# Fraction of Cycle that Diode conducts

Discussion in 'Homework Help' started by jegues, Mar 17, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
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See figure attached for problem statement as well as my attempt.

How do we go about calculating the fracion of a cycle for which each diode conducts?

Thanks again

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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In this case you don't need to calculate anything to work out the diode conduction cycle, since this an ideal full-wave bridge rectifier with a purely resistive load and without any capacitive or inductive filtering applied. Each diode will conduct for one half the AC cycle or 180° electrical.

3. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
But there will still be a small timeframe in which all the diodes are nonconducting(Using constant voltage drop model), i.e. when 0 < Vs < 1.4V and 0 > Vs > -1.4V, correct?

How do I account for this?

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
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If the diodes are non-ideal then that is correct. I now see that's stated in the question.

On any half cycle the supply must overcome a total drop of 1.4V.

The secondary supply is 20Vrms or 28.28V peak.

Solve for the time at which 28.28sinωt=1.4V

sinωt=1.4/28.28=.0495

ωt=asin(.0495)=.0495 radians

t=.0495/ω=.0495/377=131usec

The diodes will conduct from 131usec after the AC zero crossing to 131usec before the subsequent AC zero crossing.

One half AC cycle is 8.33msec.

The total conduction time per diode is then
16.67/2msec-2*131usec=8.07msec

5. ### PerunaPete New Member

Mar 17, 2011
22
1
Just model the voltage across the diodes with a sine function, and then use an inverse sine to find the value in your calculator. Do note that rms for a sine wave is amplitude/sqrt(2), and don't forget to account for the voltage drop across *both* diodes.

In the actual output of a regulator, this will look like gaps between the peaks- normally not a concern because a capacitor is typically used.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Sorry the secondary would be 12V rms not 20V rms. So you would then re-do my solution using 17V peak rather than 28.28V peak. I'll leave that task to you.

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
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As a check - I then get a total conduction time per diode of 7.895msec.

8. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
Wouldn't the RMS for a sine wave be amplitude*sqrt(2)?

Anyways,

$\theta = sin^{-1}(\frac{1.4}{12\sqrt{2}})$

I'll do both half cycles in one shot,

$2\left( \frac{\pi - 2\theta}{2\pi} \right) * 100 = 94.8% \text{ conduction interval}$

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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The question actually asks for what fraction of a cycle each diode conducts.

A cycle presumably means a complete AC cycle. Therefore I would say the fraction is in relation to a full AC cycle period rather than a half period.

The correct answer would then be ~47.37%.

10. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
D1, and D2 conduct in the first half cycle, D3 and D4 in the second half cycle, it doesn't specify which diodes in particular so if we consider all the diodes they are conducting 2(47.37%) = 94.74%, right?

Or is it more like 47.37% each half cycle?

11. ### t_n_k AAC Fanatic!

Mar 6, 2009
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At the risk of being accused of pedantry the question asks "For what fraction of a cycle does each diode conduct?" Each diode (not the totality of diodes) conducts only for 47.37% of the AC cycle.

This is "nitpicking" in the extreme. As long as you can convey your understanding of the physical behavior I wouldn't make a big deal out of which answer is more correct. Consult with your teacher and come to some agreement.

12. ### justtrying Active Member

Mar 9, 2011
329
324
knowing the teachers that i have, the correct answer would be 47.37% of the cycle statinging which pair of diodes is conducting during positive half and which pair is conducting during negative half. If you say 94.74%, that implies that all diodes are conducting for full AC cycle, which is not true.

I'm getting used to the nitpicking, ha ha...