# Half Wave Peak Rectifier

Discussion in 'Homework Help' started by jegues, Apr 22, 2011.

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

Sep 13, 2010
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I've got some questions with regards to what he's doing when he's answering this question.

See figure attached.

My textbook gives the equation,

$v_{o} = V_{p} - V_{r}$ where Vr is the peak-to-peak value of the ripple voltage, and Vp is the peak value of the input sinusoid.

I write one KVL around the loop giving me,

$v_{s} = 0.7 + v_{o}$

So I ask myself what the biggest Vo will be? Well 15V with the 1V ripple so 16V.

With the diode added as well,
$v_{s} = 16.7$

This should be the peak value of my input sinusoid, Vp = 16.7V

Now the equation I should be using to find the desired capictance value is given as,

$V_{r} = \frac{V_{p}}{fCR}$

Why does he insist on using Vp-VD?

Thanks again!

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Yes in part (b) the writer has made a transcription error of some sort. Presumably they intended to use the equation referenced as Eq(3.28). Is that an equation from your text?

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

Sep 13, 2010
735
45
That was the equation I had given in the OP.

$V_{r} = \frac{V_{p}}{fCR}$

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

Sep 13, 2010
735
45
He also uses the same thing for part c) and d). (See figure)

Can anyone explain why it's Vp-Vdo?

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5. ### t_n_k AAC Fanatic!

Mar 6, 2009
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On reflection, I see Vdo is the diode drop and Vp is the peak transformer secondary voltage.

I was thinking Vdo was the DC output but with a closer look at the solution diagram it makes more sense.

So Vp=16.7 V

Vdo=0.7V

Then the peak voltage at the load is 16.7-0.7=16V

So your equation Vr=Vp/fRC simply reflects a difference in circuit annotation.

Vp in your text is probably intended to mean the peak load voltage, rather than the peak secondary voltage behind the diode. Or perhaps the text disregards the diode drop. The solution is simply elaborating the "subtle" distinction that the diode drop introduces to the analysis.

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

Sep 13, 2010
735
45
Ah I see! Here the authour is using Vp as the peak voltage of the secondary winding, and Vp-Vdo as the peak voltage at the load!

So does the equation,

Vr = Vp/fRC use Vp as the peak voltage at the load?

Also, for part d) and e), is there an eaiser intuitive way to obtain these equations without memorizing them? Also, how would these equations vary for a full wave peak rectifier?

Thanks again!

Last edited: Apr 23, 2011
7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
790
Yes indeed, that is correct - because it is based on the analysis of the discharge of the load filter capacitor. In higher voltage applications the difference due to the diode drop becomes less important with increasing voltage.

That analysis is somewhat oversimplified with respect to the physical reality. However, it produces a result which is conservative in terms of the filter capacitor value requirement for a specified ripple voltage. So it leads to a slight overestimate of the required capacitance - which isn't really an issue.

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

Sep 13, 2010
735
45
Is there an easier way to generate those formulas for part c) and d) rather than memorizing them? They also change for a half wave/full wave peak rectifier, correct?

9. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I would take issue with the solution provided for parts d & e - particularly the average diode current solution.

The rectifier is effectively supplying 15V DC to a 150Ω. That means the mean load current is 100mA. How then could the mean diode current be 2.61A. That makes no sense.

The peak diode current looks about right. I use another approach which doesn't look like what the formula is based on.

This is my approach.

The diode conducts with short approximately "triangular" current impulses. With a 16V peak and a ripple of 2V the diode will conduct over an interval

Δt=T/4-(1/ω)*θ

Also (1/ω)*θ is the time after the AC zero crossing at which the diode begins to conduct

At that time the rate of change of the AC voltage

[ΔV/Δt]=ω*Vp*cosθ=2922 volts per second

The peak capacitor current

Ic_peak=C[ΔV/Δt]=889uF*2922=2.6A

Add the resistor current = 0.1A

Peak diode current Id_peak=2.7A

The diode conduction interval

Δt=T/4-(1/ω)*1.065=1.34 msec

I would calculate the mean diode current as
Iav=(2.7/2)*Δt/T =1.35*(1.34ms/16.67ms)=0.109A=109mA

Edit:

I note the question does actually ask for the average current during conduction so that might be where I come unstuck. I would tend to calculate the average diode current overall. If that's what the question requires then that's what one must do.

If I assume the mean capacitor charging current is given by Ic=C*Vr/Δt=889uF*2/1.34msec=1.33A and add 0.1A for the resistor current then the mean diode current during conduction would be ~1.43A which is close enough to your attached solution. Shows me that one has to read questions carefully.

In short, I can't suggest any quick alternative ways of doing the analysis without remembering the formulas. As you can see my approach still requires several steps which must be remembered with some basic formulas thrown in for good measure.

Last edited: Apr 24, 2011