Many integrated instrumentation amplifiers have architectures that permit offset compensation. The reference terminal’s voltage, V_REF, adds in phase to the output to yield a gain of one. As a result, you can reset the output offset voltage by applying to the V_REF input a correction voltage of equal value but of opposite polarity. If the instrumentation amp operates from a dual-supply voltage, you can easily provide both positive- and negative-correction voltage. However, some instrumentation amps operate from a single supply—for example, in a battery-powered application—to amplify a signal source or a sensor that introduces a positive offset voltage. A sensor such as the AD590 from Analog Devices, for example, produces an output current proportional to absolute temperature, and you should calibrate it at the lower reference temperature. In this case, the output swing of the instrumentation amp decreases, especially with high gain. To prevent this effect, you must apply a negative-correction voltage, which you generate from the positive power supply. In precision applications, the application of such a voltage may cause a problem.

This Design Idea shows you how to build an instrumentation amp operating from a single supply that permits you to reset the system offset by applying a positive-correction voltage to the V_REF input. The circuit in Figure 1 employs the dual high-precision OPA2333 op amp from Texas Instruments. This op amp can operate from a 1.8 to 5.5V supply and uses a proprietary autocalibration technique to simultaneously provide a maximum offset voltage of 10 μV and near-zero drift over time and temperature. It also offers high-impedance inputs that have a common-mode range 100 mV beyond the supply rails and rail-to-rail output that swings within 50 mV of the rails. Applying the superposition of the effects to the circuit in Figure 1 yields the following equation:

To achieve equal gain for both the V_B and the V_A inputs, resistors R₂, R₃, R₄, and R₅ must have equal values that are double the value of R₁. Using the resistor values in Figure 1, you obtain the following simplified equation:

The amplifier’s differential gain is 3+(92.8 kΩ/R_G), and the reference voltage is added, inverted together with the output signal. Resistor R_G sets the gain, and, if you do not connect R_G, the gain assumes the minimum value, which is three; decreasing the value of R_G to 93Ω increases the gain to 1000.

The V_REF input requires a low-impedance connection to preserve a good CMRR (common-mode-rejection ratio); otherwise, you can use an op-amp buffer for better CMRR, which depends mainly on resistor-ratio matching. In this implementation, to preserve an acceptable CMRR, you must use precision film resistors. Analyzing the circuit, you can calculate the worst-case CMRR at low frequency. With R₂, R₃, R₄, and R₅ all of equal value and double that of R₁ and with all the resistors having equal tolerance, you obtain:

where ΔR/R is the resistor’s tolerance. If the tolerance is 0.1% and with the minimum differential gain, which is three, you obtain a CMRR of at least 54 dB. With a differential gain of 100, you obtain a CMRR of at least 84 dB.

The V_REF input can reduce the system offset to the lower output-swing limit but does not reset it completely because, in that case, the output voltage would be unable to reach the single-supply ground. If you want instead to reset the output offset, you can subtract this value using an ADC with differential inputs (Reference 1).