Time constant, coil collapse

Thread Starter

vandaycalta

Joined Mar 22, 2016
53
Hello all!
I was thinking earlier about a npn darlington switching the primary side of a coil and wondered about the following:

If you have a coil with voltage of 10v, resistance of 10 ohms and current of 1 amp that is switched using a npn darlington transistor and you charge the coil for the 5 time constants to fully reach the 1 amp, will it take 5 time constants for the current to reach zero using a npn darlington? I assume if you were to use breaker points there would be more resistance but with a transistor darlington I would thing that there would be no additional resistance and it would take the same 5 time constants to reach 0 amps.

Am I kinda on the right track here?

Val
 

mcasale

Joined Jul 18, 2011
210
What do you mean by "breaker points"?

Charging the coil will take 5 time constants (L/R), but the Darlington will also have a voltage drop, so you won't reach 1 amp.

Discharging the coil is the same deal, except I don't understand how you want to do it. Be sure you protect your transistor with a flyback diode.

Switching is better done with a MOSFET --- lower drain-source voltage when it's on.
 

wayneh

Joined Sep 9, 2010
16,399
Why would the breaker points offer more resistance? I would think that would be the low-resistance solution. They worked great for a long time before electronic ignitions came along. A MOSFET might be superior but as noted, you have to protect it.
 
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