Is it possible to make a non-inverting difference amplifier?

Question

For a crude current sense application I am wondering whether it is possible to build a non-inverting difference amplifier without common mode offset on the output.

Reason: I would like to avoid having to create a negative supply voltage just for this function.

I.e. I am looking for the following output: Vout = gain * (V2 - V1) where V2 > V1

Answer 1

I don't see any problem. Connect your V1 and V2 as shown here

and make

gain\$=\dfrac{R_f}{R_1}=\dfrac{R_g}{R_2}\$.

Use precision resistors, to have an acceptable common-mode rejection ratio. If you need a high CMRR, use an integrated difference amplifier.

Answer 2

If what you need is just high CMRR, an instrumentation amplifier may be the best choice. Despite the name, you can use it everywhere and it behaves like a fully differential amplifier (usable also in cases where the common mode input is hundred times bigger than the signal).

The conceptual schematic of that amplifier is this:

$$ V_{OUT} = \left(1 + \dfrac{2 \cdot R_1}{R_{GAIN}}\right) \dfrac{R_3}{R_2} (V_2 - V_1) $$

As you can see, the symmetric structure gives a great performance in rejecting common mode, and the offset of the third op-amp is made less important by the gain in the previous stages.

Most (if not all) the amplifiers have offset correction, which can improved also using an external resistor; they have also programmable gain, also using a resistor.

Here you have a table from Analog Devices where you can choose the proper one.

Update

For measuring currents, you have also the choice of dedicated current sense amplifiers or, even more appropriate, high-side current monitors (like markrages suggested in the comment and here).

Answer 3

I agree with clabacchio's instrumentation amplifier solution to get a good CMRR. You can roll your own, but integrated InAmps are available at reasonable prices. Integrates ones will also have better resistor matching than you get with discrete components.

A word of warning, though, and for that we need the schematic:

Chances are that you use a high side current sensor, where \$V_2\$ is connected to \$V_+\$. If that same \$V_+\$ is the positive supply for your InAmp there's a problem. If \$V_2\$ > \$V_1\$ then there will flow a current through \$R_1\$/\$R_{GAIN}\$/\$R_1\$ from the bottom opamp to the top opamp. To make the inverting input of the bottom opamp equal to the non-inverting input (\$V_2\$) the output has to to go higher than that, and if \$V_2\$ = \$V_+\$ it can't do that. So the InAmp won't accept input voltages all the way to the rails.

The common difference amplifier from Telaclavo's answer doesn't have this limitation since the resistor divider \$R_2\$/\$R_g\$ will bring \$V_2\$ down from \$V_+\$.
For minimum offset error choose \$R_1\$ = \$R_2\$ and \$R_f\$ = \$R_g\$.

The AD820 is rail-to-rail output and has offset null inputs.