I have trouble with transformers probably because I really don't understand how they work yet. I bought a center-tapped transformer with 12.6V, 450mA secondary out. I used a bridge rectifier and taped off the center-tap on the transformer. I then put a smoothing capacitor. I calculated the results I should be getting on my schematic but I got different results on my meter. Does anything seem wrong?

-Peter

 

You might get some guidance from our Ebook - http://www.allaboutcircuits.com/vol_3/chpt_3/4.html

Pay attention to the "dual polarity center tapped" rectifier. Notice that substituting filter capacitors for the load resistors is what is normally done before feeding voltage regulators.

 

You might find the voltage drops a bit under load. I imagine you connected an ammeter straight across the output of the rectifier. It's not a very good idea, for a while the output will supply more than it's rated current, then overheat and die.
Much better to find a load that should use around 400ma then measure the current in series with that load.

 

Markd77- I didn't use the ammeter across the output of the rectifier. I took a 180k ohms resistor and then I measured the current in series to ground. But for some reason I got 0 current when I did that...I will try again.

 

If you measure current in series with a 180k resistor it will be tiny - around 0.007mA.

 

The transformer is rated for 12.6VAC when it has a 450mA load on it. R=E/I, so 12.6/0.45A = 28 Ohm load.

If there is no load, it will have a higher output voltage.

 

The transformer is rated for 12.6VAC when it has a 450mA load on it. R=E/I, so 12.6/0.45A = 28 Ohm load.

If there is no load, it will have a higher output voltage.

SgtWookie always a pleasure simple straight forward information and now its working fine. Now if only my 33 ohm resistor could absorb the power being driven into it.

 

The center tap on a transformer is a great way to make an economical ± power supply. Some circuits, like op amps, it is really handy to have.