hello guys,

The following circuit is of an opamp ac amplifier..

The text says:

If only ac signals are being amplified, it is often a good idea to "roll off' the gain to unity at dc, especially if the amplifier has large voltage gain, in order to reduce the effects of finite "input off- set voltage.

How exactly the capacitor help in putting the gain of amplifier to unity at dc inputs(input bias currents)..i mean..look the capacitor is attached in parallel acting as low pass filter i.e shorting all AC to ground..which means attenuated dc output voltage will pass to the inverting input of the opamp causing dc amplification(opamp action)..which is against what is said in the text...

please help....

 

hi,H
The C1 cap is in series with the R1 resistor, what do you calculate the impedance of the C1 cap to be close to DC.
Note: Gain = R2/(R1 +Xc1)
E

Calc: Xc1 for say 2Hz

 

How exactly the capacitor help in putting the gain of amplifier to unity at dc inputs(input bias currents)..i mean..look the capacitor is attached in parallel acting as low pass filter i.e shorting all AC to ground..which means attenuated dc output voltage will pass to the inverting input of the opamp causing dc amplification(opamp action)..which is against what is said in the text...

The way to think of circuits like this without getting yourself hopelessly confused, is to consider that at DC and sufficiently low AC frequencies, a capacitor acts as an open circuit; that is, like it simply is not there. In contrast, at sufficiently high AC frequencies, a capacitor acts like a short circuit; that is, as if it were simply a piece of wire.

So to see how your circuit behaves at DC, simply remove C1; the resulting circuit has unity gain, just as the text says.

For AC at frequencies above f = 1 / (2π ⋅ R1 ⋅ C1), C1 acts effectively as a short circuit, and the amplifier gain becomes Av = (R1 + R2) / R1.

 

The way to think of circuits like this without getting yourself hopelessly confused, is to consider that at DC and sufficiently low AC frequencies, a capacitor acts as an open circuit; that is, like it simply is not there. In contrast, at sufficiently high AC frequencies, a capacitor acts like a short circuit; that is, as if it were simply a piece of wire.

Ok thinking it this way....
The capacitor blocks DC therefore the only parts left (at DC) to consider are the op amp and R2. R1 is in series with an "infinite" impedance caused by C1 and is therefore not part of the DC analysis.
With only R2 present (at DC) the op-amp gain is unity.
But the the value of R2 plays no role since we assume (for an ideal opamP) that there is no current into the inv. input terminal (no voltage drop across R2)...and if there is no role of R2 there is no feedback ...which alternatively means that the circuit is unstable due to very large open loop gain of the opamp..

So how come the gain is unity....???

 

..So how come the gain is unity....???

Because the amplifier acts as a voltage follower.

It is easy to see how the amplifier approaches a voltage follower at low frequencies on a Bode Plot. Also note that any practical amplifier will also have a roll-off at high frequencies because of the finite Gain Band Width of the amplifier itself... (1Mhz in this example)

 

But the the value of R2 plays no role since we assume (for an ideal opamP) that there is no current into the inv. input terminal (no voltage drop across R2)...and if there is no role of R2 there is no feedback ...which alternatively means that the circuit is unstable due to very large open loop gain of the opamp..

You are totally wrong. So if no current is flowing through R2 there is no voltage drop across is. And this means that inv. input voltage is equal to the output voltage. So the feedback loop is closed and the circuit act just like a voltage follower.

 

But the the value of R2 plays no role since we assume (for an ideal opamP) that there is no current into the inv. input terminal (no voltage drop across R2)...and if there is no role of R2 there is no feedback ...which alternatively means that the circuit is unstable due to very large open loop gain of the opamp..

Stop confusing yourself with faulty "logic", Himanshoo!

Jony130 is absolutely right: obviously, if no current is flowing through R2, the voltage across it is zero; and thus, the voltage at the opamp inverting input is identical to the voltage at the opamp output, and the circuit functions as a unity-gain amplifier-- a simple voltage follower.

Stop confusing yourself with nonsense phrases like "plays no role."

Instead, stick to fundamentals. Voltage equals current times resistance. Capacitive reactance equals the reciprocal of two times pi times frequency times C.

All else flows from those two facts.

 

This thread may interest you also:
http://forum.allaboutcircuits.com/t...e-if-my-theory-is-correct.104247/#post-790259

 

You are totally wrong. So if no current is flowing through R2 there is no voltage drop across is. And this means that inv. input voltage is equal to the output voltage. So the feedback loop is closed and the circuit act just like a voltage follower.

Can it also be thought as ......R2 would attenuate the DC output voltage". But the attenuation is R2/(input impedance) = R2/infinity = ... 0. As there is no attenuation of the feedback voltage, this is a voltage follower.

 

Can it also be thought as ......R2 would attenuate the DC output voltage". But the attenuation is R2/(input impedance) = R2/infinity = ... 0. As there is no attenuation of the feedback voltage, this is a voltage follower.

Well yes you can, but it is a very strange way of thinking. And to be more precise the attenuation is (input impedance)/( R2 + input impedance).

 

Can it also be thought as ......R2 would attenuate the DC output voltage". But the attenuation is R2/(input impedance) = R2/infinity = ... 0. As there is no attenuation of the feedback voltage, this is a voltage follower.

As Jony130 noted, that's kind of a counter-intuitive way of thinking.
Viewing the feedback network as an attenuator may be technically accurate but I don't think it's that conducive to understanding what's happening.
If you want to use an intuitive approach to op amps then start with the usual premise that the (-) input is the (current) summing junction (if the op amp is operating in the linear region) and you should reference your calculations to that.
All current into that junction must sum to zero and with the (-) terminal voltage always equal to the (+) terminal voltage.
Thus if no current can flow to ground through the capacitor, then no current can flow through the feedback resistor and the voltage at the summing junction must equal the output voltage.
And since the (-) input must follow the (+) input, you have a voltage follower.

 

thanks for making my day ..guys