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I have built the instrumentation amp from the datasheet (http://www.ti.com/lit/ds/symlink/lm124-n.pdf, page 20) of the LM324. On the datasheet, they have built the amp with the 124, and I have simulated it and built it with the 324. However, on the datasheet, they state the gain of the amp should be 101. They calculate:

Vo = 1 + ( 2 * (feedback R of first amp stages) / adjustable gain resistor ) * (difference between inpunt voltage and ground)

However, I have found in my simulation and breadboard that the adjustable gain resistor has no effect. I always receive a gain of slightly larger than 2. I am not sure why the gain resistor does not work.

Attached is an image of the circuit I have simulated. I have tested the gain by inputting a DC bias into the "Vin" node; I have grounded the other input node. For instance, if I input 2VDC into the Vin, ground the other In Amp + terminal, Vout will be slightly larger than 4, for a gain of slightly greater than 2. If I adjust the gain resistor, this does not affect the gain. I am not sure why.

Can anyone explain why the gain resistor (R1 in the picture attached) does not work?

LM 324 Instrumentation Amplifier from Tex. Inst. Datasheet

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    \$\begingroup\$ with an input of 2v, and a gain of 101, what output voltage would you predict? What rail do the amplifiers use? \$\endgroup\$ – Neil_UK Aug 27 '16 at 19:59
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    \$\begingroup\$ Both Input Op-amps would saturate and give same signal on the diff. amplifier. I hope this is correct analysis. \$\endgroup\$ – ammar.cma Aug 27 '16 at 20:02
  • \$\begingroup\$ The amp is powered between 12V and ground, and actually, on the board I am using an LMC660, which has the same pinout as the 324, but it can swing very nearly to the top rail. So, with an input of 2V, I'd expect an output of 12V, as 2*101 = 204 and the amp is limited at 12V. \$\endgroup\$ – Thomas Wilk Aug 27 '16 at 20:04
  • \$\begingroup\$ With the 324 I get the same result on the breadboard, but the upper limit of the 324 on my board is just over 10V when powered from 12V. \$\endgroup\$ – Thomas Wilk Aug 27 '16 at 20:04
  • \$\begingroup\$ @ammar.cma Why would both input op amp inputs saturate? \$\endgroup\$ – Thomas Wilk Aug 27 '16 at 20:11
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There are three op-amps in this circuit. Any of the three output nodes can saturate depending on the two input voltages. All of them must be within the output range of the amplifiers for the circuit to function as an instrumentation amplifier. There is also an input common-mode range for the LM324 that extends up to about 10V, but let's ignore that for now- it's only a factor on U1A/U1B anyway.

The maximum range of inputs (assuming ideal op-amps with rail-to-rail outputs) will be achieved when the input common mode voltage is half the unipolar supply voltage (or in the middle of a bipolar range), so 6V in this case.

If one of the inputs is at either ground or +12, even with ideal op-amps (that limit at the rails) the only input possible on the other input is exactly the same, the slightest difference will cause one or the other of U1A/U1B to saturate. Reality isn't quite that nice so the amplifiers won't work with one input grounded, period. For U1A input to be grounded and any positive voltage on U1B input, the output of U1A has to go below ground, which it canna do. Any negative voltage on U1B input would mean then output of U1B would have to go negative- impossible again.

The output amplifier (ideally) can't saturate at the positive rail, since the worst case is +12 out (U1B out = 12V, U1A out = 0V), but it will saturate if the output of U1A is greater than U1B, since it cannot go below zero, even ideally.

If you have 1.0V at the input of U1A, then the minimum acceptable voltage at U1B is 1.0V (output saturation). The maximum acceptable will be set by the saturation of U1A at 0V, which will occur (ideally) at +1.02V on U1B (the current through R2 is 10uA so the voltage at U1B must be 1.0V+10uA*2K = 1.02V. Note that the output voltage can thus only go to a maximum of 2.02V = 101 * (1.02V - 1.00V).

You can repeat this analysis at various input voltages and substitute the real output range etc. if you want a complete answer, but you must determine the max/min voltage at which none of the amplifiers are saturated.

From a datasheet of a 3-amplifier style instrumentation amplifier you can see what I am talking about (presented a bit differently, in terms of common-mode voltage and output voltage rather than the two input voltages). Your case is more like the blue line.

enter image description here

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EDITED TO CORRECT U1A ERROR

Let's assume that U1A and U1B are operating "normally" - that is, their resistor networks have appropriate voltages. U1A has the + input grounded, so the - input must be grounded as well.

Now look at U2. The upper end of R1 is held at ground (well, virtual ground), so U1B acts as a standard non-inverting op amp, with gain of 51. With an input of 2 volts, the output of U1B must therefor be at 102 volts. I'm sure you'll agree that this will be difficult to do with a 12 volt power supply.

Since the voltage across R1 is 2 volts, 1 mA must flow through it. 1 mA through R2's 100k will drive the output of U1A to -100 volts. Again, I'm sure you'll agree that this will be difficult given the power supply.

So the difference amplifier U1C will take the difference of 102 and -100, for a nominal output voltage of 202 volts. Again, I'm sure you'll etc, etc, etc. While this is exactly what is predicted for circuit (a gain of 101 times 2 volts is 202 volts), none of the op amps is able to perform as their resistors require due to the limitations caused by your choice of power supply and op amp. If you had a power supply of +/- 250 volts and op amps to match, you'd be fine. Instead, each op amp is limited to much lower voltages, and changing R1 just doesn't help. Although if you increase R1 to about 100k you'll get something sort of right, or at least you'll be able to see changes in the output with varying R1. Since U1A cannot produce negative voltages the circuit still won't work properly, but at least things will change enough to see them change. If you connect pin 11 to a -12 volt supply you'll be able to play with R1 at 100k.

So try your simulation with U1A's + input tied to 2 volts, and U1B's input at 2.05 volts. This will, if the theory is correct, give you an output of about 5 volts, or halfway between V+ and ground, which is a handy place to start. Now you can play with R1.

But be careful. Remember those offsets I mentioned? For a 324 this can be as much as 3 mV. This won't matter a whole lot for U1A, but it represents more than 50% of the difference between the two inputs, so your nominal 5 volt output could be +/- 3 volts. You'll have to take this into account, unless your simulator allows you to set the offset to zero.

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  • \$\begingroup\$ Ua1s output should not be at ground. \$\endgroup\$ – Scott Seidman Aug 27 '16 at 21:52
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    \$\begingroup\$ R1, R2, and R3 all have the same current through them. If that current is not zero, UA1's output is not near ground \$\endgroup\$ – Scott Seidman Aug 27 '16 at 21:54
  • \$\begingroup\$ @ScottSeidman - Thanks. I had a brain fart when I wrote the first version, but I've edited to make more sense. Thanks for not downvoting on an obviously messed-up answer. \$\endgroup\$ – WhatRoughBeast Aug 28 '16 at 0:01

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