Fuse/supply capacitor calculations

[Richard Rasker]

Hello,

The problem seems simple: I have a circuit on 24V DC I'm trying to protect
with a 250mA SMD fuse(*) in the supply line. The circuit itself needs some
50mA at most, and several PCB tracks and components would constitute a fire
hazard at substantially more than a few hundred mA, so a 250mA fuse would
be ideal.

*: See
http://nl.farnell.com/1123250/electrical/product.us0?sku=BUSSMANN-3216FF250-R

The trouble is that the fuse often blows upon switching on the supply
voltage, probably due to the inrush current into a 220uF supply capacitor
behind the fuse.

The fuse has a typical I^2t value of 0.000084, and a cold series resistance
of 4 ohms (including wiring etcetera), so one would expect a worst case
inrush current of 24/4=6A at the moment of switching on; at a constant 6A,
charge time is some 0.9 milliseconds -- and as the datasheet specifies a
hold time of 1 millisecond at only 1.3A, it's not surprising that the fuse
should blow.

The point is, however, that this 6A very rapidly diminishes as the voltage
over the capacitor rises, following an e-power curve, so the actual I^2t
load on the capacitor is lower.
First the theory question: Is there an easy calculation providing a link
between a diminishing inrush current and an I^2t value?

The practical question is of course how to solve this particular problem.
There appear to be no slow fuses in 1206 SMD cases, so I guess I'll have to
resort to using a bigger type of fuse. I could also use a smaller capacitor
(down to some 22uF) -- but the inrush current would stay the same, only
with a correspondingly shorter charge time. So I'd need really need to know
(calculate?) what happens to the fuse in these cases.
But perhaps someone can come up with another good suggestion?

And oh: I have done some experiments with SMD resistors, but these turn out
to be amazingly resilient to excess power -- I had a 1.5 ohm 1206 resistor
desoldering itself after a minute or so of carrying almost 1 ampere of
current, with only a minor increase in resistance. And at 2 amperes, it
glowed red for several seconds before failing -- but by then I guess the
charred PCB could take over the role of conductor.

Thanks in advance,

Richard Rasker

 

[John O'Flaherty]

Hello,

The problem seems simple: I have a circuit on 24V DC I'm trying to protect
with a 250mA SMD fuse(*) in the supply line. The circuit itself needs some
50mA at most, and several PCB tracks and components would constitute a fire
hazard at substantially more than a few hundred mA, so a 250mA fuse would
be ideal.

*: Seehttp://nl.farnell.com/1123250/electrical/product.us0?sku=BUSSMANN-321...

The trouble is that the fuse often blows upon switching on the supply
voltage, probably due to the inrush current into a 220uF supply capacitor
behind the fuse.

The fuse has a typical I^2t value of 0.000084, and a cold series resistance
of 4 ohms (including wiring etcetera), so one would expect a worst case
inrush current of 24/4=6A at the moment of switching on; at a constant 6A,
charge time is some 0.9 milliseconds -- and as the datasheet specifies a
hold time of 1 millisecond at only 1.3A, it's not surprising that the fuse
should blow.

The point is, however, that this 6A very rapidly diminishes as the voltage
over the capacitor rises, following an e-power curve, so the actual I^2t
load on the capacitor is lower.
First the theory question: Is there an easy calculation providing a link
between a diminishing inrush current and an I^2t value?

The practical question is of course how to solve this particular problem.
There appear to be no slow fuses in 1206 SMD cases, so I guess I'll have to
resort to using a bigger type of fuse. I could also use a smaller capacitor
(down to some 22uF) -- but the inrush current would stay the same, only
with a correspondingly shorter charge time. So I'd need really need to know
(calculate?) what happens to the fuse in these cases.
But perhaps someone can come up with another good suggestion?

And oh: I have done some experiments with SMD resistors, but these turn out
to be amazingly resilient to excess power -- I had a 1.5 ohm 1206 resistor
desoldering itself after a minute or so of carrying almost 1 ampere of
current, with only a minor increase in resistance. And at 2 amperes, it
glowed red for several seconds before failing -- but by then I guess the
charred PCB could take over the role of conductor.

At only 50 mA, maybe you could use a resistor _and_ a fuse, along with
cutting the capacitor by a factor of ten, if you can actually stand
that. Seems like a much smaller capacitor could also have a higher
ESR, and so a lower peak current.

 

[John O'Flaherty]

At only 50 mA, maybe you could use a resistor _and_ a fuse, along with
cutting the capacitor by a factor of ten, if you can actually stand
that. Seems like a much smaller capacitor could also have a higher
ESR, and so a lower peak current.

About the theoretical part, it's an RC circuit, and the resistor
(fuse) voltage is
V * e^(-t/(RC)). (The original 24V times a factor that goes from 1 to
zero as time goes from zero to infinity).
The current is that divided by the resistance R. The I^2t should be
integral ( (V/R*e^(-t/RC))^2 dt, t = 0 to infinity).

With V = 24, R = 4ohms, C = 220uF, this is
int( (24V/4ohms * e^(-t/(4ohm*220uF)) ^2 dt, t = 0 to infinity),
which my TI-89 says is .01584 amps squared * seconds. Or square amps?
With 22 uF, it's .001584. With 220 uF and an additional 4 ohm resistor
(total 8 ohms), you get .00792 for I^2T. With 22 uF and 8ohms total,
you get 0.000792. The value scales linearly with capacitance, and non-
linearly with resistance.

This ignores any change in resistance of the parts due to heating,
which would depend on their physical description.

 

[John O'Flaherty]

About the theoretical part, it's an RC circuit, and the resistor
(fuse) voltage is
V * e^(-t/(RC)). (The original 24V times a factor that goes from 1 to
zero as time goes from zero to infinity).
The current is that divided by the resistance R. The I^2t should be
integral ( (V/R*e^(-t/RC))^2 dt, t = 0 to infinity).

With V = 24, R = 4ohms, C = 220uF, this is
int( (24V/4ohms * e^(-t/(4ohm*220uF)) ^2 dt, t = 0 to infinity),
which my TI-89 says is .01584 amps squared * seconds. Or square amps?
With 22 uF, it's .001584. With 220 uF and an additional 4 ohm resistor
(total 8 ohms), you get .00792 for I^2T. With 22 uF and 8ohms total,
you get 0.000792. The value scales linearly with capacitance, and non-
linearly with resistance.

This ignores any change in resistance of the parts due to heating,
which would depend on their physical description.

Sorry to stutter. It scales linearly up with capacitance, and down
with resistance, unless I've done something wrong.

 

[[email protected]]

Hello,

The problem seems simple: I have a circuit on 24V DC I'm trying to protect
with a 250mA SMD fuse(*) in the supply line. The circuit itself needs some
50mA at most, and several PCB tracks and components would constitute a fire
hazard at substantially more than a few hundred mA, so a 250mA fuse would
be ideal.

*: Seehttp://nl.farnell.com/1123250/electrical/product.us0?sku=BUSSMANN-321...

The trouble is that the fuse often blows upon switching on the supply
voltage, probably due to the inrush current into a 220uF supply capacitor
behind the fuse.

So put the fuse after the cap.

 

[John O'Flaherty]

Sorry to stutter. It scales linearly up with capacitance, and down
with resistance, unless I've done something wrong.

And I did something wrong - I forgot to multiply by t in the integral
of I^2T. The result is independent of resistance, and depends only on
the capacitance, which makes sense, since the capacitance charges as
current times time.. The higher the R, the lower the current but the
longer the time. The value of I^2T for 24 V and 220 uF is 7*10^-6.
So, with the same capacitor but a larger series resistance, the fuse
is able to dissipate the heat, but the energy passed through is the
same.
The only thing worse than being wrong is staying wrong. Sorry I can't
help on the theoretical question!

 

[Wimpie]

Hello,

The problem seems simple: I have a circuit on 24V DC I'm trying to protect
with a 250mA SMD fuse(*) in the supply line. The circuit itself needs some
50mA at most, and several PCB tracks and components would constitute a fire
hazard at substantially more than a few hundred mA, so a 250mA fuse would
be ideal.

*: Seehttp://nl.farnell.com/1123250/electrical/product.us0?sku=BUSSMANN-321...

The trouble is that the fuse often blows upon switching on the supply
voltage, probably due to the inrush current into a 220uF supply capacitor
behind the fuse.

The fuse has a typical I^2t value of 0.000084, and a cold series resistance
of 4 ohms (including wiring etcetera), so one would expect a worst case
inrush current of 24/4=6A at the moment of switching on; at a constant 6A,
charge time is some 0.9 milliseconds -- and as the datasheet specifies a
hold time of 1 millisecond at only 1.3A, it's not surprising that the fuse
should blow.

The point is, however, that this 6A very rapidly diminishes as the voltage
over the capacitor rises, following an e-power curve, so the actual I^2t
load on the capacitor is lower.
First the theory question: Is there an easy calculation providing a link
between a diminishing inrush current and an I^2t value?

The practical question is of course how to solve this particular problem.
There appear to be no slow fuses in 1206 SMD cases, so I guess I'll have to
resort to using a bigger type of fuse. I could also use a smaller capacitor
(down to some 22uF) -- but the inrush current would stay the same, only
with a correspondingly shorter charge time. So I'd need really need to know
(calculate?) what happens to the fuse in these cases.
But perhaps someone can come up with another good suggestion?

And oh: I have done some experiments with SMD resistors, but these turn out
to be amazingly resilient to excess power -- I had a 1.5 ohm 1206 resistor
desoldering itself after a minute or so of carrying almost 1 ampere of
current, with only a minor increase in resistance. And at 2 amperes, it
glowed red for several seconds before failing -- but by then I guess the
charred PCB could take over the role of conductor.

Thanks in advance,

Richard Rasker
--http://www.linetec.nl/

Hi Richard,

I2t = 0.5*(Ipeak)^2*(36% decay time).

So faster charging reduces the charging time, but in the end
increases I2t. I assume that you switch from a low output impedance
source. Is there any room/space/budget to add a current limiter (that
"consumes" just 100mV under normal use), or a dV/dt limitation?

Best regards,

Wim
PA3DJS
www.tetech.nl

 

[John O'Flaherty]

And I did something wrong - I forgot to multiply by t in the integral
of I^2T. The result is independent of resistance, and depends only on
the capacitance, which makes sense, since the capacitance charges as
current times time.. The higher the R, the lower the current but the
longer the time. The value of I^2T for 24 V and 220 uF is 7*10^-6.
So, with the same capacitor but a larger series resistance, the fuse
is able to dissipate the heat, but the energy passed through is the
same.
The only thing worse than being wrong is staying wrong. Sorry I can't
help on the theoretical question!

On the off chance that anyone is still following my string of self-
replies, I have to correct myself yet another time. Integration across
time is already multiplying by time, and my original calculation of
I2T, and the formulas, were correct. The total I*t to charge a
capacitor is fixed, but the I^2*t isn't.
I apologize to everyone for my dithering.

 

[Dave Platt]

The practical question is of course how to solve this particular problem.
There appear to be no slow fuses in 1206 SMD cases, so I guess I'll have to
resort to using a bigger type of fuse. I could also use a smaller capacitor
(down to some 22uF) -- but the inrush current would stay the same, only
with a correspondingly shorter charge time. So I'd need really need to know
(calculate?) what happens to the fuse in these cases.
But perhaps someone can come up with another good suggestion?

How about a Polyswitch self-resetting thermal fuse?

Tyco Electronics makes some of these in a 1206 size. As one example,
their NANOSMDC020F-2 part will sustain a current flow 200 mA
indefinitely, will trip at 420 mA, and works at up to 24 VDC.

Other vendors of polymer self-resetting PTC overcurrent protectors may
have equivalent or similar parts.

 

[Richard Rasker]

[snip]

At only 50 mA, maybe you could use a resistor _and_ a fuse, along with
cutting the capacitor by a factor of ten, if you can actually stand
that. Seems like a much smaller capacitor could also have a higher
ESR, and so a lower peak current.

I especially chose a low ESR cap of a relatively high value because it's
hooked up to an SMPS, but I think I can get away with 22uF.

About the theoretical part, it's an RC circuit, and the resistor
(fuse) voltage is
V * e^(-t/(RC)). (The original 24V times a factor that goes from 1 to
zero as time goes from zero to infinity).
The current is that divided by the resistance R. The I^2t should be
integral ( (V/R*e^(-t/RC))^2 dt, t = 0 to infinity).

I suspected out so much, although I hadn't started on calculations yet.

With V = 24, R = 4ohms, C = 220uF, this is
int( (24V/4ohms * e^(-t/(4ohm*220uF)) ^2 dt, t = 0 to infinity),
which my TI-89 says is .01584 amps squared * seconds. Or square amps?

I'd say amps squared seconds. I think I'll do the math myself as well, if
only to brush up on my math skills, but this 0.016 is indeed two hundred
times higher than the specified I^2t value of the fuse. This, by the way,
is at least an order of magnitude lower than the I^2t of regular glass
fuses, even the fast ones (which also explains why I never ran into this
problem before, having built hundreds of fuse-protected circuits ...)

With 22 uF, it's .001584. With 220 uF and an additional 4 ohm resistor
(total 8 ohms), you get .00792 for I^2T. With 22 uF and 8ohms total,
you get 0.000792. The value scales linearly with capacitance, and non-
linearly with resistance.

This ignores any change in resistance of the parts due to heating,
which would depend on their physical description.

Well, thanks a lot for your elaborate reply (and all the followups .
I'll probably just take a slower type of fuse, perhaps combined with a
smaller capacitor.

Thanks again, best regards,

Richard Rasker

 

[Richard Rasker]

How about a Polyswitch self-resetting thermal fuse?

I have looked at those, but their popularity (and thus availability) seems
to be waning -- all the types offered by Farnell, for instance, are no
longer in production. But thanks for the suggestion anyway.

Richard Rasker

 

[Richard Rasker]

So put the fuse after the cap.

Thanks, but that won't do. The techical requirements for this circuit are
that the fuse must be placed before anything else, e.g. to prevent trouble
if the supply is hooked up the wrong way round.

Richard Rasker

 

[John O'Flaherty]

Hello,
The problem seems simple: I have a circuit on 24V DC I'm trying to
protect with a 250mA SMD fuse(*) in the supply line. The circuit itself
needs some 50mA at most, and several PCB tracks and components would
constitute a fire hazard at substantially more than a few hundred mA,
so a 250mA fuse would be ideal. ...
The trouble is that the fuse often blows upon switching on the supply
voltage, probably due to the inrush current into a 220uF supply
capacitor behind the fuse.
The fuse has a typical I^2t value of 0.000084, and a cold series
resistance of 4 ohms

[snip]

At only 50 mA, maybe you could use a resistor _and_ a fuse, along with
cutting the capacitor by a factor of ten, if you can actually stand
that. Seems like a much smaller capacitor could also have a higher
ESR, and so a lower peak current.

I especially chose a low ESR cap of a relatively high value because it's
hooked up to an SMPS, but I think I can get away with 22uF.

About the theoretical part, it's an RC circuit, and the resistor
(fuse) voltage is
V * e^(-t/(RC)). (The original 24V times a factor that goes from 1 to
zero as time goes from zero to infinity).
The current is that divided by the resistance R. The I^2t should be
integral ( (V/R*e^(-t/RC))^2 dt, t = 0 to infinity).

I suspected out so much, although I hadn't started on calculations yet.

With V = 24, R = 4ohms, C = 220uF, this is
int( (24V/4ohms * e^(-t/(4ohm*220uF)) ^2 dt, t = 0 to infinity),
which my TI-89 says is .01584 amps squared * seconds. Or square amps?

I'd say amps squared seconds. I think I'll do the math myself as well, if
only to brush up on my math skills, but this 0.016 is indeed two hundred
times higher than the specified I^2t value of the fuse. This, by the way,
is at least an order of magnitude lower than the I^2t of regular glass
fuses, even the fast ones (which also explains why I never ran into this
problem before, having built hundreds of fuse-protected circuits ...)

With 22 uF, it's .001584. With 220 uF and an additional 4 ohm resistor
(total 8 ohms), you get .00792 for I^2T. With 22 uF and 8ohms total,
you get 0.000792. The value scales linearly with capacitance, and non-
linearly with resistance.

This ignores any change in resistance of the parts due to heating,
which would depend on their physical description.

Well, thanks a lot for your elaborate reply (and all the followups .
I'll probably just take a slower type of fuse, perhaps combined with a
smaller capacitor.

Thanks again, best regards,

You're welcome. I note that Wimpie's reply is exactly right (and
simpler), with R*C as the 36% discharge time - it's exactly what that
integral evaluates to. It can also be expressed as
..5* V^2 * C/R, which shows the direct linear dependence on C and
inverse on R.

 

[Richard Rasker]

Hello,

The problem seems simple: I have a circuit on 24V DC I'm trying to
protect with a 250mA SMD fuse(*) in the supply line.

[snip]

Hi Richard,

I2t = 0.5*(Ipeak)^2*(36% decay time).

So faster charging reduces the charging time, but in the end
increases I2t.

With the current's influence squared and the time factor linear, this means
that an increased current is far worse than a longer charging time.

I assume that you switch from a low output impedance source.

The 24 volts supply voltage is from a big ship's batteries, capable of
delivering a few thousand amps, so yes, I'd think we can call that a low
impedance source

Is there any room/space/budget to add a current limiter (that
"consumes" just 100mV under normal use), or a dV/dt limitation?

There is room for a current limiter, but I think the low-tech solution of
simply using a slower fuse is better.
I think I'll also run some tests with several types of fuses and several
capacitor values, having a relay switch between charging and discharging
the capacitor through the fuse once a second for an hour or so. This should
give some more information on a practical combination of values. If
anything interesting comes out of this, I'll post it here.

Thanks for your reply, best regards,

Richard Rasker

 

[Wimpie]

Hello,
The problem seems simple: I have a circuit on 24V DC I'm trying to
protect with a 250mA SMD fuse(*) in the supply line.

[snip]

Hi Richard,

I2t = 0.5*(Ipeak)^2*(36% decay time).

So faster charging reduces the charging time, but in the end
increases I2t.

With the current's influence squared and the time factor linear, this means
that an increased current is far worse than a longer charging time.

I assume that you switch from a low output impedance source.

The 24 volts supply voltage is from a big ship's batteries, capable of
delivering a few thousand amps, so yes, I'd think we can call that a low
impedance source

Is there any room/space/budget to add a current limiter (that
"consumes" just 100mV under normal use), or a dV/dt limitation?

There is room for a current limiter, but I think the low-tech solution of
simply using a slower fuse is better.
I think I'll also run some tests with several types of fuses and several
capacitor values, having a relay switch between charging and discharging
the capacitor through the fuse once a second for an hour or so. This should
give some more information on a practical combination of values. If
anything interesting comes out of this, I'll post it here.

Thanks for your reply, best regards,

Richard Rasker
--http://www.linetec.nl/

Hi Richard,

Did you also take into account that when the switching action occurs
frequently the expected life of your electrolytic capacitor may reduce
significantly.

Depending on the inductance of the feed line and resistive losses
(probably low), you may also get a voltage spike across the input
electrolytic capacitor (the better the capacitor, the higher the
spike). You might run a simplified (spice) simulation to see the
effect for yourself.

Some months ago a client had a similar fuse problem. He also wanted to
use an SMT fuse (space problem). I2t rating of the SMT fuses were far
too low (so they use a glass fuse now).

Best regards,

Wim
PA3DJS
www.tetech.nl

 

[Richard Rasker]

Hello,

The problem seems simple: I have a circuit on 24V DC I'm trying to
protect with a 250mA SMD fuse(*) in the supply line.
[snip]

Hi Richard,

I2t = 0.5*(Ipeak)^2*(36% decay time).

So faster charging reduces the charging time, but in the end
increases I2t.

Hi Richard,

Did you also take into account that when the switching action occurs
frequently the expected life of your electrolytic capacitor may reduce
significantly.

The circuit is switched on and off a few times a week at most, so I don't
think this will be a serious problem.

Depending on the inductance of the feed line and resistive losses
(probably low), you may also get a voltage spike across the input
electrolytic capacitor (the better the capacitor, the higher the
spike). You might run a simplified (spice) simulation to see the
effect for yourself.

Some months ago a client had a similar fuse problem. He also wanted to
use an SMT fuse (space problem). I2t rating of the SMT fuses were far
too low (so they use a glass fuse now).

I couldn't fit in a glass fuse, but inserting a series resistor of a dozen
ohms fixed it for me -- the resulting extra voltage drop is no problem, and
even after multiple thousand on/off cycles (with a 0.5Hz oscillator and a
relay clicking away for well over an hour) nothing failed.

Anyway, I've learned a lot about (SMT) fuses, and in particular that the
specification "fast" doesn't say anything, as the I^2t may vary orders of
magnitude among different types of "fast" fuses with the same current
rating.

Thanks once again, best regards,

Richard Rasker

 

[Paul Hovnanian P.E.]

Hello,

The problem seems simple: I have a circuit on 24V DC I'm trying to protect
with a 250mA SMD fuse(*) in the supply line. The circuit itself needs some
50mA at most, and several PCB tracks and components would constitute a fire
hazard at substantially more than a few hundred mA, so a 250mA fuse would
be ideal.

*: See
http://nl.farnell.com/1123250/electrical/product.us0?sku=BUSSMANN-3216FF250-R

The trouble is that the fuse often blows upon switching on the supply
voltage, probably due to the inrush current into a 220uF supply capacitor
behind the fuse.

The fuse has a typical I^2t value of 0.000084, and a cold series resistance
of 4 ohms (including wiring etcetera), so one would expect a worst case
inrush current of 24/4=6A at the moment of switching on; at a constant 6A,
charge time is some 0.9 milliseconds -- and as the datasheet specifies a
hold time of 1 millisecond at only 1.3A, it's not surprising that the fuse
should blow.

The point is, however, that this 6A very rapidly diminishes as the voltage
over the capacitor rises, following an e-power curve, so the actual I^2t
load on the capacitor is lower.
First the theory question: Is there an easy calculation providing a link
between a diminishing inrush current and an I^2t value?

I^2t is proportional to the capacitor's stored energy, 0.5QV^2, so this
theory suggests that the only solution is to reduce the cap size. In
reality, a fuse will pass infinite energy (given a long enough time), so
as you have concluded, one solution is to stretch out the charge time
somehow to get beyond the fuse I^2t region.

Is the device load constant enough that you could put a resistor ahead
of the cap and load? Then, bump up the supply voltage to compensate for
the IR drop. Or place a regulator downstream of the cap.

If you can't put up with the power loss through the resistor, there may
be another way. How is this device powered up? Is it possible to bypass
the fuse and power switch* with a high value resistance that will
pre-charge the cap? The resistor can be selected such that a short to
ground downstream of it will draw a low enough current so that
protecting it with a fuse is not necessary.

*Could be a series MOSFET delayed to power up once the cap voltage has
risen.