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.

(snip)

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.