# Non-inverting current to voltage converter

Using a single opamp with a resistor it is very easy to make a I to V converter, as shown in the diagram below. Note that if current is flowing from I_in towards the op-amp (let's call that a positive current), then V_out will be negative and vice versa. We can easily make V_out have the same sign as I_in by adding another op-amp in the inverting configuration.

However, is it possible to make an I to V converter that preserves the sign using a single op-amp?

(Of course this is possibly by building another opamp from discrete components but this is not what I mean, I just wondered whether there is some simple circuit that can achieve this with one op-amp.)

EDIT: Of course one could just take a resistor to ground for that (and optionally buffer the voltage). But this means the "input" is at the output voltage. But I'd be interested to have a convert for whicht he input is also kept at 0V just like in the inverting configuration.

simulate this circuit – Schematic created using CircuitLab

simulate this circuit – Schematic created using CircuitLab

Figure 1. A non-inverting I to V converter (a) buffered and (b) unbuffered.

• Thanks for the answer. The advantage of the inverting converter I posted above is that the input is always at 0V, which is a drawback of the circuit you suggested, that makes it less suitable for some applications. Is there a way to modify it to to also have the input at 0V? Jun 6 at 15:29
• @flawr, any current source should have some compliance, the range of output voltage of a constant current power supply, over which the load regulation is within certain limits. It represents the maximum voltage a current source will reach as it attempts to produce the desired current. You need to specify your current source's compliance voltage, I suspect. Jun 6 at 15:44
• @flawr The input common-mode being non-zero is a well known, and pretty much unavoidable, flaw of non-inverting op-amp circuits. Jun 6 at 15:44

If you want an I-V converter which keeps its input at 0V, then it needs to sink a current equal to the input current. In the opamp IV case, with positive input current (current going into the input), the opamp's output goes negative, and a current equal to the input current flows into the resistor.

However, there is no component that can make a current source with zero dropout voltage. To sink current from a node which is at 0V, the component will require a negative power supply.

So you can't get rid of the negative power supply (but you could if you decided to hold the input voltage at say a bit above 0V).

And... with a single opamp, it has to be inverting. So you're stuck.

You could use a resistor and an opamp as Transistor suggested, a very low value resistor and a high gain opamp would give an input "pretty close" to 0V, but this introduces more offset errors of course.