# Centre Tap Transformer Question

Discussion in 'Homework Help' started by Zaraphrax, Sep 3, 2009.

1. ### Zaraphrax Thread Starter Active Member

Mar 21, 2009
47
3
Hi there,

I'd appreciate if you could check my answer. I only require a simple "right" or "wrong". Thanks.

I need to determine the ripple peak to peak voltage and the average voltage for this transformer, that has a 100 ohm load connected. Voltages supplied are RMS values.

Here's what I've done:

Time constant on capacitor: CR= (100 x 10^-6) x 100 = 10ms

Since RMS voltages are supplied, 12 * (2^0.5) = ~17V

Plotting the output of the circuit (and the capacitor discharge curve), I determined that the average voltage was about 14.5V (~12V minimum to 17V), with a ripple of 5 V.

Is this correct?

Thanks.

2. ### t06afre AAC Fanatic!

May 11, 2009
5,939
1,227
I have not checked your numbers. But did you use 50 or 100 Hz then you calculated the ripple? You use a full wave rectifier and then you have to use mains frequencyX2

3. ### mik3 Senior Member

Feb 4, 2008
4,846
69
If you assume that the ripple voltage is small then it can be approximated by a triangular wave.

Thus the ripple frequency is:

Vripple=Vaverage/(2fCR)

Vaverage=Vmax-Vr/2

by using the two above equations you get:

Vaverage=Vmax/[1+1/(4fCR)]

f=mains frequency

4. ### t06afre AAC Fanatic!

May 11, 2009
5,939
1,227

Apr 5, 2008
18,579
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Hello,

The link does not work. The forum does not work with links that end on ")".
The "(" can be replaced with %28
The ")" can be replaced with %29

Here is the link that works:
http://en.wikipedia.org/wiki/Ripple_(electrical)

Greetings,
Bertus

6. ### t06afre AAC Fanatic!

May 11, 2009
5,939
1,227
Ouch so much for copy and paste Well anyway the link is now correct. So read it and learn

7. ### Zaraphrax Thread Starter Active Member

Mar 21, 2009
47
3
Thanks guys. The question actually doesn't give a mains frequency, so I'll just assume 50hz since thats what our main is over here. Thanks.

8. ### Audioguru Expert

Dec 20, 2007
10,577
1,178
The output voltage will be higher than is calculated since the load current is so low.