# Calculating Resistor Wattage for Voltage Divider?

Background:

I am monitoring the voltage of a 4 cell LiPo battery (max voltage 16.8V, maximum advertised current output = 60A) using a sensor with a maximum input of 3.2V.

I am using a 2-resistor voltage divider with R1 = 22KΩ and R2 = 4.7KΩ. This maps the maximum battery voltage to approximately 2.95V.

Vout = (R2 / (R1 + R2)) * Vin = (4.7KΩ / 26.7KΩ) * 16.8V = 2.95V

Question:

Once I have these values, how do I calculate the required wattage of the resistors R1 and R2?

As a practical matter I see many people using 1/4 watt resistors, but I'm interested in understanding the calculation myself.

• Calculate how much current flows through them, calculate how much power dissipation this causes, then add a margin. – PlasmaHH Feb 19 '18 at 21:52
• Many people use 1/4W leaded resistors because that was the most common (and probably cheapest) size. 1.8W resistors were more expensive, and a little harder to get, so would not normally be used except when the smaller size was required. Now, with surface-mount parts, lower power resistors are more readily available, so we may have to put a little more thought to specifying power ratings. – Peter Bennett Feb 19 '18 at 21:59
• @PeterBennett through hole 1/8W resistors break easy too, or they used to, though they may have improved over the years. – Trevor_G Feb 19 '18 at 22:16

Let's say the maximum input voltage is 20 V, just to have a round number and give some margin from the expected 16.8 V.

Now the maximum current through the divider is $$\frac{20\ {\rm V}}{26.7\ {\rm k\Omega}} \approx 0.75\ {\rm mA}$$

Now you can use the $I^2R$ form for resistor power to find the power used in each of your resistors. Start with the higher value one, since that one will consume more power:

$$(0.75\ {\rm mA})^2(22\ {\rm k\Omega})=12.3\ {\rm mW}$$

Since this is well below the power limit of even a 1/16th W SMT resistor, you won't really need to worry about power specs on these resistors. .

Once I have these values, how do I calculate the required wattage of the resistors R1 and R2?

Use the power equations.

$$P = VI$$

and by various substitutions from Ohm's Law, $V = IR$ we can generate alternatives:

$$P = VI = I^2 R = \frac {V^2}{R}$$

If you have any two values of V, I and R you can calculate the power.