Single Supply Op - Amp circuit not working

I have made a circuit like this

I am attempting to convert the -2.5V to +2.5V. I have got this circuit running using +12V and -12V but now intend to run the op amp in single supply mode. But my output is always 11.01 voltage whatever be the input. What am I doing wrong?

• Are you referring to U2.6 as your output? Commented Feb 11, 2013 at 10:52
• @abdullahkahraman yes Commented Feb 11, 2013 at 10:53
• Whether this can possibly work or not will depend on the opamp, and you don't tell us that. You need a type of opamp whose common-mode input voltage can equal or exceed the negative supply.
– user16324
Commented Feb 11, 2013 at 12:06
• Key point: "Ground" is at +6V relative to your ground signal. All else follows from that ||| Detail: Your effective "ground" point is now at +6V. IF your -2.5V signal is relative to this +6V signal then your circuit will work. BUT if the -2.5V is relative to the ground signal = 0V on the 12V supply , then U2 is "seeing" -8.5V wrt to the 6V on U2.3, and is trying to set U2 output to +6V + +8.5V = 14.5V. 11.01V is max possible with a 12V supply. || NB the 741 was retired years ago. Use an LM324 or LM358 or better. They will produce the same result here BUT are happier in single supply mode. Commented Feb 11, 2013 at 13:16
• You only tell us the output you actually get, but what do you expect to see? And when running in single supply mode, what happens to the -2.5V at R4? Commented Feb 12, 2013 at 19:09

An op-amp amplifies the difference between its inputs, not the difference between an input and whatever happens to have the ground symbol attached to it in your schematic. Look at the dual-supply version of the inverting amplifier:

It amplifies voltage differences from ground because the non-inverting input (+) is attached to ground. It could just as easily be any other voltage, and this is why single-supply op-amp circuits can work.

In your circuit, the non-inverting input isn't attached to ground. It's attached to 6V. An op-amp with negative feedback will strive to make its inputs equal through the feedback path. Your input is -2.5V, and it wants the inputs to be equal, so the voltage drop over R4 should be $6V - (-2.5V) = 8.5V$. The voltage drop over R3 should be equal, since they are the same value, and the op-amp bias current is assumed to be none. Thus, the output would have to be $6V + 8.5V = 14.5V$. Of course, this is impossible for an op-amp with a 12V supply, so it does the best it can manage, 11.01V.

You can decrease the voltage at the non-inverting output by making R1 bigger and R2 smaller to move the virtual ground to a lower voltage, but you won't be able to move it all the way to 0V relative to ground because an op-amp generally can't operate all the way to the rails. Read more about common-mode input range to learn more. You could use a rail-to-rail op-amp to get closer. Or, you could continue to use a negative supply. Or, you could add a voltage to your input with a zener diode in series with it. Or, you may be able to AC couple the inputs. Which option is best depends on what the desired behavior when the input isn't -2.5V.

Here are some ideas for converting -2.5V to a positive voltage, without an op-amp:

The core of the problem is that -2.5V is actually -8.5V relative to the 6V bias of your op-amps, and so the circuit tries to change -8.5V to 8.5V which is really 14.5V, as others have pointed out.

You must do all your calculations relative to the 6V virtual earth. The output that you want, 2.5V, is actually -3.5V below the virtual earth. The circuit must be designed in the same way as a dual supply circuit which converts -8.5V to -3.5V.

An amplifier that goes from -8.5 to -3.5 is noninverting. But noninverting amplifiers cannot have a gain of less than unity. What you can simply do is attenuate the input voltage and buffer it.

With the inverting stage, you have a voltage divider from the input to the output of U2. Why don't you make a voltage divider from the -2.5V input to the 6V output of U1 instead: a voltage divider whose resistors are chosen such that it puts out 2.5V. Then U2 can be a noninverting stage which just buffers the 2.5 volts.