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Analog Signal
Conditioning for Accurate Measurements

Reference
Number: AA-00615 Views:
1181 Created: 10/29/2014 11:18
am Last Updated: 10/29/2014 05:30
pm

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Q: Should I put some sort of circuit between my
sensor and an analog-to-digital
converter?

A: Yes. You probably need some signal conditioning.
The explanation below goes on for a bit, but stay with it and you'll
understand what you need and why you need it.

Before you make any connections, get the electrical
specifications for the analog-to-digital converter (ADC) and
for the sensor or sensors in your system. Let's
assume the data-acquisition module uses a Maxim Integrated
Products MAX197 12-bit ADC. This device can accept eight
differential (2-wire) inputs or 16 single-ended (1-wire)
inputs. Maxim's specifications show an input impedance of 21 kohms for single-ended
inputs and 16 kohms for differential inputs.

Now let's assume your system uses an Analog Devices AD590 temperature sensor in a simple circuit (Figure
1) to produces a 1-mV/K output. The sensor serves as a variable- current
source that passes 1 µA per Kelvin (K), or per degree Celsius
(°C). The Kelvin scale uses the same increments as the Celsius
scale, but it starts at absolute zero: 0K or -273°C. At 0°C,
the Kelvin scale reads 273K and at 25°C it reads 298K. The
Kelvin symbol does not include a degree mark. (For a
temperature-scale comparison, visit: http://www.magnet.fsu.edu/education/tutorials/magnetminute/kelvin-transcript.html.)

Figure 1. The current from this AD590 sensor passes through a precision
1000-ohm resistor. The resulting voltage represents temperature at a rate
of 1 mV/K. So at 20°C (293K), you would measure 293 mV at the
circuit output.

Now what happens when you connect the AD590 to a
single-ended input on the MAX197 ADC (Figure 2)? Note
this arrangement places the 1000-ohm resistor at the sensor in parallel with the ADC's input
resistance; 21,000 ohms.

Figure 2. The connection of an AD590 sensor and MAC197 ADC creates a circuit
with two resistances in parallel.

Could the ADC input resistance affect temperature readings?
Ohm's Law lets us calculate the equivalent resistance for these two resistors:

As a result, the complete circuit looks like that shown in
Figure 3. If a 1000-ohm resistor provides a 293 mV output for a
20°C temperature, what happens when the circuit has a 955-ohm
resistor instead?

Figure 3. The equivalent circuit for the AD590
sensor and the MAX197 ADC.

Once again, Ohm's Law lets us calculate the voltage from the known current through
the sensor and the parallel resistance. At 293K, the AD590 passes 293
µA, but with the 955-ohm resistor, what voltage goes to the ADC input?

The voltage output does not match what we
expected. In other words, the input impedance of the ADC will cause an
8-Kelvin error in the ADC reading. So you should not connect
the AD590 sensor-circuit output directly to a MAX197
input.

Keep in mind the AD590 controls the small current flow and
it--not the 1000-ohm resistor--creates a high-impedance
output.

Q: So how do I overcome this impedance problem?

A: You need an amplifier with a low-impedance, or
low-resistance, output that will provide sufficient current so
the ADC input resistance does not affect the
amplifier's output signal. In other words, if the amplifier
produces a 2.20-volt output, for example, the 21,000-ohm input
resistance of the MAX197 ADC should not
affect the 2.20-volt signal. (Or if a slight difference
occurs, it should be small enough that the ADC will not detect
it.)

The data sheet for the AD590 sensor includes a circuit that uses an
OP177 operational amplifier (op amp) to boost the sensor signal (Figure 4). The
OP177 circuit produces a 100-mV/°C, or 100-mV/K, signal. The
OP177 data sheet lists a typical open-loop output resistance of 60 ohms. This resistance in parallel with the
21,000-ohm input resistance of the MAX197 ADC yields a
combined resistance of 59.8 ohms--a change of less
than one percent.

Figure 4. Example circuit for an AD590 sensor that uses an OP177 operational
amplifier (op amp) to provide higher voltages to an ADC
or other measurement circuit. (Courtesy of Analog Devices.)

If an op-amp data sheet lacks output-resistance
information, you can perform an experiment to determine it.
Let's assume an op amp produces a 2.20-volt output with
nothing connected to the output--an open circuit. Place a
100-ohm precision resistor between the output and ground (Figure 5) and measure the
op-amp output voltage and the current through the
resistor. You can use a different
resistance value if you wish. Caution:
Use a digital meter with a high input impedance or the meter itself will alter
the measurements.

Figure 5. Test circuit for an op-amp output impedance, or resistance. In this circuit the
non-inverting op-amp configuration amplifies the 1.10-volt
signal by two.

Say the voltage drops from 2.20 volts to 2.17
volts for a current of 20 mA (0.02A) through the 100-ohm
resistor. Use Ohm's Law to calculate the
op-amp's output resistance. You know the current from the
op-amp's output and you know the voltage difference between an
open-circuit and a loaded-circuit condition:

That's a sufficiently low output impedance that the ADC input resistance will not load the amplifier
enough to alter the 2.21-volt output.

To learn more about op amps I recommend the book, "Op Amps for Everyone," by Bruce Carter and
Ron Mancini. Elsevier-Newnes, 2013. ISBN-13: 978-0123914958.
You can find a lot of useful information on the Internet,
too.

Q: What if I don't have the time or experience to
build and test op-amp circuits?

A: Use an instrumentation amplifier instead. This
type of amplifier, often called an "in amp" or INA, provides a "gain block" that
simplifies connections between sensors and ADCs. The INA827
amplifier from Texas Instruments provides a good example. One
external resistor sets the gain or amplification
factor. Op-amps on the other hand, require several carefully
matched resistances. The INA827 offers amplification factors
of five and as high as 100. Companies such as Analog Devices, Linear Technology,
Microchip Technology, and Texas Instruments sell many types of
in-amp integrated circuits (ICs), which make it easy to design
and construct an in-amp "front end" signal-conditioning
circuit for your ADC.

Unfortunately, only a few companies sell in-amp modules
that let you set the gain with an external resistor you select. Most commercial
in-amp modules offer only preset gains of, say, 1, 10, 100 and
1000. These modules also constrain input voltages to a preset
range and produce an output voltage within a preset range. In other
words, they don't offer much flexibility. For examples of
modules that provide more flexibility, visit Omega Engineering or Ocean Controls (Australia).

If you decide to design and build an in-amp circuit, the
evaluation boards and modules available from the in-amp IC
manufacturers let you experiment with the ICs in working
circuits. Digikey lists 13 in-amp evaluation boards,
most for under $100. If you decide to use an in-amp IC, I
might choose a prototype board that fits in a DIN-compatible
enclosure. The "Electronics Salon" eBay store sells several
such boards and housings that clip onto a standard DIN rail
used in industrial-control equipment.

Q: Does an instrumentation amplifier offer
advantages over an op amp?

A: Yes. In addition to using only one resistor to set the gain, or
amplification factor, in-amps have a high input impedance, so they don't affect signals
from high-impedance sensors such as an AD590. In amps also
have a high common-mode AC/DC rejection ratio (CMRR). That
means a signal common to both the positive and negative
amplifier inputs gets greatly attenuated before it reaches the
in-amp output. The INA827 in amp has an 88 dB CMRR for DC signals. The
CMRR in amplifiers decreases at the frequency of input signals
increases. An in-amp can remove common-mode AC signals such as
60- or 50-Hz electrical noise caused by nearby equipment. For
more information about CMRR, visit: http://en.wikipedia.org/wiki/Common-mode_rejection_ratio.

The circuit in Figure 6 illustrates an in amp connected in a bridge circuit that
could measure strain, pressure, humidity, or another physical
phenomena with a resistive sensor substituted for the Rs component.
One side of the bridge connects to the in-amp positive input
and the other side connects to the in-amp negative input. To
start, assume all resistors in the bridge have the same
value.

Figure 6. This bridge circuit indicates a resistance change in the sensor element (Rs) as a voltage difference between the two output
signals applied to the in amp. The amplifier removes any
common-mode voltage from the excitation source and
puts out only the voltage difference between the two
inputs. Resistor Rg determines the in-amp's
gain.

As the sensor's resistance changes, the bridge circuit
produces a small voltage difference between the two
connections to the in amp. Both inputs also receive a part of
the bridge's excitation voltage, say +3 volts. Thus,
in this example the in-amp minus input "sees," a 2.000V
signal, while the positive input "sees" 2.015V. The 0.015V
difference between the two inputs provides the useful sensor information. Note that the in-amp
circuit converted a differential signal to a
single-ended signal referenced to ground. In-amps have a reference
input, so if necessary you can offset the in-amp's output.
I'll talk more about amplification and offset in my next
column.

For more information about in amps, I recommend the
following:

"A Designer's Guide to Instrumentation
Amplifiers," 3rd edition, Analog Devices

"Getting the Most out of Your Instrumentation
Amplifier Design," Texas Instruments

"Ultra-precise Instrumentation Amplifier Makes
Robust Thermocouple Interface," (PDF)
Design Note 302, Linear Technology.

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