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Dec. 12th, 2007 06:51 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
nibotSuppose we have an opamp connected in the inverting amplifier configuration:
The gain of this thing is just -R2/R1, i.e., if R2 is 10000 ohms and R1 is 5000 ohms, then the output will be twice as large as the input, and with its sign flipped. That is well and good, but we are concerned about the noise introduced to the circuit. Noise is introduced both by the opamp itself, which, in our abstract view, is intrinsic to the amplifier; and by Johnson noise in the two resistors in the feedback network.
Horowitz and Hill give an expression for the noise of this circuit on page 447 of Art of Electronics, but they dedicate remarkably little space to it. Only about 8 3/4 square inches, actually, and their answer is somewhat confusing.
First, the amplifier noise. The amplifier noise is specified as an equivalent voltage noise e_a and current noise i_a seen at one (or both, you never can be sure what the datasheet is quoting) of the opamp terminals. We can pretend that the voltage noise e_a is all clumped at the noninverting input to the opamp. We then multiply that by the "noise gain," which is the transfer function from the opamp's noninverting input to the output of the circuit: 1+R2/R1. Done.
in R1 R2
o----+--/\/\/\------------+-----/\/\/\-------------+----o out
| |
+--|-\ |
| \__________________|
R_BIAS | /
+------/\/\/\-----------|+/ OPAMP
|
GNDThe gain of this thing is just -R2/R1, i.e., if R2 is 10000 ohms and R1 is 5000 ohms, then the output will be twice as large as the input, and with its sign flipped. That is well and good, but we are concerned about the noise introduced to the circuit. Noise is introduced both by the opamp itself, which, in our abstract view, is intrinsic to the amplifier; and by Johnson noise in the two resistors in the feedback network.
Horowitz and Hill give an expression for the noise of this circuit on page 447 of Art of Electronics, but they dedicate remarkably little space to it. Only about 8 3/4 square inches, actually, and their answer is somewhat confusing.
First, the amplifier noise. The amplifier noise is specified as an equivalent voltage noise e_a and current noise i_a seen at one (or both, you never can be sure what the datasheet is quoting) of the opamp terminals. We can pretend that the voltage noise e_a is all clumped at the noninverting input to the opamp. We then multiply that by the "noise gain," which is the transfer function from the opamp's noninverting input to the output of the circuit: 1+R2/R1. Done.
