EDN Blog Postings - August 2012
Instrumentation amplifier input-circuit strategies
Bonnie Baker
8/20/2012 4:30 PM EDT
Many industrial and medical applications use instrumentation
amplifiers to condition small signals in the presence
of large common-mode voltages. INAs have complete
closed-loop operational amplifiers with feedback components
included. Under normal conditions, an INA is easy
to use, as long as you pay attention to the input stage and
the output of the first stage. The quick and inexpensive solutions described
here can help you avoid the INA circuit’s input-stage pitfalls.
In a classic, three-op-amp INA (Figure 1), the input stage has two op amps in an adjustable-gain configuration and provides high input impedance on both the inverting (VIN−) and noninverting (VIN+) inputs. The output stage has four matched resistors around a single op amp. When the circuit designer exercises proper precautions with the input pins, this configuration rejects external common-mode voltage and noise. The output stage has a reference voltage that level-shifts the output with respect to ground. The level shift is convenient in single-supply applications. As with many INAs, you can program the gain found in Figure 1 with a single resistor, RG.
Depending on the INA’s silicon process, VIN+ and VIN− connect to a bipolar transistor base, FET gate, or CMOS gate. All inputs to the INA require a current-return path to ground and a bias-voltage reference. Without the current-return path and the bias-voltage reference, the INA input stage saturates or floats to an undesirable voltage. Either condition creates an invalid output voltage.
The floating thermocouple circuit
in Figure 2a does not provide a
current path to ground or a bias-voltage
reference for the INA’s input pins. Thus,
the input-current leakages are not dissipated,
and the two inputs can float
to any undefined voltage. That situation,
in turn, causes the INA’s output
to change to an invalid output voltage,
usually in the middle of the INA’s output
range. The invalid voltage can appear
to be legitimate, making it difficult to
detect the correctness of the INA input
implementation. Figure 2b illustrates
a correct thermocouple connection to
an INA; both inputs have a path to
ground—in this case, through a 10-kΩ
resistor, biased to a voltage within the
INA’s input range, or ground.
Circuit designers often misapply a thermocouple or even a two-wire microphone to the INA’s input circuit; the problem arises when the INA inputs are connected without proper consideration to current paths or biasing. Heed those considerations for the INA’s input stage, and you can be confident that the INA’s output voltage is representative of the thermocouple’s voltage.
But wait—there may be a problem with the output values of A1 and A2 (Figure 1). How would you solve this problem? Comment below. I’d really like to hear from you!
References
- Kugelstadt, Thomas, “Getting themost out of your instrumentationamplifier design,” Analog Applications Journal, Texas Instruments, 4Q 2005.
- Trump, Bruce, “Instrumentation amplifiers—avoiding a common pitfall,” TI E2E Community, March 27, 2012.
- Trump, Bruce, “Instrumentation amplifiers—they’re not op amps butwhat are they?” TI E2E Community, April 2, 2012.
Bonnie Baker is a senior applications engineer at Texas Instruments.
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gkoko
8/27/2012 8:50 AM EDT
Probably we could connect two resistors (e.g. 20K) on both inputs (Vin+ & Vin-) in order to be balanced.
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Bonnie Baker Texas Instruments
8/28/2012 2:01 PM EDT
gkoko,
If you conned tow resistors to the tow inputs and then to ground or power you would provide a path for the input bias current of the IA inputs. The bottom line is that you can not connect either or both inputs to an extremely high impedance or worse yet, leave them open. I am not sure what you mean by the word "balance".
Bonnie
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WKetel
8/28/2012 9:29 PM EDT
Since the resistance of most thermocouples is much less than one ohm, it is not likely that the unbalance caused by using a single resistor to ground would cause any detectable unbalance, and it certainly would not introduce any detectable errors if the 10K value shown were used. If the device were an RTD, then there might be a concern, except that the RTD implementation uses a bias source, which holds the amplifier's input voltage in the required range. The first time that I ran into this problem, the resistor that I used was 100K, partly because it was handy, but also because I did not understand the problem completely. That was a lot closer to the start of my career.
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Bonnie Baker Texas Instruments
8/30/2012 1:26 PM EDT
WKetel
Your comment about higher impedances is completely right. This particular circuit only applies to a low impedance input, such as a thermocouple. On the other hand an RTD would require a different strategy, even though RTD resistances can be very low. The most important thing to understand about this article is that the inputs do need a termination current path. It is completely useless to tie the two inputs together. These inputs are high impedance and the terminals have no capability to source or sink predictable currents.
Bonnie
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cigrbox
9/8/2012 12:27 AM EDT
I think your problem is that the gain of A1 and A2 is also applied to the common mode voltage, so these stages will clip when you have high gain and high common mode. I'd be interested in a work-around for a monolithic INA, but in discrete circuits I've moved more of the gain to the output stage.
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Antonio Rodrigues
9/11/2012 3:15 PM EDT
ACarlos
I would connect two resistors (e.g. 20K or 100K) being one from Vin- to ground and another from Vin+ to ground. Besides that, we should be aware of thy offset errors. I mean that the sum of the offset errors of A1 and A2 will be multiplied by the gain and this resulting voltage will be present in the output of the first stage.
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