# Measure small currents without adding resistive insertion loss

**Figure 1**shows two typical methods of making the conversion. In one method, you insert a probing resistor, R

_{P}, in series with the current path and use differential amplifier IC

_{1}to measure the resulting voltage drop (

**Figure 1a**). A second method is a widely known operational amplifier current-to-voltage converter in which inverted IC

_{1}’s output sinks the incoming current through the feedback resistor (

**Figure 1b**).

In the first method, the same current that flows into one node flows from the second node, but a significant voltage drop occurs across probing resistor R_{P}. In the second method, the voltage drop is on the order of tens of microvolts to millivolts, depending on IC_{1}’s quality, but the measured current flows only into the sensing node with no return to the circuit. You can measure only currents flowing to ground.

The circuit in **Figure 2** operates in a somewhat similar manner to the one in **Figure 1b** in that an op amp’s output sinks the incoming measured current. However, the other op amp’s output sources an equal outgoing current back to the circuit under test.

In **Figure 2**, input current I flows through R_{1} into the output of IC_{2}, which reduces its voltage by the amount of IR_{1} relative to the input node. That voltage equals the voltage mean of the op amp’s outputs, which R_{3} and R_{4} set at the op amp’s inverting inputs. Consequently, the output of IC_{1} must rise to a voltage of IR_{2} relative to the inverting inputs and the equal-voltage noninverting input node of IC_{2}. IC_{1} sources this current, which returns through R_{2} to the circuit under test. R_{1}=R_{2}, so the output current is the same as the input current. Because the op amp’s outputs maintain their inputs at equal voltages, the circuit under test has virtually no resistance.

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