Is there any practical difference as to how a unity gain buffer using an op-amp might be configured? For example, the input signal can be connected to the noninverting input and the feedback to the inverting input; but it can also be configured the other way, with the inverting input as the signal and the noninverting as the feedback. Practically speaking (noise, stability, frequency response, offset error and so on), is there a difference between these configurations?
Electronic – Difference between configurations of unity gain buffer using op-amp
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It's best to think of this circuit in two sections. Everything before R4 belongs in the first section. R4, and everything after it, belongs in the second section. Now the first section looks like a standard op-amp noninverting voltage amplifier. As you probably know, the gain of such an amplifier is equal to \$1 + Z_f/Z_{in}\$, where \$Z_f\$ is the feedback impedance and \$Z_{in}\$ is the input impedance. The second section is an inverting voltage amplifier, with gain equal to \$-Z_f/Z_{in}\$.
The gain of the first section (with the LMC6001 op-amp) is therefore equal to 1 + (R2+R_V)/R1, where R_V is the variable resistance setting of the R3 potentiometer. The simplest way to change this to a unity-gain buffer is actually just to remove R1 and short the V- terminal to the output. The op-amp forces V- to equal V+, so if V- is shorted to the output, then the output must also equal V+.
The gain of the second section is -R9/R4 = -1. You correctly observe that the network connected to the positive input of the LMC6041 is supposed to add a DC offset. So this section is already at unity gain, but it's inverting.
The capacitor C1, in combination with R9, acts as a low-pass filter with a cutoff frequency of 0.7 Hz. That makes sure that the output voltage changes slowly, so you can read the output with a multimeter.
How about this? You do give up very high impedance input, but the resistors do not have to all be 1K.
Best Answer
A normal opamp has an infinite gain, practically [factor] x 10^5. The difference between + and - terminal determines its output:
Vout = (V+ - V-) * A_ol
For an opamp you will have 2 rules:
When you make a real circuit, you reduce the open loop gain to a closed loop gain. However, the 2 rules stated only work for negative feedback. If you use positive feedback, they do not apply.
So, if the rule of no input voltage difference doesn't apply, the opamp basically becomes an comperator. An inverting situation would try to get the difference to 0V because of its feedback. Now it will be become a simple comperator with Vout=H if V+ > V-, Vout=L if V+ < V-. In an wrong unity gain buffer, you'll see Vout=L because V+ is lower then the signal you're feeding it with.
Because I couldn't believe both situations would simulate the same, I did it myself:
Just 2 opamps which are internally fed to +/-15V. They follow a 1kHz 10Vpp source. The results are:
(Note: Colors are inverted, so green = purple, cyan = red)
Oh so they do amplify correctly. But the ideal opamp has an infinite gain, no offset voltages, no input bias currents, no bandwith limitations (however, we wont notice much of that at 1kHz) etc. If we look at a real opamp, I picked one randomly (TL031):
An now it suddenly clips, because the opamp doesn't have the correct feedback.