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Old 23rd June 2009, 11:34 AM   #1
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Default Calculating current capacity?

Here are my 2 PT's. They both have 3.75"x3" laminations. The bigger one on the right (called "Venti" from now on) has a 1.75" thick stack, and the smaller one on the left (called "Tall") has a 1.5" stack.

Click the image to open in full size.

Here are some measurements taken with no load on the secondaries. HV winding resistances were closely matched on either side of the center tap.

House mains: 126VAC

Tall:
Primary, black-black 4 ohms
Secondary 1, red-red/yellow-red: 370-0-370V, 183 ohms
Secondary 2, brown-brown: 17VA ~1 ohm
Secondary 3, yellow-yellow: 5.4V, ~.2 ohms
Secondary 4, green-green: 6.6V, ~.1 ohms

Venti:
Primary, black-black: 1.6 ohms
Secondary 1, red-red/yellow/red: 380-0-380V, 90 ohms
Secondary 2, green-green: 6.5V ~.1 ohms
Secondary 3, yellow-yellow: 5.4V ~.1 ohms
Secondary 4, brown-brown: 6.5V ~.1 ohms

Any ideas how I might go about guesstimating what current capacities I have? My idea would be to divide the HV windings by 126, multiply by 115 (probably the intended primary voltage), then fiddle around in PSUD II until I could get the right secondary resistances. Any other (better) ideas? Thanks!
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Old 23rd June 2009, 12:29 PM   #2
SY is offline SY  United States
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Johan Potgeiter posted some nice rules of thumb some time back. It might be worth doing a search for them.
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Old 23rd June 2009, 02:20 PM   #3
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I'm not having the best of luck with the search. Can you perhaps point me in the right direction?
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Old 23rd June 2009, 04:11 PM   #4
ChrisA is offline ChrisA  United States
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Default Re: Calculating current capacity?

Quote:

Tall:
Primary, black-black 4 ohms
Secondary 1, red-red/yellow-red: 370-0-370V, 183 ohms
Secondary 2, brown-brown: 17VA ~1 ohm
Secondary 3, yellow-yellow: 5.4V, ~.2 ohms
Secondary 4, green-green: 6.6V, ~.1 ohms
I'm kind of in the same way, I can figure out the turns ratio and the DC resistance but that's not enough.

You can apply a rule of thumb and guess based on weight. (10 pounds is about 200VA.) Or you can try and match up what you have with the specs on the Hammond web site.

I think the only 100% positive, fool proof method is to test the transformers on a dummy load. That sounds easy until you figure out the specs on the load and see that you need something like a 5K 200W resistor. And it would have to be adjustable so you could use it for more than just one test.

I think I want to do this. Even if I have the specs in hand. Before I invest time in a 35 year old salvaged PT I want to be 100% certain the PT is not defective and the only way to know is to perform a burn in test. Let it run for hours. then lower the resistance of the load and see what this does to (1) the voltage and (2) transformers temperature.

I thought about buying 100 ceramic, wire wound power resistors . For $20 I could build a dummy load board that could fully test a 500VA transformer (futurlec.com has 5W parts for 20 cents each)

I'd like to design the "load board" so that it could be re-configured to test OPTs as well. For that you need an 4, 8 or 16 ohms and 250+ watts. I could wire it with some terminal strips and crip-on ring terminals

I thought about 100W light bulbs. But I think all of the bulbs and ceramic sockets and the plywood to mount them on would cost more and be less practical than the 100 resistors.
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Old 23rd June 2009, 04:23 PM   #5
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I've used light bulbs. A 100W incandescent light bulb draws ~800mA. A 40W draws ~300mA. Put 6 in series and you have a cheap (and bright) 720V load. Put an ammeter in series to get the real current load, since the above loads are wild guestimmates.
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Old 23rd June 2009, 04:37 PM   #6
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Hm, so if I manage to get a 16-watt light bulb, I should only be drawing around 135mA? 150mA for an 18-watter? Then the issue would be to get 3 of them and put them in series. I'll check my local hardware store, but I'm not so sure I'll be able to find such low wattage lamps..
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Old 23rd June 2009, 04:53 PM   #7
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I read someplace that transformer current ratings are based on, or partially based on temp rise over a period of time. But the whole transformer rating system seems to be a lot of black magic no one will talk about! At least i have never found a clear definitive way to measure an unknown transformer as say for sure it is X-Number of amps or whatever.

I would think....if you used an adjustable load and loaded down the transformer until you got some % of Voltage drop. then measured the temp rise over a period of 1 hour. you could find some ratio of load VS temp rise that would give you an answer.

So there are 2 questions that need to be answered there first. 1- what is an acceptable or normal amount of Voltage drop? 2- what is a normal or acceptable rise in temp after 1 hour. I wonder if there are any sort of standards written as such?

There is an eBay seller that list's some very detailed ratings of there transformers. it seems to be an excellent guide as such.

here is the specs for one of there Toroid transformers. from the specs i think this would give a person a pretty good idea of how to test.


Open Circuit Test (core loss test):
TEST CONDITION: Apply variable voltage to primary coil (115V terminals) from 115-180VAC at 60Hz. No load on secondary coils.
1. Primary V = 115VAC, Primary I = .02A or core loss is 2.3Watt.
2. Primary V = 140VAC, Primary I = .03A
3. Primary V = 160VAC, Primary I = .03A
4. Primary V = 180VAC, Primary I = .07A

Short Circuit Test (copper loss test):
TEST CONDITION: Apply variable voltage to primary coil (115V terminals) from 1-20VAC at 60Hz and short circuit on secondary coils. The test result is 10.25V and 3.48A or 35.6Watt which is the maximum rated copper loss

Dielectric Test:
TEST CONDITION: Apply dielectric meter between primary coils and secondary coils; and increase voltage up to 3500VAC. No initiate any spark.

Load Test Result:
TEST CONDITION: Input 115VAC 60Hz to the primary coil; parallel 2 secondary coils, 5 ohm 250W resistors as dummy load and room temperature at 25 degree C. Temperature rises 5 degree C after an hour test at load test #4.
1. Primary I = 0.02A, Secondary V = 18.4VAC at 0.0A, Power = 0 Watt
2. Primary I = 0.65A, Secondary V = 18.1VAC at 3.61A, Power = 65 Watt
3. Primary I = 1.20A, Secondary V = 17.7VAC at 7.08A, Power = 125 Watt
4. Primary I = 2.77A, Secondary V = 16.6VAC at 16.59A, Power = 275 Watt
5. Primary I = 4.00A, Secondary V = 16.4VAC at 24.11A, Power = 400 Watt



Now i don't quite understand the copper loss test. how the specs are measured...

But the load test is quite interesting. only a 5 deg C rise after 1 hour.


I look forward to comments from those more in the know then I!
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Old 23rd June 2009, 05:07 PM   #8
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Quote:
Originally posted by sorenj07
Hm, so if I manage to get a 16-watt light bulb, I should only be drawing around 135mA? 150mA for an 18-watter? Then the issue would be to get 3 of them and put them in series. I'll check my local hardware store, but I'm not so sure I'll be able to find such low wattage lamps..
Start with 25W bulbs or whatever is the cheapest and add another to the chain. You'll be increasing the "hot" resistance of your load which will reduce the current. It's easier if you know the actual load you will be applying in your project, try to reproduce that, and then run the transformer for a while and see how it does. If you get about the voltage you want and the transformer can run for an hour or two without getting too hot, then you are probably well within spec. Of course, that approach only works within reason...if you grossly overload the thing it may die before the core gets hot enough to let you know.

I think the rule of thumb for the amount of voltage drop you will see for unloaded versus loaded is 10%. So if you are seeing a much greater drop than that with the load, you are loading it pretty hard. Not necessarily out of spec, but given that you don't know anything about it....
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Old 23rd June 2009, 05:39 PM   #9
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Quote:
Originally posted by rknize
I've used light bulbs. A 100W incandescent light bulb draws ~800mA. A 40W draws ~300mA. Put 6 in series and you have a cheap (and bright) 720V load. Put an ammeter in series to get the real current load, since the above loads are wild guestimmates.
The resistance across a light bulb varies depending on the voltage (or really with temperature of the filiment.). Try putting an ohm meter cross a 100W bulb, then light it up with 120VAC and measure the voltage drop. Ohms law does not exactly apply because lights, unlike resistors are design to run at very high temperature.

Whenever I start working this out I find I need more bulbs then my initial guess. Say I have a 500V transformers I think might supply 250ma. That is 125VA. I think four 40W bulbs in a series. That is about 1778 ohms or 280ma or about 10% over current. maybe "close enough" But what if I guesse wrong and it's only a 100ma transformer? Better start with five 25W bulbs. And then you think doubling the power is to large of a step. I'd prefer to jump in units of about 20%. So you end up with about 10 lamp sockets. At about $3 each plus a $20 stash of 25W, 40W and 60W bulbs. So then I go back to thinking about buying 100 resistors again.

The problem here is that you don't know, not even within a factor of two what your transformers can do and you might have several transformers some at 350V and another 750V. and you want to make small 25ma or 50ma jumps.
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Old 23rd June 2009, 05:56 PM   #10
rknize is offline rknize  United States
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Yes, that's what I meant about "hot" resistance. As far as cost, well I have a large box full of old bulbs from the many that have been switched over to CFLs. Sockets? What sockets?

Edit: I don't recommend that!
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