diff-ampdifferentialoperational-amplifier
Is there any reason why the gain of a single ended input BJT differential amplifier, with one Rc, is a good approximation of the value of Adm for a common mode input.
The simulation shows the frequency response for a single ended input and the circuit below shows th common mode input circuit
You cannot just connect the probe across inputs to the inamp and expect good results.
Operational amplifiers need a path to ground from each input. So the second circuit is better.
the 741 is totally unsuitable, but the LMC6041 in the notes should work well for this circuit. (2 fA is typical rating though, use the guaranteed value of 4 pA for design unless you want your circuit to only mostly work... 4pA through 300Mohm sensor is 1.6mV offset.)
LMC6041 will be happy with a single supply, so if the -5V supply is inconvenient have a look at this circuit: What is the purpose of this op amp?
I suggest you stop simulating and actually build the circuit. LMC6041 is $2 from a distributor or ask National, they'll give free samples for school projects. You don't need to hook up the ADC yet, just get the probe and signal conditioning amplifier working. Read the output with a voltmeter. Just like software, breaking the problem into small pieces is the way to make progress.
does cmrr value make sense for a differential amplifier if its inputs are separate and totally different voltages?
Yes, it makes sense. Consider two independent voltage sources as the inputs to a differential amplifier:
simulate this circuit – Schematic created using CircuitLab
This is equivalent to:
simulate this circuit Where we've defined the differential and common mode voltages to be:
$$V_d = V_{in1} - V_{in2}$$
$$V_{cm} = \dfrac{V_{in1} + V_{in2}}{2}$$
Best Answer
When you apply no AC signal to Q1's base, the shared emitter connection is still largely held at a small signal value and the gain of the right-side input to the left-side output is the same formula as the full differential mode amplifier gain: -
$$\dfrac{R_C}{2\cdot r_E}$$
The "2" factor in the above equation is because we are applying a differential signal. This means that the voltage gain is halved because each base only receives 50% of the full differential signal.
When applying a single ended signal to a differential amplifier the gain is twice as high but given we might only be applying half the full differential signal, it becomes the same formula.