| Author |
Message |
Mike
Guest
|
 Posted:
Tue Nov 08, 2005 1:35 am Post subject:
How to calculate the peak inductor current? |
|
|
Could someone clairify these equations to calculate the peak current
reached in an inductor when a capacitor is discharged into it
The equations can be seen at
http://home.san.rr.com/nessengr/techdata/rlc/rlc.html
In particular I cannot make out the character that is part of the
exponent of e in the i(t) equations for all 3 cases of damping. It
looks like a t, but I'm not sure. Maybe e^((-R/2L)*t) ?
Also what does the small symbol that looks like a sinewave over the
omega symbol mean that is in the i(t) equation for the overdamped
case?
Thanks
Mike
|
|
| Back to top |
|
 |
John Popelish
Guest
|
| Posted:
Tue Nov 08, 2005 8:51 am Post subject:
Re: How to calculate the peak inductor current? |
|
|
Mike wrote:
| Quote: | Could someone clairify these equations to calculate the peak current
reached in an inductor when a capacitor is discharged into it
The equations can be seen at
http://home.san.rr.com/nessengr/techdata/rlc/rlc.html
In particular I cannot make out the character that is part of the
exponent of e in the i(t) equations for all 3 cases of damping. It
looks like a t, but I'm not sure. Maybe e^((-R/2L)*t) ?
|
Yes. t, for time in seconds. These are expressions for some value in
terms of time.
| Quote: | Also what does the small symbol that looks like a sinewave over the
omega symbol mean that is in the i(t) equation for the overdamped
case?
|
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. |
|
| Back to top |
|
|
The Phantom
Guest
|
| Posted:
Tue Nov 08, 2005 4:52 pm Post subject:
Re: How to calculate the peak inductor current? |
|
|
On Mon, 07 Nov 2005 21:51:50 -0500, John Popelish <jpopelish@rica.net>
wrote:
| Quote: | Mike wrote:
Could someone clairify these equations to calculate the peak current
reached in an inductor when a capacitor is discharged into it
The equations can be seen at
http://home.san.rr.com/nessengr/techdata/rlc/rlc.html
In particular I cannot make out the character that is part of the
exponent of e in the i(t) equations for all 3 cases of damping. It
looks like a t, but I'm not sure. Maybe e^((-R/2L)*t) ?
Yes. t, for time in seconds. These are expressions for some value in
terms of time.
Also what does the small symbol that looks like a sinewave over the
omega symbol mean that is in the i(t) equation for the overdamped
case?
It is defined as a function of R, L and C in the previous equation.
|
Looks to me like ~wo = wo * SQRT(-1)
| Quote: | It is the sine component of the decay
|
Wouldn't it be better to call it the "sinh" component?
| Quote: | , even though the real
exponential dominates the result, because it decays so much in a small
part of a cycle. |
|
|
| Back to top |
|
|
John Popelish
Guest
|
| Posted:
Tue Nov 08, 2005 5:35 pm Post subject:
Re: How to calculate the peak inductor current? |
|
|
The Phantom wrote:
| Quote: | On Mon, 07 Nov 2005 21:51:50 -0500, John Popelish <jpopelish@rica.net
(snip)
Also what does the small symbol that looks like a sinewave over the
omega symbol mean that is in the i(t) equation for the overdamped
case?
It is defined as a function of R, L and C in the previous equation.
Looks to me like ~wo = wo * SQRT(-1)
It is the sine component of the decay
Wouldn't it be better to call it the "sinh" component?
(snip) |
Yes. Thanks. |
|
| Back to top |
|
|
Mike
Guest
|
| Posted:
Wed Nov 09, 2005 12:25 am Post subject:
Re: How to calculate the peak inductor current? |
|
|
On Tue, 08 Nov 2005 10:31:20 -0500, John Popelish <jpopelish@rica.net>
wrote:
| Quote: | The Phantom wrote:
On Mon, 07 Nov 2005 21:51:50 -0500, John Popelish <jpopelish@rica.net
(snip)
Also what does the small symbol that looks like a sinewave over the
omega symbol mean that is in the i(t) equation for the overdamped
case?
It is defined as a function of R, L and C in the previous equation.
Looks to me like ~wo = wo * SQRT(-1)
It is the sine component of the decay
Wouldn't it be better to call it the "sinh" component?
(snip)
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 |
|
| Back to top |
|
|
John Popelish
Guest
|
| Posted:
Wed Nov 09, 2005 1:11 am Post subject:
Re: How to calculate the peak inductor current? |
|
|
Mike wrote:
| Quote: | Thanks, for the clarifications.
Is there a reason why there is no equation to calculate the peak
current for the overdamped case?
|
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. |
|
| Back to top |
|
|
The Phantom
Guest
|
| Posted:
Wed Nov 09, 2005 1:35 am Post subject:
Re: How to calculate the peak inductor current? |
|
|
On Tue, 08 Nov 2005 13:25:30 -0500, Mike <nomtrxspam@comcast.net> wrote:
| Quote: | On Tue, 08 Nov 2005 10:31:20 -0500, John Popelish <jpopelish@rica.net
wrote:
The Phantom wrote:
On Mon, 07 Nov 2005 21:51:50 -0500, John Popelish <jpopelish@rica.net
(snip)
Also what does the small symbol that looks like a sinewave over the
omega symbol mean that is in the i(t) equation for the overdamped
case?
It is defined as a function of R, L and C in the previous equation.
Looks to me like ~wo = wo * SQRT(-1)
It is the sine component of the decay
Wouldn't it be better to call it the "sinh" component?
(snip)
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
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)
-----------------------------
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. |
|
| Back to top |
|
|
John Popelish
Guest
|
| Posted:
Wed Nov 09, 2005 8:52 am Post subject:
Re: How to calculate the peak inductor current? |
|
|
The Phantom wrote:
| Quote: | 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?
|
I read it as pi.
| Quote: | 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)
-----------------------------
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.
|
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? |
|
| Back to top |
|
|
The Phantom
Guest
|
| Posted:
Wed Nov 09, 2005 9:35 am Post subject:
Re: How to calculate the peak inductor current? |
|
|
On Tue, 08 Nov 2005 21:52:52 -0500, John Popelish <jpopelish@rica.net>
wrote:
| Quote: | The Phantom wrote:
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?
I read it as pi.
|
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.
| Quote: |
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)
-----------------------------
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.
I tested the current versus time for the under damped and critical
cases, with the inductance nearly the same,
|
For this test, you should leave the inductance at 1 uH and vary the
resistance.
| Quote: | and got 4 times higher
current for the critically damped case.
|
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.
| Quote: |
Could a multiplier of 4 be missing from the underdamped case? |
|
|
| Back to top |
|
|
John Popelish
Guest
|
| Posted:
Wed Nov 09, 2005 11:41 pm Post subject:
Re: How to calculate the peak inductor current? |
|
|
The Phantom wrote:
| Quote: | John Popelish wrote:
The Phantom wrote:
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?
I read it as pi.
So does the complete expression look like:
Ipeak = (Vo/wo/L)*Exp(-R*Pi/(L*wo)) ?
|
That is what I think I am seeing.
| Quote: | When I use the values L = 1 uH, C = 1 uF, R = .2 ohms,
|
And Vo = 10 volts, I assume.
| Quote: | 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.
|
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)
| Quote: | I tested the current versus time for the under damped and critical
cases, with the inductance nearly the same,
For this test, you should leave the inductance at 1 uH and vary the
resistance.
and got 4 times higher
current for the critically damped case.
How did you test? Using the web page expressions?
|
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.
| Quote: | Using a simulator?
|
Using Mathcad to evaluate the given expressions for i(t).
| Quote: | 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?
|
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.
| Quote: | Did you try the values of R that I suggested?
|
I've lost track.
| Quote: | 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.
|
So does this lead you to suspect the formula for i(t) for the under
damped case is in error? |
|
| Back to top |
|
|
Mike
Guest
|
| Posted:
Thu Nov 10, 2005 1:35 am Post subject:
Re: How to calculate the peak inductor current? |
|
|
<snip>
| Quote: |
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)
-----------------------------
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.
|
I found this online simulator
http://www.oz.net/~coilgun/mark2/rlcsim.htm and it seems to agree
pretty well with the above equations.
Mike |
|
| Back to top |
|
|
The Phantom
Guest
|
| Posted:
Thu Nov 10, 2005 1:35 am Post subject:
Re: How to calculate the peak inductor current? |
|
|
On Wed, 09 Nov 2005 12:41:08 -0500, John Popelish <jpopelish@rica.net>
wrote:
| Quote: | The Phantom wrote:
John Popelish wrote:
The Phantom wrote:
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?
I read it as pi.
So does the complete expression look like:
Ipeak = (Vo/wo/L)*Exp(-R*Pi/(L*wo)) ?
That is what I think I am seeing.
When I use the values L = 1 uH, C = 1 uF, R = .2 ohms,
And Vo = 10 volts, I assume.
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.
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)
I tested the current versus time for the under damped and critical
cases, with the inductance nearly the same,
For this test, you should leave the inductance at 1 uH and vary the
resistance.
and got 4 times higher
current for the critically damped case.
How did you test? Using the web page expressions?
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 a simulator?
Using Mathcad to evaluate the given expressions for i(t).
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?
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.
Did you try the values of R that I suggested?
I've lost track.
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.
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. |
|
| Back to top |
|
|
John Popelish
Guest
|
| Posted:
Thu Nov 10, 2005 9:35 am Post subject:
Re: How to calculate the peak inductor current? |
|
|
John Popelish wrote:
| Quote: | Mike wrote:
Thanks, for the clarifications.
Is there a reason why there is no equation to calculate the peak
current for the overdamped case?
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.
|
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. |
|
| Back to top |
|
|
|
|
|
|
FAQ
Memberlist
Register
Profile
Log in to check your private messages
Log in