# Determine resistor power rating using data sheet?

Info:

I am having trouble properly calculating the power rating for a resistor on a robotics design.

I have a circuit diagram which I need to use for my power calculations.

Can someone please clarify what is the correct method to calculate the power rating for the resistor? (diagram below)

My calculations are giving me an extremely high power rating which I really hope is wrong.

What I have tried:

The motor requires 3.17A but will draw more during start up, so the current drawn from the power source has to be at least:

3.17A + Iu + In = Itot.

(In goes to the 10n and Iu goes to the 1000u).

PVDD = 48V.

The current charging the capacitor is

Vs/R so 48/3.3 = 14.545454...

So according to my calculations, the resistor needs to have a power rating of ((48/3.3) * 48) = 698W?

Extra information: Here's a link to the datasheet (the circuit in question can be found on page 20)

Note: in the diagram (below) [PVDD] is the motor power source rated at 48V and the branch that goes off screen leads to the motor.

Edit: I was hoping to use a small SMD resistor for this. I also believe 50W is a physically large resistor, at least for my application.

[CIRCUIT DIAGRAM BELOW]

• I think what you're missing is that the current will only be flowing for a few microseconds. Even with a massively under-sized resistor, it's not going to get hot. Dec 17, 2020 at 8:38
• Hi, Thanks for your speedy reply! Can you kindly expand on what "massively under sized" means? I was hoping to use a small SMD resistor for this. I also believe 50W is a physically large resistor, at least for my application. Dec 17, 2020 at 9:00
• Look for a resistor that has some kind of pulse rating data in the datasheet. It is very possible that a quarter, half or one Watt resistor will do the job. A non pulse rated resistor will probably work, too, but to be safe you can pick a pulse rated resistor. Dec 17, 2020 at 9:04
• Take a look at this: koaspeer.com/pdfs/SG73.pdf Dec 17, 2020 at 9:08
• Thank you for the input everyone. I think you are correct though @mkeith, thank you for that PDF also. If you would like to make this as an answer I would be happy to mark it as solved by you. Dec 17, 2020 at 9:11