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I'm using a U18 latching Hall sensor (datasheet) to toggle a LED. I'm driving it off a 6V L1016 battery.

My issue is that the LED is very dim.

My understanding is that the U18 limits the output current to 20~ish mA. In theory my LED wants 20mA forward current.

So I am not sure why it is not very bright. Testing the LED directly with the same battery both 150 & 300 ohms resistance it's much brighter.

schematic

simulate this circuit – Schematic created using CircuitLab

Original and tidied-up schematic.

How can I make this brighter? Would the best solution would be an op-amp? Is there something else I've overlooked (in terms of setting it up incorrectly)?

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  • \$\begingroup\$ The output of the sensor is open collector. Examine your datasheet, you need to sink current to switch on the LED. \$\endgroup\$ Sep 7, 2018 at 4:54

3 Answers 3

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The hall sensor is not designed to have enough current to drive an LED.

You need to use a small mosfet or transistor so that you can have the right current on the LED.

enter image description here

If the output is an open collector output, you can do it with a PNP (supposedly you want the led on when it activates.)

enter image description here

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  • \$\begingroup\$ Your PNP schematic is "upside down". D1 and 220 Ω should be below the PNP. \$\endgroup\$ Sep 7, 2018 at 6:40
  • \$\begingroup\$ It works either way \$\endgroup\$
    – Damien
    Sep 7, 2018 at 6:45
  • \$\begingroup\$ (1) The datasheet says that the sensor can switch 20 mA. This is enough to drive an LED. (2) The PNP transistor would have to be on the high side of the load to be able to be switched by an NPN output. Harry is correct. \$\endgroup\$
    – Transistor
    Sep 7, 2018 at 6:48
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    \$\begingroup\$ You are right since the OUT is not at 0 but 0.7V. However, 20mA is the maximum limit of the sensor to which point it may be damaged. Also this would likely increase the Vce of the output and may also explain why the LED does not light properly. I still believe the last schematic is the best solution, with as you suggested the PNP above the led and resistor. \$\endgroup\$
    – Damien
    Sep 7, 2018 at 6:54
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Output current at 5V is 5.5 mA. It's also an open collector output, so it pulls to ground when triggered. Your schematic as drawn would not work. At minimum the led is backwards.

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That device has an open-collector output - the LED and its current-limiting resistor must be connected between the positive supply and the device output terminal - like so:

schematic

simulate this circuit – Schematic created using CircuitLab

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