What is the Amplitude of the Repetitive Current in a Full Wave Rectifier?

Click For Summary
SUMMARY

The discussion focuses on calculating the amplitude of the repetitive current in a full-wave rectifier operating at 60Hz with a transformer output voltage of 15V and a diode voltage drop of 1V. The calculated values include a DC output voltage of 20.2V, a minimum capacitance of 1.3475F to maintain ripple voltage below 0.25V, and a PIV rating of 42.4V. The surge current is calculated at 10,777A, while the amplitude of the repetitive current through the diode is confirmed to be approximately 1,650A, despite initial confusion over the calculations.

PREREQUISITES
  • Understanding of full-wave rectifier circuits
  • Knowledge of electrical engineering concepts such as voltage drop and ripple voltage
  • Familiarity with the equations for DC output voltage and surge current
  • Basic proficiency in calculus for understanding current variations over time
NEXT STEPS
  • Study the derivation of the surge current formula: I = C * dV/dt
  • Learn about ripple voltage calculations in capacitor-filtered circuits
  • Explore the implications of diode voltage drop on rectifier performance
  • Investigate the effects of frequency on current amplitude in rectifier circuits
USEFUL FOR

Electrical engineers, students studying power electronics, and anyone involved in designing or analyzing rectifier circuits will benefit from this discussion.

shalzuth
Messages
2
Reaction score
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?
 
Physics news on Phys.org
shalzuth said:

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
 
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.
 

Similar threads

Replies
1
Views
7K
  • · Replies 11 ·
Replies
11
Views
6K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
10K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 12 ·
Replies
12
Views
5K