The cheap and convenient hall effect liquid flow sensors available for a few bucks online nearly all cite the device output as being a square pulse train whose frequency is a linear function of the flow rate.

My application is a water flow indicator. No measurement of flow rate, no microprocessors needed - just an output that indicates whether water is flowing or not flowing.

All the devices have a hall sensor and a pinwheel and a conditioning/amplifying/shaping circuit internal to the assembly, but I have not seen anywhere information that says if the output remains high (or low) when flow stops.

If the pinwheel magnet(s) are aligned away from the hall effect sensor when rotation stops then I would expect the output to go to logical 0. But what happens if the wheel stops when the magnet is aligned with the sensor? Does the output now stay high, or is the internal circuitry such that there has to be changing flux for there to be any pulse train at all, and there is a defined state for "zero flow"?

Anyone know from experience before I start breadboarding and prodding the pinwheel?



closed as unclear what you're asking by Andy aka, Voltage Spike, PeterJ, Sparky256, Dmitry Grigoryev Mar 2 '18 at 20:54

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  • \$\begingroup\$ I would guess they have a retriggerable 1shot with edge trigger and common emitter output \$\endgroup\$ – Sunnyskyguy EE75 Feb 14 '18 at 4:46
  • \$\begingroup\$ most likely it stays high when the magnet stops next to the hall-effect. \$\endgroup\$ – Jasen Feb 14 '18 at 8:22
  • \$\begingroup\$ Hard to tell anything in general. Pick a sensor, post the relevant part of the datasheet and ask about that. \$\endgroup\$ – Dmitry Grigoryev Mar 2 '18 at 20:55

high-pass filter the pulse signal and then rectify that.


simulate this circuit – Schematic created using CircuitLab


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