Resistor power rating

[dincz]

I'm about to build a Faultline and, as the pdf doesn't specify power ratings, I've chosen 1/4W resistors. I'm sure they'll be fine but I wonder about the 47ohm resistor in series with the 9V power coming into the PCB. I guess this is to limit current through the polarity protection diode. To be safe, the 47ohm would need to be rated at around 2W - or is it better to use a lower rating and have the resistor function as a fuse?

[kothoma]

http://www.catalinbread.com/SFT states 4mA draw.
P = I^2 x R = 0.004^2 x 47 = 0.000752 W

[dincz]

I meant that the 47ohm would need to be 2W to not burn up if reverse polarity power is accidentally connected. In normal use 1/4W would be more than enough.

[RobA]

You could tie a 1N5817 in series with the 47Ω resistor (solder the leads together and partially stand them up in the holes for the resistor). That way you would block the majority of reverse current and a 1/4W resistor should be able to handle any reverse current that gets through. It's safer for the power supply that way too.

[dincz]

No, there is already a diode to protect against reverse polarity but current is limited in this case by a 47ohm resistor. If the power is connected with reverse polarity, the current through that resistor will be about 200mA. That means the resistor will have to dissipate around 2 watts.

So my original question - is it better to use a 1/4 watt resistor that will act as a fuse, or up the rating to 2 watts?   

[RobA]

Yes there is already a diode there in a crowbar type configuration going from ground to the input rail. I meant to put a Schottky (1N5817) in series with the power input and the 47Ω resistor. This arrangement blocks the reverse flow of voltage. It doesn't short the voltage to ground, it blocks it. You use a Schottky because it has less voltage loss (~300mV). There is some current that will still leak from a reverse biased Schottky, so it is good to still have the crowbar diode in place. But, the voltage that gets through is much lower and the 1/4 watt 47Ω resistor should be fine in this arrangement.

[stecykmi]

a 2W resistor is significantly larger than a 1/4W so it may not fit the pad.

i would just use the 1/4W and hope whatever power supply you're using has over current protection that kicks in before the 47R burns out (it can withstand short surge as the failure is heat related).

[kothoma]

You could use a Schottky diode instead of R24.

Or better use something like this instead of the typcial D1/R24 hack: