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Consider an ideal differential operational amplifier whose circuit looks like this,

enter image description here

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??

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Assuming the op amp is ideal, the virtual ground will be maintained. The ideal op amp has infinite gain and is infinitely fast.

However, any real op amp will not have infinite gain. It will not be infinitely fast. And more than likely it will have some limited output slew rate. EDIT: Another important thing to consider is that a practical op amp has limited output voltage range. You will likely see the output signal clipping terribly for high-frequencies.

So, using a real op amp, the op amp will not be able to keep its negative terminal at 0 V at high frequencies. You can see this by connecting a square wave at the input of the differentiator. Just after the rising and falling edges, you will likely see the op amp's negative terminal jump in the direction of the edge of the input square wave. I.e. if the square wave is transitioning from low to high, you will see a positive spike at the inverting terminal of your real op amp, and a negative spike on the high to low transition.

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The concept of "virtual ground" implies that the Opamp must be capable of maintaining the "virtual ground". In the situation that you sketch: high frequency, high value of the capacitor, as long as the opamp is able to keep it's - input at ground potential, it will be a virtual ground.

So assuming the opamp is ideal: the - input will always be virtually grounded, no matter what (ads I'm also assuming the R and C to be ideal).

But in practice, you're right, the opamp will have limitations so under certain conditions it will not be able to keep up and the - input will not be at 0V anymore.

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Please realize that "virtual ground" does NOT mean "ground (0 volts)". The term "virtual" is just an indication that the voltage at the inverting terminal is so small (Vout/Aol) that it can be neglected (set to 0 volts) during calculation. That means: This voltage is never 0 volts but we are setting it (during analyses) to zero.

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  • \$\begingroup\$ I knew that! But the voltage at the inverting input will be almost Vin now. If I set Vin at an instant to be 5V.The inverting input will be almost at 5v.Can this value be neglected!This is not at virtual gnd now . \$\endgroup\$ – Andrew Flemming Aug 19 '15 at 7:42
  • \$\begingroup\$ OK - do you simulate or is at a real circuit (measurement)? In any case, the shown circuit is unstable if you apply a real opmap (model). That`s a typical problem for this circuit and this may be the explanation for your observation. Place a resistor in series to the input capacitor. Try different values and use the lowest possible value. For more exact calculations we need the open-loop function of the opamp. \$\endgroup\$ – LvW Aug 19 '15 at 9:32
  • \$\begingroup\$ You mean, from Vin = 0V you instantly make Vin = 5V, so now the - input cannot be kept at 0V level so it will not be a virtual ground. Well, if the opamp is ideal, it will keep up and it will supply an infinite amount of current to keep the virtual ground at 0V. In practice, ideal opamps do not exist so then the answer is no, the virtual ground point will not be at 0V. But after a while it will return to 0V when the capacitor has been charged. \$\endgroup\$ – Bimpelrekkie Aug 19 '15 at 9:33
  • \$\begingroup\$ Do you use an ideal opamp model or a realistic device? In the latter case, the circuit - most probably - cannot work as intended and is unstable (oscillations). \$\endgroup\$ – LvW Aug 19 '15 at 9:38
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The op-amp is able to make that pin a virtual ground because of the feedback configuration you have. He will drive a signal through that resistor to try to cancel out whatever signal comes in the negative pin. You can check this quite long video out if you have some time, probably the first half of it will be enought for you to anwser your questions.

Video Link

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