Electronic – Full-wave rectified sine wave representation

accapacitorrectifierwave

Let us suppose I have an input voltage \$V=V_0 \sin(\omega t)\$ connected to a full-wave rectifier bridge which changes the voltage to \$V=V_0|\sin(\omega t)|\$. Now, if I want to work out the voltage drops, currents and the impedances in the following circuit, how do I represent the rectified waveform in terms of a superposition of elementary sinusoidal signals?

schematic

simulate this circuit – Schematic created using CircuitLab

I tried to use wolframalpha to give me a Fourier decomposition of the wave (which I don't know how to compute analytically). But now I do not know how to find the final output voltage across the load resistor. I know the basic methods of finding impedances of capacitors and resistors, but since there is a superposition of many waves here,I am unable to find the answer. The impedance of capacitor is \$\frac{1}{i\omega C}\$, I don't know the ω here.

Best Answer

The problem with analysing it formulaicly is that the diodes are very non-linear devices and this is implied in the diagram below (red curve is output): -

enter image description here

The blue curve shows a full-wave rectified signal with no capacitor on the output, just a load resistor. When you add a capacitor it charges very quickly due to the forward conduction of one of the diodes but as soon as it attains near-peak voltage (and the blue waveform turns back down towards zero volts), the capacitor remains charged thus reverse biasing the diode and no further diode conduction is seen until the next peak.

You have to think along these lines to adequately "model" what happens. The red curve shows a discharging element between peaks and this is the effect of drawing current into the load resistor. It's a reasonable approximation to regard this as a time-linear discharge but in fact it's an exponential discharge. However, for most decently designed (and used) bridge circuits, using a linear approximation is fine and the ripple voltage formula works pretty good.

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