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Center Tap Transformer

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Member level 3
Jun 30, 2008
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Hello All,

I buy a center tap transformer, it has a sticker attached where it is written:
6V 0.5A and a part number. I have to question:

1- 6V is the rms voltage, right? and it means that the transformer is 6V-0V-6V, so if I only use the 2 terminal of the secondary (without the center tap) I will get 12V rms,right?

2- 0.5A: does this mean that I can get 1A load current in a rectifier circuit where I use the center tap transformer with two diode connecter to the two terminals and connect the center tap to ground?

Thanks in advance,


If you double specified winding current, magnectic flow may saturate, and overheat transformer.
( and voltage ratio also not works properly like expected ).


The answer to your first question is positive, voltages add up (6+6=12) ..
The answer to your second question is negative, current stays as per transformer's rating (0.5A) as each of the halves of the CT winding conducts current only over half of the cycle ..


i think if you use both 6 volt and the center tap ,at a time you get only one half cycle so the output is 6volt.
Also the transformer only produce 0.5 ampere ,how it produce 1 ampere.

I like to add a comparison of the transformer dissipation due to current in two popular configurations as rectifiers.

The given data:
To simplify the analysis, the smoothing capacitor is removed, and the rectifier load will be just a resistor.
The voltage drop of the forward diode is neglested (assumed 0V).
The transformer input is 0-Vp and Rp is its internal primary resistance.
The transformer output is Vs-0-Vs and Rs1 is the secondary resistance of one coil.
N= Vs/Vp for one coil, therefore 2N is voltage ratio if the two coils are used.
Each coil sees Rp as Rsp1=Rp*N^2
The two coils in series sees Rp as Rsp2 = Rp*(2N)^2 = 4*Rp*N^2 = 4Rsp1
R (of the load) >> the internal resistances of the transformer

(1) Full wave using 2 diodes and the center tap:
Let us assume the load Ro1 will require the current Is1 , that is Is1 = Vs/Ro1
One half-cycle:
Power dissipation of one coil = (Is1^2)*(Rs1+Rsp1)/2 (/2 because it occurs in one half cycle only)
For the two cycles:
P_dis1 = (Is1^2)*(Rs1+Rsp1)

(2) Full wave using 4 diodes over the two secondary coils in series:
Let us assume the load Ro2 will require the current Is2 , that is Is2 = 2*Vs/Ro2
P_dis2 (of the two coils) = (Is2^2)*(2*Rs1+4*Rsp1)

For a typical transformer of this type Rsp1=Rs1 therefore:
P_dis1 = (Is1^2)*(2*Rs1)
P_dis2 = (Is2^2)*(6*Rs1)

So if we let P_dis1=P_dis2
Is1 = Is2 * √3

Is1 = Is2 * 1.732
Is2 = Is1 / 1.732

This result gives a rough idea that the maximum current in the second case has to be lower than of the first one to allow the same copper dissipation in the transformer.

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