How to calculate the peak inductor current?

Yes. t, for time in seconds. These are expressions for some value in terms of time.

It is defined as a function of R, L and C in the previous equation. It is the sine component of the decay, even though the real exponential dominates the result, because it decays so much in a small part of a cycle.

Looks to me like ~wo = wo * SQRT(-1)

Wouldn't it be better to call it the "sinh" component?

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Yes. Thanks.

Thanks, for the clarifications. Is there a reason why there is no equation to calculate the peak current for the overdamped case?

Thanks Mike

I can't see one, except that it is messy.

I solved for the peak current (with lots of help from Mathcad). I'm posting the formula as a graphic on A.B.S.E.

But it varies little. The extreme values are Vo/R as L approaches zero, and about .736 times that for the just barely over damped case.

The equation for the peak current in the underdamped case is hard to read, and I'm not sure it's correct anyway. Can anybody make out the expression, and does it work?

All you have to do is have an expression for the time of the first current peak and substitute it in the expression for the current as a function of time.

The correct equations are:

Underdamped case

(set your calculator to radians mode, of course)

Time to first current peak is tp = (1/wo)Arctan(2*L*wo/R)

Magnitude of peak current = (Vo/L/wo)*Exp(-R*tp/2/L)*Sin(wo*tp)

Overdamped case

Time to only current peak is tp = (1/~wo)Arctanh(2*L*~wo/R)

Magnitude of peak current = (Vo/L/~wo)*Exp(-R*tp/2/L)*Sinh(~wo*tp)

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For the critically damped case, his schematic shows a 2 ohm resistor, but his text says it's 20 ohms.

A good test of these expressions is to set the resistor value to 1.99 ohms (barely underdamped) and compute the peak current. Then set the resistor to 2.01 ohms (barely overdamped) and compute the peak current. The two peak currents should be nearly the same. I get 3.691096 and

3.66657 amps for the two cases.

So does the complete expression look like:

Ipeak = (Vo/wo/L)*Exp(-R*Pi/(L*wo)) ?

When I use the values L = 1 uH, C = 1 uF, R = .2 ohms, I get a peak current of 5.3448 amps from this expression. But if you look at the graphic from the simulation on the web page, the current peak is obviously up around 8 amps. I get a peak current of 8.626 amps using the expressions given below.

For this test, you should leave the inductance at 1 uH and vary the resistance.

How did you test? Using the web page expressions? Using a simulator? Did you use the L and C from the web page, and if so, what value of R did you use for the underdamped case? Did you try the overdamped case too? Did you try the values of R that I suggested? And if you did, what peak currents did you get for the various cases?

The expressions for the critically damped case are really simple:

Time to current peak is tp = 2*L/R

Magnitude of peak current is Exp(-1)*Vo*SQRT(C/L)

Using the web page values of L = 1 uH, C = 1 uF and R=2, I get a peak current of 3.67879 amps.

The expression on the web page for the peak current in the critically damped case is just the same as above, but with 2/R substituted for SQRT(C/L), since the critically damped case is obtained when SQRT(C/L)=2/R.

The peak currents when R = 1.99 and when R = 2.01 must be very close to the 3.67879 amps of the critically damped case.

I read it as pi.

I tested the current versus time for the under damped and critical cases, with the inductance nearly the same, and got 4 times higher current for the critically damped case.

Could a multiplier of 4 be missing from the underdamped case?

That is what I think I am seeing.

And Vo = 10 volts, I assume.

The graphs are so poor that I have been able to make little sense of them.

I also get a peak value of 8.626 at t=1.478, found by solving for when the derivative i(t) = 0 and applying that time to the i(t) formula, but 5.345 from the i peak formula. At least one of them is wrong.

If I solve produce a formula for peak current, based on the i(t) formula, I get:

ipeak=(Vo/(L*w0))*exp((-R/(2*L^2*w0))*atan(2*L^2*w/R))*...

sin(atan(2*L^2*w)/R)

and this gives a value of 8.626, which just verifies that I did the differentiation and simplification right, since the i(t) it was based on looks like it has the same peak value. It doesn't prove that i(t) was right to start with.

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Yes. I used values of Vo=1, C=1, R=1 and L=R^2*C/4 and a very slightly higher value (1.0001 times) for L, for the under damped case.

Using Mathcad to evaluate the given expressions for i(t).

I experimented with the over damped case, earlier, (deriving the i peak formula I pasted to A.B.S.E) to solve for i peak) and verified that i peak based on the the given over damped i(t) converged to i peak given for the critical damped case as critical damping was approached from the over damped side.

So I think the over damped i(t) formula is at least consistent with the critical i peak formula. But I haven't yet derived the critical i peak formula from the critical damped i(t) formula.

I've lost track.

So does this lead you to suspect the formula for i(t) for the under damped case is in error?

Yes, indeed. You get the same numerical value from his web page formula as I do for the peak current in the underdamped case, and I think this formula, at least as I am able to read it, is wrong.

The formulas I posted mostly came from an application note published by Sprague concerning their energy storage capacitors.

------------------------------------------------------------------------ I also solved the problem by writing an expression for the current as:

I(s) = (Vo/L)/(s^2 + s*R/L + 1/L/C)

and looking up the appropriate Laplace transform. The result is:

i(t) = (Vo/L/(a-b))*(Exp(a*t)-Exp(b*t))]

where a and b are the roots of the denominator of the I(s) expression.

This can be whipped into the form of the expressions for the under and overdamped cases from the Sprague app note that I posted yesterday. The critically damped case requires taking limits or explicitly making the roots equal and then transforming.

tpeak can be found as: tp = (Ln(a/b) - 2*Pi*j*n)/(b-a), j=SQRT(-1)

What is neat about that expression is that if you have a calculator or PC program that can deal with complex numbers (including taking logs and exponentials of complex arguments), you succesively set the variable n to

0,1,2,... and this expression will give the times for the successive peaks (positive and negative) in the underdamped case; and with n=0, it will give the peak time for the overdamped case. Then you just substitute those times back into the expression for i(t) and you get the peak currents. So you can get numerical results without having to manipulate the expression for i(t) and possibly making a mistake in the algebra.

I've done all this and verified that the formulas from the Sprague app note that I posted yesterday are correct (or made minor corrections as necessary). The numerical results that you have gotten so far are the same as mine, and I'm confident the formulas I posted yesterday are correct.

I found this online simulator

and it seems to agree pretty well with the above equations. Mike

I just posted another version of the formula for the peak current in the over damped series RLC case on A.B.S.E. Here is a tip of the hat to John Woodgate for the suggestions.