Secondary half wave rectification

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191017-1041 EDT

mbrooke:

Run an experiment.

Since there is no primary DC current, only secondary, you can use an electronic RMS current meter on the primary. Load the secondary with the sing;e diode, and raise the secondary load until the primary RMS current reaches the transformer rating. Do this at maximum expected primary voltage to pick the worst case.

After this determination you could do a transformer temperature rise measurement at this loading.

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gar said:
191017-1041 EDT

mbrooke:

Run an experiment.

Since there is no primary DC current, only secondary, you can use an electronic RMS current meter on the primary. Load the secondary with the sing;e diode, and raise the secondary load until the primary RMS current reaches the transformer rating. Do this at maximum expected primary voltage to pick the worst case.

After this determination you could do a transformer temperature rise measurement at this loading.

.
Click to expand...

Sure, but first you must formulate a hypothesis before hitting the lab :p
 
steve66 said:
It directly uses the half wave rectified output.

I'm not sure how the flux people explain that.
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In an ideal transformer, ampere turns on the secondary are directly balanced by ampere turns on the primary, so that the core flux is not changed by changes in loading.

Think about it: put a perfectly balanced AC load on the secondary. The reflected primary current might be many times the transformer magnetizing current, but the peak core flux does not _increase_ because of the increased ampere turns.

So in an ideal transformer with no impedance in the primary circuit feeding the transformer, a rectifier and pulsed DC output on the _secondary_ would not cause any net DC core flux.

Real transformers have resistance in the coils and are fed with circuits that have impedance, so a rectifier load will be reflected back to the primary and cause a DC bias of the primary, and thus net DC core flux.

Back to the OP: my hypothesis is that the bulk of the increased transformer heating will not be caused by DC bias, but rather by the poor crest factor of the rectified and filtered load.

-Jon
 
191018-1010 EDT

mbrooke:

You created a hypothesis in your first post. Design some experiments and go play.

Have you run any experiments?

rlundsrud ask a good question. You night define your circuit more clearly for us to know where you are headed with your original question.

In post #21 I suggested two experiments.

.
 
Besoeker3 said:
I don't see why it would but I've never tried it. What do you want half wave for anyway?
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A correction to this. We made hexaphase rectifiers for high current rectifiers, typically 10kA. Each secondary winding was effectively half wave with just one semiconductor (SCR) in each leg. I don't recall that causing problems with winding saturation.
 
191018-1319 EDT

Besoeker3:

I do not believe you have a half wave rectifier. Rather I would guess it is really a full wave center tapped rectifier, which balances the DC components.

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gar said:
191018-1319 EDT

Besoeker3:

I do not believe you have a half wave rectifier. Rather I would guess it is really a full wave center tapped rectifier, which balances the DC components.

.
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Each of the six secondary windings carries DC.
 
GoldDigger said:
But does each winding share a common core with windings of a different phasing? That would allow for a net 0 flux (ampere turn) integral overtime in the core.

Sent from my XT1585 using Tapatalk
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The primary. The set up had four 10kA rectifiers feeding a 40kA output. There was a 15 degree phase shift between primaries to cancel harmonics.
 
191018-1328 EDT

mbrooke:

Since you don't seem to be able to run an experiment I did the following quick crude, not well controlled, test,

Circuit: Signal Transformer A41-175-24 transformer, roughly a 120 V primary, 24 V secondary, calculated ratio 5 to 1, 175 VA rating, on secondary a large enough diode as a half wave rectifier, in series with a 2 ohm Ohmite 50 W resistor. The transformer primary is fed from a Variac, and I used the dial calibration for voltage. Primary AC current read with a true RMS DVM. Primary waveform observed on a scope. Secondary DC current measured with a Fluke 27.

Results:

No load primary input --- 122.8 V, 0.25 A, 4.6 W, 29 VA, 0.16 PF.

24 V secondary load --- diode in series with 2 ohm power resistor. Note that the Variac output is somewhat greater than the dial reading.

AC V --- Primary Input --- DC Output

30 V --- 0.43 A, 15 VA --- 1.1 A, 2.2 W
50 V --- 0.85 A, 43 VA --- 1.9 A, 7.2 W
70 V --- 1.36 A, 95 VA --- 2.8 A, 15.7 W
90 V --- 2.00 A, 180 VA -- 3.7 A, 37.4 W
110 V -- 2.60 A, 285 VA -- 4.5 A, 40.5 W

On the primary waveform there was a good replication of the half wave current pulse on its half of the cycle. On the other half of the cycle there was a large peaking of the current toward the zero crossing. Although there has to be an equal area under each half cycle on the primary side, no DC on primary side, I did not have a way to measure the areas, but an eyeball estimate would verify this.

Note: I did not go to to full line voltage. I do not have a large enough assortment of resistors in this range to perform a variable load test at full line voltage. As viewed at the transformer primary I was overloading the transformer before full rated voltage, and less than 1/4 of the VA rating.

.
 
191018-2024 EDT

In rereading my last post my last statement --- "I was overloading the transformer before full rated voltage, and less than 1/4 of the VA rating"

This means at both 90 and 110 V input I was exceeding the transformer input current rating which is 175/120 = 1.46 A, and that the DC output power was less than 1/4 of the transformer VA rating when at the 110 V input.

.
 
191018-2137 EDT

Ran another experiment, more like what mbrooke probably wants. Found some other resistors.

AC V --- Primary Input --- DC Output

120 V --- 0.62 A, 74 VA --- 1.04 A, 10.8 W --- 10 ohms load
120 V --- 1.40 A, 169 VA -- 2.47 A, 24.4 W --- 4 ohms load

At about full primary load current rating, 175/120 = 1.459 A, the DC output power is 24.4/175 = 14 % of transformer VA rating.


Next is no diode and just the 10 ohm resistive load.

120 V --- 0.65 A, 76.7 VA, 73.3 W, 0.96 PF --- 2.56 A AC, 67.9 W calculated, 26.5 V measured.


I did not have the power rating for resistors to more heavily load the no diode case. You can extrapolate from this to full primary current and get 152 W output.

The results show a huge difference as a result of the load DC current compared to AC loading.

mbrooke go run some of your own experiments and report back.

.
 
191019-1047 EDT

mbrooke:

Pictures would provide no useful information. My setup for the experiment is a totally scrambled mess.

Basically I used a 7.5 A Variac (really a Powerstat) to provide input voltage. A true RMS current meter in series with the Variac output to the transformer primary. Also when feasible monitored power input to the under test transformer with a Kill-A-Watt-EZ V-I-P-VA-Hz-PF meter. Only good for 120 V and up to 15 A.

On the secondary side was a series circuit of a DC ammeter, diode, and various load power resistors. Used the marked value on the Ohmite resistors for the resistance value.

This was not a high accuracy measurement. Not needed for the kind of variations that exist.

I don't believe you have mentioned the details of your application, or why you have to use a half wave rectifier.

You list yourself as an Electrical Engineer. Thus, you should be able to easily draw a schematic for the test circuit from what I have described so you can visually see the circuit.

It would be useful if you indicated the transformer input and output voltages, I believe these are 480 and 120. What is the VA rating? Why the half wave rectifier? What is the rectifier load? What other loads on the transformer?

.
 
I might have missed this in the answers? A Transformer is a Alternating voltage component on the primary it would get hot even under no load. Half wave voltage would be pulsating DC.
It needs a full AC voltage to work on the primary side or the input side. The Transformer would get hot with half wave and if it was filtered it would burn the Transformer out.

Ronaldt.
 
191020-2106 EDT

ronaldrc:

In the real world a transformer takes a changing voltage or current signal into one coil (primary), and via the changing magnetic flux field from said one coil coupling to another coil (secondary) within the first coil's field induces a voltage or current into the another coil. The another coil must have a load for any secondary current to flow.

Apply a changing signal to any coil, and a current will flow, and a changing magnetic field will be produced around that coil. Because a coil has resistance the current will produce heat in the coil. Heat is also produced even if the input to the coil is DC. The heating of the coil will result from the current flow.

For pure DC applied to the coil the current is defined by the coil resistance and applied voltage. Transformers are wound with relatively low resistance coils.

When you look at a coil with an AC component applied, and no DC component, then the current that flows is a function of both the DC (really AC) resistance, and the inductance of the coil. This current will be less than the DC current where the AC current measurement is an RMS value.

Next couple the coil to a high permeability magnetic circuit, and the inductance greatly increases. This means the AC impedance greatly increases, and the current flow becomes quite small. This is the condition of a transformer with no secondary load.

All known high permeability magnetic materials have what is known as a hysteresis curve. As the flux level gets high the permeability drops which means the inductance drops, and thus the instantaneous impedance drops. This is called the saturation region. There is a great difference between different materials in how sharp is this saturation curve. Old transformer magnetic materials were relatively soft, and newer transformers may be sharper. Some materials are very square (sharp), expensive, and usually not used for most power applications.

The general design criteria for power transformers is to design for slight entry into the saturation region on each half cycle, opposite flux levels. This produces a peaking in the excitation current near the voltage zero crossing.

If you look at an unloaded transformer primary current with a scope, and gradually increase the excitation voltage, then as you get around the rated voltage for the primary you will see a rapid increase in this current peaking as you go beyond the rated voltage.

If you add a DC current component to your AC input, then you bias the hysteresis curve in one direction, and cause a peaking on 1/2 cycle at a lower AC input voltage than with a balanced AC signal.

Put some other coil on this magnetic circuit, and apply a DC current to that coil, and the result can be the same on the input current as if the DC current was on the primary.

.
 
gar said:
191020-2106 EDT



If you look at an unloaded transformer primary current with a scope, and gradually increase the excitation voltage, then as you get around the rated voltage for the primary you will see a rapid increase in this current peaking as you go beyond the rated voltage.



.
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In over simplified terms, is this because the core can no longer hold any more magnetic field, and as a result it can't push back (counter) the extra field from the winding? Is that extra field in the winding, or lack there of in the core, proportional to the current?

Do the EMFs around a transformer go up during saturation? Is it from the core or wingdings?
 
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