operational-amplifier
I want to use an op-amp as an inverting amplifier to convert negative voltages to positive voltages,
simulate this circuit – Schematic created using CircuitLab
If my input varies from 0 to -5V and I want from 0 to +5V as output, is it Ok to connect V+ to 5V and V- to ground?
From what I know, the rail voltages just saturate the output between the V+ and V- values but I'm not sure about how it sources and sinks the functions internally.
This should work, but usually there's also a resistor from the non-inverting input to ground:
If \$\dfrac{R4}{R2} = \dfrac{R3}{R1}\$ then:
\$V_{OUT} = \dfrac{R_{3}}{R_{1}} \times (V_{IN+} - V_{IN-})\$
To minimize offset error both inputs have to see the same impedance, and therefore \$R1 = R2\$ and \$R3 = R4\$.
Omitting R4 will only give a scaling factor for the voltage on the non-inverting input, namely:
\$V_{OUT} = 2 \times V_{IN+} - V_{IN-}\$
but changing the setting for the lower potmeter should definitely have a result in the output. Did you measure the voltages on both inputs?
What happens if you set the lower voltage to 2.5V and the upper to 1V? The inverting input should also be 2.5V, and the output 4V. What do you measure?
Note: especially the lower voltage follower is not necessary in your version; the opamp's input current is low enough to be negligible for most uses, and by the way, you're connecting the potmeter's wiper to an exactly same input!
Further reading
differential amplifier tutorial (interesting site overall!)
The op-amp can't do anything useful in this configuration. You need to give it more supply voltage than the desired output voltage, even though it's "rail to rail".
If you ask it to output 8V or even 11V with a 12V supply, it may be able to do that, provided the load is not too much for the op-amp to supply. In this case, that might be 10mA or so.
I don't know exactly why you're only getting 4V at the output, but I suspect it's because the load is something like a couple hundred ohms or equivalent, so the output current is limiting at 20-30mA.
Even if you put a lighter load on there, my first comment applies- this circuit can not do anything useful without some changes (more supply voltage compared to desired output voltage).
Best Answer
A couple of things.. the op-amp won't get all the way to the positive rail, how far it gets will be 'splained by the datasheet. Note that so long as the amplifier is balanced, the inverting input is a virtual ground so the feedback resistor acts as a load on the amplifier output, just like a resistor to ground. Maximizing the resistor value will allow the output to get closer to the rail, but that difference might be 10mV or it might be 2V, depending on the op-amp. A 'rail-to-rail' type will usually get to within some hundreds of mV or better with a 'reasonable' load.
When the amplifier output is very close to either rail, the characteristics change somewhat and you might find micro-oscillation going on or other strange stuff that requires some thought to compensation. Increasing the feedback resistors generally decreases the phase margin and increases the likelihood of strange stuff happening, as well as increasing errors due to input bias current and leakage, so it's a trade-off.
If you're going into an ADC with a 0-5V range you might choose to use a 0.5~4.5V output (requiring an offset so a couple more resistors) to make sure you can actually measure 0V in and -5V in. The cost of that is a loss of 20% of your resolution or about 1/3 of a bit.
The input common mode range must include ground for this application but it does not have to be R-R input, a device advertised as "single supply" with "rail-to-rail output" is probably going to be your best choice.