1
Electronic Circuits and Logic Design Laboratory
List of Experiments
Part A
Sl.No Experiment Name
1 a. To study the working of positive clipper, double –ended clipper and positive clamper
using diodes.
OR
b. To build and simulate the above circuits using a simulation package
2 a. To determine the frequency response, input impedance, output impedance, and bandwidth
of a CE amplifier.
OR
b. To build the CE amplifier circuit using a simulation package and determine the voltage
gain for two different values of supply voltage and for two different values of emitter
resistance.
3 a.To determine the drain characteristics and transconductance characteristics of an
enhancement-mode MOSFET.
OR
b. To implement a CMOS inverter using a simulation package and verify its truthtable.
4 a. To design and implement a Schmitt trigger using Op-Amp for given UTP and LTP values.
OR
b. To implement a Schmitt trigger using Op-Amp using a simulation package for two sets of
UTP and LTP values.
5 a. To design and implement a rectangular waveform generator (Op-Amp relaxation
oscillator) for given frequency.
OR
b. To implement a rectangular waveform generator (Op-Amp relaxation oscillator) using a
simulation package and observe the change in frequency when all resistor values are
doubled.
6 To design and implement an astable multivibrator circuit using 555 timer for a given
frequency and duty cycle.
7 To implement a +5V regulated power supply using full-wave rectifier and 7805 IC regulator
in simulation package. Find the output ripple for different values of load current.
Part B
1 a. Given any 4-variable logic expression, simplify using Entered Variable Map and realize
the simplified logic expression using 8:1 multiplexer IC.
b. Write the Verilog /VHDL code for an 8:1 multiplexer. Simulate and verify its working.
2 a. Realize a full adder using 3-to-8 decoder IC and 4 input NAND gates.
b. Write the Verilog/VHDL code for a full adder. Simulate and verify its working.
3 a. Realize a J-K Master/Slave Flip-Flop using NAND gates and verify its truth table.
OR
2
b. Write the Verilog/VHDL code for D Flip-Flop with positive-edge triggering. Simulate
and verify its working.
4 a. Design and implement a mod-n (n<8) synchronous up counter using J-K Flip-Flop ICs.
OR
b. Write the Verilog/VHDL code for mod-8 up counter. Simulate and verify its working.
5 a. Design and implement a ring counter using 4-bit shift register.
b. Write the Verilog/VHDL code for switched tail counter. Simulate and verify its working.
6 Design and implement an asynchronous counter using decade counter IC to count up from 0
to n (n<=9).
7 Design a 4-bit R-2R ladder D/A converter using Op-Amp. Determine its accuracy and
resolution.
GENERAL INTRODUCTION TO PSPICE
Why simulate circuits?
In order to ensure a successful circuit design and mitigate costly and potentially dangerous
design flaws, careful planning and evaluation must occur at every stage of the process.
Circuit simulation provides a cost-effective and efficient method for identifying faults before
moving to the more expensive and time-consuming prototyping stage. Including simulation
in the design process reduces design errors and speeds the design cycle by allowing you to
predict and better understand circuit behavior.
The main purpose of simulation is to predict and understand the behavior of electronic
circuits. Experiment with ―What-if‖ Scenarios.
Limitations of simulation
Simulation is never meant to replace prototyping. Certain real-world effects, like cross-talk,
electromagnetic noise, and spurious line noise, can be too difficult, time-consuming, or
costly to model in simulation.
Note: While a prototype helps you to verify and validate your design in the real world,
simulation helps you catch design errors before spending money and time on prototyping.
SPICE is program that simulates electronic circuits on your PC.
WHY USE SPICE?
SPICE is a great tool for learning electronics. You can increase your understanding of
circuits as you play and tinker with them. Modify the circuit and see what happens. Change
a component value and see the effect on a circuit in seconds.
PSPICE by OrCAD is one of the most popular circuit simulators.
Orcad 9.2 Lite Edition Installation
1. Insert Cadence CD into CD-ROM drive
2. Select Products to install
Capture – Schematic entry application – it must be installed
3
Capture CIS – Should be grayed out.
PSpice – For conducting mixed-signal analog and digital simulations
Layout– For creating PC Board layouts from schematics
Then follow the instructions as it appears on the monitor and complete the installation
The steps to simulation:
1. Create a simulation project
2. Draw schematic to simulate
3. Establish a simulation profile
4. Set up simulation type
5. Simulate circuit
6. Analyze results in Probe
To start a simulation session
1. On the start menu select ORCAD FAMILY RELEASE 9.2 LITE EDITION - CAPTURE LITE
EDITION
2. Once the capture window appears, select FILE - NEW - PROJECT
The following window appears:
3. Give the project a descriptive name (spaces can be included).
4. Select ANALOG OR MIXED-SIGNAL CIRCUIT WIZARD.
5. Specify a location where the project is to be stored
a. Click Browse to change the location the files will be stored
6. Click OK.
4
The following window appears:
7. Select Create a blank project and click OK. The following window appears:
8. To place parts, click PLACE PART (Shift+P) The following window appears:
5
If Libraries are not appearing in the window then click Add Library. The
library files will generally be available in the following path by default C:\Program Files\Orcad_Demo\Capture\Library\Pspice.Select all Library files
by pressing Ctrl A and then press Open. All the Library files will appear in the window.
9. Select the part you wish to place in the schematic. Insert as many as
needed
10. Right Click and select END MODE to stop inserting parts
6
11. To wire parts together, click Place Wire Icon from the Right hand side
vertical Icons list (Shift+W). Place cursor over boxes at ends of parts and draw wires connecting parts.·When done, right click and select End Wire.
12. To insert a ground node, click Place – Ground Icon·Window appears with caption Place Ground with only ground nodes available for selection.
7
Altenative method:Click on place Net alias Icon N1 on the Right hand side
vertical Icons list.
Always select 0/SOURCE for the ground node of an analog circuit
– every analog circuit must contain atleast one 0 ground. This is not a requirement for digital circuits.
13. To change component values that are displayed·Double click the displayed value· Change the desired value in the dialog box that appears.
To set up a simulation profile
Select PSPICE ---NEW SIMULATION PROFILE from the menu
8
Give a descriptive name to the type of simulation .
Select the desired parameters for the particular circuit and then click OK
Place voltage, current, and power markers from the PSPICE --- MARKERS menu where needed
Click PSPICE -- RUN.
Simulation Experiments
General procedure for all experiments:
1. Select the required components from the menu.
2. Place all the required components in the schematic. 3. Simulate the circuit using RUN option from the menu.
4. Observe the waveforms form the output
1 a. To study the working of positive clipper, double –ended clipper and positive
clamper using diodes.
COMPONENTS REQUIRED: Diode (BY-127 / IN4007), Resistors-10 K & 3.3k , DC
regulated power supply (for Vref), Signal generator (for Vi) and CRO.
CIRCUIT DIAGRAM OF POSITIVE CLIPPER
9
Fig.1 a Positive clipper Circuit b. Transfer Characteristics
Clippers clip off a portion of the input signal without distorting the remaining part of the
waveform. In the positive clipper shown above the input waveform above Vref is clipped off.
If Vref = 0V, the entire positive half of the input waveform is clipped off.
Plot of input Vi (along X-axis) versus output Vo (along Y-axis) called transfer characteristics
of the circuit can also be used to study the working of the clippers.
Choose R value such that rf RRR where Rf = 100 and Rr = 100k are the resistances
of the forward and reverse diode respectively. Hence R = 3.3k .
Let the output resistance RL = 10k . The simulations can be done even without RL.
WAVEFORMS
Fig. 2. Input and output waveform for positive Clipper
DOUBLE ENDED CLIPPER
Fig.3 Double ended clipper Circuit b. Transfer Characteristics
10
Fig.4. Input and output waveform for double-ended clipping circuit
Note: The above clipper circuits are realized using the diodes in parallel with the load (at the
output), hence they are called shunt clippers. The positive (and negative) clippers can also
be realized in the series configuration wherein the diode is in series with the load. These
circuits are called series clippers.
POSITIVE CLAMPER
COMPONENTS REQUIRED: Diode (BY-127), Resistor of 200 K , Capacitor - 0.1 F, DC
regulated power supply, Signal generator, CRO
Fig. 5 Positive Clamper
The clamping network is one that will “clamp” a signal to a different DC level. The network
must have a capacitor, a diode and a resistive element, but it can also employ an
independent DC supply (Vref) to introduce an additional shift. The magnitude of R and C
must be chosen such that time constant ζ=RC is large enough to ensure the voltage across
capacitor does not discharge significantly during the interval of the diode is non-conducting.
Aim: To simulate clipper circuit
POSITIVE CLIPPER
Components to be placed in the schematic: ac voltage source, diode IN 4002,
Resistor 10k and 3.3k
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Type of analysis: TIME DOMAIN (TRANSIENT) Run to time: 5msec step size:0.01msec
POSITIVE CLIPPER WITH REFERENCE VOLTAGE
Components to be placed in the schematic: ac voltage source, diode IN 4002,
Resistor 10k, Resistor 3.3k,dc voltage source
R1
3.3kV
R2
10k
v 1
FREQ = 1khzVAMPL = 5vVOFF = 0
V
0
D1
D1N4002
Time
0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms
V(V1:+) V(R2:2)
-5.0V
0V
5.0V
12
Type of analysis: TIME DOMAIN (TRANSIENT) Run to time: 5msec step size:0.01msec
NEGETIVE CLIPPER
Components to be placed in the schematic: ac voltage source, diode IN 4002,
Resistor 10k, Resistor 3.3k
V
2V
R1
3.3k
v 1
FREQ = 1khzVAMPL = 5vVOFF = 0
0
V
D1
D1N4002
R2
10k
Time
0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms
V(V1:+) V(R2:2)
-5.0V
0V
5.0V
13
Type of analysis: TIME DOMAIN (TRANSIENT) Run to time: 5msec step size:0.01msec
NEGETIVE CLIPPER WITH REFERENCE VOLTAGE
Components to be placed in the schematic: ac voltage source, diode IN 4002,
Resistor 10k,Resistor 3.3k, dc voltage source
R1
3.3kV
R2
10k
D1
D1N4002
0
Vv 1
FREQ = 1khzVAMPL = 5vVOFF = 0
Time
0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms
V(V1:+) V(R2:2)
-5.0V
0V
5.0V
14
Type of analysis: TIME DOMAIN (TRANSIENT) Run to time: 5msec step size:0.01msec
DOUBLE ENDED CLIPPER
V
R2
10k
v 1
FREQ = 1khzVAMPL = 5vVOFF = 0
2V
R1
3.3k
D1
D1N4002
0
V
Time
0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms
V(V1:+) V(R2:2)
-5.0V
0V
5.0V
15
Type of analysis: TIME DOMAIN (TRANSIENT)
Run to time: 5msec step size:0.01msec
DOUBLE ENDED CLIPPER WITH REFERENCE VOLTAGE
V
D2
D1N4002R2
10k
R1
3.3k
D1D1N4002
v 1
FREQ = 1khzVAMPL = 5vVOFF = 0
0
V
Time
0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms
V(V1:+) V(R1:2)
-5.0V
0V
5.0V
V12V
R210k
0
V22V
V
R1
3.3k
V3
FREQ = 1khzVAMPL = 5vVOFF = 0 D2
D1N4002V
D1D1N4002
16
CLAMPER
Type of analysis: TIME DOMAIN (TRANSIENT)
Run to time: 5msec step size:0.01msec
Time
0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms
V(V1:+) V(R1:2)
-5.0V
0V
5.0V
V5
FREQ = 1khzVAMPL = 5vVOFF = 0
V
R2250k
0
V
C2
.1uf
D1D1N4002
17
CLAMPER WITH REFERENCE VOLTAGE
Type of analysis: TIME DOMAIN (TRANSIENT)
Run to time: 5msec step size:0.01msec
Time
0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms
V(V1:+) V(C1:2)
-5V
0V
5V
10V
V5
FREQ = 1khzVAMPL = 5vVOFF = 0
R2250k
V
C2
.1uf
0
V
D1D1N4002
V6
1v
Time
0s 0.5ms 1.0ms 1.5ms 2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms
V(V1:+) V(C1:2)
-5V
0V
5V
10V
15V
18
2b. To build the CE amplifier circuit using a simulation package and determine the voltage
gain for two different values of supply voltage and for two different values of emitter
resistance. Components required:- transistor(QN3904),resistor,VAC,VDC,capacitor
The single stage common emitter amplifier circuit shown above uses what is commonly called
"Voltage Divider Biasing". This type of biasing arrangement uses two resistors as a potential
divider network and is commonly used in the design of bipolar
transistor amplifier circuits. This method of biasing the transistor
greatly reduces the effects of varying Beta, ( β ) by holding the Base
bias at a constant steady voltage level allowing for best stability. The
quiescent Base voltage (Vb) is determined by the potential divider
network formed by the two resistors, R1, R2 and the power supply
voltage Vcc as shown with the current flowing through both
resistors. Then the total resistance RT will be equal to R1 + R2
giving the current as I = Vcc/RT. The voltage level generated at the
junction of resistors R1 and R2 holds the Base voltage (Vb) constant
R5
1k
R2
2.2k
0V
0
R1
10k
C2
.47uf
0
V2
10v
V
C1
2.2uf
0
C4
10uf
R4
3.6k
0
R3
600V7
1mv
0Vdc
V
Q1
Q2N3904
0
R6
10k
19
at a value below the supply voltage. Then the potential divider network used in the common
emitter amplifier circuit divides the input signal in proportion to the resistance. This bias
reference voltage can be easily calculated using the simple voltage divider formula below:
The same supply voltage, (Vcc) also determines the maximum Collector current, Ic when the
transistor is switched fully "ON" (saturation), Vce = 0. The Base current Ib for the transistor is
found from the Collector current, Ic and the DC current gain Beta, β of the transistor.
The voltage gain of an electronic circuit as a function of frequency.
Av = Vout/Vin
Analysis Type: Ac Sweep
Start Frequency: 10hz
End Frequency:100meg
Points/Decade: 40
When the supply voltage is 1mv
Voltage gain Av= 80mv
Frequency
10Hz 100Hz 1.0KHz 10KHz 100KHz 1.0MHz 10MHz 100MHz
V(R6:2) V(C2:2)
0V
40mV
80mV
20
When the supply voltage is changed to 2mv
Voltage gain Av= 150mv
When the supply voltage is 1mv and the emitter resistor value is 1k
Voltage gain Av= 80mv
When the supply voltage is 1mv and the emitter resistor value is 2.2k
Voltage gain Av= 40mv
Frequency
10Hz 100Hz 1.0KHz 10KHz 100KHz 1.0MHz 10MHz 100MHz
V(R6:2) V(C2:2)
0V
50mV
100mV
150mV
Frequency
10Hz 100Hz 1.0KHz 10KHz 100KHz 1.0MHz 10MHz 100MHz
V(R6:2) V(C2:2)
0V
40mV
80mV
Frequency
10Hz 100Hz 1.0KHz 10KHz 100KHz 1.0MHz 10MHz 100MHz
V(R6:2) V(C2:2)
0V
20mV
40mV
21
3.b To implement a CMOS inverter using a simulation package and verify its
truthtable.
Type of analysis: TIME DOMAIN (TRANSIENT)
Run to time: 100usec step size:0.1usec
Time
0s 10us 20us 30us 40us 50us 60us 70us 80us 90us 100us
V(V1:+) V(R1:2)
0V
2.5V
5.0V
V(R1:2)
0V
2.5V
5.0V
V(V1:+)
0V
2.5V
5.0V
SEL>>
22
4b. To implement a Schmitt trigger using Op-Amp using a simulation package
for two sets of UTP and LTP values. COMPONENTS REQUIRED: IC μA 741, Resistor of 10KΩ, 90KΩ, DC regulated power
supply, Signal generator, CRO
Theory: Open loop gain of op amp is defined as:
AOL = Vo / VD
where VD = Vin – Vref
• Open loop gain of op amp is very high (ideally infinite).
• Any small difference between VNI and VINV results into saturation of output voltage
±VSAT
Value of VSAT is limited by the supply voltage of op amp
DESIGN:
From theory of Schmitt trigger circuit using op-amp, we have the trip points,
VR
RRLTPUTPV
KRKR
RRV
RR
VRLTPUTP
RR
VRLTPUTP
VRR
RR
VR
RR
VRLTP
VVRR
VR
RR
VRUTP
ref
sat
sat
ref
ref
satref
ccsatsatref
33.32
))(( have we(1)equation From
90 then ,10Let
9 yields (2)equation then 2V, LTP & 4V UTP10V,Let
-(2)----- 2
-(1)----- 2
used. isdesign following the values,&, thefind to values UTP& LTP given the Hence
&
of 90% opamp theof saturation positive theis where
1
21
12
21
21
1
21
1
21
21
2
21
1
21
2
21
1
UTP=-1 V LTP=-3v
Vref=-2.2v
23
Type of analysis: TIME DOMAIN (TRANSIENT)
Run to time: 40msec step size:0.1msec
UTP=4v LTP=2v
Vref=3.3v
Hysteresis Curve
U1
uA741
3
2
74
6
1
5+
-
V+
V-
OUT
OS1
OS2
V
0
V4
10v
V1
FREQ = 50hzVAMPL = 5vVOFF = 0
R2
90k
V2
3.3v
0
R1
10k
0
V
V3
10v
0
Time
0s 5ms 10ms 15ms 20ms 25ms 30ms 35ms 40ms
V(V1:+) V(U1:OUT)
-10V
-5V
0V
5V
10V
24
To get the Hysterisis curve
1. Choose Axis settings from Plot menu
2. Under X-axis select axis variable
3. Select the input wave
4. Click Apply and ok
5. From the output check the UTP and LTP values
6. When Vin >UTP The o/p changes from +vsat(10v) to –vsat(-10v)
7. When vin< LTP the o/p changes from -vsat(10v) to +vsat(-10v)
UTP=-1 V LTP=-3v
Vref=-2.2v
Hysteresis Curve
V(V1:+)
-5.0V -4.0V -3.0V -2.0V -1.0V 0.0V 1.0V 2.0V 3.0V 4.0V 5.0V
V(V1:+) V(U1:OUT)
-10V
-5V
0V
5V
10V
Time
0s 5ms 10ms 15ms 20ms 25ms 30ms 35ms 40ms
V(V1:+) V(U1:OUT)
-10V
-5V
0V
5V
10V
25
V(V1:+)
-5.0V -4.0V -3.0V -2.0V -1.0V 0.0V 1.0V 2.0V 3.0V 4.0V 5.0V
V(V1:+) V(U1:OUT)
-10V
-5V
0V
5V
10V
26
5.b. To implement a rectangular waveform generator (Op-Amp relaxation oscillator) using
a simulation package and observe the change in frequency when all resistor values are
doubled.
TYPE OF ANALYSIS : TIME DOMAIN
RUN TO TIME : 4ms
MAXIMUM STEP SIZE : 0.01ms
Time period =.3ms
Frequency=3.3 KHz
Resistor values Doubled
V1
12v
C1
.1uf
U1
uA741
3
2
74
6
1
5+
-
V+
V-
OUT
OS1
OS2
R3
10k
V
0
V2
12v
RELAXATION OSCILLATOR
0R1
1k
R2
10k
0
0
V
Time
2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms
V(R3:1) V(R3:2)
-20V
-10V
0V
10V
20V
27
Time period =.5ms
Frequency= 2khz
Conclusion:As the resistor values are increased Time period increases and so the frequency
decreases
Time
2.0ms 2.5ms 3.0ms 3.5ms 4.0ms 4.5ms 5.0ms
V(R3:1) V(R3:2)
-20V
-10V
0V
10V
20V
28
6. To design and implement an astable multivibrator using 555 Timer for a given frequency
and duty cycle.
COMPONENTS REQUIRED: 555 Timer IC, Resistors of 3.3KΩ, 6.8KΩ, Capacitors of 0.1
μF, Regulated power supply, CRO
Capacitance
This is a measure of a capacitor's ability to store charge. A large capacitance means that more
charge can be stored. Capacitance is measured in farads, symbol F.
Three prefixes (multipliers) are used, µ (micro), n (nano) and p (pico):
µ means 10-6
n means 10-9
p means 10-12
Capacitor Number Code for Electrolytic capacitors
A number code is often used on small capacitors where printing is difficult:
the 1st number is the 1st digit,
the 2nd number is the 2nd digit,
the 3rd number is the number of zeros to give the capacitance in pF.
Ignore any letters - they just indicate tolerance and voltage rating.
For example: 102 means 1000pF = 1nF (not 102pF!)
101 means 100pF=.1nF
Colour Coding For Resistors
Resistor values are always coded in ohms ( symbol Ω), capacitors in picofarads (pF)
A 100 kΩ, 5%
band A is first significant figure of component value
band B is the second significant figure
band C is the decimal multiplier
band D if present, indicates tolerance of value in percent (no color means 20%)
29
Color Significant
figures Multiplier Tolerance
Temp. Coefficient
(ppm/K)
Black 0 ×100 – 250 U
Brown 1 ×101 ±1% F 100 S
Red 2 ×102 ±2% G 50 R
Orange 3 ×103 – 15 P
Yellow 4 ×104 – 25 Q
Green 5 ×105 ±0.5% D 20 Z
Blue 6 ×106 ±0.25% C 10 Z
Violet 7 ×107 ±0.1% B 5 M
Gray 8 ×108 ±0.05% A 1 K
White 9 ×109 – –
Gold – ×10-1
±5% J –
Silver – ×10-2
±10% K –
None – – ±20% M –
DESIGN: Given frequency (f) = 1KHz and duty cycle = 60% (=0.6)
The time period T =1/f = 1ms = tH + tL
Where tH is the time the output is high and tL is the time the output is low.
For an astable multivibrator using 555 Timer we have
tL = 0.693 RB C ------(1)
tH = 0.693 (RA + RB)C ------(2)
T = tH + tL = 0.693 (RA +2 RB) C
Duty cycle = tH / T = 0.6. Hence tH = 0.6 T = 0.6ms and tL = T – tH = 0.4ms.
Let C=0.1μF and substituting in the above equations,
So RB= tL/0.693 x C =0.4 x 10-3
/0.693 x .1x10-6
=5.8kΩ
RA = tH - 0.693 x Cx RB / 0.693 x C
=0.6 x 10-3
x0.693 x5.8 x 103 0.1x x10
-6/0.693 x .1x10
-6=2.9kΩ
RB = 5.8KΩ (from equation 1) and RA = 2.9KΩ (from equation 2 & RB values).
The Vcc determines the upper and lower threshold voltages (observed from the capacitor voltage
waveform) as CCLTCCUT VVVV3
1&
3
2.
30
Note: The duty cycle determined by RA & RB can vary only between 50 & 100%. If RA is much
smaller than RB, the duty cycle approaches 50%.
Circuit Diagram
Connect the pin 2 to the CRO to get the capacitor waveform check the amplitude from the
waveform to get the UTP and LTP values
Connect pin 3 to CRO to get the output. Find out the TH and TL values
PROCEDURE:
1. Before making the connections, check the components using multimeter.
2. Make the connections as shown in figure and switch on the power supply.
3. Observe the capacitor voltage waveform at 6th
pin of 555 timer on CRO.
4. Observe the output waveform at 3rd
pin of 555 timer on CRO (shown below).
5. Note down the amplitude levels, time period and hence calculate duty cycle.
Example 2: frequency = 1kHz and duty cycle =75%, RA = 7.2kΩ & RB =3.6kΩ,
Duty cycle = tH / T = 0.75. Hence tH = 0.75T = 0.75ms and tL = T – tH = 0.25ms.
Let C=0.1μF and substituting in the above equations,
So RB= tL/0.693 x C =0.25 x 10-3
/0.693 x .1x10-6
=3.6kΩ
RA = (tH - 0.693 x RB x C) /0.693 x C
=0.75 x 10-3
x 0.693 x 3.6 x 103 x 0.1x 10
-6 / 0.693 x .1x10
-6=7.2kΩ
Choose RA = 6.8kΩ and RB = 3.3kΩ.
RESULT:
The frequency of the oscillations = _1KHz.
WAVEFORMS
31
THEORY:
Multivibrator is a form of oscillator, which has a non-sinusoidal output. The output
waveform is rectangular. The multivibrators are classified as: Astable or free running
multivibrator: It alternates automatically between two states (low and high for a rectangular
output) and remains in each state for a time dependent upon the circuit constants. It is just an
oscillator as it requires no external pulse for its operation. Monostable or one shot
multivibrator: It has one stable state and one quasi stable. The application of an input pulse
triggers the circuit time constants. After a period of time determined by the time constant, the
circuit returns to its initial stable state. The process is repeated upon the application of each
trigger pulse. Bistable Multivibrators: It has both stable states. It requires the application of an
external triggering pulse to change the output from one state to other. After the output has
changed its state, it remains in that state until the application of next trigger pulse. Flip flop is an
example.
Threshold voltage VUT VLT
theoretical (2/3 x Vcc)=3.3V (1/3 x Vcc)=1.6V
practical
Each division in oscilloscope is 0.2
Time=no of div in x-axis x time base
Amplitude= no of div in y-axis x volt/div
Duty cycle= (Ton/Ton +Toff) *100
Duty cycle Duty cycle Ton Toff
Theoretical 75% .6ms .4ms
Practical
32
Type of analysis: TIME DOMAIN (TRANSIENT)
Run to time: 100msec step size:0.01msec
33
Waveforms:Load Resistor RL=1k
RL=500
RL=100
Conclusion: when The resistor values are decreases Load current increases and the ripple
increases
Time
0s 10ms 20ms 30ms 40ms 50ms 60ms 70ms 80ms 90ms 100ms
V(D2:2) V(R1:2)
-5V
0V
5V
10V
Time
0s 10ms 20ms 30ms 40ms 50ms 60ms 70ms 80ms 90ms 100ms
V(D2:2) V(R1:2)
-5V
0V
5V
10V
Time
0s 10ms 20ms 30ms 40ms 50ms 60ms 70ms 80ms 90ms 100ms
V(D2:2) V(R1:2)
-5V
0V
5V
10V
34
1.a. Given a four variable expression, simplify using Entered Variable Map (EVM) and
realize the simplified logic using 8:1 MUX.
E.g., Simplify the function using MEV technique
f(a,b,c,d)=∑m(2,3,4,5,13,15)+dc(8,9,10,11)
Components used: IC 74 LS151, patch chords, power chords, trainer kit.
Theory
The term multiplex means “many to one”. A multiplexer (MUX) has n inputs. Each line is
used to shift digital data serially. There is a single output line.
Pin diagram of IC 74LS151
Decimal LSB f MEV map entry
00
1
12
3
24
5
36
7
48
9
510
11
612
13
714
15
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111
0
0
1
1
1
1
0
0
X
X
X
X
0
1
0
1
0------Do
1------D1
1-----D2
0-----D3
X-----D4
X-----D5
d----D6
d----D7
( Solution is given
below)
35
D—Map Entered Variable
000 001 010 011 100 101 110 111
0000
0
0010
1
0100
1
0110
0
1000
X
1010
X
1100
0
1110
0
0001
0
0011
1
0101
1
0111
0
1001
X
1011
X
1101
1
1111
1
D0=0 D1=1 D2=1 D3=0 D4=X D5=X D6=d D7=d
Circuit Diagram
0
1
O/p
`
D
Low
A B C
ABC
0
1
D
04 D0 Vcc 16
03 D1 GND 8
02 D2
01 D3
15 D4
14 D5 Y5
13 D6
12 D7
9 10 11
7 STROBE S2 S1 S0
36
Procedure:
1. Verify all components & patch chords whether they are in good condition or not.
2. Make connections as shown in the circuit diagram.
3. Give supply to the trainer kit.
4. Provide input data to circuit via switches.
5. Verify truth table sequence & observe outputs.
37
VHDL VHDL stands for Very High Speed Integrated Circuit Hardware Description
Language. It describes the behavior of an electronic circuit or system, from which the
physical circuit or system can then be implemented
VHDL was originally intended to serve 2 main purposes-
• It was used as a documentation language for describing the structure of complex
digital circuits.
• VHDL provides features for modeling the behavior of a digital circuit.
General Features of VHDL
• The language can be used as an exchange medium between chip vendors and CAD
tool user and can be used as communication medium between CAD and CAE tools.
• It supports hierarchy.
• It is not a case sensitive language.
• It is strongly type checked language.
• It provides design portability and flexible design methodologies: top down, bottom
up or mixed
• It supports both synchronous and asynchronous timing models.
• Nominal Propagation delays, min-max delays, setup and hold timing constraint and
spike detection can be described in this language.
Usage of the Tool
It is one of most popular software tool used to synthesize VHDL code. This tool Includes
many steps. To make user feel comfortable with the tool the steps are given below:-
Select NEW PROJECT in FILE MENU.
Enter following details as per your convenience
Project name : sample (should be same as the entity name in
your VHDL code
Project location : C:\example( As per convenience use default
Top level module : HDL
In NEW PROJECT dropdown Dialog box, Choose your appropriate device
specification. Example is given below:
Device family : cyclone
Device : EP1C6Q240
Package : PQFP
Pincount :240
Speed grade 8
On File Drop down menu choose new Vhdl file
Type the Vhdl code
Under the Processing Drop down menu choose Start compilation
If there are errors go back to the VHDL code and correct it. Once the
compilation is successfull
38
Under the processing drop down box select simulator tool select the
simulator mode to functional and click on generate functional simulation
netlist. we Get the success message.
Under simulator tool click on open. In the empty location right click . Click on
insert on NODE or BUS. Then click on NODE finder. In the window opened
select pins to unassigned. Click on List . IT will list all the Net list seloct all
and click ok.
The input and output appears give appropriate input.
On the simulator tool click on start simulation. Simulation success message
will be prompted
on simulator tool click on report to see the output.
Create a new project for every new VHDL code
The primary data type std_ulogic (standard unresolved logic) consists of nine character literals in
the following order:
'U' - uninitialized
'X' - strong drive, unknown logic value
'0' - strong drive, logic zero
'1' - strong drive, logic one
'Z' - high impedance
'W' - weak drive, unknown logic value
'L' - weak drive, logic zero
'H' - weak drive, logic one
'-' - don't care
39
1.b. Write the verilog /VHDL code for 8:1 MULTIPLEXER. Simulate
and verify its working.
TruthTable
INPUTS OUTPUTS
SEL (2) SEL (1) SEL (0) Zout
0 0 0 I(0)
0 0 1 I(1)
0 1 0 I(2)
0 1 1 I(3)
1 0 0 I(4)
1 0 1 I(5)
0 1 1 I(6)
1 1 1 I(7)
VHDL code for 8 to 1 mux (Behavioral modeling).
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
entity mux1 is
Port ( I : in std_logic_vector(7 downto 0);
sel : in std_logic_vector(2 downto 0);
zout : out std_logic);
end mux1;
architecture Behavioral of mux1 is
begin
zout<= I(0) when sel="000" else
I(1) when sel="001" else
I(2) when sel="010" else
I(3) when sel="011" else
I(4) when sel="100" else
I(5) when sel="101" else
I(6) when sel="110" else
I(7);
end Behavioral;
MULTIPLEXER
8 TO 1
8
Zout I
SEL
3
40
41
2.a. Realize a full adder using 3-8 decoder IC and 4 input NAND. Components used: IC 74 LS138, IC 74LS20, patch chords, power chords, trainer kit.
Pin diagram of ICs used:
42
Theory:
The simplest Binary adder is a half adder. It has 2 inputs and 2 output bits. One is the sum
and the other is carry.
A half adder has no provision to add carry of lower order bits when binary numbers are
added. When two input bits and a carry are to be added, the number of input bits become 3
and input combination increases to 8. For this a full adder is used. Like half adder, it has 2
outputs. One is sum and the other is carry. New carry generated is denoted as Cn and carry
generated from addition of previous lower order bits is denoted as Cn-1
Function table
Circuit
Diagram
A B C S C
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
0
1
1
0
1
0
0
1
0
0
0
1
0
1
1
1
43
Procedure:
1.Verify all components & patch chords whether they are in good condition or not.
2. Make connections as shown in the circuit diagram.
3. Give supply to the trainer kit.
4. Provide input data to circuit via switches.
5. Verify truth table sequence & observe outputs.
44
2.b. Write the verilog/ VHDL code for FULL ADDER. Simulate and verify its working.
Truth Table
INPUTS OUTPUTS
X Y Z SUM CARRY
0 0 0 0 0
0 0 1 1 0
0 1 0 1 0
0 1 1 0 1
1 0 0 1 0
1 0 1 0 1
1 1 0 0 1
1 1 1 1 1
Expression: Sum(S)=X Y Z
Carry(Cout) = XY +YZ + ZX
FULL
ADDER
S
Cout
X
Y
Z
45
VHDL code for Full adder using data flow
library IEEE;
use IEEE.STD_LOGIC_1164.ALL
use IEEE.STD_LOGIC_ARITH.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;
ENTITY fa IS
Port (X, Y, Z : in STD_LOGIC;
Cout, S : out STD_LOGIC);
End fa;
ARCHITECTURE df OF fa IS
Begin
Cout<= (X AND Y) OR( Y AND Z) OR (X AND Z);
S<= (X XOR Y) XOR Z;
End df;
SUM CARRY
46
3.a.Realize a J-K Master/Slave FF using NAND gates and verify its
truth table.
Components used: IC 74 LS00, IC 74LS10, patch chords, trainer kit.
Pin diagram of ICs used:
7400
7410
Clk J K Q ---
Q
comment
0
0
1
1
0
1
0
1
Q0
0
1
Q0
----
Q0
1
0
Q0
No change
Reset
Set
toggle
47
Theory:
The circuit below shows the solution. To the RS flip-flop we have added two new
connections from the Q and Q' outputs back to the original input gates. Remember that
a NAND gate may have any number of inputs, so this causes no trouble. To show that
we have done this, we change the designations of the logic inputs and of the flip-flop
itself. The inputs are now designated J (instead of S) and K (instead of R). The entire
circuit is known as a JK flip-flop.
In most ways, the JK flip-flop behaves just like the RS flip-flop. The Q and Q' outputs
will only change state on the falling edge of the CLK signal, and the J and K inputs will
control the future output state pretty much as before. However, there are some important differences.
Since one of the two logic inputs is always disabled according to the output state of the
overall flip-flop, the master latch cannot change state back and forth while the CLK input is at logic 1. Instead, the enabled input can change the state
of the master latch once, after which this latch will not change again. This was not true
of the RS flip-flop.
If both the J and K inputs are held at logic 1 and the CLK signal continues to change,
the Q and Q' outputs will simply change state with each falling edge of the CLK signal.
(The master latch circuit will change state with each rising edge of CLK.) We can use
this characteristic to advantage in a number of ways. A flip-flop built specifically to
operate this way is typically designated as a T (for Toggle) flip-flop. The lone T input is in fact the CLK input for other types of flip-flops.
The JK flip-flop must be edge triggered in this manner. Any level-triggered JK latch
circuit will oscillate rapidly if all three inputs are held at logic 1. This is not very useful.
For the same reason, the T flip-flop must also be edge triggered. For both types, this is
48
the only way to ensure that the flip-flop will change state only once on any given clock pulse.
Because the behavior of the JK flip-flop is completely predictable under all conditions,
this is the preferred type of flip-flop for most logic circuit designs. The RS flip-flop is
only used in applications where it can be guaranteed that both R and S cannot be logic 1 at the same time.
At the same time, there are some additional useful configurations of both latches and
flip-flops. In the next pages, we will look first at the major configurations and note their
properties. Then we will see how multiple flip-flops or latches can be combined to perform useful functions and operations.
Master Slave Flip Flop:
The control inputs to a clocked flip flop will be making a transition at approximately the
same times as triggering edge of the clock input occurs. This can lead to unpredictable
triggering.
A JK master flip flop is positive edge triggered, where as slave is negative edge triggered.
Therefore master first responds to J and K inputs and then slave. If J=0 and K=1, master
resets on arrival of positive clock edge. High output of the master drives the K input of the
slave. For the trailing edge of the clock pulse the slave is forced to reset. If both the inputs
are high, it changes the state or toggles on the arrival of thepositive clock edge and the
slave toggles on the negative clock edge. The slave does exactly what the master does.
Procedure:
1Verify all components & patch chords whether they are in good condition or not.
1. Make connections as shown in the circuit diagram.
2. Give supply to the trainer kit.
3. Provide input data to circuit via switches.
5. Verify truth table sequence & observe outputs.
3.b. Write the verilog/ VHDL code for D Flip-Flop with positive- edge
triggering. Simulate and verify its working.
Truth table
clk D(input) Q(output) QBar
- 0 No change
49
- 1 No change
0 0 1
1 1 0
VHDL code for D Flip Flop Counter.
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
entity dff is
Port ( D,Clk : in std_logic; Q : out std_logic;
Qbar : out std_logic:='1'); end dff;
architecture Behavioral of dff is
begin
process(clk) begin
if rising_edge(clk) then Q<= D;
Qbar<= not D; end if;
end process;
end Behavioral;
D-ff Simulation Results
50
4.a. Design and implement a mod n (a<8)synchronous up counter
using JK FF. Components used: IC 74 LS76, IC 74LS08, patch chords, trainer kit.
PIN DIAGRAM OF JK FLIP FLOP 74LS76 ICS ( TWO JK FLIP FLOP IN ONE ICS))
FOR INITIALIZATION EITHER FOR 1S OR FOR 0S
PR1,PR2-Preset Used To Store 1 By Making PR=LOW AND CLR=HIGH
CLR1,CLR2-Clear Used To Store 0 By Making CLR= LOW AND PR=HIGH
FOR NORMAL OPERATION THAT IS ,DEPENDS ON INPUT:-
PR1 ,PR2,CLR1,CLR2=HIGH
for initialization
for normal
51
PIN DIAGRAM OF 74LS08 ICS( 2 INPUT AND GATE( FOUR
AND IN ONE ICS))
FUNCTIONAL TRUTH TABLE FOR J-K FLIP FLOP
J K Qn Qn+1
0 0 0
1
0
1
0 1 0
1
0
0
1 0 0
1
1
1
1 1 0
1
1
0
STATE SYNTHESIS TABLE FOR JK FLIP FLOP
Present state Next state J K
0 0 0 X
0 1 1 X
1 0 X 1
1 1 X 0
52
Theory:
The ripple counter requires a finite amount of time for each flip flop to change state. This
problem can be solved by using a synchronous parallel counter where every flip flop is
triggered in synchronism with the clock, and all the output which are scheduled to change
do so simultaneously.
The counter progresses counting upwards in a natural binary sequence from count 000 to
count 100 advancing count with every negative clock transition and get back to 000 after
this cycle.
Synchronous counter design
To successfully design synchronous counters we may employ the following six basic steps:
1. Create the state transition diagram.
2. Create a present state-next state table (often referred to as the next state table).
3. Expand the table to form the transition table for each flip-flop in the circuit. The
transition table shows the flip-flop inputs required to make the counter go from
present state to the desired next state. This is also referred to as the excitation
table.
4. Determine the logic functions of the J and K inputs as a function of the present
states.
5. Analyse the counter to verify the design. 6. Construct and test the counter.
Let us employ these techniques to design a MOD-8 counter to count in the following sequence: 0, 1, 2, 3, 4, 5, 6, 7.
Step1: Creating state transition diagram.
53
Step 2: Creating present state-next state table
Present State Next State
Qc Qb Qa Qc Qb Qa
0 0 0 0 0 1
0 0 1 0 1 0
0 1 0 0 1 1
0 1 1 1 0 0
1 0 0 1 0 1
1 0 1 1 1 0
1 1 0 1 1 1
1 1 1 0 0 0
Step 3: Expand the present state-next state table to form the transition table.
Present
State
Next State Present inputs
Qc QB QA QC QB QA JC Kc JB KB JA KA
0 0 0 0 0 1 0 X 0 X 1 X
0 0 1 0 1 0 0 X 1 X X 1
0 1 0 0 1 1 0 X X 0 1 X
0 1 1 1 0 0 1 X X 1 X 1
1 0 0 1 0 1 X 0 0 X 1 X
1 0 1 1 1 0 X 0 1 x X 1
1 1 0 1 1 1 X 0 X 0 1 X
1 1 1 0 0 0 X 1 X 1 X 1
„X‟ indicates a "don‟t care" condition.
54
Step 4: Use Karnaugh maps to identify the present state logic functions for each of the inputs.
BA
00 01 11 10
1
X X X X
Jc=QaQb
BA
00 01 11 10
X X X X
1
Kc=QaQb
BA
00 01 11 10
0 1 X X
0 1 X X
JB=Qa
AB
00 01 11 10
X X 1 0
X X 1 0
Kb=Qa
AB
00 01 11 10
1 X X 1
1 X X 1
Ja=1
C
c 0
1
C
0
1
C
0
1
C
0
1
C
0
1
55
BA
00 01 11 10
X 1 1 X
X 1 1 X
Ka=1
Step 5: Trace through indicates circuit should work correctly.
Step 6: Constructing Circuit
A three-bit synchronous counter
*NOTE:-CONNECT PRESET AND CLEAR TO HIGH TO WORK ON NORMAL OPERATION
4.b. Write the verilog/ VHDL code for mod-8 up counter. Simulate
and verify its working.
Truth table
rst clock Q2 Q1 Q0
1 x 0 0 0
0 0 0 1
0 0 1 0
0 0 1 1
0 1 0 0
C
0
1
56
0 1 0 1
0 1 1 0
0 1 1 1
0 0 0 0
VHDL code for Mod-8 Counter.
library IEEE;
use IEEE.STD_LOGIC_1164.ALL; use IEEE.STD_LOGIC_ARITH.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;
entity mod8 is
Port ( rst,clk: in std_logic; q : buffer std_logic_vector(2 downto 0):="000");
end mod8;
architecture Behavioral of mod8 is
begin
process(clk,rst) is begin
IF (CLK'event and clk='0') then If(rst='1') then
q<="000"; Else
q <= q+1; End if;
End if; End process;
End Behavioral;
57
Mod 8 Counter Simulation Results
Output
58
5.a. Design and Implement a Ring Counter using 4 bit Shift Register.
Components used: IC 74 LS95, patch chords, trainer kit.
Pin diagram of ICs used:
Pin Diagram of IC: 7495
Mode Control Input : If Mc= 1 Parallel input Mc=0 Serial input
Theory:
Ring Counter is a basic register with direct feedback such that contents of the register
simply circulate around the register when the clock is running. Here last output Qd in a shift
register is connected back to the serial input.
Function Table:
Clk Qa Qb Qc Qd
0
1
2
3
1
0
0
0
0
1
0
0
0
0
1
0
0
0
0
1
59
Circuit Diagram: CLK(Mono pulse)
Qd
Vcc Qa Qb Qc
A B C D MC gnd
1. Make Mc=1 Give parallel input 1000 through ABCD
2. Then make Mc=0 for shift operation
Procedure: 1 Verify all components & patch chords whether they are in good condition or not.
2 Make connections as shown in the circuit diagram.
3 Give supply to the trainer kit.
2 Provide input data to circuit via switches .
4. Verify truth table sequence & observe outputs.
14 13 12 11 10 9 8
IC 7495
1 2 3 4 5 6 7
60
5.b. Write the verilog/ VHDL code for switched tail counter. Simulate and
verify its working.
Truth table
clk Q3 Q2 Q1 Q0
0 0 0 0 0
1 0 0 0 1
1 0 0 1 1
1 0 1 1 1
1 1 1 1 1
1 1 1 1 0
1 1 1 0 0
1 1 0 0 0
1 0 0 0 0
1 0 0 0 1
1 0 0 1 1
1 0 1 1 1
1 1 1 1 1
61
VHDL code for Johnson counter.
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
use IEEE.STD_LOGIC_ARITH.ALL;
use IEEE.STD_LOGIC_UNSIGNED.ALL;
entity jc is
Port ( clk : in std_logic;
Q : buffer std_logic_vector(3 downto 0));
end jc;
architecture Behavioral of jc is
begin
process(clk)
begin
IF (CLK'event and clk='1') then
Q(3 downto 1)<=Q(2 downto 0);
Q(0)<=not Q(3);
end if;
end process;
end behavioral;
Johnson Counter Simulation Results
62
OUTPUT
63
6) Design and implement asynchronous counter using decade counter IC to count
up from 0 to n (n≤9).
Components used: IC 74 LS90, patch chords, trainer kit.
Pin Diagram of 7490
Theory:
Asynchronous counter is a counter in which the clock signal is connected to the clock input
of only first stage flip flop. The clock input of the second stage flip flop is triggered by the
output of the first stage flip flop and so on. This introduces an inherent propagation delay
time through a flip flop. A transition of input clock pulse and a transition of the output of a
flip flop can never occur exactly at the same time. Therefore, the two flip flops are never
simultaneously triggered, which results in asynchronous counter operation.
1.Cp0(pin 14) to be connected to clock
2.Cp1(pin1) to be connected to Q0.( The output of the first flip flop drives the second clock)
64
For decade counter(mod 10)
MR1 AND MR2-CLEAR ALL FILPFLOP (HIGH ACTIVE AND LOW FOR NOT
ACTIVE)
MS1 AND MS2- SET ALL FLIP FLOP (HIGH ACTIVE AND LOW FOR NOT
ACTIVE)
CPO-CLOCK PULSE TO FIRST FLIP FLOP
CP1-CLOCK PULSE TO SECOND FLIP FLOP (OUTPUT OF FIRST FLIP FLOP
CLOCK FOR SECOND FILP FLOP)
For mod 9
connect Q0 and Q3 to reset(clear) through an AND gate.
Note :reset should not be connected to the switch
For mod8
Connect Q3 to reset
For mod7
Connect Q2, Q1,Q0 to reset through an And Gate
For Mod 6
Connect Q2 and Q1 to reset through an AND gate
65
For mod 5
Connect Q0 and Q2 to reset through an AND gate
For Mod 4
Connect Q2 to reset through an AND gate
For mod 3
Connect Q1 and Q0 to reset through an AND gate
For mod 2
Connect Q1 to reset
Function Table:
Clock Q3 Q2 Q1 Q0
0 0 0 0 0
1 0 0 0 1
2 0 0 1 0
3 0 0 1 1
4 0 1 0 0
5 0 1 0 1
6 0 1 1 0
8 0 1 1 1
9 1 0 0 0
10 1 0 0 1
Procedure:
1 Verify all components & patch chords whether they are in good condition or not.
2. Make connections as shown in the circuit diagram.
3. Give supply to the trainer kit.
4. Provide input data to circuit via switches.
5. Verify truth table sequence & observe outputs.
66
7.Design a 4-bit R-2R ladder D/A converter using Op-Amp. Determine its accuracy
and resolution.
AIM:
EQUIPMENTS AND COMPONENTS REQUIRED:
Op-amp 741, Resistors , dual power supply, VRPS ,CRO, Bread board.
PROCEDURE:
1. The circuit is connected as per the diagram
2. Give digital inputs from 0000 (2) to 1111(2) in steps using switch arrangement and note down
the corresponding analog voltage.
3. For each case verify the theoretical value with practical value.
4. To find linearity plot the practical analog output values Vs digital inputs on a graph sheet. A
straight line connecting input and output values is drawn.
DESIGN:
Analog output voltage Vo for 4-bit DAC is given by
Vo = (8 D3 + 4 D2+2 D1+1 D0)*V
Where V = Vref. (2R) /24 . (3R)
= Vref/ 24 =5/24=0.20833 Example:
For Digital input D1D,D,D0 1 011
Vo=(8+0+2± 1 ) 0.20833
Vo= 2.291 volts.
•Resolution: This is the number of possible output levels the DAC is designed to reproduce.
This is usually stated as the number of bits it uses, which is the base two logarithm of the number
of levels. Resolution is related to the effective number of bits (ENOB) which is a measurement
of the actual resolution attained by the DAC.
67
Example:-
1 bit DAC is designed to reproduce 2 (21) levels while an 8 bit DAC is designed for 256 (2
8)
levels.
(a)Resolution for 4bit DAC is designed to reproduce 16(24) levels
(b)Accuracy formula= 100-(Theoretical value- Practical value)
Example:-
100-(3.12-3.00)=100-.12=99.88%
Decimal
Equivalent
Digital Input Analog Output voltage
D3 D2 D1 D0 Theoretical Practical
0 1 2 3 4 5 6 7 8 9 10
11
12
13
14
15
68