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I have to design an amplifier that has $$V_{out} = 2 V_1 + 3 V_2 + 5 V_3$$ supposed to use two op-amps and as many resistors as needed. But i cant figure out resistors |
November 25, 2016 |
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How about this: There are two stages: the first stage (OA1 and R1,R2,R3,R4) does the weighted summing. The second stage (OA2 and R5,R6) is simply an inverting gain of -1. The resistors R1,R2,R3 are chosen so as to set the individual gains for each input. Because node A is a virtual ground, each input voltage source V1,V2,V3 contributes a current proportional to its voltage: $i_1 = \frac {V_1} {R_1}$ and so on. These currents are added together, $i_{total} = i_1 + i_2 + i_3$ and flow through resistor R4. The resulting voltage $$V_{INVSUM} = -R_4\big(\frac {V_1} {R_1} + \frac {V_2} {R_2} + \frac {V_3} {R_3}\big)$$ So arbitrarily picking $R_4 = 10k\Omega$ then we have $2=\frac {10k\Omega} {R_1} $ or $R_1 = \frac {10k\Omega} {2}$ -- and so on for the other two inputs, $R_2 = \frac {10k\Omega} {3}$, $R_3 = \frac {10k\Omega} {5}$. In practice you could adjust all of these to find real resistor values that you could actually buy or adjust with a potentiometer. The second stage just inverts $V_{INVSUM}$, so $$V_{OUT} = R_4\big(\frac {V_1} {R_1} + \frac {V_2} {R_2} + \frac {V_3} {R_3}\big)$$ Click the circuit & try the simulation! |
by mrobbins November 27, 2016 |
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