Here I can't understand how adding a capacitor reduces the dc gain to 1?

 

I can't understand how adding a capacitor reduces the signal gain to 1?

Hello there
The inverting amplifier circuit has the function of amplifying the input signal and inverting output, which is a negative feedback technique. Negative feedback returns a portion of the output signal to the input. The reason why the inverting amplifier can only connect the signal to the inverting input is because the negative feedback can be formed only in this way, otherwise it will not work in the linear amplification region
certain criteria must be met when using an inverting amplifier compared to a non-inverting amplifier and its feedback network
for stability the phase shift induced by the op amp and the feedback network must be less than 180 degrees when Avβ = 1
Feedback can exist in an amplifier whether it's intentional or not. Parasitic capacitance can couple outputs back to inputs, as can magnetic coupling. If there happens to be a frequency at which these un-intentional feedback paths can result in a loop gain over 1, a circuit that is intended as an amplifier can quickly turn into an oscillator at that frequency. It's often the case that a feedback network is added to the amplifier just to counteract the effects of parasitic feedback

 

Here I can't understand how adding a capacitor reduces the dc gain to 1?

At DC the capacitor acts as an open connection. Therefore R1 is effectively taken out of the equation. The remaining circuit is very similar to a voltage follower configuration with R2 in its feedback path. A voltage follower as you probably know has a gain of 1.

 

The remaining circuit is very similar to a voltage follower configuration with R2 in its feedback path.

And with no current through R1, then R2 also has no effect on the DC gain, so the circuit becomes a follower.

It has a gain of 1 only at DC, of course.
For an AC input, the gain will be a function of the frequency, as determined by the values of R1 and C1.

 

The title of this thread is wrong.
1) The capacitor is not added, it is always there. Then the ratio of R1 and R2 sets the signal gain.
2) The capacitor does not reduce the signal gain, instead it does not pass DC so the DC gain is 1.

 

At DC, C1 is open, thus a gain of 1.

then yiou can use Fc=1/(2*3.14*2K*4.7uF). That's the -3db corner frequency or when the output is down 70.7%

 

Not exactly. Just think of the case where the AC gain is less than 3dB. Even when gain is more than 3dB, you're not dealing with a simple dipole. The formula you wrote is valid only for a dipole. The NFB path is a quadripole.