A question regarding a transformer

Thread Starter

AlexK

Joined May 23, 2007
34
Suppose I want to create some sort of an "inductive kick" using a transformer, however instead of disconnecting a battery I prefer using a very fast discharge RL circuit.

Suppose I have a constant current source, I,connected to the primary winding. After a long time a switch disconnects the primary from the source and connects it to a parallel resistor R (thus forming an RL circuit).
So now the current equation is IL=Iexp(-t/tau).
My question is this: If tau is very small (meaning the resistor R is large), so the discharge time is very small, will this create a pulse on the secondary winding (similar to the inductive kick, where the current goes rappidly from constant to zero)?

Thanks.
 

beenthere

Joined Apr 20, 2004
15,819
You will get a kick, but the resistive path will pull current and make the inductive kick smaller. The kick is powered by the collapse in the magnetic field surrounding the transformer winding.
 

Thread Starter

AlexK

Joined May 23, 2007
34
What do you mean by "pull current"? Won't the current behave according to the equation I wrote above? And how much smaller will it make the kick?

Suppose I make the following (very) approximate calculation.
magnetic field=B=N1*u*IL/2R; R-radius ; u-permeability

flux=B*A; A- cross section area
d(flux)/dt=k*dIL/dt=k*I0*exp(-t/tau)*(-1/tau)

where 'k' is the product of all the constants.
So the e.m.f induced in the secondary winding at the moment t=0 is:

E=N2*d(flux)/dt=N2*k*I0*(-1/tau);

So by choosing tau very small won't I be able to get a high E value according to the above equation?
 

beenthere

Joined Apr 20, 2004
15,819
The resistor in parallel with the winding will make a current path. You get your inductive kick when the induced magnetic field collapses in a shot period and inpuces a much higher voltage in the winding. Allowing a resistor to make a path for current outside of the transformer is going to reduce the inductive kick.
 
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