# Ripple in a rectifier circuit

Discussion in 'Homework Help' started by AD633, Jul 12, 2013.

Jun 22, 2013
96
1
Calculate the value of the capacitor so that the average voltage on the load is 9.75 V.The current on R2 equals 0,2 A

If the input voltage has an amplitude of 10 V,therefore the ripple equals to 0,25V

$

I*delta T=C*Delta V

0,2A*10ms=C*0,25V

C=8 mF

$

Is this right?

Thanks

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2. ### WBahn Moderator

Mar 31, 2012
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Is this last part a given part of the problem, or are you assuming it, or what?

If you came up with the 0.25V ripple voltage, how are you defining ripple voltage (there are a couple of different definitions that get used).

Are both sources 10V in amplitude? Are they in phase?

If so, which one can effectively be removed because it's diode will never be forward biased?

Jun 22, 2013
96
1
No,the values of Iload,and Vomed are given and the waveforms of the sources.

$
Vomax=Vomed-Vripple

10V=9,75 V-Vripple

Vripple=0.25 V
$

The both have 10 V amplitude and they are in opposition of phase.When V1=10 V,V2=-10V.

They also say that the conduction time of D2 is approximately 0 that is the capacitor charges instantaneously.

4. ### WBahn Moderator

Mar 31, 2012
18,087
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It's pretty important that we know that they are in opposition, don't you think? We are not mind readers.

Don't your equations yield a ripple voltage of -0.25V.

You really need to slow down and take more time to review your work and make sure you at least have it reflecting what you think.

If the sources are 10V amplitude, how can the load possibly have an average voltage of 9.75V? Consider what would happen if you replaced both sources by 10V batteries. What would the voltage across the load be?

Jun 22, 2013
96
1
My equation was wrong.It should be $Vomax=VoDC+Vripple$

The average voltage of 9,75 V, wasn't calculated by me.It is said in the question that they want to have a VoDC=9,75 V with those sources.

6. ### WBahn Moderator

Mar 31, 2012
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But how is that achievable? Answer the question I asked: Consider what would happen if you replaced both sources by 10V batteries. What would the voltage across the load be?

Hint: What is the voltage drop across a forward-biased 1N4148 diodel?

That represents the MAXIMUM voltage that can appear across the load.

Jun 22, 2013
96
1
My mistake ,i forgot to mention that they also say to consider that the drop voltage when the diodes are foward-biased is 0 V(ideal diodes).
Considering this is the value of the capacitor i mentioned correct?

Thanks

8. ### WBahn Moderator

Mar 31, 2012
18,087
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What simplifying assumptions are you making regarding your waveforms.

Jun 22, 2013
96
1
Here is a sketch of the graphical representation of the waveforms(both have f=50 Hz)(The first one is suposed to be a sine wave with 10 V of amplitude
$
V1(wt)=10Vsin(wt)
$

The second one is a square wave with 10 V of amplitude and in oposition of phade with v1(wt)

The other aproximation to be consider is the fact that the capacitor charges instantaneously.

Thanks

Last edited: Jul 13, 2013
10. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
The graph that will help you most would be one that has these two waveforms AND the load voltage all on the same diagram. Then you can easily identify one key approximation that you are making, but not seeing, and also how this circuit differs from the case where you just have a sinusoidal source in terms of the relationship between the average voltage and the min/max voltages.

Jun 22, 2013
96
1

The aproximation that i am making is considering that the discharge time of the capacitor is equal to half of the period( that is 10ms)...but when we have square wave aplied to the load the capacitor does not charge right?

So the discharge time to be used in calculations should be 5 ms?

In terms of the relationship between the average and min/max voltage when we have a square wave on the load the capacitor won't have any effect on the value of the voltage right?

Thanks

12. ### WBahn Moderator

Mar 31, 2012
18,087
4,917
I think you are seeing the points, but it will be a lot easier to discuss if you draw a sketch.

Jun 22, 2013
96
1
That means that the ripple value i mentioned was incorrect right?I think the value of the capacitor is equal to:

$

C=(0,2A)*(5ms)/(0,125V)=8mF$
,right?

Thanks

Mar 31, 2012
18,087
4,917