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I'm building an autonomous boat which will contain solar panels for its energy needs. The current prototype contains a 20W solar panel. All the electronics work great so it's time to scale up. The next prototype I'm working on will contain a 100W solar panel and the final version 200W of solar panels.

I'm currently using this INA219 power sensor:

enter image description here

It works great and I would like to continue to use it. However I don't think the sensor will be able to handle the increased amounts of energy I'm going to receive from the solar panels. I just tested running about 50W through the sensor (12V x 4A) for about 20 seconds and it heated up really quick.

Over on this page at Adafruit I read that their version "can measure up to ±3.2 Amps". I presume the version I have will be able to handle the same or less. Furthermore, the page at adafruit says:

Advanced hackers can remove the 0.1 ohm current sense resistor and replace it with their own to change the range (say a 0.01 ohm to measure up 32 Amps with a resolution of 8mA)

Although I wouldn't call myself an "advanced hacker" (which should be obvious from me asking this question) I guess this is what I need to do. So first things first. Is this the part that they're talking about replacing (the current sense resistor)?

enter image description here

So as far as I understand this R100 means it's 0.1 ohm resistance. So I need one which says R010 like for example one of these: https://www.aliexpress.com/item/32847907759.html

Would one of those be enough and can I simply replace it using my soldering iron? Do I need a certain specific size or do these always have a "standard size"? And most importantly; will changing this out with the R010 prevent the sensor from heating up like it does now? (it actually hurt my finger)

Are there other things I'm missing here? All tips are welcome!

kramer65
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3 Answers3

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You are two for two:

  1. Yes, that is the sense or shunt resistor they are talking about replacing.

  2. Yes, it is a 0.1 ohm resistor. In resistor code, the letter describes the range and its location is the decimal point. For example, 2K7 is a 2.7K resistor. R for ones, K for thousands, M for millions. There is no letter for the 0.001 to 0.999 ohm range, so the R is used with more zeros. There usually is not a leading 0. Thus, a 0.01 ohm resistor is R01 or R010.

For any given current, both the voltage across it (Ohm's Law) and the power dissipation (Joule's Law) are directly proportional to the resistance value. You really want to stay with the same size for maximum power dissipation capability. Here is a chart:

enter image description here

AnalogKid
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  • Thank you for the information. This helps a lot. One more question. With 50W the sensor heats up really quick (becomes untouchable hot within 20 seconds). Will this other resistor solve that, or do I also need to do something about the heating? If so; what can I do? Or do I simply need a different sensor if I want to avoid the heat? – kramer65 Feb 18 '21 at 21:12
  • For the same current, power dissipation is proportional to the resistance. 0.1x the resistance, 0.1x the power dissipation - at the same current. The tradeoff is that the voltage across the resistor also will be less, possibly affecting the accuracy of the sensor. – AnalogKid Feb 18 '21 at 22:21
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Keep in mind that the resistor is not the only factor that limits the current you can pass through the sensor. At some point the traces on the PCB will exceed their maximum reliable current and fail.

And I would absolutely not assume that a device you bought on aliexpress (or ebay or even amazon) works as well, works exactly the same, and is as reliable as a similar looking unit from a reliable vendor.

Elliot Alderson
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  • Thanks for the tips. And about your mention of quality. I'm fully aware that electronics on AliExpress can be of pretty bad quality. But the price which is often one tenth of "regular" electronics is on my level of hobbyism definitely worth it. If I would have to buy everything of "branded" items I wouldn't be able to build the autonomous boat that I'm building. Sometimes it takes a few tries to find a good part on AliExpress. But my main goal; learning about electronics is definitely met.. :-) – kramer65 Feb 19 '21 at 11:44
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You are on the right track on all counts. Calculate the power dissipation of the resistor as follows:

P = I^2 * R

Where P is power in Watts, I is maximum sustained current (sustained for more than like 1 second) and R is the resistance of the resistor you intend to put in. You want P to be less than the rated P of the resistor you use. Ideally, P would be like only half of the rated resistance for a reliable solution.

You also have to consider the voltage swing at the ADC. It will scale with the resistor vale. If you get 1V for each 1A now, that will be reduced to 0.1V for each 1A if you go from 0.1 ohms to 0.01 ohms.

user57037
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