OpAmp's Input Confusion

Hi

When analyzing an OpAmp circuit using KVL/KCL/Nodal, how do the + and - signs designating non-inverting and inverting OpAmp inputs affect the analysis? How do we apply them?

Thanks.

vo = A ( v(+) - v(-) )
that is output voltage vo is equal to the difference in voltage between + and - inputs multiplied by the Open loop Gain of the op. amp.
In an ideal op. amp the current at the inputs is zero (infinite input impedance) and A is very large (--> infinite). Output resistance is zero.

albert22 said:

vo = A ( v(+) - v(-) )
that is output voltage vo is equal to the difference in voltage between + and - inputs multiplied by the Open loop Gain of the op. amp.
In an ideal op. amp the current at the inputs is zero (infinite input impedance) and A is very large (--> infinite). Output resistance is zero.

Click to expand...

So after finding the voltages at the terminals using any of the analysis methods, I use A ( v(+) - v(-) ) with the gain A always > 0?

Frankly it's not detailed in any of the supper expensive textbooks.

Thanks.

To take you back to basics, if you do the current analysis of an op amp circuit, the polarity of the circuit will naturally affect the output, just make sure you maintain the same current direction you chose for your input impedance across your feedback impedance.
Ex: for inverting resistive feedback op amp, you will have the amp pos input grounded, and the neg input tied to the input via resistor. make a directional assumption( in this case I will say vin > 0), so current of R1 = Vin/R1, meaning current is flowing from Vin toward the output. so the next step would be (0-Vout)/R2 because this is the same direction(right side node subtracted from left side node). combine to get Vin/R1 = -Vout/R2 solve for Vout you get -R2/R1 * Vin, note the sign has followed through. if you do the same analysis for a non inverting amp you will end up with (Vout - Vin)/R2 = Vin/R1, which when solving for Vout you get Vout=R2/R1*Vin + Vin => (R2/R1+1)Vin, note the polarity worked itself out, the trick is to make sure you dont change the direction of your current from your initial guess. Sorry if this was too much back to basics, just didnt want to make any assumptions.

Added quick note, if youre asking about the difference of the positive and negative inputs of the amplifier and why it matters which you connect to, its simply because of the internal workings of the amplifier, a pos change on the pos input will have a positive change on the output, while a pos change on the neg input will have a negative change on the output, due from the current steering in a diff pair(because a single ended output must be pulled off on 1 side of the diff pair ota, assuming single ended output)

Is this equivalent to assigning the same polarities of the input terminals to the elements ends connected to the terminals, except for the power sources, and this is regardless of the assumed current directions?

Yes if making a voltage subtraction circuit etc you would keep your input + on the + side of the amp and input - on the - side of the opamp, but this is also concluded from doing the current equations. fully differential amplifiers are a special case though , where you would have a Negative input on the in- side of an opamp via resistor or cap and Positive input on in+ side of opamp via resistor or cap, since it is fully differential, it is perfectly symmetric and therefor the +,- inputs outputs are swappable, as long as you swap the outputs around as well. The plus and neg notation on these are merely a point of reference.