# Opamp doubt

1. May 29, 2012

### reddvoid

This is the snap shot i took from Pspice simulation

I have a doubt here
***why is voltage at non inverting terminal is 13.77V even though i connected 5V dc source to it and why the voltages at inverting and non inverting terminals are not almost same ( which we consider while solving opamp circuits)

thank you .

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2. May 29, 2012

### Staff: Mentor

Normally, when someone says they connected a 5v source to a point they mean they connected a voltage source between that point and ground. But in the schematic, it shows the 5v source connected between that (strange) 8.769V point and the non-inverting input. So naturally it is going to make the non-inverting input 8.769 + 5.00 = 13.769V.

Is this a schematic you thought up? What is it supposed to do or show?

3. May 29, 2012

### Kholdstare

I have exactly resimulated your circuit.

I've also opened "eval.lib" and found the macromodel

Code (Text):
*-----------------------------------------------------------------------------
* connections:   non-inverting input
*                |  inverting input
*                |  |  positive power supply
*                |  |  |  negative power supply
*                |  |  |  |  output
*                |  |  |  |  |
.subckt uA741    1 2 3 4 5
*
c1   11 12 8.661E-12
c2    6  7 30.00E-12
dc    5 53 dx
de   54  5 dx
dlp  90 91 dx
dln  92 90 dx
dp    4  3 dx
egnd 99  0 poly(2) (3,0) (4,0) 0 .5 .5
fb    7 99 poly(5) vb vc ve vlp vln 0 10.61E6 -10E6 10E6 10E6 -10E6
ga    6  0 11 12 188.5E-6
gcm   0  6 10 99 5.961E-9
iee  10  4 dc 15.16E-6
hlim 90  0 vlim 1K
q1   11  2 13 qx
q2   12  1 14 qx
r2    6  9 100.0E3
rc1   3 11 5.305E3
rc2   3 12 5.305E3
re1  13 10 1.836E3
re2  14 10 1.836E3
ree  10 99 13.19E6
ro1   8  5 50
ro2   7 99 100
rp    3  4 18.16E3
vb    9  0 dc 0
vc    3 53 dc 1
ve   54  4 dc 1
vlim  7  8 dc 0
vlp  91  0 dc 40
vln   0 92 dc 40
.model dx D(Is=800.0E-18 Rs=1)
.model qx NPN(Is=800.0E-18 Bf=93.75)
.ends
That thing has two transistor in it and hence its gain is not infinity. Rather in the datasheet it mentions 200 V/mV (min 50 or 75 V/mV)

Anyway, 200x1000x(13.77-11.69) = 416000 ! looks like the thing got saturated and its placing +VCC - some Vce drops at the output. Now do a consistency check and you'll find out all nodes/currents has correct values to make it 13.77V and 11.69V at the input.

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4. May 29, 2012

### Kholdstare

Also this is the output file of the simulation

Look how most of the solution voltages inside opamp are close to 15V

Code (Text):
**** 05/29/12 20:45:20 ************** PSpice Lite (Mar 2000) *****************

** Profile: "SCHEMATIC1-bias"  [ D:\test-SCHEMATIC1-bias.sim ]

****     CIRCUIT DESCRIPTION

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

** Creating circuit file "test-SCHEMATIC1-bias.sim.cir"
** WARNING: THIS AUTOMATICALLY GENERATED FILE MAY BE OVERWRITTEN BY SUBSEQUENT SIMULATIONS

*Libraries:
* Local Libraries :
* From [PSPICE NETLIST] section of C:\Program Files\OrcadLite\PSpice\PSpice.ini file:
.lib "nom.lib"

*Analysis directives:
.OP
.PROBE V(*) I(*) W(*) D(*) NOISE(*)
.INC ".\test-SCHEMATIC1.net"

**** INCLUDING test-SCHEMATIC1.net ****
* source TEST
X_U1         N00089 N00379 +VCC -VCC N00517 uA741
V_V1         N00089 N00241 5Vdc
R_R1         N00241 N00379  1k
R_R2         0 N00241  1k
R_R3         N00517 N00241  1k
V_V2         +VCC 0 15Vdc
R_R4         N00517 N00379  1k
V_V3         -VCC 0 -15Vdc

**** RESUMING test-SCHEMATIC1-bias.sim.cir ****
.END

**** 05/29/12 20:45:20 ************** PSpice Lite (Mar 2000) *****************

** Profile: "SCHEMATIC1-bias"  [ D:\test-SCHEMATIC1-bias.sim ]

****     Diode MODEL PARAMETERS

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

X_U1.dx
IS  800.000000E-18
RS    1

**** 05/29/12 20:45:20 ************** PSpice Lite (Mar 2000) *****************

** Profile: "SCHEMATIC1-bias"  [ D:\test-SCHEMATIC1-bias.sim ]

****     BJT MODEL PARAMETERS

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

X_U1.qx
NPN
IS  800.000000E-18
BF   93.75
NF    1
BR    1
NR    1
CN    2.42
D     .87

**** 05/29/12 20:45:20 ************** PSpice Lite (Mar 2000) *****************

** Profile: "SCHEMATIC1-bias"  [ D:\test-SCHEMATIC1-bias.sim ]

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

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

NODE   VOLTAGE     NODE   VOLTAGE     NODE   VOLTAGE     NODE   VOLTAGE

( +VCC)   15.0000  ( -VCC)  -15.0000  (N00089)   13.7690 (N00241)    8.7688

(N00379)   11.6920 (N00517)   14.6150 (X_U1.6)   -1.5906 (X_U1.7)   15.0540

(X_U1.8)   15.0540 (X_U1.9)    0.0000 (X_U1.10)   13.1260

(X_U1.11)   15.0000                   (X_U1.12)   14.9150

(X_U1.13)   13.1260                   (X_U1.14)   13.1550

(X_U1.53)   14.0000                   (X_U1.54)  -14.0000

(X_U1.90)    8.7859                   (X_U1.91)   40.0000

(X_U1.92)  -40.0000                   (X_U1.99)    0.0000

VOLTAGE SOURCE CURRENTS
NAME         CURRENT

V_V1        -1.705E-07
V_V2        -1.651E-03
V_V3         1.667E-03
X_U1.vb     -1.591E-05
X_U1.vc     -1.686E-05
X_U1.ve      2.862E-11
X_U1.vlim    8.786E-03
X_U1.vlp    -3.121E-11
X_U1.vln    -4.879E-11

TOTAL POWER DISSIPATION   4.98E-02  WATTS

**** 05/29/12 20:45:20 ************** PSpice Lite (Mar 2000) *****************

** Profile: "SCHEMATIC1-bias"  [ D:\test-SCHEMATIC1-bias.sim ]

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

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

**** VOLTAGE-CONTROLLED CURRENT SOURCES

NAME         X_U1.ga     X_U1.gcm
I-SOURCE     1.598E-05   7.824E-08

**** VOLTAGE-CONTROLLED VOLTAGE SOURCES

NAME         X_U1.egnd
V-SOURCE     0.000E+00
I-SOURCE    -8.785E-03

**** CURRENT-CONTROLLED CURRENT SOURCES

NAME         X_U1.fb
I-SOURCE    -1.597E-01

**** CURRENT-CONTROLLED VOLTAGE SOURCES

NAME         X_U1.hlim
V-SOURCE     8.786E+00
I-SOURCE    -1.757E-11

**** DIODES

NAME         X_U1.dc     X_U1.de     X_U1.dlp    X_U1.dln    X_U1.dp
MODEL        X_U1.dx     X_U1.dx     X_U1.dx     X_U1.dx     X_U1.dx
ID           1.69E-05   -2.86E-11   -3.12E-11   -4.88E-11   -3.00E-11
VD           6.15E-01   -2.86E+01   -3.12E+01   -4.88E+01   -3.00E+01
REQ          1.53E+03    1.00E+12    1.00E+12    1.00E+12    1.00E+12
CAP          0.00E+00    0.00E+00    0.00E+00    0.00E+00    0.00E+00

**** BIPOLAR JUNCTION TRANSISTORS

NAME         X_U1.q1     X_U1.q2
MODEL        X_U1.qx     X_U1.qx
IB          -3.32E-12    1.71E-07
IC           5.18E-12    1.60E-05
VBE         -1.43E+00    6.13E-01
VBC         -3.31E+00   -1.15E+00
VCE          1.87E+00    1.76E+00
GM           0.00E+00    6.18E-04
RPI          9.38E+13    1.52E+05
RX           0.00E+00    0.00E+00
RO           1.00E+12    1.00E+12
CBE          0.00E+00    0.00E+00
CBC          0.00E+00    0.00E+00
CJS          0.00E+00    0.00E+00
BETAAC       0.00E+00    9.38E+01
CBX/CBX2     0.00E+00    0.00E+00
FT/FT2       0.00E+00    9.84E+15

JOB CONCLUDED

TOTAL JOB TIME             .05

5. May 29, 2012

### reddvoid

I was trying to solve this
i got ans as B

in the simulation we get Ia/Vg=1.75e-3
but according to schematic we get 3/R=3e-3
not matching !

why is this ?

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6. May 29, 2012

### reddvoid

got it :) and Is there a way in Pspice to simulate using all ideal Opamp conditions so that we get almost same voltage at inverting and non-inverting input terminals on simulation ?

7. May 29, 2012

### Kholdstare

Yes. You can get an ideal opamp by selecting "OPAMP" from "ANALOG" library. Its has a gain of 1E6 with default voltage limits of 15V.

Code (Text):
**** INCLUDING test-SCHEMATIC2.net ****
* source TEST
V_V1         N00089 N00241 5Vdc
R_R1         N00241 N00379  1k
R_R2         0 N00241  1k
R_R3         N00517 N00241  1k
R_R4         N00517 N00379  1k
E_U2         N00517 0 VALUE {LIMIT(V(N00089,N00379)*1E6,-15V,+15V)}

When you first simulate it the voltage saturate.
Notice the voltage difference b/w positive and negative pins.

Change the opamp settings from properties to make voltage limit 30V.
Then you get the correct answer.

Yes your answer is correct. Ia/Vg = 15mA/5V = 3mS = 3/1kOhm = 3/R

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8. May 30, 2012

### reddvoid

https://www.physicsforums.com/attachment.php?attachmentid=47783&stc=1&d=1338382293
still not getting
i think i should use newer version of pspice
I'm using V9.1

9. May 30, 2012

### Kholdstare

still not getting what?
I use v9.2

10. May 30, 2012

### Jony130

Why you use the simulation instead of proper analysis? For example by using nodal analysis ?

11. May 31, 2012

### Staff: Mentor

I agree with (B).

12. May 31, 2012

### Kholdstare

For Sanity Check.

13. May 31, 2012

### jim hardy

Basic rule for opamps: It is the duty of the circuit designer to surround the opamp with a circuit that allows it to balance its inputs.

Q: In order for the opamp to balance its inputs, what must it be able to do? How can it drive them to same voltage?
A: It must be able to push five milliamps down through R4. That'll make inverting and noninverting inputs equal at 5 volts > than the node at top of R1

Which means it must raise its output pin to ten volts> whatever voltage is at top of R1, in order to push those same five milliamps through R3 as well.
Which in turn means it will also push ten milliamps down through R2.
Which means those fifteen milliamps will need to flow down through R1.

Which means voltage at top of R1 will be fifteen ,
at top of R4 twenty volts, that's inverting input and note noninverting would be same,,,
and at output pin twenty-five volts.
That's hard for the opamp to do with just a fifteen volt supply.
So it goes as far as it can and gives up.
EDIT Which i think Kholdstare figured out in post #7.

If i missed something please advise, this was from a quick look.

But it looks to me like the circuit designer blew his chance.
Kirchoff is your friend - let him help.

Last edited: May 31, 2012