Circuits to help verify matched resistors

Texas Instruments LM336-2.5 OPA2197

Analog designers often need matched resistors for their circuits [1]. The best solution is to buy integrated resistor networks [2], but what can you do if the parts vendors do not offer the desired values or matching grade?

The circuit in Figure 1 can help. It is made of two voltage dividers (a Wheatstone bridge) followed by an instrumentation amplifier, IA, with a gain of 160. R3 is the reference resistor, and R4 is its match. The circuit subtracts the voltages coming out of the two dividers and amplifies the difference.

The intuitive solution is a circuit made of a Wheatstone bridge and an instrumentation amplifier.
Figure 1. The intuitive solution is a circuit made of a Wheatstone bridge
and an instrumentation amplifier.

Calculations show that the circuit provides a perfectly linear response between output voltage and resistor mismatch (see Figure 2). The slope of the line is 1 V per 1% of resistor mismatch; for example, a VOUT of –1 V means -1% deviation between R3 and R4.

Circuit response is perfectly linear with a 1:1 ratio between output voltage and resistor mismatch.
Figure 2. Circuit response is perfectly linear with a 1:1 ratio between output voltage
and resistor mismatch.

A possible drawback is the price: instrumentation amplifiers with a power supply of ±5 V and more start at about 6.20 USD. Figure 3 shows another circuit using a dual op-amp, which is 2.6 times cheaper than the cheapest instrumentation amplifier.

This circuit also provides a perfect 1:1 response, but at a lower cost.
Figure 3. This circuit also provides a perfect 1:1 response, but at a lower cost.

The transfer function is:

  (1)

Assuming,

  (2)

converts the transfer function into the form,

  (3)

If the term within the brackets equals unity and R5 equals R6, the transfer function becomes

  (4)

In other words, the output voltage equals the percentage deviation of R4 with respect to R3. This voltage can be positive, negative, or, in the case of a perfect match between R3 and R4, zero.

The circuit is tested for R3 = 10.001 kΩ and R4 = 10 kΩ ±1%. As Figure 4 shows, the transfer function is perfectly linear (the R2 factor equals unity) and provides a one-to-one relation between output voltage and resistor mismatch. The slope of the line is adjusted to unity using potentiometer R2 and the two end values of R4. A minor offset is present due to the imperfect match between R5 and R6 and the offset voltage VIO of the op-amps.

The transfer function provides a convenient one-to-one reading.
Figure 4. The transfer function provides a convenient one-to-one reading.

A funny detail is that the circuit can be used to find a pair of matched resistors, R5 and R6, for itself. As mentioned before, it is better to buy a network of matched resistors. It may look expensive, but it is worth the money.

Equation 3 shows that circuit sensitivity can be increased by increasing R7 and/or VREF. For example, if R7 goes up to 402 kΩ, the slope of the response line will increase to 10 V per 1% of resistor mismatch. A mismatch of 0.01% will generate an output voltage of 100 mV, which can be measured with high confidence.

Watch the current capacity of VREF and op-amps when you deal with small resistors. A reference resistor of 100 Ω, for example, will draw 25 mA from VREF into the output of the first op-amp. Another 2.5 mA will flow through R5.

References

  1. Bill Schweber. The why and how of matched resistors (a two part series).
  2. Art Kay. Should you use discrete resistors or a resistor network?

Materials on the topic

  1. Datasheet Texas Instruments LM336-2.5
  2. Datasheet Texas Instruments OPA2197

EDN