ELECTRICAL ENGINEERING in a Nutshell Dr. Robert A. Durham, PhD, PE Dr. Marcus O. Durham, PhD, PE A Quick Reference The book is designed with four purposes: A Study Guide to provide assistance with preparation for professional examinations. A Reference for investigating those occasional problems outside your regular specialty. A Curriculum for professional classes for maintaining your license. A Synopsis for university students in electrical engineering and other technical disciplines.
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ELECTRICAL ENGINEERING in a Nutshell
Dr. Robert A. Durham, PhD, PE
Dr. Marcus O. Durham, PhD, PE
A Quick Reference
The book is designed with four purposes:
A Study Guide to provide assistance with preparation for
professional examinations.
A Reference for investigating those occasional problems outside your
regular specialty.
A Curriculum for professional classes for maintaining your license.
A Synopsis for university students in electrical engineering and other
Contents Chapter 1 - Circuit Elements and Analysis ................................................................................................. 1-1
2.6 Fourier Series ................................................................................................................................. 2-11
2.6 Fourier Series ................................................................................................................................. 2-11
3.2 Power Definitions ............................................................................................................................. 3-2
3.3 Three-Phase AC ................................................................................................................................ 3-5
3.4 Power Transfer across a Reactance ................................................................................................. 3-8
3.5 Power One-Line ................................................................................................................................ 3-9
3.6 Power Problem Plan ....................................................................................................................... 3-10
3.7 Mechanical Power .......................................................................................................................... 3-11
3.8 Electric Machinery ......................................................................................................................... 3-13
4.5.5 FET Transistor Models ............................................................................................................. 4-29
4.6 State Space ..................................................................................................................................... 4-35
4.6.1 The 6-Minute Approach .......................................................................................................... 4-35
5.3.3 Routh-Hurwitz – Special Cases .................................................................................................. 5-6
5.4 Root Locus ..................................................................................................................................... 5-11
5.6 Analog Filters ................................................................................................................................. 5-25
Problem 5-1 ...................................................................................................................................... 5-30
Chapter 6 Digital ........................................................................................................................................ 6-1
7.2 Time and Interest ............................................................................................................................. 7-2
7.2.1 Uniform series ........................................................................................................................... 7-3
9.2 NEC Synopsis .................................................................................................................................... 9-1
9.2 NEC Synopsis .................................................................................................................................... 9-1
Practice Problem 3-4 (old style – similar to new style)
SITUATION:
In order to accommodate the needs of larger plant power systems, a switchgear manufacturer is
offering a new line of vacuum circuit breakers rated in accordance with ANSI C 37.04 “Rating Structure
for A-C High Voltage Circuit Breakers.
The following data apply:
Nominal voltage: 7.2kV
Nominal 3-phase MVA class: 700 MVA
Rated maximum voltage: 8.25kV
Rated voltage range factor (k): 1.3
Rated insulation level – low frequency: 36kV
Rated continuous current: 3000 A
Rated short circuit current: 46kA
Rated interrupting time: 5 cycles
REQUIREMENTS:
Determine the following related required capabilities (to three significant figures)
The symmetrical interrupting capability with the prefault voltage @ 7.2kV
The symmetrical interrupting capability with the prefault voltage @ 6.1kV
The symmetrical interrupting capability with the prefault voltage @ 8.5kV
Closing and latching capability operating voltage at 7.2kV
Closing and latching capability operating voltage at 6.1kV
Closing and latching capability operating voltage at 8.5kV
Three-second short time current carrying capability operating @ 7.2kV
If the circuit breaker is applied at a point where the system impedance is 0.08Ω/phase, what is the
change in margin of symmetrical interrupting capability over short circuit current available when the
operating voltage is changed from 7.2kV to 7.4kV?
Page | 3-66
SOLUTION:
Symmetrical interrupting capability at 7.2kV
Max kVA remains the same, so, as voltage goes down, current goes up, as long as voltage is within
voltage range (1.3)
8.25
1.157.2
, so keep constant kVA
3 3max 8.2546 10 52.71 10
7.2SC rated
nom
VI I Amp
V
Symmetrical interrupting capability at 6.1kV
Max kVA remains the same, so, as voltage goes down, current goes up, as long as voltage is within
voltage range (1.3)
8.25
1.356.1
, so limit voltage ratio to 1.3
3 3*1.3 46 10 *1.3 59.8 10SC ratedI I Amp
Equipment is rated at 8.25kV, do not use at 8.5kV
Max interrupting current: 3 31.3*46 10 59.8 10mI Amp
Refer to ANSY C 37.04 standards
Closing & latching is 1.6 times the max current (at any voltage below rated)
3& *1.6 95.68 10c l mI I Amp
Closing & latching is 1.6 times the max current (at any voltage below rated)
3& *1.6 95.68 10c l mI I Amp
Equipment is rated at 8.25kV, do not use at 8.5k
Three second current capability = max interrupting current (3s < 5s rating)
3 31.3*46 10 59.8 10mI Amp
7.2kV is L-L voltage, need L-N (phase) voltage
Page | 3-67
7200
4156.93 3
LLLN
VV V
System Z is 0.08,
3720051.96 10
3 *0.08scI Amp
From a) above, Max interrupting current at 7.2kV is 52.71x103 Amp, so margin is 748 Amp
at 7.4kV, short circuit current is
3740053.406 10
3 *0.08scI Amp
Max short circuit current is
3 3maxmax
8.2546 10 51.283 10
7.4SC rated
nom
VI I Amp
V
So, equipment is overrated by 2,122 Amps. Change in margin is 2,870 Amps.
Page | 3-68
Practice Problem 3-5(old style)
SITUATION:
An industrial plant has four generators connected to a 13.8kV bus as shown in Figure 433 below. The
neutral of each generator is connected through a neutral circuit breaker to a common neutral bus. This
bus is connected through a neutral resistor to ground.
G
G1
G
G2
G
G3
G
G3
Individual generator reactances in pu:
2
0
0.16
0.16
0.08
dX j
X j
X j
REQUIREMENTS:
a) Determine the value of the neutral resistor required to limit a line-to-ground fault to that of
a three-phase fault when only one of the units is operating.
b) If a 0.1 pu neutral resistor is used, determine the ground fault current in pu when all units
are operating, but with only one neutral circuit breaker closed.
c) Same as b., except all neutral circuit breakers closed.
SOLUTION:
Estimate positive, negative and zero sequence impedances from data given.
1
2 2
0 0
0.16
0.16
0.08
dZ X j
Z X j
Z X j
a) f resistorZ Z
Page | 3-69
3
1
f
ph
f
vI
z z
1
0 1 2
3
3
f
ph
f
vI
Z Z Z Z
Operating at 100% voltage (1.0 pu)
3
1.06.25
0.16 0phI j
j
1
3*1.0 3
0.16 0.08 0.16 3 0.4 3ph
resistor resistor
Ij j j Z j Z
Set 1phI = 3 phI and solve for resistorZ
36.25
0.4 3
30.4 3
6.25
3 0.48 0.4
0.026667
resistor
r
R
R
jj Z
j Zj
Z j j
Z j
b)
1
0 1 2
12 2 2 2
3 3*1
3 0.08 0.04 0.04 3* 0.1
3 3
( 0.16) (0.3)
8.8235
f
ph
f
ph
vI
Z Z Z Z j j j j
Ij X R
pu
c) With all breakers closed, Zo=0.02j
1
0 1 2
12 2 2 2
3 3*1
3 0.08 0.04 0.02 3* 0.1
3 3
( 0.1) (0.3)
9.487
f
ph
f
ph
vI
Z Z Z Z j j j j
Ij X R
pu
3 x 0.1
j0.16
j0.16
j0.08
j0.16
j0.16
j0.08
j0.16
j0.16
j0.08
j0.16
j0.16
j0.08
Positive sequence
impedance
Negative sequence
impedance
Zero sequence
impedance
One breaker
closed
Page | 3-70
Problem 3-6 (old)
SITUATION:
A 1000/1250 kVA, OA/FA, 13.2kV:4160V single phase transformer is part of a 3000/3750 kVA Y-Δ
bank.
Factory tests are made on this transformer at 25°C and the following data recorded.
DC Resistance: r1 = 0.40 Ω r2 = 0.035 Ω
With secondary open and 13.2kV applied to the primary: I1 = 10A, Pin = 5500W
With secondary shorted and 800V applied to the primary: I1 = 75.76A , Pin = 5800W
Assume the three single phase transformers are equal.
REQUIREMENTS:
For the operating temperature of 75°C, determine:
The percent effective resistance on the self-cooled rating base
The percent reactance on the self-cooled rating base.
The percent impedance on the self-cooled rating base
The no-load loss of the three-phase bank (kW)
The total loss of the three-phase bank (kW) with the transformer operating at its force cooled rating.
The efficiency of the bank carrying 3750 kVA at 85% pf
Background
1000/1250 kVA OA/FA
13.2kV/4.16 kV
DC Resistance: r1=0.40Ω r2=0.035Ω
Open Circuit Test: V1=13.2kV I1=10A Pin=5500W
Short Circuit Test: I1=75.76A Pin=5800W
Fan Load = 750W
Sbase=1,000 kVA Vbase=13.2kV Zbase=2 2(13,200)
174.241,0000,000
base
base
V
S
Turns Ratio: 13.2
3.1734.16
p
s
Va
V
Page | 3-71
Solution:
(a)Percent effective (ac) resistance on the self-cooled rating base
ac dc core mechr =r +r + r (rmech is 0 for transformer)
Equivalent dc resistance referred to primary:
2 21 2(25 ) 1.4 3.173 *0.035 0.7524dcr C r a r
Effective resistance from short circuit test
2 21
5800(25 ) 1.0105
(75.76)
inac e
Pr r C
I
The components of the ac resistance at test temperature
ac dc core
core
core
r (25 C)=r (25 C)+r (25 C) 1.0105
1.0105 0.7524 r (25 C)
r (25 C) 1.0105 0.7524
0.2581
Resistance changes with temperature.
rdc increases with temp (positive temp coeff)
rcore resistance decreases with temp (negative temp coeff)
∆R = α T0
∆T
or
R = R0 [ 1 – α (T – T0)]
For copper, the inferred absolute zero coefficient is -234.4.
So the equation reverts to
R = 234.4 + T
R0 234.4 + T0
Apply to both the copper and the core resistance.
ac
234.5 75 234.5 25r (75 C)=0.7524 0.2581
234.5 25 234.5 75
1.1138
Convert to per unit.
1.1138
( 75 ) 0.006392 0.6392%174.24
acr pu C
Page | 3-72
(b) Percent reactance on the self-cooled rating base
Impedance
Z = V = √R2 + X2
2 2
2 2
80010.56
75.76
10.56 1.1138
10.501
10.501( ) 0.060267 6.0267%
174.24
scac
sc
ac ac ac
ac
ac
VZ
I
X Z r
X
X pu
(c) Percent impedance on the self-cooled rating base
10.56( ) 0.0606 6.06%
174.24acZ pu
(d) No-load loss of 3 phase bank (from open circuit test)
Pno-load=3*Pin = 3*5500 = 16.5kW
(e) Total loss of 3 phase bank operating at FA rating
S = VI* → I = S / V
2
2
125094.697
13.2
3*( )
3*(94.697 *1.1138 5500)
46.46
FA
lossFA FA ac no load
kVAI A
kV
P I r P
kW
(f) Efficiency
3750 *0.85
3,187.5
3,187.598.54%
(3,187.5 46.46 0.75 )
out
out
in
P kVA pf
kW
P kWeff
P kW kW kW
Page | 3-73
Problem 3-7 (old) 436
SITUATION:
A generating station is connected as shown in Figure Problem 2-7 below. Transformer T2 was destroyed
and must be replaced; however, no records exist of the nameplate, and the proper phase relations must
be determined so that a new transformer can be specified.
REQUIREMENTS:
Neatly sketch and label phasors A'B'C', and state sequence A'B'C' or C'B'A'.
Neatly sketch and label phasors A''B''C'' and state sequence A''B''C'' or C''B''A''
Complete the nameplate for T2 – ratings not required.
H0 H1 H2 H3
X1 X2 X3
GENT2
H1 H2 H3
X0 X1 X2 X3
T1
A’B’
C’
ABC
N
B’’C’’N
A’’
T3 H0 H1 H2 H3
X0 X1 X2 X3
A
C
B
N
H3
H1H2
H0
X3
X1
X2
H3
H1
H2
H3
H1H2
H0
X3
X1X2
X0
T1
T2
T3
Page | 3-74
SOLUTION:
This is a problem about phase sequences. It illustrates the phase shifting between (1) wye and delta,
(2) between line-line and line-ground, and (3) between line and phase. Although these are obviously
related, the actual connections can be quite different.
Delta Delta Wye Wye
Phase Line Phase Line
L-L L-L L-N L-L
Phase sequence is drawn from the perspective of looking down the x-axis to the left. The phasors
rotate CCW. Record the phase sequence AN, BN, CN or CN, BN, AN or record the line sequence AB, BC,
CA or CA, BC, AB. Select every other letter. The sequence is positive ABC or negative CBA.
For a transformer the terminals are labeled on the primary and secondary.
Neutral
Primary H1 H2 H3 H0
Secondary Additive X1 X2 X3 X0
Transformers in a wye-delta configuration are shown. Note the
corresponding orientation that does not result in a phase shift. AN-
XY, BN-YZ, CN-ZX
Steps for determining transformer connection. Make a table of
the line connections and the transformer connections. Fill in the
rows of unknowns. Note the order that data is filled.
Order Action Options
1 Reference phase AN BN CN or AB BC CA
AN BN CN
2 Transformer primary connection H1 H2 H3 H0
3 Primary actual phase/line connection AN BN CN or AB BC CA
4 Transformer secondary connection X1 X2 X3 X0
5 Secondary actual phase/line connection AN BN CN or AB BC CA
6 Orientation of primary & secondary draw sketch
0o 120o 240o or 90o 210o 330o
7 Sequence ABC or CBA
A B
N
C
X
Y
Z
Page | 3-75
Transformer T3 is a wye-wye. The primary and secondary are aligned in phase.
Order Action Connection
1 Reference phase AN BN CN or AB BC CA
AN BN CN
2 Transformer primary connection H1 H2 H3 H0
H1H0 H2H0 H3H0
3 Primary actual phase/line connection AN BN CN or AB BC CA
AN BN CN
4 Transformer secondary connection X1 X2 X3 X0
X1X0 X2X0 X3X0
5 Secondary actual phase/line connection AN BN CN or AB BC CA
C”N B”N A”N
6 Orientation of primary & secondary draw sketch
0o 120o 240o or 90o 210o 330o
0 o 120 o 240o
7 Sequence ABC or CBA CBA
Transformer T1 is a wye-delta. The primary and secondary are shifted in phase.
Order Action Connection
1 Reference phase AN BN CN or AB BC CA
AN BN CN
2 Transformer primary connection H1 H2 H3 H0
H3H0 H2H0 H1H0
3 Primary actual phase/line connection AN BN CN or AB BC CA
AN BN CN
4 Transformer secondary connection X1 X2 X3 X0
X3X1 X2X3 X1X2
5 Secondary actual phase/line connection AN BN CN or AB BC CA
A’C’ B’A’ C’B’
6 Orientation of primary & secondary draw sketch
0o 120o 240o or 90o 210o 330o
0 o 120 o 240o
7 Sequence ABC or CBA ABC
Page | 3-76
Transformer T2 is a delta-wye. The primary & secondary are shifted in phase. The secondary
orientation is unknown
Order Action Options
1 Reference phase AN BN CN or AB BC CA
AN BN CN
4 Transformer primary connection H1 H2 H3 H0
H1H3 H2H1 H3H2
2 Primary actual phase/line connection AN BN CN or AB BC CA
A’C’ B’A’ C’B’
5 Transformer secondary connection X1 X2 X3 X0
X1X0 X2X0 X3X0
3 Secondary actual phase/line connection AN BN CN or AB BC CA
C”N B”N A”N
6 Orientation of primary & secondary draw sketch
0o 120o 240o or 90o 210o 330o
90o 210o 330o
7 Sequence ABC or CBA CBA
Page | 4-1
Chapter 4 - Electronics
4.1 Introduction Electronics involves a small signal operating on active devices to produce a different signal. Active
devices are non-linear energy amplifiers. Passive devices are simply modeled as RLC components.
Electronics is predominantly small signal variations that exist around a larger operating point. This fits
the standard response or solution, without the exponent, or essentially the cos( )t portion.
( ) cost
y t F I F e t
The constant part of the signal is the power supply, called the bias. The dc voltage and current
define an operating point , quiescent point (q-point), or bias. The bias dictates what portion of the curve
the device is operating on.
The small signal is the AC component. Generally AC analysis involves investigating a transfer
function operating on the input signal. The function is a two port model of the circuit.
The signal voltage and current cause slight variations around the operating point defined on the dc
bias line.
0 1 cosi I I t
ln1
1ln
new old
spacingX X
GMR
oI and oV are the DC operating point. The cos ωt term is the signal varying about the operating
point.
Electronic functions are by definition non-linear systems. In order to simplify the solution, it is
necessary to use assumptions.
Linearize with time varying about operating point is appropriate if
1 0cos( )I t I
For AC small signal, the DC supply voltage appears as a short
Page | 4-2
4.1.1 Solid State Device Characteristics
Electronics often use semiconductors. The conductivity of a semiconductor material is a physics
property.
n pq n p
electron mobilityn
hole mobilityp
electron concentrationn
hole concentrationp
-19charge on electron (1.6 10 )q C
The silicon material is doped to change the number of electrons (negative) or holes (positive).
p-type material; p ap N
n-type material; n dn N
Carrier concentrations at equilibrium are the product of the p and n concentration.
2( )( ) ip n n
in = intrinsic concentration
Built in potential or contact potential of a p-n junction depends on Boltzmann’s constant,
temperature and the electrical charge.
2
ln a do
i
N NkTV
q n
Thermal voltage is the constant, temperature, and charge components.
0.026T
kTV V
q at 300°K
Na = acceptor concentration
Nd = donor concentration
T = temperature (K)
k = Boltzmann’s Constant = 1.38x10-23 J/K
Page | 4-3
Capacitance of abrupt p-n junction diode is a function of the voltage.
( )1
o
bi
CC V
VV
Co = junction capacitance at V=0
V = potential of anode with respect to cathode
Vbi = junction contact potential
Resistance of a diffused layer is based on a sheet of material.
s
LR R
W
Rs = sheet resistance = d
in ohms per square
= resistivity
d = thickness
L = length of diffusion
W = width of diffusion
Page | 4-4
4.2 Boundary Conditions The response to a signal is investigated at three different conditions: the midpoint, the upper limit
or boundary, and the lower limit or boundary. Frequency changes the impedance and the time constant.
As a result, the voltage and gain of the circuit is changed.
The upper and lower boundary analysis determines the frequency of the transition.
High Frequency Cutoff – Shunt Capacitance dominates
f
Series capacitor is conducting 1
02
se
se
XfC
Frequency is determined by shunt capacitance& parallel resistors.
1 2
1 1 1 1c shC C
RC R R R
Low Frequency Cutoff – Series resistance and capacitance dominates
0f
Shunt capacitor is open 1
2sh
sh
XfC
Frequency is determined by series (coupling) capacitor and resistance in series.
1 2
1c serR R R C C
RC
3 dB Frequency
The response is the voltage that changes with frequency. It has an exponential rise to the low cutoff,
transitions to a bandwidth that is relatively flat, then transitions to an exponential decay past the upper
cutoff. The cut-off frequencies are called 3dB because the voltage has decayed to that value.
Vmax
.707 Vmax
ω ωoω3db
Page | 4-5
3 2 between +3db and -3dbndb
20log o
in
Vdb
V
30.15 log
20
o
in
Vdb
V
0.1510 2 o
in
V
V
3db point @ 2o inV V
Attenuation
Attenuation is the reduction in signal. It is influence by the cut-off of each stage in an amplifier.
ATF (Attenuation Factor) = 1
2 1n
n= # of stages
*high onef f ATF
fone =one stage frequency
fhigh = all stage high frequency
flow = all stage low frequency
onelow
ff
ATF
Page | 4-6
4.3 Diodes & Rectifiers Diodes are the simplest semi-conductor device that consists of a single p-n junction. Functionally,
these are rectifiers.
4.3.1 Diodes
Diodes are non-linear as noted in the waveforms. They are used for power supplies, wave shaping,
and logic. For most circuit analysis, the ideal model of the diode can be used. The assumptions for an
ideal diode are zero current in reverse direction and zero voltage drop in forward direction. Real diodes
are somewhat less idyllic.
Ideal Diode
I=0+ -
V=0- +
Real Diode
(Piecewise Linear)
+ - - +
The Shockley equation defines the V/I relationship in a real diode.
1D
T
vV
D si I e
sI = saturation current
η = emission coefficient, typically 1 for Si
TV = thermal voltage = kT
q
Page | 4-7
4.3.2 Rectifiers & Clippers
The application or performance of diodes is a rectifier. Representative circuits and their output are
shown. The input voltage Vin is a sinusoid.
Input Waveform
Half Wave Rectifier
RL VoAC
Full Wave Rectifier
RLVin
Vo
Voltage Clipper
V1
RLVo
AC
Page | 4-8
4.4 Operational Amplifiers
Ideal
An ideal op-am has infinite input impedance, zero output impedance, and almost infinite gain. The
gain, A is described by the relationship.
1 2( )ov A v v
A is large (>104) and v1 and v2 are small enough so the amplifier is not
saturated. For the ideal operational amplifier, assume that the input
currents are zero and that the gain A is infinite. Then, when operating
linearly, the voltage difference is 2 1 0v v .
Differential Amplifier
Two ungrounded input terminals create a
differential amplifier. The device amplifies the
voltage difference between the inputs.
The general form is shown. The significant
relationships are calculated.
out
in
VA
V
The op amp has a high Z input. That implies the
voltage between the terminals is essentially 0. So,
there is no current flow.
V- = V+
Compare the current on the negative and positive input branches. Use those functions to calculate
the output voltage.
01
1 2
V VV VI
Z Z
2
3 4
0V V VI
Z Z
The circuit is a voltage divider between the feedback side and balancing side.
2 4 1 20 1 2
1 1 3 4
Z Z Z ZV V V
Z Z Z Z
(Voltage divider)
-
+
v2
v1
vo
V1
Z1
V2
Z3
Z2
Z4
-
+
Vo
Page | 4-9
The connection of the op-amp changes the gain. Complex reactance creates mathematics functions
of integrator and differentiator.
Inverting Amplifier
Voltage on – terminal
2 2
1 1
Z RA
Z R
-
+
R1 R2
ViA
Non-Inverting Amplifier
Voltage on + terminal
1 2
1
R RA
R
-
+
R1 R2A
Vi
Integrator
Capacitor in feedback
2
1 1 1
1 1 1ZA
Z R sC sCR
-
+
R1 AVi
C
Differentiator
Capacitor on input
22 2
1
( )( )Z
A R sC sCRZ
-
+
AVi
R2
C
Page | 4-10
4.5 Transistors Transistors are the next level of complexity above diodes for semi-conductor devices. A transistor
consists of two junctions. Functionally, transistors are amplifiers. When operating at a boundary
condition, such as in digital circuits, a transistor is a switch.
I
vbe
Q-point
time
time
Vbe
cI
Input Signal
Output Signal
4.5.1 Bias vs. Small Signal
Bias is the DC voltage that determines the operating point. Small signal is the AC message that is
superpositioned on the DC.
Transistor Bias – External
Transistor biasing is a process of external connections similar to those in the figure. Because of the
connection, we know certain items.
Know:
Voltage @ ground = 0
Voltage @ Rc = Vcc
Want to find:
Voltage @ base. It is a divider of Vcc
The Thevenin equivalent input is a voltage divider of the power supply. If Rb
>>10R2, then the base current can be ignored and VBG can be easily determined.
1
1 2
CCBG Thevenin
V RV V
R R
R1
R2
Rc
RE
VL
VccB
C
E
Page | 4-11
The Thevenin resistance is obtained by shorting the power supply (VCC). The equivalent
circuit is shown in the figure.
where
Next, find the current.
Current @ base (input)
Current @ emitter (output)
The analysis must include loop elements. In essence do per unit conversion. The current through the
transistor has gain. IB is the base current. If Rbase (=hfeRE) is not >> than 210* R , then VBG must be
determined from this circuit.
The operating point of the transistor is referred to as the Quiescent point (Q point). This is the point
that the circuit will operate at when only the DC signal is in place. The Q-point is defined by the Ic
current at the Q-point (ICQ) and VCE at the Q-point (VCEQ). For an existing circuit, the Q-point can be
determined by using the Thevenin equivalents.
2
1 2
1 2
( )
TH CC
TH
TH BECQ
THE
FE
CEQ CC CQ C E
RV V
R R
R R R
V VI
RR
h
V V I R R
Often when designing an amplifier
circuit, a graphical approach is easier. The
Q-point is the intersection of the DC load
line and the AC load line. The DC load line is
the line drawn between the
1 2/ /THR R R
1CC
THCC TH
VR R
V V
2CC
THTH
VR R
V
( 1)
( 1)
E B C B
TH BE TH B E E
TH BEB
TH E
I I I I
V V R I R I
V VI
R R
R1
R2
VBGVcc
RTH
RE
IE
VTH
VBE
IB
DC
CCV
R
CQI
CEQVCCV
Page | 4-12
Collector/Emitter current when the transistor is conducting (shorted) and the Collector voltage (VCC).
Collector/Emitter current is VCC divided by RDC where
The load line is shown superimposed on the transistor characteristic curves in the figure to the right.
The next step is to select a Q-Point. The goal for most linear (Class A) amplifiers is to select a Q-point
near the middle of the load line. This allows for an equal amount of swing above and below the Q-point.
Once the Q-point is selected, VCEQ and ICQ can be determined.
The AC load line is the line that represents the small signal that is superimposed on the DC bias
circuit. AC signals bypass (short) VCC; it also includes the load impedance (RL). The endpoints of the ac
load line are found using the following
formulae
( )
( )
( )
( )
CEQ CE off
c sat CQ
C ac
CE off CEQ CQ C
ac C L E
V vi I
R R
v V I R
R R R R
The ideal situation for a Class A
amplifier is for the AC load line is for the Q-
point to be in the middle of the AC load
line. This can be adjusted by modifying RC.
(no AC coupling)dc C ER R R
DC
CCV
R
V
CQI
CEQV
( )CE off
ac
v
R
( )CE offvCCV
Page | 4-13
4.5.2 Transistor Mathematical Relationships
Name and Schematic Symbol Mathematical Relationships
NPN Bipolar Junction Transistor (BJT)
Relationships valid in active mode of operation.
PNP Bipolar Junction Transistor (BJT)
Relationships valid in active mode of operation.
N-Channel Junction Field Effect Transistor (JFET)
P-Channel Junction Field Effect Transistor (JFET)
Cutoff Region:
Triode Region:
Saturation Region:
C
E
B
iC
iE
iB
E B Ci i i
C Bi i
C Ei i
1
BE
T
VV
C si I e
emitter saturation currentSI
thermal voltageTV
C
E
B
iC
iE
iB
E B Ci i i
C Bi i
C Ei i
1
BE
T
VV
C si I e
emitter saturation currentSI
thermal voltageTV
D
S
G
iD
iS
D
S
G
iD
iS
GS pv V
0Di
GS Pv V
GD Pv V
2
22DSS
D DS GS P DS
p
Ii v v V v
V
GS Pv V
GD Pv V2
1 GSD DSS
P
vi I
V
Page | 4-14
N-Channel Depletion MosFET (NMOS)
IDSS = drain current with (in the saturation region
K = conductivity factor VP = pinch off voltage
P-Channel Depletion MosFET(NMOS)
Same as N-Channel, with current directions and voltage polarities reversed
D
S
G
iD
iS
B
N-Channel Enhancement MosFET (NMOS)
Cutoff Region:
GSv Vt
0Di
Triode Region:
GSv Vt
GD tv V
22D DS GS t DSi K v v V v
Saturation Region:
GS tv V
GD tv V
2
D GS ti K v V
IDSS = drain current with 0GSv (in the
saturation region K = conductivity factor Vt = pinch off voltage
D
S
G
iD
iS
B
P-Channel Enhancement MosFET(NMOS)
Same as N-Channel, with current directions and voltage polarities reversed
D
S
G
iD
iS
B
0GSv
2
DSS PI KV
D
S
G
iD
iS
B
Page | 4-15
4.5.3 General Two-port Models
Two-port models have external measurements which take three forms; impedance, admittance, or
hybrid. Impedance models have voltage in terms of current. Admittance models have current in terms of
voltage. Hybrid models have both voltage and current as outputs. The internal connections have three
arrangements: T or wye, Π or delta, and L.
4.5.4 BJT Transistor Models
Bipolar Junction Transistor – Use Common emitter as general example
The complete model is shown on the left, while the simplified version is on the right.
The hybrid two-port network is the generic form for the BJT transfer function and includes the
current, voltage, and hybrid parameters, h.
1 1 2
2 1 2
i r
f o
v h i h v
i h i h v
The definition of the hybrid parameters are in terms in input (1) , and output (2).
General Form Specific to Common Emitter
2
1
1 0
i in
v
vh R
i
(1 )ie ib b ich h i h
2
2
1 0
f
v
ih
i
1
cfe
b
ih
i
1
1
2 0
r
i
vh
v
= gain of circuit
1
2
2 0
1o
i
ih
v out
= gain of transistor
hi
hrV2 hfi11/ho
i1 i2B
E
C
V1 V2
hfe
ib ib
hfeib
B
E
C
Page | 4-16
The second subscript is determined by which terminal connects to the common lead for the two-
port model.
For common base – b input – e, output – c
For common emitter – e input – b, output – c
For common collector – c input – b, output – e
Common collector is also called an emitter follower.
Common collector circuits permit assumptions which greatly simplify the circuit.
2 0reh v
1
0 ignoreoe
o
hh
H- parameters are real at low frequencies and complex a high frequencies. At high frequencies, the
capacitance has more effect.
Since this is a small signal investigation on internal parameters, the DC supply appears as a short.
Page | 4-17
Common Emitter – Hybrid Parameters
The common emitter is the most common configuration for
transistor amplifiers.
The h-parameters for a common emitter model are listed.
4
Q point
10coe
ce
ih
v
Q point
cfe
b
ih
i
3
Q point
(25 10 ) febe Tie fe
b EQ EQ
hv Vh h
i I I
i feA h
The small signal circuit removes the DC components.
The equivalent circuit input and output impedance are found.
b iei ie
b ie
R hZ h
R h
If the base bias resistor Rb>>hie, then the output is greatly
simplified.
1
o
oe
Zh
Conditions Small Signal Equivalent Model
Complete
hi
hrV2 hfi11/ho
i1 i2B
E
C
V1 V2
As hre0
hie hfeib1/hoe
ib icB
E
C
vbe vce
iB
ic
+
V
be
-
+
Vce
-
iL
Rb
VBB Re
RL
Ce
VCC
i1
RbRL
ib
ic
ie
+
Vbe-
+
Vce-
iL
Page | 4-18
As hre0 and hoe e0
hie hfeib
ib icB
E
C
vbe vce
Complete common-
emitter amplifier
hie hfeib
ib icB
E
C
vbe vceRbRLii iL
ZiZo
Page | 4-19
Hybrid Model – Common Emitter Model at High Frequency
The hybrid is a two port network defined in terms of both h-parameters and circuit parameters.
As a result, it incorporates both the non-linear transistor and a two-port model as shown below.
B
C
E
B C
E
rbb’rb’e Cb’e
rb’crce
Cb’c
Eb’
e
Ebe Ece
b’
Because of the capacitance, the frequency has a major impact on impedance. The Miller effect
resolves the situation where the voltages at both ends of a capacitor change at the same time. Rather
than consider the actual effect, the model uses a large capacitor, CMiller . The Miller replaces Cb’c. At low
freq assume Cm is included in Cb’c. The circuit assumptions are noted.
Cmiller '(1 )Lb c
d
RC
r
' ' '
0.025@ 300
fe
ie bb b e bb
EQ
hh r r r T K
I
β = common emitter low frequency current gain = 1
feh
Ebe = diode drop forward biased b-e voltage 0.6V
The equivalent circuit has the input side coupled to the output by the input voltage operating on the
transconductance, gm. By using superposition, the input is calculated first. Then the results are used on
the output side.
Page | 4-20
Rbb’
ReVbe Zc
gmVbe
B
E
C
Input Side
Calculated 1st
Output Side
Use input to
calculate
rπ
gmVbe
B
E
C
Rce
The frequency effects are calculated. At the cutoff frequency, the gain is down 3dB.
Cutoff frequency = ' ' ' ' '
1 1
2 ( ) 2b e b e b c b e b e
fr C C r C
3 db bandwidth = fef h f 1
fb
i
ve
hA
jh
The internal circuit parameters have the following relationships.
'
40 300fe
m EQ
b e
hg I T K
r
' 10 50
2 5 (hi freq caps)
bbr
'
'
fe mb e
T b e T
h gC
r
'
0.26
40
fe
b e
EQ E
hr
I I
1 1' 2 3
5 pb c ob cbC C v pf p
Page | 4-21
Common Base – Hybrid Parameters
A common base is used for high frequency applications because the base separates the input and
output, minimizing oscillations at high frequency. It has a high voltage gain, relatively low input
impedance and high output impedance compared to the common collector.
1 0
11
e
cob
o cb i i
ih
Z v
0
1
cb
cfb
e v
ih
i
0
1cb
eb ieTib i
e EQ fev
v hVh Z
i I h
i fbA h
A simplified circuit model is developed.
Complete Circuit
ri
vi
VCC
iE iLRL
R2R1Cb
iC
Small Signal Circuit
ri
vi
iE
iL
RLiC
Page | 4-22
Small Signal Equivalent Model
ri
E
vi hrbvcb
hib
hebi1 1/hob
RL
B
C
iE
iL
iC
i1
Simplified Small Signal Equivalent
Model
0rbh
1fbh
0obh
ri
E
vi
hib
i1 RL
B
C
iE
iL
iC
i1
Page | 4-23
T Model – Common Base Model at High Frequency
The T-model is simply a configuration that uses conventional circuit analysis without special
parameters.
re
Ce
rb
Zc
IcIee
b
c
At low frequency, capacitors are high impedance and appear as open circuits.
e eZ R
'' is close to unity(1)b e
e m b e
e
VI i g V
r
c cZ R
At high frequency, capacitors are low impedance and appear as a short circuit which passes signals.
1 1
2e e e
e t e
Z sC CR f r
1 tc b
c
fZ I I
sC f
c E cBOI I I
The variables have the following meanings.
α = fraction of emitter current collected
Icbo = reverse current characteristics of collector base junction
IE = dc value of emitter current
ft = high frequency performance (where hfe decreases to unity)
The emitter resistance is the base to emitter diode resistance. It is also called rb’e & rd.
26 T
E
E e
Vr
I r at room temperature
The typical ranges of values are noted.
Rb range 10’s to 100’s ohms – base spreading resistor
α range 0.9 to 0.999
rc range hundreds megohms
Vb’e 0.6v (0.5 – 0.7v) which is diode drop forward bias voltage.
Page | 4-24
Page | 4-25
Common Collector (Emitter Follower)– Hybrid Parameters
The common collector amplifier is also called an emitter follower because the output is taken from
the emitter resistor. The connection has a gain of near unit. It can be an impedance matching device
since its input impedance is much higher than its output impedance. It is used as a buffer in digital
circuits.
'
1
io ib
fe
rZ h
h
( 1)i ie fe eZ h h R
1vA
Complete Circuit
ri
vi
iE
ib
+
vE
-
BC
E
RE
RB
VCC
Small Signal Circuit ri
viiE
Rb
ib
+
vE
-
BC
E
RE
+
vb
-
Small Signal Equivalent Model
Looking into base
ri
viiE
hib
Rb +
vE
-
B
C
E
RE
+ vbe -
Page | 4-26
Small Signal Equivalent Model
Looking into emitter
v’i
iE
hib
+
vE
-
B
C
E
RE
+ vbe -
'
1
i
fe
r
h
Page | 4-27
Emitter Follower (common collector) Model at High Frequency
High frequency performance is very different because of the capacitors created in the device and in
the circuit. Therefore, alternative analysis assumptions are needed.
Circuit
Ri
vi Cb’c
Cb’e
Re
i1 ib
ie
vb
iCb’e
iRe
Equivalent Circuit
Ri
vi Cb’c
Cb’e
Ze
i1
vb
rb’e
ve
The frequency impact on the impedance is readily observable.
1
( 1)1
Te fe
j
Z hj
Assumptions reduce the complexity of the problem.
Cb’c is relatively small, and '1fe e b e ih R r R
e
1' R
1( )
1 1
T
i
T
j
ev R
i
vA
v j
Upper cutoff frequency 11 i
e
Th R
R
ff
If Cb’c is large enough for its impedance to be equal to Ri at fh2<<fh1, then the gain is in terms of h-
parameters.
Page | 4-28
''
1 1( )
11
fe e
v
i b cb e fe e
h RA
j R Cr h R
Upper cutoff frequency 2
'
1
2h
i b c
f fR C
If 1 2h hf f the gain changes.
2 2
1
111
T
i
e
j
v RR
h T T h
Aj j
j
Page | 4-29
4.5.5 FET Transistor Models
Field Effect Transistors (FET) are devices that are less affected by frequency. They can also handle
more power, in the range of amps. They are also more expensive.
Circuit
Vdd
RdId
Drain
SourceGate
RsRs CsVin
Equivalent circuit
Rg
Rd
rd
gmVqVqin
CsRs
The following constraints apply.
There may be capacitor coupling which includes Cqd Cqs & Cds in the small signal.
Vp is the pinch off voltage, which is negative. Pinch-off is the condition that stops conducting ID but
still has leakage current IDSS.
VDS is the breakdown voltage where the transistor fails.
Keep the supply voltage, VDD, below the breakdown voltage, VDS.
Igs is the gate leakage current in nanoamperes. DI is the drain current.
For a JFET, the drain current is dependent on the leakage current and drain voltage.
2
D gs
D DSS
D
V VI I
V
Page | 4-30
Example
A FET circuit with self bias is desired. The FET limits
and the supply consumption are provided.
Vp = -3 volts with current of 6ma
VBD = 30 volts
Required: Determine the bias for 10V Drain to
Source and channel current of 4ma.
Given:
ID = 4 mA
IDSS = 6 mA
|VD| = 3V
Solution:
2 2
34 6
3
0.55
D gs gs
D DSS
D
gs
V V VI I
V
V V
Assumptions:
Select Rg = 1MΩ
This makes voltage drop across Rg0 (Input resistance >10X source)
Calculations:
Write Kirchhoff current law @ FET
0 0D g S g
D S
I I I I
I I
For 30BDV , use 30DDV = 24 volts (80%)
Write KVL around loop for Bias
0D SDD R DS RV V V V
Substitute using Ohm’s law & values
24 10 0D D D SI R I R
Vgs is determined by ID & RS
RdId
Drain
SourceGate
RgRsVin
VDD
Page | 4-31
Vgs = IDRS 3
0.55140
4 10
gs
S
D
VR
I
Substitute in KVL
3
14 0
14 4 10 ( ) 0
3500
D D D S
D S
D S
I R I R
R R
R R
Use RS = 140
3500 3.3kD SR R
Page | 4-32
Basic FET Models
The FET has several different models that are used for design purposes.
Current Source Model
rds
G
S
D
gmvgs
Voltage Source Model
rds
G
S
D
μvgs
Complete Circuit
Common source voltage amplifier
ri
vi
cc1
R1
R2
R3
Rd
Rs Cs
Cc2
RL
DC
Small Signal Equivalent ri
vi R3+(R1||R2) Rdrds RL+
vL
-
gmvgs
Page | 4-33
The equations for the FET are the normal big three: input impedance, output impedance, and gain.
3 1 2( || )iZ R R R
||o d dsZ R r
1 3 1 2
1( || )
1 / ( || )v m L oA g R Z
r R R R
Voltage Gain
usually 1 3 1 2( || )r R R R
If RL<<Zo then
v m LA g R
Page | 4-34
Source Follower (Common Drain) Amplifier
The FET source follower is similar to the BJT emitter follower configuration.
Complete Circuit
vi
cc R1
Rs1
Rs2
VDD
+
vs
-
ii
Small Signal analysis
vi
Rs=Rs1+Rs2
VDD
+
vs
-
ii
Small Signal Equivalent
rds
vi
μvgs
+
vs
- Rs=Rs1+Rs2
ii
G
S
D
Thevenin Equivalent Circuit
Zo
vi
Zi μvgs+
vs
-
Rsii
GS
D
Assumptions assist in solving for the coupling voltage.
1R
s gs o ds sv v i r v
The big three are then calculated.
1
1
dso
m
rZ
g
2
1 2
11
1 ( )
( 1)1 s
s s
g
i Ri R R
v RZ R
i
'1
vA
Page | 4-35
4.6 State Space State space models can be used for a wide range of problems from controls to communications and
computer modeling. State space is a technique to represent differential equations when applied to the
LaPlace s-domain.
4.6.1 The 6-Minute Approach
Definitions
The state equation has the form ( )
( ) ( )dx t
Ax t Bu tdt
The output equation has the form ( ) ( )y t Cx t
The state variables can be defined as 1 2( ) ( ), ( ) / ( )x t x t dx t dt x t
A, B, and C are matrices while u is input, y is output, and x is the transfer variable.
The transfer function, Y(s)/U(s) can be broken into parts by ( ) ( )
( ) ( )
X s Y s
U s X s. Then Y(s)/X(s) is the
numerator of the transfer function and X(s)/U(s) is 1 over the denominator.
The s-operator is defined as the first derivative. ' /s x dx dt
S-domain to Differential
Convert the numerator Y(s)/X(s) to ( ) ( )* ( )Y s numerator X s .
Substitute to obtain y(t)= f(x(t)), using #6 definition.
( ) / ( ) 4
( ) ' 4
Y s X s s
y t x x
Develop an equation for the highest derivative of x to be the same power as the highest power of s
in the denominator.
2
1
3 6s s
''x
Replace the highest power with the input.
'' ...x u
Replace the remaining s-values using #6 definitions.
'' 3 ' 6x u x x
The result is two differential equations.
Page | 4-36
( ) ' 4y t x x
'' 3 ' 6x u x x
Substitute 1 2( ) ( ) and ( ) / ( ) ( )x t x t d x d t x t into the matrix form.
'1 1
'22
1
2
0 1 0
6 3 1
4 1
x xu
xx
xy
x
Page | 4-37
4.6.2 Description
State space is used to solve ordinary differential equations. The process uses op-amps to create
summers and integrators. Note, differentiation produces noise. The output is limited to maximum and
input is limited to minimum because of noise. The values are selected so the constant coefficient has a
range of 0.1 to 10.0. Because of the op-amp circuit, each operation is always negative or inverting
Multiplier Summer Integrator
KVi Vo
V1
Vn
Vo
V1V2
V3
Vn
vo
o iV KV and 0 1K 1 1( )o n nV a V a V 1 1
0( )
t
o n nV a v a v
Third order differential equation
2 1 0 ( ) 0x a x a x a x f t
1 11x
x x x
a2
a1
a0
1
f(t)
Page | 4-38
State Space method
òS S ò òS
a2
a1
a0
- - -
f(t)x
Scaling and Sampling Time
Time – T=ht
h = scaling factor
T = simulation time
t = time between samples
3 2 2 1 0 ( ) 0Th x h a x ha x a x fh
Simply replace 3 by x h x etc.
Time scaling factor of 1
hreplaces 1 on the input of each integrator.
( ) ( )
( ) ( )( )
t kT
y t Fx t
y kT Fx kt
k = which sample
T = interval between samples
Magnitude scaling
Used to get full linear range from an op-amp.
X=kx
X = simulation variable
x = dependent variable
k = scaling factor
max
max
Xk
x
k is chosen and incorporated in each term so that it does not change the equation.
Page | 4-39
3 0 02 2 1 1
3 2 1 0
( )0
m
m
k x a k x f f ta k x a k x
k x k k k f
Simplify equations but keep terms in parentheses together. This is scaled variable X.
Example
0.82 18 250 400 0tx x x e
Initial conditions (0) 35 (0) 10x x
Maximum values max max max35 250 2500x x x
Maximum simulation V = 10 volts
Solution:
Find scaling factors
max2
max
100.004
2500
vk
x
max
max1
max
100.04
250
vk
x
max0
max
100.2
35
vk
x
maxmax
100.025
400
vf
f
Make differential w/ unity coefficient on USB
Divide by coefficient (example = 2)
0.89 125 200 0tx x x e
Substitute scaling factors
0.8200 0.0250.004 9 0.04 125 0.2
00.004 0.04 0.2 0.025
tex x x
Make differential w/ unity coefficient on USB
0.80.9 0.04 2.5 0.2 0.8 0tx x x e
Multiply scaling factor time initial value to obtain scaled initial values
1
0
(0) (0.04)(10) 0.4
(0) (0.2)(35) 7
k x
k x
Page | 4-40
Adders (summers) use round values (0.1, 1, 10)
coefficients are obtained by using multiplier k and
summer value
Example: 2.5(0.2x)
Adjust scale coefficients by integrator coefficient
Example:
Draw differential equation (from above)
100.004x 0.04x
5
0.9
0.25
0.8e-0.8t = force
0.2x
-0.4 7
0.25 10(0.2x)
summer
100.004x 0.04x
Page | 4-41
Integrator Realization – Draw it Out
Locate input x(t) and output y(t) along a line
Place as many integrators as order (n) of system
Place a summer before the output and one before each integrator. Total number of summers = n+1,
or the maximum number of y terms
Use summer to bring signals together. Use connectors for departing signals
Draw forward, x-input signals. Use scalars for gain on each signal. Terminate the signal after as many
integrators as the order. Feedback will prove the differentiation.
Draw feedback, y-output signals. Use scalars for gain on the signal. Terminate the signal after as
many of the integrators as the order. This represents the differentiation.
Draw the main line. Remove all operators that are null. Use as many feed-forward and feedback taps
as necessary to complete the system. If the previous two steps are consistent, this will be one line.
Page | 4-42
Chapter 4 Problems
Problem 4-1 (Old Style)
SITUATION:
The operational amplifier circuit shown below is connected to provide a variable output resistance
at terminals a-b.
-
+
VinRo
Ra
LOAD
a
b
REQUIREMENTS:
Find a Thevenin equivalent circuit as a function of the potentiometer setting α at terminals a-b.
Assume the op-amp to be ideal and that R is much larger than RA.
SOLUTION:
Circuit is a non-inverting amplifier (voltage on + terminal).
Redraw into standard form
Find Thevenin Equivalent voltage (VTH).
Open circuit terminals, leave sources active, calculate voltage across open terminals
( (1 ) )
TH ab
ab A
A
V v
v I R R
I R
abA
vI
R
-
+(1-a)R
RA Vin
aR
b Vo a
Page | 4-43
( (1 ) )
( (1 ) )
( (1 ) )
in A A
abA
inab A
V I R R
vR R
R
Vv R R
R
Find Thevenin Equivalent Impedance (ZTH)
Short voltage sources, open current sources, then calculate series/parallel resistances.
Alternately, short circuit the terminals, leave sources active, then find ISC, TH
TH
SC
VZ
I
inSC
A
VI
R
( (1 ) )
( (1 ) )
in
ATHTH
SC in
A
A
A
RVR RV
RI V
R
RR
R R
Page | 4-44
Problem 4-2 (Old Style)
A FET amplifier composed of three identical stages is shown in the Figure below. It may be assumed
that Cs adequately bypasses Rs. It may also be assumed that there is 40 pF of stray shunt capacitance per
stage. Determine a value of RD such that the mid-frequency amplification is 60dB if
32.5 10mg 8dr k 100gR k
Cb
Rg
Rd
CsRs
Cb
Rg
Rd
CsRs
Cb
Rg
Rd
CsRs
Cb
Rg
REQUIREMENTS:
Determine a value of RD such that the mid-frequency amplification is 60dB.
SOLUTION:
FET Model
ReId
Drain
SourceGate
RgRg CSVin
Cb
Rgout Cstray/stage
Page | 4-45
Rg
VDD
Rd
rd
gmVgsVgin
CsRs
Rgout
Assumptions:
Gain of FET is across rd. If CS shunts RS, then we can simply connect rd to ground.
At the signal frequency, VDD is a short.
Once input and output stages are isolated using model, each can be solved individually.
Rg
Rd
rd
gmVq
Vqin
Cb
Rgout
a
Given:
32.5 10mg 8dr k 100gR k 40strayC pf shunt
Gain relates the output and input voltage.
20logo o
in in
V VGain A db
V V
For gain of 60db overall:
3
60 20log
60log 3
20
10
o
in
o
in
o
in
V
V
V
V
V
V
For 3 stage amplifier, gain for each stage is 10 (10*10*10=103)
Page | 4-46
Circuit analysis:
For an AC signal, the Capacitor bC is a short, which means it conducts.
Calculate the nodal equation at “a”.
31 1 1( 0)( ) 2.5 10N out in
d D g
I V Vr R R
Use the gain relationship to find the resistance.
3
3
1 1 1
1 1 110 ( 2.5 10 )
8 100
18.7
1.15 10
outm
in d D g
D
D
vA g
v r R R
k R k
R k
Page | 4-47
Problem 4-3
In the figure of Problem 4-2 above determine the high frequency cutoff for the entire amplifier if
32.5 10mg 8dr k 100gR k 8.7DR k
SOLUTION
Cutoff for each stage is reciprocal of the time constant. The capacitance is given as the stray:
1
2hi cutoff
sh sh
fC R
40shC stray pf
The resistors are in parallel.
1 1 1 1
4
SH d D g
SH
R r R R
R k
Calculate the frequency.
12 3
10.995
2 (40 10 )(4 10 )hi cutofff MHz
Cutoff for entire amplifier is a root containing the number of stages.
1
1
36 6
2 1
0.995 10 2 1 0.995 10 0.51 0.501
nhi all hi one one
hi all
f f ATF f
f MHz
Page | 4-48
Problem 4-4
In the figure of Problem 4-2 above determine a value of Cb such that the low frequency cutoff for
the entire amplifier is not greater than 200Hz (indicate any further assumptions made) if
32.5 10mg 8dr k 100gR k 8.7DR k
SOLUTION:
Low frequency cutoff for each stage:
1
lo cutoff
series seriesC R ?seriesC
Calculate the equivalent resistance.
511.042 10
1 1series g
d D
R R
r R
The frequency range is reduced by multiple stages.
1
1
13
2 1
2 1
200 2 1 102
n
n
lo onelo all lo one lo all
ff f f
Hz Hz
Cut-off frequency is time constant dependent.
5
1 10.015
2 (102)(1.042 10 )series
lo series
C fR
Page | 4-49
Problem 4-5
A voltage amplifier is shown in the figure below. Determine the high-frequency cutoff. Use
appropriate approximations.
' 20b br ' 500b er ' 0.05b ec equiv F 0.004mg
100kΩ
50kΩ
5kΩ
5kΩ Xc=0
Vo
a a’
SOLUTION:
The circuit is a common emitter BJT amp, draw equivalent hybrid small signal model
Rbb’
Recbe
gmVbe
B
E
C
Input Side
Calculated 1st
Output Side
Use input to
calculate
+
Vbe
-
Ignore R’s which are used for Bias.
Large C will pass anything.
Resistors rb’e & rb’b are in parallel on input.
20*500
2020 500
inr
The input circuit to the transistor is shown at the right.
20
cbe
B +
Vbe
-
Page | 4-50
'
11
1 1b e in
sCV VsCRR
sC
The cutoff freq is inversely related to the time constant.
8
8
1 110 / sec
20 5 10c rad
RC
810
1592 2
f kHz
Page | 4-51
Problem 4-6
In the voltage amplifier shown in Problem 4-6, a 10μH coil is put in place between the a and a'.
Determine the upper frequency cutoff with this coil in place.
' 20b br ' 500b er ' 0.05b ec equiv F 0.004mg
SOLUTION:
The circuit is a common emitter BJT amp, draw equivalent hybrid small signal model.
Rbb’
Recbe
gmVbe
B
E
C
Input Side
Calculated 1st
Output Side
Use input to
calculate
+
Vbe
-
Ignore R’s, they are used for Bias.
Large C – pass anything
Resistors rb’e & rb’b are in parallel on input.
20*50020
20 500inr
The input circuit of the transistor is shown at right.
Use the standard bandpass or resonance equation for a second order
system. The equation can be stated in terms of circuit elements or in terms of
frequency as noted in chapter 2.
2
/( )
1
s LY s
Rs s
L LC
2 2
0
/( )
s LY s
s s
From the admittance, calculate the voltage drop across the capacitor.
12
' 2 6 122
1 2 10
1 2 10 2 10
in in
b e
V VLCV
R s ss sL LC
20
cbe
B
E
+
Vbe
-
10-5H
Page | 4-52
The denominator is the characteristic equation when the input is set to zero. Equate the frequency
form and the circuit element forms to find the frequency.
2 22 n ns s
2 12 61 2 10 1.414 10n nLC
6
0.7072 10
o nR QL Q
1
2n
R
Q L
The 3db cutoff frequency is 225.2n kHz
Page | 4-53
Problem 4-7
The circuit shown in the figure below is a cascade connection of a FET and a junction transistor.
100ieh 10DR k 10dr k 100feh 5s cR R k 310 (FET)mg
You can neglect hoe, hre and all capacitance.
Determine the overall small signal amplification out
in
VV
for this cascode connection. A cascode is a
high frequency connection with an input as a common source ant the output is a common gate.
+
Vin
-
RD
RS
Rc
+
Vout
-
SOLUTION
Draw the small signal models and solve.
Rg
Rd
rd
gmVqVqin
Rs
hie
Rc
hfeib
Page | 4-54
Simplify:
Since d ier h and D ieR h , in a parallel arrangement, you can drop both and leave only hie.
Determine the output voltage.
( )
(1 ( ))
o s s C C
m g fe m g s fe m g c
m g fe s c
v i R i R
g V h g V R h g V R
g V h R R
Assume the forward parameter is not a contribution.
For hfe>>1
( ( ))o m g fe s cv g v h R R
Next determine the input voltage so the gain relationship can be determined.
(1 )
(1 )
1
( )
1
g i m g fe s
in fe m g s
in fe m s g
ing
m fe s
fe m c sout
in m fe s
v v g V h R
V h g V R
V h g R V
VV
g h R
h g R RV
V g h R
Substitute values to obtain the gain.
3
3
( )
1
100 10 (5 5 )
1 10 100 5
2
fe m c soutv
in m fe s
h g R RVA
V g h R
k k
k
RSgmVg
hie
hfeib=-hfegmVg
vo
Rc
Page | 5-1
Chapter 5 - Controls
5.1 Introduction
Controls are predominantly about using transfer functions to describe the response of a system to signal
inputs.
The transfer function is usually a two-port network. The most complex circuit is a second order, which
has been discussed in everything from circuits and power to electronics. By having multiple transfer
functions which result in multiplication, the system transfer function can become very long. Controls
analysis is the process of deciphering these complex functions into a physically realizable performance in
terms of real and angular components.
The standard waveforms for signals are the big three: dc or step, exponential, and sinusoidal.
Special cases of the exponential are also considered. These are impulse and ramp.
Page | 5-2
5.2 Controls Basics
5.2.1 Introduction
Control basics set the profile. The profile includes manipulation of the one-line or block diagram, the
errors, and the overshoot. A characteristic equation is used for stability analysis. The roots of the
numerator are called zeroes while the roots of the denominator are called poles. The gain of the
characteristic equation can be manipulated to improve stability.
5.2.2 Block Diagrams
A block diagram is used as a one-line representation of the process. It consists of three components:
input, R; output, Y; and the system transfer function. For analog systems, the diagram is in the LaPlace
transformed or s-domain. Obviously, this can be transformed directly into the frequency domain.
s j
Transfer Function – Output / Input = F=Y/R
Forward Gain – 1 2G G G
Feedback Gain – H
Unity Feedback – H = 1
Error signal – E R HY
Output – Y GE
Open Loop Transfer – ( )
( )
K s zGH
s p
Closed Loop Transfer – 1
Y GF
R GH
Characteristic Equation – 1 0GH ; 1GH
Denominator – 0
Open Loop 1 1 180
Roots of Characteristic Equation are the eigenvalues
eigenvalues – poles of YR
- closed loop poles
If a system is stable, all roots are in the Left Half Plane
The system is marginally stable if 1 root at origin, or 1 complex conjugate on jω axis
For frequency questions - s j
S G2Gain
G1Plant
HFeedbac
k
EError
RInput
YOutput
+
-
Page | 5-3
Low frequency gain - K (DC, ω=0)
00
( )
( )
Y K j zF
R j p
Bandwidth ( BW ) is at amplitude equal to 12
* low frequency gain
0
1
2BW
BW
Y YF
R R
- Solve for BW
Page | 5-4
5.2.3 Steady State Errors
Steady state errors are the response of the system to inputs. A second order system will have three
error responses; position (t=0), velocity (t), and acceleration (t2). The inputs that cause these responses
are step (DC), ramp (constant slope), and parabolic (exponential).
Use unity feedback & transfer function
Use final value theorem
0 0
( )lim ( ) lim
1 ( )s s
sR sess sE s
G s
The signal must be stable – not oscillating.
Error constants are the results of stability.
Position Velocity Acceleration
constant 0
lim ( )ps
k G s
0
lim ( )vs
k S s
2
0lim ( )s
ka S s
Laplace ds
dt
22
2
ds
dt
Steady state calculation depends on the order of the input.
step ramp parabolic
Time ( ) 1r t t 212
t
s 1( )R ss
21
s 3
1s
ess 0
1
lim1 ( )s
ss
G s
2
0
1
lim1 ( )s
ss
G s
3
0
1
lim1 ( )s
ss
G s
0
1
1 lim ( )s
G s
0
1lim
( )s sG s
20
1lim
( )s s G s
1
1 pk
1
vk
1
ak
0 no erroress
no control, explodesess
0ess as add pure integrators (increase system type)
System type = power of s in denominator
Page | 5-5
5.2.4 Time Response
The time response of a system is the familiar three component DC, exponential, sinusoid.
From this, the time domain positions can be calculated. The waveform chapter in the section RLC
System Response has additional relationships.
Percent overshoot exists in under-damped systems only.
For second order system
211 ( )
% 100( )
e yos
y
( ) 1 for normalized systemy
21(max) 1 for normalized systemy e
Rise time is the amount of time to rise from 10% to 90% of final value
Settling time is the amount of time necessary to reach and stay within a band around the final value
The damping frequency is a shift from the natural or resonant where the wave is dropping off.
2 2d n
( ) ( )% 100 max
( )
y t yos
y
max ( )y t
p
d
nt
( )y
Page | 5-6
5.3 Routh-Hurwitz Criteria
5.3.1 Introduction
Controls problems are the result of multiplications which often yield functions that are higher than
second order. Consequently, traditional solutions become unworkable and alternative approaches are
necessary. Factoring is a possibility, but can be very tedious. The Routh-Hurwitz criteria were developed
to eliminate factoring. The coefficients of a function are used to find stability.
5.3.2 Rules
Determine the function to be analyzed for stability.
Put coefficients of the function in 1st two rows of array – as even/odd powers
Next row – use determinant of 1st column & column to right of location being evaluated
Complete the row (use -1 on determinants)
If have a zero as last element in row, simply move last element of previous row to present row
Interpretation
# of roots in right half plane = # sign changes in first column
Example
4 3 2( ) 6 13 12 4f s s s s s
Create Routh table.
4
3
22,1 2,2
11,1
00,1
1 13 4
6 12 0
11 4
9.8 0
4
s
s
s P P
s P
s P
2,1
1 13
6 121 11
6P 2,2
1 4
6 01 4
6P
6 12
11 41,1 1 9.8
11P
2,1
(1*12 6*13)1 11
6P
2,2
(1*0 6*4)1 4
6P
1,1
(6*4 11*12)1 9.81
11P
5.3.3 Routh-Hurwitz – Special Cases
Page | 5-7
A zero in the first column of the array implies division by zero – not kosher.
Two cases
First Case – zero in first column of row & some non-zero elements in the row.
Replace the zero by a small number (ε) and proceed.
Second Case – all elements in row are zero – This occurs if roots are on jω axis or roots are symmetrical
on axis about origin – ( p root ).
Form auxiliary polynomial fa using coefficients of row before zero row.
Take derivative of auxiliary polynomial and use as substitute for zero row.
Derivative gives max/min when zero.
Example 1
Determine the stability of the function.
3 2( ) 2 2f s s s s
Develop the Routh array from the coefficients.
3
2
11,1
0
1 1
2 2
4 0
2 0
s
s
s p
s
1,1
1 1
2 21 0
2p - would give zero row for s1
Instead, make fa using row above
2
1,1
2 2
4
4
a
a
f s
dfs
ds
P
Interpretation – first column positive – no sign change
no poles in RH plane, - stable
Example 2
Page | 5-8
Determine the value of k for stability.
ΣRInput
Y(s)Output
+
-
2
2
1
( 1)( 2 2)
k s
s s s
Closed Loop Characteristic Equation - 1 0GH
In this problem, H=1.
2
2
( 1)1 1 [1] 0
( 1)( 2 2)
k sGH
s s s
Expand to obtain the stability function.
RInput
Y(s)Output
1
G
GH
2 2
3 2
( 1)( 2 2) ( 1)
( 3) (4 2 ) ( 2)
f s s s s k s
s k s k s k
Create Routh table.
3
2
11,1
00,1
1 4 2
3 2
s k
s k k
s P
s P
To be stable, the first column must have no sign changes. In this case the signs remain positive.
2,1 0 3 0P k 3k
0,1 0 2 0P k 2k
2
1,1
2 3 100 0
3
k kP
k
Denominator will be positive if ( 3k ). For the coefficient to be positive, the numerator must be >0
22 3 10 0k k
Solve using quadratic equation.
0,1 2P k
2
1,1
1 4 2
3 2 2 3 10( 1)
3 3
k
k K k kP
k k
Page | 5-9
( 3) 9 (4)( 2)(10)
2( 2)
3 89
4
3.11&1.61
3.11 1.61
k
k
Combine all the constraints.
2 1.61k
If k = limits, have poles on jω @ the limits
If k outside limits, have instability
Page | 5-10
Example 3
For the problem above, what are the poles on the jω axis?
ΣRInput
Y(s)Output
+
-
2
2
1
( 1)( 2 2)
k s
s s s
Use Routh table with the k used to calculate the limits.
When k=-2 when k=1.61
3
2
1
0
1 8
1 0
8
0
s
s
s
s
3
2
1
0
1 1.68
4.61 3.61
0
3.61
s
s
s
s
If any rows are zero, develop an alternative function.
s0 row = zero row s1 row – zero row
8
@ 0
8( ) 0
0
a
a
f s
s j f
j
2
*
* 2
4.61 3.61
@ 0
4.61( ) 3.61 0
a
a
f s
s j f
j
The frequencies of the roots are available.
@ origin, 0s j 2 3.61
4.61j
3.61
4.61
0.7831
0.885s j j
Page | 5-11
5.4 Root Locus
5.4.1 Introduction
Root Locus is the process of determining the location of all the roots to a feedback function. The
location of the roots will change as the value of the gain, k changes. By adjusting k, the stability can be
improved or degraded. The function that is investigated is the characteristic equation, which is the
denominator of the transfer function.
Closed Loop Transfer Function 1
Y G
R GH
Open Loop Transfer Function – GH
When H=1. the open loop is the plant, G
Characteristic Equation
1 0GH
Solving characteristic equation implies 1GH
1GH
,( 1,3,5,7)
(2 1),( 0, 1, 2)
GH m m
or
GH k k
Adjust the gain, k, so the open loop transfer is one. Substitute the actual function values for GH and set
the equation to one.
1 2
1 2
( )( )1
( )( )
k s z s zGH
s p s p
Calculate the system gain, K.
product of vector length of poles
product of vector length of zeroesK
2( 1)GH s zi s pi k
Page | 5-12
5.4.2 Object of Root Locus
The object of the root locus process is to determine the movement of the roots as k changes. The
analysis goes through this development.
Make plot of open loop function for 0k to k .
See where the function is stable & unstable. Avoid or fix unstable points.
Make sketch of significant points to draw.
Use the six rules to aid in the construction.
Page | 5-13
5.4.3 Rules for Root Locus Construction
Each rule provides a significant factor to assist in the sketch of the roots or eigenvalues. The sketch then
shows the stability as the roots change due to the gain, k being adjusted.
1) Starting and Ending Points Root Locus plots start ( 0k ) on the open-loop poles and
end ( k or k ) on the open-loop zeros
2) Root Locus Segments on the Real Axis Root loci occur on a particular segment of the real axis if and only if there are an odd number of total poles and zeros of the open-loop transfer function lying to right of that segment.
3) Imaginary Axis Intersections Use the Routh-Hurwitz criterion to determine j axis crossings of the root locus plots. Both the
gain k and the value of ω* may be found from the Routh table.
4) Asymptotes (for p z )
Root locus plots are asymptotic to straight lines with angles given by
(2 1)
( )A
k
p z
as s approaches infinity. These straight lines intersect at a point 1 on the real axis specified by:
1
poles of ( ) zeros of ( )GH s GH s
p z
Where p is the number of finite poles of GH(s) and z is the number of finites zeros of GH(s).
5) Angles of Departure and Arrival Assume a point s arbitrarily near the pole (for departure) or the zero (for arrival) and then apply
the fundamental angel relationship
angles of zeros of ( ) angles of poles of ( ) (2 1)GH s GH s k
6) Breakaway Points Breakaway points may be determined by expressing the characteristic equation for the gain k as a function of s and then solving for the breakaway points sB from
( )
0
Bs s
dk s
ds
Page | 5-14
Example 1
Sketch the root locus plot for the system in the diagram.
ΣRInput
Y(s)Output
+
-
( 1)( 2)
( 4)
k s s
s s
Set up the open loop function.
( 1)( 2)
( 4)
k s sGH G
s s
Plot the locus from poles to zeroes.
X – poles – denominator
O – zeros – numerator
Use rule 1, 2
Stay on axis only when have odd # of p&z to right
Example 2
Plot the root locus from the plant and feedback functions.
( 2) ( 3)
( 10) ( 1)
k s sG H
s s
The location of the poles and zeroes are obvious. However, the path is somewhat more tedious.
Obviously the path needs to cross from the right to
the left plane. Use Routh with the characteristic
equation which contains the open loop transfer
function.
1 0GH
Substitute the values.
Set up the Routh table.
2 2
2
( 1)( 10) ( 2)( 3) 0
9 10 5 6 0
( 1) (5 9) (6 10) 0
s s k s s
s s ks ks k
k s k s k
k=0k=k=k=0
-1-2-4
k=0
-1-2-3 10
2
1
0
1 6 10
5 9
6 10
s k k
s k
s k
Page | 5-15
Make s1=0 with 95
k .
Develop the auxiliary function fa from the row s2
2( 1) (6 10)af k s k
Determine the frequency for the crossing the axis, jω*
*( 1)( ) (6 10) 0k j k
A complex conjugate results in complementary frequencies. When 95
k , the root locus crosses the
vertical axis.
2
95
6 10
1 k
k
k
0.54
Make s0 =0 for 106
k
When 106
k , the root locus crosses jω @ 0 on its path to the right half plane, RHP.
The root locus moves from one pole in the LHP, crosses into the RHP at 0 . It collides with a pole in
the RHP. The path then comes back @ 95
k in complex conjugate pairs. It stays close to the origin.
The track should be circle, since make polynomial is s2. If the roots are s3 or higher, the path becomes an
ellipse.
k=0
-1-2-3
Example 3
Develop the root locus for an open loop transfer function.
Page | 5-16
2( 2)( 2 2)
kGH
s s s
The second order poles can be factored.
( 2)( 1 )( 1 )
kGH
s s j s j
Observations can be made about the expected results.
Not all poles are real. jω
p>z asymptotes
complex angle of departure
Follow the rules to develop all the plotting parameters.
Rule 2 – on real axis
Rule 3 – not all real, use Routh to find k & jω
21 ( 2)( 2 2) 0GH s s s k
3 21 4 6 (4 ) 0GH s s s k
Make the Routh table.
3
2
1
0
1 6
4 4
20
4
4
s
s k
ks
s k
1,0
1 6
4 4 20
4 4
k kP
Determine the frequency to make a row zero.
1 0 @ 20s k
24 (4 )af s k
* 24( ) 24 0j
6 @ 20k
Rule 4 - p z 3 0 3 3zeros @ p z
Page | 5-17
, ,3 33 0
A
Intersection
( 2) ( 1 ) ( 1 ) 0 4
3 0 3i
poles zeros j j
p z
Rule 5 - The complex conjugates need to 180 to all other poles & zeros
Pole @ 1 902
s j
zeros of GH - poles of GH=(2 )k
from -1-j = 902
from -2= 454
All these are s from poles, from local pole = x
(0) ( ) (2 )2 4x k
4x
, the θ for conjugate is complex
Plot the calculated points and draw the root locus from poles to zeroes.
-1- j
-4/3-2
-1+ j
Page | 5-18
5.5 Frequency Response Plots
5.5.1 Introduction
Frequency plots indicate the performance of a system over a range of frequencies. The plots can be
linear, semi-log (Bode), log-log, or polar plots (Nyquist). All these tools have the same information, and
are used to determine relative stability.
In contrast, the transition from the RHP using Routh- Hurwitz gives absolute stability in a limited system
without time shift.
A comparison of the costs and benefits of each response plot assists in selecting the one most
appropriate for a particular problem.
Linear – direct, but cumbersome math
Bode – easiest. Use log10 of open loop frequency response
Magnitude values in decibels (db) are plotted on the vertical
1020log ( )G j
Perform for each pole & zero @ K, then add the sketches.
Phase angle is plotted under the magnitude plot
Start w/ ( )G j in standard form
2
2
1( ) ( )
( ) (1 2 ( )
n
n n
j
ms j j
kG j G s
j
Plot the asymptotes.
These differ from actual values by 3db @ limits (ωn is corner frequency)
Slope of asymptotes = 20*m on the db/decade
Slope is negative for poles and positive for zeros.
m is the exponent power of s associated with the root.
Determine the phase angle @ the values.
1tann
90 *m for all ω
At the root, cross the 45o angle on the way from 0o two decades before the root in route to 90o two
decades after the root.
Page | 5-19
5.5.2 Bode Plots – Basic rules
The basic rules for Bode are quite short. The examples are better illustrations than a wordy explanation.
1) Magnitude – make break @ ωn, plot asymptote 20*m db/decade, where m = power of s.
2) Phase angle – make 45 break @ ωn plot from 2 decades before to 2 after the root.
3) Break down for poles and up for zeros
4) Plot 2 decades before to two decades after
5) Σ plots to get effective
Example 1
Make a frequency plot of a plant function.
1000( 1)
( )( 2)( 10)( 20)
s j
sG s
s s s
Convert the form to a magnitude and phase angle.
2
2 2 2 2 2 2
1000 1( )
2 10 20G j
1 1 1 1( ) tan tan tan tan1 2 10 20
G j
The above equations are used for the linear plots.
Example 2
Make a frequency plot of a plant function above.
Convert the function to standard form for Bode & log-log.
1000(1)1
(2)(10)(20) 1( )
1 1 12 10 20
j
G jj j j
1000(1)
2.5(2)(10)(20)
K
Page | 5-20
20 log(2.5) = 8, DC portion – normalized
Magnitude is plotted for each root.
Break up + 20db for zero @ 1
Break down -20db for poles @ 2, 10, 20
@ DC, response is8 0 , start phase diagram at 0°
Phase angle is plotted for each root.
+45˚ break points for zero @ ω=1
-45˚ break points for poles @ ω = 2, 10, 20
Combine the magnitude with 3 db smoothing @ corners (1,2,10,20)
1
rad/sec0 dB
-40 dB
20 dB
rad/sec
0 dB
-180°
1 100 10k
+45°
10 10000.1
10 1000
1000100
Combine the angles with a 5.7o smoothing at each break.
Page | 5-21
Page | 5-22
5.5.3 – Phase Margin and Gain Margin
The phase margin and gain margin of a system are indicators of the relative stability of the system. The
phase margin basis of the system is the absolute value of the difference between the actual phase angle
and -180° when the gain is 0dB.
phasemargin ( 180 ) @0dB
The gain margin basis of the system is the absolute value of the difference between the actual gain and
0dB when the phase angle is -180°.
gain margin ( ) 0G j dB @-180°
Example:
For the system response shown below, determine the phase margin and the gain margin.
0.01
rad/sec0 dB
-80 dB
40 dB
0.1 100.001 1001
-160 dB
0.01
rad/sec0 dB
-90°
0.1 100.001 1001
-180°
At approx 0.01 rad/sec, the gain crosses 0dB. At that frequency, the phase angle is -90°
0.01
phasemargin 90 ( 180 ) 90
At approx 10 rad/sec. the phase angle crosses -180°. At that frequency, the gain is approx -70dB.
10
gain margin 70 (0 ) 70dB dB dB
Page | 5-23
5.5.3 Polar - Nyquist Plot.
Re ( )G j vs Im ( )G j
Use the Bode Plot from the previous example to determine points on the Plot. Only a couple of points
are needed to determine the overall curve
The following steps outline the process.
@ ω=0
(8/20)10 2.5
0
G
G
Plot the point (2.5,0)
At the point that the phase = 0 real axis, called ωpc (phase crossing)
@= 0
(12/20)10 3.98
0
G
G
Plot the point (3.98,0)
Page | 5-24
At the point when 1G , this is called ωgc , arc from origin
@ 1G
01 10
125
G
Plot the point (-0.574,-0.819)
If zeros of |G| in left HP, called minimum phase
Plot the mirror image for complex conjugate pairs
Page | 5-25
5.6 Analog Filters
Filters are classified according to the order of the transfer function, and according to function. The most
common functions are low-pass, high-pass, band-pass and band-reject filters. Other types are phase-
lead and phase-lag. A chart of filter information, including transfer functions, system response and
circuits are shown below for some of the most common type filters.
First Order Low-Pass Filters
Frequency Response
ωc0 ω
H
1( ) (0)
2cH j H
RC Voltage Source (Thevenin)
v1
R1
C R2
+
v2
-
2
1 1
1( )
1
eq
eq
RvH s
v R sR C
1 2
1 2
eq
R RR
R R
1c
eqR C
RL
v1
R1
R2
+
v2
-
L
2 2
1
1( )
1eqeq
v RH s
Lv R sR
1 2eqR R R eq
c
R
L
RC Current Source (Norton)
R1 R2C
i2
i1
2
1 1
1( )
1
eq
eq
RiH s
i R sR C
1 2
1 2
eq
R RR
R R
1c
eqR C
Page | 5-26
First Order High-Pass Filters
Frequency Response
1( ) ( )
2cH j H j
ωc0 ω
H
RC
v1
R1
R2
+
v2
-
C
2 2
1
( )1
eq
eq eq
sR Cv RH s
v R sR C
1 2eqR R R 1
c
eqR C
RL Voltage Source (Thevenin)
v1
R1
R2
+
v2
-L
2
1 1
( )1
eq eq
eq
LsR Rv
H sLv R s
R
1 2
1 2
eq
R RR
R R
eq
c
R
L
RL Current Source (Thevenin)
R1 R2L
i2
i1
2
1 1
1( )
1
eq
p
RvH s
v R sR C
1 2
1 2
eq
R RR
R R
1c
pR C
Page | 5-27
Band-Pass Filters
Frequency Response
ωo0 ω
H
ωL ωU
0
1( ) ( ) ( )
2L UH j H j H j
3-dB Bandwidth = u LBW
Series
v1
R1
R2
+
v2
-
CL
2
21 1
1( )
1eq
v sH s
Rv R Cs s
L LC
1 2eqR R R 1
0LC
2 2
1 2
( )o
eq
R RH j
R R R
eqRBW
L
Parallel
v1
R1
R2
+
v2
-
L C
2
21 1
1( )
1eq
v sH s
v R C s sR CLC
1 2
1 2
eq
R RR
R R
10
LC
2
1 2 1
( )eq
o
RRH j
R R R
1
eq
BWR C
Page | 5-28
Band-Reject Filters
Frequency Response
ωo0 ω
H
ωL ωU
1( ) ( ) 1 (0)
2L UH j H j H
3-dB Bandwidth = u LBW
Series
v1
R1
R2
+
v2
-
C
L
2
2 2
21
1
( )1
eqeq
sv R LCH ssv R s
R C LC
1 2eqR R R 1
0LC
2 2
1 2
(0)eq
R RH
R R R
eqRBW
L
Parallel
v1
R1
R2
+
v2
-
L
C
2
2
21 1
1
( )1
eq
eq
sRv LCH sRv R
s sL LC
1 2
1 2
eq
R RR
R R
10
LC
2
1 2 1
(0)eqRR
HR R R
1
eq
BWR C
Page | 5-29
Phase-Lead Filters Phase-Lag Filters
ω1 ωm ω2
ω
log|H(jω)|
ωm0 ω
φm
ω1 ω2
ω1 ωm ω2
ω
log|H(jω
)|
ωm0 ω
φm
ω1 ω2
v1
R2
+
v2
-C
R1
2 1 1 1
1 1 22
11
( )1 1
eq
eq
sRv sR C
H ssv R sR C
1 2
1 2
eq
R RR
R R
1
1
1
R C 2
1
eqR C
1 2m max ( )m mH j
2 1 2 1
1 2
arctan arctan arctan2
m
m
1
1 2
(0)eqR
HR
1
2
mH j
( ) 1H j
v1
R1
R2
+
v2
-
C
2 2 2
1 1
1 1( )
1 1
eq
eq eq
Rv sR C sR CH s
v R sR C sR C
1 2eqR R R 1
1
eqR C 2
2
1
R C
1 2m min ( )m mH j
1 2 1 2
2 1
arctan arctan arctan2
m
m
(0) 1H 1
2
mH j
2 1
2
( )eq
RH j
R
Page | 5-30
Chapter 5 Problems
Problem 5-1
The circuit shown below is constructed
from an ideal op-amp (infinite gain
bandwidth product)
RC1 -
+
R
C2
Vout
Vin
The phase shift between input and output signals, as used
below, does not include the inversion (180°) from the
operational amplifier circuit.
20R k
1 100C pF
2 0.01C F
5.1.1 Which of the following is most
correct?
A) This is not a high-pass active filter.
B) The input circuit time constant is 2 μS.
C) The feedback circuit time constant is
0.2mS.
D) The phase of the output lags the phase
of
the input signal
5.1.2 The poles and the zeros of the transfer function
are approximately
A) One pole at 5 kHz and one zero at 500 kHz
B) Two zeros, one at 800 Hz and one at 80 kHz
C) Two poles, one at 5 krad/sec and one at
500krad/sec
D) A paired pole and zero at 55 krad/sec
E) One pole at 800 Hz and one zero at
500krad/sec
5.1.3 The dc response (dB) is
A) 20
B) 0
C) -20
D) -30
E) -40
5.1.4 The response (dB) at 10 MHz is most nearly
A) 10
B) 0
C) -10
D) -20
E) -40
Page | 5-31
5.1.5 The frequency at which the response
is -20dB is most nearly:
A) 0 Hz
B) 800 HZ
C) 500 kHz
D) 50 krad/sec
E) 80 krad/sec
5.1.6 The maximum phase shift between input and
output signals occurs under approximately which of the
following conditions?
A) When the response is minimum
B) When the response is equal to the square root of the
ratio of the capacitances
C) When the response is equal to the ratio of the time
constants
D) When the response is maximum
E) At a frequency that is twice that of a pole
5.1.7 The phase shift is approximately half
of its maximum value under which of the
following conditions?
A) When the response is -20 dB
B) When the response is equal to the
square root of the ratio of the capacitances
C) When the response is equal to the ratio
of the time constants
D) At the frequencies of a zero and a pole
E) At a frequency that is half that of a zero
5.1.8 The phase shift is 90° under which of the following
conditions?
A) When the response is -10dB
B) When the response is equal to the square root of the
ratio of the capacitances
C) When the response is equal to the ratio of the time
constants
D) At the frequencies of 800Hz and 80kHz
E) Never
5.1.9 The phase shift is approximately 5%
of its maximum value under which of the
following conditions?
A) When the response is -10dB
B) When the response is equal to the ratio
of the capacitances
C) When the response is a maximum
D) At the frequencies of 500 rad/sec and 5
Mrad/sec
E) At a frequency that is half that of a zero
5.1.10 Which of the following is true of this circuit?
A) Its sinusoidal response is maximum at the pole and
minimum at its zero
B) Its response is maximum when it is equal to the ratio
of the square of the capacitances
C) Its response is maximum at dc and minimum at very
high frequency
D) Its response increases with increasing frequency from
its value at dc
E) Its output phase leads the input signal phase
Page | 5-32
SOLUTION:
Calculate the gain function.
2 1
2
1
11
2
2
5
3
11
11
1
1
5 100.01
5 10
out
in
RRsCV RsC
V RsCRRsC
SRCC
CS
RC
s
s
Convert to standard form in terms of frequency.
5) 5 5
3
3 3
1 10.01(5 10 5 10 5 10
15 10
1 15 10 5 10
out
in
j j
V
j jV
Observe poles and zeros.
Pole at 5x103 rad/sec = 796Hz – Break Down
Zero at 5x105 rad/sec = 7.96kHz – Break Up
Determine the DC valued from K.
1; 20log 0K K , DC response is 0dB
@ DC, the gain is 1 1 180 . The starting point for the Phase diagram is 180°.
Plot gain and phase angle asymptotes.
Page | 5-33
500 5x103 5x104 5x105 5x106
500 5x103 5x104 5x105 5x106
79.6 796 7.96k 79.6k 796k
rad/sec
Hz
0 dB
-40 dB
180°
90˚
-
3dB
-37dB
-5.7°
101.4°
5.1.1
This is not a high-pass active filter. The correct answer is A)
5.1.2
Zero at 5x105 rad/sec (79.6kHz)
Pole at 5x103 rad/sec (796 Hz)
The correct answer is E)
5.1.3
By inspection @ DC, the gain is 0 dB. The correct answer is B)
5.1.4
At 10 MHz, by inspection, the gain is -40dB. The correct answer is E)
5.1.5
By inspection, -20dB occurs at 50krad/sec (7.96kHz). The correct answer is D)
5.1.6
By inspection, max phase shift occurs at -20dB (0.1), which is 1 2C C . The correct answer is B)
Page | 5-34
5.1.7
By inspection, the phase shift is approx. half its max value at the zero and the pole. The correct answer is
D).
5.1.8
The phase shift never reaches 90°. The correct answer is E)
5.1.9
The phase shift is approx 5% of its maximum value at one decade above the zero frequency (5x106
krad/sec) and one decade below the pole frequency (500 rad/sec). The correct answer is D)
5.1.10
By inspection, the response is maximum at DC and minimum at very high frequencies. The correct
answer is C).
Page | 5-35
Problem 5-2
The response to a unit step test is shown in the
figure below.
0.58152
0.5
0.9069
Data List
Peak value of unit step response – 0.58152
Steady-state value of unit step response – 0.5
Time to peak value of unit step response – 0.9069
sec
5.2.1 Data from the data list that are necessary
and sufficient to determine the percent
overshoot are:
A) 1 only
B) 2 only
C) 1 and 2 only
D) 1 and 3 only
E) 2 and 3 only
5.2.3 The expression for the time to peak in terms
of damping ratio and natural frequency of
oscillation is
A) 21
p
n
t
B) 1
p
n
t
C) 1
p
n
t
D) 2
1
1p
n
t
E) p
n
t
5.2.2 The percent overshoot is most nearly
A) 0%
B) 0.163%
C) 16.3%
D) 58.2%
5.2.4 The dc gain is most nearly
A) 0
B) 0.5
C) 1
D) 2
Page | 5-36
E) 100% E) 8
5.2.5 Data from the data list that are necessary
and sufficient to determine the damping ratio are
A) 1 only
B) 2 only
C) 1 and 2 only
D) 1 and 3 only
E) 2 and 3 only
5.2.7 The natural frequency of oscillation n is
most nearly
A) 1
B) 1.414
C) 2
D) 4
E) 16
5.2.6 The damping ratio is most nearly
A) 0
B) 0.5
C) 1
D) 2
E) 8
5.2.8 The transfer function of the open loop system
is
A) 2
8
8 16s s
B) 2
0.5
8 16s s
C) 2
1
4 16s s
D) 2
8
4 16s s
E) 2
0.5
4 16s s
Page | 5-37
SOLUTION
5.2.1 max ( ) ( )
% overshoot = 100( )
y t y
y
Need 1 and 2 only. The correct answer is C)
5.2.2 0.58152 0.5
% overshoot = 100 16.304%0.5
. The correct answer is C)
5.2.3 21
p
n
t
. The correct answer is A)
5.2.4 The dc gain is the steady state value divided by the input (unit step), or 0.5. The correct answer is
B
5.2.5 The damping ratio may be found from the equation 21
% overshoot = 100e
. The data needed
are 1 and 2 only. The correct answer is C)
5.2.6 210.58152 0.5
% overshoot = 100 16.304% 1000.5
e
.
21
2
2 2 2
2 2
2
1006.135
16.304
ln 6.135 1.8141
3.29(1 )
( 3.29) 3.29
0.25; 0.5
e
The correct answer is B)
5.2.7
2
2
1
1
4.0 / sec0.9069 0.75
p
n
n
p
n
t
t
rad
The correct answer is D)
5.2.8 Form of the equation is 2
2 22
n
n n
Ks
2
160.5
4 16s s
The correct answer is D)
Page | 5-38
Problem 5-3
A control system has an open loop response of 2
8
4 16s s . A tachometer feedback system is installed
in the control system in order to make the system critically damped.
5.3.1 The rate gain needed for the tachometer
feedback to achieve a critically damped system is
most nearly
A) 0.125
B) 0.5
C) 1
D) 2
E) 4
5.3.2 The location of the critically damped poles is
most nearly
A) -16
B) -4
C) 0
D) 4
E) 16
SOLUTION
The block diagram, with the tachometer (rate) feedback is shown
5.3.1
The transfer function is developed in the circuit
element form.
.
The frequency form of the second order bandpass or resonance equation is 2
2 22
n
n n
Ks
.
Set the denominator characteristic equation in circuit form to the equivalent frequency form. Then
calculate the frequency variables.
.
For critical damping, ζ=1. Therefore, the gain b can be calculated.
The correct answer is B)
2 2
( ) 8 8
( ) 1 4 16 8 (4 8 ) 16T T
Y s G
X s GH s s K s s K s
2 16 4n n
4 8 2T nK
4 8 2 2(1)(4) 0.5T n TK K
X(s)Input
Y(s)Output
+
-2
8
4 16s s
TK s
S
Page | 5-39
5.3.2
The critical damped transfer function is
2 2
8 8
8 16 4s s s
. Therefore the poles are -4, -4.
The correct answer is B)
Page | 5-40
Problem 5.4
A control system is described by the block diagram
X(s)Input
Y(s)Output
+
-2
4
6 13
s
s s
Σ ( )cG s
controller
Data List:
Gs(s)=2
4
6 13
s
s s
Gc(s)=( )
( 1)
K s z
s s
Possible values of z = -3 or -5
5.4.1 The open loop transfer function is
A) K
B) ( )
( 1)
K s z
s s
C) 2
( )( 4)
( 1)( 6 13)
K s z s
s s s s
D) 2( 1)( 6 13)s s s s
E) 2
4
6 13
s
s s
5.4.4 The angles the asymptotes of the root loci
makes with the real axis of the s plane are
A) ±0°
B) ±90°
C) ±180°
D) ±270°
E) ±360°
5.4.2 The poles of the open loop transfer function
are:
A) z, -4 only
B) 0, -1 only
C) -3±j2 only
D) 0,-1, -3±j2 only
E) z, 0, -1, -3±j2, -4
5.4.5 Data from the data list that are necessary
and sufficient to determine the intersection of
the root loci with the real axis of the s plane are
A)1 only
B) 1 and 2 only
C) 1 and 3 only
D) 2 and 3 only
E) 1, 2, and 3
Page | 5-41
5.4.3 The zeros of the open loop transfer function
are:
A) z, -4 only
B) 0, -1 only
C) -3±j2 only
D) 0,-1, -3±j2 only
E) z, 0, -1, -3±j2, -4
5.4.6 The intersections of the root loci with the
real axis of the s plane, if 3 5z and are
A) 0 and 0
B) 0 and 1
C) 0 and +1 and -1
D) 1 and 1
E) -3 and -5
5.4.7 The characteristic function if z = -3 is
A) 4 3 27 19 13s s s s
B) 4 3 27 20 17s s s s
C) 4 3 27 20 20 12s s s s
D) 4 3 27 34 118 180s s s s
E) 4 3 27 (19 ) (13 7 ) 12s s K s K s K
5.4.9 What is the first term in the third row
Routhian array if z=-3?
A) (156+12K2)
7
B) 120
7
C) 19 K
D) 13 K
E) 12K
5.4.8 Which of the following values of z provide for
a stable system for any value of K?
A) -3 only
B) any z>-5
C) -3 and -5 only
D) any z>-3
E) Neither -3 nor -5
5.4.10 For with values of K is the system stable is
z = -3
A) All K>0
B) All K<0
C) All large K
D) All small K
E) All K<15
Page | 5-42
SOLUTION:
5.4.1
The open-loop transfer function ( )
( ) ( )( )
c s
Y sG s G s
X s
2
( )( 4)
( 1)( 6 13)
K s z s
s s s s
.
The correct answer is C)
5.4.2
The poles of the open lop transfer function are the roots of 2 6 13 1s s s s
Roots are 0, -1, 3 9 13 =0, -1, 3 2j
The correct answer is D)
5.4.3
The zeros of the open loop transfer function are the roots of ( )( 4)K s z s . Roots are z= -4
The correct answer is A)
5.4.4
Asymptotes – angle = (180 360 )
p z
k
n n
0, , p zk n n ; 4; 2p zn n . Therefore the angle = 90
The correct answer is B)
5.45
- c
p z
poles zeros
n n
H(s), Gc(s) and the value of z are needed.
The correct answer is E)
5.4.6
Page | 5-43
0 ( 1) ( 3 2) ( 4) - 1.5 0.5
4 2c
p z
j zpoles zerosz
n n
For 3, 0cz . For 5, 1cz
The correct answer is B)
5.4.7
2
2
( 1)( 6 13) ( )( 4)1 ( )
( 1)( 6 13)c
s s s s K s z sG s H
s s s s
The Characteristic equation = 4 3 27 (19 ) 13 (4 ) 4s s K s K z s Kz
For 4 3 23, 7 19 13 (4 ) 12z s s K s K z s K
The correct answer is E)
5.4.8
0 ( 1) ( 3 2) ( 4) - 1.5 0.5
4 2c
p z
j zpoles zerosz
n n
For 0c , there are poles in the right half of the s-plane – unstable.
The correct answer is D)
5.4.9
For z=-3, the Routh array is
2,1
19 (13 7 ) 120
7 7
K KP
2,2 12P K
2,2
1,1
2,1
13 7 713 7 4.9 13 2.1
K PP K K K
P
0,1 2,2 12P P K
The correct answer is B)
5.4.10
4
3
22,1 2,2
11,1
00,1
1 19 12
7 13 7 0
s K K
s K
s P P
s P
s P
Page | 5-44
From the Routh array above, 2,1 0P , 1,1 0 6.19P for K , 0,1 0 0P for K .
Therefore, the function is stable for all 0K .
The correct answer is A)
Page | 6-1
Chapter 6 Digital
6.1 Introduction Digital systems are the language of computers and microprocessor hardware. A basic understanding of
the principals associated with digital systems are presented.
6.2 Binary (Digital) Systems
6.2.1 Number Systems
Decimal numbers
When using numbers, we usually write only the coefficient and let the position indication the power of
10. The coefficient range = 0-9 for base 10
Binary numbers
The coefficient range = 0-1 for base 2
3 2 1 02 101010 1 2 0 2 1 2 0 2 10
Octal numbers
Octals are simply binary numbers combined into groups of 3 ( base 8=23)
388 2 1| 001 12
The coefficient range for octal numbers is 0-7
3 2 1 0
107392 7 10 3 10 9 10 2 10
Page | 6-2
Hexadecimal numbers
Hexadecimal numbers are binary numbers that are combined into groups of 4 (base 16 = 24)
41616 2 0 |1010 0A
The range for hexadecimal numbers is 0-F.
Hex # Decimal Equiv Hex # Decimal Equiv
0 1 8 9
1 2 9 10
2 3 A 11
3 4 B 12
4 5 C 13
5 6 D 14
6 7 E 15
7 8 F 16
Page | 6-3
Arithmetic
Arithmetic rules for all bases have the same basic set of rules.
Addition
111
101
1100
Subtraction
111
101
010
Multiplication
111
101
111
000
111
100011
Decimal to Binary Equivalence
0 0000 5 0101
1 0001 6 01101
2 0010 7 0111
3 0011 8 1000
4 0100 9 1001
Page | 6-4
6.2.2 Binary System Wiring
Equivalent Names
5V = 1 = high = True = on = closed
0V = 0 = low = False = off = open
Resistors
Pull-up – from power source –
2.2KΩ
Current limiting – in series with load
330Ω
Switch
When switch = open, output = 1 (True)
When switch = closed, output = 0 (False)
Called an inverting switch
2.2kΩ
output
5 Volts
switch
Transistor
Simply a digital switch with an electrical
input (inverter)
When base switch = off, output = 1
(True)
When base switch = on, output = 0
(False)
2.2kΩ
output
330Ω
5V
Page | 6-5
Output – LED
Generally connected to ground
Use current limiting resistor
LED Long leg is the ground
This is a common cathode arrangement.
A common anode would have inverted
diodes
330Ω
5 Volts
330Ω
TTL
Standard Arrangement
5V Power, VCC, on upper right pin
Ground on lower left pin
Notch at top
a11
a22
3a3
4a4
b1
b2
b3
b4
14
13
12
11
a15
a26
7a3
b1
b2
b3
10
9
8
5 volts
Problem:
Wire a switch circuit with an LED connected to the output pin.
Page | 6-6
6.2.3 The Huntington Postulates
Required entities to define an algebra: A defined group of coefficients (R, N, Q, Z [0,1]), Real Natural, Complex Binary
A defined group of operators and a table which defines how each operator works.
A defined group of axioms or postulates (unproven theorems) from which new theorems,
lemmas, corollaries, and propositions may be constructed.
[0,1] = B (coefficients)
OR AND NOT
For all x.y elements of group B: for each operator
P1 (a) x y B (b) *x y B (closure) only 1 or 0
P2 (a) 0x x (add Identity, I=0) (b) *1x x (multiply Identity, I=1) P3 (a) x y y x (b) * *x y y x (commutative)
P4 (a) *( ) ( * ) ( * )x y z x y x z (b) ( * ) ( )*( )x y z x y x z
(distributive)
P5 (a) 1x x (b) * 0x x (complement, unique) P6 (a) B contains at least 2 distinct elements
Useful theorems: listed in pairs that have correspondence – duality – every expression is valid if operator and identity elements are changed. To find the dual, exchange + to * and 1 to 0.
T1 (a) x x x (b) *x x x T2 (a) 1 1x (b) *0 0x
T3 (a) ( )x x (involution)
T4 (a) ( ) ( )x y z x y z (b) ( ) ( )x yz xy z (associative)
DeMorgan’s Theorem – complement function by interchanging AND & OR operators and complementing each literal
T5 (a) ( )x y x y (b) ( * )x y x y
T6 (a) x xy x (b) ( )x x y x (absorption)
T7 (a) x x y x y (b) ( )x x y xy
Operator Precedence
Parentheses () Not AND OR
Complement – take dual and complement each literal
+ 0 1
0 0 1 1 1 1
* 0 1
0 0 0 1 0 1
~
0 1 1 0
Page | 6-7
6.2.4 Basic Digital Gates
NAND
X
Y
Z
F=(xyz)’
And Invert
XOR
F=x’+y’+z’ = (xyz)’xyz
Invert Or
NOR
X
Y
Z
F=(x+y+z)’
Or Invert
XAND
F=x’y’z’=(x+y+z)’x
yz
Invert And
A one input gate acts like an inverter
To use NAND requires Sum of Products (SOP) form
To use NOR requires Product of Sums (POS) form
Page | 6-8
6.3 Karnaugh Maps
Definitions: Literal a variable of the problem Don’t Care Output values for which no value is necessary (situations that never occur, or if
they do, can be ignored SOP Abbreviation for Sum of Product POS Abbreviation for Product of Sum Reflected Code A binary code which has the property that the codes following every 2n-1 codes
are the reflection of the first 2n-1 codes in all bits except the most significant. The MSB of the first 2n-1
is a zero, and the MSB of the next 2n-1 codes is a 1.
Example Binary Code Reflected Code
00 00
01 01
10 11
11 10
PURPOSE:
A method of realizing a function in either of the two standard forms such that the number and
complexity of terms in the function is minimal.
Construction:
Draw a square or rectangular figure allowing 2n squares (for a problem with n variables).
The table should be drawn so that there are 2x rows and 2y columns where x y n .
Label the rows and columns with reflected code from left to right and top to bottom.
Fill in each square of the map with its corresponding truth table function value.
Simplification:
Circle the largest contiguous binary (2w, w n ) group of 1’s for SOP (0’s for POS) which is rectangular or square
Consider all the edges of the map to be physically adjacent.
Each circled block of 1’s (0’s) corresponds to one SOP (POS) term. The term is extracted by observing which literals do not change for the block. These can be complemented to create the POS terms using 0’s. The literals are then AND’ed (OR’ed) together to realize the function.
Finally, each term, which corresponds to each rectangular block, is OR’ed (AND’ed) together to realize the function
Each 1 (0) in the map must be circled at least once to realize the function. (Sometimes there is more than one way to do it).
Additional Rule for Don’t Cares: It is not required that don’t cares be circled, but treat them as 1 (0) if it will help with minimization.
Page | 6-9
6.3.1 Construction / Simplification of Karnaugh Maps:
Draw K-map for FF input @ this present state. Where is the T FF input required from excitation table to get next state. d. Use PS & Input to yield FF input
Make transition table with added columns of current sequence and desired sequence. Table columns are current sequence, desired sequence, input, present state, next state, output, FF input
Label current state & desired state
Write desired sequence under column for all rows
Place sequence up to this point under correct sequence
Draw line through as many as useable of current sequence to next state in desired sequence.
Same number of states must be crossed in both
Output = 1 when all states are crossed for Mealy. For Moore, add one more state when all states are crossed
Else, go to 7.
Note: it is easier to make input
Page | 7-1
Chapter 7 - Economics – Time Value of Money
7.1 Introduction Does money have the same value today that it did 10 years ago? Would you rather have your money
now or later? If you can get it sooner, would you be willing to give a discount on the amount. If you are
willing to get it later, how much interest do you need?
In the process of creating projects, it is necessary to balance time, money, and quality. Other chapters
look at the time and quality issues. This section discusses the relative value of money. The study of the
value of money is called economics.
Economics is often divided into two segments: macro and micro. Macroeconomics deals with large-scale
money manipulation that is often controlled by government. It attempts to look at an economy as a
whole. Microeconomics is local scale money that is controlled by a project, corporation, or other entity.
This chapter will focus on project economics.
Economics has been called the science of allocating scare resources. That definition is limiting. It
assumes there is a fixed sized pie. If one group gets a larger slice, then someone else must necessarily
receive a smaller share. That is a negative feedback analysis, and conflicts with the fundamental premise
of positive feedback.
Positive feedback philosophy is the basis for developing new technology. It assumes something that has
little value can have value added through technology, thereby creating more value or wealth. This
comes without reducing the wealth of others.
Consider a common example. Sand has little value. The most common component of sand, however is
silicon oxide; it can be modified to create the silicon wafers used in microelectronics. A previously
limited value material now has great value.
Page | 7-2
7.2 Time and Interest At the present time money has a known value, called the present value or purchase price, PV. At some
time in the future, the same money will have a different value called the future value, FV.
Two things relate the values. Interest rate, i, is the percentage of a sum of money charged for its use.
This is also called the discount rate. Interest is based on a period of time. In most cases it is one year, but
it can be any interval. The number of interest periods is n.
It is assumed that all money transfer is at the end of a period. If it is at the beginning, the present value
must be calculated from one period in the future.
The simplest calculation is for a single payment exemplified in the cashflow diagram. A value below the
line is negative or paid out. A value above the line is positive or received in. Hash marks without arrows
simply identify a time period.
FV
PV
Equations for calculating the values are shown below. In addition, many texts give an abbreviation for
the factor that relates the value to be calculated to the known value. In addition, we will show the
equivalent spreadsheet command, (Microsoft Excel 2003), since that is the most common way of
calculating value of money. [Excel] The Excel formulae start with an “=” sign
Find the amount of money that it would take in the future, FV, which has an equivalent value to the
present worth, PV.
*(1 )nFV PV i
( , ,0, ,)FV i n PV
Conversely, when the value in the future is known, determine the present value.
*(1 ) nPV FV i
( , ,0, ,)PV i n FV =PV(i,n,0,FV,)
Page | 7-3
7.2.1 Uniform series
Rather than single payments, a much more common practice is to have progress payment amounts, A. If
these are at regular time intervals, the process is called a uniform series. When uniform amounts are
applied to the future value and present value, four new relationships are established. Some authors
treat these as equivalent uniform annual cost (EUAC).
Future value and uniform amounts are illustrated in a cashflow diagram.
FV
A A A A A A A
Series compound amount is the future value found from the uniform amount.
(1 ) 1ni
FV Ai
( , , ,,)FV i n A
Sinking fund is the uniform payment found from the future value.
(1 ) 1n
iA FV
i
( , ,0, ,)PMT i n FV
The cashflow diagram demonstrates the present worth and the uniform payment amounts.
PV
A A A A A A A
Series present worth is the present value found from the uniform payment.
(1 ) 1
(1 )
n
n
iPV A
i i
( , , ,,)PV i n A
Page | 7-4
Capital recovery is the uniform payment found from the present value.
(1 )
(1 ) 1
n
n
i iA PV
i
( , , ,,)PMT i n PV
Capital recovery on salvage is a modification of the capital recovery relationship. The future value of the
salvage, S, is converted to a uniform payment and subtracted from the present value convert to a
uniform series. Alternately the salvage value is transferred through the present worth as developed
above.
(1 )
( )(1 ) 1
n
n
i iA PV S iS
i
( , , , )PMT i n PV S
Page | 7-5
7.2.2 Gradient
A gradient is a value that increases or decreases at a rate for each time interval. An arithmetic gradient
is an integer multiple during each interval. It can have either a uniform amount or a present value, PV.
G
A A A A
PV
AA
2G
3G
4G
(n-1)G
Arithmetic gradient uniform series is calculated from the gradient value, G, after the first interval.
1
(1 )n
nA G
i i i
In a similar manner the present value is derived from the first gradient, G.
(1 ) 1
(1 )
n
n
i inPV G
i i i
A geometric gradient grows by a rate, g, during each interval. Therefore the change in amount varies
according to an exponential curve.
A1
PV
A2
A3
A4
An
The present value depends on two rates, the interest and the gradient, as well as the first amount in the
series.
1 (1 ) (1 )n ng i
PV Ai g
Frequently, the gradient rate is equal to the interest rate. Then the equation reduces significantly.
11 (1 )PV A n i
Page | 7-6
7.2.3 Nominal Interest or APR
Nominal interest rate, r, is also called the annual percentage rate (APR). It is the reported rate that is
compounded multiple times M, per year. It may be stated as 18% APR or 1.5% per month. These are
equivalent. However, they are not the effective interest rate.
Effective annual interest rate, ia is also called annual effective yield (AYR). It is substantially greater than
the nominal rate.
1 1
m
a
Ri
M
The most insidious effect is continuous compounding at a nominal rate. This creates an exponential
growth.
For a single payment, the relationships are simple, but expensive.
( )rnFV PV e
( )rnPV FV e
Page | 7-7
7.2.4 Perpetual and Rule of 72
Present value of perpetuities is a common way to determine the worth of an ongoing, perpetual uniform
payment. this is valuable in evaluating annuities, insurance, and lotteries.
A
PVi
Rule of 72 is a quick technique to determine how long it takes for the present value of money to double.
The number Of periods is determined from 72 divided by the interest rate.
72
ni
Page | 7-8
7.3 Rate of Return
Project economics are based on recovery of money. A variation of present worth has been used in all
the calculations to this point. Projects typically can be implemented with a number of different
alternatives. Therefore, the economic analysis is a comparison of alternatives.
Internal rate of return (IRR) is the break-even interest rate i* at which the present worth of a project is
zero. With a present worth of zero, we are ambivalent and do not particularly care about the project.
( *) PV cas in - PV cash out =0PV i
The calculation of IRR is an iterative process. Begin with the number of time intervals, n, and the
present, future, or uniform amounts. Guess at an interest rate and calculate the present worth. If it is
positive increase the interest rate. If the present worth is negative decrease the interest rate. Continue
estimating interests until the present value is adequately close to zero.
It is not necessary to laboriously perform this calculation. Many calculators and spreadsheets have the i*
function built into the spreadsheet package (Microsoft Excel 2003).
( , )IRR values guess
Investors have a minimum acceptable rate of return, MARR. This is the interest that must be exceeded if
the project is to be funded. Compare the calculated rate of return with the MARR. If the IRR is less than
the MARR, then the project should not be done.
Typically the analyst will go back and tweak the costs associated with the project until an acceptable IRR
is obtained. This is a very important process since it forces the analyst to have a better handle on the
project, and, as a result, the project is more likely to perform as expected.
Page | 7-9
7.3.1 Incremental Analysis
There is an anomaly with simple IRR analysis. For projects with a short term and low investment, the IRR
may be very high. However, that project may not have the greatest present worth. This is because
present worth is an absolute dollar measurement while IRR is a relative measurement.
Incremental investment analysis is used to compare the difference in investments for projects.
Begin with a table for the number of years. In year zero, place the present worth or purchase price
under each project. In year one, place the future value or uniform amount under each project. Continue
for all remaining years.
Next create a column that is the difference in the cashflow between the two projects at each time
interval. This gives a sequence of incremental values.
n Project1 Project2 2 –1
0 P10 P20 P20 – P10
1 F11 F21 F21 – F11
2 F12 F22 F22 – F21
IRR1 IRR2 IRR2-1
Next calculate the rate of return on the differences. If the rate of return on the difference exceeds the
MARR, then the incremental investment should be made to do the project.
Page | 7-10
7.3.2 Payback
Payback is simply how long it takes for net receipts to equal investment outlays. Conventional payback is
calculated ignoring the time value of money.
Discounted payback includes the interest on the money. For a project with the same cash inflows each
year,
cost
Payback=uniform annual benefit
For projects with varying cash inflows, these inflows are added until the cost is equaled.
For payback analysis, all costs and all revenue prior to the payback are included without considering
differences in their timing. All consequences after the payback time are completely ignored.
This is obviously a simple technique. It also has some serious constraints. Nevertheless, it is still a
popular indicator since it gives the time to recover the investment. It is important to note again that the
simple payback ignores the time value of money, which may be very high for an expensive, or lengthy,
project.
Page | 7-11
7.3.3 Benefit Cost Ratio
Benefit is simply another way a describing the positive cash flow or value. Cost describes the negative
cash flow or value. When comparing projects, these are calculated using incremental analysis. The ratio
is the benefit to cost ratio. If the ratio is greater than 1, the incremental project should be done.
(benefits)
(costs)
PVBC
PV
Page | 7-12
7.3.4 Tax Implications
The taxes on income and investments will influence the value of the project. Calculate the value of the
project before income tax (BFIT). Financial types refer to this as EBITDA (Earnings Before Interest, Tax,
Depreciation and Amortization). Calculate the value of the project after income tax (AFIT). The process
can be used with all the techniques including present worth, incremental analysis IRR, and payback.
The value of the analysis is it reflects the real net worth, since different investments have alternative tax
implications. BFIT may be a better indicator of projects, but AFIT is the better indicator of value to the
organization. After all, the purpose of projects is not for the projects sake, but to increase the value to
the organization.
Page | 7-13
7.4 Table of Terminology
Numerous symbols are used for economic comparisons. These are shown in a table.
Symbol Definition I Interest rate per interest period, often one year N Number of interest periods PV Present value of money FV Future value of money A Amount of money at end of period in a uniform series for n periods G Arithmetic gradient, uniform period change in amount of money G Geometric gradient, uniform rate of cash flow change. R Nominal interest rate per interest period M Number of compounding subperiods S Salvage value is cash recoverable at the end of a project. PV*, FV* Amount of money flowing continuously during one period
7.5 Commentary
Determining which project is the best choice is less than straightforward. Many issues enter the
evaluation, besides the pure economic calculations.
There are biases and preferences from the management and the project team. These are difficult to
eliminate, and the biases may even be desirable. With practice, there can come that ‘gut feel’ that this is
right. However, gut feel can also be wrong.
Most organizations place some criteria on selecting projects. Traditionally over many years, an IRR of
15% AFIT has been acceptable, assuming the project was well defined. For higher risk ventures, the
MARR often moves toward 30%.
Other companies place great emphasis on payback. Traditionally, a 3-year term has been acceptable. As
risk increases, the term is frequently shortened toward 1 year.
With these more stringent criteria, one must wonder if any projects will ever be built. That is not the
point. The objective is to compensate for higher than normal risk. There is no future in investing money
in ventures that do not increase the wealth of the stakeholders.
7.6 Review
Page | 7-14
Money has different values, depending on the time it is obtained. The worth of the money is dependent
on present, future, and uniform amounts. There are multiple projects on which to invest money. The
preferred project is determined by incremental analysis that compares the projects. The methods are
investor rate of return, payback, and benefit to cost ratio. The return after taxes is a better indicator of
worth.
Page | 7-15
Chapter 7 Problems
1. Retirement savings. Given: Uninflated growth of capital is 3% per year. You begin paying into a retirement program when you turn 25. After25 years, you begin collecting $50,000 annually. You collect for 40 years. a. What is the present value of the annuity? b. What are your annual contributions to have that amount in the account? c. How much will be in the account when you begin collecting? d. Is the annual contribution a reasonable amount to expect you can contribute after you pay taxes
on a $50,000 salary? e. Wait 35 years to begin collection and receive for 30 years, do you contribute more or less each
month? f. To retire at 50, what is the best approach?
2. Auto Purchase. Given: A new university graduate purchases a new Whizmobile for $25,000. She borrows the money at an interest of 1%/mo. Loan period is 60 months. a. Find, future value, if a single balloon payment is made at 60 months. b. Monthly payments if made at the end of each month. c. Total amount paid (cash-flow) d. Total Interest paid e. Amount unpaid after 12 months a. Assume the same monthly payments were deposited in savings program. The account pays 1%
per month. How many months would it before you had $25,000?
3. Rule of 72: Given: $1000 to invest and the interest is compounded 10% per year. a. How long will it take to double your money? b. What is the effective monthly interest rate?
4. House Purchase. Given: Engineer with a salary of $40,000 per year. Federal and state income tax of 25%. Mortgage company will lend 2.5 times the annual salary for a house. Interest rate is 7% per year. Loan period is 30 years. Use uniform series of annual payments procedure.
Create a spreadsheet with the following columns. Calculate the values for the end of each year.
a. Principal remaining at start of year b. Interest owed for year c. Annual payment d. Amount of payment that is principal e. Total payments to date
Page | 7-16
f. Total interest to date g. Double the payment. h. Apply the extra amount to principal only. i. Tabulate the lower principal remaining. j. How long does it take to pay off the loan, with double payments?
Calculate the following items for a 15-year loan.
a. Total cash flow (amount paid). b. Total interest paid.
The next sections should be added to the spreadsheet after taxes.
c. Effect on income tax for interest paid d. Cash flow after taxes
Bibliography
Anthony Tarquin, Anthony Engineering Economy, McGraw-Hill Education, 1989
A 0–3.14 B 3.15–3.54 C 3.55–3.99 D 4.0–4.49 E 4.5–4.99 F 5.0–5.59 G 5.6–6.29 H 6.3–7.09 J 7.1–7.99 K 8.0–8.99
Code Letter
Kilovolt-Amperes per Horsepower
with Locked Rotor
L 9.0–9.99 M 10.0–11.19 N 11.2–12.49 P 12.5–13.99 R 14.0–15.99 S 16.0–17.99 T 18.0–19.99 U 20.0–22.39 V 22.4 and up
Table 430.52 Maximum Rating or Setting of Motor Branch-Circuit Short-
Circuit and Ground-Fault Protective Devices
Percentage of Full-Load Current Type of Motor Non-time
Delay Fuse Dual Element (Time-Delay) Fuse
Instantaneous Trip Breaker
Inverse Time Breaker
Single-phase motors
300 175 800 250
AC polyphase motors other than wound-rotor Squirrel cage other than Design B energy-efficient
300 175 800 250
Design B energy-efficient
300 175 1100 250
Synchronous 300 175 800 250 Wound rotor 150 150 800 150 Direct current (constant voltage)
150 150 250 150
Page | 9-9
Table 430.91 Motor Controller Enclosure Selection
For Outdoor Use
Enclosure Type Number1
Provides a Degree of Protection Against the Following Environmental Conditions
3 3R 3S 3X 3RX 3SX 4 4X 6 6P
Incidental contact with the enclosed equipment
X X X X X X X X X X
Rain, snow, and sleet X X X X X X X X X X
Sleet2 — — X — — X — — — —
Windblown dust X — X X — X X X X X
Hosedown — — — — — — X X X X
Corrosive agents — — — X X X — X — X
Temporary submersion — — — — — — — — X X
Prolonged submersion — — — — — — — — — X
For Indoor Use
Enclosure Type Number1
Provides a Degree of Protection Against the Following Environmental Conditions
1 2 4 4X 5 6 6P 12 12K 13
Incidental contact with the enclosed equipment
X X X X X X X X X X
Falling dirt X X X X X X X X X X
Falling liquids and light splashing
— X X X X X X X X X
Circulating dust, lint, fibers, and flyings
— — X X — X X X X X
Settling airborne dust, lint, fibers, and flyings
— — X X X X X X X X
Hosedown and splashing water
— — X X — X X — — —
Oil and coolant seepage — — — — — — — X X X
Oil or coolant spraying and splashing
— — — — — — — — — X
Corrosive agents — — — X — — X — — —
Temporary submersion — — — — — X X — — —
Prolonged submersion — — — — — — X — — —
Page | 9-10
1Enclosure type number shall be marked on the motor controller enclosure. 2Mechanism shall be operable when ice covered. FPN: The term raintight is typically used in conjunction with Enclosure Types 3, 3S, 3SX, 3X, 4, 4X, 6, 6P. The term rainproof is typically used in conjunction with Enclosure Type 3R, 3RX. The term watertight is typically used in conjunction with Enclosure Types 4, 4X, 6, 6P. The term driptight is typically used in conjunction with Enclosure Types 2, 5, 12, 12K, 13. The term dusttight is typically used in conjunction with Enclosure Types 3, 3S, 3SX, 3X, 5, 12, 12K, 13.
Notes: 1. These resistance values are valid only for the parameters as given. Using conductors having coated strands, different stranding type, and, especially, other temperatures changes the resistance. 2. Formula for temperature change: R2 = R1 [1 + α (T 2 - 75)] where α cu = 0.00323, α AL = 0.00330 at 75°C. 3. Conductors with compact and compressed stranding have about 9 percent and 3 percent, respectively, smaller bare conductor diameters than those shown. See Table 5A for actual compact cable dimensions. 4. The IACS conductivities used: bare copper = 100%, aluminum = 61%. 5. Class B stranding is listed as well as solid for some sizes. Its overall diameter and area is that of its circumscribing circle.
Page | 9-16
Table 9 Alternating-Current Resistance and Reactance for 600-Volt Cables, 3-Phase, 60 Hz, 75°C (167°F)
— Three Single Conductors in Conduit
Ohms to Neutral per Kilometer Ohms to Neutral per 1000 Feet6
Size
XL (Reactance) for All Wires
Alternating-Current Resistance for Uncoated Copper Wires
Notes: 1. These values are based on the following constants: UL-Type RHH wires with Class B stranding, in cradled configuration. Wire conductivities are 100 percent IACS copper and 61 percent IACS aluminum, and aluminum conduit is 45 percent IACS. Capacitive reactance is ignored, since it is negligible at these voltages. These resistance values are valid only at 75°C (167°F) and for the parameters as given, but are representative for 600-volt wire types operating at 60 Hz. 2. Effective Z is defined as R cos(θ) + X sin(θ), where θ is the power factor angle of the circuit. Multiplying current by effective impedance gives a good approximation for line-to-neutral voltage drop. Effective impedance values shown in this table are valid only at 0.85 power factor. For another circuit power factor (PF), effective impedance (Ze) can be calculated from R and XL values given in this table as follows: Ze = R × PF + XL sin[arccos(PF)].
Page | 9-18
Table C.4 Maximum Number of Conductors or Fixture Wires in Intermediate Metal Conduit
NEMA Controller Size for Motors, Transformers, & Capacitors
Load Voltage
Continuous Current
Service Limit Current
Motor Maximum
Motor Maximum
Transformer Primary Switching
Transformer Primary Switching
Capacitor Switching
Circuit Closing Maximum
Non-plugging and Non-jogging Duty
Plugging and Jogging Duty
Inrush Current < = 20 times Continuous Amp
Inrush Current = 20 to 40 times Continuous Amp
Inrush Current Peak Including Offset
NEMA
V Amp Amp HP HP HP HP kVA kVA kVA kVA kVAR Amp
Size 1φ 3 φ 1 φ 3 φ 1 φ 3 φ 1 φ 3 φ 3 φ 3 φ
00 115 200 230 380 460 575
9 11 1/3 — 1 — — —
— 1-1/2 1-1/2 1-1/2 2 2
1/4 — 1/2 — — —
— 1 1 1 1-1/2 1-1/2
— — — — — —
— — — — — —
— — — — — —
— — — — — —
— — — — — —
87
0 115 200 230 380 460 575
18 21 1 — 2 — — —
— 3 3 5 5 5
1/2 — 1 — — —
— 1-1/2 1-1/2 1-1/2 2 2
0.6 — 1.2 — 2.4 3
— 1.8 2.1 — 4.2 5.2
0.3 — 0.6 — 1.2 1.5
— 0.9 1 — 2.1 2.6
— — — — — —
140
1 115 200 230 380 460 575
27 32 2 — 3 — — —
— 7-1/2 7-1/2 10 10 10
1 — 2 — — —
— 3 3 5 5 5
1.2 — 2.4 — 4.9 6.2
— 3.6 4.3 — 8.5 11
0.6 — 1.2 — 2.5 3.1
— 1.8 2.1 — 4.3 5.3
— — 6 — 13.5 17
288
1P 115 230
36 42 3 5
— —
1-1/2 3
— —
— —
— —
— —
— —
— —
— —
2 115 200 230 380 460 575
45 52 3 — 7-1/2 — — —
— 10 15 25 25 25
2 — 5 — — —
— 7-1/2 10 15 15 15
2.1 — 4.1 — 8.3 10
— 6.3 7.2 — 14 18
1 — 2.1 — 4.2 5.2
— 3.1 3.6 — 7.2 8.9
— — 12 — 25 31
483
3 115 200 230 380 460 575
90 104 7-1/2 — 15 — — —
— 25 30 50 50 50
7-1/2 — 15 — — —
— 15 20 30 30 30
4.1 — 8.1 — 16 20
— 12 14 — 28 35
2 — 4.1 — 8.1 10
— 6.1 7.0 — 14 18
— — 27 — 53 67
947
4 115 200 230 380 460 575
135 156 — — — — — —
— 40 50 75 100 100
— — — — — —
— 25 30 50 60 60
6.8 — 14 — 27 34
— 20 23 — 47 59
3.4 — 6.8 — 14 17
— 10 12 — 23 29
— — 40 — 80 100
1581
5 115 200 230 380 460 575
270 311 — — — — — —
— 75 100 150 200 200
— — — — — —
— 60 75 125 150 150
14 — 27 — 54 68
— 41 47 — 94 117
6.8 — 14 — 27 34
— 20 24 — 47 59
— — 80 — 160 200
3163
6 115 200
540 621 — —
— 150
— —
— 125
27 —
— 81
14 —
— 41
— —
6326
Page | 9-22
230 380 460 575
— — — —
200 300 400 400
— — — —
150 250 300 300
54 — 108 135
94 — 188 234
27 — 54 68
47 — 94 117
160 — 320 400
7 230 460 575
810 932 — — —
300 600 600
— — —
— — —
— — —
— — —
— — —
— — —
240 480 600
9470
8 230 460 575
1215 1400 — — —
450 900 900
— — —
— — —
— — —
— — —
— — —
— — —
360 720 900
14205
9 230 460 575
2250 2590 — — —
800 1600 1600
— — —
— — —
— — —
— — —
— — —
— — —
665 1325 1670
25380
Service-Limit Current Ratings - The service-limit current ratings shown represent the maximum rms current, in amperes, which the controller
shall be permitted to carry for protracted periods in normal service. At service-limit current ratings, temperature rises shall be permitted to exceed
those obtained by testing the controller at its continuous current rating. The current rating of overload relays or the trip current of other motor
protective devices used shall not exceed the service-limit current rating of the controller.
Plugging or Jogging Service - The listed horsepower ratings are recommended for those applications requiring repeated interruption of stalled
motor current encountered in rapid motor reversal in excess of five openings or closings per minute and shall not be more than ten in a ten minute
period.
Capacitor terminals - If maximum available current is greater than 3,000 amperes, consult NEMA ICS-2 Standard.
Page | 9-23
NEMA Table 11 Typical Characteristics and Applications of Fixed Frequency Small and Medium AC
Squirrel-Cage Induction Motors
Design Letter Locked Rotor Torque
Pull- up Torque
Break-down Torque
Locked Rotor Current
Slip Typical Applications Relative Efficiency
Polyphase Characteristics
Percent Rated Load Torque*
Percent Rated Load Torque
Percent Rated Load Torque
Percent Rated Load Current
Percent Sync Speed
Design A
High locked rotor torque High locked rotor current
70-275 65-190 175-300 Not defined
0.5-5% Fans, blowers, centrifugal pumps and compressors, motor-generator sets, etc., where starting torque requirements are relatively low.
Medium or high
Design B
Normal locked rotor torque Normal locked rotor current
70-275 65-190 175-300 600-700 0.5-5% Fans, blowers, centrifugal pumps and compressors, motor-generator sets, etc., where starting torque requirements are relatively low.
Medium or high
Design C
High locked rotor torque Normal locked rotor current
200-285 140-195 190-225 600-700 1-5% Conveyors, crushers, stirring motors, agitators, reciprocating pumps and compressors, etc., where starting under load is required
Medium
Design D
High locked rotor torque High slip
275 NA 275 600-700 5-8% High peak loads with or without flywheels such as punch presses, shears, elevators, extractors, winches, hoists, oil-well pumping and wire-drawing motors
Low
Design N
Small motor
- NA - - - Centrifugal loads where starting torque requirements are relatively low.
Low
Design 0
Small motor - NA - - NA
Design L
Medium motor
- 100% - - NA Fans, blowers, centrifugal pumps and compressors, motor-generator sets, etc., where starting torque requirements are relatively low.
Medium or low
Design M
Medium motor
- 100% - - NA Fans, blowers, centrifugal pumps and compressors, motor-generator sets, etc., where starting torque requirements are relatively low.
Medium or low
*Higher values are for motors having lower horsepower ratings.
Frame Size Information Suffix letters after the NEMA frame size indicates that the frame differs in some way from the standard frame. Below is a list of suffixes that may be found after the frame size and their definition.
A DC Motor or Generator
C ―C‖ flange mounting on drive end **
D ―D‖ flange mounting on drive end **
E Shaft dimensions for elevator motors in frames larger than the 326U frame
H Frame with an ―F‖ dimension larger than a frame without (small framed motors)
J Jet pump motors
JM ―C‖-face mounted close coupling pump with mechanical seal
JP ―C‖-face mounted close coupling pump -packed pump
K Sump pump motor LP & LPH ―P‖ flange mounting vertical solid shaft pump
P & PH ―P‖ flange mounting vertical hollow shaft pump
S Standard short shaft T Included as part of a frame number-standard dimension
U Included as part of a frame number-standard dimension V Vertical mounting
Y Special mounting dimensions -manufactured specified Z Special shaft dimensions -manufactured specified
** If the face mounting is on the end opposite the drive, the suffix will be as follows: ―FC‖ or ―FD‖
Assembly Position F-1 VS. F-2
F-1 F-2
Notes From Front: -When a ―C‖ flange has been added to a NEMA motor the ―BA‖ dimensions are:
1 Motor horsepower NEC 430.250 - 20 2 Full load Amps - FLA NEC 430.250 - 27 3 Lock letter code & kVA/hp NEC 430.7(B) F 5.59 112 4 Lock rotor amp calculate kVA*1000/1.732 V 112,000/1.732*460 141 5 Lock rotor amp for disconnect NEC 430.251(B) - 145 6 Wire rating:1.25*largest+other NEC 430.24 1.25*27 + 0 34 7 Insulation type NEC 310-16 - THHN 8 Insulation temperature NEC 310-16 - 90C 9 AWG / kcmil NEC 310-16 - 10 AWG 10 Temperature correction amp NEC 310-16 0.91 36 11 Max breaker rating & type NEC 430.52 800 instant 216 12 Actual breaker size NEC 240.6(A) - 200 13 Controller enclosure NEC 430.91 - 3R 14 Controller size NEMA Controller - 2 15 Controller max closing amp NEMA Controller - 483 16 Overload setting % - Amp - 105 28.4 17 Motor enclosure NEMA Enclosure TEFC 18 Motor NEMA design NEMA 11 - B 19 Motor sync speed 120 * freq / poles 120*60/4 1800 20 Motor slip - shaft speed NEMA 11 2% 1764 21 Motor frame NEMA Dimension 2 - 256T 22 Shaft diameter NEMA Dimensions U 1-5/8 23 Sheave diameter P(n*dia) = M(n*dia) 300*18/1800 3‖ 24 # wires - 3Φ + N 4 25 Conduit type & size NEC C4 et al 1/2 3/4 26 Total kVA 1.732 V * I / 1000 1.732*480*27/1000 22.5 26 Transformer kVA size NEMA - 3 – 7.5 28 Secondary volt & Y-Δ - 277 / 480 Y 29 Primary volt & Y-Δ - 12470 Δ 30 Transformers taps two 2-1/2 + 31 Transformer impedance PU