# Low current sensor INA219 **URL:** \ **Category:** General Guidance\ **Created:** [April 28, 2025, 12:01am UTC](https://forum.arduino.cc/t/low-current-sensor-ina219/1377106 "2025-04-28T00:01:49Z")\ **Posts on this page:** 10\ **Page:** 1
**Author:** ![alandinho10](https://dub1.discourse-cdn.com/arduino/user_avatar/forum.arduino.cc/alandinho10/32/1098598_2.png) [@alandinho10](https://forum.arduino.cc/u/alandinho10)\ **Post date:** [April 28, 2025, 12:01am UTC](https://forum.arduino.cc/t/low-current-sensor-ina219/1377106/1 "2025-04-28T00:01:49Z")
I have a battery going to my esp32 xaio module. I have the xaio module when it sleeps it is currently using 47 micro amps measured from a multimeter. I want to monitor this value and the voltage from the battery over 10 days and print the values to an SD card. What module can I use? I dont think the INA219 can go this low. Any guidance will help. This is one I have looked at: > **[HiLetgo INA219 I2C IIC Bi-Directional DC Current Power Supply Sensor Breakout...](https://www.amazon.com/HiLetgo-INA219-Bi-directional-Current-Breakout/dp/B01ICN5OAM)** > > Description: INA219 can be achieved within the maximum error accuracy 85℃, temperature range from -40℃ to + 1%, the maximum offset is 100 uV. The high-precision products combine the advantages of 12-bit resolution. This module can be sensed bus... ---
**Author:** ![LarryD](https://dub1.discourse-cdn.com/arduino/user_avatar/forum.arduino.cc/larryd/32/215550_2.png) [@LarryD](https://forum.arduino.cc/u/LarryD)\ **Post date:** [April 28, 2025, 12:20am UTC](https://forum.arduino.cc/t/low-current-sensor-ina219/1377106/2 "2025-04-28T00:20:38Z")
> [@alandinho10](#): > > [HiLetgo INA219 I2C IIC Bi-Directional DC Current Power Supply Sensor Breakout Module Power Monitoring Sensor: Amazon.com: Industrial & Scientific](https://www.amazon.com/HiLetgo-INA219-Bi-directional-Current-Breakout/dp/B01ICN5OAM) - I believe the INA219 uses a 0.1 ohm shunt, you could raise this to 1 or more ohms. ---
**Author:** ![alandinho10](https://dub1.discourse-cdn.com/arduino/user_avatar/forum.arduino.cc/alandinho10/32/1098598_2.png) [@alandinho10](https://forum.arduino.cc/u/alandinho10)\ **Post date:** [April 28, 2025, 12:28am UTC](https://forum.arduino.cc/t/low-current-sensor-ina219/1377106/3 "2025-04-28T00:28:37Z")
Do I have to desolder the on board shunt? Or what would be the best course of action? ---
**Author:** ![LarryD](https://dub1.discourse-cdn.com/arduino/user_avatar/forum.arduino.cc/larryd/32/215550_2.png) [@LarryD](https://forum.arduino.cc/u/LarryD)\ **Post date:** [April 28, 2025, 12:38am UTC](https://forum.arduino.cc/t/low-current-sensor-ina219/1377106/4 "2025-04-28T00:38:07Z")
- Yes, the existing 0.1 ohm would need to be removed and a similar foot print 1Ω or 10Ω soldered in its place. Use 1% or better if you want accuracy. - Remember, if you used 10Ω if the normal current went to 10mA, there is a 0.1v drop across the shunt . . . ---
**Author:** ![alandinho10](https://dub1.discourse-cdn.com/arduino/user_avatar/forum.arduino.cc/alandinho10/32/1098598_2.png) [@alandinho10](https://forum.arduino.cc/u/alandinho10)\ **Post date:** [April 28, 2025, 12:42am UTC](https://forum.arduino.cc/t/low-current-sensor-ina219/1377106/5 "2025-04-28T00:42:23Z")
1% tolerance you mean right? ---
**Author:** ![LarryD](https://dub1.discourse-cdn.com/arduino/user_avatar/forum.arduino.cc/larryd/32/215550_2.png) [@LarryD](https://forum.arduino.cc/u/LarryD)\ **Post date:** [April 28, 2025, 12:44am UTC](https://forum.arduino.cc/t/low-current-sensor-ina219/1377106/6 "2025-04-28T00:44:05Z")
- Yes, if accuracy is needed. ---
**Author:** ![alandinho10](https://dub1.discourse-cdn.com/arduino/user_avatar/forum.arduino.cc/alandinho10/32/1098598_2.png) [@alandinho10](https://forum.arduino.cc/u/alandinho10)\ **Post date:** [April 28, 2025, 1:00am UTC](https://forum.arduino.cc/t/low-current-sensor-ina219/1377106/7 "2025-04-28T01:00:53Z")
Great thank you! I expect 50 micro amps the whole day for 23 hours and 59 minutes a day. For only a minute a day it will go to 100mA. Will 1 ohm still be good? ---
**Author:** ![belmont1591](https://dub1.discourse-cdn.com/arduino/user_avatar/forum.arduino.cc/belmont1591/32/1133217_2.png) [@belmont1591](https://forum.arduino.cc/u/belmont1591)\ **Post date:** [April 28, 2025, 5:18am UTC](https://forum.arduino.cc/t/low-current-sensor-ina219/1377106/8 "2025-04-28T05:18:47Z")
If you want to measure current in the micro-amp range then you need to choose a low-input bias current amplifier such as the INA190. THIS IS A ANALOG OUTPUT CURRENT-SENSE AMPLIFIER. According to the [INA190](https://www.ti.com/lit/ds/symlink/ina190.pdf?ts=1745789294067) datasheet (page 21), the maximum R\_sense should be: ![imagen](https://europe1.discourse-cdn.com/arduino/original/4X/d/1/8/d180fc43f086a7da76762a1a08de66e9dc98958e.png) In the same page you can see the definition of each parameter, but quickly: - I\_max is the maximum current you want to measure - GAIN is the GAIN of the current-sense amplifier (you have 5 different options) - Vsp is the positive output swing, e.g., if you amplifier is fed with 5V, the Vsp is: Vsp = 5V - 20mV = 4.98V ![imagen](https://europe1.discourse-cdn.com/arduino/original/4X/a/5/5/a55ad88bb4223bfaed74ad80ae557d9c34d2e522.png) - V\_ref is the externally applied voltage on the REF pin, you can tie this pin to GND, so V\_ref = 0V On the other hand, the minimum value of your sense resistor must be: ![imagen](https://europe1.discourse-cdn.com/arduino/original/4X/5/4/6/54695e41756fd66669c55edcf339682a0b2c8982.png) Where - I\_min is the minimum current you want to measure - Vsn is the negative output swing of the device, i.e., Vsn = Vgnd + 0.05mV = 0V + 0.05mV = 0.05mV ![imagen](https://europe1.discourse-cdn.com/arduino/original/4X/7/9/c/79c9f9ff4f514b4df2d42ebb825ba7c8cf57b5be.png) This can be rearranged as ![imagen](https://europe1.discourse-cdn.com/arduino/original/4X/6/8/0/680fa45a1395f28c1faa66f39d666cbd8eca12c6.png) The output voltage is given by ![imagen](https://europe1.discourse-cdn.com/arduino/original/4X/b/9/0/b90b2fc50a7708d4581b2acee18531a8a7a9f80f.png) If you want to know the value of the current, then you can easily solve for I: ![imagen](https://europe1.discourse-cdn.com/arduino/original/4X/6/e/f/6efc5c814bcedfb46e28479615af654a2aaa45a3.png) For example, if you want to measure a minimum current of 1uA and a maximum current of 100 uA, and you use a current-sense amplifier with a GAIN = 25 V/V, then the current resistor must be: ![imagen](https://europe1.discourse-cdn.com/arduino/original/4X/d/a/8/da8a4ccb3a70dc8a9e2e5efbf70efc9cf80089eb.png) We can use a 1.5 kOhm resistor. We can choose a precision 0.1% tolerance shunt resistor to limit gain error at higher currents. This is a simulation result using the INA190 with a maximum current of 100uA and a shunt resistor of 1.5k ![imagen](https://europe1.discourse-cdn.com/arduino/original/4X/0/3/d/03d97f38ebd56870e155969dd4797dd598d275b2.png) ![imagen](https://europe1.discourse-cdn.com/arduino/original/4X/1/f/8/1f835f01ad7e2f613599d6869e91908d4923295f.png) ---
**Author:** ![srnet](https://dub1.discourse-cdn.com/arduino/user_avatar/forum.arduino.cc/srnet/32/238868_2.png) [@srnet](https://forum.arduino.cc/u/srnet)\ **Post date:** [April 28, 2025, 8:21am UTC](https://forum.arduino.cc/t/low-current-sensor-ina219/1377106/9 "2025-04-28T08:21:29Z")
> [@alandinho10](#): > > I have the xaio module when it sleeps it is currently using 47 micro amps measured from a multimeter. I want to monitor this value and the voltage from the battery over 10 days and print the values to an SD card. You get a 47uA sleep current, including the SD card ? Which XIAO module are you using ? ---
**Author:** ![system](https://dub1.discourse-cdn.com/arduino/user_avatar/forum.arduino.cc/system/32/1140315_2.png) [@system](https://forum.arduino.cc/u/system)\ **Post date:** [October 25, 2025, 8:21am UTC](https://forum.arduino.cc/t/low-current-sensor-ina219/1377106/10 "2025-10-25T08:21:40Z")
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