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There are basically two questions I have for this circuit.

1) Which capacitor configuration is good? before fuse or after fuse

2) How do I limit my inrush current?

My thoughts: I am confused between weather to put filtering capacitor before the fuse or after the fuse? Putting a capacitor at the IN (Voltage in) pin of the voltage regulator(LM1085) is recommended by LTC. According to me, if I put my cap after the fuse, it will act like a short when I connect my 12V PC supply to my board. The inrush current could kill my fuse every-time I connect PC-supply. To me putting a cap before the fuse make sense but still, I have a feeling it could still act like a short during inrush and harm my PC-supply. What I want is a nice filtering at the IN pin of my regulator while limiting my inrush current. The circuit powered by my regulator (U2) has max 300 mA loading. Any suggestion of what I need to do here?

Here is the datasheet link for my:

capacitor-> http://industrial.panasonic.com/lecs/www-data/pdf/ABA0000/ABA0000CE2.pdf

fuse-> http://www.littelfuse.com/~/media/electronics/datasheets/fuses/littelfuse_fuse_154_154t_154l_154tl_datasheet.pdf.pdf

schematic

simulate this circuit – Schematic created using CircuitLab

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The 10uF cap can be near the regulator where it does the most good. The inrush current with that size of capacitor should be no problem for a 3.5A fuse.

The fuse you have selected is a fast acting fuse but the thermal mass of the fuse will unlikely respond in the short time that it takes to charge a 10uF capacitor.

You will always have a certain amount of series resistance in the wiring, connectors, PCB traces that also will help to limit inrush current. Plus under normal circumstances I suspect that you would cycle the mains power switch of the 12V supply to power down your circuit instead of direct connecting the +12V. In this case the supply will have a fairly lengthy rise time at its output that limits the inrush current to same levels.

For grins I ran a simulation assuming a wiring resistance of 0.1 ohm and the 12V supply coming up to full voltage in 10us. Under these conditions the inrush current is ~12A for the 10usec rise time of the supply (linear rise used). My estimation that under such conditions the fuse material may over heat and blow only if the switching duty cycle was faster than the fuse can cool down from a 10usec pulse.

Do note that in the past I have had first hand experience of seeing fuses crystallize and fail after years of service being subjected to inrush currents. That was on the rectified DC lines of a Cromemco S100 chassis that had enormous capacitors. Your 10uF caps would look like specks in comparison. The Cromemco fuse in question was the 30A fuse on the 8V rail as shown (in the photo here). Inside the back of the unit the associated capacitor was the large soup can in the (closer part of this image). That capacitor was a 130,000uF / 15V unit.

Now if you had something like a 4700uF capacitor then there may be more concern. In that case you may want to select a time delay SloBlo type fuse.

In some electronics devices where there is indeed a very large inrush current possible the equipment is designed with a low ohms resistive device in series with the input. As the device comes up a special circuit either detects when the input caps are charged or just waits some nominal delay and then activates a relay that shorts out the low resistance device.

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  • \$\begingroup\$ I understand that 10 µF cap will have much shorter time to charge up and during that very short period, my fuse does get expose to some what high current than it might be rated at. Do you think that this will degrade the fuse performance over the years although the fuse may not necessarily burn out? \$\endgroup\$ – dr3patel Jun 15 '15 at 6:02
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    \$\begingroup\$ I cannot say what the long term effect would be for your fuse. Fuse manufacturers usually do not provide this type of data either so if you want to know then I would encourage you to run some accelerated characterization tests. If you were to setup a test circuit that consists of a configuration equivalent to the input section of your product and with worst case current loading in the "My Circuit" part you can test to see what happens. Try driving the circuit with a FET switched input of the +12V that you think represents the worst case speed at which the supply would come on. (continued) \$\endgroup\$ – Michael Karas Jun 15 '15 at 11:04
  • \$\begingroup\$ (continued from above) Switch the circuit at say four times a second using an MCU board (such as a PI or Arduino). Make sure that when the FET goes off that input capacitors drain back to zero volts. It may take a slower switching time to achieve this. Lets say you find that once per second does the trick. Let this run for 30 days during which you will accumulate over 2.5 million cycles. After the test run take the fuse out of the test circuit and do comparative measurements on it compared to a new fresh fuse. Look at parameters such as resistance, inductance, visual appearance (continued) \$\endgroup\$ – Michael Karas Jun 15 '15 at 11:15
  • \$\begingroup\$ (continued from above) of the fuse element and other such things. You could also choose to add fuse blow comparisons as well. If you see any changes in the test fuse that leads to concern you would then want to retest with a larger sample size to conform the results. If you see little or no effects on the test fuse then you can rest assured that your fuse is OK for someone turning your product on an off millions of times. \$\endgroup\$ – Michael Karas Jun 15 '15 at 11:21
  • \$\begingroup\$ If you add a FET in front of the fuse, wouldn't the Rds act as a series resistor, limiting the current ever so slightly? I suppose you'd have to be picky with which one you choose, since 12V / 3.5A gives 3.4 ohm and any larger resistance than that would therefore create a false test? Depending on how accurate the 3.5A specification is, that is. Or maybe it is not an issue at all, given how low the Rds is on modern MOSFETs. \$\endgroup\$ – Lundin Jun 15 '15 at 11:40
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Ask yourself - if the capacitor fails short circuit, how will the fuse protect against overcurrent (fire, etc)?

That dictates the correct placement of fuse and capacitor.

(I have no problem with Michael's excellent answer, but I wanted to highlight the most important point more clearly. Not just for this example, but thinking about circuits in general.)

If the capacitor is large enough for inrush current to be a problem, then a slower blowing fuse would be needed - or some alternative soft-start circuit (e.g. a current limit in the 12V supply.)

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  • \$\begingroup\$ this is the correct answer. The fuse has to protect the whole circuit. \$\endgroup\$ – dmb May 31 '18 at 21:04
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I would also feel that at a capacitor of only 10uf would have a negletable effect upon your fuse in any way. Having the cap after the fuse in my opinion is also smart, because if it was to ever become leaky or directly shorted to ground it would pop your fuse, thus protecting your 12V supply from a possible overload. If a larger cap situation was ever needed for proper filtering in another application you could always place a NTC Thermistor in series just before the fuse. Usage and proper selection of such a device would require doing some math. Here's a link "if interested" to the basics of how a NTC Thermistor functions. Good luck.

https://en.m.wikipedia.org/wiki/Inrush_current_limiter

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1) Which capacitor configuration is good? before fuse or after fuse

After fuse, so that the cap provides a better bypass path for the regulator and so that the fuse provides protection in-case the capacitor fails short.

2) How do I limit my inrush current?

For a cap that small I wouldn't bother.

Lets try and put some numbers to the inrush problem.

Lets model the fuse as a resistor, the supply as an ideal voltage source and the capacitor as an ideal capacitor and assume that the supply is applied instantaneously. This should give us a worst case figure for the energy dissipated in the fuse.

The energy drawn from the supply is \$CV^2\$ the energy stored in the capacitor is \$\frac{1}{2}CV^2\$ so the energy dissipated in the fuse is \$\frac{1}{2}CV^2\$ which works out to 0.7 millijoules.

The part number for your actual fuse is 045303.5 (015403.5 is the fuse/fuseholder combo) the datasheet for the fuse is at https://www.mouser.co.uk/datasheet/2/240/Littelfuse_Fuse_451_453_Datasheet.pdf-468354.pdf

Unfortunately the datasheet doesn't directly tell us the minimum energy needed to blow the fuse but we can get a ballpark estimate of it from the time/current curves and the nominal resistance.

unfortunately the time/current curves for the different ratings are overlaid and thus hard to read but it seems that to blow the fuse in 1ms takes about 100A and that the resistance of the fuse is about 0.02 ohms.

\$E = I^2Rt\$

Thus the minimum energy required to blow the fuse is approximately 0.2 Joules. That is approximately 285 times higher than the energy your capacitor can possibly deliver to the fuse.

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