39.4k views
3 votes
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

User Aogan
by
7.0k points

1 Answer

2 votes

Answer:

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)

User Azamat Abdullaev
by
8.1k points