If the forward voltage drop of each conducting diode is Vf, which expression gives the peak output voltage of a full-wave bridge rectifier in terms of the input peak voltage of the transformer secondary Vp(in)?
Vp(out) = Vp(in) − 2Vf
In a full-wave bridge rectifier, during each half-cycle current flows through two diodes in series with the load (D1 and D2 in one half-cycle, D3 and D4 in the other).
Each conducting diode drops Vf, so two drops appear in series in the current path during every half-cycle.
Applying KVL at the peak: \(V_{p(in)} = V_{p(out)} + 2V_f\), so \(V_{p(out)} = V_{p(in)} - 2V_f\) (with silicon diodes, \(V_{p(out)} \approx V_{p(in)} - 1.4\) V).
The other expressions fail: subtracting only one Vf applies to a half-wave or centre-tapped rectifier, where one diode conducts at a time; diodes can only reduce the output, so adding 2Vf is impossible; and the bridge does not double the secondary peak, so 2Vp(in) is wrong.
Hence, the answer is Vp(out) = Vp(in) − 2Vf.
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