Secondary half wave rectification

mbrooke

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Can secondary half way rectification over heat an industrial control transformer? And is so what conservative multiplier is typically used for de-rating? 480:120
 

Besoeker3

Senior Member
Location
UK
Can secondary half way rectification over heat an industrial control transformer? And is so what conservative multiplier is typically used for de-rating? 480:120
I don't see why it would but I've never tried it. What do you want half wave for anyway?
 

gar

Senior Member
191016-0815 EDT

mbrooke:

Yes. Don't run much DC in either the primary or secondary. This is why you must use special phase shift dimmers into a transformer.

Unbalanced DC thru any coil on a transformer unbalances the hysteresis curve increasing saturation in one direction, and increasing magnetizing peak current in one direction. Thus, increasing RMS current. A full wave center tapped rectifier balances the two half wave rectified waveform DC components.

Run experiments. You need an RMS meter that also simultaneously measures the DC component along with the AC component.

Do temperature rise experiments.

.
 
Last edited:

mbrooke

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191016-0815 EDT

mbrooke:

Yes. Don't run much DC in either the primary or secondary. This is why you must use special phase shift dimmers into a transformer.

Unbalanced DC thru any coil on a transformer unbalances the hysteresis curve increasing saturation in one direction, and increasing magnetizing peak current in one direction. Thus, increasing RMS current. A full wave center tapped rectifier balances the two half wave rectified waveform DC components.

Run experiments. You need an RMS meter that also simultaneously measures the DC component along with the AC component.

Do temperature rise experiments.

.
This is what I was thinking.

I know when the diode is on the primary the thing gets HOT (had to unplug it)- tried that experiment long time ago. But I'd imagine the secondary would not be as bad... right?
 

gar

Senior Member
191016-1021 EDT

mbrooke:

Generally I would judge just about as bad, possibly worse. It will depend on what the primary and secondary currents are, and how all circuit elements balance out.

I think if the ampere-turns of DC are the same, that secondary loading maybe worse. The DC ampere-turns produces the flux unbalance, and if both primary and secondary have current flow, then power is dissipated in two places.

.
 

Besoeker3

Senior Member
Location
UK
This is what I was thinking.

I know when the diode is on the primary the thing gets HOT (had to unplug it)- tried that experiment long time ago. But I'd imagine the secondary would not be as bad... right?
On the primary, it will cause saturation. Don't do it.
 

mbrooke

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191016-1021 EDT

mbrooke:

Generally I would judge just about as bad, possibly worse. It will depend on what the primary and secondary currents are, and how all circuit elements balance out.

I think if the ampere-turns of DC are the same, that secondary loading maybe worse. The DC ampere-turns produces the flux unbalance, and if both primary and secondary have current flow, then power is dissipated in two places.

.
50% va

I know things like hair dryers and heater regulate power through half wave rectification. So at some power level it must be fine.
 

GoldDigger

Moderator
Staff member
The key difference is that a diode on the secondary cannot cause saturation in the same way as on the primary because the secondary voltage is dependent on the rate of change of the flux in the core. A load on the secondary can never increase the core flux, just the non-magnetizing current in the primary. You will see a drop in output voltage on the secondary before you see saturation of the core. It is still not a good idea because the DC component will definitely cause additional heating of the windings.
 

gar

Senior Member
191016-1342 EDT

mbrooke:

There is not just some level at which a problem occurs, it is a proportional thing. The greater the DC current the greater is the problem. But how sharply this occurs is a function of the shape of the hysteresis curve. Thus, the core material alloy is important. This is the basis of magnetic amplifiers.

.
 

steve66

Senior Member
One problem with half wave is that all the power that needs to be delivered to the load is taken in very brief pulses that only happen once every cycle. So the peak current through the transformer is much higher than the output current of the rectifier/filter.

The better the filtering (assuming you are using a capacitor across the output), the shorter and larger the spikes are. So a 80 ma current may flow for 2 ms, and that allows the filter to supply a 10 mA load for a full 16 ms. In both cases the energy is the same 80*2 = 16*10.

With no filtering at all, the diode is allowed to conduct for a full half cycle, and the output current also only flows for one half cycle. So 10 mA out is only 10 mA in. That is probably your hair dryer.

An inductor at the input of the filter changes everything, and can get rid of all those nasty harmonics and spikes.
 

gar

Senior Member
191016-1452 EDT

steve66:

What you say is true, but it is only part of the story.

The DC flux bias is probably the most important factor, and is most easily observed with an unloaded secondary and a series diode in the primary.

All flux sources in a closed core add.

.

.
 

winnie

Senior Member
Location
Springfield, MA, USA
Occupation
Electric motor research
GoldDigger's point is key. A half wave rectifier on the _primary_ blocks half of the magnetizing current causing huge saturation problems.

A half wave rectifier on the _secondary_ blocks half of the _load_ current. The magnetizing current is still AC. Thus DC saturation effects must be indirect, eg by increased voltage drop on one half of the primary AC cycle.

-Jon
 

gar

Senior Member
191016-1646 EDT

winnie:

No matter where the DC ampere-turns come from on the transformer, the same number of DC ampere-turns produce the same saturation effect. This assumes a single magnetic core path.

.
 

mbrooke

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One problem with half wave is that all the power that needs to be delivered to the load is taken in very brief pulses that only happen once every cycle. So the peak current through the transformer is much higher than the output current of the rectifier/filter.

The better the filtering (assuming you are using a capacitor across the output), the shorter and larger the spikes are. So a 80 ma current may flow for 2 ms, and that allows the filter to supply a 10 mA load for a full 16 ms. In both cases the energy is the same 80*2 = 16*10.

With no filtering at all, the diode is allowed to conduct for a full half cycle, and the output current also only flows for one half cycle. So 10 mA out is only 10 mA in. That is probably your hair dryer.

An inductor at the input of the filter changes everything, and can get rid of all those nasty harmonics and spikes.
Not 10ma, but literally 15 amps. Hair dryers have a massive diode that will literally rectify everything when set to low.


https://jimogaoshou.en.made-in-china.com/product/gCFQOAGhEWco/China-High-Temperature-Mica-Electric-Hair-Dryer-Heating-Element.html


I mean a 25kv pole pig is obviously fine with it. But what about a 3000 va transformer converter as an example?


I gets what being said here, but some posts downplay others overplay the hysteresis curve when all else is adjusted for. IMHO
 

GoldDigger

Moderator
Staff member
191016-1646 EDT

winnie:

No matter where the DC ampere-turns come from on the transformer, the same number of DC ampere-turns produce the same saturation effect. This assumes a single magnetic core path.

.
Quite true, but please keep in mind that because the magnetizing current comes from only one side, a diode on the primary will not create the same DC ampere turns in the core as a diode on the secondary. And, of course, the magnitude of the diode load current will depend on the load resistance, which we have not really considered so far except to tacitly assume that the transformer will be close to fully loaded.
 

mbrooke

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Quite true, but please keep in mind that because the magnetizing current comes from only one side, a diode on the primary will not create the same DC ampere turns in the core as a diode on the secondary. And, of course, the magnitude of the diode load current will depend on the load resistance, which we have not really considered so far except to tacitly assume that the transformer will be close to fully loaded.
Yup- agree

Now, can we find a way to determine how much single wave rectified load a transformer can take without over heating?
 

steve66

Senior Member
191016-1452 EDT

steve66:

What you say is true, but it is only part of the story.

The DC flux bias is probably the most important factor, and is most easily observed with an unloaded secondary and a series diode in the primary.

All flux sources in a closed core add.
Yes, others had mentioned the flux effect, so I wasn't going to repeat what had already been said.

As far as which is more important, I think that would depend on all the factors including transformer parameters, filtering, and load. As I mentioned, the current spikes get worse with better filtering because the diodes only conduct when the input voltage is higher than the output voltage. Less droop on the output means a shorter conduction time and higher current spikes.

With the right (or should I say wrong?) filtering, the current spikes could easily be 10X, 100X, or even 1000X the output current.

So what happens to your flux when the output current spikes are 10 times the transformers rated current?
 

steve66

Senior Member
Not 10ma, but literally 15 amps. Hair dryers have a massive diode that will literally rectify everything when set to low.


https://jimogaoshou.en.made-in-china.com/product/gCFQOAGhEWco/China-High-Temperature-Mica-Electric-Hair-Dryer-Heating-Element.html


I mean a 25kv pole pig is obviously fine with it. But what about a 3000 va transformer converter as an example?


I gets what being said here, but some posts downplay others overplay the hysteresis curve when all else is adjusted for. IMHO
Yes, this works for millions of hair dryers because there is no filtering of the output with capacitors.

It directly uses the half wave rectified output.

I'm not sure how the flux people explain that.
 
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