Continue to Site

# fuse resistor calculation

Status
Not open for further replies.

#### zuzu

##### Member level 3
Hello,

The fuse resistors are rated by "fuse power". How to translate this in real world, protecting a PSU by limiting inrush current?
So far, my assumption is (10uF)

Rlim = Vpeak/Iinrush = 240*sqrt(2)/38A = 10R

But how to compute required power of resistor?

What maximum operating current ?
Are you trying to reduce inrush current? If so use an ICL.
or protect it from surge voltage or what else are you concerned about?

Fuses are rated by Amp-seconds , Holding current rating and Trip current rating.
ICL's are NTC resistors called Inrush Current Limiters that are rated in R@25'C, Current - Steady State Max, and sometimes R@ current
They are opposite to Polyfuses (PTC) which quickly trip when excess current is reached after a thermal rise to 85'C

zuzu

### zuzu

Points: 2
The normal current is low, about 10mA. I want a fused resistor since is much cheaper, it will current limit and both fuse the protected device.

The only problem is to determine it's required power properly.

The normal current is low, about 10mA.

The only problem is to determine it's required power properly.

RC time constant is τ=10Ω*10µF=100µs . For an approximation, we keep the initial voltage constant (which is a worst case scenario). The anti-log decay from I0=38A to I=10mA takes T=-τ*ln(I/I0)=-τ*ln(10mA/38A) ≈ 800µs . Taking a linear current decay vs. time instead of the anti-log decay provides a worst case electrical energy of (I0/2)2*R*T ≦ 3Ws. Any resistor ≧ ½W can stand this energy occurring during less than a millisecond - more important are sufficiently thick connection wires which can support the initial current of 38A. I'd use a 2W resistor.

Points: 2

### nguyenxuansy

Points: 2
The term "fuse resistor" normally refers to a safety feature, a resistor that will fail open circuit under certain overlload conditions without catching fire (or burning the PCB).

I'm not sure if this is what you want?

zuzu

### zuzu

Points: 2
We want just a cheaper solution than NTC plus fuse.

RC time constant is τ=10Ω*10µF=100µs . For an approximation, we keep the initial voltage constant (which is a worst case scenario). The anti-log decay from I0=38A to I=10mA takes T=-τ*ln(I/I0)=-τ*ln(10mA/38A) ≈ 800µs . Taking a linear current decay vs. time instead of the anti-log decay provides a worst case electrical energy of (I0/2)2*R*T ≦ 3Ws. Any resistor ≧ ½W can stand this energy occurring during less than a millisecond - more important are sufficiently thick connection wires which can support the initial current of 38A. I'd use a 2W resistor.

This is the problem The 2.5W resistor we use just blows immediately.

Initially, I estimate the power based on inrush energy for charging capacitor ~ 0.57 Joules. Compared to equivalent energy deduced from resistor graph (power vs. time blow) take some margin and I still was in 10% range.

Then, I tried more accurate calculation. I plotted resistor power starting from 0 (max. current 38A) and take the time when power decays to 2W, so under resistor withstand capability. The medium energy was computed as averaged power on this time 440uS and guess what, is ~0.528 Joules so almost the same energy as initial calculation.

We used Tyco FCR series, 1206 model, 2.5W "fuse power"
https://www.farnell.com/datasheets/1770680.pdf

But datasheet isn't clear about "fuse power" as impulse capability. I just assume that if at 2W "fuse" will blow in 8 sec., we can apply 20W for 0.8 sec. for the same result.

Other explanation can be over voltage. In other manufacturer 1206 specs, this resistors are rated for 200V (DC or RMS) so when apply 240 (or 260 with tolerance) it can blow.

- - - Updated - - -

Hmm.... erikl, I suspect you're right!
The 1206 contacts just cannot handle the 38 Amps....

I'll ask my colleague to see at microscope the blown one.

I didn't expect that you are using a small 1206 resistor.

It simply can't handle the multi kW inrush power during the first microseconds. You can review datasheets of similar fusible resistors and see that they have an absolute pulse power limit, e.g. https://www.vishay.com/docs/20031/m25_si.pdf

... I just assume that if at 2W "fuse" will blow in 8 sec., we can apply 20W for 0.8 sec. for the same result.
It simply can't handle the multi kW inrush power during the first microseconds.

With my above (overestimated) approximation, the resistor will receive a medium power of P=(I0/2)2*R*T/T = 3.6kW for 0.8ms !

I think a linear projection (20kW for 0.8ms) is not valid.

The energy dissipated in a resistor in series with a capacitor dissipates energy equal to the charged capacitor energy or 1/2 CV^2. Using the RC charging time constant you can then determine the average power dissipated in the resistor over this charging time (using approximately two time-constants for the time period).

A problem comes when you try to extrapolate to timescales
shorter than the resistor body mass's thermal time constant.
In short time, only the resistor film itself (a much smaller
substrate and encapsulation (if any) slower, and possibly
too slow for the film to survive the heat spike.

You might like a look at thick film vs thin film resistors.
Perhaps you will find a useful difference in short-pulse
power handling.

There are also metal-only surface mount resistors
(used for current sense on high power stuff) which
would be much more robust (if you need very low
resistance - like sub-ohm.

You don't want a fuse as an inrush current limiter
because fuses only work once while inrush current
happens at every cold start.

0.45 Ws can be easily absorbed by a 1206 chip resistor if it's allowed to heat the ceramic body uniformly. The problem is that according to 100 µs time constant, only the resistance film and a thin ceramic layer must handle the energy in first instance. Apparently it can't.

A wirewound resistor can be expected to have sufficient thermal capcity in the resistive material itself and absorb the energy.

Status
Not open for further replies.