I have an amplifier in a closed loop, the diagram looks as follows:
The DC gain is simple which \$ -Rf/R1 \$. If I make \$ -Rf=2\cdot R1 \$ the the DC gain is two. This is verified using AC analysis in Cadence.
However there seems to be a complex conjugate pole also, which leads to unstable behavior.
Hence I modify the gain equation as:
\$Av= \frac{-Rf}{R1} \cdot \frac{1}{(s+\sigma + j\omega)(s+\sigma – j\omega)} \$. The notation \$ + \sigma \$ is because the phase goes negative. I observed that if I put a capacitor across Rf as shown in this figure then I can get rid of the complex conjugate poles. The value of capacitor is 1pF.
So now the gain equation becomes:
\$ Av= \frac{-Rf}{R1 \cdot C} \cdot \frac{1}{(s+\sigma + j\omega)(s+\sigma – j\omega)(s+p1)} \$ where \$ p1= 1/Rf\cdot C \$ This equation checks out because at DC, the gain is \$ -Rf/R1 \$.
My question is, from this equation how is the pole being compensated? Is my analysis even correct?
To add more information the open loop gain of the amplifier is a single pole system as shown here:
Best Answer
One common cause for the pole is: capacitance on the Virtual Ground of the opamp.
With high_value resistors, the DELAY on Virtual Ground, becomes large, and you get ringing or oscillation.
The feedback capacitor can be viewed as part of a voltage_divider with the VirtualGround capacitance. Thus at high frequencies, the delay is removed.