0V Dropout Shunt Using Instrumentation Amplifier

Passive shunts measure the current flowing through a relatively small resistor, typically with a full-scale voltage drop of 60 mV for higher power devices and 200 mV for electronic instruments. Similarly, current-to-voltage converters measure the current flowing through a sense resistor, which typically has a higher voltage drop. However, in some cases, the voltage drop between the input and ground must be as low as possible, with 0V being the ideal value (regardless of the current being measured). If your application requires this characteristic, you can use the current-to-voltage converter shown in Figure 1. In this circuit, resistor R1 acts as a classic current sense resistor, and the instrumentation amplifier senses the current being measured across it, thereby obtaining a voltage drop. The instrumentation amplifier and R1 not only act as an inverting current-to-voltage converter, but also establish a voltage through a resistor network at Point B. The magnitude of this voltage is equal to the voltage drop across R1, and the polarity is opposite to ΔVR1. The end result is that the voltage at input A is an ideal 0V, regardless of the magnitude and polarity of the current flowing into the input.

The design uses the Analog Devices AD8223 instrumentation amplifier because it has a default voltage gain of 5, which is close to the ideal value with high accuracy. The typical gain error at default gain is 0.03%, and the worst-case error for B-grade ICs is 0.1% (Reference 1). When the gain is 5 and R1 and R2 have the same value, it can be concluded that for a 0V voltage drop at input A, the value of R3 is twice that of R2 (Figure 2). R1, R2, and R3 in Figure 1 should all be high-precision, low-temperature-coefficient resistors. In the experimental circuit with R1 and R2 values ​​of 20Ω, there is a zero-drift current of 0.8 µA referenced to the input terminal, and the voltage drop at input A changes by 0.27 mV at 1 mA input current. For negative input currents, a similar negative voltage change will appear at input A. The transfer constant (or mutual resistance) of the circuit is: (ΔVOUT)/(ΔIIN)=–5R.

Therefore, if the input current is 1 mA, -100 mV appears at the output. Because the AD8223 output can source about 2.5 times the current it can source, the input range for positive current is also 2.5 times greater. You can further increase the positive and negative current ranges by increasing the supplies from ±5V to ±12V; you can also use 12V and -5V. If your design requires higher input currents, place a precision voltage buffer with a correspondingly high output current capability between the in-amp output and resistor R3.

References
1. “Single-Supply, Low-Cost Instrumentation Amplifier, AD8223,” Analog Devices Inc, 2008.