[EE] Input fuse protection circuit

[Rob]

Hello Group,
  Hate to display my ignorance but...
  I have a small PIC based device and have sold a small number of them
and have not had any problems until possibly now. I received a report of
a blown fuse and that has made me want to understand why that happened
since there was not an apparent obvious reason for this, or so I
thought. My project uses about 20mA while idle and can jump up to maybe
350mA for brief moments when an audio amplifier becomes active and set
to full volume but typically runs around 150mA under normal peak spikes.
Currently, the input goes directly through a 750mA fast blow soldered
pico fuse to a 1N4007 for reverse voltage protection. Then into a low
pass pi filter consisting of 2 0.1uF caps and a 2.2uH inductor to
hopefully filter out any RF. This then goes into a 100uF/50v cap for
filtering. This then is fed into an LM2940 5 volt regulator with 47uF
and 0.1 caps on the output. I typically feed this device with a
9vDC/800mA switching wall adapter but can also be connected to a high
current 13.8vDC supply or even a car battery at times. I selected the
750mA fuse because it is about double maximum current draw and just
under the wall wart capability so I don't know if that was the right
choice. Also, I never considered the potential for high inrush current
to charge the 100uF cap when connected to a high current power supply.
Also when I first built this, I was also wanting to keep the input
resistance low in case it was being used out in the field on a small
capacity battery so wanted to keep the input voltage drop relatively low
which of course after considering it now, increases the inrush current
capability.
  I don't have a digital storage scope so can't easily measure the
inrush current and I don't know if a min/max function on my Fluke 189 is
fast enough for capturing an event like this. So if you don't mind, I
have a few questions regarding this "design".
  Initially until the blown fuse report, I wanted to know if there is a
better choice for the reverse diode that would work for this circuit
with a lower forward voltage drop? Probably some schottky diode but now
realize that would also contribute to  a higher inrush current so should
probably stick with the 1N4007.
  The current fuse is a soldered infuse so replacement is a pain. I
tried looking into a PTC resettable type fuse with approximately the
same parameters but all I could find that could handle a sustained
current of around 500mA has an opening current rating of around 1.5
amps. That seems a bit high for this device since the wall wart
allegedly won't even supply that much current. Plus the higher series
resistance will add to the input voltage drop.
  I'm hoping I can just replace the fast fuse with an equivalent slow
blow version but I don't know if that will survive in all circumstances.
At a minimum, I'm thinking I'll make room on the PCB for a blade fuse
holder so it can be more easily replaced by the user. But a quick look
didn't turn up many automotive blade style fuses with a 750mA slow blow
rating.
  Other thoughts are to use a smaller input filter cap where I was just
thinking bigger is better. Maybe use a 47uF on the input as well as what
I'm using on the output of the regulator.
  I've also considered moving the fuse to the other side of the input
capacitor but don't know the wisdom of that location.
  I realize not knowing the actual inrush current I'm dealing with at
this point makes this difficult to accurately design for but I'm sure
you guys have delft with this many times already and can offer some
advice as far a just better general design principles. And if
recommending any parts, they need to be through hole.
Thanks in advance,
Rob

[Richard Prosser]

You could try something like theMAX17608 - other manufacturers also produce
similar units.
Reverse Protection, programmable Inrush limit, +/-60V max input.

I'd still use a fuse though (slow blow) or fusable tracks. If the fuse
fails, the unit's probably a write-off anyway.

RP

Post by Rob
Hello Group,
Hate to display my ignorance but...
I have a small PIC based device and have sold a small number of them
and have not had any problems until possibly now. I received a report of
a blown fuse and that has made me want to understand why that happened
since there was not an apparent obvious reason for this, or so I
thought. My project uses about 20mA while idle and can jump up to maybe
350mA for brief moments when an audio amplifier becomes active and set
to full volume but typically runs around 150mA under normal peak spikes.
Currently, the input goes directly through a 750mA fast blow soldered
pico fuse to a 1N4007 for reverse voltage protection. Then into a low
pass pi filter consisting of 2 0.1uF caps and a 2.2uH inductor to
hopefully filter out any RF. This then goes into a 100uF/50v cap for
filtering. This then is fed into an LM2940 5 volt regulator with 47uF
and 0.1 caps on the output. I typically feed this device with a
9vDC/800mA switching wall adapter but can also be connected to a high
current 13.8vDC supply or even a car battery at times. I selected the
750mA fuse because it is about double maximum current draw and just
under the wall wart capability so I don't know if that was the right
choice. Also, I never considered the potential for high inrush current
to charge the 100uF cap when connected to a high current power supply.
Also when I first built this, I was also wanting to keep the input
resistance low in case it was being used out in the field on a small
capacity battery so wanted to keep the input voltage drop relatively low
which of course after considering it now, increases the inrush current
capability.
I don't have a digital storage scope so can't easily measure the
inrush current and I don't know if a min/max function on my Fluke 189 is
fast enough for capturing an event like this. So if you don't mind, I
have a few questions regarding this "design".
Initially until the blown fuse report, I wanted to know if there is a
better choice for the reverse diode that would work for this circuit
with a lower forward voltage drop? Probably some schottky diode but now
realize that would also contribute to a higher inrush current so should
probably stick with the 1N4007.
The current fuse is a soldered infuse so replacement is a pain. I
tried looking into a PTC resettable type fuse with approximately the
same parameters but all I could find that could handle a sustained
current of around 500mA has an opening current rating of around 1.5
amps. That seems a bit high for this device since the wall wart
allegedly won't even supply that much current. Plus the higher series
resistance will add to the input voltage drop.
I'm hoping I can just replace the fast fuse with an equivalent slow
blow version but I don't know if that will survive in all circumstances.
At a minimum, I'm thinking I'll make room on the PCB for a blade fuse
holder so it can be more easily replaced by the user. But a quick look
didn't turn up many automotive blade style fuses with a 750mA slow blow
rating.
Other thoughts are to use a smaller input filter cap where I was just
thinking bigger is better. Maybe use a 47uF on the input as well as what
I'm using on the output of the regulator.
I've also considered moving the fuse to the other side of the input
capacitor but don't know the wisdom of that location.
I realize not knowing the actual inrush current I'm dealing with at
this point makes this difficult to accurately design for but I'm sure
you guys have delft with this many times already and can offer some
advice as far a just better general design principles. And if
recommending any parts, they need to be through hole.
Thanks in advance,
Rob
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[RussellMc]

I'd not have thought a 100 uF would do that.
Energy content at say 15V is very small so inrush time very limited so
absolute energy in fuse small.

But, based on ESR alone, peak inrush current can be large.

ESR of eg Panasonic FT V SMD 100 uF, 15V cap is 0.34 Ohm.

https://industrial.panasonic.com/cdbs/www-data/pdf/RDE0000/ABA0000C1240.pdf

Peak inrush current from a suitable source can notionally be tens of Amps -
for a very short period.
A power supply rated at say 1A MAY have an output capacitor able to source
much higher currents very briefly.

If you can tolerate the drop from a 1N400x diode then maybe a small amount
of added series resistance may help.
Changing to a Schottky diode recovers part of the drop - but odds are a
small resistor that may make the difference will not add enough drop to
affect normal operation.

Using say 3R3 with about 350 mA usual peak adds about 1V drop - but
increases your effective ESR by a factor of about 10 - and drops your peak
inrush to around 4A at 12V. All very approximate but perhaps worth looking
at the several contributors with a scope.
How much instantaneous current will the supply source?
What is the peak worst-case real world inrush current?
Does the fuse have a peak current duration versus blow time graph or
formula?


Russell

[Sean Breheny]

It seems that my earlier post did not make it to the server - I apologize
if this is a duplicate (but from the archive it appears not)

Hi Rob,

A few things to consider:

1) Most fuses have I2t (or current squared times time) ratings. There
should be a melting I2t and a clearing I2t. This is the integral of current
squared over time - which for a roughly fixed fuse internal resistance
represents energy. The melting one is roughly the amount needed to begin
melting the fuse element (which would be permanent damage if you are trying
NOT to blow the fuse) and the clearing one is the amount required to not
only melt the fuse element completely but also to blow the molten metal out
of the gap and result in a truly open fuse.

When charging a cap through a resistor from a constant voltage source, the
I2t is equal to 0.5*(1/R)*C*V^2 where R is the resistance in ohms, C is the
capacitance in Farads, and V is the source voltage. For your case of
charging a 100uF cap through 0.34 ohms (as Russel found) from 15V, the I2t
is 0.0331 amps^2 seconds. Looking on Digikey at 750mA fast picofuses, I see
two values for melting I2t: 0.041 and 0.15. Notice that both of these are
higher than the value in your case. However, the 0.041 value is
uncomfortably close, considering that you also have to charge the cap on
the output side of the regulator (although that will contribute somewhat
less than its value would suggest because of the voltage drop of the
regulator).

To be comfortable that the inrush is not damaging the fuse, I would want to
see the actual I2t be less than 1/3rd of the melting value.

2) The current rating of a fuse is the highest current at which it is
guaranteed NOT to blow at 25 deg C ambient temperature. Fuses have a "no
man's land" from the rating up to about 3 times the rating. The exact
factor depends on fuse design and testing and can be as low as 1.7 and as
high as 3.5 from what I've seen. Current in this range should not be
applied to the fuse for more than a fraction of a second (maybe longer for
a slow blow fuse). It may or may not blow the fuse. It may fail to blow it
but it may leave it damaged with different characteristics. It could make
the fuse get so hot that it melts the fuseholder or even ignites something
in the vicinity of the fuse. It may blow the fuse but not clear it
completely, leaving it to act like a resistor - even as high as 10k ohms or
more.

3) Fuses also have a maximum interrupting current rating - if the circuit
can source more than this then the fuse may arc over or explode in a short
circuit. The AC and DC values of this current may differ. The value may
also depend on the maximum voltage applied.

4) The voltage rating of the fuse is often very different for AC vs DC
because DC will sustain a continuous arc while AC has automatic current
zero-crossings to help interrupt an arc

Points number 1 and 2 are the most relevant to your case. I included 3 and
4 because I was on my soapbox ;)

In summary, your design challenge is to keep your worst-case inrush I2t
well below the fuse's melting I2t AND also to ensure that you are not
relying on the fuse to interrupt current below about 3X the rated fuse
current AND that you will not subject the fuse to current above the rating
but below 3X the rating (other than inrush or other rare brief events).

One more caveat I've been bitten by - resistors also have a maximum I2t
which they can absorb without damage. It can be quite small for SMD
resistors. If you are relying on a resistor to limit inrush current AND the
peak inrush would be significantly more than the resistor could handle
continuously THEN you should consult the resistor datasheet to see if there
are pulse ratings (either a curve or an I2t value). If not, choose a
different resistor which has pulse withstanding ratings. I have been bitten
by this - it is real. Your resistor may handle 100 inrushes and fail open
on the 101st.

Sean

Post by Rob
Hello Group,
Hate to display my ignorance but...
I have a small PIC based device and have sold a small number of them
and have not had any problems until possibly now. I received a report of
a blown fuse and that has made me want to understand why that happened
since there was not an apparent obvious reason for this, or so I
thought. My project uses about 20mA while idle and can jump up to maybe
350mA for brief moments when an audio amplifier becomes active and set
to full volume but typically runs around 150mA under normal peak spikes.
[SNIP]

[RussellMc]

On Wed, 29 Aug 2018 at 16:17, Sean Breheny <***@cornell.edu> wrote:

...
When charging a cap through a resistor from a constant voltage source, the
I2t is equal to 0.5*(1/R)*C*V^2 where R is the resistance in ohms, C is the
capacitance in Farads, and V is the source voltage. For your case of
charging a 100uF cap through 0.34 ohms (as Russel found) from 15V, the I2t
is 0.0331 amps^2 seconds. Looking on Digikey at 750mA fast picofuses, I see
two values for melting I2t: 0.041 and 0.15. Notice that both of these are
higher than the value in your case. However, the 0.041 value is
uncomfortably close,

Adding a say 1 Ohm series input resistor (as suggested in my prior post)
would give you the i^2t margin that Sean recommends and add only about 0.3
V drop under your typical peak current. Adding slightly more (say 1.5 Ohm)
would give you in excess of that margin of error for caps with any ESR
(including 0 Ohms).

Russell

[Sean Breheny]

I agree - but said resistor should be a pulse-withstanding type :)

Post by Sean Breheny
...
When charging a cap through a resistor from a constant voltage source, the
I2t is equal to 0.5*(1/R)*C*V^2 where R is the resistance in ohms, C is the
capacitance in Farads, and V is the source voltage. For your case of
charging a 100uF cap through 0.34 ohms (as Russel found) from 15V, the I2t
is 0.0331 amps^2 seconds. Looking on Digikey at 750mA fast picofuses, I see
two values for melting I2t: 0.041 and 0.15. Notice that both of these are
higher than the value in your case. However, the 0.041 value is
uncomfortably close,
Adding a say 1 Ohm series input resistor (as suggested in my prior post)
would give you the i^2t margin that Sean recommends and add only about 0.3
V drop under your typical peak current. Adding slightly more (say 1.5 Ohm)
would give you in excess of that margin of error for caps with any ESR
(including 0 Ohms).
Russell
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[AB Pearce - UKRI STFC]

Post by Sean Breheny
It seems that my earlier post did not make it to the server - I apologize
if this is a duplicate (but from the archive it appears not)
Hi Rob,

Well, I'm glad you reposted that, as it is one of the most succinct treatises on fuses I have seen. Got some designing to do where this will be useful instead of using "rule of thumb".

Thanks.

[Ruben Jönsson]

+1

/Ruben


Den Wed, 29 Aug 2018 09:43:51 +0000, AB Pearce - UKRI STFC skrev:

&gt; It seems that my earlier post did not make it to the server - I apologize
&gt; if this is a duplicate (but from the archive it appears not)
&gt;
&gt; Hi Rob,
&gt;
&gt; A few things to consider:

Well, I'm glad you reposted that, as it is one of the most succinct treatises on fuses I have seen. Got some designing to do where this will be useful instead of using "rule of thumb".

Thanks.



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