HSpice Essential Examples

HSPICE Essential Examples

Dariush Naseh

Shahid Beheshti University of Tehran

Voltage Divider

** Voltage Divider **R1 in out 1KR2 out 0 1kV1 in 0 DC 4V.OP.END

R1

1k

R2

1k

V1

4v

0

outin

Dariush Naseh - Shahid Beheshti University of Tehran

10/2/2016

Subcircuit

** Subcircuit Voltage Divider**.SUBCKT BLK1 input outputR1 input 1 1kR2 1 output 1K.ENDSX1 in out BLK1X2 out 0 BLK1V1 in 0 DC 4v.OP.END

V1

4v

BLK1

input output

BLK1

input

outp

ut

0


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To Call a Subcircuit

** Call subcircuit **.INCLUDE "z:\Voltage Divider.sp"X1 in out BLKX2 out 0 BLKVDC in 0 DC 4V.OP.END


10/2/2016

Switch tclose

** RC switch **R1 2 3 1kC1 3 0 1u IC=0V1 1 0 DC 1vG1 1 2 VCR PWL(1) 4 5 0 100MEG 1 0.1mV2 4 5 PWL 0 0 4.9999m 0 5m 1.TRAN 1n 10m 0.01m.END

R1

1k

C1

1u

IC = 0

U1

TCLOSE = 5m

1 2

V1

1v

0

1 32


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Other SW tclose & Creating Step Function

** RC Switch 1 **I1 0 V1 PWL 0 0 0.0000001 1C1 V1 0 1 IC=0R1 V1 0 1G1 V1 V2 VCR PWL(1) v4 0 0 100000meg 1 0.001mR2 V2 0 1G2 V2 v3 VCR PWL(1) v5 0 0 100000meg 1 0.001mC2 V3 0 1 IC=0V1 V4 0 PWL 0 0 10 0 10.000001 1V2 V5 0 PWL 0 0 15 0 15.000001 1.TRAN 10m 20 .END

I1

1A

C1

1IC = 0

R1

1

U1

TCLOSE = 10s

1 2

R2

1

U2

TCLOSE = 15s

1 2

C2

1IC = 0

0

V1 V2 V3


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Pulse Generator

.TITLE pulse waveVin in 0 PULSE 1 -1 0 1n 1n 1m 2mR1 in out 1kC1 out 0 100n.TRAN 10u 10m.PROBE output=par('V(out)').PROBE input=par('V(in)').END

C2

100n

outR2

1k

in

V3

TD = 0

TF = 1nPW = 1mPER = 2m

V1 = 1

TR = 1n

V2 = -1

0


10/2/2016

Triangle Pulse Generator

.TITLE triangle waveV1 in 0 PWL 0ms 0 5ms 15v 15ms -15 20ms 0R1 in out 1kD1 A out d1n4734D2 A 0 d1n4734.MODEL d1N4734 D Is=1.085f Rs=.7945 N=1 Xti=3 Eg=1.11 Cjo=157p M=.2966+ Vj=.75 Fc=.5 Bv=5.6 Ibv=.37157 Nbv=.64726.TRAN 20us 20ms.PROBE vin=par('v(in)').PROBE Vout=par('v(out)').END

V1

R1

1k

D1

D1N4734

D2

D1N4734

in

A

out

0


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Dependent Sources

** VCCS **G1 0 1 VCCS 0 2 2R1 1 2 1I1 0 2 DC 1R2 2 3 1 V1 3 0 DC 2.OP.END

V1

2v

R1

1

R2

1

+

-

G1

GAIN = 2I1

1A

0

0

1 32

02


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Dependent sourse , another Ex.(VCCS & CCVS)

** Other example in VCCS and CCCS **I1 V1 V3 DC 1R3 A V2 1R1 V2 V3 1R2 V3 0 1R4 v1 V4 1V1 V2 0 2H1 V4 0 CCVS Vc 4Vc V1 A DC 0G1 V2 V4 VCCS V3 V2 2.OP.END

V1

2V+

-

G1

GAIN = 2

R1

1

R2

1

V3

I1

1AR3

1

V2

R4

1

V4

H1GAIN = 4 0

V1 A

V2V3


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Ideal Transformer

** Rectifire with RC Filter **E1 3 4 TRANSFORMER 1 2 10Vin 1 2 SIN 0 318 50 0 0 0D1 3 out d1n4007D2 0 3 d1n4007D3 4 out d1n4007D4 0 4 d1n4007.MODEL d1n4007 D IS=1.09774E-008 n=1.78309 Rs=0.0414388

+ Eg=1.11 Xti=3 Cjo=2.8173E-011 Vj=0.50772+ m=0.318974 Fc=0.5 TT=9.85376E-006 Bv=1100 Ibv=0.1 Kf=0

R1 Out 0 1kC1 out 0 220u.TRAN 0.1m 1000m START=900m.PROBE IDiode=par('I(D1)').PROBE Vtrans=par('V(3)-V(4)').END

V1

FREQ = 50HzVAMPL = 318v

TX1 D1

D1N4007

D2

D1N4007

D3

D1N4007

D4

D1N4007

R1

1k

C1

220uF

Out3

4

1

2


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Linear mutual inductor

.Title Linear mutual inductorR1 1 2 2L1 2 0 8L2 3 0 200K1 L1 L2 1R3 3 0 200Vs 1 0 AC 50 0.AC DEC 200 100 100k .PROBE vtarget=par('V(2)').PROBE Phase=par('VP(2)').END

V150v

ACPHASE = 0

R1

2

L1

8

L2200 R2

200

0

1 2 3

K

COUPLING=

K1

COUPLING = 1


10/2/2016

Ideal Op-Amp

.TITLE Ideal Op-Amp

.INCLUDE "Z:\7805.sp"Vi in 0 DC 15vx1 in 2 out LM7805R1 3 0 1kR2 1 0 1kR3 out 1 2kE1 2 3 OPAMP 1 2.TRAN 0.1m 10m .END

V1

15v

U1

OPAMP

+

-

OUT

R1

1k

R2

1k

R3

2k

U2

LM7805C

IN1

OUT2

GN

D3

0

in out

1

2

00

3


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Identifying Rin & Rout & Av in OPAMP

.TITLE op-ampV1 2 0 DC 1V2 1 0 DC 100mR1 2 4 500R2 1 3 5kR3 3 0 10kR4 4 3 100K **Rin=100k**R5 4 6 1kR6 5 6 1k ** Rout=1k**R7 6 0 1kE1 5 0 OPAMP 3 4 100K **Av=100K**.DC V1 LIN 499 1m 500m.PROBE Vout=par('v(6)').END

V1

1v

V2

100m

U1

OPAMP

+

-

OUT

0

0

R1

500

R2

5k

R3

10k

0

R5

1k

R7

1k

0

6

31

42


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Plotting characteristic curve of Doid(source sweeping)

** characteristic curve of Diod **R1 in out 1kD1 out 0 d1n4148V1 in 0 DC 1v.MODEL d1n4148 D Is=2.682n N=1.836 Rs=.5664 Xti=3 Eg=1.11 Cjo=4p+ M=.3333 Vj=.5 Fc=.5 Bv=100 Ibv=100u Tt=11.54n.DC V1 LIN 300 0 30 SWEEP TEMP LIN 2 25 50.PROBE IDiode=par('I(D1)').END

D1

D1N4148

V1

1v

R1

1k

0

outin


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Maximum Power Transfer(elements sweeping)

** Maxium Power Transfer**.PARAM r=1R1 in out r R2 out 0 10kV1 in 0 DC 10v.DC r LIN 400 8k 12k.PLOT P(R1).END

V1

10v

R1

{r}

0

outin

PARAMETERS:

r = 1k

R2

10k


10/2/2016

Thevenin Equivqlent & IF-ELSE Structure

.TITLE Thevenin equivqlentR1 1 2 1R2 2 3 1R3 2 4 1VSENSE 4 0 DC 0H1 3 0 CCVS VSENSE 0.5.PARAM k=0.IF(K==1)

{Vin 1 0 DC 1v.TRAN 0.1m 10m .PROBE Rin=par('-V(1)/I(Vin)')}

.ELSEIF(K==0){Vin 1 0 DC 0.TRAN 0.1m 10m.PROBE ISC=par('I(Vin)')}

.ENDIF

.END

R1

1

R2

1

R3

1

H1

GAIN = 0.5

0

2 31

0

VSENSE4


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.SENS CommandSensitivity of element’s variation

.TITLE Self Bias Sensivity ro VccVCC VCC 0 DC 15vRB1 VCC B 3.9kRB2 B 0 1.5kRC VCC C 470RE E 0 180Q1 C B E BC107A.MODEL BC107A NPN Is=7.049f Xti=3 Eg=1.11 Vaf=116.3 Bf=375.5 Ise=7.049f+ Ne=1.281 Ikf=4.589 Nk=.5 Xtb=1.5 Br=2.611 Isc=121.7p Nc=1.865+ Ikr=5.313 Rc=1.464 Cjc=5.38p Mjc=.329 Vjc=.6218 Fc=.5 Cje=11.5p+ Mje=.2717 Vje=.5 Tr=10n Tf=451p Itf=6.194 Xtf=17.43 Vtf=10.OP.SENS IC(Q1).END

VCC

15v

Q1

BC107A

RC

470

RB1

3.9k

RE

180

RB2

1.5k

0

00

dc sensitivities of output ic(q1)

element element element normalizedname value sensitivity sensitivity

(amps/unit) (amps/percent)

0:rb1 3.9000k -4.1551u -162.0503u0:rb2 1.5000k 10.4832u 157.2486u0:rc 470.0000 -101.2284n -475.7735n0:re 180.0000 -99.2755u -178.6959u0:vcc 15.0000 1.4890m 223.3519u


10/2/2016

Sweeping the specifications of models

.TITLE DC sweep

.PARAM k=5.6Vi in 0 1vR1 in out 1kD1 0 out d1n4734.MODEL d1N4734 D Is=1.085f Rs=.7945 N=1 Xti=3 Eg=1.11 Cjo=157p

+ M=.2966 Vj=.75 Fc=.5 Bv=k Ibv=.37157 Nbv=.64726

.DC Vi LIN 300 -15 +15 SWEEP k LIN 6 1.6 6.6

.PROBE target=par('V(out)')

.END

D1

D1N4734

R1

1kV1

1v

0

outin


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Output characteristic curve of BJT

.TITLE CurveIDC 0 B DC 1mVDC C 0 DC 5VV1 E 0 DC 0Q1 C B E BC107A.MODEL BC107A NPN Is=7.049f Xti=3 Eg=1.11 Vaf=116.3 Bf=375.5 Ise=7.049f+ Ne=1.281 Ikf=4.589 Nk=.5 Xtb=1.5 Br=2.611 Isc=121.7p Nc=1.865+ Ikr=5.313 Rc=1.464 Cjc=5.38p Mjc=.329 Vjc=.6218 Fc=.5 Cje=11.5p+ Mje=.2717 Vje=.5 Tr=10n Tf=451p Itf=6.194 Xtf=17.43 Vtf=10.DC VDC LIN 200 0 20 SWEEP IDC LIN 5 0 40u.PROBE I=par('IC(Q1)').END

IDC

1m

Q1

BC107A VDC

5v

00 0


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ICQ-B Curve in BJT

.TITLE BetaIDC 0 B DC 1mVDC C 0 DC 5vV1 E 0 DC 0Q1 C B E BC107A.MODEL BC107A NPN Is=7.049f Xti=3 Eg=1.11 Vaf=116.3 Bf=375.5 Ise=7.049f+ Ne=1.281 Ikf=4.589 Nk=.5 Xtb=1.5 Br=2.611 Isc=121.7p Nc=1.865+ Ikr=5.313 Rc=1.464 Cjc=5.38p Mjc=.329 Vjc=.6218 Fc=.5 Cje=11.5p+ Mje=.2717 Vje=.5 Tr=10n Tf=451p Itf=6.194 Xtf=17.43 Vtf=10.DC IDC LIN 10000 1u 10m SWEEP VDC LIN 3 5 15.PROBE Beta=par('IC(Q1)/IB(Q1)').PROBE ICollector=par('(IC(Q1))').END

IDC

1m

Q1

BC107A VDC

5v

00 0


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HFE & Transition Frequency of BJT

** HFE and transition frequency **.PARAM k=1uVDC C 0 DC 5vIDC 0 B DC k AC 1 0V1 E 0 DC 0Q1 C B E BC107A .MODEL BC107A NPN Is=7.049f Xti=3 Eg=1.11 Vaf=116.3 Bf=375.5 Ise=7.049f+ Ne=1.281 Ikf=4.589 Nk=.5 Xtb=1.5 Br=2.611 Isc=121.7p Nc=1.865+ Ikr=5.313 Rc=1.464 Cjc=5.38p Mjc=.329 Vjc=.6218 Fc=.5 Cje=11.5p+ Mje=.2717 Vje=.5 Tr=10n Tf=451p Itf=6.194 Xtf=17.43 Vtf=10.AC DEC 200 1 1G SWEEP K LIN 3 5u 15u .probe beta=par('IC(Q1)/IB(Q1)').end

IDC

1u

Q1

BC107A VDC

5v

00 0


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Wien Bridge Oscillator

.TITLE wien brigde oscillator

.INCLUDE "Z:\LM741.sp"Vcc Vcc 0 DC 12vVee 0 Vee DC 12vX1 3 2 Vcc Vee 1 LM741R1 5 2 18kR2 2 0 10kR3 3 4 1.6kR4 3 0 1.6kR5 1 5 4.7kC1 4 1 10nC2 3 0 10nD1 5 1 d1n4148D2 1 5 d1n4148I1 0 3 PWL 0 0 1us 1mA 4us 1mA 4.0001us 0.MODEL d1n4148 D Is=2.682n N=1.836 Rs=.5664 Ikf=44.17m Xti=3 Eg=1.11 Cjo=4p+ M=.3333 Vj=.5 Fc=.5 Isr=1.565n Nr=2 Bv=100 Ibv=100u Tt=11.54n.TRAN 1n 2m.PROBE vout=par('V(1)').END

U2

LM741/NS

+3

-2

V+7

V-4

OUT6

OS25

OS11

R1

18k

R2

10k

R3

1.6k

R4

1.6k

C1

10n

C2

10n

0

00

2

3

4

1

V1

12v

V2

12v

0

Vcc

Vee

Vee

Vcc

R5

4.7k

D1D1N4148

D2D1N4148

5

I1

0


10/2/2016

Power Amplifiers in B-Class and Fourier Analysis

.TITLE Power amplifire in B-ClassVCC VCC 0 12vVee 0 Vee 12vQ1 Vcc B out BD135Q2 Vee B out BD136RL out 0 100R1 in B 50Vin in 0 Sin 0 8V 10KHz 0 0 0.MODEL BD135 NPN IS = 4.815E-14 NF = 0.9897 ISE = 1.389E-14 NE = 1.6 BF = 124.2 + IKF = 1.6 VAF = 222 NR = 0.9895 ISC = 1.295E-13 NC = 1.183 + BR = 13.26 IKR = 0.29 VAR = 81.4 RB = 0.5 IRB = 1E-06 RBM = 0.5 + RE = 0.165 RC = 0.096 XTB = 0 EG = 1.11 XTI = 3 CJE = 1.243E-10 + VJE = 0.7313 MJE = 0.3476 TF = 6.478E-10 XTF = 29 VTF = 2.648 + ITF = 3.35 PTF = 0 CJC = 3.04E-11 VJC = 0.5642 MJC = 0.4371 + XCJC = 0.15 TR = 1E-32 CJS = 0 VJS = 0.75 MJS = 0.333 FC = 0.9359 .MODEL BD136 PNP IS = 7.401E-14 NF = 0.9938 ISE = 4.104E-16 NE = 1.054 BF = 336.5+ IKF = 0.1689 VAF = 22.47 NR = 0.9913 ISC = 1.290E-14 NC = 1.100+ BR = 13.91 IKR = 9.888E-2 VAR = 30.00 RB = 0.500 IRB = 1E-06+ RBM = 0.500 RE = 0.208 RC = 5.526E-02 XTB = 0 EG = 1.11 XTI = 3+ CJE = 1.066E-10 VJE = 0.6900 MJE = 0.3676 TF = 2.578E-10 XTF = 13.56+ VTF = 2.366 ITF = 1.3040 PTF = 0 CJC = 5.234E-11 VJC = 0.6431+ MJC = 0.4436 XCJC = 0.440 TR = 1E-25 CJS = 0 VJS = 0.75 MJS = 0.333+ FC = 0.990 .TRAN 0.1u 1m START=0.8m.PROBE output=par('v(out)').FOUR 10k v(out).END

Q1

BD135

Q2

BD136

Vcc

12

Vee

12RL

100

R1

50

B

Vcc

Vee 0

out

Vee

Vcc

0V3

FREQ = 10kVAMPL = 8vVOFF = 0

0

in


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.MEASURE Command (Calculation of output DC Level)

.TLTEL CEVCC VCC 0 DC 12vRB1 VCC B 100KRB2 B 0 27kRC VCC C 4.7kRE1 E 1 47RE2 1 0 3.9kRS in 2 50CB 1 0 470uCC 2 B 10uQ1 C B E BC107A.MODEL BC107A NPN Is=7.049f Xti=3 Eg=1.11 Vaf=116.3 Bf=375.5 Ise=7.049f+ Ne=1.281 Ikf=4.589 Nk=.5 Xtb=1.5 Br=2.611 Isc=121.7p Nc=1.865+ Ikr=5.313 Rc=1.464 Cjc=5.38p Mjc=.329 Vjc=.6218 Fc=.5 Cje=11.5p+ Mje=.2717 Vje=.5 Tr=10n Tf=451p Itf=6.194 Xtf=17.43 Vtf=10

Vin in 0 0 AC 1 Sin 0 5mV 10kHz.op.AC DEC 200 1 100MEG .TRAN 0.1us 1ms.PROBE AC Gain=par('VDB(c)').PROBE AC Phase=par('-ABS(VP(c))').MEASURE TRAN val avg v(c) from=0 to=0.1ms.PROBE TRAN output=par('V(c)-val').PROBE AC Rin=par('V(in)/I(Rs)').END

VCC

12vRC

4.7k

RB1

4.7k

RB2

27k

RE1

47

Q1

BC107A

RE2

3.9k

RS

50

C1

10uV2

FREQ = 10kVAMPL = 5mVOFF = 0

AC = 1

00

B

C2

470u

0

1

0in 2

Vcc

0


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Example of .MEASURE CommandWHEN-FIND

TITLE measureVin in 0 PWL 0 0 1ns 1v 10ms 1 10.0000001m 0R1 in out 1kC1 out 0 1u.TRAN 20u 20ms.MEASURE TRAN val1 FIND I(R1) AT=1m.MEASURE TRAN Val2 WHEN V(out)='0.5' CROSS=2.PROBE input=par('V(in)').PROBE output=par('V(out)').END

R1

1k

C1

1u

0

V1

outin


10/2/2016

Square wave, MEASURMET of RISE TIME(MIN,MAX,PP &TRIG… TARG… )

.TITLE Rise TimeVi in 0 PULSE 1 -1 0 1p 1p 1m 2mR1 in out 1kC1 out 0 100n.TRAN 0.1u 10m.MEASURE MAX MAX V(out) FROM=6.1m TO=8m.MEASURE MIN MIN V(out) FROM=6.1m To=8m.MEASURE PP PP V(out) FROM=6.1m TO=8m.MEASURE TRAN T1 TRIG V(in) val=1 TD=6.1ms RISE=1 TARG V(out) val=0.1*PP+MIN TD=6.1ms RISE=1.MEASURE TRAN T2 TRIG V(in) val=1 TD=6.1ms RISE=1 TARG V(out) val=0.9*PP+MIN TD=6.1ms RISE=1.MEASURE TRAN Tr=param'T2-T1'.PROBE in=par('v(in)').PROBE OUT=par('V(out)').END


10/2/2016

.PZ CommandPole & Zero

.TITLE Pole and zeroV1 in 0 DC 0 AC 1 R1 in out 1kC1 out 0 100n.AC DEC 200 1 100k.PROBE gain=par('VDB(out)').PROBE pahse=par('VP(out)').PZ v(out) v1.END

V21Vac

0Vdc

out

0

R1

1k

C1

100n

in


10/2/2016

.TF Command(Transfer Function)

.TITLE Diffrentional AmplifireVCC VCC 0 12vVEE 0 VEE 12vV1 in 0 DC 0Q1 1 1 VCC BC177Q2 2 1 VCC BC177Q3 1 in 3 BC107AQ4 2 0 3 BC107AQ5 3 4 6 BC107AR1 4 VEE 1.8kR2 4 0 22kR3 6 VEE 100.MODEL BC107A NPN Is=7.049f Xti=3 Eg=1.11 Vaf=116.3 Bf=375.5 Ise=7.049f+ Ne=1.281 Ikf=4.589 Nk=.5 Xtb=1.5 Br=2.611 Isc=121.7p Nc=1.865+ Ikr=5.313 Rc=1.464 Cjc=5.38p Mjc=.329 Vjc=.6218 Fc=.5 Cje=11.5p+ Mje=.2717 Vje=.5 Tr=10n Tf=451p Itf=6.194 Xtf=17.43 Vtf=10.MODEL BC177 PNP Is=336.7f Xti=3 Eg=1.11 Vaf=55.46 Bf=154.4 Ise=412.1f+ Ne=1.429 Ikf=.2994 Nk=.7028 Xtb=1.5 Br=3.99 Isc=1.03n Nc=1.958+ Ikr=9.726 Rc=1.833 Cjc=11p Mjc=.2223 Vjc=.5 Fc=.5 Cje=33p+ Mje=.3333 Vje=.5 Tr=10n Tf=847.7p Itf=2.198 Xtf=23.26 Vtf=10.OP.TF v(2) v1.END

Q1

BC177

Q2

BC177

Q3

BC107A

Q4

BC107A

Q5

BC107A

V1

12

V2

12

VEE

VCC

R1

100

R2

1.8k

R3

22k

VEE

0

VCC

0

V3VAC=1VDC=0

0

Vin


10/2/2016

LAPLACE Transform Simulation

TITLE LAPLACEERC 2 0 LAPLACE 1 0 1/1,1E-3Vin 1 0 DC 0 AC 1 sin 0 1 159.AC DEC 200 0.1 100MEG.TRAN 0.6u 60m.PROBE TRAN out=par('V(2)').PROBE AC ampel=par('VDB(2)').PROBE AC phase=par('VP(2)').MEASURE AC MAX FIND VDB(2) AT=1.MEASURE AC cutoff WHEN vDB(2)='MAX-3'.END

V2

R1

1k

0

C1

1u


10/2/2016

HSpice Essential Examples

Engineering

Transcript of HSpice Essential Examples