0
\$\begingroup\$

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.

\$\endgroup\$
3
  • \$\begingroup\$ Calculate how much current flows through them, calculate how much power dissipation this causes, then add a margin. \$\endgroup\$ – PlasmaHH Feb 19 '18 at 21:52
  • \$\begingroup\$ 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. \$\endgroup\$ – Peter Bennett Feb 19 '18 at 21:59
  • \$\begingroup\$ @PeterBennett through hole 1/8W resistors break easy too, or they used to, though they may have improved over the years. \$\endgroup\$ – Trevor_G Feb 19 '18 at 22:16
3
\$\begingroup\$

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. .

\$\endgroup\$
4
\$\begingroup\$

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.