Self-heated transistors used as thermal air movement sensors are a selected (obsessive?) curiosity of mine, and over time I should have designed dozens of variations on this theme. Determine 1 illustrates one such topology seen right here earlier than. It connects two transistors in a Darlington pair with Q2 serving as an unheated ambient thermometer and Q1 because the self-heated airflow sensor. Reference amplifier A1 and present sense resistor R3 regulate a relentless 67 mA = heating present = 333 mW @ 5 V heating energy.
Determine 1 Typical self-heated transistor thermal airflow sensor.
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This warmth enter raises Q1’s temperature above ambient by 64oC at 0 fpm air velocity, cooling to 24oC at 1000 fpm as proven in Determine 2.
Determine 2 Thermal sensor temperature versus air velocity.
As proven in Determine 2, the connection between the airspeed and cooling of the self-heated transistor sensor is extremely nonlinear. That is an inherent attribute of such sensors and causes the sensor temperature versus air velocity sign to be equally nonlinear. Consequently, even comparatively small energy provide instabilities, that translate % for % into instability in sensor temperature rise, can create surprisingly giant airspeed measurement errors.
Clearly, something lower than good energy provide stability could make this an issue.
However Determine 3 gives a surprisingly easy and cheap repair consisting of simply two added resistors: R7 and R8.
Determine 3 Added R7 and R8 set up an instability-cancelling relationship between heating voltage V and heating present I.
The added Rs sum suggestions from present sensing R3 with heating voltage supply V. Summation occurs in a ratio such {that a} proportion enhance in V produces an equal and reverse proportion lower in present I, and vice versa. The result’s proven graphically in Determine 4.
Observe the null (inflection) level at 5 V the place heating is completely unbiased of voltage.
Determine 4: Sensor temperature versus provide voltage the place: Blue = heating voltage V and (uncorrected) energy; Pink = heating present I; and Black = I*V = heating energy / temperature.
Right here’s the identical factor in easy nullification math:
I = (0.2 – V*R8/R7)/R3 = (0.2 – 0.02V)/R3
H = I*V = (0.2V – 0.02V2)/R3
dH/dV = (0.2 – 0.04V)/R3 = (0.2 – 0.2)/R3 = 0 @ V = 5 volts
dH = -0.01% @ V = 5 volts ±1%
Observe the 200:1 stability enchancment that attenuates a ±1% variation in V right down to solely -0.01% variation in heating energy and subsequently temperature.
Drawback solved. Cheaply!
Stephen Woodward’s relationship with EDN’s DI column goes again fairly a good distance. Over 100 submissions have been accepted since his first contribution again in 1974.
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