Skip navigation
Electronic Design

Linear pitot-tube air-speed indicator

Among the many methods used for measuring air speed, one approach excels in applications related to aerodynamics and wind power: the Pitot-tube impact-pressure air-speed indicator. The so-called “impact” or “stagnation” pressure exerted by airflow striking a surface is given by P = (D × V2)/2, where D = air density and V is air speed. Pitot tube anemometry uses this relationship to produce: V = √—2P/D.

Although it might seem that this dependence of the air-speed estimate upon density is an undesirable complication, it’s actually advantageous in those applications to which Pitot-tube anemometers are uniquely suited. This is true because, in these applications, the forces of primary interest (e.g., lift from an airplane’s wing or propulsion from a boat’s sail) are themselves related directly to impact pressure and therefore to air density. The air speed measured by a Pitot tube is automatically compensated for variations in air density caused by changes in temperature, barometric pressure, or altitude. It’s therefore exactly the air-speed information most wanted by the pilot or sailor.

The heart of the Pitot tube airspeed indicator is piezoresistive differential pressure sensor G1. Op amp A1, in combination with VR1 and R1, R2, and R3 controls Q1 and Q2 to generate constant-current-bridge bias I1 = 500 µA. In response, G1 produces a differential voltage (V1 − V2) equal, after zero compensation via R4, R5, R6, and R7 to approximately 8 mV/psig of impact pressure. The A2, A3 differential pair controls Q3 so as to convert (V1 − V2) to impact-pressure-proportional current I2 ≈ 160 µA/psig.

Calculating √—I2 works as follows: Due to the logarithmic behavior of silicon transistors, base-emitter voltages Vbe2 = Alog(I1/2) + B, and Vbe3 = Alog(I2) + B, where A and B are constants common to all five transistors in the LM3046 array. Because addition of logs is equivalent to multiplication, (Vbe2 + Vbe3) = (Alog(I1 × I2/2) + 2B). Series/parallel connection makes (Vbe2 + Vbe3) = (Vbe4 + Vbe5) and, because of the implicit matching between transistors in a monolithic array, Vbe4 = Vbe5. Therefore, Vbe4 = Vbe5 = (Alog(I1 I2/2)/2 + B). Division of logs is equivalent to inverse exponentiation. Hence, Vbe5 = log(√—I1× I2/2) + B.

This makes log(I3) = log(√—I1 × I2/2). I1/2 = 250 µA, so substitution and exponentiation yields I3 = √—I2 × 250mA.

With the components and circuit values shown, I3 ≈ 1 µA/knot. With suitable adjustment of R8, a wide range of full-scale air speeds can be calibrated for. If R8 ≈ 10k, for example, Vout = 10 mV/knot for a 1-V fullscale output at 100 knots. This is just about right for duty as a primary airspeed indicator in an ultralight aircraft. Of course, suitable adjustment of R8 can accommodate different preferences in air-speed measurement units, such as meters per second or statute miles per hour.

Supply-voltage regulation for this circuit isn’t critical and power demand is modest, easily allowing battery operation. The simple power supply shown will provide more than 100 hours of operation from a 9-V alkaline battery.

See associated figure

Hide comments


  • Allowed HTML tags: <em> <strong> <blockquote> <br> <p>

Plain text

  • No HTML tags allowed.
  • Web page addresses and e-mail addresses turn into links automatically.
  • Lines and paragraphs break automatically.