Transformers

[gar]

090311-1623 EST

The question of using a step down transformer in the reverse direction seems to pop up fairly often. Today I ran a quick check on one transformer. This is a Sq-D SKC B-4 0.5 KVA.

The primary can be wired 480 or 240. I wired it 240.
The secondary can be wired 240 or 120. I wired 240.
Thus a nominal 1 to 1 transformer. I ran this at 120 V instead of 240 because that was easy. Also ran the transformer unloaded.

The resistance of the normal primary = 2.7 ohms.
The resistance of the normal secondary = 3.3 ohms.
Obviously the design of the primary is to reduce power dissipation in the primary from the magnetizing current.

Used in the normal direction
120 V input
125 V output
Apparent turns ratio 125/120 = 1.042

Used in the reverse direction
120 V input
114.8 V output
Apparent turns ratio 120/114.8 = 1.045

This is a good correlation.

Used in the reverse direction I believe the KVA and KW capability will be slightly reduced because the power dissipation will be greater in the 3.3 ohms than in the 2.7 ohms.

Clearly the transformer is wound to provide slightly higher open circuit voltage on the secondary than one would select just on the basis of desired output voltage.

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[jim dungar]

In many, transformers below 3kVA it is not uncommon to find inherent 'compensating turns' on the primary windings in order to overcome some of the internal losses of the transformer.

 

[glene77is]

Gar,
Very interesting.
Good, simple experiment.
Clear documentation.
Even something that seems obvious can be interesting when described in numbers.

 

[LarryFine]

That's one reason we have to separate the theoretical from the real-world when discussing and answering questions.

"Welcome, Neo, to the real world."

 

[gar]

090211-1808 EST

I will add some load to the transformer in the normal direction. My guess is that somewhere between 75% and 100% of rated load KVA the load voltage will approximately equal the input. I would expect the transformer to be designed this way. The voltage was about 4% high unloaded and that implies a 4% impedance.

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[gar]

090311-2052 EST

I have now run the load test and I guessed wrong.

My two loads were a 100 W and a 250 W bulbs.

Used in the normal direction no load
120.3 V input
125.8 V output
Apparent turns ratio 125.8/120.3 = 1.046

With 100 W load
120.8 V input
121.2 V @ 0.81 A = 98.2 W
internal impedance 125.8*120.8/120.3 = 126.3
126.3-121.2 = 5.1, 5.1/0.81 = 6.29 ohms

With 250 W load
120.4 V input
113.2 V @ 2.03 A = 229.8 W
internal impedance 125.8*120.4/120.3 = 125.9
125.9-113.2 = 12.7, 12.7/2.03 = 6.26 ohms

Add the two DC resistances and we get 2.7 + 3.3 = 6.0 ohms .
This implies that a great part of the impedance is the resistive component. However, if I was at full excitation the leakage inductance might be somewhat higher.

Note that 250 W with the transformer at half voltage is the same current as the full load rating of the transformer. Assuming the internal impedance does not change much with doubling the voltage, then the % voltage change from no load to full load is about 12.7/251.8 = about 5% at full voltage.

Note output voltage = to input voltage was at about 0.81/2.03 = 40% of full load. So I was far off in my guess. More iron and less copper would probably increase this point.

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