# Transformer VA

Discussion in 'General Electronics Discussion' started by komalbarun, May 20, 2013.

1. ### komalbarun

67
0
Nov 25, 2011
Step down 20:1 transformer
240V input.
Transformer VA = 3.6 Watts (12V @ 0.3A)

If i want to use it as a step up 1:20 with 12V as input... will 0.3A the maximum safe current which can flow through it?

also...
if I decreased the input voltage applied to 1.8V will the maximum safe current be 2A since P=IV = 1.8 x 2 = 3.6W ?

pls help

Last edited: May 20, 2013
2. ### duke37

5,364
771
Jan 9, 2011
If you use a transformer backwards you get a low voltage out. You may only get 180 to 200V with a low power transformer.

You should not exceed the rated current. The heating in the transformer = P = I * I * R where R is the resistance of the winding wire. This does not change with the voltage.

3. ### komalbarun

67
0
Nov 25, 2011
True..but since V = I*R , P = I * [ I * R] = I * V, right? But I will go with your advice . Thnx.

4. ### komalbarun

67
0
Nov 25, 2011
It seems that P = IV CANNOT be used for wires since very low V-drop on them right?

Using P = I x I x R :

3.6 = (0.3x0.3) x R
0.09 R = 3.6
R = 3.6 / 0.09 = 40 ohms

applying 2A current:

P = 4 * 40 = 160W !!!!

using P = IV :

p = 2 * 1.8 = 3.6W --> would have been applicable if V-drop on transformer input coil was 1.8V, right?

I think I will stick to the P = I x I x R formula .

Thnx dude, you helped me save my transformer

Last edited: May 21, 2013
5. ### duke37

5,364
771
Jan 9, 2011
P = I * V is the output power of the transformer and should be much more than the power lost in the windings.

P = I * I * R is the power lost in the transformer winding where R is the winding resistance NOT the load resistance.
Low resistances can be measured with various techniques. In a transformer you have two windings and both need to be taken into account. The current and resistance in each winding will differ.

The transformer is rated by how much voltage will be dropped under load and how hot it will get. Modern transformers have good insulation and can run hot, old transformers should run only just warm so will have a lower rating for the same size.

6. ### komalbarun

67
0
Nov 25, 2011
So basically what I have here as 40 ohms is supposed to be the load resistance since am using 3.6W which is actually the transformer's output power?

Last edited: May 21, 2013
7. ### duke37

5,364
771
Jan 9, 2011

The resistance of the transformer secondary will be much less than this so that the power lost in the winding will be low. A transformer will normally be designed so that the loss in the primary and secondary is equal. The primary resistance will be much higher than the secondary, by the square of the turns ratio.

I guess that your transformer secondary plus the referred primary would be of the order of 4 ohms.

8. ### komalbarun

67
0
Nov 25, 2011
The question is "What is the maximum safe current I can apply to my step up transformer? Is it more that 0.3A or is 0.3A the maximum safe current?"

Last edited: May 22, 2013
9. ### komalbarun

67
0
Nov 25, 2011
The question is "What is the maximum safe current I can apply to my step up transformer? Is it more that 0.3A or is 0.3A the maximum safe current?"

10. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,496
2,837
Jan 21, 2010
If it is a 12V 3.6VA transformer then 0.3A RMS represents a safe current. It is the maximum recommended power.

Sure, it might handle a little higher power for a short time, but it will be operating beyond its specs and may be damaged if the ambient temperature is high, if there is insufficient ventilation, etc., etc.

11. ### komalbarun

67
0
Nov 25, 2011
But if I appplied 9V @ 0.4A to the primary winding, then even though rating is 3.6VA, this might still damage the transformer due to current being higher than the maximum safe current? [as explained by duke ]

12. ### duke37

5,364
771
Jan 9, 2011
P=I*I*R
P1=0.3*0.3*R = 0.09R
P2=0.4*0.4*R = 0.16R

So going up from 0.3A to 0.4A will almost double the heating of the transformer. You may get away with this if, as *steve* says, you run for a short time and then let the transformer cool or you may be able to cool it with a fan.

13. ### komalbarun

67
0
Nov 25, 2011
Thnx for that one duke, sure helps a lot

I never thought of using P = I*I*R in that way before(P=0.09R and 0.16R) lol...

thnx to steve too