*by Crutschow, Electro-Tech-Online community member*A Wheatstone bridge is sometimes used to convert a variable resistance, such as a thermistor or strain-gage, to a voltage with zero volts out at some particular resistance (such as at 0°C for the thermistor). A bridge circuit has the advantage of being able to generate a 0V null output for any arbitrary resistance, with minimal sensitivity of the null to the supply voltage. The disadvantages of the Wheatstone bridge are that the output voltage is differential across the bridge, and that voltage is a nonlinear function of any change in the bridge sense resistor.

The differential to single-end conversion, required for most applications, is typically done with a relatively expensive differential instrumentation amp or with a cheaper op amp connected as a differential amp, which requires some additional resistors, in addition, the bridge. And if a linear output is desired, then additional correction circuitry is required.

Discussed here is a circuit using only one op amp with no added resistors to provide both those functions for a bridge type circuit — differential to single-ended conversion with a linear output. It’s basically a variation of the typical 4-resistor differential op-amp circuit which uses the bridge for those 4-resistors, forming a sort of quasi bridge configuration.

By using the bridge sense resistor as the op amp feedback resistor and lifting that resistor from ground, the circuit generates a constant current through the sensor. The output is thus directly (linearly) proportional to the resistance change, not to the ratio between the resistance change and the other series bridge resistor, as in a standard bridge, which gives the nonlinear change of voltage versus the change. Although this is not a standard bridge, since one of the resistors is not connected to ground or the bridge supply voltage, it has characteristics similar to a bridge but with a linear, single-ended output.

This circuit is shown below:

The bridge voltage and resistor values were selected to give an output going from 0V @ 20°C to 1V @ 30°C.As previously noted, the sensor (thermistor) is placed in the op amp feedback loop so that its current is independent of its resistance. For

**, the thermistor (sensor) current is**

*R1=R2**.*

**It = (1/2 Vb) / (R3 + R_U2)**(Notice the current is independent of the thermistor resistance as

**is not in the equation).**

*Rt***Vout = 1/2 Vb – (It * Rt)**.

Pot U2 is for zero/offset adjustment.

The circuit output gain sensitivity to a change in sensor resistance is determined by sensor current ** (ΔVout = ΔRt * It)**, which is a function of the bridge voltage,

*. To adjust the sensitivity without affecting the zero setting, you can vary this bridge voltage. (For stability, this voltage should be derived from a voltage reference.)*

**Vb**In the example circuit, * Vb* was selected to give a

**of ≈1V for 10°C temperature change (999Ω), so the value of**

*ΔVout**to give a thermistor*

**Vb***current of 1mA is*

**It***, where*

**Vb = 2It*Rt = 2mA*2814Ω = 5.63V****is the thermistor resistance at 0V output.**

*Rt*Note that many single supply op amps don’t pull the output down close to zero with a single supply voltage (perhaps 50-100mV minimum), thus may need a pull-down load. (In general CMOS op amps, such as the one in the simulation, may be better at this than bipolar amps since MOSFETs don’t have a saturation voltage). This pulldown can be done with a resistor, but its value may need to be quite low to get near zero.

An alternative is to use a current mirror to ground drawing a current at slightly more than the sensor design current,** It**. That will pull the output down to near the saturation voltage of the current-mirror transistor without drawing additional current as the output voltage increases.