# How does this rectifier work?

In a car battery charger, I found a strange rectifier. Can someone explain to me how does it work?

Here I have a transformer which is unmarked. By measuring resistance between its output terminals, I determined that the plus cable is connected to the center of the transformer. The two outer connections both have a diode connected as shown on the image. To me it looks similar to this:

rectifier, but the diodes are turned backwards. I checked their orientation several times with two different multimeters and I'm sure that I've drawn them correctly.

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For this circuit, the orientation of the diodes does not matter, as long as both are the same way round, common cathodes or anodes. - the only difference is the polarity of the output - the bottom circuit - the diodes are positive, and the centre tap is negative. – Wonko The Sane Mar 26 '11 at 9:20

Maybe these illustrations will help:

Let's assume that the start and end of the primary and secondary windings are such that the 'starts' are at the top of the picture and the 'finishes' are at the bottom.

When primary current flows from the top of the picture to the bottom, the top of the primary winding is at higher voltage than the bottom. This will induce a voltage in the secondary winding with the highest potential at the top of the winding and the lowest at at the bottom (and a potential somewhere in between, at the center-tap).

You can tell quickly that D1 will be reverse-biased, because the voltage at the top is higher than at the center-tap. Current will flow however it can, which will be out of the center-tap, through D2 and into the bottom of the winding.

When primary current flows from the bottom of the picture to the top, the reverse condition holds true in the secondary: D2 will be reverse-biased, and current will flow from the center-tap through the load and through D1 back to the winding.

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This drawing is so simple and perfect I don't even need to read the answers to understand what's going on. – Alain Mar 25 '11 at 13:29
@Alain, got my +1. I would agree. I thought the same thing but did not have time or tools to make it. – Kortuk Mar 25 '11 at 14:58
Ms paint is a wonderful tool : ) – Adam Lawrence Mar 26 '11 at 11:29
I just feel lazy now :) – MikeJ-UK Mar 27 '11 at 12:12

By center tapping you have just picked a reference that has two different sides that at any one point will oscillate in reference to it.

When the top side goes negative in reference to the center tap it will conduct from the load. The same applies for the bottom side. The number of turns on each side will determine what voltage peak you can receive. The center tap just makes it like you have 2 transformers each with have of the loops of the larger transformer it could be by ignoring the center tap. With a capacitive load and regulator this is easily turned into a DC voltage.

### Why does diode direction matter?

It is just going to change which of the pins is the negative reference. The graph you loaded in shows the rectified hills which are all positive, because of the diode direction they are all negative. They need a negative reference for something in your car.

I hope this helps. Let me know if there is something in this answer I can expand upon to help.

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I do not understand your answer despite understanding the question and the circuits... ;] – jpc Mar 25 '11 at 12:18
@jpc, I am sorry to hear that. I was trying to focus on the connections and what was happening. I could have done the full wave rectifier approach, this just seemed like what the OP may be missunderstanding. – Kortuk Mar 25 '11 at 12:54
OTOH maybe it is just me since you get only positive votes for this. :) – jpc Mar 25 '11 at 13:15

It's the same, although I've never seen a full-wave rectifier this way round before. Electriclly it makes no difference which way it's done, but it seems more 'natural' to connect the centre-tap of the transformer to the circuit's reference rail - which is usually the negative.

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he did an edit and we both answered in a 1.5 minute time span. – Kortuk Mar 25 '11 at 11:15
@Kortuk - These answers are like London buses. You wait 20 minutes and three turn-up together :o} – MikeJ-UK Mar 25 '11 at 11:59
haha. Mainly on the common routes though. – Kortuk Mar 25 '11 at 12:03

The diodes conduct on alternate half cycles (the two ends of the secondary winding give a phase reversal), so that the circuit behaves like a full-wave rectifier with four diodes. It's a popular circuit, because it's cheaper to manufacture than one using a single winding secondary with a four diode bridge rectifier.

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and you were 30 seconds after Mike. I need to just give others a chance on these questions. – Kortuk Mar 25 '11 at 11:26
Cheaper and much more efficient at low voltages since you have only 1 diode drop instead of 2. – jpc Mar 25 '11 at 12:17
@jpc, but you do "lose" half your number of loops because you center tap. There is a tradeoff there. – Kortuk Mar 26 '11 at 2:24

From a practical point of view this half bridge arrangement is exellent if you do not want the trouble to isolate the nut of a power diode. This speeds up assembly and cools the diode more effectively because of the metal-to-metal thermal coupling with nothing in between. Downside is that a ground referenced positive supply needs reverse direction diodes. A negative ground referenced supply can however use standard direction diodes where the diode nut is the cathode.

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One thing I haven't seen in the other answers:

This is a centre-tapped rectifier. It uses a centre-tapped transformer to create a full-wave rectifier.

Its biggest positive is that you only need 2 rectifiers to do this. Because of this, these were used when solid state rectifiers were still very pricey or you had to use valve rectifiers. Its not used often today because the cost of creating two secondary windings in a transformer is much higher than adding 2 more rectifiers to create a full-wave rectified signal from a simple transformer.

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