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According to this application note from Fairchild (btw, I find Fairchild's application notes to be of immense value - thanks to all who have taken the time to write one), the load resistor value for a phototransistor in a common emitter configuration, to obtain switch-like performance, can be calculated with:

VCE = VCC – ICE • RL RL = (VCC - VCE)/ICE

Their example conditions results with a 68kΩ resistor being selected. My question is: is this a minimum value? What affect would increasing this value have on operation? If my values resulted in a 15kΩ RL value, and I used 100kΩ, what behavior might be different?

Background: I'd like to use a GPIO IC that has internal 100kΩ pull-up resistors and am wondering if the pull-ups will work as load resistors. At VCC=3.3V, VCE=0.4V, and ICE probably somewhere around 500µA (LTR-301), my load resistor (according to Fairchild's formula) should be ~15kΩ.

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A larger resistor means that there will be more margin to assure operation if the opto ages, the optical path gets dirty (for interruptors) or the temperature changes. If you go nuts with a really high value you might have to consider dark current and leakage.

It also means that turn-off will be slower- see this graph in the LTR-301 datasheet:

enter image description here

As you can see the turn-off time may be in the 1.5 millisecond range with a 100K load.

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  • \$\begingroup\$ So, voltage drop across the resistor would be V=IR=(10E-7)*(1E5)=0.1. Therefore, at 3.3V Vcc the input would be seeing 3.2V with no light. That's almost certainly going to be a logic high. Thanks! \$\endgroup\$ Commented Jan 23, 2015 at 3:25

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