Assignment Simulasi

ASSIGNMENT 1

OPERATING POINT ANALYSIS

ELECTRONIC CIRCUIT SIMULATION

NAME : AHMAD FAKHRURRAZI BIN AHMAD NOORDEN

I/C N0. : 901109-03-5695

Lecturer’s Name: P.M Dr. Ahmad Radzi Bin Mat Isa

Soalan 1

Buat kajian literature mengenai beberapa model diod yang digunakan dalam perisian SPICE.

Soalan 2

Buat kajian literature mengenai beberapa model transistor bipolar (BJT) yang digunakan dalam perisian SPICE. Sertakan persamaan matematik dan graf yang berkaitan dengan model-model yang diberikan itu

Soalan 3

Tuliskan fail input ( tak perlu jalankan simulasi ) bagi litar dalam Rajah 1 bagi menjalankan simulasi titik operasi.

Soalan 4

Tuliskan fail input bagi litar dalam Rajah 2 untuk menjalankan simulasi analisi titik operasi. Sahkan hasil simulasi dengan pengiraan ke atas litar tersebut.

Soalan 5

Tuliskan fail input bagi litar dalam Rajah 3 untuk menjalankan simulasi analisi titik operasi. Sahkan hasil simulasi dengan pengiraan ke atas litar tersebut.

1.0 LITERATURE REVIEW ON SPICE

1.1 Diode Model

Diode circuit models

The SPICE circuit simulation program provides for modeling diodes in circuit

simulations. The diode model is based on characterization of individual devices as described

in a product data sheet and manufacturing process characteristics not listed. Some

information has been extracted from a 1N4004 data sheet in Figure below.

Data sheet 1N4004 excerpt, after [DI4].

The diode statement begins with a diode element name which must begin with “d” plus

optional characters. Example diode element names include: d1, d2, dtest, da, db, d101. Two

node numbers specify the connection of the anode and cathode, respectively, to other

components. The node numbers are followed by a model name, referring to a subsequent

“.model” statement.

The model statement line begins with “.model,” followed by the model name matching one or

more diode statements. Next, a “d” indicates a diode is being modeled. The remainder of the

model statement is a list of optional diode parameters of the form

ParameterName=ParameterValue. None are used in Example below. Example2 has some

parameters defined. For a list of diode parameters, see Table below.

General form: d[name] [anode] [cathode] [modelname]

.model ([modelname] d [parmtr1=x] [parmtr2=y] . . .)

Example: d1 1 2 mod1

.model mod1 d

Example2: D2 1 2 Da1N4004

.model Da1N4004 D (IS=18.8n RS=0 BV=400 IBV=5.00u CJO=30 M=0.333

N=2)

The easiest approach to take for a SPICE model is the same as for a data sheet: consult the

manufacturer's web site. Table below lists the model parameters for some selected diodes. A

fallback strategy is to build a SPICE model from those parameters listed on the data sheet. A

third strategy, not considered here, is to take measurements of an actual device. Then,

calculate, compare and adjust the SPICE parameters to the measurements.

Diode SPICE parameters

If diode parameters are not specified as in “Example” model above, the parameters

take on the default values listed in Table above and Table below. These defaults model

integrated circuit diodes. These are certainly adequate for preliminary work with discrete

devices For more critical work, use SPICE models supplied by the manufacturer [DIn],

SPICE vendors, and other sources. [smi]

SPICE parameters for selected diodes; sk=schottky Ge=germanium; else silicon.

Otherwise, derive some of the parameters from the data sheet. First select a value for

spice parameter N between 1 and 2. It is required for the diode equation (n). Massobrio

[PAGM] pp 9, recommends ".. n, the emission coefficient is usually about 2." In Table above,

we see that power rectifiers 1N3891 (12 A), and 10A04 (10 A) both use about 2. The first

four in the table are not relevant because they are schottky, schottky, germanium, and silicon

small signal, respectively. The saturation current, IS, is derived from the diode equation, a

value of (VD, ID) on the graph in Figure above, and N=2 (n in the diode equation).

ID = IS(eVD/nVT -1)

VT = 26 mV at 25oC n = 2.0 VD = 0.925 V at 1 A from graph

1 A = IS(e(0.925 V)/(2)(26 mV) -1)

IS = 18.8E-9

The numerical values of IS=18.8n and N=2 are entered in last line of Table above for

comparison to the manufacturers model for 1N4004, which is considerably different. RS

defaults to 0 for now. It will be estimated later. The important DC static parameters are N, IS,

and RS.

Rashid [MHR] suggests that TT, τD, the transit time, be approximated from the

reverse recovery stored charge QRR, a data sheet parameter (not available on our data sheet)

and IF, forward current.

ID = IS(eVD/nVT -1)

τD = QRR/IF

Take the TT=0 default for lack of QRR. Though it would be reasonable to take TT for

a similar rectifier like the 10A04 at 4.32u. The 1N3891 TT is not a valid choice because it is

a fast recovery rectifier. CJO, the zero bias junction capacitance is estimated from the VR vs

CJ graph in Figure above. The capacitance at the nearest to zero voltage on the graph is 30 pF

at 1 V. If simulating high speed transient response, as in switching regulator power supplies,

TT and CJO parameters must be provided.

The junction grading coefficient M is related to the doping profile of the junction.

This is not a data sheet item. The default is 0.5 for an abrupt junction. We opt for M=0.333

corresponding to a linearly graded junction. The power rectifiers in Table above use lower

values for M than 0.5.

Take the default values for VJ and EG. Many more diodes use VJ=0.6 than shown in

Table above. However the 10A04 rectifier uses the default, which we use for our 1N4004

model (Da1N4001 in Table above). Use the default EG=1.11 for silicon diodes and rectifiers.

Table above lists values for schottky and germanium diodes. Take the XTI=3, the default IS

temperature coefficient for silicon devices. See Table above for XTI for schottky diodes.

The abbreviated data sheet, Figure above, lists IR = 5 µA @ VR = 400 V,

corresponding to IBV=5u and BV=400 respectively. The 1n4004 SPICE parameters derived

from the data sheet are listed in the last line of Table above for comparison to the

manufacturer's model listed above it. BV is only necessary if the simulation exceeds the

reverse breakdown voltage of the diode, as is the case for zener diodes. IBV, reverse

breakdown current, is frequently omitted, but may be entered if provided with BV.

Figure below shows a circuit to compare the manufacturers model, the model derived

from the datasheet, and the default model using default parameters. The three dummy 0 V

sources are necessary for diode current measurement. The 1 V source is swept from 0 to 1.4

V in 0.2 mV steps. See .DC statement in the netlist in Table below. DI1N4004 is the

manufacturer's diode model, Da1N4004 is our derived diode model.

SPICE circuit for comparison of manufacturer model (D1), calculated datasheet model (D2),

and default model (D3).

SPICE netlist parameters: (D1) DI1N4004 manufacturer's model, (D2) Da1N40004 datasheet

derived, (D3) default diode model.

*SPICE circuit <03468.eps> from XCircuit v3.20

D1 1 5 DI1N4004

V1 5 0 0

D2 1 3 Da1N4004

V2 3 0 0

D3 1 4 Default

V3 4 0 0

V4 1 0 1

.DC V4 0 1400mV 0.2m

.model Da1N4004 D (IS=18.8n RS=0 BV=400 IBV=5.00u CJO=30

+M=0.333 N=2.0 TT=0)

.MODEL DI1N4004 D (IS=76.9n RS=42.0m BV=400 IBV=5.00u CJO=39.8p

+M=0.333 N=1.45 TT=4.32u)

.MODEL Default D

.end

Compare the three models in Figure below. and to the datasheet graph data in Table

below. VD is the diode voltage versus the diode currents for the manufacturer's model,

calculated datasheet model and the default diode model. The last column “1N4004 graph” is

from the datasheet voltage versus current curve in Figure above which we attempt to match.

Comparison of the currents for the three model to the last column shows that the default

model is good at low currents, the manufacturer's model is good at high currents, and our

calculated datasheet model is best of all up to 1 A. Agreement is almost perfect at 1 A

because the IS calculation is based on diode voltage at 1 A. The model grossly over states

current above 1 A.

First trial of manufacturer model, calculated datasheet model, and default model.

Comparison of manufacturer model, calculated datasheet model, and default model to

1N4004 datasheet graph of V vs I.

model model model 1N4004

index VD manufacturer datasheet default graph

3500 7.000000e-01 1.612924e+00 1.416211e-02 5.674683e-03 0.01

4001 8.002000e-01 3.346832e+00 9.825960e-02 2.731709e-01 0.13

4500 9.000000e-01 5.310740e+00 6.764928e-01 1.294824e+01 0.7

4625 9.250000e-01 5.823654e+00 1.096870e+00 3.404037e+01 1.0

5000 1.000000e-00 7.395953e+00 4.675526e+00 6.185078e+02 2.0

5500 1.100000e+00 9.548779e+00 3.231452e+01 2.954471e+04 3.3

6000 1.200000e+00 1.174489e+01 2.233392e+02 1.411283e+06 5.3

6500 1.300000e+00 1.397087e+01 1.543591e+03 6.741379e+07 8.0

7000 1.400000e+00 1.621861e+01 1.066840e+04 3.220203e+09 12.

The solution is to increase RS from the default RS=0. Changing RS from 0 to 8m in

the datasheet model causes the curve to intersect 10 A (not shown) at the same voltage as the

manufacturer's model. Increasing RS to 28.6m shifts the curve further to the right as shown in

Figure below. This has the effect of more closely matching our datasheet model to the

datasheet graph (Figure above). Table below shows that the current 1.224470e+01 A at 1.4 V

matches the graph at 12 A. However, the current at 0.925 V has degraded from 1.096870e+00

above to 7.318536e-01.

Second trial to improve calculated datasheet model compared with manufacturer model and

default model.

Changing Da1N4004 model statement RS=0 to RS=28.6m decreases the current at VD=1.4 V

to 12.2 A.

.model Da1N4004 D (IS=18.8n RS=28.6m BV=400 IBV=5.00u CJO=30

+M=0.333 N=2.0 TT=0)

model model 1N4001

index VD manufacturer datasheet graph

3505 7.010000e-01 1.628276e+00 1.432463e-02 0.01

4000 8.000000e-01 3.343072e+00 9.297594e-02 0.13

4500 9.000000e-01 5.310740e+00 5.102139e-01 0.7

4625 9.250000e-01 5.823654e+00 7.318536e-01 1.0

5000 1.000000e-00 7.395953e+00 1.763520e+00 2.0

5500 1.100000e+00 9.548779e+00 3.848553e+00 3.3

6000 1.200000e+00 1.174489e+01 6.419621e+00 5.3

6500 1.300000e+00 1.397087e+01 9.254581e+00 8.0

7000 1.400000e+00 1.621861e+01 1.224470e+01 12.

Suggested reader exercise: decrease N so that the current at VD=0.925 V is restored to 1 A.

This may increase the current (12.2 A) at VD=1.4 V requiring an increase of RS to decrease

current to 12 A.

Zener diode: There are two approaches to modeling a zener diode: set the BV parameter to

the zener voltage in the model statement, or model the zener with a subcircuit containing a

diode clamper set to the zener voltage. An example of the first approach sets the breakdown

voltage BV to 15 for the 1n4469 15 V zener diode model (IBV optional):

.model D1N4469 D ( BV=15 IBV=17m )

The second approach models the zener with a subcircuit. Clamper D1 and VZ in Figure

below models the 15 V reverse breakdown voltage of a 1N4477A zener diode. Diode DR

accounts for the forward conduction of the zener in the subcircuit.

<="" a="">

Zener diode subcircuit uses clamper (D1 and VZ) to model zener.

<="" a=""> Tunnel diode: A tunnel diode may be modeled by a pair of field effect transistors

(JFET) in a SPICE subcircuit. [KHM] An oscillator circuit is also shown in this reference.

Gunn diode: A Gunn diode may also be modeled by a pair of JFET's. [ISG] This reference

shows a microwave relaxation oscillator.

• REVIEW:

• Diodes are described in SPICE by a diode component statement referring to .model

statement. The .model statement contains parameters describing the diode. If parameters are

not provided, the model takes on default values.

• Static DC parameters include N, IS, and RS. Reverse breakdown parameters: BV,

IBV.

• Accurate dynamic timing requires TT and CJO parameters

• Models provided by the manufacturer are highly recommended.

1.2 Bipolar Junction Transistors

BJT circuit models

A large variety of bipolar junction transistor models have been developed. One

distinguishes between small signal and large signal models. We will discuss here first the

hybrid pi model, a small signal model, which lends itself well to small signal design and

analysis. The next model is the charge control model, which is particularly well suited to

analyze the large-signal transient behavior of a bipolar transistor. And we conclude with the

derivation of the SPICE model parameters.

1.2.1. Small signal model (hybrid pi model)

The hybrid pi model of a BJT is a small signal model, named after the “”-like equivalent

circuit for a bipolar junction transistor. The model is shown in Figure 1.1. It consists of an

input impedance, r, an output impedance r0, and a voltage controlled current source described

by the transconductance, gm. In addition it contains the base-emitter capacitances, the junction

capacitance, Cj,BE, and the diffusion capacitance, Cd,BE, and the base-collector junction

capacitance, Cj,BC, also referred to as the Miller capacitance.

Figure 1.1 :Small signal model (hybrid pi model) of a bipolar junction transistor.

The transconductance, gm, of a bipolar transistor is defined as the change in the collector

http://ecee.colorado.edu/~bart/book/book/chapter5/ch5_6.htm#fig5_6_1


current divided by the change of the base-emitter voltage.

(1.1)

The base input resistance, r, is defined as the change of the emitter-base voltage divided by

the change of the base current.

(1.2)

The output resistance, ro, is defined as:

(1.3)

The base-emitter and base-collector junction capacitances are given by:

(1.4)

(1.5)

for the case where the base-emitter and base-collector junctions are abrupt. Since the base-

emitter is strongly forward biased in the forward active mode of operation, one has to also

include the diffusion capacitance of the base:

(1.6)

Based on the small signal model shown in Figure 1.1, we can now calculate the small

signal current gain versus frequency, hfe, of a BJT biased in the forward active mode and

connected in a common emitter configuration. The maximum current gain is calculated while

shorting the output, resulting in:

(1.7)

The unity gain frequency, fT, also called the transit frequency is obtained by setting the small


signal current gain, hfe, equals to one, resulting in:

(1.8)

This transit frequency can be expressed as a function of the transit time, :

(1.9)

Where the transit time, , equals:

(1.10)

The circuit model therefore includes the charging time of the base-emitter capacitance,

E, as well as the base transit time, B, but not the transit time of the carriers through the base-

collector depletion region, C.

The collector transit time:

(1.10a)

The total transit time then becomes:

(1.11)

The corresponding transit frequency, fT, can still be calculated using (5.6.9).

While the unity gain frequency, fT, is an important figure of merit of a bipolar transistor,

another even more important figure of merit is the maximum oscillation frequency, fMAX. This

figure of merit predicts the unity power gain frequency and as a result indicates the maximum

frequency at which useful power gain can be expected from a device. The maximum

oscillation frequency, fMAX, is linked to the transit frequency, fT, and is obtained from:

(1.12)

Where RB is the total base resistance and Cj,BC is the base-collector capacitance. The total

base resistance consists of the series connection of metal-semiconductor contact resistance,

the resistance between the base contact metal and the emitter and the intrinsic base resistance.

http://ecee.colorado.edu/~bart/book/book/chapter5/ch5_eq.htm#eq5_6_9

Assuming a base contact, which is longer than the penetration depth this base resistance

equals

(1.13)

for a one-sided base contact, where Rs,c, Rs,BE and Rs are the sheet resistances under the base

contact, between the base contact and the emitter and underneath the emitter respectively. Ls,E

is the emitter stripe length of the emitter, Ws,E is the emitter stripe width of the emitter and L

is the alignment distance between the base contact and emitter. For a double-sided base

contact, the total base resistance equals

(1.14)

The base-collector capacitance equals:

(1.15)

Where AC is the base-collector area.

5.6.2. Large signal model (Charge control model)

The charge control model of a bipolar transistor is an extension of the charge control

model of a p-n diode. Assuming the “short” diode model to be valid, one can express the

device currents as a function of the charges in each region, divided by the corresponding

transit or lifetime. In the general case one considers the forward bias charges as well as the

reverse bias charges. This results in:

(1.16)

(1.17)

(1.18)

Under forward active mode of operation, this model can be simplified since the reverse

mode components can be ignored. A transient model can be obtained by adding the rate of

change of the charges over time. To further simplify the model, we also ignore the minority

carrier charge, Qp,E, in the emitter. This results in the following equations:

(1.19)

(1.20)

(5.6.21)

As an example we now apply this charge control model to the abrupt switching of a bipolar

transistor. Consider the circuit as one applies a positive voltage to the base, the base-emitter

junction will become forward biased so that the collector current will start to rise. The input is

then connected to a negative supply voltage, VR. This reverses the base current and the base-

emitter junction capacitance is discharged. After this transient, the transistor is eventually

turned off and the collector current reduces back to zero. A full analysis would require solving

the charge control model equations simultaneously, while adding the external circuit

equations. Such approach requires numeric simulation tools.

To simplify this analysis and provide insight, we now assume that the base current is constant

before and after switching. This approximation is very good under forward bias since the

base-emitter voltage is almost constant. Under reverse bias, the base current will vary as the

base-emitter voltage varies, but conceivably one could design a circuit that does provide a

constant reverse current.

The turn-on of the BJT consists of an initial delay time, td,1, during which the base-emitter

junction capacitance is charged. This delay is followed by the increase of the collector current,

quantified by the rise time, trise. This rise time is obtained by applying the charge control

equation for the base current, while applying a base current IBB with the voltage source VBB:

(1.22)

where:

(1.23)

This differential equation can be solved resulting in:

(1.24)

If the device does not reach saturation, the charge reaches its steady state value with a time

constant tr,B, which equals the base transit time of the BJT. The corresponding collector

current will be proportional to the excess minority carrier charge until the device reaches

saturation or:

(1.25)

A larger base voltage, VBB, will therefore result in a larger charging current, IBB, which in turn

decreases the rise time and causes the BJT to saturate more quickly. There also will be more

excess minority carrier charge stored in the base region after the BJT is turned on. The rise

time, trise, is then obtained by finding the time when the saturation current is reached or:

(1.26)

While switching back to the negative power supply, VR, the base current is reversed. As long

as significant charge is still stored in the base region, the collector current will continue to

exist. Only after this excess charge is removed, will the base-emitter junction capacitor be

discharged and the BJT be turned off. The removal of the excess charge can take a significant

delay time labeled as td,2 on the figure. Again we can calculate the time evolution of the excess

charge and calculate the collector current from it. To first order the delay time, td,2, equals:

(1.27)

This delay time can be significantly larger than the rise time trise. Also note that a higher base

turn-on current IBB results in a larger turn-off delay as more minority carrier charge is stored in

the base.

The actual fall time, tf, depends on the remaining storage charge at the onset of saturation as

well as the charge stored by the base-emitter junction capacitance.

Figure 1.2. :Switching behavior of a BJT: a) bias circuit used to explain the

switching behavior. b) Applied voltage and resulting collector current.


The SPICE model of a bipolar transistor includes a variety of parasitic circuit elements and

some process related parameters in addition to the elements previously discussed in this

chapter. The syntax of a bipolar transistor incorporates the parameters a circuit designer can

change as shown below:

BJT syntax

Q []

+ []

.MODEL NPN(BF= BR= IS= CJE=

+ CJC= VJE= VJC= VAF= VAR=

+ NF= NR=)

BJT Parameters

BF Forward active current gain

BR Reverse active current gain

IS Transport saturation current

CJE Base-emitter zero-bias junction capacitance

CJC Base-collector zero-bias Junction capacitance

VJE Base-emitter built-in potential

VJC Base-collector built-in potential

VAF Forward mode Early voltage

VAR Reverse mode Early voltage

NF Forward mode ideality factor

NR Reverse mode ideality factor

Example:

Q1 3 2 1 BJTNAME

.MODEL BJTNAME NPN(BF=100 CJC=20pf CJE=20pf IS=1E-16)

where Q1 is one specific transistor in the circuit, while the transistor model "BJTNAME" uses

the built-in model NPN to specify the process and technology related parameters of the BJT.

The built-in model PNP is used for p-n-p bipolar transistors. A list of SPICE parameters and

their relation to the parameters discussed in this text is provided in the table below.

Table 1.1 : Selected SPICE parameters of a BJT.

In addition, there are several parameters, which can be specified to further enhance the

accuracy of the model, such as:

RB base resistance

RE emitter resistance

RC collector resistance

MJE base-emitter capacitance exponent

http://ecee.colorado.edu/~bart/book/book/chapter5/ch5_6.htm#tab5_6_1

MJC base-collector capacitance exponent

EG energy gap for temperature effect on IS

The exponents NJE and MJC are used to calculate the voltage dependence of the base-emitter

and base-collector junction capacitances using:

(1.28)

(1.29)

This exponent allows the choice between a uniformly doped junction (m = ½), a linearly

graded junction (m = 1/3) or an arbitrarily graded junction for which the exponent must be

independently determined.

The temperature dependence of the transport saturation current is calculated from the energy

bandgap, since the primary temperature dependence is due to the temperature dependence of

the intrinsic carrier density, which results in:

(1.30)

The corresponding equivalent circuit is provided in Figure 5.6.3. The output resistance, ro, was

added to represent the Early effect, which is included in the BJT model by specifying VAF

and VAR.


Figure 5.6.3. : Large signal model of a BJT including the junction

capacitances.

Reference of : Boulder, December 2004


2.0 NESTLIST FILE

Question 3

SOALAN 3

V1 1 0 80V

R14 2 1 470

R12 3 1 390

R23 4 3 560

R45 4 2 220

R56 5 4 180

R49 9 2 1.2K

R58 7 4 680

R67 6 5 120

R78 7 6 91

R910 9 0 820

R89 9 7 120

R811 8 7 68

R110 10 8 390

R100 10 0 220

.OP

.END

Question 4

SOALAN 4

V1 1 0 DC 24

V2 3 0 DC 15

R1 1 2 10K

R2 2 3 8.1K

R3 2 0 4.7K

.OP

.END

Question 5

SOALAN 5

V1 1 0 DC 10V

V2 3 0 DC -10V

R1 1 2 4K

R2 2 0 8K

R3 2 3 2K

.OP

.END

3.0 OUTPUT FILE

Question 3

**** 10/16/11 22:55:20 *********** Evaluation PSpice (Nov 1999) ************** SOALAN 3

**** CIRCUIT DESCRIPTION

V1 1 0 80V

R14 2 1 470

R12 3 1 390

R23 4 3 560

R45 4 2 220

R56 5 4 180

R49 9 2 1.2K

R58 7 4 680

R67 6 5 120

R78 7 6 91

R910 9 0 820

R89 9 7 120

R811 8 7 68

R110 10 8 390

R100 10 0 220

.OP

.END

**** 10/16/11 22:55:20 *********** Evaluation PSpice (Nov 1999) **************

SOALAN 3

**** SMALL SIGNAL BIAS SOLUTION TEMPERATURE = 27.000 DEG C

******************************************************************************

NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE

( 1) 80.0000 ( 2) 55.4750 ( 3) 67.0940 ( 4) 48.5620

( 5) 41.1880 ( 6) 36.2730 ( 7) 32.5450 ( 8) 29.2810

( 9) 30.5640 ( 10) 10.5600

VOLTAGE SOURCE CURRENTS

NAME CURRENT

V1 -8.527E-02

TOTAL POWER DISSIPATION 6.82E+00 WATTS**** 10/16/11 22:55:20 *********** Evaluation PSpice (Nov 1999) **************

SOALAN 3

**** OPERATING POINT INFORMATION TEMPERATURE = 27.000 DEG C

*****************************************************************************

JOB CONCLUDED

TOTAL JOB TIME .02

Question 4

SOALAN 4 -TITLE


******************************************************************************

V1 1 0 DC 24

V2 3 0 DC 15

R1 1 2 10K

R2 2 3 8.1K

R3 2 0 4.7K

.OP

.END


SOALAN 4


******************************************************************************


( 1) 24.0000 ( 2) 9.7470 ( 3) 15.0000


NAME CURRENT

V1 -1.425E-03

V2 -6.485E-04

TOTAL POWER DISSIPATION 4.39E-02 WATTS


SOALAN 4

**** OPERATING POINT INFORMATION TEMPERATURE = 27.000 DEG C

*****************************************************************************

JOB CONCLUDED

TOTAL JOB TIME 0.00

Question 5

** 10/21/11 10:05:53 *********** Evaluation PSpice (Nov 1999) **************

SOALAN 5


******************************************************************************

V1 1 0 DC 10V

V2 3 0 DC -10V

R1 1 2 4K

R2 2 0 8K

R3 2 3 2K

.OP

.END


SOALAN 5


******************************************************************************


( 1) 10.0000 ( 2) -2.8571 ( 3) -10.0000


NAME CURRENT

V1 -3.214E-03

V2 3.571E-03

TOTAL POWER DISSIPATION 6.79E-02 WATTS


SOALAN 5

**** OPERATING POINT INFORMATION TEMPERATURE = 27.000 DEG

******************************************************************************

JOB CONCLUDED

TOTAL JOB TIME .05

4.0 DESCRIPTION

Question 3

This circuit consist of 10 node respect to the ground ‘0’.

The words “SOALAN 3” in the first line is the title of the simulation analysis. Then the second and third line refer to the DC voltage source which regarding to its arrangement in between the node 1 “ 1 0 ” and ground of the circuit. At the line 3 until R100 , this command describe about the resistor place between each state node “R14 2 1” means the resistor between those 2 node and 10 k is its value. “.OP” command is refer to operating point analysis. .END is the end statement.

At words “( 1) 80.0000” it is describing about voltage at node 1 respect to the ground at the node 0. The current of V1 is state at the lower part of the file.

Question 4


The words “SOALAN 4” in the first line is the title of the simulation analysis. Then the second and third line refer to the DC voltage source which regarding to its arrangement in the node 1 “ 1 0 ”and node 3 “ 3 0 ” of the circuit. At the line 3 until 6 , this command describe about the resistor place between each state node “R1 1 2” means the resistor between those 2 node and 10 k is its value. “.OP” command is refer to operating point analysis.

Output File

At words “( 1) 24.0000” it is describing about voltage at node 1 respect to the ground at the node 0. The current of V1 and V2 is state at the lower part of the file.

Question 5


The words “SOALAN 5” in the first line is the title of the simulation analysis. Then the second and third line refer to the DC voltage source which regarding to its arrangement in between the node 1 “ V1 1 0 ” and ground of the circuit. At the line 2 until line 5 , this command describe about the resistor place between each state node

“R 1 2” means the resistor between those 2 node and 4 k is its value. “.OP” command is refer to operating point analysis. .END is the end statement.

At words “( 1) 10.0000” it is describing about voltage at node 1 respect to the ground at the node 0. The current of V1 and V2 is state at the lower part of the file.

5.0 ANALYSIS

Question 4

Value from the simulation SPICE: V at node 2 to node 0

V = 9.7470 V

USING SUPERPSITION METHOD :

V1 = 24V V2 = 15V

Short V2

V20’ at node 2 to 0

=V1(R2//R3) / ((R2//R3)+R1)

=24 (8.1K//4.7K)/ ((8.1K//4.7K)+10K)

=24(2.97K)/(2.97K+10K)

=5.5V

Short V1

V20’’ at node 2 to 0

=V1(R1//R3) / ((R1//R3)+R2)

=15 (4.7K//10K)/ ((4.7K//10K)+8.1K)

=15(3.2K)/(3.2K+8.1K)

=4.25V

V20 = V20’+ V20’’ = 5.5 + 4.25 = 9 .745V ~ 9.747V

Question 5

Value from the simulation SPICE: V at node 2 to node 0

V = -2.8571V

CALCULATION ON CIRCUIT USING KIRCHOFF’S LAW:

V= IR

Loop 1

10V = I 1 R1 + I 3 R3 - 10V ----------

Loop 2

10V = - I 2 R2 + I 3 R3 ----------

I 1 = I 2+¿ I3¿

10V + 10V = (I 2+ I 3)R1 + I 3 R3

(I 2 + I 3)R1 + I 3 R3 = 20

4KI 2 + 4KI 3 + 2KI 3= 20

4KI 2 + 6KI 3 = 20

4KI 2= 20 - 6KI 3---------

I 2 = 20−6 K I 3

4 K

10V = - I 2 R2 + I 3 R3 ----------

Insert into

10V = -(20−6 K I 3

4 K¿ R2 + I 3 R3

10V = -40 + 12K I3 + 2KI3

10V = -40 + 14KI 3

I 3 = 50

14 K

I 3 = 3.571 X 10−3A

Insert I 3 into equation 1

10V = I1 (4K) + (3.571 X 10-3)(2K)

I1 = 7.145 X 10-4 A

I 1 = I 2+¿ I3¿

I2 = I1 – I3

= ( 7.145 X 10-4) – ( 3.571 X 10-3)

= - 2.857 X 10-3 A

Voltage node between 0 and 2

V = I3R3 - 10V

= (3.571 X 10-3) ( 2K) – 10V

= 7.142 V – 10V

= -2.858 V ~ -2.8571V

6.0 DISCUSSION

This simulation mainly focuses on DC circuit, with nodes and ground. SPICE is a

program which can simulate circuit with texting a several command to its program. By using

spice its easily to determine the value of the parameters in electric signal such as voltage,

current, power and temperature. One of the methods in using SPICE is operating point

analysis which had been used in this simulation.

In an operating point analysis, a power supply is connected to the circuit to be

analyzed. At this stage, no stimulus is applied to the input. The operating point analysis

provides voltage information on every node of the circuit with respect to ground.

In this report, there are 3 circuits that we need to simulate and confirming the

simulation with manual calculate. As the analysis in question 4 we determine the value of the

voltage in node 2 and node 0. By simulate, 9.7470V is the value of the voltage that we

obtained. in order to prove it we used manual calculation of electronic concept which is

superposition method. In the calculation we compute the value of V20’’ and V20’ of each

short DC voltage source value respectively. By the calculation the value that I get is 9.745

which is only slightly difference from the simulation.

In question 5, the tabulated result from the simulation is -2.8571V which is the value

between node 2 to node 0. In order to prove the simulation is correctly simulate the circuit I

used the KIRCHOFF’S LAW method to solve the value manually and -2.858V is the result.

Those of result shows the simulation is working properly.