Choosing a Shunt Resistor for Low Current Applications

Anyway, Ohm's law states V=IR. V is voltage, I is current through your sense resistor, and R is the resistance of your sense resistor. If we re-arrange it, we get:

R = V/I.

Let us say you want to have 100 mV when I = 10 mA. Plug that in and you will get:

R = 100 mV / 10 mA = 10 Ohms

If your load is 1 Meg, 10 Ohms will not affect it. Now you just need to measure the Voltage across your 10 Ohm sense resistor. If you can arrange to put the sense resistor on the low voltage side of the load, you can sense the voltage with a volt meter.

Just re-arrange Ohms law to understand the relationship between sense voltage and current:

I = V/R (where V is meter voltage and R is value of sense resistor) I = V/10 = 0.1 * V

Addendum The power dissipated by the current sense resistor can be calculated using one of three forumulas: P = I^2 * R P = I * V P = V^2 / R

If the peak current through the resistor is definitely going to be 100 mA, then the peak power will be: P = 100 mA * 100 mA * 10 P = 0.1 * 0.1 * 10 = 0.1W or 100 mW

So you can use a resistor rated for 1/8 W or more. If you change the resistor value, or change your estimate of the maximum current, you should recalculate the power dissipation of the resistor.