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Center Tapped Transformer Question

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OK, I might be over thinking this but I'm going to ask to check. I'm looking at a center tapped transformer (yes, from All Electronics) that is listed as 12.6VAC center tapped. The photo shows 6.3 - 0 - 6.3. If I connect from one 6.3 to the other 6.3 I'll have 12.6VAC, and if from either 6.3 to 0, then 6.3VAC. I need both the 12.6VAC and 6.3VAC converted to DC with a result of 12VDC and 5VDC. From the 12VDC regulator I will have 12VDC and a 12VDC GND. From the 6.3-0 connections to a 5VDC regulator, I have 5VDC and another GND. Are both the 5VDC GND and the 12VDC GND at the same potential? ie., are they the same ground?

The circuit being used to rectify the voltage is a single diode between the transformer and the voltage regulator (from Velleman kit K2639). It appears to me that the 6.3V terminal on the transformer that is common to both would be the ground. So the transformer would be better labelled 0 - 6.3 - 12.6 in my situation.
 
Kind of hard to tell from your word salad, but if you wire the power supply thus:

**broken link removed**

then the two outputs will have a common ground. (Filter capacitors, etc., have been omitted for simplicity.)

And yes, you're correct, in this configuration it's really a 0-6.3-12.6V transformer. Center-tapped transformers aren't generally used this way, hence the 6.3-0-6.3 designation.
 
I must say you can read "salad" very well. That's what I was trying to describe, the question I was trying to ask, and the answer I was hoping for. I was afraid there was some "trick" to center tapped transformers I was forgetting about.
 
Voltage is relative.
You could use the transformer as normally used. (+6,0,-6) Four diodes and two caps. Then connect the -6 to ground (0 volts). This makes the old -6=0, the old 0 is now at +6 and the old +6 is 12V.
 
I would just place a full wave bridge across the transformer secondary, omit the center tap and filter the bridge out. 12.6 VAC rectified will deliver about 11.2 VDC (12.6 less the forward drop of two diodes conducting on alternate half cycles). Once a good filter capacitor is placed across that the cap will charge to E peak which is 1.414 * 11.2 = 15.8 VDC which is fine for a LM78XX series 12 volt regulator allowing for the dropout voltage. Additionally the 5 volt regulator could be driven.

The downside to using the 12.6 CT transformer as described is you get two DC outputs but each is only half wave rectified to drive the regulator. This is why the Velleman circuit likely used a 1,000 uF cap at the regulator input (C 10). With that design the load needs to be kept very light (low current). Remember in this configuration you are only getting 1/2 wave rectification. The filter cap will only charge on alternate half cycles and the load will discharge it on alternate half cycles. If the load is great enough the circuit won't work.

Ron
 
Here's a neat and simple connection that does what you want with full-wave rectification for both voltages with one bridge.

Connect a full-wave bridge circuit between the two 6.3V windings. Ground the negative node of the bridge. You will now get a full-wave rectified +12.6V from the bridge. Nothing unusual there.

But what's interesting is that there will also appear a full-wave rectified +6.3V at the center-tap of the transformer. The reason is that the two diodes going to ground in the bridge act as a 2-diode full-wave center tap rectifier. It appears a little odd since the rectifiers are in the ground leg instead of in the hot leg as normally seen, but it works just the same.

So with the addition of just two diodes you go from two half-wave rectified voltages to two full-wave rectified voltages.

One consideration is that the two bridge rectifier diodes going to ground will have to carry the total current drawn from both voltages.

Edit: Below is a simulation of the circuit.

Bridge.gif
 
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If you use half wave rectification as in your original post (schematic provided in post #2) you won't be able to get the rated current. The transformer would buzz a lot and overheat if loaded to rated current. The DC in the secondary winding will cause the transformer core to saturate. With a half wave load the current you would be able to draw without overheating the transformer would be less than half the (rectifier) rated current.

If you use the circuit described in post #6 you will be able to get the rated current for rectifier use. Note that the transformer's rated current is probably for a pure resistive load. To get the rating for rectified output, divide by 1.5
 
So what are you proposing, rectifiying each of the 6.3VAC taps and "stacking" the bridge rectifiers by connecting V- of the top one to V+ of the bottom. Sort of a center-tap fulll wave rectifier? I've never tried that. FYI, the transformer is only 1A and I don't expect to need more than 1/2 that (or 1/2A).
 
So what are you proposing, rectifiying each of the 6.3VAC taps and "stacking" the bridge rectifiers by connecting V- of the top one to V+ of the bottom. Sort of a center-tap full wave rectifier? I've never tried that....
Sort of. But it's all done with one bridge. See my edited post circuit simulation.
 
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I would just place a full wave bridge across the transformer secondary, omit the center tap.
The output of the 12.6v AC will be over 19v DC after the bridge, on no load. This will give you plenty of voltage to put a 12v regulator. Once you have the 12v DC output, you can put a 5v regulator on the 12v line, to get 5v.
This arrangement will produce the best regulation and prevent putting "DC on the transformer."
 
The output of the 12.6v AC will be over 19v DC after the bridge, on no load. This will give you plenty of voltage to put a 12v regulator. Once you have the 12v DC output, you can put a 5v regulator on the 12v line, to get 5v.
This arrangement will produce the best regulation and prevent putting "DC on the transformer."
The peak output of a 12.6V AC RMS waveform is 17.7V not including diode drop. Or are you including the normal overvoltage that most power transformers exhibit under no load?

The scheme I proposed generates both voltages which minimizes power dissipation in the linear regulators. It also causes no net DC current from the transformer.
 
Carl, yours is a good design. I did not see your posting before posting mine. There is always a 2v - 6v "over-voltage" provided by small transformers due to the poor regulation of these devices and this has to be taken into account as it adds to the heating of the regulator.
 
That was my first thought. But as both regulators are TO220 78xx series, they have the same power capacity so I can't draw from either at 100%. The 5V supply with be loaded with quite la large "surge" load.
 
Wow. I didn't expect that.

If the transformer is rated to 1A, then I shouldn't be able to draw more than 1A through the bridge, so I only need a bridge for 1A (aim for 2A for safety margin/heat dissipation)
 
Would I be better off with individual diodes or a single bridge rectifier in this design?
Functionally they are the same as long as both are rated for the maximum current.

The main reasons to use a bridge is they're cheaper and there's less wiring.
 
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