R

#### Richard Rasker

- Jan 1, 1970

- 0

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