*SENSOR BRIDGE*
CIRCUIT
SBRIDGE1.CIR
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SPICE file
Suppose you have a sensor that converts a physical
phenomenon like temperature or pressure into a resistance change. How do you
convert the resistance change into a useful voltage you can measure?
SENSOR BRIDGE
A sensor bridge gives you a differential
output voltage given a sensor's resistance change. The resistive sensor XSENSOR is placed in one of four branches of the bridge.
XSENSOR is a subcircuit that models a voltage controlled resistor. Resistors R1, R2, and R4 are chosen to be equal to the nominal value of XSENSOR.
The voltage at node 2 is developed from voltage divider R1 and XSENSOR. The voltage at node 3 comes from voltage divider R2 and R4. When all resistors are equal, the voltage
at nodes 2 and 3 are equal. Therefore, the difference between 2 and 3 is zero. However, a resistance change
of XSENSOR unbalances the two voltage dividers producing a
useful voltage change between 2 and 3.
TEMPERATURE SENSOR
Suppose you’ve selected a temperature sensor whose
resistance is 1k ohms at 25 deg C and changes
ΔRsensor = +/- 100 ohms at temperatures
50 and 0 deg C.
__ T deg
C R (ohms) __
0
900
25 1000
50 1100
Choose RB1, RB2 and RB4 = 1k to produce 0 volts output
at 25 deg C. But, what kind of voltage change can you expect from
ΔRsensor=100 ohm? For small resistance changes the differential output can
be approximated by
CIRCUIT ANALYSIS
For Rsensor = 1k,
ΔRsensor=100 and VBIAS = 10 V, calculate the
output voltage swing to be Vout = (10/4) x (100 / 1k) = 0.25 V. The voltage
ramp from VCONT controls the resistance of the XSENSOR changing it from 900 to
1100 ohms
over 10ms. Run a simulation and plot the voltage
at V(2) and V(3). Do you see V(2) producing the expected +/- 0.25V change
from the 5V bias point?
SENSOR BRIDGE KEY FUNCTION
What’s interesting is that the small voltage change is
centered around 5V! This doesn’t help us especially if we want to amplify
this signal at a later stage. We would end up amplifying the 5V too -
saturating the amplifier stage!
It would be great if we could subtract off the 5V to
produce a more useful output. That’s where the other leg of the bridge comes
in. Note that V(3) is conveniently sitting at 5V. So now by taking the
differential voltage at Vout, you get a much more useful result. View the difference
between nodes 2 and 3 by the plotting the variable V(2,3) to see the +/- 0.25V
output centered
around 0V. ( Here is a great application for a differential or
instrumentation amplifier to measure the difference voltage V(2) - V(3). )
NON-LINEAR OPERATION
What are the limitations of a single sensor in a four
legged bridge? The output voltage versus resistance becomes non-linear for
large resistance changes.
CIRCUIT INSIGHT
Increase the resistance change from +/-100 to a greater value
like +/-500 ohms, for
example. You can do this by changing the control voltage to look like
VCONT 10 0
PWL(0MS 0.5 10MS
1.5)
VCONT now generates a ramp from 0.5 to 1.5V that
effectively sweeps the nominal 1000 resistor of XSENSOR from 500 to 1500 ohms.
Run a simulation and plot V(2,3). Can you see a bend in the output voltage?
SIMULATION NOTE
Although SPICE does not directly have a model for
voltage controlled resistor, you can put one in your circuit using the
subcircuit XSENSOR. Change the value of 1K in the VALUE expression to
the nominal resistance value you desire. This resistance will appear at the
output terminals (1) and (2) when the control voltage at nodes (4) and (5)
is 1.0V. The output resistance will then scale according to the control
voltage value. ( For more info, see
Voltage Controlled Resistor. )
SPICE FILE
Download the file
or copy this netlist into a text file with the *.cir
extention.
SBRIDGER.CIR - INSTRUMENTATION BRIDGE
*
VBIAS 1 0 DC 10
*
* INSTRUMENTATION BRIDGE
RB2 1 3 1K
RB4 3 0 1K
RB1 1 2 1K
*
* RESISTIVE SENSOR (VOLTAGE CONTROLLED)
XSENSOR 2 0 10 0 V_CONT_R
*
* CONTROL VOLTAGE
VCONT 10 0 PWL(0MS 0.9 10MS 1.1)
RD1 10 0 1G
*
* SUBCIRCUIT FOR VOLTAGE CONTROLLED RESISTOR
* RESISTOR - 1,2 CONTROL - 4,5
*
.SUBCKT V_CONT_R 1 2 4 5
ERES 1 3 VALUE = { I(VSENSE)*1K*V(4,5) }
VSENSE 3 2 DC 0
.ENDS
*
* ANALYSIS
.TRAN 1MS 10MS
* VIEW RESULTS
.PLOT TRAN V(2) V(3)
.PRINT TRAN V(2) V(3)
.PROBE
.END
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