How to calculate gain?
2 Attachment(s)
The gain is determined by a bridge of R1R4 that is intentionally unbalanced by R5.
How come the gain becomes 30X? How to calculate the gain for this "intentionally unbalanced R1R4 bridge"? 
R2 and R3 are shorted at lower frequencies, so the resulting impedance is R2/R3 = 1K
Same goes for R4 and R5 (but for all frequencies), so R4/R5 = 909R Makes the calculation a bit easier ;) 
Hi, ACD,
I still don't understand this. To get gain 30x there should be a resistor about 29k or 30k value somewhere in the network. But in here, all the resistors are only 2k and 1k, how come the gain becomes 30x? 
Just notised a mistake :xeye:
R2 and R3 doesn't get shorted as R2 is connected to  Input and R3 to + Input... Sorry for the confusion :rolleyes: 
Hi,
I can't see it either. Stick the front end numbers in a sim and see what it spits out. 
The key is "balanced bridge" of 4 resistors. And the references for 2 of them is to output node. But how to calculate gain?

2 Attachment(s)
How to merge this 2 equations, so the result is Vout/Vin in terms of R1,R2 and R3?

The gain can be determined by solving the circuit equations. Neglect the effects of L1 and R13, and assume an ideal op amp.
The equations are: (VinVplus)/R1 = (VplusVout)/R2 Vplus=Vout*Rp/(Rp+R3) Where Vplus is the voltage at the op amp input terminals. (For an ideal op amp the voltages are the same at both input terminals.) Rp is the parallel combination of R4 and R5. Solving the above equations gives: Vout/Vin=R2*(Rp+R3)/(Rp*R2R1*R3) The gain for this circuit is 32. 
how about the usual way?  do the voltage divider algebra with the ideal op amp assumption that the opamp +, inputs are held exactly equal by the loop gain

Hi, Sawreyrw,
Thanks for the equation :D One dumb question, the gain has () sign infront of it, is this circuit is an inverting amplifier? 
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