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Basic Electronics » INA118 -- Instrumentation Amplifier

November 28, 2010
by lcruz007
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Greetings!

I've connected an instrumentation amplifier (INA118) with a gain of 5000 in order to multiply that gain with the input voltage difference. The gain resistor is set to 10 ohms, according to the datasheet: G = 1+(50kΩ/Rg), ∴​ G = 5001.

I'm using a virtual ground as the reference voltage, it's set to 2.5V. The positive voltage is 1.4 above that reference and the negative input is 0.5V below that difference.

Correct me if I'm wrong, but the output voltage shouldn't be (V+) - (V-) * G? So, 5000(1.4V-0.5V)? If I measure the output voltage with a multimeter and measure the voltage reference (with respect to ground - 0V), the measure I receive is exactly the same. It is not amplifying, Vref = Vout.

What am I doing wrong? Here are my schematics:

http://img228.imageshack.us/img228/9056/ina118.jpg

November 28, 2010
by BobaMosfet
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You don't have it wired right. Check saturation, bias current, feedback loop. supply voltage could be larger; the gain you've chosen can't be supported with what you're now doing. try starting with a smaller gain, and just get it to work first.

BM

December 01, 2010
by lcruz007
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Well unfortunately lowering the gain didn't work for me. You can check this other example for instance:

http://img547.imageshack.us/img547/8508/circuit.png

The oscilloscope measure states that the signal is just 200mV, which doesn't coincides with my calculations. That's a gain of 50. There's a small voltage difference with respect to the voltage reference, which could perfectly be amplified.

December 01, 2010
by mongo
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Swap the resistors 3 and 4. The voltage there is 4.5 V, not .5 V, If I am reading the power supply and the other divider networks correctly.

December 01, 2010
by lcruz007
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Well according to my schematics ( http://img403.imageshack.us/img403/8508/circuit.png ), resistors R3 and R4 would create a voltage divider of 0.4545V because: [R2/(R1+R2)] * 5V = Vout

100/(100+1000) * 5 = 0.454545455

The other input should be 0.05V

December 01, 2010
by lcruz007
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Is that right though?

December 02, 2010
by mongo
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Yeah, I misread the values. For some reason I saw a divider that would give about 4.5 V.

That's why I mentioned my possible error in reading the supply.

December 02, 2010
by BobaMosfet
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lcruz007-

I'm looking at: http://img403.imageshack.us/img403/8508/circuit.png

Let's re-orient this a little to help you:

You're not calculating your input voltage or current correctly. You're overloading the common input range because of this, and your amp is flatlining. You're dealing with both parallel and series paths here-- think about that, because that is something you're missing in your math.

Read your datasheet and make sure everything is within specification, or the amp will just try to protect itself. I'm not guessing either, I did the mathematical analysis of your entire circuit, and matched that against the spec's in the datasheet to see if you were within spec. As the above circuit stands, you have -0.41V differential between inputs, and are applying way too much current and voltage to them as well. IF your inputs were within range, you'd be seeing about -20V on the output.

BM

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