Capacitance on the output of an op amp

I was looking at this post and I totally understand except the answer said that the series resistance at the output of the op amp is creating a pole and a zero. I am thinking pole is something that would attenuate the gain at 20dB/dec and zero is something increasing the gain by 20dB/dec. But here I cannot see that there is a zero being created. Shouldn't adding a capacitance always add a pole?

Also, doesn't zero means there is some frequency at which the output = 0? I am always confused of shouldn't this be a negative infinity in the gain of the bode plot since log(0) = -infinite?

  • \$\begingroup\$ In the circuit that you refer to, there is a pole but no zero. The pole is given by the inverse of the time constant involving \$C_{10}\$ determined when the stimulus, \$V_{in}\$ is zeroed. So, in the given circuit, the pole should be close to \$\omega_p=\frac{1}{C_{10}(R_{20}||R_{19})}\$. There would be a zero if the author had included the equivalent series resistance (ESR) of capacitor \$C_{10}\$. \$\endgroup\$ May 27, 2023 at 7:42

1 Answer 1


Consider a transconductance amplifier output stage where you connect the series resistor and output cap.

enter image description here

If you draw the bode plot for magnitude at the Vout node, it will look like the following

enter image description here

So, clearly there is a pole and zero. The pole was always there even without Rseries. Addition of the Rseries changes the pole frequency and adds a zero.

Consider the transfer function H1(s) = (1+s/wz)
The magnitude goes to zero only at s = -wz. In the bode plot, we deal the case where s = jw and for real frequencies w, the magnitude does not become 0. For eg., in the case where w = wz, magnitude is not 0 as shown below.
|H1(w)| = mag(1 + j) = sqrt(2)

  • \$\begingroup\$ Thank you so much! How does H1 goes to zero at zero frequency? Shouldn't H1 = (1-0/wp) which equals 1 at zero ferquency? \$\endgroup\$
    – Wu Eric
    May 30, 2023 at 15:10
  • \$\begingroup\$ I have updated the last few statements as they were ambiguous. I think it will be clear now. \$\endgroup\$
    – sai
    May 30, 2023 at 16:12

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