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} \$
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.
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.
Does that look reasonable? Or is there is better way of calculating the lowpass response of the smoothing capacitor?
Best Answer
No because the diode only provides charging current during the part of the AC cycle that it conducts: -
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,
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: -
Picture source