finding the value of C

Discussion in 'Homework Help' started by led23, Apr 27, 2009.

1. led23 Thread Starter New Member

Mar 10, 2009
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0

using this diagram i can't figure out how to find the value of C. Can anyone help to lead me to a formula to solve this problem?

2. KL7AJ AAC Fanatic!

Nov 4, 2008
2,040
287
No answer is possible, unless the question specifies the amount of ripple.

eric

3. led23 Thread Starter New Member

Mar 10, 2009
6
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I forgot to mention the value of the capacitor c , to have a DC voltage across the load equal to 9.5v

4. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
I don't think this is possible.

Even with zero capacitance the average DC value will be at least 12.3 V assuming some realistic forward voltage drop in the rectifier diodes.

5. led23 Thread Starter New Member

Mar 10, 2009
6
0
the only other part of this I am to assume that they are ideal diodes.

6. t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
OK. I did originally assume the primary to secondary ratio was 5:1 for each winding. That's probably incorrect so ...

Let me go through the circuit behavior in a little detail.

The primary is connected to a 60Hz voltage supply with a peak of 100V.

The primary to secondary ratio is stated as 5:1. Assuming this means the total secondary winding is 1/5th of the primary or 20V peak, then the secondary half voltage will be 10V peak. That will then "work" for the expected 9.5V DC load voltage. So on that basis disregard my earlier post.

To solve your problem requires several assumptions - given there is no other data such as KL7AJ suggests.

I would then try this approach.

1. Assume ideal diodes. Hence rectified wave Vmax = 10V.
2. The DC voltage = (Vmax+Vmin)/2 = 9.5V
3. Vmin= 9V
4. The rectified 60Hz sine wave goes from zero to Vmin [= 9V] in 2.97ms
5. So the capacitor discharge time Δt = 8.33/2 + 2.97 = 7.14ms
6. Assuming I load is 'constant' = 9.5/330 = 28.8mA
7. Capacitor voltage droop ΔVc= 1v {from Vmax to Vmin}
8. For this ΔVc at 28.8mA drain for Δt=7.14ms we require C=IxΔt/ΔVc
9. C=205.5uF

Hope that makes sense.

7. led23 Thread Starter New Member

Mar 10, 2009
6
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I have finally had time to look at this and there is something that bothers me.
the Time period of a 60 Hz sin wave is 1/60= 1.6 mS, so I am confused how you came up with 2.93? then you say the capacitor discharge time = 8.33 how am I to come up with the number.Sorry for the questions, i am just trying to understand.

8. Papabravo Expert

Feb 24, 2006
10,148
1,791
Off by a decimal place. 1/60 th of a second is 16.6666... milliseconds

9. jvjtech Member

Jan 26, 2008
23
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From: http://en.wikipedia.org/wiki/Rectifier

For single-phase AC, if the transformer is center-tapped, then two diodes back-to-back (i.e. anodes-to-anode or cathode-to-cathode) can form a full-wave rectifier.

My text follows: Refering to the graphic of the recifier using a center tapped transformer on the wiki page you will see the waveform of rectified AC. Without a load the capacitor will charge up to the peak voltage and remain at that level. The load will load down both the transformer (through the diodes) and the capacitor. The capacitor will discharge through the load during the time the voltage less than the AC peak. This discharge formula is given by the RC time constant of RL and C. As the frequency is 60 Hz and the rectifier is a full wave rectifier the time we are dealing is 1/2 of 1/60 or 8.3 ms. The ripple, or the amount the capacitor is allowed to discharge, is given by the target of 9.5 V at the load. t_n_k showed that Vmax is 10V and calculated Vmin at 9V to get Vavg at 9.5V. Regards. JJ

Apr 5, 2008
15,648
2,347
Hello,

Read the attached PDF. That will make things more clear.

Greetings,
Bertus

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