My Question

Why is my amplifier circuit amplifying more than I expect, and what can I do to fix it?

What I want to accomplish

I want to amplify an input that, at most, is 1.5[V] to, at most, be 2[V].

What I have tried

I have the below circuit set up. When I measure the voltage of OUT against GND, I get values that are 7 times higher than the value IN.

I used the following formula:

$$V_o = V_i * (1 + \frac{R_2}{R_1})$$

Plugging in \$2\$ for \$V_o\$ and \$1.5\$ for \$V_i\$ evaluates to:

$$R_1 = 3R_2$$

I tried using 300[Ω] and 100[Ω] for \$R_1\$ and \$R_2\$ respectively, which yielded a different, but also undesirable, gain. I recall that it was a bit lower.

What I got

Measuring the voltage at IN and OUT against GND using a multimeter gives me about 0.5[V] for IN and 3.5[V] for OUT.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Are you sure that's how you built it? That shouldn't work properly at all. No negative feedback, and your positive input is connected to ground through R1. What is the supply voltage for the LM358? \$\endgroup\$
    – JRE
    Commented Aug 18, 2016 at 8:40
  • \$\begingroup\$ @JRE sorry, the poles were flipped on the op amp. The supply voltage is 5[V], as seen on the left. \$\endgroup\$ Commented Aug 18, 2016 at 8:42
  • \$\begingroup\$ Since you are using positive feedback, the output likely saturates to bias voltage. Why not use negative feedback? Also the feedback should be connected to the other side of R1 not to GND. \$\endgroup\$
    – Durmus
    Commented Aug 18, 2016 at 8:45
  • 3
    \$\begingroup\$ That amplifier's open loop, its gain will be more like 100000 than 7, at least until it saturates. Follow the schematic for a non-inverting amplifier next time. \$\endgroup\$
    – user16324
    Commented Aug 18, 2016 at 10:41
  • 1
    \$\begingroup\$ The bottom part of R2 should go on top of the R1 resistor ... As it is the output will just go to the rail since anything is higher than ground (with V+ is tied to through the resistor R1) \$\endgroup\$
    – jbord39
    Commented Aug 19, 2016 at 1:13

3 Answers 3


Change to the following to get a non-inverting amplifier with gain = \$1 + R_2/R_1\$


The difference is that \$R_2\$ is connected to the op amp's inverting input instead of ground.

Please see Scott Seidman's answer for an explanation of what the incorrect circuit was doing.

  • 2
    \$\begingroup\$ @abligh Please take a realy close look and find the (important) difference ... \$\endgroup\$
    – Batuu
    Commented Aug 18, 2016 at 11:23
  • 4
    \$\begingroup\$ For those who like me thought "huh, that's the same!", R2 on the original diagram is connected to ground, as opposed to be connected to the negative input of the OpAmp. \$\endgroup\$
    – abligh
    Commented Aug 18, 2016 at 11:33

Your circuit is wrong. Try to connect the lower terminal of R2 to the upper terminal of R1 (negative input of the opamp). This would give you an non-inverting amplifier with Vo = Vi * (1 + R2/R1)

For proper simulation add a ground symbol.


To add to the other answers, because the inverting terminal voltage is FIXED AT GROUND (we know this because no current, in the ideal, can enter the input terminals, thus there is zero voltage drop across the resistor R1), you have no negative feedback. Because of the lack of negative feedback, the gain of your amplifier is the open-loop gain of the op amp (maybe \$10^5\$ or so), and the output is simply determined by how close the output can get to the rail.

  • \$\begingroup\$ Maybe it's good to know for the O.P. that his [original] circuit behaves like a comparator not a [linear] amplifier... \$\endgroup\$
    – user59864
    Commented Aug 19, 2016 at 8:26
  • \$\begingroup\$ @Nasha I prefer to avoid answers that might encourage people to use op amps as comparators. \$\endgroup\$ Commented Aug 19, 2016 at 10:32
  • 2
    \$\begingroup\$ Hmmm... I beg to differ. It should not be a pretext not to teach nor to learn in any way. After all, it's up to whom decides to use an opamp as a comparator (if at all). And you also might as well explain why you don't recommend this. \$\endgroup\$
    – user59864
    Commented Aug 19, 2016 at 13:20
  • \$\begingroup\$ @Nasha -- I've done so repeatedly. If you look around this stack, you'll find a trail of people who have problems with a circuit because they expect an op amp to behave like a comparator when used open loop, and almost no problems from people who start with an actual comparator optimized to function as a comparator. \$\endgroup\$ Commented Aug 19, 2016 at 14:35
  • \$\begingroup\$ Well, providing answers to such questions on a site with thousands of hits a day makes it bound to happen, I suppose... \$\endgroup\$
    – user59864
    Commented Aug 19, 2016 at 14:46

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