Finding shunt resistance

Thread Starter

engr_david_ee

Joined Mar 10, 2023
176
I would like to use INA226 to monitor the current.

https://www.ti.com/product/INA226

The nominal voltage and the nominal current through the load is 4 A and 5 V. The maximum current through the load is 5 A.

I started calculating shunt resistance by defining the peak power dissipation across the shunt resistor which is 200 mW.

Let's calculate the shunt voltage given the power is 200 mA and the max current is 5 A.

P = V x I

V = P / I

V = 200 mW / 5 A

V = 40 mV

This shunt voltage 40 mV is well within the input shunt voltage range according to datasheet INA226.

Now calculate the value of the shunt resistance given the maximum current (5 A) and the shunt voltage (40 mV)

V = I x R

R = V / I

R = 40 mV / 5 A

R = 8 m Ohm

The value of the shunt resistance is calculated as 8 m Ohm.

Now we need to calculate the Current LSB according to equation (2) on page 15/39 of the datasheet.

Current LSB = Maximum Current / 2^15

Current LSB = 5 A / 2^15

Current LSB = 152 uA/bit

Assuming the last two bits are noisy and not very stable. Or we do not want very high precision. Let's ignore the last two LSBs. We multiply by 4 to get new Current LSB.

Current LSB = 610 uA/bit

Is my calculation are ok until now ?


Now we calculate the Calibration Register value using equation (1) of the datasheet. The values of Current LSB and the R_Shunt need to be converted to A/bit and Ohm before using in equation (1)

CAL = 0.00512 / (Current_LSB x R_Shunt)

CAL = 0.00512 / (610 /1000/1000 x 8 /1000)

CAL = 1049 decimal

Is that number also correct to be written in Calibration Register ?

Two more questions.

The resolution of Shunt Voltage Register is 15 bit ? The LSB is 2.5 uV ?

Does it means that full scale reading will be 2^15 x 25 uV = 81 mV approx ?
 

Jerry-Hat-Trick

Joined Aug 31, 2022
579
I think your calculations are correct. How about trying it to check? I wonder if a higher shunt resistor value would make sense, is there a reason to limit the power loss to 200mW?

I’ve not come across this device before and I think it’s a fairly complicated way of measuring current, there are devices which simply amplify the voltage across the shunt resistor which can be converted to digital with an A/D processor input pin. And I agree with your suggestion to ignore the lowest 2 bits, 2^15 sounds incredibly optimistic!
 

Thread Starter

engr_david_ee

Joined Mar 10, 2023
176
I was assuming the the power dissipation across the shunt resistor will be used in the calculation of junction temperature.

If the power dissipation across the shunt resistor is 200 mWatt for example. Do we need to use 200 mWatt multiplied by junction-to-ambient thermal resistance and add to ambient temperature to calculate the junction temperature ?

For example junction-to-ambient thermal resistance is 171.4 degC/Watt. The power dissipation across the shunt resistor is 200 mWatt.

The junction temperature will be

Tj = Tambient + junction-to-ambient thermal resistance x Pd

Tj = 40 + 171.4 x 0.2

Tj = 74.2 degC

Is that correct ?
 

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Thread Starter

engr_david_ee

Joined Mar 10, 2023
176
On the other hand, increasing the value of the shunt resistor in m Ohms will increase the shunt voltage which will decrease the error in current measurement but will increase the power dissipation. What are the consequence of higher power dissipation across the shunt resistor ?
 

Jerry-Hat-Trick

Joined Aug 31, 2022
579
Metal film resistors typically have a temperature coefficient of less than 100ppm/°C so even with 100°C change you are looking at a resistance change of less than 1%. A series/parallel arrangement of multiple resistors would allow heat to be lost from the copper track - you could use strip board for this. Maybe the odd thermistor touching the copper to correct for temperature. It all depends on the accuracy you are looking for - as already discussed, 2^15 resolution is a tad optimistic, what accuracy do you need? It will always be a compromise between temperature sensitivity and full scale mV.
 

Thread Starter

engr_david_ee

Joined Mar 10, 2023
176
There is still one thing I am not sure about it. I understand that increasing the value of shunt resistor will cause more heat dissipation across the shunt resistance and also crease the shunt voltage that may improve the current measurement accuracy but at the of more heat dissipation across the shunt resistor.

Will it also increase the junction temperature of the INA226 ? I am asking this question because the shunt resistor is located physically outside the INA226. So the question is, increasing the shunt resistor value will also increase the junction temperature of the INA226 ?
 

Jerry-Hat-Trick

Joined Aug 31, 2022
579
There is still one thing I am not sure about it. So the question is, increasing the shunt resistor value will also increase the junction temperature of the INA226 ?
No, whilst the internals of the IC are not described it will be high impedance, like an op amp to amplify the voltage across the shunt resistor.

Note that figure 21 of the data sheet shows a 10 ohm resistor in each of the two input lines, at most a very small current will be flowing into the IC.
 

Thread Starter

engr_david_ee

Joined Mar 10, 2023
176
I guess this 10 ohm is the resistor to be used for filtering together with capacitor. But this resistor is physically outside the chip. How do we calculate the current flowing into the chip ?
 

Jerry-Hat-Trick

Joined Aug 31, 2022
579
Consider the shunt resistor to be 20 mOhm, just to make the maths easy. Now think of the internal resistance between the input pins of the IC to be zero. In that case, the 5A flowing through the shunt resistor in parallel with the 20 ohm input resistance would be about 5mA through the IC. But the actual internal resistance of the IC is likely to be in the region of 200 kOhm, maybe an order of magnitude larger which would mean a current of less than 0.5uA.

If you really want to know the resistance between the input pins of the IC try putting a 20K resistor in place of one of the 10 ohm resistors and see how much the 'measured' current falls. To get familiar with the IC it would be a good idea to simulate the 5A current by making the shunt resistor say 8 ohms instead of 8 mOhms and simply pass a current of 5mA through it to get the same result
 

Thread Starter

engr_david_ee

Joined Mar 10, 2023
176
I think I am getting it now. I just use the current divider formula, see attachment.

Total current through the load is 5 A.
Load voltage = 5 V

Shunt resistor = 20 m Ohm
Input resistance of the IC = 10 + 10 = 20 Ohm

Current running through shunt resistor = 4.995 A
Current running through the IC = 4.995 mA

This shows the current through the IC is just 5 mA.

Now in order to calculate the junction temperature of the INA 226. How do I calculate the power dissipation in the IC.

If current through the IC is 5 mA for example, then which internal resistance should I consider, 20 Ohm ?

Pd = V x I
Pd = 5 x 5/1000
Pd = 25 mWatt

Is that looks correct ?

Junction Temperature = Ambient Temperature + Pd * thermal resistance
Junction Temperature = 50 + 25/1000 * 171
Junction Temperature = 54.27 degC

The question still is the internal resistance that will tell us the power dissipation through the IC which will be used to calculate the junction temperature to determine whether we need heat sink or not.

Untitled 626.png
 

Attachments

I think you should not overthink this, my example was to show how little current will actually flow through the IC. The power loss in it will be tiny - I really wouln't worry about it. As I said, if you use a large resistor instead of one of the 10 ohm input resistors the change in measured value will give you the internal resistance of the IC.
 
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