Re: 1 KVA 240 v transformer output 6v,8v,12 v AC how to test if it can supply 83 amps?

From: The Phantom (phantom_at_aol.com)
Date: 03/07/05 

Date: Sun, 06 Mar 2005 16:24:24 -0800

On Sun, 06 Mar 2005 12:09:30 -0600, John Fields <jfields@austininstruments.com> wrote:

>On 5 Mar 2005 21:29:04 -0600, The Phantom <phantom@aol.com> wrote:
>
>>On Fri, 04 Mar 2005 10:47:06 +0000, rob <sasman@sasman.com> wrote:
>>
>>>Hi
>>>
>>>I am a total novice in electrics and am not sure were to post this
>>>question?.. i hope some one can help me...I need a transformer that
>>>can supply aprrox 6 Volts Dc at 75 amps..
>>
>>Rob,
>> Do you have any instruments that can measure currents of ~75 amps
>>AC? Of 5 amps AC?
>>
>>Do you have an AC voltmeter?
>>
>>The tests that others have proposed so far require you to dissipate
>>power on the order of .5 kw.
>>
>>The standard method for testing the capability of a transformer
>>without actually dissipating a power equivalent to the rating of the
>>transformer is the short circuit test.
>
>---
>AFAIK, the short test is used to measure the copper losses, not the
>"capability" of a transformer, whatever that means.

  I think, as I explained to Fritz Schlunder, that for small transformers like this, the
copper loss *is* what, to a great extent, determines the (power handling) "capability" of
the transformer.

>---
>
>>You need a variac for this. You dead short the 6 volt output of the
>>transfomrer.
>
>---
>NO!!! If you do you will exceed the current rating of the secondary
>and, possibly, damage the transformer. The transformer is rated for
>1kVA out of the _entire_ secondary which, at 12VRMS out comes to the
>83.3 amps noted on the transformer's faceplate. That is, the
>transformer secondary is wound with wire which is designed to carry
>83.3 amps no matter which voltage tap is used.

   I think you are quite right. I didn't read your first response to Rob where you
interpreted the label, which I also didn't bother to look at. :-( I was thinking that
the secondary had multiple windings which could be paralleled, which, after looking at the
label, I would agree that it doesn't. The 83 amps times 12 volts gives 996 VA, the rating
of the transformer. Mea Culpa. He should limit the primary current to 1.875 amps as I
mention later.

>---
>
>>You connect a variac capable of about 5 amps output to
>>the power line. Turn the output of the variac all the way down to
>>zero output (THIS IS VERY IMPORTANT). Connect the variac output to
>>the 240 volt winding of the transformer with an ammeter in series to
>>monitor the current. SLOWLY turn up the variac until the ammeter
>>reads the rated current for the 240 volt winding (about 4 amps AC in
>>this case). You may need to put a (power) resistor of a few ohms in
>>series with the 240 volt winding of the transformer if the current
>>comes up too fast (which it undoubtedly will) when you SLOWLY turn up
>>the variac. (If you don't have any suitable power resistors, use an
>>electric heater or some light bulbs as resistors). (An even better
>>alternative would be to follow the variac with a small step-down
>>transformer. Something with about 5 amps output at 6 or 12 volts.
>>The variac would power the line-voltage winding and the low voltage
>>winding would power the 240 volt winding of the 1kva transformer you
>>are testing.)
>>
>>When you have the current into the 240 volt winding set to 4 amps AC,
>>you just wait around and see how hot the transfomer gets.
>
>---
>That's not really a valid criterion, since the core losses are only
>going to be a fraction of what they would normally be with 240VAC on
>the primary. Moreover, if you're going to do it properly you need to
>monitor the temperature rise of the transformer over ambient and make
>sure it doesn't exceed the spec.

  I'm not sure what instrumentation Rob has access to; that's why I asked about meters.
If you're really going to do it right, monitoring the temperature rise of the
"transformer" over ambient isn't enough. You need to know the temperature of the
windings, and the rating of the insulation. Since the transformer doesn't appear to have
thermocouples buried in the windings, what is what I usually do with a new design, one
would have to use the method of measuring the resistance of the windings at room temp, and
then applying current for several hours, followed by a measurement of the winding
resistances when hot. From this the temperature rise of the windings can be inferred.
One starts out with less than rated current and makes a measurement, then if temperatures
are under allowable values, increase the current and take some more measurements.

Given a starting temp (room temp), designated Tcold, and measured resistance of a winding
at that temp, designated Rcold, and a measured resistance Rhot after allowing the
transformer to reach equilibrium with heating from the current in the windings; the temp,
Thot, is given by:

           Rhot
   Thot = ------ (234.5 + Tcold) - 234.5
           Rcold

 in degrees centigrade.

Since we don't know whether his transformer has Class C, Class H, or whatever, insulation,
we can't really tell what temperature rise is acceptable. One way to find out would be to
assume (there I go again) that the transformer was designed properly, and will exhibit a
temp rise at full load of just about the rating of the insulation system. Rob could short
the 12 volt winding, and apply rated current to the primary, wait a few hours and measure
the temperature rises of the windings. This would presumably be what the transformer was
designed for. He could then measure the temp rises with his rectifier load and see if
they are <= to what he got with the full load test I just described.

I'm sure you know all this, John, but I'm going into detail for Rob's benefit.

>---
>
>>The current in the shorted 6 volt winding should be about 160 amps.

  This is too much, as John points out. Don't do this.

>>Ideally you should short the 6 volt winding into a suitably
>>rated shunt and then you could directly measure the current there.
>
>---
>The shunt's an OK idea, but it needs to short out the _entire_
>secondary, for the reason given earlier.

  Or limit the current to less than 83 amps in the shunt; this would be acceptable for the
6 volt tap. The shunt could also be used to short the rectifier output so that the actual
current there could be measured. To assess the heating in the 6 volt winding, a true RMS
meter would need to be used. The electrochemical effect of the current is proportional to
coulombs/sec, so an average responding meter would be appropriate to assess this.

Rob's original question was whether the transformer could supply 75 amps at 6 volts. It
would be reasonable to assume that the transformer can meet its nameplate rating. So if
he has access to a high current shunt, he can just put it in series with the 6 volt output
tap and, with a true RMS meter, verify that the current is less than 83 amps, and feel
safe without all this fooling around we've all been telling him to do! :-)
 
I somehow doubt that Rob has any high current shunts around. Maybe he has access to some,
but I got the impression that he might be short on instrumentation.

So, what we all need to help him with is the measurement of the secondary current without
having to buy a high current shunt. One possibility is to make a current transformer
somehow. Or, how about this? Ten guage copper wire (here in the states, anyway) has very
close to .001 oms/ft. If he were to cut a 1.5-foot length of 10 guage wire, and solder on
a couple of smaller wires exactly 1 foot apart, he would have a shunt of 1 milliohm, and
with a current of 75 amps, he would need to measure 75 millivolts AC, and even the
cheapest Radio Shack DVM can do that. The problem is that the piece of wire would be
dissipating 5.6 watts and would get hot. The resistance would change enough to inspire a
vote of no confidence. He could parallel two such wires and have only 2.8 watts
dissipation, and finally, he could put the put the homemade shunt in a plastic tray of
water (at 20 degrees C), which would probably limit the temperature rise to an acceptable
value. (He can find out the resistance/foot of available wire of about this size and
figure out exactly how far apart to put the meter taps.) (And, he will need a true RMS
meter)
>---
>
>
>>Without knowing the details of the transformer insulation system, I
>>can't say what the maximum allowable temperature rise is, but if the
>>transformer doesn't start stinking too much, you're probably ok. So
>>far, we would be testing the transformer at its full rated power.
>
>---
>No, you'd only be exciting the copper, not the core. To determine the
>core losses, you'd need to do the open circuit test.

   I'm aware of this, but as I explained in another reply to Fritz Schlunder, typically
small transformers like this aren't optimized for continuous operation at full load. It
could be, but I think it unlikely. Anyway, to do it right (measure the core loss) would
require a wattmeter designed for measurements under low power factor conditions.

But, he could get an idea of the relative magnitude of the core loss by measuring the room
temp resistances of the windings as mentioned earlier and then applying 240 volts to the
primary with the secondaries unloaded, waiting several hours, measuring the warm winding
resistances and using the formula given earlier.

>---
>
>>Since you only want 75 amps from the 6 volt winding, you could set the
>>primary current in this test to 1.875 amps AC. This would correspond
>>to 75 amps in the 6 volt secondary. Just let it run for a few hours
>>while monitoring the temperature rise. This will provide a test of
>>the transformer at the actual current which you will be using.
>
>---
>This isn't a good test either, because the core losses will be even
>less than they were previously and won't reflect the conditions under
>which the transformer will be expected to function in real life.
>---
>
>>Rectifying the output of the transformer will cause the current to be
>>non-sinusoidal which will increase the copper (I^2*R) losses.
>
>---
>Yes. Conversion efficiency is about 81% for a full-wave bridge.

And the distortion of the current waveform will make the the copper losses even more
significant compared to the core loss than with a pure resistive load.

>---
>
>>You could also connect your rectifiers in whatever configuration you
>>will use, and short the output from the rectifiers. This will give
>>you a method of testing the heatsinking of the rectifiers. As before,
>>turn the variac which is supplying the 240 volt winding ALL THE WAY
>>DOWN before you start the test. Turn up the variac SLOWLY while
>>monitoring the current in the 240 volt winding until you reach 1.875
>>amps AC. You should use a true RMS ammeter when the rectifiers are in
>>circuit for best results. I wouldn't turn the current up to 1.875
>>amps right away with the rectifiers in circuit; turn it up gradually
>>while monitoring the temperature of the rectifiers. If the rectifiers
>>sizzle when touched with a wet finger, you probably should use some
>>more heat sinking.
>>
>>The advantage of the short circuit test method is that you only
>>dissipate a power equal to the losses in the transformer which should
>>be under 100 watts for a transformer of this size. (light bulbs or
>>electric heaters extra) It does not include the core loss, but that
>>shoud be a negligible part of total losses at nearly full power out of
>>the transformer.
>
>---
>As Fritz Schundler pointed out earlier, an optimum transformer design
>will yield core losses equal to copper losses.

As I explained to Fritz, this is only true for a transformer used nearly all the time at
full rated load.

In some special applications the disparity between core and copper loss is even greater
than the example I cited to Fritz. Inverters for off grid use are designed for very low
core loss because the customer wants to be able to leave the inverter running for extended
periods without a large drain on the battery. A transformer I designed for a 2400 watt
inverter had about 12 watts of core loss and 347 watts of copper loss (full load) on EI200
laminations, 2.75" stack height. I think inverter transformers represent the extreme of
this disparity.

A good example of a transformer where the core loss is prominent is the transformer in a
"wall wart". Those transformers are usually blazing hot even without a load. They use
cheap iron and run 'em hot. These transformers are very small, and the resistance of the
primary is so high that the volt-seconds seen by the core is greatly reduced by the IR
loss in the primary when the transformer is loaded. This reduction in core loss
compensates for the increase in copper loss due to load so that the transformer is within
ratings (barely, I think).

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