Rectifier with cap and zener

Discussion in 'Homework Help' started by scottfarcuz, Dec 9, 2012.

1. scottfarcuz Thread Starter New Member

Dec 9, 2012
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Im lost as to how to calculate the Vripple (pk-pk) and Vaverage (Including Vripple) of the attached circuit.

For Vsec I used
75Vrms *sqrt(2) = 106Vpk then divided it by the turns ratio of 5 coming up with 21.21Vpk

For the voltage at the top of the cap I just subtracted the 0.7V drop from the diode leaving me with 20.51Vpk

Any help or links would be very appreciated.

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2. #12 Expert

Nov 30, 2010
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It seems you have misunderstood the question. The ripple voltage is at the top of C1. What you have found is the peak voltage of the transformer, and that is a good thing to do.

There are 2 diodes in the path of each wave, so your .7 volts is a wrong belief.

You need to find the amount of (mostly) dc on C1. Then calculate the current of the circuit in the DC section. Then calculate the p-p ripple voltage and describe that combined with the DC voltage on C1.

3. scottfarcuz Thread Starter New Member

Dec 9, 2012
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0
I think I may have this problem solved, but would really appreciate anyones input.

4. MrChips Moderator

Oct 2, 2009
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I don't think anyone would want to help by looking at random equations on a piece of paper.

Dec 9, 2012
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6. #12 Expert

Nov 30, 2010
18,076
9,676
Your third equation is wrong. The 200 ohm resistor doesn't go to ground. It goes to a 6 volt zener so the voltage is 19.8 minus 6.

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7. scottfarcuz Thread Starter New Member

Dec 9, 2012
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0
Thank you. Not sure how I missed that, but it changed almost every eqatiion below it.

8. The Electrician AAC Fanatic!

Oct 9, 2007
2,644
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You've got wrong units on Vripple2; they should be mV, not mA.

Also, the voltage Vc1 can't be anywhere near 6 volts. It should be more like 19 or so volts.

The current in the transformer secondary consists of very narrow, large amplitude pulses. These large pulses cause a larger voltage drop across the diodes than .7 volts each. This will cause the peak voltage across the capacitor to be somewhat less than it would otherwise be.