# Re: L/R response time

J. Eric Bracken ([email protected])
Tue, 18 Feb 1997 09:15:46 -0500

Greg,

You didn't mention whether you take the output across the R or across
the L; I assume you meant R.

It's not hard to show that the response of the circuit to an applied
ramp with rise time tr is

t/tr + (L/R) * 1/tr * [ exp( -t / (L/R) ) - 1 ], t < tr

The first term is just the original ramp; the second term contains an
exponential that decays with time constant L/R.

If the L/R time constant is a lot shorter than the signal rise time,
then we can neglect the exp() and simplify the response to

(t - L/R) / tr

This shows that the LR circuit just *delays* the ramp; it doesn't change
its slope (i.e. the risetime is not affected.)

On the other hand, if the L/R time constant is much larger than the
input rise time, then the output rise time is greatly increased. The
10-90% rise time is just 2.2 * (L/R), because it's a simple exponential
up-cay to the final value.

Summary: for small L/R, rise time is not affected; for large L/R,
rise time is independent of input signal and purely controlled by L/R.
For values of L/R in between, expect a smooth transition from the
one behavior to the other.

```--
J. Eric Bracken, Ph.D.                          Tel: 1.412.261.3200
Ansoft Corp. Pittsburgh, PA USA                 Fax: 1.412.471.9427
http://www.ece.cmu.edu/afs/ece/usr/bracken/.home-page.html
```