Full Wave Rectifier, Ripple Voltage?

In summary, the student is trying to find the maximum voltage and the minimum ripple current for a Full Wave Bridge Rectifier circuit. They found that the maximum voltage is 5.5 volts and the minimum ripple current is 0.008 seconds.
  • #1
Naptime
2
0

Homework Statement



Full Wave Bridge Rectifier circuit

four-diode-full-wave%20bridge-rectifier_7-3.jpg


Basically this circuit, but ZL is r//c not simply r as it is in this circuit (also, disregard any values... just the model is relevant)
Vs=6.5cost(2pi60t)
CL=1μF
RL=500Ω

I'm trying to find Vmax and Vripple.

Homework Equations



Vmax = Vdc-2Von
Vr = (Vmax Tau) / (2RC)

Tau = 1/f <-- This must be wrong but it's the only Time constant equation I have.



The Attempt at a Solution



Vmax = Vdc-2Von = 6.5 - 2(.5) = 5.5v
Tau = 1/f = 1/60 = .0166
Vr = (Vmax Tau) / (2RC) = 5.5 * Tau / (2(500)(1x10-6) = 91.6

My output graph (generated using SPICE) shows a sinusoid from around .2ish to 5ish Vo so clearly 91 Vripple is wrong. There should be a Vripple of about 5v to achieve this specific Vo. I'm 95% that my error is in the Tau calculation but I just can't find any info in my book/web about it.

Any help is appreciated, ty.
 
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  • #2
First off 0.5 is a bit low for Vf of the diodes.

As for the ripple. I use..

Q=VC

differentiate to give

I=CdV/dt

rearrange to give

dV = Idt/C

So then need to find approx I and dt...

if V=5V and R = 500 Ohms

I=V/R
= 0.01Amps

dt is approx
1/(60*2) = 0.008s

so
dV = 0.01 * 0.008 / 1*10-6
= 80V

Except it can't be 80V as that's >> 5V.

In other words the capacitor is not big enough to remove any significant ripple. The ripple is therefore roughly same as the AC waveform. That appears to be what Spice is telling you.
 
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  • #3
Hi Naptime. http://img96.imageshack.us/img96/5725/red5e5etimes5e5e45e5e25.gif
Naptime said:
CL=1μF
RL=500Ω

I'm trying to find Vmax and Vripple.
Check that you have the correct value for CL.

If CL is not large enough to produce small ripple, then the ripple will be large and the approximations involved in deriving Vr = (Vmax Tau) / (2RC) do not apply, hence that formula does not apply. CL of 1 μF is way too small to do anything useful here.
 
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  • #4
ps: The above method I used for calculating the ripple is only very approximate. It assumes the load is roughly constant current. That's a valid approximation when the ripple is small. If the ripple voltage is large it's not an accurate method.

However in this case the ripple appears so large and larger than the supply voltage that accuracy isn't that important.
 
  • #5
With a 500 ohm load you can effectively disregard the capacitor. Just assume the transformer is ideal and away you go. You won't get "constant load current" or anything near it with Cload = 1μF and Rload = 500Ω.

Different story, and much more complicated, if Rload = 12K as shown in the figure. There the rolloff is at ~ 13Hz and you're running 60Hz input voltage.
 
  • #6
OK I figured it out. Basically I was confusing Tau for T. Time constant and the period were the same thing in my calculations which threw everything off. This was basically a design where the output was designed poorly (on purpose so that we could learn to fix it). When Vo is ~ Vin the rectifier wasn't doing it's job. Basically the purpose is the flip the bottom half of the sinusoid to "mirror" the positive side and to do so the filter needed to be above a specific range (using the Vr equation in the first post, recalculate for C). If anyone needs me to fully rework this let me know, but with the help of those who responded and my TA I was able to redesign the circuit given the criteria. Ty.
 
  • #7
Naptime said:
OK I figured it out. Basically I was confusing Tau for T. Time constant and the period were the same thing in my calculations which threw everything off. This was basically a design where the output was designed poorly (on purpose so that we could learn to fix it). When Vo is ~ Vin the rectifier wasn't doing it's job. Basically the purpose is the flip the bottom half of the sinusoid to "mirror" the positive side and to do so the filter needed to be above a specific range (using the Vr equation in the first post, recalculate for C). If anyone needs me to fully rework this let me know, but with the help of those who responded and my TA I was able to redesign the circuit given the criteria. Ty.

Good job. have a nice nap, Ty! :smile:
 

Related to Full Wave Rectifier, Ripple Voltage?

1. What is a full wave rectifier?

A full wave rectifier is an electronic circuit that converts an alternating current (AC) input signal into a direct current (DC) output signal. It uses a combination of diodes to only allow the positive portion of the AC signal to pass through, resulting in a smoother DC output compared to a half wave rectifier.

2. How does a full wave rectifier work?

A full wave rectifier works by using two diodes in a bridge configuration to convert the AC input signal into a DC output signal. During the positive half of the AC signal, one diode conducts and allows current to flow through, while the other diode is reverse biased and blocks current. During the negative half of the AC signal, the roles of the diodes are reversed, resulting in a continuous DC output.

3. What is ripple voltage?

Ripple voltage is the small amount of AC voltage that remains in the DC output of a rectifier circuit. It is caused by imperfections in the rectifier circuit and can be reduced by using smoothing capacitors or by using a full wave rectifier instead of a half wave rectifier.

4. How is ripple voltage calculated?

Ripple voltage can be calculated by subtracting the minimum voltage from the maximum voltage of the DC output and dividing by two. This value is then divided by the average DC output voltage and multiplied by 100 to get the percentage of ripple voltage.

5. How can ripple voltage be minimized?

Ripple voltage can be minimized by using smoothing capacitors in the circuit, which help to filter out the AC component of the output signal. Additionally, using a full wave rectifier instead of a half wave rectifier can also reduce ripple voltage as it allows for a more continuous DC output. Other methods, such as using a voltage regulator, can also be used to further reduce ripple voltage.

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