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I was wondering if someone could help me understand why this instrumentation amp isn't working the way I think it should. The amp is an INA155.

My intention is that at if a voltage (positive or negative) is applied across the input pins, I should see the same value, multiplied by 10 (since this amp has a gain of 10) at the output pin, relative to the ref pin. I have a virtual ground using a 2.5V reference at the ref pin. I have some low pass RC filters on the input. Eventually I want to read the output voltage using an ADC.

I have the circuit in this picture bread-boarded:

enter image description here

I have the ref and output pin connected to a multi meter. I am powering it with a bench top supply. I haven't applied a voltage to the input pins yet, because I thought I would first test the offset voltage by measuring the output when the input pins are shorted together. It should show a voltage close to 0, with a small offset right???

What I'm seeing is the output voltage start at close to the rail (so I measure exactly 2.5V on the meter since the ref pin is tied to 2.5V),then it slowy starts to drift, over the course of minutes, until the voltage on the meter hovers somewhere around 200 to 300mV between the output and the ref pin. The stability increased since I added the low pass filter, but the output doesn't make sense (at least to me).

Can anyone explain why I'm seeing this? There is obviously some gap in my understanding of these amps. So far all I've worked with is a unity gain buffer. I'm just trying to design a simple buffer for measuring a signal, mainly DC, that can be bipolar.

Looking for any tips or help with this, or even a different approach to achieve the same thing. I am forced to work with a single 5V supply, hence the 2.5V virtual ground, but I need to measure both positive and negative voltage. I've tried this with added resistors from the input to 2.5V for a return path, but the output didn't change.

Thanks!!

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  • \$\begingroup\$ One a different note, I have reservations about your circuit which drives the REF pin. See here and here. \$\endgroup\$ – Nick Alexeev Jan 27 '18 at 1:19
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The problem is you are not applying voltage to the inputs, which makes them 'float'. Instrumentation amplifiers have a very low input bias current in the pico volt range normally (meaning it takes pico volts for them to make a measurement). A lot of times this is close to currents flowing through the air.

Odds are there is not a lot of resistance through the caps and you are seeing the leakage through them.

Apply some voltage to the inputs (or in your case the resistors before the inputs) and the problem should go away.

Edit: Yep, the INA155 has a very low input bias current:

LOW INPUT BIAS CURRENT: 0.2pA

If you didn't have the filter could wave your hand over the inputs and the value of the amplifier would change from charge of your hand.

Tie the inputs to something that isn't floating like ground or a voltage reverence. (A 1.5V battery is sometimes fun for a voltage source) I forgot you had a gain of ten, you'd have to divide down your 1.5V battery with a 10 to 1 resistor attenuator.

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  • \$\begingroup\$ So I had tried doing just as you said, and connecting my inputs (before the filter) to the 2.5V reference with 47k resistors. But I still noticed the same effect. I also tried connecting them after the filter, and the output was the same. Am I just dealing with too low of an bias current to practically breadboard? \$\endgroup\$ – Michael Jan 26 '18 at 20:34
  • \$\begingroup\$ If you connect both your inputs to 2.5, then the result will be near zero, The INA155 can't output zero because of the common mode range (which is ~0.2 to 0.3V). So you can either move the grounded rail to a negative rail (even -1V would work) OR find a voltage source that is less than 0.45V and more than 0.03V. \$\endgroup\$ – Voltage Spike Jan 26 '18 at 20:42
  • \$\begingroup\$ When I connect both inputs directly to 2.5 I measure about 150mV although it moves around a lot. I don't have a negative rail available. When you say a voltage source <0.45 and >0.03, are you talking about the reference voltage? Can you elaborate on this, or point to a page on the datasheet? Sorry for my complete ignorance on this!! \$\endgroup\$ – Michael Jan 26 '18 at 20:50
  • \$\begingroup\$ Yeah look at the input common-mode range vs reference voltage graph, the equation states that you would be 0.25 to 4.75 (by substituting 0 for V-, 5 for V+ and 2.5 for Ref.) This is also with a gain of 10. The common mode range is for the output of the amplifier, if you want to find the range at the input divide by 10 \$\endgroup\$ – Voltage Spike Jan 26 '18 at 21:31
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    \$\begingroup\$ Actually, with both inputs tied to 2.5, your output voltage should be 2.5 volts. You are in the middle of your common mode range, and the output voltage is 0 volts (differential) x 10 but referenced the 2.5 volts. I would check for a loose connection. \$\endgroup\$ – John Birckhead Jan 26 '18 at 21:50
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The input common-mode range with a 5.5V supply and gain of 10 is as follows, according to the datasheet:

enter image description here

It will be a bit tighter on the top end with a 5.0V supply.

So with a 2.5V reference your inputs have to be within about 0.5V to 4.5V. You need to keep both inputs within that range, whatever the differential voltage you are applying. This is actually a pretty wide range, and you will find some in-amps are not as accommodating.

Right now you show it as floating, so there is no guarantee that it will be within that range and if it's misbehaving, it probably isn't.

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  • \$\begingroup\$ I didn't show it, but my inputs are connected to another voltage source for testing purposes. Right now it's set to 0mV. When I check the common mode voltages of the inputs, I measure 1.57 volts on each with a meter. \$\endgroup\$ – Michael Jan 26 '18 at 20:42
  • \$\begingroup\$ So you have two voltage sources? One for differential and one for common mode? \$\endgroup\$ – Spehro Pefhany Jan 26 '18 at 21:24
  • \$\begingroup\$ I have the voltage reference (2.5V) in the circuit, and a precision voltage calibrator as the differential input \$\endgroup\$ – Michael Jan 26 '18 at 21:55
  • \$\begingroup\$ So you connected one side of the calibrator to the reference?? \$\endgroup\$ – Spehro Pefhany Jan 27 '18 at 2:53
  • \$\begingroup\$ not quite, its all in the image \$\endgroup\$ – Michael Jan 27 '18 at 22:36

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