Lowering current output from supply transformer

[sdickson]

Our current supply transformer at work has a range of 146A-7000A, in order to preform lower current tests on CTs I need to lower the output current to around 30A.

For example, the CT's resistance is around 2.46m?, to achieve 30A output I need to set the voltage of my supply transformer to around .00246*30=.0738V, which is lower than the minimum .49V output of the transformer. I've thought of a couple of ways to do this, the easiest of which would be installing a very low resistance, high current resistor on the output. One that I've spec'd out is a .209?, 200A, 8300W rated ribbon resistor. So using this resistor to try and achieve 30A output I should have .209+.00246=.21146? which would give me 30A at .21146*30=6.34V which the transformer can handle. Does this sound like it would work?

Another option I considered is installing another buck transformer in series with the one already installed and using that to bump down the transformer's output voltage. I'd rather not pursue this option as it would involve taking apart the supply and installing new hardware.

The only reason I'm not simply reducing my input voltage is because I think the relays seen at the bottom left of the schematic won't be able to operate otherwise.

Let me know what you guys think and if I'm totally off base with all this, please let me know. I've also included schematics for both of the solutions described above.

Thanks!

 

[gar]

141120-1627 EST

Your drawings are too small to read easily, and I don't want to wade thru all the detail either.

So to the basics.

You need a transformer with a rating of possibly 50 A on the secondary, and a low secondary open circuit voltage. The transformer should be capable of a short circuit of possibly 100 A ( this is simply the relationship of open circuit voltage to internal impedance ).

Suppose secondary open circuit is 2.5 V ( an old filament voltage standard voltage ). Pick a resistance of 2.5/100 = 0.025 ohms. This resistance plus internal impedance of transformer should be selected to provide at least 50 A short sircuit with full line voltage to the transformer primary. 100 A at 2.5 V is a VA rating of 250 VA. Supply the transformer primary with a Variac to adjust voltage and thus current. If you really wanted a 100 A capability, then the resistor (0.025 ohms) needs a power rating of 10,000 * 0.025 = 250 W, but at 30 A the power dissipated would be about 900 * 0.025 = 22.5 W. Thus, a small 25 W resistor would work. Ohmite power resistors can tolerate large overload for short times like 1 second.

#14 copper wire is 2.5 ohms per 1000 ft at 20 C. Thus, 10 ft of #14 copper would be about the correct resistance. 30 A in #14 is in excess of its rating and there will be greater heating of the wire, but that is adjusted for by the Variac. Better to use #10 which is 1 ohm / 1000 ft. Thus, you need 25 ft of #10.
.

 

[sdickson]

Thanks for your help gar, I really appreciate the response. I'll try and keep my question more succinct next time and actually upload images correctly. And if you are interested in the schematic, here's a full sized copy.

 

[mgookin]

Can you just get another power supply for the specific need?

 

[gar]

141120-2033 EST

sdickson:

I do not know how your circuit works.

I assume the current transformer at the top of the last photo is the one under test.

The transformer on the right has nothing connected to its secondary. Thus, it is simply a tap adjustable inductor. The current transformer on the right has no current thru it.

The selectable capacitors form a parallel resonant circuit with the tapped transformer on the right.

The Powerstat (Variac) provides an adjustable voltage to the series circuit including the current transformer at the top.

So far the circuit makes no sense. What is it supposed to do? Where is there any reference for whatever is being tested?

Something is missing including the purpose and how it is used.

.

 

[GoldDigger]

Without the instructions, the function is hard to decipher. What I see is that an output current is taken from one of the taps of transformer on the right.
The powerstat and the buck/boost transformer along with the R contacts controlled by the switches on the lower left control the voltage (and therefore the short circuit current.) The relay switched capacitors allow the phase of the output voltage and current to be varied with respect to the input voltage.
The purpose appears to be to generate a test current output for testing and calibrating CTs. (And maybe the wattmeters they attach to?)