I want to use specifically the SFH 309 NPN phototransistor in order to create a pulse sensor that works as seen below:


The IR diode emits IR light to the fingertip from the one side and on the other side the phototransistor is picking up the light "converting" it to voltage values. When there is a pulse, the excess blood "blurs" further the fingertip and the transistor is picking a weaker light beam.

The problem statement here is: How do I find the right resistor to get the best result? (interested more in the procedure of what to look for in the data sheet and how to calculate it)

This is the kind of signal that I am looking for (here the diff is 50mV, and I am hoping for something better than this).


The VCC is 5V.

  • \$\begingroup\$ Where on the circuit did you probe to measure the signal in the scope image? \$\endgroup\$ – tokamak Apr 3 '16 at 20:13
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    \$\begingroup\$ Because it's so hard to define the nature of this system, I would suggest simply using a potentiometer or decade box and experiment. \$\endgroup\$ – uint128_t Apr 3 '16 at 20:55
  • \$\begingroup\$ The system shown will not produce pulses, since Vcc is a 5 volt DC supply and the LED current (and light output) will therefor be DC as well. You need to define your pulse amplitude, LED resistor and LED. Then you need to define the distance from the LED to the diode. Only then can you get a grip on the resistor needed. Oh yes, and a scope trace is completely useless without identifying the vertical sensitivity AND the timebase setting. \$\endgroup\$ – WhatRoughBeast Apr 3 '16 at 21:14
  • \$\begingroup\$ 1) I measure the voltage between the resistor and the phototransistor. 2)Yes the LED current is steady and the LED emits IR light on the finger and on the other side there is the phototransistor receiving the light signals. Then I measure between the resistor and the phototransistor the voltage. The voltage changes because the impedance of the transistor changes according to the light intensity. 3) The vertical sensitivity is 50mV and the freq is between 1 and 2 Hz... because it is a heartbeat. but in this case it is just to give an idea of what to expect. \$\endgroup\$ – Dimitrios Torssøn Apr 4 '16 at 8:09

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