Hello. I have a question about load current in a rectifier.
I've thought about it a lot, but it just isn't clear to me. When you are trying to figure out the current through the load, in a center tapped full-wave rectifier, do you use the full secondary voltage in the calculation, or do you use just half of it? Also, how do the diode voltage drops figure into this?(silicon diodes with 0.7V barrier potential)

 

A center-tapped full wave rectifier will have only one diode drop, while a non-center tapped full wave rectifier will have two diodes drop.

http://www.allaboutcircuits.com/vol_3/chpt_3/4.html

Do you understand why this is? The current in a non-center tapped FW rectifier must travel through two dioes to get back to return to the transformer. A center-tapped transformer lets the current bypass the other diode and take the center connection to return to the transformer. ANd of course, a center-tapped transformer needs less diodes also...but you need a center-tapped transformer.

You use the secondary voltage that appears *across the load* to calculate the current across the load. NOw that I have said this blatantly obvious thing that might not have been so obvious before I said it, are things more obvious? The load is effectively always connected between the center-tap and one end of the transformer (though the specific end alternates due to the diode's rectifying action). THerefore you use the voltage between center and end, or half the secondary voltage.

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Thanks!

Okay I understand this.


Yeah.

Makes sense.

That's what I needed.
let's say the total secondary voltage is 10V across the coil.
*So in a center tapped rectifier,would I do 5V/R(load) to calculate the load current? Or would I take 4.3V/R(load) to calculate the load current?
*And would it be proper to call the result the "forward" current, or is that something else?
*Would the load current be the same as the total current in the whole circuit in this case?
Thanks

 

It's kind of I = V/R, except you will need to account for the diode drops, commutation overlap of finite diode switching time, and the fact the transformer output is not steady (rectified sinusoid if you have no caps, or a DC voltage with ripple if you have capacitors to smooth the voltage).

THe load current is pretty much the same, just be aware it's alternating through the diodes every half cycle.

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Sounds good. Thanks man