Hello again,
We would all like to know why the current mirrors are being used. It could be as simple as extending the usable output voltage range (because it will do that), but i dont want to guess at this point what the original designer had in mind.
To clarify again, when i say the first stage i mean both the first op amp on the left and the (now) four transistors. The output of the first stage then appears across R1 because the right side of R1 is at virtual ground, and virtual ground is very close to zero volts.
As i am sure you know, a current mirror takes an input current of some level and outputs a current at that same level in the output branch. The two current mirrors are connected to the supply lines of the left op amp, so they mirror the positive supply current and negative supply current. Since their output junctions are connected together and to R1 that has the other end connected to virtual ground, the current through R1 is the difference in currents that the op amp draws from the positive and negative supply rails. So if the op amp draws 1.1ma from the positive supply rail and the op amp draws 1ma from the negative supply rail then the current through R1 is the difference 1.1ma-1.0ma which equals 100ua. So the current through R1 is:
iR1=Ip-In
where Ip is the positive rail current and In is the negative supply rail current.
Now when the left op amp has zero input, there is zero output (assuming zero offset for now). Zero output voltage produced zero current in R2, so the positive rail current is the same as the negative supply rail current which would be the quiescent current draw of the op amp. If this was 1ma then that would appear in both rails, so iR1=0 also.
When a positive voltage is applied to the left op amp, the output goes higher by the same amount. So for 1v input the output of the left op amp goes to 1v also. This produces a current equal to:
Iout1=Vout1/R2
This current must come from the positive rail, so now the positive rail current is higher than the negative rail current so we get a net output current from the two current mirrors of:
iR1=Ip-In
and since the current in R2 is now 1/10000=100ua we have iR1=100ua also.
We could now state the transconductance of the first stage, which would be:
gm=100ua/1v=0.0001/1=100uS
but it's also a simple matter to state the voltage output of the first stage because the current iR1 produces a voltage Vout1=iR1*R1 so the voltage is:
Vout1=Vin/R2*R1
so the voltage gain of the first stage is:
Vout1/Vin=R1/R2
The second stage (the right side op amp) has a gain R3/R1, so the overall voltage gain of the whole circuit is:
Av=(R1/R2)*(R3/R1)=R3/R2
We might be able to call the two current mirrors used that way a "Norton Amplifier" because it works on the input difference of two currents rather than the difference of two voltages, for what it is worth. More strictly speaking though we could probably call the two current mirrors combined with the output op amp stage a Norton Amplifier because it takes the difference of two input currents and amplifies that difference into a voltage output. We could look into this more.