# Don't understand this op amp's circuit design

I'm trying to understand op amps and, to do so, I'm reading a manual. For now I was understanding everything but I have reached a point that I can't get.

The manual I'm reading (linked below) presents the following schematics:

And says:

[...] The positive op amp input lead is at a voltage of approximately 65 mV, and normal op amp operation keeps the inverting op amp input lead at the same voltage because of the as- sumption that the error voltage is zero.

I understand the part in which it is stated that the negative lead is at the same voltage as the possitive one, because that is one of the op amp's assumptions.

However, I can't see why the positive lead is at 65mV. I only see a voltage divider formed by Vcc, R2 and R1. How can it be 65mv?

Op amps manual: http://web.mit.edu/6.101/www/reference/op_amps_everyone.pdf

See page 69.

• If the resistors in the input voltage divider on the non-inverting input were swapped, the statement would be accurate Feb 4, 2021 at 15:06
• $$V_{+} = V_{CC} \times \frac{R_1}{R_1 + R_2} = 10V \times \frac{2k\Omega}{302k\Omega} = 66.225mV$$ Which means that there is an error in the schematic, R1 and R2 resistors need to be swapped.
– G36
Feb 4, 2021 at 15:08
• Si it seems to be an erra Feb 4, 2021 at 15:09
• Looks like it is Feb 4, 2021 at 15:10
• @Martel - it would seem so. Simply swap the values of R1 and R2 and it becomes accurate (there is no such thing as an error free text). Feb 4, 2021 at 15:10

It looks like a mistake : the positive end of the opamp is at potential $$\ V_+ = \frac{R_1}{R_1 + R_2}.V_{cc} \approx 9.934 V \$$.
Exchanging the values for $$\ R_1 \$$ and $$\ R_2 \$$ gives : $$\ V_+ = \frac{R_2}{R_1 + R_2}.V_{cc} \approx 66 mV \$$.
Hence it is most likely a typo, and the $$\ R_1 \$$ and $$\ R_2 \$$ values needs to be exchanged.