Why is RC circuit considered closed loop system with negative feedback? Is it because it has a differential equation due to which current 'I' through capacitor 'C' will be C*(dVo/dt) = C*{Vo(t)-Vo(t-h)}/h, where 'h' tends to Zero. Since a past output 'Vo(t-h)', appears in the equation, does it makes it a closed loop control system with negative feedback?

  • Vo is Voltage across Capacitor. enter image description here
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    \$\begingroup\$ As the capacitor voltage changes, the voltage across the resistor continues to be (Vbat - Vcap), thus the change in current is controlled to be (-)delta_Vcap/R. \$\endgroup\$ – analogsystemsrf Nov 18 '19 at 11:24
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    \$\begingroup\$ Strictly speaking, to have a negative feedback system there is a need of some gain... but it is interesting to look at the humble RC circuits from this perspective. The "vicious circle" is: the output voltage Vo across the capacitor subtracts from the input voltage... the difference between the two voltages determines the current that charges the capacitor... thus determining the output voltage... and so on... up to infinity... Such processes lead to the exponent. \$\endgroup\$ – Circuit fantasist Nov 18 '19 at 12:19
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    \$\begingroup\$ I assisted in debugging a quadrature-generator circuit, needed for sideband suppression and carrier suppression. The suppression was poor (about 20dB). OK. WHY? We started analyzing the circuit via simulator, finding as others had warned "There is a sweet spot." . I sat and wrote the G/(1 + G * H) and found the G*H was only 1.6X Thus the servo loop behavior was horrid. FIrst thing we did was to greatly slow the input edges, to ensure a change in slicing threshold would actually alter the duty cycle. Lesson: don't trust somebody else's circuit. You must fully understand it. \$\endgroup\$ – analogsystemsrf Nov 18 '19 at 12:49

Never thought of it that way, but whatever floats your boat, I guess. A closed loop system has a form like this.

closed loop


The closed loop transfer function is

$$\frac{Y}{X} = \frac{G}{1 + GH}.$$

The RC system has a transfer function

$$\frac{1 / (C s)}{R + 1 / (C s)} = \frac{1}{1 + R C s}.$$

This is the closed loop transfer function with \$G \equiv 1\$ and \$H \equiv RCs\$.

If you are interested in a more qualitative analysis, note that the input to the feedback transfer function \$H\$ is \$Y\$, which is the capacitor voltage. Hence

$$YH = RCs Y = Ri = V_\mathrm{R}.$$

So what you are feeding back is the resistor voltage.

  • \$\begingroup\$ Thanks for the insight, but I would like to ask, 1) That for any system whose transfer function can be reduced to the form G/(1+GH) , is a negative feedback system? 2)Does the presence of exponential term ensures that the system is a feedback system? \$\endgroup\$ – Vasu Goyal Nov 18 '19 at 15:59
  • \$\begingroup\$ @VasuGoyal I would say yes to 1 and no to 2. Keep in mind the exponential form is there only to a step input. Put a sine wave input and the output is completely different. \$\endgroup\$ – user110971 Nov 18 '19 at 18:07

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