Full wave Rectifier Question...

[Projectile]

I'm confused about a specific aspect of how a full wave rectifier works (not a bridge rectifier, but the one that consists of 2 half-wave rectifiers). Specifically, I'm interested in how a reversed biased diode in the secondary circuit is reflected in the primary circuit. Here is a schematic of a full wave rectifier:

http://www.electronics-tutorials.ws/diode/diode18.gif

My confusion lies in the fact that a load in the secondary gets reflected into the primary as described here:

http://www.wisc-online.com/objects/ViewObject.aspx?ID=ACE12607

Now, we take the case of a diode. When a diode is reversed biased, effectively no current flows, making it's resistance close to infinite ohms as long as the voltage level doesn't exceed it's breakdown voltage.

Well, in a full wave rectifier we have a transformer center tapped with a diode in the upper secondary circuit preventing current from flowing through the upper circuit on the negative half cycle of a sinusoid, and we have a bottom secondary circuit with a diode preventing current from flowing in the bottom circuit on a positive half-cycle.

So my question is, why isn't the impedance of the reverse biased diode, which is effectively infinite ohms, reflected in the primary circuit? Why does it not limit the current in the primary?   

I hope that makes sense. I'm really confused on this one and have been scouring the internet looking for answers. Any help would be greatly appreciated. Thanks.

[R.G.]

That's because all secondary loads are reflected into the primary in parallel. So the second diode being reverse biased appears as an open circuit in parallel with the conducting one.

This even makes sense with no conducting secondaries, as they are then infinite impedances in parallel with the primary inductance, so they cannot stop the current going into the primary inductance.

[Projectile]

That's because all secondary loads are reflected into the primary in parallel. So the second diode being reverse biased appears as an open circuit in parallel with the conducting one.

Thanks R.G! That's exactly the answer I was looking for.

I posted this same question on several different message boards and got about 20 different in-depth responses back, but nobody simply said that reflected loads add in parallel, which is what was causing my confusion.  I was starting to think that I had poorly worded my question or something. I just waded through post after post of people trying to explain to me how diodes work and trying to break it down into a simpler circuit and nobody was answering my question. This was such a breath of fresh air. You solved the entire problem in one sentence.  Thank you!

I now realize that my confusion was in that I was imagining transformers as two air core inductors coupled by proximity, but they're usually not. The fact that they are coupled through a magnetic core means that you will never have 1/2 of the primary decouple in a center tapped situation. It's always coupled. It makes so much sense now. I don't know why I didn't realize that before.

This even makes sense with no conducting secondaries, as they are then infinite impedances in parallel with the primary inductance, so they cannot stop the current going into the primary inductance.

Now I'm confused. I thought that if you only have one secondary, and the number of windings in the primary and secondary are the same, then the load will be directly reflected in series with the primary circuit. So therefore, no AC current will flow through the transformer if the secondary is disconnected. I assumed this was because the transformer essentially just becomes a very large inductor, with a very large reactance, when there is a very large resistance in the secondary. Is this not correct?

[R.G.]

This was such a breath of fresh air. You solved the entire problem in one sentence.  Thank you!

Thanks! Sometimes I get lucky. Lots of times I just blather on about stuff too. 

I now realize that my confusion was in that I was imagining transformers as two air core inductors coupled by proximity, but they're usually not. The fact that they are coupled through a magnetic core means that you will never have 1/2 of the primary decouple in a center tapped situation. It's always coupled. It makes so much sense now. I don't know why I didn't realize that before.

It actually goes a little deeper. Two air core inductors are coupled by proximity, and that does constitute a transformer if you want the coupling or crosstalk if you don't want it. Same thing. Two ANY core inductors where the flux from the driven inductor (that is, the primary, where power comes in)  gets into the non-driven core (the secondary) constitute a transformer.

It's just that unless the coefficient of coupling is nearly unity, the amount of power you can transfer is severely limited. The magnetic field from the driven/primary coil that does not couple through the secondary coil looks like an inductance in series with the primary. This is leakage inductance - literally, the magnetic field that leaks outside the secondary. Power and signal transformers do everything anybody can think of to get 100% of the flux from the driven/primary coil confined into the secondary coil. That's the whole point of iron cores and concentric windings - force the flux from the primary to flow inside the hole in the middle of the secondary coil. The iron sucks it in by being a better magnetic field "conductor" than free space, and the intermixed concentric windings try to make the hole in the middle of each coil be literally the same volume of three-space.

Now I'm confused. I thought that if you only have one secondary, and the number of windings in the primary and secondary are the same, then the load will be directly reflected in series with the primary circuit. So therefore, no AC current will flow through the transformer if the secondary is disconnected. I assumed this was because the transformer essentially just becomes a very large inductor, with a very large reactance, when there is a very large resistance in the secondary. Is this not correct?

No, and it's the same misconception that got you on the diodes and rectifiers. The way to think of this is that the primary always has a big inductance across it - this is the magnetizing inductance that lets the magnetizing current flow, and charges the core up with magnetic field so it all works. The secondary loads always appear in parallel with the primary inductance, not in series.

With no load, the (ideally very small) magnetizing current flows, and that's all there is. If you load a secondary, the reflected secondary load appears in parallel with the primary inductance.

Example: A transformer with 120Vac input, 31.8H of primary inductance, a turns ratio of 1:1 is loaded with 120 ohms of resistance which has a switch in series with it. The switch is initially open. What current flows?

The impedance of the primary inductance is Xl = 2*pi*F*L = 2*3.14159*60*31.8 = 11988 ohms. The current that flows is I = 120Vac/11988 = 0.010009753... OK, call it 10milliamperes. (and I obviously made up neatly fitting numbers  )

What is the current when we close the switch?

Neglecting resistive losses and other imperfections, the voltage on the secondary is 120Vac as well. The current is 120Vac/120 ohms, or 1A.

The current in the primary is the secondary 1A transformed by the 1:1 ratio, so that's 1A too; and the magnetizing current still flows, so the total primary current is now 1.00A + 0.01A = 1.01A.

If we change the secondary resistor to 240 ohms, the primary current drops to 0.510A, with half an amp of transformed secondary current plus the still-active 10ma of magnetizing current.

This is a dramatically simplified example, because the primary inductance is not really constant, and because there are eddy current losses in the core that are best modeled as a nonlinear resistance that's always there, and there are winding resistance losses as well as leakage inductance losses, and... well, you see where that goes.

But to understand the basics, the magnetizing current is always there. The reflected secondary loads are in parallel with it. You can't stop primary magnetizing current from the secondary.

[Projectile]

That makes sense.

When I said "no current", I meant in more of an simplified first approximation where the magnetizing inductance and other losses are ignored for convenience. Your explanation makes it a lot clearer though. It's really hard to find good study material on this stuff. The text I am using as a reference explains most of the elemental stuff okay, but then just glosses over the whole part about multiple loads, and instead just gives a whole bunch of complicated equations. I can understand and manipulate the equations alright, but it's really hard to connect them back to the circuit I'm looking at and understand how they relate to electrons moving in the wire. I really wish I could find source material that describes this stuff better, but most of the material I find is either too simple or overly complex. There's not much in between.

Thanks for taking the time to explain things. I've read a lot of your posts on these forums and they always help a ton. If you ever wonder if your wasting your time rambing on about this stuff on internet forums, I guarantee that you're not. I'm sure there are plenty of other guys like me that lurk through back posts on this forum and find these in-depth explanations golden. Thanks again.

[R.G.]

You sound like you're on a knowledge quest. There is a way to look at transformers that even most EEs that work with transformers either don't understand or remember.

Left to itself, the primary inductance and primary voltage allow a current to flow, this being the magnetizing current. The magnetizing current causes a changing magnetic field to flow in the core.

The changing field in the core causes the turns of the primary winding to have a voltage induced in them too. This reverse voltage is in opposition to the voltage driving the primary. The resulting total current in the primary and hence the magnetization of the core, is actually the result of this difference between driving voltage and the reverse EMF, as it's called. The core magnetization is a result of the balance of these two. It's a simple feedback/offset mechanism.

If you now pull energy out of the core by connecting a load to a secondary, this pulls energy out of the core field. That reduces the back EMF and lets a matching increase of primary current flow until the primary current offsets the losses to the secondary and brings the core field back into balance. It instantly rebalances itself. The core conditions remain essentially constant (ignoring losses again) while energy flows through the magnetic coupling but do not affect the field that remains in the core to any great extent. It's very much like the core field is holding off current flow from the primary voltage source by means of the back EMF. Secondary current reduces the back EMF and lets current flow to just balance what the secondary takes out, so the core conditions remain almost constant.

This is also why you can't saturate a power-line AC-only transformer from the secondary. Saturation is a total volt-time issue. As long as the primary volt-time integral remains limited to the values provided by the incoming AC-only voltage and back EMF, the core does not saturate. It may overheat if you pull so much current through the transformer that the ohmic heating in the winding resistances generate heat, but it can't saturate. Half wave rectifiers would be severely limited if this were not true - they'd always be saturating the core. Instead, they work only on "flow-through" energy, not affecting the core to any huge extent.

[PRR]

> got about 20 different in-depth responses back

Lots of people can post "answers".

Not as many answerers can READ the question.

R.G. does. He may blather-on, but the core question gets answered.


Here's a different take.

The CT winding is, of course, just a convenience and a penny-saver.

You could wire the 2-diode rectifier with a dual-winding transformer.

You could wire the 2-diode rectifier with two *separate* transformers.

What is the half-wave situation for the transformer with the reverse (open-circuit) diode?

Essentially the same as a transformer powered-up but with nothing connected to the leads.

This is very common. Doorbell transformers run open-load 99.9% of the time (after the silly light-bulb in the switch dies). Your utility transformer at the street may run 110% of load for an hour at Thanksgiving Supper (electric stove), at 1% of load when lights and TV are off and it's just clocks and vampire taps, and is happy to sit at zero load if you don't pay the bill and they disconnect you.

That's just a different illustration of the "parallel" concept. There's good economic reasons we don't ever do a full-wave rig with two half-wave loaded transformers. Averaged over the full wave that's more iron and steel than we need to get the job done. But leave that for post-grad or on-the-job learning.

> magnetizing inductance and other losses are ignored for convenience

Right. You ignore that until ideal transformers seem simple.

In practice, it may matter. In small transformers there's good economic reason to tolerate quite low inductance and large inductive energy. "Most" of that energy returns to the power company and (on residential rates) you may not pay. But there's copper loss through the winding, some of the imaginary energy becomes real (and billable) heat. Very small transformers sometimes run hot at no load. Utility companies have LOTS of transformers and while they have low idle loss, it adds up. When there's real money in the deal, you balance transformer cost against idle loss and capital costs. But when you buy a $5.98 wall-wart, you have no say in idle losses, and first-cost demands as high a loss as won't smoke in 30 days.