Measurement Technology › Op-Amp Circuits › Further Amplifier Types

Current-to-Voltage Converter

Some sensors provide a current as their electrical output quantity. One example is the photodiode, which outputs a current that depends on the light. In biology there are processes that emit light, and its intensity can be measured with photodiodes. We are looking for a solution that converts a current into a suitable voltage for the ADC in the analogue signal processing. This is done by the current-to-voltage converter, also called a transimpedance amplifier (TIA). I will use this abbreviation from now on.

The TIA is based on the inverting amplifier.

Inverting amplifier with currents I_R1 and I_R2

\[ \begin{gathered} \text{Mesh with the input voltage:} \\[6pt] -U_{\mathrm{Ein,OP}} + U_{R1} - U_d = 0\,\mathrm{V} \rightarrow U_{R1} = U_{\mathrm{Ein,OP}} \text{ with } U_d = 0\,\mathrm{V} \\[6pt] \text{At resistor } R_1\text{: } I_{R1} = \frac{U_{R1}}{R_1} = \frac{U_{\mathrm{Ein,OP}}}{R_1} \\[6pt] \text{Node equation: } I_{R2} = I_{R1}\text{, no current flows into the op-amp} \\[6pt] \text{At resistor } R_2\text{: } U_{R2} = R_2 \cdot I_{R2} \\[6pt] \text{Mesh with the output voltage:} \\[6pt] U_d + U_{R2} + U_{\mathrm{Aus,OP}} = 0\,\mathrm{V} \rightarrow U_{\mathrm{Aus,OP}} = -U_{R2} \text{ with } U_d = 0\,\mathrm{V} \\[6pt] U_{\mathrm{Aus,OP}} = -U_{R2} = -R_2 \cdot I_{R2} = -R_2 \cdot I_{R1} = -\frac{R_2}{R_1} \cdot U_{\mathrm{Ein,OP}} \\[6pt] \text{Key equation for the TIA: } U_{\mathrm{Aus,OP}} = -R_2 \cdot I_{R1} \end{gathered} \]

The inverting amplifier outputs the current in the input mesh multiplied by R₂. If a sensor feeds this current directly into the node between R₁ and R₂, the output voltage of the op-amp is already proportional to the input current. A circuit for this looks as follows:

Current-to-voltage converter (TIA): current source I_S at the minus input, feedback via R_2

\[ \begin{gathered} U_{\mathrm{Aus,OP}} = -R_2 \cdot I_S \\[6pt] H = \frac{U_{\mathrm{Aus,OP}}}{I_S} = -R_2 \end{gathered} \]

The sensor current is modelled as an ideal current source. So you connect a photodiode directly to the op-amp input.

A real sensor is always a real current source. It therefore has an output resistance. So let us change the model of the sensor in the circuit:

Current-to-voltage converter with internal resistance R_i of the current source

\[ \begin{gathered} \text{Mesh with the current source:} \\[6pt] -U_{Ri} - U_d = 0\,\mathrm{V} \rightarrow U_{Ri} = 0\,\mathrm{V} \text{ with } U_d = 0\,\mathrm{V} \end{gathered} \]

The voltage U_d between the two op-amp inputs is always 0 V. The current source and its internal resistance are in parallel with the voltage U_d. So there is no voltage across either the current source or the internal resistance. The current through the internal resistance – which would actually falsify the measurement – is always 0 in this circuit. That is why it is popular for current measurement.