amplifiercircuit analysisdiff-ampinstrumentation-amplifieroperational-amplifier
I am simulating a difference amplifier circuit using LT1215 Opamp. I'm using LT1215 since I need supply voltage around 3V. The input voltage for both the terminals of the opamp is 200 μV and 100 μV. The output should have been around 100 μV but I'm getting output 6.24 mV to 6.2mV. The closed loop gain of the opamp is 1.
Why don't you get an ac signal source, attenuate it 500:1 with a 499 ohm resistor and a 1 ohm resistor and insert this into one grounded element of the bridge. DC offset introduced can be countered with another 1 ohm resistor in the bottom leg of the other grounded element.
Also shown is a method that doesn't break-into the bridge - it applies a high value resistor across one of the grounded bridge elements in order to simulate a shift in bridge output. This can be controlled with a fet fed with a squarewave as shown. It's even possible to use an opto-coupler in this type of application should the return line of the bridge be very sensitive.
Now you can inject (say) 1kHz at a known measurable amplitude and see what comes out from the instrumentation amplifier. If the AC gain is fine then you have to look at the dc issues that may be causing your problem BUT until you rid yourself of the doubts about ac gain you might keep going round in circles.
Each opamp is a high-gain differential amplifier. Therefore, the unit can do nothing else than to amplify the voltage difference between both input nodes - independent if one of the input nodes is grounded or not. This voltage difference is in the micro-volt range and does not allow real applications.
However, because of the large open-loop gain (Aol) of the opamp unit we always use negative signal feedback. For simple amplifier purposes the feedback network consists of resistors only. Referred to the applied signal voltage, the so-called closed-loop gain Acl is reduced correspondingly and the applied input voltage now can assume realistic values in the millivolt or even volt range.
However, the voltage difference directly BETWEEN the opamps input nodes is still in the micro-volt range. During calculation of opamp amplifiers (with feedback) this tiny voltage can be neglected in comparison with the two other voltages within the circuit (applied signal input voltage, opamp output voltage). This approach leads to the concept of "virtual ground" (for inverting opamp applications).
Best Answer
In the circuit shown, you are simply out of luck. Assuming the two inputs are at different frequencies, your expected output range is +/- 300 uV. With a minus power input of zero volts, you can never get less than 0 volts out, so the zero to -300 uV portion of your output range is simply impossible. Even the with a different op amp, the positive portion of your range is not likely to work, either. Op amps with rail-to-rail outputs do have some limits, and most don't do well when operating within a few millivolts of the rails. Since you want sub-millivolt outputs, I'm afraid you're doomed to disappointment.
What you need to do (whether you like it or not) is to provide something like -3 volts to the minus power input.