The op-amp tries to keep its two input terminals at the same voltage, so as the non-inverting (+) input changes, the op-amp tries to output whatever voltage it can to get the inverting (-) input to the same voltage. Since the voltage at the inverting input is set to 1/10th of V(out) by the resistor voltage divider ratio R1/(R2+R1), V(out)=R1/(R2+R1)*V(in).
You can vary the input over the range -1 to +1 volts and see how the output reacts, just as you might if you were playing around with this circuit on a breadboard.
V1 is currently configured to output a 1kHz sine wave, and you can see the output as well.
Try the Bode simulation mode. At low frequencies, we get a gain of +20dB (a factor of 10^(20/20) = 10 in voltage) with a phase of 0 degrees. You'll see that at higher frequencies, the circuit no longer provides a gain of 10! You're seeing the op-amp's Gain-Bandwidth product at work. If you change R2 to be 99K instead of 9K, you'll notice that the amplifier's low-frequency gain increases by a factor of 10, but its bandwidth (the frequency range before this gain starts to fall off) drops by a factor of 10.
Op-amp inverting amplifier:
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How did you get that op-amp? |
May 27, 2013 |
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from components list. |
May 27, 2013 |
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Obviously the formula for V(out) V(out)=R1/(R2+R1)*V(in). is wrong. Just enter R1 = 1 kOhm and R2 = 9 kOhm and you'll end up with a Gain of 1/10 (instead of 10). Correct is "V(out)=(R2+R1)/R1*V(in)" |
May 07, 2024 |
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