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Help with Instrumentation Amplifier

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The only way that circuit could work is if the opamp had a seperate isolated ground from the input, otherwise the input is beyond the power supply rails of the opamp.
 
1) you say you need gain of 100, but then give example where gain must be 100,000 (G=1V / 0.01mV). Gain of this amp, if properly biased, is 1+(50K/Rg) and you have Rg=10K max, so you have max gain of 6.
2) the input common mode range of this part, with 5V supply, has an extremely narrow range of operation between 2 and 3 volts (varies with output voltage). Inputs beyond this range will do nothing but hold the amp in saturation with output at either 0.6 or 4.4 Volts.
3) the input bias configuration puts the inputs at 0V DC which is outside the input common mode range of the amp the way you have it powered. So it will not work. You need to raise the bias level of the inputs to mid-Vcc to get anything to work at all. Mid-Vcc is 2.5V.
4) If my eyes don't deceive me, it looks like you have an input voltage of 12V feeding a 5ohm resistor, which means there is a flow of 2.4A of current. The 5 ohm resistor will dissipate 28.8 watts and will burn up. However, perhaps I misunderstand the label on the resistor. What is "5E"?
5) what is the purpose of this circuit? Is the input only 12V DC, or does it vary, or does it have an AC signal on it? The way it is drawn now, you have a differential voltage of 12V at the input and you have the amp configured for a mid-adj gain of about 3 which means the amp wants to have an output of 36V. It is impossible to reach that output when the power supply to it is only 5V. Don't forget that this amplifier will amplify the DC at the input as well as any AC.
6) I think you should increase the power supply voltage by removing the ground from pin 4 and putting -5V there instead.
 
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Why do some people say, "5E", other people say, "5R" and other people say, "5 ohms"?
Isn't the resistor 5 ohms? A heater?
 
There's a lot of inconsistencies. 5E can't be 5 ohms, if that were the case then it would draw almost 30watts, you'd be smoking the resistor in seconds. In order to draw 10 watts at 12 volts you'd need a 15 ohm resistor.

Ian could you explain how a diode on the VREF would allow this circuit to function without either the sense element or the sense circuit having isolated grounds, or did you mean the diode would be required as well as isolation? The only way the voltage drop of the sense resistor would reach bellow 5 volts is if the power supplies output impedance was near 25 ohms if the sense resistor/circuit were at the same ground potential, that's also assuming the sanity adjusted 15ohm resistor. Even with the diodes to raise the ground the high side would always be off scale high (possibly terminally)
 
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