# DC amps on the output of a regulated PSU

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#### akis

I should be building and trying this but it's too cold in my workshop at the moment so I am just simulating it.

Supposing we have a transformer and we want to find out what is the maximum DC current we can get regulated.

My (elementary) theory says this:

The power of the transformer's secondary will be expended on the bridge, regulator pass transistors and on the actual load. The DC current available will be the VA of the secondary divided by the peak rectified voltage (this voltage will be then divided on the regulator and the load).

So for a 150VA/25V/6A transformer we can get a maximum DC current of 150 / (25*sqrt(2)-2) = 4.5 A DC according to the theory above.

My simulator is less optimistic depending on the size of the filter capacitor : I get something like 1.8A to 2.2 A out of the 6A RMS of the secondary.

My theory has not accounted for the filter capacitor which influences the results negatively - although the bigger it is the smoother the output.

The above suggests that with a secondary of 25V/6A I am looking at about 1.8A-2.2A at most out of a regulated supply with a filter cap of 4.7mF to 30mF.

Have I done this right? It sounds so wrong to me.

#### akis

Ahh the current provided by the secondary winding is not a sinusoidal so you cannot simply measure it with an RMS multimeter. This is what it looks like (attached). The real energy spent is the area under the curve.

I have done some basic measurements and we can chop up the area under the curve bits into sections to fit next to the square section. This makes the square section wider but it still leaves a gap until the next square section. I have empirically measured this to be 72.2% "on" and the rest "off".

The "RMS" current is 72.2% of the level of the square section which works out as 6A RMS for a 4.94A DC output.

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#### AndrewT

you have omitted the manufacturer's de-rating when using a transformer feeding a capacitor input filter.

Each manufacturer will show how to de-rate, but most seem to say that 70% of rating is available from the capacitor output filter.

The reason for this is the heating effect of I^2R. For the short term high pulse currents that flow to charge up the capacitor the heating effect in the transformer windings is much greater than if sinusoidal current were flowing.

Combining your rectified rating factor with the capacitor input de-rating you find that the continuous DC current available from a transformer is ~50% of the AC current rating.

If you use a 2A rated transformer and draw a continuous 1Adc from it then it will heat up to roughly the same extent as drawing a continuous 2Aac from it. Most transformers are VA rated for maximum temperature operation. This means that for a temperature limited VA rated transformer it will run hot using the 50%*AC rating.
Many, myself included, suggest that for cool running and long life and reliability of the transformer, that your continuous DC current be limited to 1/4 of the AC current rating.

If you care to search you will find I have discussed this at least a dozen times on this Forum. Many others also discuss this continuous DC rating after a capacitor input filter.

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#### akis

My 72.2% by counting squares on the oscilloscope screen is not very bad then if the manufacturers say 70%?

I am still trying to work it all out in my head, slowly

#### DF96

You can measure it with an RMS meter, provided it really is a true RMS meter. Most meters cheat: they rectify it, measure the average (or the peak), then apply a fiddle factor which happens to be correct for a pure sine wave. So when a DMM says 1V RMS, it means if this is a sine wave then it is 1V RMS but if it is any other waveform I can't help you so I will tell lies instead. True RMS meters are usually expensive.

Think of your charging pulses as being rectangular in shape (not true, but not as daft as it sounds). If the charging pulses occupy 25% of the possible time, then a 1A DC current needs a 4A pulse. Heating effect, which is the main limit in a mains transformer, goes as I-squared. 4^2=16, but this only happens for 25% of the time so our 1A DC is equivalent to 4A RMS.

#### sawreyrw

akis,

The current rating of a transformer is basically determined by the power loss in the windings. The power loss in the secondary is Irms^2*R. Now the RMS current will be higher than the average current. You need to determine the RMS current for various loads and filter cap values. LTspice will numerically do this for you, but I don't know if your simulator will or not. Note that you must use an integer number of cycle for this to work correctly, and all transient must have died out.

BTW, your current traaes look strange. The triangle like part is ok, but what is causing the square portion of the waveform?

Rick

#### megajocke

If you want to simulate it and get meaningful results you will have to include the effect of transformer winding resistances.

A simulation without it, where the transformer is modeled as a perfect voltage source, will give you overly pessimistic transformer RMS current and overly optimistic output voltage under load.

For a 10% regulation transformer with low leakage inductance the AC RMS current is about 1.8 times larger than the DC current. For a 6% regulation unit, it is about 2.0 times higher.

The above suggests that with a secondary of 25V/6A I am looking at about 1.8A-2.2A at most out of a regulated supply with a filter cap of 4.7mF to 30mF.

I'd expect about 3A to be deliverable and the on-load voltage to be about 30V DC if the transformer is rated 25V at full resistive load. It is because you have omitted the transformer resistance the simulation is pessimistic. The transformer resistance actually dominates the charging cycle and you'll find that the capacitor value does not affect the current significantly!

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