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Why does the choke input filter produce a DC output voltage equal to the average value of the rectified voltage given as input to this filter, while the capacitor input filter produces a DC output voltage equal to the peak value of the rectified voltage given as input to the filter?
Why is it so when both are supplied from the same rectifier?
Why does the choke input filter produce a DC output voltage equal to the average value of the rectified voltage given as input to this filter,
There is a property of inductors that, once you understand it, makes analysis of circuits with inductors much easier. This property is known as inductor voltage balance (or inductor volt-second balance). If a circuit is in a periodic steady state, that is, the same cycle of states occurs over and over periodically, then the average voltage on one side of an ideal inductor is equal to the average voltage on the other. (Real inductors have resistance, and there is a voltage drop due to this resistance, so the inductor voltage balance described above is not exact for real inductors).
This is the derivation.
For an ideal inductor
$$V_L = -I_L'L$$
so over a cycle,
$$\int_0^T V_L dt = -\Delta I_L L = -L \Delta I_L$$
Since we have assumed the circuit is in steady state, \$I\$ returns to its original value, and so \$\Delta I = 0\$.
Thus
$$\int_0^T V_L dt = 0$$
which says that the average of \$V_L\$ over a cycle is 0.
Since the average voltage across the inductor is 0, the average voltage on one side of the inductor must equal the average voltage on the other side of the inductor.
And that means that in any filter of the form
simulate this circuit – Schematic created using CircuitLab
the average voltage at \$V_{in}\$ will equal the average voltage at \$V_{out}\$ if we assume that L1 is an ideal inductor (i.e. has no internal resistance).
the capacitor input filter produces a DC output voltage equal to the peak value of the rectified voltage given as input to the filter?
Consider this unregulated supply circuit:
Here are the input voltage, the absolute value of the input voltage, and output voltage.
And here is the current through the rectifier.
Notice that current only flows through the rectifier when the absolute value of the input voltage is greater than the output capacitor voltage. Put another way, the capacitor charges through the rectifier, but discharges through the load.
It is often a good approximation to say that a diode has a voltage drop of some fixed amount, say 0.65 V, when it is conducting. The voltage dropped by a bridge rectifier would then be 2 diode drops, or \$\approx\$ 1.3V.
Since the voltage drop through the rectifier is "fixed" when it is conducting, the maximum voltage of the capacitor will be this fixed amount less than the maximum (absolute value of the) voltage applied to the input of the rectifier. And that is why it is (more or less) correct to say that the peak output voltage of the unregulated power supply is equal to the peak input voltage (minus the rectifier voltage drop).
