I've got a pneumatic cylinder reed switch sensor which has the internal layout as per the attached diagram: Reed switch sensor

If I put 9v directly to it (without the load in the circuit diagram) then I get a strong LED illumination, however, if I wire it into the vehicle I'm using it on then the illumination isn't as bright and is illuminated for a shorter duration in the stroke of the cylinder (the magnet the activates the reed switch sensor is around 20mm).

I've tested the voltage on the vehicle at the terminals the sensor is connected to and it's 12v, so I would expect the illumination of the LED to be stronger - why isn't it?

  • \$\begingroup\$ How many volts does the load drop? \$\endgroup\$ – Dampmaskin May 13 '17 at 18:28
  • \$\begingroup\$ Maximum voltage drop across the sensor unit (enclosed in the rectangle in the circuit diagram) is listed as 3.5V in the technical datasheet \$\endgroup\$ – Andy May 13 '17 at 18:36
  • \$\begingroup\$ Then how did it survive you putting 9V to it? \$\endgroup\$ – Dampmaskin May 13 '17 at 18:36

The light is weaker because the load limits the current into the LED.

This should not be wired without a load. You have a high chance of burning the (5) Zener diode. A 9V battery has a high internal resistance, so the current into the diode was limited. If you connected 12V from the car battery, it would likely be destroyed or the fuse would blow.

  • \$\begingroup\$ Thanks for your answer - what is the purpose of the Zener diode in parallel with the resistor in this particular circuit? \$\endgroup\$ – Andy May 13 '17 at 19:11
  • \$\begingroup\$ the zener protects the LED max current and something else , not shown like a polyfuse , must protect the zener. Is it a 5V zener or a 12V zener or else? and whatever is the intended series load regulates the brightness. like a 4-20mA current loop etc. where are the specs? with them, the solution will become obvious. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 13 '17 at 21:21
  • \$\begingroup\$ The zener is likely in a range of 4-5V. It enables the current flow into the load and together with the resistor limits the current into the LED. \$\endgroup\$ – pkuhar May 15 '17 at 0:35

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