Download The Modern Devices wind speed sensor

The Modern Devices wind speed sensor
©2012 H Schiretz, BEng (Hons) - Deakin University Australia
The circuit for the Modern Device thermistor (hot-wire) wind sensor is shown in Figure 3. In our
application, the output for the wind sensor, designated as V_RAW, is taken at the junction of NPN
transistor Q1 and the top of a Wheatstone bridge. Modern Devices advise that the lowest voltage at
this point is ~1.8 V. At the time of writing, the circuit diagram and the lowest voltage point at the
emitter of Q1 are the only specifications presented by the manufacturer on their website however,
Modern Device support provided details of the thermistors used in personal communications which
made it possible to analyse the circuit response to temperature variation and determine its
relationship with air velocity.
Figure 1 The circuit diagram for the Modern Device wind-speed sensor (moderndevice.com).
R4 is the ‘wind speed’ sensing thermistor, Murata 0402 surface mount device, NCP15XQ471E03RC
470 Ω @ 25°C, with a B-constant (slope of the thermistor resistance, RT, to the inverse of
temperature, 1/T characteristic) which is 3650 +/- 2%. The significance of the B-constant is that it
can be used to determine the thermistor resistance for any specific temperature.
R4 is in one arm of a Wheatstone bridge consisting of voltage divider, R3 which is 1 Ω and the
thermistor. The other arm of the bridge consists of the voltage divider circuit, R5 (2 X 103 Ω) and R6
(165 X 103 Ω). When the bridge is balanced the voltage at the junction of the voltage dividers of each
the arms of bridge will be equal so that the voltage difference between these junctions will be 0 V.
Theoretically, the bridge will be balanced when;
= ×
×
= .
(10)
Thus, using the Murata B-constant specification for the thermistor, the bridge will be in a balanced
condition at ~ 74.4 °C, and hence under a normal Wheatstone bridge configuration there would
always be a voltage difference between the voltage divider junctions of the two arms of the bridge.
However, operational amplifier (op-amp) U1A is configured as a differential comparator that outputs
a voltage indicative of the voltage difference between the voltage divider junctions of the two arms
of the Wheatstone bridge. R8 (10 X 103 Ω) serves as a voltage pull-up resistor that acts as a current
source for U1A so as to not to draw current from the thermistor arm of the bridge but the voltage at
that point will be that of the thermistor voltage divider junction.
Because the op-amp has high gain the difference between its inputs (= op-amp output / gain) will be
negligible. As shown in equation (10), for the bridge to be in a balanced condition the thermistor
resistance must be 82.5 Ω. The op-amp output voltage will go to whatever voltage is required to
produce sufficient current through Q1 to self-heat the thermistor to 74.4 °C and maintain the
thermistor resistance at 82.5 Ω.
If the thermistor resistance is less than 82.5 Ω, the net voltage to the op-amp will be tending
towards negative, reducing the op-amp output voltage, reducing the current, reducing the selfheating, reducing the temperature and, increasing the thermistor resistance. Conversely, if the
thermistor is more than 82.5 Ω, the net voltage to the op-amp will be tending towards positive,
increasing the op-amp output voltage and, decreasing the thermistor resistance.
We can conclude then that the feedback is negative (necessary for the feedback to be stable) and
furthermore, the feedback regulates the thermistor resistance to 82.5 Ω and hence its temperature
to 74.4 °C.
The whole purpose of the exercise is to regulate the temperature of the thermistor to a fixed value
considerably above ambient. The current (actually power) in the thermistor depends on the heat
loss which depends on (a) the air speed (b) the temperature difference between the thermistor and
ambient and (c) surface area and other fixed factors. Increased wind speed tends to cool the
thermistor (increased convection) and the circuit responds by heating it more in order to regulate
the thermistor temperature to 74.4 °C.
V_RAW will increase proportional to wind speed in a non-linear fashion, similar to that of Figure 2,
with V_RAW described by;
V_RAW= ⁄82.5 = ⁄82.5 × 83.5
(11)
where, IT is be thermistor current and, PT is the thermistor power. Restating equation 7;
∝ × ⁄∆ ( !""#/℃)
(12)
As, Tth = 74.4 degC and Ta is the value obtained from the R1 and R2 voltage divider circuit for the 10
kΩ thermistor (R2) for the ambient (wind) temperature at TEMP_OUT on the sensor circuit board –
as described later - therefore ∆T= 74.4 – Ta. This then accounts for any variation in the temperature
of the air passing over the wind speed sensing thermistor.
Thus for the wind speed sensor,
∝ ⁄(74.4 − *_,) ( !""#/℃)
(13)
where;
such that:
= -_./0 ⁄82.5
!12 = -_./0
∝ (-_./0 ⁄82.5)/(74.4 − *_,)
(watts/°C)
(14)
Now the important thing to notice here is that wind speed is proportional to Watts/°C which is
where equation (9) comes in which essentially states that;
= / + 4(-_./0 ⁄82.5)/(74.4 − *_,)56 (m s-1)
(15)
The constants A and B may be found in a wind tunnel test from known wind speeds, U, versus the
right hand side of equation (14).