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Topic: Indicator LED doubling as pull-up circuit (Read 1 time) previous topic - next topic


May 26, 2017, 06:13 pm Last Edit: May 26, 2017, 06:19 pm by GilchristT

Just a quick question on best practices. I'm designing a simple hall effect sensor based circuit as shown below.

The output requires a pull-up resistor and I'm also adding a LED to indicate when the sensor is triggered.

The circuit got me thinking, is the 10K pull-up resistor redundant as the LED circuit is carrying out the same function?

It's more a just-wondering-keen-to-learn thing, obviously removing a single resistor from the BOM isn't a big issue but I'd be interested to know if there are any good reasons not to use the indicator circuit as the pull-up as well?




May 26, 2017, 06:25 pm Last Edit: May 26, 2017, 06:25 pm by jremington
Yes, in that circuit the pullup resistor is redundant.

It is required if the output goes instead to a digital input with no pullup.


It is required if the output goes instead to a digital input with no pullup.
I'm not sure I understand that last statement?

Surely a pull-up resistor is redundant in any situation where the digital input has a pull-up resistor enabled (setting aside arguments about the relative strenghts of integrated pull-up resistors)?


The pull up resistor is a requirement, whether it is part of the circuit that you show, or part of the circuit that it is connected to.
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If the output is On/Off the 10K is optional.

If the output is analog, I recommend using the 10K as the LED is essentially open circuit until conduction occurs, but this could be argued otherwise. 

Adding a parallel 10K resistor costs little and consumes little current.

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May 26, 2017, 08:00 pm Last Edit: May 26, 2017, 08:01 pm by jremington
If the sensor has an open collector or drain output, a connection from the output to Vcc via a resistor or other circuitry is absolutely required to determine the sensor state.

Which sensor do  you have?

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