Full Wave Rectifier (2 diodes)

In summary, a full-wave rectifier operating at 60Hz and with an rms value of 15 V has a dc output voltage of 20.2 V, a minimum required capacitance of 1.3475 F to maintain a ripple voltage of less than .25 V with a resistance of .5 ohm, a PIV rating of 42.4 V, and a surge current of 10777 A when power is first applied. The amplitude of the repetitive current in the diode is 1650 A.
  • #1
shalzuth
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?
 
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  • #2
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
 
  • #3
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 to Full Wave Rectifier (2 diodes)

1. What is a Full Wave Rectifier?

A Full Wave Rectifier is an electronic circuit that is used to convert an alternating current (AC) into a direct current (DC). It is made up of two diodes that work together to allow current to flow in one direction only.

2. How does a Full Wave Rectifier work?

The two diodes in a Full Wave Rectifier are arranged in a way that they only allow current to flow in one direction. When the AC input signal is positive, one diode conducts and when it is negative, the other diode conducts. This results in a DC output signal with no negative cycles.

3. What are the advantages of using a Full Wave Rectifier?

The main advantage of a Full Wave Rectifier is that it provides a more efficient conversion of AC to DC compared to a half wave rectifier. This is because it utilizes both the positive and negative cycles of the AC input signal, resulting in a smoother DC output. It also has a higher output voltage and less ripple.

4. What are the applications of a Full Wave Rectifier?

Full Wave Rectifiers are commonly used in power supplies for electronic devices such as laptops, TVs, and phone chargers. They are also used in audio amplifiers, motor control circuits, and battery charging circuits.

5. Can a Full Wave Rectifier be used for high voltage applications?

Yes, a Full Wave Rectifier can handle high voltage applications as long as the diodes used have a high peak inverse voltage (PIV) rating. The PIV rating should be at least twice the peak AC input voltage to ensure the diodes do not get damaged.

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