Question

For the circuit below with the source peak Vs-10V and diode voltage Vpo-0.7V, the average output voltage (Vo, avrthe peak current in the diode (omax! and PNV are: - 0.7 V + Rikos

Answer 1

The voltage source shown in th figure is a dc voltage source and the diode is reverse biased so no current flows through the circuit at all. And output voltage will be zero always. PIV is the peak reverse voltage acting across the diode during reverse bias so in this case entire Vs is acting across the diode and therefore PIV is equal to Vs

$V_{avg}=0$

$I_{peak}=0$

PIV = 10V

Now let the case be the source is a sinusoidal source with peak voltage 10V then the case will be as shown

The circuit will act as half wave rectifier which only pass the negative half cycle and block positive half cycle. For positive half cycle the diode is reverse biased and $V_{0}=0$ and no current will flow through the circuit. In negtaive half cycle the diode will conduct but the diode drop will be there and therefore $V_{0}=V_{s}-0.7$ (only when this value is less than 0 the diode conducts )

$\therefore V_{0}=9.3\sin \omega t$(only during negative half cycle)

So by the equation of average voltage of a half wave rectifier

$\therefore V_{0avg}=V_{m}/\pi$

$V_{m}=9.3V$

$\therefore V_{0avg}=9.3/\pi$

$\therefore V_{0avg}=2.96V$

Now coming to peak current the max current will flow when max voltage come across the resistor and its magnitude is 9.3V(in negative half cycle )

$\therefore I_{peak}=9.3/1k\Omega$

$\therefore I_{peak}=9.3mA$

Now coming to peak inverse voltage it happens in positive half cycle when diode is reverse biased the source voltage come across the diode and it is equal to 10V

PIV=10V

:) :) :) :) :)