INA326: Minimum measurable current range

Hi Anand.

Q: With the same reference we designed the circuit which can measure current in the range 1uA - 200uA. 

The transfer function for the part is: Vout = Gain*(V+ - V-) + Vos, where Gain = 2(110k/20k) = 11 V/V

Say the constant I=1uA, the voltage delta across the INA326 is only 1.5Ω, Vout = 11*1.5uV + Vos = 16.5uV + Vos. Since Vos is up to 124uV, the voltage offset will dominate the Vout, you do not see the correct Vout. 

I increased the resistor from 1.5Ω to 1kΩ, the Vout = 11*1uA*1kΩ=11mV, and the simulation is much closer (I used dual supply rails here). 

Here is what you need to do.

1. increase the resistor value or gains in the circuit, so Vout >> Vos. 

2. Make sure the Vcm is operating within the allowed input voltage range for a single supply rail (ideally in the mid of Vcc/2).

3. You may need to remove the Vos, see Figure 6 or use MCU to null out the Vos. As you can see, Vp or Vout is linear to 1uA -200uA (including Vos). 

4. You may also use other Instrumentation Amplifier, such as INA333 or similar, but Vos will be an issue with low delta input voltage at IA. 

INA326 1u-200uA 03242021B.TSC

There may be some issues with pspice model with single supply rail, and I will find out why. 

If you have additional questions, please let me know. 

Best,

Raymond

Hi Raymond,

Thanks for your feedback.

Actually in my design i have two different gains. ie Max gain = (6600K/20K) 660 v/v & Min gain = (110K/20K) 11 v/v.
So for 1uA current i should get output as 660*1.5uV = 0.99mV. But i am getting different value.

And i couldn't find any formula to calculate output voltage if the circuit has two different gains like the one in this design note https://www.ti.com/lit/an/sboa210a/sboa210a.pdf .

As you said, if I increase shunt resistance or gain my current measuring range will be less (in few uA).