So I have a load that needs constant power to maintain a specific temperature and this load changes over time. For this, I am using the design outlined in the following article:


Now to design the error amplifier shown in the article I want to use a difference amplifier with a closed loop gain of 5 V/V along with a 100uF shunt capacitor at the output of the amplifier to create (at least what I think is) a PI controller.

Now my question is, is the capacitor necessary and does adding it actually create the behaviour of a PI controller? I added it because I wanted to prevent crazy oscillation when I turn on the system. I would also like to avoid feeding too much power to the load as well since it could damage the system. However, I am not too sure if adding that helps at all.

  • \$\begingroup\$ The typical PI controller design I know puts the capacitor in series with the feedback resistor of the Op-Amp in inverting amplifier mode. As the gain is proportional to the impedance of the feedback resistor, you get a low gain for high-frequency fluctuations (you want the proportional action to counter them), but a high gain for low-frequency errors (up to infinite gain at DC), which will get cancelled by the integral action. \$\endgroup\$ – Michael Karcher Sep 30 '16 at 19:02
  • \$\begingroup\$ That makes sense but since I am using a difference amplifier I should be adding the capacitors in series with the two feedback resistors correct? \$\endgroup\$ – Biraj Kapadia Sep 30 '16 at 19:19
  • \$\begingroup\$ upload.wikimedia.org/wikipedia/commons/a/a2/… only has one feedback resistor, $R_f$. $R_g$ is not connected to the output and thus no feedback resistor. So only one capacitor in series with $R_f$, and no capacitor in series with $R_g$. \$\endgroup\$ – Michael Karcher Sep 30 '16 at 19:44

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