# Voltage at inverting input of OPAMP

Discussion in 'Homework Help' started by xxxyyyba, Sep 30, 2015.

1. ### xxxyyyba Thread Starter Active Member

Aug 7, 2012
283
6
Hi!

Here is my circuit:

(v3=vul(t)=sin(2*pi*100*t))

My task is to find expression for viz(t) and expression for voltage at inverting input (- input). I found using superposition expression for viz(t):
viz(t)=-R3*Vcc/R2-R3*vul(t)/R1.

Here is sketch of viz(t):

But how to find expression for voltage at inverting output?

2. ### shteii01 AAC Fanatic!

Feb 19, 2010
4,440
704
You don't. It is zero.

3. ### xxxyyyba Thread Starter Active Member

Aug 7, 2012
283
6
It is zero when OPAMP works in linear mode, but when it goes to saturation, $V^{+}\neq V^{-}$ (in that case it is also $V^{+} = 0V$). That's how they told us in school, If I got them

Last edited: Sep 30, 2015
4. ### xxxyyyba Thread Starter Active Member

Aug 7, 2012
283
6
Here is voltage diagram from Multisim.
Red - viz(t), green vin(t) ( = vul(t) in original scheme), blue - voltage at inverting input), OPAMP - LT1097CN8

5. ### WBahn Moderator

Mar 31, 2012
23,194
6,989
What is Viz when the opamp is saturated?

Given that value and the value of Vin, can you not calculate the value at the inverting input of the opamp? After all, isn't it just a problem with three resistors sharing a common node but each resistor being held at a different potential?

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6. ### shteii01 AAC Fanatic!

Feb 19, 2010
4,440
704
Like this one?

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7. ### WBahn Moderator

Mar 31, 2012
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Something like that, though for the range of operation we are interested in Viz is effectively a DC source.

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Oct 1, 2015
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1

9. ### vinodh2308 New Member

Oct 1, 2015
3
1
The voltage at the inverting terminal
can be found from the graph that is present above in the following way.
Op amp will be in virtual short condition only if Vcc < Vout(output voltage of opamp) < -Vee
if op amp is in virtual short then voltage at inverting terminal is zero.since inverting terminal is at zero potential.
but in our problem Vee is 12 volts...k...
our output graph is exceeding -12 volts at some point of time.
Op amp is not capable of providing a voltage more than -12v. it will saturated at -12 volts.
so now output voltage = -12 is fixed
and Vcc = 12 is also fixed (input to circuit)
and Vul = sin(2*pi*100t) is also fixed.
using the above values and node equations you just find the equation for voltage at inverting input.
if you want i can upload my solution....

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10. ### xxxyyyba Thread Starter Active Member

Aug 7, 2012
283
6
So after solving circuit in post #6 (Viz=-12V), I get this expression for voltage at inverting node:

$v^{-}(t)=\frac{\frac{Vcc}{R_2}+\frac{Viz}{R_3}}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}}+vul(t)(1-\frac{R_1}{R_1+\frac{R_2R_3}{R_2+R_3}})$.
Now I need to find point in time when $v^{-}(t)$ start (marked point):

I found it solving this equation (for t):
viz(t)=-R3*Vcc/R2-R3*vul(t)/R1=-12V,
-R3*Vcc/R2-R3*vul(t)/R1=-12V=>t=0.001s.
Is it ok?

11. ### WBahn Moderator

Mar 31, 2012
23,194
6,989

But do look at your simulation and see if it is reasonable to use -12 V as the saturated output voltage of the opamp. Sure looks closer to -11 V to me.

12. ### xxxyyyba Thread Starter Active Member

Aug 7, 2012
283
6
I didn't say at beginning that OPAMP is considered to be ideal with exactly +-12V at Viz when OPAMP works in saturation, so that is why I used exactly -12V for Viz. We were told to compare results we got by hand with results from simulator.

13. ### WBahn Moderator

Mar 31, 2012
23,194
6,989
Sounds reasonable. You might mention in your analysis that the simulation (or, better yet, the data sheet) show that this opamp's saturation voltage is about 1V from the rail and that, with that information, more accurate hand results are easily obtainable. But that is going a bit above and beyond -- but THAT is something that good engineers typically do.

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14. ### crutschow Expert

Mar 14, 2008
20,263
5,737
That's only for an ideal op amp with infinite gain.
For the non-ideal device shown, the voltage at the (-) input is the (inverse) of the output voltage divided by the op amp open-loop gain (ignoring any DC offsets).

15. ### WBahn Moderator

Mar 31, 2012
23,194
6,989
For this problem, this isn't the case, either. The whole point of this exercise would appear to be to recognize that when the opamp saturates that the voltage difference between the inputs is no longer constrained to be zero (for an ideal opamp) or the output voltage divided by the open-loop gain (for a real opamp) at all.

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