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#1 Feb 10, 2021
sourav
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Joined: 2017-07-29 06:48

Getting wrong simulation results for op-amp circuit

Hello everyone,

I was trying to simulate the below circuit in .CIR file

As Rs< Rfn * (Ro/Rfp) {here Rs=0 ohms ; i.e., resistance between Vin and inv-opamp terminal};

the expected result is unstable operation as positive feedback dominates here.

but i am getting a stable output voltage( i.e, V(5,0)= 2.81 volts)

Below is my Source code:

Open circuit stable circuit
.lib "C:\Cadence\SPB_16.6\tools\pspice\library\opamp.lib"
Vin  1 0 DC 1.6v
Vccp 3 0 DC 12
Vccn 0 4 DC 12
XOP    6 1 3 4 5 uA741
Rfn 1 5 2k
Rfp 5 6 3k
Ro 6 0 4k
.tran 0.1m 1 
.Probe V(Vin),V(5,0)
.end

Simulation Result:

So, what changes should i make to get the correct results?

Thu, 2021-02-11 08:33
retiredEE
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Joined: 2018-03-21 12:19

This circuit has two stable states.  Even though you have positive feedback, during the initial bias solution the algorithm incrementally reaches a point where all the voltages/currents fall within the solution error criteria.  V(5) reaches a level where it equals [V(6)-1.6] times the amp gain.  In the absence of noise, no further transition occurs.  Add the statement .NODESET V(6)=1.62 to get the solution started above that point and V(5) will rise to a saturation level of 11.8120 volts and the circuit will stay at that point.

Fri, 2021-02-12 00:55
sourav
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Last seen: 23 hours 25 min ago
Joined: 2017-07-29 06:48

Hi retiredEE

Thanks for your kind information;

but when i set Rs=1 kohm ,then still we expect saturated output as positive feedback still dominates,

but i am getting V(5,0)= 4.52 volts ,

so how to solve this?

below i have attached my modified circuit:

.CIR code:

Open circuit stable circuit
.lib "C:\Cadence\SPB_16.6\tools\pspice\library\opamp.lib"
Vin  1 0 DC 1.6V
.param RsValue=1k
Rs 1 2 {RsValue}
.param Vcc=12
Vccp 3 0 DC {Vcc}
Vccn 0 4 DC {Vcc}
XOP    6 2 3 4 5 uA741
Rfn 2 5 2k
Rfp 5 6 3k
Ro 6 0 4k
.NODESET V(6)=2.7v
.tran 0.1m 1 
.probe V(Vin) V(5,0) V(2,0)
.end

Simulation result:

Sun, 2021-02-14 11:47
retiredEE
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Joined: 2018-03-21 12:19

It's the same explanation as before except now you've changed the gain by introducing some negative feedback so the unsaturated null point is different.  By starting the bias point algorithm at different points using .NODESET you can latch the output at the positive or negative rail also.

Mon, 2021-02-15 02:23
sourav
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Last seen: 23 hours 25 min ago
Joined: 2017-07-29 06:48

Previously (when Rs=0) , voltage at inverting terminal of the opamp was 1.6V , so we are setting the initial bias point as V(6)=1.62V;

this time (Rs=1 kohm) ,voltage at inverting terminal of the opamp is 2.6V , so I am setting the initial bias point as V(6)=2.7V;

so what else should we choose as the initial bias point this time (.i.e., .NODESET V(6)=? ) and why?

Tue, 2021-02-16 10:14
retiredEE
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Last seen: 1 week 2 days ago
Joined: 2018-03-21 12:19

I went back to your original transient simulation and got the same results.  However, when I shortened it significantly things changed.
.TRAN 0 40u tripped off a ramp in the negative direction
.TRAN 0 40u 0 10n tripped off a ramp in the positive direction
I can only guess that changes in the maximum time step may have something to do with this. Give it a try on your system.

Thu, 2021-02-18 09:27
sourav
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Last seen: 23 hours 25 min ago
Joined: 2017-07-29 06:48

Thanks a lot retiredEE

I am getting the desired results on my system

thanks for your kind information

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