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I have a Hall-effect sensor with an open-collector output rated for max. 25 mA. This sensor will be used to switch an NPN transistor powering a 12 V LED strip. The load is 300 mA.

Could I get away with a single pull-up resistor here? Assume the transistor has a gain of 100 and a saturation voltage of 1 V. If I use a (12-1) V / 3 mA = 3.6 kΩ pull-up resistor, would the transistor switch on the LED strip when the sensor output goes high?

Secondly, could I use the same single-resistor configuration with a PNP transistor instead if I wanted the LED to switch on with a low signal from the sensor?

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  • \$\begingroup\$ For npn version seems ok to me. For pnp you need two resistors, one pull-up and one for base current limit. But pull-up can be something like 22k or 47k just to dicharge the base. \$\endgroup\$ Commented Jan 3 at 21:44

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Could I get away with a single pull-up resistor here?

Yes, but not the way you describe it.

What you describe is called "dangle biasing" and is not recommended. It relies on the transistor's beta being constant, stable, and predictable. In reality it is none of those things. A transistor's beta changes with temperature, aging, collector-emitter voltage, collector current, base current, manufacturer, lot-to-lot variations - just about everything except the phase of the moon. This is why the long-time rule of thumb for driving a transistor as a saturated switch is to assume a gain of 10. That's 1 mA of base current for every 10 mA of collector current.

That rule is from the 1950's, back when transistors (and especially power transistors) sucked. Today's are much better, and I'm comfortable with a 20:1 ratio. Since the sensor output is rated for 25 mA, use your same math to design for 15 to 20 mA of base current.

For a PNP transistor, the sensor output now is sourcing the base current rather than bypassing it. Either way, the base current-limiting resistor is the same value. Add a 100K resistor from the base to the emitter to assure a fast and crisp turn-off.

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For a saturated switch the rule of thumb is to use a base current 1/10th of the collector current. This would mean 30 mA, so you would need more than one stage. This could be done a number of ways, Darlington, Sziklai, or a cascade arrangement.

Here's a circuit using cascaded amplifiers, the sensor drives a common collector stage with 4-5 mA, which drives a common emitter stage with 35-40 mA, this switches 300 mA through the LED array. The D44H11 is a relatively inexpensive high current transistor with low \$V_{CE_{sat}}\$, other transistors could be used though, just make sure they'll handle the current with a good safety margin. Hall Effect sensor to LED driver

This is low side drive using NPNs, high side drive and/or PNP arrangements are also possible. The output can also be inverted depending on if you want the load on with the sensor output low or high.

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