Author Topic: Current monitoring circuit using INA226  (Read 474 times)

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Offline joniengr081Topic starter

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Current monitoring circuit using INA226
« on: March 11, 2024, 01:22:19 pm »
I would like to monitor the current using INA226 in am application.

I am wondering about the value of the resistor to be used as shunt resistor. I am not sure if it is ok to ask here because this is a design parameter.

We have a DC to DC converter that provide 4 Amp current with the output voltage of 5V. There are some LDOs connected to the output of LDOs which supply the power to the several ICs. We actually want to continuously measure the total current at the output of the DC to DC converter. We can accept few tens of mV drop in the supply voltage to the LDOs which is not a problem. Any suggestion on the value of the shunt resistor and of what power rating ?   
 

Offline BennoG

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Re: Current monitoring circuit using INA226
« Reply #1 on: March 11, 2024, 01:31:10 pm »
The specs says -81 to 81 mv on the current shunt.
So you can not have more than 81 mv drop on the shunt before saturation of the ADC
You mention few tens of mV drop so say 20 mV drop gives  R = U / I  ==  0.020 / 4 = 0,005 Ohm shunt.
Then you have 20mV drop at 4 Amp.

Benno
 

Offline Leuams

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Re: Current monitoring circuit using INA226
« Reply #2 on: March 11, 2024, 01:33:33 pm »
TI offers a wealth of advice for this on the product page. I suggest looking through the technical documents along with the datasheet.
 

Offline joniengr081Topic starter

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Re: Current monitoring circuit using INA226
« Reply #3 on: March 11, 2024, 02:08:30 pm »
I understand that for 20 mV drop across the shunt resistor. The value of the shunt resistor for 4 A current will be:

R = V / I  = 0.020 / 4 = 0.005 Ohm = 5 m Ohm

This 5 m Ohm resistor is very small resistance value. Do we have such resistors available with resistances in m Ohm range ?

I am also wondering about the power rating of the shunt resistor.

P = V x I
P = 0.020 x 4
P = 0.080 Watt
P = 80 mWatt

I am not sure if that is correct.

 

Offline BennoG

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Re: Current monitoring circuit using INA226
« Reply #4 on: March 11, 2024, 02:16:23 pm »
You have a lot of options to choose from.

https://www.digikey.nl/short/npznmdh4

these are probably a better option.

https://www.digikey.nl/short/c8mjhrmp

Be aware that even the solder joints have resistance.

Benno
« Last Edit: March 11, 2024, 02:18:57 pm by BennoG »
 

Offline joniengr081Topic starter

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Re: Current monitoring circuit using INA226
« Reply #5 on: March 12, 2024, 10:11:41 am »
I am still in to this.

I am not sure if my calculation of shunt resistor are correct. I calculate 5 m Ohm.

I also calculated power for the shunt resistor which is 80 m Watt.

Does it means that I can use a resistor 5 m Ohm of power rating 250 mW to be on the safe side. the 250 mW resistor are very common to be use on the PCBs.

 

Offline BennoG

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Re: Current monitoring circuit using INA226
« Reply #6 on: March 12, 2024, 10:27:00 am »
Yes you can. But remember the pcb traces have resistance too and even the solder joints.
So your PCB traces need to be something like below.

 

Offline joniengr081Topic starter

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Re: Current monitoring circuit using INA226
« Reply #7 on: March 12, 2024, 11:54:15 am »
Yes, I will include and consider the resistance of PCB traces and also the solder joint.

Simultaneously, I am wondering about the power rating of 1206 resistors. I found them in two power ratings. Are they available in 250 mWatt and also in 500 mWatt ?

https://uk.farnell.com/bourns/crl1206-fu-r020elf/res-thick-film-0r02-1-0-25w-1206/dp/2328117RL
https://uk.farnell.com/yageo-phycomp/rl1206fr-7w0r022l/res-thick-film-0r022-1-0-5w-1206/dp/8067546

 


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