INA225: Output range too small

hwswdesign

Part Number: INA225

I'm measuring current on a 4S Li-Ion battery, so the voltage is about 12V to 14.8V. This is the nominal range, deeply discharged would be less). I'm currently powering the INA225 with 3.3V.

The nominal battery current is from -10 amps (charging) to 100 amps (discharging), my sense resistor is 0.001 ohms.. A the min gain that's a range of (110*25*0.001)=2.75V.

My power supply could be as low as 3.2V (3%), the INA225 max output could be as low as VDD-0.2V. Since I'm bidirectional, it looks like the INA225 can output as low as 5 mV.

I also want to add some margin on top of that, and I end up with the INA225 not being able to output signals over my entire desired range.

To rectify this, I have two ideas:

1. Use higher values for the input resistors. I'm afraid this way relies on precisely knowing the values of the INA225 input resistor (like 50K or so?).

2. Use the battery (10V-14.8V) as Vs to the INA225, instead of 3.3V. Then I probably need to divide the output down a little bit to get it to fit within my ADC's range, and add protection on the ADC against high voltage output from the INA225 (just in case).

Any other ideas?

over 9 years ago

0 Carolus Andrews over 9 years ago

TI__Mastermind 26375 points

Dear hwswdesign,

I would definitely not recommend the first solution. These resistors would be placed in series with the gain network internal to the part, and severely alter your gain error. As such, we typically recommend that you use resistors no larger than 10 ohms for input filtration.

As for the second option, that does seem a bit tedious. In addition to this, the voltage division would add additional noise to the measurement, so that doesn't seem ideal either.

Looking at the design, some thoughts I had that might help are:

Could you change the size of your resistor from 1mohm to a smaller value, such as, for example, say 500uohm? If you were to do this, while the math would not be quite as nice, it would change your total range to ((110*25*.0005)= 1.375V. With this maximum, you could bias the reference at midsupply (1.65V), and have plenty of margin in both directions to ensure total linear operation. You could also look at shunts between 500uohm and 1mohm to find something that brings the operating range into your desired ideal.

Also, have you thought about placing your reference lower than midsupply, since you have non-uniform directions? For example, If you maintain your current setup (Vcc = 3.3V, Rshunt = .001ohm, total range = 2.75 V), then if you placed the reference at .4V, the downswing on charging would drop your measurement by .25V to 150mV, which, unless I missed a parameter, should keep you well inside linear range, and a decent distance from the offset voltage. On the upswing, maximum current would place your measurement at 2.9V, which is tight, but should fall within the Vdd-.2 requirement, even with the potential built-in 3% error you've listed from your source. If you wanted, you could also examine other values that might provide a better margin in line with your needs. Keep in mind, though, that we recommend that you use a low impedance source to drive the reference pin, so if you plan to divide off of Vcc, you may want to buffer the division.

Hope this helps, and let me know your thoughts on this.

Carolus

0 hwswdesign over 9 years ago in reply to Carolus Andrews

Intellectual 350 points

Thanks. I did not mention that I was planning on having the reference less than mid-supply.

If I design for 3V (3.3*0.97-0.2)=3V to be the highest guaranteed output voltage, and then 10mV to be the lowest guaranteed output voltage (from the datasheet when Vref is not GND), using a reference of 0.38V centers the 2.75 range within that 10mV-3V range.

The issue then is that only provides for a guaranteed 4.8 amps overhead at each end. I kind of want something more.

Thank you for your suggestions.

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INA225: Output range too small