# Phototransistor resistor value help

I have a rowing machine that I'm trying to replace the computer on. It uses a phototransistor switch to detect flywheel rotation speed. The phototransistor is an Optek OPB365T55.

Here is the datasheet.

TT electronics provides an App Bulletin showing a sample circuit and formulas for calculating the values for R1 and R2. Here is the link.

When I examine the existing computer's circuit board it seems they are using a value of 5.6K for R2. This seems to be in line with all of my research on this type of circuit. Usually, I find most similar circuits are using values from 4.7K to 10K.

My application uses the circuit config from Figure 1 in the App Bulletin. The formula given for R2 is R2 = (VCC - VCEsat)/Ic(on). Using the values from the datasheet (Ic(on) 3.5mA to 14mA) and with VCC = to 3.3V I calculate a value of between 207 Ohm and 828 Ohm. This doesn't seem in line with what the observed circuit is using or with any of my research. Please educate me on what I am missing. This is fairly new to me.

By the way, my calculation for R1 would seem to be 100 Ohm. Does this seem correct?

Thank you for your help.

## 1 Answer

The App bulletin is likely using a OPB860/870/865/875 part which has a lower Ic(on) rating (0.5ma). Your OPB36xT55 part has a higher min Ic(on) so it makes sense that the R2 value can be lower. (Also note that the exact p/n OPB365T55 is not actually included in the data sheet.)

Your calculations for R2 and R1 are correct assuming a Vce(sat) of 0.4v, and an LED current of 20ma.