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Kenneth43

11/16/2012 12:43 PM EST

The circuit uses two amplifiers. In the first case, the wanted gain G is shared ...

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BOB.prasatik

1/19/2010 11:23 AM EST

I'm no math wizard so can ...

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# Resistor compensates for instrumentation-amp gain drift

## 10/8/2009 10:00 AM EDT

Some instrumentation amplifiers use external resistors to set their gain. Unfortunately, the lack of temperature-coefficient matching between the external and the internal resistors results in a high gain drift. If, however, another on-chip resistor is available, you can use it to compensate for gain drift as a result of temperature.

As an example, Analog Devices’ AD8295 has a drift of as much as –50 ppm/°C, even if you use a zero-drift gain-setting resistor. In this Design Idea, you can compensate this drift with an extra zero-drift resistor in combination with an internal chip resistor.

The gain-set equation from the data sheet (Reference 1) is

From this gain-set equation, you can assume that the chip uses two 24.7-kΩ resistors with the external gain resistor, RG, to set the amplifier’s gain. The chip has two more 20-kΩ resistors. Because all of these chip resistors are of the same magnitude, they probably will have good temperature-coefficient matching, and you can use this matching for compensation. If the amplifier resistance, RA,and the gain resistor are zero-drift resistors (Figure 1), then

where Δ is the drift of the internal matched resistors. If

then the first-order drift of the gain cancels, and the gain splits equally between the instrumentation amplifier and A1. Solving for RG and RA yields

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For gain greater than 100, the amplifier resistance becomes greater than 90 kΩ, which can be problematic. In this case, you can use A1 in an inverting configuration with a gain of –1 (Figure 2). With an amplifier resistance of 10 kΩ,

This case sizes RG using a value from the data-sheet formula. If the gain is 50, the internal matching and the negative drift compensate the “49” part of the gain, and the “one” part is just the drift divided by 50 in the total gain, resulting in a typical figure of –1 ppm/°C. In both cases, the resulting gain temperature coefficient can be less than 5 ppm/°C, which is 10 times better than the original outcome.

 Reference

BOB.prasatik

1/19/2010 11:23 AM EST

I'm no math wizard so can someone explain the slight-of-hand that gets you from the gain set equation from the data sheet to the gain set resistor equation involving the square root of gain?

Kenneth43

11/16/2012 12:43 PM EST

The circuit uses two amplifiers. In the first case, the wanted gain G is shared so each stage takes a gain of square root of G. In the second case the gain of the second stage is 1, so the first stage takes the whole gain G.