HV transformer sc current measurements / High Voltage / Forums

Forums

4hv.org :: Forums :: High Voltage

« Previous topic | Next topic »

HV transformer sc current measurements

1 2 next

Move Thread

LAN_403

Tom R.

Thu Oct 15 2009, 06:57PM

Registered Member #2437 Joined: Thu Oct 15 2009, 05:58PM
Location:
Posts: 3

Hello group.

This is my first post, i'm just beginning with HV so please bear with me :)

I have a home built HV transformer of the following parameters:
input 230V, output around 10 kV
winding ratio is 500 on the primary side and around 23 000 on secondary giving a turn ratio of 46

Now i am wondering what are the max. currents on the primary and secondary side once the secondary is shorted.
How can i measure this safely without burning the ammeter and myself? :)

I'm thinking of applying 230V to the secondary terminals, then shorting the primary through an ammeter (since it will have only 5V) and simply rescaling the result obtained.

Is that proper thinking?

Sulaiman

Thu Oct 15 2009, 07:06PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141

Please don't take this badly, I really want to help.
If you really are new to high voltage stuff - please do not make this transformer.
There are many ways to fail technically
If you succed the result will be lethal - start safer.

Tom R.

Thu Oct 15 2009, 07:47PM

Registered Member #2437 Joined: Thu Oct 15 2009, 05:58PM
Location:
Posts: 3

This transformer is already done, it has been tested and seems to operate correct drawing around 2-3 cm long arc.
I'm not taking any unnecessary risks.

Can you tell where did i make a mistake in my reasoning and how would you do it?

Dr. Dark Current

Thu Oct 15 2009, 09:18PM

Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384

I'd do it with a multimeter set to the 10(20) A range. Do you think your transformer puts out more than 10A?

Or you can do it like you said (run in reverse), but the s/c current calculated this way could be a little off (lower) than the real one.

Sulaiman

Thu Oct 15 2009, 09:57PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141

Assuming that you've made a reasonably good transformer, the short circuit current can be assumed to be dominated by resistance and transformer action.
e.g 230 Vac to primary of 1 ohm, turns ratio 500:23,000 secondary resistance 2500ohms.
the primary resistance 'transformd' to the secondary is 1 x 46^2 = 2116 ohms
Added to the actual secondary resistance of 2500 ohms gives 4616 ohms effective.
So the output of your transformer behaves like 230 v 46 = 10500 volts rms
with a source resistance of 4616 ohms.
So the short circuit current should be 10500/4616 = 2.3 Amps rms.

The primary current should be 105.4 Arms - if a circuit breaker or fuse doesn't prevent it.

Hazmatt_(The Underdog)

Fri Oct 16 2009, 07:06AM

Registered Member #135 Joined: Sat Feb 11 2006, 12:06AM
Location: Anywhere is fine
Posts: 1735

Here's some more info on transformers, hope this helps a bit:

Dr. Dark Current

Fri Oct 16 2009, 08:44AM

Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384

Sulaiman: The leakage inductance often plays a big role too.

Sulaiman

Sat Oct 17 2009, 06:53AM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141

True, flux leakage will reduce the short-circuit secondary current,
but not enough to protect against overload, which was my main point - SAFETY.

Tom, you need a ballast, an inductor wold be most efficient, a mains lamp in series with the priary would be easy/cheap - or an electric heater element for more current etc.

Steve Conner

Sat Oct 17 2009, 09:44AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

Running the transformer backwards will kind of work. Here's what a commercial transformer maker will do:

The secondary winding is shorted, and a variable voltage is applied to the primary using a variac, through amp, volt and wattmeters. The voltage is carefully brought up from zero until the rated primary current is flowing, and the amount of voltage needed to flow this current is noted.

The impedance of the transformer can then be calculated by Ohm's law. Typically it's stated as a percentage. For instance, if your transformer has a 240V primary, and on the short-circuit test it only takes 12V to make rated current flow, then the transformer has an impedance of 5%.

This implies that if you apply 240V to the primary and short the secondary, then 100/5 = 20 times the rated current will flow. Bam!

If you don't have a variac, and don't know what the "rated" current of your transformer is because you made it yourself, you can't really do this, but you can still estimate what the short circuit current would be. Just do the calculations in ohms instead of percent.

A transformer maker would also do an open-circuit test, where he applies rated voltage to the primary with the secondary open, and measures the magnetizing current (on the ammeter) and core losses (on the wattmeter) On the short-circuit test, the wattmeter reads copper losses, so the transformer is now completely characterized by these two tests. The method really comes into it's own when you're making a 750MVA transformer but don't have a city to use as a dummy load.

Antonio

Sat Oct 17 2009, 12:32PM

Registered Member #834 Joined: Tue Jun 12 2007, 10:57PM
Location: Brazil
Posts: 644

The scaled down measurement can be easily done with a lamp, or better, a linear resistor, in series with the primary coil. Measure the short-circuit secondary current and the primary voltage, that will be both low enough. Assuming that the transformer is linear (what is quite optimistic), the obtained transconductance will be valid with the normal input voltage too. Measure also the input impedance with the secondary coil open, with the same setup, and repeat both measurements with the transformer inverted. The four parameters obtained characterize completely a linear model of the transformer at the measurement frequency. The contributions of resistances and reactances can be obtained by looking at the phase differences between the signals in the measurements.

1 2 next

Moderator(s): Chris Russell, Noelle, Alex, Tesladownunder, Dave Marshall, Dave Billington, Bjørn, Steve Conner, Wolfram, Kizmo, Mads Barnkob