How do I calculate the required wattage for a current sense resistor?

Question

I'm choosing a current sense resistor for a current sense IC. I have a few considerations for the IC (e.g. ZXCT1107, INA169, INA219, etc) which all require an external current sense resistor.

I see that the Adafruit INA169 module comes with an 0.1 Ω SMD current sense resistor, and can handle 5 A according to the specs—how was that current capacity calculated? Some tutorials talk about replacing the resistor on the Adafruit module to increase the current capacity, but the guides are usually vague about what resistor to choose.

My circuit needs to handle 20A at 12V, so what current sense resistor should I use, and what wattage should it have? I am considering how much power dissipation it should need to be capable of.

Answer 1

The Adafruit shunt is \$0.1\Omega\$. A current of 5A will produce a voltage of \$0.5V\$.

For \$20A\$ of current a the shunt will be: $$R_{shunt}=\frac{0.5V}{20A}=0.025\Omega$$

Power dissipation is 10W so double it to 20W.

Addition:

If 10% of the Adafruit ADC range can be tolerated, then only 50mV need be used. Then you can use a 2W 0.0025 ohm shunt resistor.

Otherwise, the internal Adafruit shunt would need to be removed and external circuitry would need to be used to convert the current range to the Adafruit full scale capabilities.

Answer 2

A current sense resistor drops some small voltage across its terminals when a large current is applied. Due to Ohm's law:

$$ V = iR $$

the voltage produced is equal to the current times the resistance.

The Adafruit module handles a 5 A current using a 0.1 Ω resistor -- meaning that it drops 0.5 V. Without digging too deeply into the INA169 datasheet, this at least lets us know that the amplifier can handle a differential voltage of at least 0.5 V. This seems like a good target to design for.

Since you need to handle 20 A, and drop 0.5 V, rearrange Ohm's law:

$$ R = \frac{V}{I} $$

and you get a resistor of 25 mΩ.

Finally, remember the equation for power:

$$ P = VI $$

and since your resistor drops 0.5 V with 20 A, you end up with a power dissipation of 10 W.

10 W is a lot for a resistor. At 10 W, it might be more accurately called "a heater;" this is going to be a chunky block that puts out a lot of heat. You may need to heat sink it, or choose a resistor that can handle significantly more than 10 W with ambient cooling.

This circuit works, but that amount of power loss is not good. You might want to dig deeper into the INA169 datasheet after all, and make the circuit better. Specifically, you can look at the typical application circuit and determine how \$R_L\$ sets the output gain, i.e. how you can use a smaller sense resistor that burns less power, and still effectively measure the current.

Answer 3

The ADAfruit module uses a 2W 0.1Ω sense resistor which will dissipate 2.5W at 5A and will likely overheat and possibly become desoldered, or may be damaged. The 10k load resistor provides 0-5V for 0-5A. For 20A I would recommend a 0.0025Ω 2W sense resistor which will dissipate 1.0W at the maximum 20A. To obtain the full 0-5V output, the load resistor would need to be 100kΩ. This is too high for the ADC input, and the existing 10k would give only 500 mV. So this would require either a 12W or greater 0.025Ω sense resistor, or a device like the INA198, which has a slightly different pinout, or the INA180A3, which has a 26V common mode limit.

Answer 4

According to Practical Electronics for Inventors, Fourth Edition (Paul Scherz and Simon Monk):

The wattage of the resistor must be at least 2 Ω times the square of the maximum current (expressed in amps). For example, if the maximum anticipated current is 0.5 A, the minimum wattage of the resistor should be \$2\ Ω \times (0.5\ A)^2 = \frac{1}{2}\ W\$.

Therefore, for 20 A with a 0.001 Ω current sense resistor, the equation is:

\$W = 0.001\ Ω \times (20\ A)^2 = 0.4\ W\$

In this case, I'd use a 1 W current sense resistor in case the current is higher than expected.

If the current was 10 A, I might choose a higher resistor value, such as 0.01 Ω. This would allow for more accuracy in measuring the voltage drop.

\$W = 0.01\ Ω \times (10\ A)^2 = 1\ W\$

In this case, I'd choose a current sense resistor that can dissipate a bit more heat (such as 2 W), again, in case the current is higher than expected.

Common values for SMD current sense resistors, for example, are \$\frac{1}{4}\ W\$, \$\frac{1}{2}\ W\$, \$1\ W\$, \$2\ W\$, and \$3\ W\$, but they are generally available in a range from 50 mW to 50 W.

Higher values such as 25 W might be better suited to measuring hundreds of amperes:

\$W = 0.0001\ Ω \times (500\ A)^2 = 25\ W\$

Answer 5

Page 10 of TI's shunt resistor training slides explains...

Max power dissipation W = 5 mΩ ∗ 10^2 A = 0.5 W

In other words, in this example, to calculate the power dissipation required, multiply the resistance (\$0.005 Ω\$) by the current to the power of 2 (\$10^2 = 100\$), which gives us \$0.5 W\$.

Max power dissipation \$P = R(I^2)\$

\$0.005Ω(10^2A) = 0.5W\$

To calculate the absolute maximum current that a resistor is capable of handling based on the power dissipation rating, it's the square root of the max power dissipation divided by the resistance.

Absolute max current \$I = \sqrt{\frac{P}{R}}\$

Example with a 0.001 Ω resistor that has 0.5 W power dissipation.

\$\sqrt{\frac{0.5 W}{0.001 Ω}}\$ = \$22.36 A\$

Edit: As mentioned in the comments, this is the absolute maximum current that can be passed before the resistor fails, and as current increases so does temperature and impedance which would effect the measurement. Therefore, a 1 W resistor would be preferable (usually the next rating up from 500 mW). I noticed that with 0.001 Ω SMD current sense resistors, 2W ratings are the most commonly used. For 0.001 Ω SMD, typically values up to 10 W exist which would be able to handle an absolute maximum of 100 A.

Answer 6

This simple circuit could be used with a resistor of 0.001 Ohm - 2 W.
This resistor could be done DIY with Maillechort wire (?).
It needs a preamplifier (or instrumentation amplifier) as the voltage at 30 A should be only 30 mV.
The two resistors (R6 & R5) used for gain should be "stable" with temperature or this circuit needs calibration (to be checked at the lab and by software).
Searching for circuits that don't need this (ratiometric ? or auto corrective) ...