# Question about rise time calculation?

Discussion in 'General Electronics Chat' started by wes, Oct 29, 2010.

1. ### wes Thread Starter Active Member

Aug 24, 2007
242
2
Hello, I have a question about a formula I have used many times to figure out how high the current would be at a certain point in time

uH (micro Heneries) / Voltage = RiseTime ( microseconds) to 1 amp

I can clarify more if someone isn't sure what it means exactly.

I know that when calculating the rise time, the first time constant is 63% of the total value. Since the current basically jumps to 63% then takes more and more time to reach full, Does this equation tell you the 1st time constant or the total overall value?

The reason I am asking is because if the equation tells you the overall value then that means that you could reach 63% of that total value much faster and if you just increased the voltage so that the 1st time constant (63%) is equal to that of the total current value at the lower voltage. Then once you have reached that value you can just stop trying to reach maximum since that is going to take much longer.

So does anyone know what that Equation tells you exactly, Is it the Total value or is it like just the 1st time constant? I believe it is the Total value that the current would reach in that amount of time, not the 1st time constant or 63% of it's total value.

Also I have attached a picture demonstrating what I am talking about.
You can see in the image that the voltage decreases rapidly then levels off.
You can't see the current, but the current is basically the reverse going up and then leveling off toward the top.

By the way, If anyone is interested in the program, just message me.

File size:
259.1 KB
Views:
41
2. ### DonQ Active Member

May 6, 2009
320
14
In an L-R circuit, the time constant is a function of both the inductance and the resistance. In your 'formula', you do not have a term for the resistance. Without a resistance, there is no time constant.

Once you know the time constant, a set of equations with e^(t/tc) will come into play. These will predict the voltage and current as a function of time.

Note that 63% is from (1-1/e^(t/tc)) when t=tc. In that case:

t/tc=1,
e^1 =2.71828...,
1/2.71828=.368...,
1-.368=.632 = 63%

Actually, what you will find is that in one more time constant, it will go 63% of the way that is left. Then in another tc, it will go 63% of what is left after that... starting anywhere, in one time constant, it will go 63% of the way to the steady-state condition.

With no resistance, I'm afraid the steady-state condition is infinite current. Since *that* ain't gonna happen, you need to find out what your resistance really is, even if it is just the resistance of the coil of the inductor and/or the impedance of the voltage source.

3. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
720
There will always be an R component, out of the windings of the inductor if nothing else, it may be small, but it exists (as well as a capacitive component which is smaller). Using a very low value for resistance, such as 0.01Ω for a small inductors inherent resistance is acceptable.

Think of it as the impedance of an inductor goes down the closer it is to "charged", where the impedance of a capacitor goes up the closer it is to "charged".

The "knee" is where ${dV\over dt} = 1$ (slope of curve) As you can see in your plot, which is around the 66% mark.

Last edited: Oct 30, 2010
4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
Also, I thought "rise time" had a definition along the lines ..... the time taken for a transient signal to rise from 10% to 90% of its final value.

In an LR circuit the current rise time for a step input voltage (by the aforementioned definition) would be 2.2*TC where TC is the time constant.

5. ### wes Thread Starter Active Member

Aug 24, 2007
242
2
thanks for the replies.

I was messing with the program and resistance has a BIG effect, lol. If you double the resistance then you can cut the fall time in half but the voltage spike is also doubled aswell.