Question

the diodes were ideal:a.What dc output voltage should the half-wave rectifier produce?b.What dc output voltage should the bridge rectifier produce?2.What is the peak reverse voltage impressed across the diode: a.In the half-wave rectifier with capacitor in parallel with the load resistor?b.In the bridge rectifier with capacitor in parallel with the load resist

Answer 1

Answer:

1) a)  i)  V = $\frac{V_{peak} }{\pi }$  

       ii) V = $V_{peak} [ 1 - \frac{1}{2fR_{l} C} ]$

  b) i)   V = $\frac{2V_{peak} }{\pi }$

     ii)     V = $V_{peak} [ 1 - \frac{1}{4fR_{L}C } ]$

2) a) 2Vpeak

   b) P/ V = $V_{0max} = V_{peak}$

Explanation:

1) If the diodes were ideal

a)  The dc output voltage the half-wave rectifier should produce

 ii ) without a capacitor

     V = $\frac{V_{peak} }{\pi }$  

  ii) with a capacitor filter

    V = $V_{peak} - \frac{V_{r} }{2}$      ;   where $V_{r} = \frac{V_{peak} }{fR_{L} C}$

    V = $V_{peak} [ 1 - \frac{1}{2fR_{l} C} ]$

b) The  dc output voltage the bridge rectifier should  produce

  i) without a capacitor

     V = $\frac{2V_{peak} }{\pi }$

 ii ) with a capacitor

    V = $V_{peak} [ 1 - \frac{1}{4fR_{L}C } ]$

2) a) peak reverse voltage in the half-wave rectifier with capacitor in parallel with the load resistor

during -ve half cycle

 P/V = $V_{peak} + V_{C}$

 Voltage across capacitor = Vc = Vpeak

Therefor the PIV for a half-wave rectifier with capacitor in parallel

= 2 * Vpeak = 2Vpeak

 b) Peak reverse voltage in the bridge rectifier with capacitor in parallel with the load resist

P/ V = $V_{0max} = V_{peak}$