Hello all!

Can anyone please help me with calculating the gain of this Op-Amp circuitry? The design note from which it has been picked up claims it's gain to be -9.91. However, my calculations reveal it to be the exact half.



Any guidance would be really appreciated.

 

Assume for now that Vin = +1V from this we can say that the voltage at A2 "+" input is also +1V. And voltage at A1 "-" and "+" inputs is equal to 1V * 10/110 = 90.0mV.

Hence the voltage across a lower A1 100kΩ resistor (at A1 + input) is (1V - 90.0mV) = 0.9090V so the current is 9.0909μA.
So, the voltage at A1 output is 1V + 10kΩ*9.0909μA = 1.09090V
Current through A1 feedback resistor (10kΩ) is (1.090V - 90.0mV)/10kΩ = 100μA

And finally the output voltage is Vo = ( 1V * 10/110) - 100μA*100kΩ = -9.909V

And if we renamed the 100kΩ as a R2 and 10kΩ as a R1 the gain is equal to:

Av = R1/(R2 + R1) - R2/R1 = 10/(100 + 10) - 100/10 = -9.9090 V/V

 

That is certainly helpful, Jony130. Thank you very much.

I assumed that the current through the feedback resistor would be same as that of the lower 100K resistor (at +IN of A1) as well as the 10K resistor before the feedback resistor (at +IN of A2). Which, I suppose, is not correct. Since the triangle is a blocked representation of the actual circuitry. The power ports are omitted, and so we happen to forget about them. Which effectively means that the current sourced/sinked by the output of an op-amp may not necessarily be the same as through the feedback network!

 

That is certainly helpful, Jony130. Thank you very much.

I assumed that the current through the feedback resistor would be same as that of the lower 100K resistor (at +IN of A1) as well as the 10K resistor before the feedback resistor (at +IN of A2). Which, I suppose, is not correct.

Nope, it's not correct. In order for the current in the feedback resistor to be the same as the current in the lower 100 kΩ resistor, the current in the upper 100 kΩ resistor would have to be the same as well (the feedback resistor is effectively in series with the upper 100 kΩ resistor since we assume negligible current flows into the opamp inputs). But this would require that the voltage drop across each of the 100 kΩ resistors be the same. While the voltage at the right end of each resistor is the same (because we assume the opamp is in the active region and has driven the differential input voltage to zero), there is no reason to assume that the voltages at the left ends (pins 1 and 8) are the same.

There is also no reason to believe that the current in the feedback resistor of the first opamp is the same as the current in the bridging resistor to the second opamp since they are not effectively in series. That would require that the assumption that the opamp is neither sourcing nor sinking any current at all, which would make its presence rather pointless. Instead, we assume that it will source or sink whatever current is needed in order to drive the differential voltage at its inputs to zero.

Since the triangle is a blocked representation of the actual circuitry. The power ports are omitted, and so we happen to forget about them.

Who's we?

Which effectively means that the current sourced/sinked by the output of an op-amp may not necessarily be the same as through the feedback network!

This makes no sense, particularly in the context of what you were assuming earlier. If the output of the opamp was the same as what was in the feedback resistor, then there would be NO current going through the bridging resistor, but you assumed that ALL of the current in the feedback resistor went through the bridging resistor.

Can't have it both ways.

 

Thank you very much, WBahn, for pointing that out.

This makes no sense, particularly in the context of what you were assuming earlier. If the output of the opamp was the same as what was in the feedback resistor, then there would be NO current going through the bridging resistor, but you assumed that ALL of the current in the feedback resistor went through the bridging resistor.

Seems like I am not very good with words, my apologies.

What I meant there is that the current in the feedback network may not be the same as that drawn by the load. The op-amp may source/sink any amount of current, irrespective of the current through the feedback network, needed to drive the load at the output voltage (not considering the overload condition at the moment).

Instead, we assume that it will source or sink whatever current is needed in order to drive the differential voltage at its inputs to zero.

Something very similar to what you wrote.