Hi team

I am experimenting with an INA122 In Amp with a view to measuring input voltages of fractions of a mV. I'm using two bias resistors of 330K on the inputs and gain of from 25 to 200. Ultimately I'd like to use this with sensors such as a thermopile, but currently I'm using a wheatstone bridge of 3 x 100K resistors with an 83 K resistor + 25K and 1 K pots in series as the 4th arm. By varying the 1 K pot, I am able to graph Vin vs Vout and get good linearity from about Vin = 1mV upwards. However, with anything less than Vin= 1 mV, the graph levels off and certainly does not pass through the origin. I would have thought that an in-amp should be able to give good results with input down to microvolts. Any suggestions?

 

...and yes, I have the leads and amplifier well shielded

 

What's the supply and common-mode voltage?

 

12V gel cell supply for the in-amp; separate 12 V gel cell supply for the resistor bridge. Must admit I have trouble understanding the common-mode concept except that these amps are supposed to reject noise that is common to both inputs. offset (pin 5) is directly connected to 0V (pin 4).

 

(330 K bias resistors from pins 2 & 3 to earth not shown in diagram above)

 

The bias resistors will load down the bridge, need to keep this in mind for the calculations. Are you expecting the output (pin 6) to go below 1 mV.

 

The specs. state that the bias resistors are essential; I was assuming that very high value bias resistors would minimise loading on the bridge.
If I plan for voltage amplification of 100, then Vin= 0.01 mV should give Vout = 1 mV. I'd be more than happy with that performance. I did wonder whether implementing an offset on pin 5 might get the output into the linear region.
Thank you for taking an interest!

 

The problem with bias resistors only applies when you have a floating signal, as a thermo-coupler signal. In contrast, a measuring bridge always keep in-amp inputs to the middle of the bridge supply.

This type of in-amp has not "rail two rail" output, the datasheet says 0.15V with 20 KOhm to the middle of supply. Two solutions are: either lift the Ref terminal up to at least 200 mV (where the output signal also gets the same offset) or use a similarly small negative supply on the in-amp.

 

Thank you for your help - I'll try that.

 

Any suggestions

Your circuit is quite unbalanced, having 100k vs 83/25/1k. It should produce a differential input on the order of 200mv. So if you are getting but 1nv out of it, something else is wrong.

I have used similar topology on much smaller input signal, with higher gsins and on inferior parts (ne5532 and tl0x2). So I tend to think your issue is elsewhere.

A simpler way to check would be to out the pot in between some large value resistors, like 330k or 470k x 2 and measure the voltage drop over the pot.

 

Looks like you’re on the right track. You are trying to use the device down to the negative rail when the device is not rated rail-rail.

I suggest adding a switch-capacitor supply inverter to create a negative 12V rail. Add filtering to reduce the ripple. This will provide a negative supply rail and not require offsetting the I-Amp or signal.

 

Your circuit is quite unbalanced, having 100k vs 83/25/1k. It should produce a differential input on the order of 200mv. So if you are getting but 1nv out of it, something else is wrong.

I have used similar topology on much smaller input signal, with higher gsins and on inferior parts (ne5532 and tl0x2). So I tend to think your issue is elsewhere.

A simpler way to check would be to out the pot in between some large value resistors, like 330k or 470k x 2 and measure the voltage drop over the pot.

I'll try that too, thanks

 

Looks like you’re on the right track. You are trying to use the device down to the negative rail when the device is not rated rail-rail.

I suggest adding a switch-capacitor supply inverter to create a negative 12V rail. Add filtering to reduce the ripple. This will provide a negative supply rail and not require offsetting the I-Amp or signal.

Excellent - As a newcomer to all of this I appreciate your assistance.