487451449AN589| Application Note

487451449AN589 PDF

487451449AN589| Application Note


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AN-589 APPLICATION NOTE

One Technology Way P.O. Box 9106 Norwood, MA 02062-9106 Tel: 781/329-4700 Fax: 781/326-8703 www.analog.com

Ways to Optimize the Performance of a Difference Amplifier
by Reza Moghimi
There are times when a small signal needs to be measured in the presence of a large common-mode signal. Traditional instrumentation amplifiers (in amps) that have two op amp or three op amp internal configurations are commonly used in these applications. Although in amps have good common-mode rejection ratios (CMRR), price and sometimes specifications prevent their usage in these applications. In amps may not have the right bandwidth, dc accuracy, or power consumption requirements that the user requires. Therefore, in these situations, users build their own difference amplifiers by using a single amplifier and external resistors as an alternative to instrumentation amplifiers. Unless a set of tightly matched resistors is used, CMRR of these circuits will be very low. This Application Note presents several ways to build and optimize the performance of a discrete difference amplifier. It also recommends amplifiers that will make the overall solution cost/performance competitive with monolithic instrument ampli
fiers. A typical difference amplifier using a single amplifier connected to a sensor bridge is shown in Figure 1.
VREF 4 AO1 1 AOZ 3 0 2 R3 R4 2 0
R3 3 R4 2 0 0 BR4 VREF/2 VD/2 BRIDGE R1 V+ V U9 1 VOUT

By applying the superposition principle, it can be shown that the output is a function of the difference of the two inputs. The transfer function of the circuit in Figure 1 is:

VOUT

R 2 R 2 1+ R 1 = R 3 V 2 -V 1 R1 1+ R 4

(1)

A special situation arises when
R2 R 4 = R1 R 3

and the above equation is reduced to:

R2 R 4 = R1 R 3 R2 (V 2 -V 1) VOUT = R1

(2)

The output is the difference of the two inputs times a gain factor that can be set to unity. Equation 2 holds true if the ratio of the resistors is tightly matched. Assuming perfectly matched resistors with values of R2 = R4 = 10 k, R1 = R3 = 1 k, V1 = 2.5 V, V2 = 2.6 V, then VOUT = 1 V. As stated above, one of the shortcomings of the circuit in Figure 1 is the poor CMRR, which is caused by mismatch of the resistors. In order to investigate this, the circuit is redrawn for clarity in Figure 2.

3 V+ V

U8 1 VOUT

R1

R2

VREF/2 + VD/2

Figure 1.

R2 (1-ERROR)

Figure 2.

REV. 0

Analog Devices, Inc., 2002


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