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I understand qualitatively and electrically how a smoothing capacitor works in the case of a diode-based full-bridge rectifier but what I am not too certain of is how to describe it in terms of its frequency response, since the capacitor is in parallel with the load resistor.

It makes sense for this parallel RC circuit to act as a low-pass filter to remove the harmonics and fundamental frequency of the rectified AC signal to preserve only the DC component. But all analyses of RC-based lowpass filters are series-based circuits, with the frequency response based on voltage gain Vout/Vin. i.e. \$H(\omega)=\frac{1}{1+j\omega RC} \$

schematic

simulate this circuit – Schematic created using CircuitLab

However, for the parallel RC circuit, this voltage gain is unity since the voltage across the R and C is the same.

schematic

simulate this circuit

So how does one find the frequency response of smoothing capacitor in parallel with the load resistor and show that it is a lowpass response?

My guess is that we are modelling the current gain, i.e. treating the diode rectifier circuit as a current source, so that we can transform the current source in parallel with load resistor (Norton equivalent) into a voltage source in series with the load resistor (Thevenin equivalent), which makes it into the conventional series RC circuit.

schematic

simulate this circuit

Does that look reasonable? Or is there is better way of calculating the lowpass response of the smoothing capacitor?

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Does that look reasonable?

No because the diode only provides charging current during the part of the AC cycle that it conducts: -

enter image description here

So, the diode conducts during the charging period for a short part of the cycle and, for the rest of the cycle the capacitor is free-falling due to ONLY the load resistor,

Or is there is better way of calculating the lowpass response of the smoothing capacitor?

Take into account what I've said above and develop an exponential decay formula that describes the falling capacitor voltage under the influence of the discharging resistor: -

enter image description here

Picture source

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  • \$\begingroup\$ It seems ok for the current to conduct during only part of the cycle, since that suggests it's a rectified (and therefore periodic) wave. Therefore the Fourier series of the current would consist of the DC component, a fundamental and its harmonics. So the parallel RC would act as a filter to attenuate the latter two frequency components of the current, instead of the voltage. I'm just trying to view this part of the circuit from a signals point of view and the missing link seem to be that Norton to Thevenin transformation, which I assume holds for time-varying currents/voltages as well. \$\endgroup\$ – Steve May 29 '18 at 13:27
  • \$\begingroup\$ Yes you can treat it in the frequency domain like this but if you want a time domain analysis to find ripple voltage then my answer is the more direct route. \$\endgroup\$ – Andy aka May 29 '18 at 15:28

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