Is there a virtual ground in this op-amp?

Discussion in 'General Electronics Chat' started by rkbshiva, Aug 19, 2015.

1. rkbshiva Thread Starter New Member

Aug 1, 2015
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0
Consider an ideal differential operational amplifier whose circuit looks like this,

Now in this circuit diagram it is given that, the negative terminal is at 0v.

But if I give a sine wave as input to this op amp whose frequency is high and capacitance of the capacitor is 1.I think almost can't be such a high voltage drop across the capacitor to make the negative terminal at 0v.

If so, is there a virtual ground in this circuit. If yes, please explain me how??

2. Wendy Moderator

Mar 24, 2008
20,735
2,499
Op amps with negative ffedback work by making the negative input match the positive. In this configuration the capacitor will have lower reactance at higher frequencies, requiring large signal on the output to compensate. At low frequencies the reactance will be large, so the output will be small.

G = R / (Xc) = R / (1/(2πCf) = R 2π C f

Yes, there is a virtual ground.

Last edited: Aug 19, 2015
3. dl324 Distinguished Member

Mar 30, 2015
3,114
601
The zero differential input theorem states the voltage at both inputs will be the same.

4. AnalogKid Distinguished Member

Aug 1, 2013
4,284
1,172
As long as the opamp has enough open loop gain to satisfy the feedback equation at the frequency of interest, the loop will be closed and the inverting input will be a virtual ground. Note that that is not the same as 0 Vdc. The DC potential of the virtual ground will be the DC potential of the non-inverting input. In your case that happens to be 0 Vdc, but it can be something else if the non-inverting input is boased up to some non-zero voltage.

If the non-inverting input has a varying signal on it such as audio, the inverting input will attempt to track it through the action of the feedback loop. In this case the inverting input is a virtul ground at varying voltage, and the overall circuit is a true differential amplifier.

ak

5. MrAl Well-Known Member

Jun 17, 2014
2,204
433
Hi there,

When you start to use extreme values of components you have to look carefully at the practical aspects along with the theoretical aspects. This is true in any situation really, but lets take a quick look at this circuit...

First, the transfer function is:
Vout=-(s*Vin*A*C*R)/(s*C*R+A+1)
where R is the resistor value and C the capacitor value, and
A is the open loop gain of the op amp at the desired frequency, and
Vin is the AC input voltage, and
s is the complex frequency.

If we solve for the inverting terminal node, we get:
Vx=(s*Vin*C*R)/(s*C*R+A+1)

That is the expression for the inverting terminal node regardless of any other theory such as a virtual ground.

The first thing we might notice is that the open loop gain A is in the denominator only, so the limit as A approaches infinity is zero. That's because as A gets larger the node voltage gets lower and lower until it reaches zero when A gets to be a very high value. This is where the concept of a Virtual Ground really comes from. The only time this is possible, in theory, is when A is infinite. For any other value of A it is not zero, but for most modern op amps A can be taken to be quite high, on the order of 100000 or so, so the node voltage is still very close to zero, although not quite zero.
Lets see what it is when we use some components such as those that you might be considering.

First, we calculate the amplitude of Vx:
Ampl=(Vin*w*C*R)/sqrt(w^2*C^2*R^2+(A+1)^2)

Next, if we let Vin=1vac peak, C=1 Farad and R=1000 Ohms, and A=100000 we have:
Ampl=(1000*w)/sqrt(1000000*w^2+10000200001)

Now say we have a low frequency w=1, then we get:
Ampl=0.0099994

So we see the amplitude at the inverting terminal is not zero but fairly low.
We think of this as zero though in the concept of a virtual ground.

But now the practical side: just what output does it take to achieve this result, with low frequency such as w=1 ?

Working with the equation for the output voltage we again have:
Vout=-(s*Vin*A*C*R)/(s*C*R+A+1)

the amplitude of that is:
AmplVout=(Vin*w*A*C*R)/sqrt(w^2*C^2*R^2+(A+1)^2)

and now with the same values for A,R,C and Vin we have:
AmplVout=(100000000*w)/sqrt(1000000*w^2+10000200001)

and notice the numerator is much larger for this amplitude.
Now using w=1 as before, we get:
AmplVout=999.94 volts peak AC.

Wow, that's almost 1000 volts peak, which is not possible to obtain from most op amps.

What this means is that although in theory we get 0v at the inverting terminal, in a practical circuit this would not happen because the output voltage would have to go too high to get that to work. The output would saturate and clip so we would not get zero volts we would get some bad waveshape and that means the circuit does not work as intended.

This was mainly due to the size of the capacitor. If the capacitor was smaller the reactance would be higher and thus make it easier for the circuit to force the inverting terminal to zero. If the frequency was higher that would help too as then the capacitive reactance is higher also.

We also should take into account the current requirement to make sure the output can supply enough current to force a zero voltage condition at the inverting terminal. Note that 1000 volts across a 1000 ohm resistor requires 1000 watts of power to operate, something else op amps usually can not do

There is also another consideration however. In the above we looked at what we might call the "Static AC conditions" and found some limitations, although with proper design it might work. But when we look at all the dynamic conditions that can occur, we find we still have a problem. That is because we did not yet examine the transient effects of an input wave that does not start exactly at zero volts. If the input wave does not start at zero volts (a sine wave that starts at zero volts that is) then we also have to deal with a step change in input.

A step change in input means the capacitor, for a short time, FORCES the inverting terminal to go to a non zero condition. This is pretty simple to think about because with the input side of the cap going to say 0.1v means that the inverting terminal side of the cap goes to 0.1v, and there is nothing in theory to stop it from doing that for at least some amount of time. It is then up to the output to try to compensate, which will never be able to do so even for a small input change with such a larger cap such as 1 Farad.
So for the time that it takes the op amp to respond, the inverting terminal will be non zero and this could be a long time in a practical circuit with C=1 Farad. The calculation is not that hard to do as it is just a capacitor being charged by a resistor between two voltage sources at that point:
(Vin-Vout)*e^(-t/(C*R))+Vout=0

and solve this for t, where Vout is negative because it goes negative with positive Vin:

t=-ln(-Vout/(Vin-Vout))*C*R

and with an assumed output op amp saturation at -10 volts and the values we used previously, we get:
t=95.3 seconds

This means it would take more than 95 seconds to force the inverting terminal back to zero volts with a step change input of +1v. Not very desirable

So while the AC theory seems to confirm that any RC value would work with some limitations, there are other practical concerns that make the circuit impractical with RC values too large.

Last edited: Aug 19, 2015
6. crutschow Expert

Mar 14, 2008
12,531
3,069
In summary, if the op amp has sufficient dynamic (output voltage) range, slew rate, and frequency response to maintain the (-) input voltage at (very near) 0V for a given input waveform, then it will.
If not, the input will deviate from zero.