# Centre Tap Rectifier

#### Ledwardz

Joined Oct 31, 2010
37
hi ,i am after a few things cleared up about the center tap rectifier which is in the attachment.

I am asked to work out the average value of voltage across the 100 ohm load resistor.
the power dissipated by the resistor
the minimum PIV rating of the diode

1. first off i'm told that the voltage of the primary coil is 120v

i'm then told that the coil ratio NP:NS = 4:1 where NS is half of the secondary coil. giving 30v each on half of the coil. Am i right in thinking that this is 30v from the positive and 30 from the negative AC in but because it goes through the Load resistor in the same dirction it gives a voltage of 30v minus the 2 diode voltage drops of 1.4 ( assuming they are silicone) giving a Vout of 28.6????

2.okay so then i go on to work out the current through the resistor which i multiply by the voltage to give the power.

3. okay the average voltage is 28.6v so the peak voltage is root 2 * 28.6 = 40.45 V + the peak secondary voltage = root2 *30 = 42.4 v okay so i add them together and half them to give 40.45 + 42.4 =82.9 /2 = 41.4 V

does this seem right? tbh i dnt really have any idea what i'm doing. any help is much appreciated. Thanks, Lee.

#### t_n_k

Joined Mar 6, 2009
5,455
The current only flows through one diode per half AC cycle.

Presumably the primary voltage of 120V is the RMS value so you need to convert the secondary voltages to peak values to find the diode PIV.

#### Ledwardz

Joined Oct 31, 2010
37
The current only flows through one diode per half AC cycle.

Presumably the primary voltage of 120V is the RMS value so you need to convert the secondary voltages to peak values to find the diode PIV.
thanks, so this is what ive done by multiplying it by √2 ?

#### t_n_k

Joined Mar 6, 2009
5,455

The total applied peak voltage around the loop comprising the full transformer secondary and the two diodes is 60*√2=84.85V. At the AC cycle peak one diode is conducting & the other (withstanding the PIV) isn't. Subtracting the 0.7V forward voltage drop of the conducting diode from 84.85V peak gives a PIV of 84.15V.

#### Bob Scott

Joined Apr 13, 2009
3
The turns ratio is 4:1 so the total secondary is 30VAC, or 15-0-15.

The peak reverse voltage for the rectifiers is 30 * 1.414 = 42.2VDC.

The output is not filtered, so the RMS voltage into the load is 15V.

I don't think they would take into account the diode forward voltage drop. This loss is insignificant in the real worls. Your 120VAC mains varies more than that.

Normally the secondary turns are for the entire secondary winding and expressed as center tapped if one exists. eg: 30VCT

#### t_n_k

Joined Mar 6, 2009
5,455

"i'm then told that the coil ratio NP:NS = 4:1 where NS is half of the secondary coil. giving 30v each on half of the coil."

#### JoeJester

Joined Apr 26, 2005
4,266
i'm then told that the coil ratio NP:NS = 4:1 where NS is half of the secondary coil. giving 30v each on half of the coil.
Are you sure your professor said there was 30 volts on each half of the secondary?

If this is an exercise from a book, scan the question and post it.

The only way you can get 30V on each half of a center tap transformer with a 4:1 ratio is to have dual primaries and dual secondaries. Then the primaries can be wired in parallel and the secondaries in series. Then you would have 30-0-30

The OP needs to clarify the question ... by scanning in the exact question.

Last edited:

#### Ledwardz

Joined Oct 31, 2010
37
okay the coil defnitley has Ns taking up half of the secondary coil i assumed that the output would be 30 - 0.7 (i c why we only minus one diode coz it only goes through one at a time now thanks for tht) does it not take into account the minus side of the AC wave

so

30 0 -30 ???? but it all goes throuh the load the same way given 30 out minus the diode drop?

when i get home ill upload a picture

cheers again, Lee.

#### JoeJester

Joined Apr 26, 2005
4,266
The one condition I left out is single primary and dual secondaries. Under that condition, you could have the 30 - 0 - 30 outputs.

Attached is the illustration.

#### Ledwardz

Joined Oct 31, 2010
37
Sorry for the slow reply here is the question i think its the second one down tht u posted. so mayb 2 coils in the secondary....

#### JoeJester

Joined Apr 26, 2005
4,266
You missed what was to the right of Np ...

That is a center tapped transformer. I would take the 4:1 turns ratio to be over the whole secondary, not each half of the taps.

#### djsfantasi

Joined Apr 11, 2010
6,532
I think the 4:1 ratio applies to one half of the secondary...
i'm then told that the coil ratio NP:NS = 4:1 where NS is half of the secondary coil

#### JoeJester

Joined Apr 26, 2005
4,266
So it's a 4:2 or 2:1 ratio.

The point I'm making is there is no reference for using half a coil on centertaped transformers.

Everyone I've ever seen referenced the whole coil in the secondary side when centertapped.

Of course I shouldn't worry about it as this was a classroom assignment and doesn't mean nothing till the student buys a transformer.

Joined Dec 26, 2010
2,148
The OP should state the question exactly as it was set, and answer it in those terms in order to get credit for it. The student will not get far with his instructor by protesting that commercial transformer ratios are customarily described differently.

#### JoeJester

Joined Apr 26, 2005
4,266

I agree. Answer to the standard set by the professor. Wait till they try to simulate it and then they can discuss with the instructor the error of the turns ratio

#### The Electrician

Joined Oct 9, 2007
2,778

I agree. Answer to the standard set by the professor. Wait till they try to simulate it and then they can discuss with the instructor the error of the turns ratio
Commercial transformers are not described in terms of turns ratio; see: