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

1) a) i) V =
`(V_(peak) )/(\pi )`

ii) V =
`V_(peak) [ 1 - (1)/(2fR_(l) C) ]`

b) i) V =
`(2V_(peak) )/(\pi )`

ii) V =
`V_(peak) [ 1 - (1)/(4fR_(L)C ) ]`

2) a) 2Vpeak

b) P/ V =
`V_(0max) = V_(peak)`

Step-by-step explanation:

1) If the diodes were ideal

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

ii ) without a capacitor

V =
`(V_(peak) )/(\pi )`

ii) with a capacitor filter

V =
`V_(peak) - (V_(r) )/(2)` ; where
`V_(r) = (V_(peak) )/(fR_(L) C)`

V =
`V_(peak) [ 1 - (1)/(2fR_(l) C) ]`

b) The dc output voltage the bridge rectifier should produce

i) without a capacitor

V =
`(2V_(peak) )/(\pi )`

ii ) with a capacitor

V =
`V_(peak) [ 1 - (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)`