transformer inductance 16000H??

komalbarun

Nov 25, 2011
67
Joined
Nov 25, 2011
Messages
67
Am a bit confused...

20:1 step down transformer, 240V to 12V @ 0.3 A 3.6VA.

what is the indcutor's primary coil's and secondary coil's inductance?

I get 16000H for primary and 40H for secondary coil !!?

pls help..I think my calculus is wrong or maybe it's my concept about the transformer itself which is wrong xD
 
Last edited:

BobK

Jan 5, 2010
7,682
Joined
Jan 5, 2010
Messages
7,682
16000H sounds a little high. How did you measure it?

The ratio is correct though: the square of the ratio of the windings.

Bob
 
Last edited:

komalbarun

Nov 25, 2011
67
Joined
Nov 25, 2011
Messages
67
well...
V = -L I ?

Pin = Pout, so ,
240V x I = 3.6VA
I = 3.6/240 V = 0.015A

240 = -L 0.015
-L = 240/0.015 = 16000

am I making a mistake somewhere?
 
Last edited:

duke37

Jan 9, 2011
5,364
Joined
Jan 9, 2011
Messages
5,364
Not V = I * L but V = I * Xl

Xl = 2*pi*F*L

You are too high by a factor of about 300.
 

komalbarun

Nov 25, 2011
67
Joined
Nov 25, 2011
Messages
67
Not V = I * L but V = I * Xl

Xl = 2*pi*F*L

You are too high by a factor of about 300.

how do I calculate the Xl if I don't know the L or vice-versa?

how do I use 12V @ 0,3A 3.6 VA to calculate the L (if possible)?
 

duke37

Jan 9, 2011
5,364
Joined
Jan 9, 2011
Messages
5,364
You measured Xl NOT L as I pointed out in the first equation.

Transposing the second equation (can you do this?)

L = Xl/(2*pi*F)

The 0.3A is the maximum current that the transformer can provide in the load without getting too hot or dropping too much voltage. It has nothing to do with inductance.
 

Raven Luni

Oct 15, 2011
798
Joined
Oct 15, 2011
Messages
798
Is it just me or are transformers much easier to understand than plain inductors?
 

komalbarun

Nov 25, 2011
67
Joined
Nov 25, 2011
Messages
67
You measured Xl NOT L as I pointed out in the first equation.

Transposing the second equation (can you do this?)

L = Xl/(2*pi*F)

The 0.3A is the maximum current that the transformer can provide in the load without getting too hot or dropping too much voltage. It has nothing to do with inductance.

thnx for that, its good to know ;)
 

komalbarun

Nov 25, 2011
67
Joined
Nov 25, 2011
Messages
67
Not V = I * L but V = I * Xl

Xl = 2*pi*F*L

You are too high by a factor of about 300.

lol I just understood what you meant by this xD, so if am understanding right :

16000 = 2 * pi * f * L?

16000 = 2 * pi * 50 * L

L = 16000/ approx. 314
L = 50H ?
 
Last edited:

duke37

Jan 9, 2011
5,364
Joined
Jan 9, 2011
Messages
5,364
That looks right to me.

Raven
A simple transformer is relatively easy to understand but as soon as non linear effects or energy transfers are included, my mind boggles. Inductors can be used in so many ways.
 
Top