3–286
Motorola Sensor Device Data
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Figure 3. Sensor Specific Amplifier
* NOTE: FOR MPX2010 R5 = 75 OHMS
C1
1
μ
F
I
O
G
3
1
2
B+
GND
OUT
R9 200
ZERO
CAL.
R8
1.5 k
XDCR1
MPX2000 SERIES
PRESSURE SENSOR
1
2
4
3
–
+
U1A
MC33274
1
4
3
2
TP2 +8 V
C2
0.1
μ
F
R6 7.5 k
R5
120*
–
+
U1B
MC33274
7
11
6
5
R3
820
–
R1 2 k
+
U1C
MC33274
8
10
9
R2 2 k
–
+
U1D
MC33274
14
12
13
R4 1 k
U2
MC78L08ACP
across R1, and the voltage at pin 8 is 4.0 V – 4.0 V = 0 V. In
practice, the output of U1C will not go all the way to ground,
and the voltage injected by R8 at the wiper of R9 is
approximately translated into a DC offset.
Gain is approximately equal to R6/R5(R1/R2+1), which
predicts 125 for the values shown in Figure 3. A more exact
calculation can be performed by doing a nodal analysis, which
yields 127. Cascading the gains of U1A and U1C using
standard op amp gain equations does not give an exact result,
because the sensor’s negative going differential signal at pin
4 subtracts from the DC level that is amplified by U1C. Setting
offset to 0.5 V results in an analog zero to full scale range of
0.5 to 4.5 V. For this DC output voltage to be independent of
the sensor’s common mode voltage it is necessary to satisfy
the condition that R1/R2 = (R3+R9)/R4.
This approach to interface amplifier design is an
improvement over the classic instrument amplifier in that it
uses fewer resistors, is inherently more stable, and provides
a zero pressure output voltage that can be targeted at .5 V. It
has the same tolerance problem from matching discrete
resistors that is associated with classic instrument amplifiers.
SENSOR MINI AMP
Further improvements can be made with the circuit that is
shown in Figure 4. It uses one dual op amp and several
resistors to amplify and level shift the sensor’s output. To see
how this amplifier works, let’s simplify it by grounding the
output of voltage divider R3, R5 and assuming that the divider
impedance is added to R6, such that R6 = 12.4 k. If the
common mode voltage at pins 2 and 4 of the sensor is 4.0 V,
then pin 2 of U2A and pin 6 of U2B are also at 4.0 V. This puts
4.0 V across R6, producing 323
μ
A. Assuming that the current
in R4 is equal to the current in R6, 323
μ
A
100
produces
a 32 mV drop across R4 which adds to the 4.0 V at pin 2. The
output voltage at pin 1 of U2A is, therefore, 4.032 V. This puts
4.032 – 4.0 V across R2, producing 43
μ
A. The same current
flowing through R1 again produces a voltage drop of 4.0 V,
which sets the output at zero. Substituting a divider output
greater than zero into this calculation reveals that the zero
pressure output voltage is equal to the output voltage of
divider R3, R5. For this DC output voltage to be independent
of the sensor’s common mode voltage it is necessary to satisfy
the condition that R1/R2 = R6/R4, where R6 includes the
divider impedance.
Gain can be determined by assuming a differential output at
the sensor and going through the same calculation. To do this
assume 100 mV of differential output, which puts pin 2 of U2A
at 3.95 V, and pin 6 of U2B at 4.05 V. Therefore, 3.95 V is
applied to R6, generating 319 uA. This current flowing through
R4 produces 31.9 mV, placing pin 1 of U2A at 3950 mV + 31.9
mV = 3982 mV. The voltage across R2 is then 4050 mV –
3982 mV = 68 mV, which produces a current of 91
μ
A that
flows into R1. The output voltage is then 4.05 V + (91
μ
A
93.1 k) = 12.5 V. Dividing 12.5 V by the 100 mV input yields
a gain of 125, which provides a 4 V span for 32 mV of full
scale sensor output. Setting divider R3, R5 at 0.5 V results
in a 0.5 V to 4.5 V output that is comparable to the other two
circuits.
This circuit performs the same function as the other two with
significantly fewer components and lower cost. In most cases
it is the optimum choice for a low cost interface amplifier.
F
Freescale Semiconductor, Inc.
n
.