# Need help understanding pulse response of L/R circuit

#### electronice123

Joined Oct 10, 2008
311
Thanks in advance-this question has been bothering me for years!

I have a series LR circuit:
L=100mH
R=100 Ohms
V=100V

The L/R time constant is .1H / 100 ohms = .001 seconds (which means full current will occur after 5 time constants or .001 * 5 =.005 seconds). Also, peak current is 100V/100Ohms=1A.

I apply a pulse for .002 seconds then turn it off for .001 seconds....Since the on time is longer than the off time the inductor never fully discharges so does the current increase each time the pulse is applied?

How do you do the math to calculate how long it will take for full current to flow through the inductor when the on time is greater than the off time?

My reasoning:

Pulse on for .002 seconds (two tc's - 86.4%) so current would be 864mA.
Off for .001 seconds, this is where I get confused-are we supposed to look at the inductor like its just beginning the discharge cycle-so it would drop 1 time constant (36.8%) of 864mA?
so it would be 864mA * .368 = 317mA?
Pulse on for .002 seconds again-how do I determine the current at the end of the pulse?
Pulse off for .001 second?

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#### electronice123

Joined Oct 10, 2008
311
I didn't even think of doing that.

Thanks for the great idea. I'll try it in multisim tomorrow. After some simulation I should be able to get the math close enough to understand.

My only fear is that it will lead me to calculus as the only way to do the math accurately....

#### ericgibbs

Joined Jan 29, 2010
8,873
hi 123,
Calculus is fun when you figure out how helpful it can be in solving problems.
BTW: In that sim I assumed a zero impedance Voltage source and an Inductor with no internal resistance.
E

• electronice123

#### crutschow

Joined Mar 14, 2008
23,518
Calculus is fun when you figure out how helpful it can be in solving problems.
For you perhaps, but I have never found calculus to be "fun". • electronice123