Full Wave Rectifier (2 diodes)

  • Thread starter shalzuth
  • Start date
  • #1
2
0

Homework Statement


A full-wave rectifier is operating at a frequency of 60Hz, and the rms value of the transformer output voltage is 15 V.
(a) What is the value of the dc output voltage if the diode voltage drop is 1 V?
(b) What is the minimum value of C required to maintain the ripple voltage to less than .25 V if R = .5 ohm?
(c) What is the PIV rating of the diode in the circuit?
(d) What is the surge current when power is first applied?
(e)What is the amplitude of the repetitive current in the diode?


Homework Equations


Vdc = Vp - Von
Vr = Vdc/R * T/2C
PIV = 2 * PIV
Vp = Vrms * 2^.5


The Attempt at a Solution


I've solved a-d.
(a) Vrms * .5^2 - Von (20.2 V)
(b) Rearrange Vr to solve for C (1.3475 F)
(c) PIV = 2 * Vp (42.4 V)
(d) Isc = 2 * pi * f * C * Vp (from deriving and such, 10777 A)
I know all of those are right.
(e) I have tried a bunch of number juggling, V = IR, I = C * dV/dt, a bunch of different variations, etc., and can't get the correct answer. The final answer should be about 1650 A.
Point me in the right direction?
 

Answers and Replies

  • #2
berkeman
Mentor
59,409
9,530

Homework Statement


A full-wave rectifier is operating at a frequency of 60Hz, and the rms value of the transformer output voltage is 15 V.
(a) What is the value of the dc output voltage if the diode voltage drop is 1 V?
(b) What is the minimum value of C required to maintain the ripple voltage to less than .25 V if R = .5 ohm?
(c) What is the PIV rating of the diode in the circuit?
(d) What is the surge current when power is first applied?
(e)What is the amplitude of the repetitive current in the diode?


Homework Equations


Vdc = Vp - Von
Vr = Vdc/R * T/2C
PIV = 2 * PIV
Vp = Vrms * 2^.5


The Attempt at a Solution


I've solved a-d.
(a) Vrms * .5^2 - Von (20.2 V)
(b) Rearrange Vr to solve for C (1.3475 F)
(c) PIV = 2 * Vp (42.4 V)
(d) Isc = 2 * pi * f * C * Vp (from deriving and such, 10777 A)
I know all of those are right.
(e) I have tried a bunch of number juggling, V = IR, I = C * dV/dt, a bunch of different variations, etc., and can't get the correct answer. The final answer should be about 1650 A.
Point me in the right direction?

There have got to be typos in your current numbers. Are they missing decimal points?

10777 A
1650 A
 
  • #3
2
0
Positive there are no types :/, I know the values seems large, but the first one is a peak value of the surge current. It's 2*pi*f*C*V, which is derived from...

I = C * dV/dt
= C * d(Vcos(2*pi*f*t))
= C * -V*2*pi*f * sin(2*pi*f*t)
and since it's a peak value, the sin component can only max out at 1.
therefore
= C * 2 * pi * f * V
= 1.34 * 2 * pi * 60 * 20.2

And I can't figure out how to get the amplitude of the current through the diode.
And the solutions say 1650A. The other answers have been correct, so I'd assume e is correct as well.
 

Related Threads on Full Wave Rectifier (2 diodes)

  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
2
Views
9K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
6
Views
7K
Replies
5
Views
2K
  • Last Post
Replies
11
Views
5K
Replies
3
Views
18K
Replies
11
Views
4K
W
Replies
2
Views
11K
Top