Computer Aided Design

Course 5

Transient Analyses

These time-based analyses evaluate circuit performance in response to independent input source signals varying with time. The transient, or time-domain, response is the most used analysis in PSpice simulation. Using this type of analysis means that the operation of your circuit is being simulated as time progresses and various input variables change their values, or as oscillations are developed depending on component values.  On the other hand, one must bear in mind that the transient analysis  tends to cause trouble due to the inevitable compromises which have to be made between speed and accuracy. 

    In order to set up and run Transient Analysis, for example,  you would go through the following steps:

Specifying  Time-dependent stimulus

Setting up Time response

Setting up Fourier Components

Selecting Options

1. Time-dependent stimulus

The generators of time-dependent input signals for your circuit can be classified in two categories according to the way of setting their transient behaviour parameters:

• by using Standard Symbol names at Schematics stage where parameter adjustment is possible

• by introducing the desired input waveform as a function or a graphic plot with the help of Stimulus Editor

V3

FM =

VAMPL = VOFF =

FC = MOD =

V4

TD =

TF = PW = PER =

V1 =

TR =

V2 = V2

FREQ = VAMPL = VOFF =

V5

V6

TD1 =

V1 =

TD2 = TC1 =

V2 =

TC2 =

2. Time response

The Transient response analysis allows the response of the circuit to be calculated from TIME = 0 to a specified time. Minimum circuit design requirements Circuit should contain one of the following:

•an independent source with a transient specification (if it is an input excitation) - see previous step

•an initial condition on a reactive element

•a controlled source that is a function of time

During a transient analysis, any or all of the independent sources may have time-varying values.

Minimum program setup requirements A transient analysis specification includes:  

*Detailed Bias Point: The transient analysis does its own calculation of a bias point to start with, using the same technique as described for DC analyses. This is necessary because the initial values of the sources may be different from their DC values. If you want to report the small-signal parameters for the transient bias point, you should use the Transient command and enable Detailed Bias Point. Otherwise, if all you want is the result of the transient run itself, you should enable the Transient command only.

Detailed Bias Point.

V 1

F R E Q = 1 kV A M P L = 0 . 5V O F F = 0

R 11 k

0

R 2

1 k

C 1

1 0 0 n

Q 1

Q 2 N 2 2 2 2

R 31 0 0 k

R 42 0 k

R 55 k

V C C

o u t

V 21 2 V d c

0

V C C

V

V

in

Time

0s 5ms 10msV(OUT) V(IN)

-10V

0V

10V

**** INITIAL TRANSIENT SOLUTION TEMPERATURE = 27.000 DEG C************************************************************************* NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE( IN) 0.0000 ( OUT) 5.9272 ( VCC) 12.0000 (N00101) 1.2222 (N00563) 1.8718 (N00870) 0.0000 VOLTAGE SOURCE CURRENTS NAME CURRENT V_V1 0.000E+00 V_V2 -1.316E-03 TOTAL POWER DISSIPATION 1.58E-02 WATTS

**** 03/29/07 15:33:55 ******* PSpice 10.3.0 (Jan 2004) ******* ID# 1111111111 ** Profile: "SCHEMATIC1-t1" [ C:\DOCUMENTS AND SETTINGS\OVI\MY DOCUMENTS\CARTE_CAD_V1\simulari\t1-PSpiceFiles\SCHEMATIC1\t1.sim ] **** OPERATING POINT INFORMATION TEMPERATURE = 27.000 DEG C***************************************************************************** BIPOLAR JUNCTION TRANSISTORSNAME Q_Q1 MODEL Q2N2222 IB 7.69E-06 IC 1.21E-03 VBE 6.50E-01 VBC -4.06E+00 VCE 4.71E+00 BETADC 1.58E+02 GM 4.68E-02 RPI 3.72E+03 RX 1.00E+01 RO 6.43E+04 CBE 5.57E-11 CBC 3.87E-12 CJS 0.00E+00 BETAAC 1.74E+02 CBX/CBX2 0.00E+00 FT/FT2 1.25E+08

(<tstop>-<tstart>)/50 şi 2<tstep>.

5. Fourier components

A periodic signal can be represented by a Fourier series:

)sincos()(1

021 tkbtkaatv

n

kkk

where a0/2 is the dc component, ak and bk the coefficients of the series give the magnitude of the signal Ak of kth harmonic component and the frequency k.. Fourier series is:

)cos()(1

021

k

n

kk tkAatv

where the amplitude Ak and the phase k are given by:

kakb

arctg

baA

k

2k

2kk

To perform a Fourier analysis is only possible after a transient analysis has been completed. This is due to the very simple reason that the  Fourier analysis actually calculates DC and Fourier components of the result of a transient analysis. By default, the 1st through 9th components are computed, however, more can be specified. So, first, you have to perform a transient analysis before doing a Fourier analysis. The sampling interval used during the Fourier transform is equal to the print step specified for the transient analysis. When selecting Fourier to run a harmonic decomposition analysis on a transient waveform, only a portion of the waveform is used. Using Probe, a Fast Fourier Transform (FFT) of the complete waveform can be calculated and its spectrum displayed.

A Fourier analysis setting up is performed in Transient dialog box. Specification includes:

• FOURIER COMPONENTS OF TRANSIENT RESPONSE V(OUT)• DC COMPONENT = 5.925863E+00• HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED• NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG)• 1 1.000E+03 2.264E+00 1.000E+00 -1.744E+02 0.000E+00• 2 2.000E+03 4.320E-03 1.908E-03 1.011E+02 4.498E+02• 3 3.000E+03 5.686E-04 2.511E-04 -1.611E+02 3.621E+02• 4 4.000E+03 9.627E-05 4.252E-05 -5.221E+01 6.453E+02• 5 5.000E+03 3.538E-05 1.563E-05 1.055E+01 8.825E+02• 6 6.000E+03 1.936E-05 8.552E-06 7.584E+00 1.054E+03• 7 7.000E+03 1.723E-05 7.613E-06 4.507E+00 1.225E+03• 8 8.000E+03 1.554E-05 6.866E-06 2.057E+00 1.397E+03• 9 9.000E+03 1.410E-05 6.227E-06 -3.378E+00 1.566E+03• 10 1.000E+04 1.153E-05 5.092E-06 6.167E-01 1.744E+03• 11 1.100E+04 1.089E-05 4.812E-06 1.417E+01 1.932E+03• 12 1.200E+04 1.139E-05 5.031E-06 2.265E+00 2.095E+03• 13 1.300E+04 9.579E-06 4.231E-06 -1.306E+01 2.254E+03• 14 1.400E+04 7.705E-06 3.404E-06 8.332E+00 2.450E+03• 15 1.500E+04 8.825E-06 3.898E-06 1.654E+01 2.632E+03

The program compute the THD (Total Harmonic Distorsion):

%100A

A....AA[%]THD

1

2n

23

22

The results of the Fourier Analysis are only available in the output file (.out). They cannot be viewed in Probe.

Frequency

0Hz 0.5KHz 1.0KHz 1.5KHz 2.0KHzV(OUT)

0V

5V

10V

7. Options  The Options is used to set all the options, limits, and control parameters for the simulator.

 *These options are available for modification in PSpice, but it is recommended that the program defaults be used. **For these options zero means infinity.

 Convergence problems: There are only a few remedies available for this one. Try relaxing RELTOL from 0.001 to 0.01. Try setting ITL4=40 in an OPTIONS statement. This setting will slow down the simulation, so it does not recommend using it for circuits that do not have a convergence problem in transient analysis.When using PSpice for high voltage and currents, it may be appropriate to relax VNTOL frm 1uV to 1mV and ABSTOL from 1pA to 1nA.

Example 1: A simple BJT Switching circuit

00

V C C _ C I R C L E V C C _ C I R C L E

0Q 1

Q 2 N 6 9 6

R 1

1 k

R 2

2 . 2 k

V 11 0 V

V in

Comutator cu BJT

Time

0s 0.2s 0.4s 0.6s 0.8s 1.0sV(Vin:+)

0V

2.5V

5.0V

SEL>>(9.1603m,76.675p)

V(Q1:c)0V

5V

10V

(377.099m,81.685m)(38.168m,146.230m)

Example 2: Schmitt trigger comparator

+15 -15

0 0

+15

-15 0

0

out

V2 -15V

V1 15V

U1

uA741

3

2

7

4

6

1

5 +

-

V+

V- OUT

OS1

OS2

R3

500k

R4

100k VI

.PARAM RP=500k

Time

0s 100ms 200ms 300ms 400ms 500ms 600msV(out)

-20V

0V

20V

RP=1MEG

RP=500k

RP=200k

V(VI:+)

-15V -10V -5V 0V 5V 10V 15VV(out)

-20V

0V

20V

Example 3: Astable circuit with OpAmp

0

0 0 0

+15

-15 +15 -15

C1 0.01u

R3

500k

R2

100k

R1

500k

U1

uA741

3

2

7

4

6

1

5 +

-

V+

V- OUT

OS1

OS2

V1 15V

V2 -15V

Astabil

Time

0s 5ms 10ms 15ms 20msV(R1:2) V(C1:2)

-20V

0V

20V

IC dat de PSF

IC=1V

Time

0s 5ms 10ms 15ms 20msV(R1:2) V(C1:2)

-20V

0V

20V

Example 4: Hartley oscillator

00

0

0

0

0

0

J 1B F 2 5 6 B

J 2

B F 2 5 6 B

C 1

1 0 0 n

C 2

1 0 0 nC 3

1 0 0 p

C 4

2 5 pC 5

1 n

C 61 0 p

R 12 . 2 k

R 25 6 k

R 3

1 k

R 41 k

R 5

5 6 k

L 1

1 6 u H

L 2

4 u H

V 15 V

I C =0

0

+

Time

0s 2us 4us 6us 8us 10usV(J1:s)

-5.0V

0V

5.0V

Time

18.0us 18.4us 18.8us 19.2us 19.6us 20.0usV(J1:s)

-5.0V

0V

5.0V