I am trying to use the wheel encoder mentioned in the circuit but somehow i can't manage to understand why there is a full voltage drop in the IR led and why the resistors are not limiting that voltage drop, considering that in "ohmic" terms my circuit should work.

I would like somebody to help me pin what i am doing wrong

The datasheet is http://sensing.honeywell.com/index.php?ci_id=50399

In the electric characteristics the datasheet says that the forward voltage should be 1.6 V@20mA and the supply of detector should be between 4.5 and 5.5V@7mA

So to use the resistors i have at hand the circuit is in the following attachment. For the IR led i connected it to a 5V source with a 220hm resistor and to the receiver 47 + 47 Ohm resistors.

Also i am connecting the A and B channel outputs to an atmega328 directly as the datasheet says the HOA0901 has internal pullups. I also set the atmga328p in input mode.

My design Thanks

  • \$\begingroup\$ What exactly do you mean by "full voltage drop"? What is the actual voltage and where are you measuring it between? \$\endgroup\$
    – Oli Glaser
    Jul 24, 2012 at 14:32
  • \$\begingroup\$ By full voltage drop i mean that the voltage measured across the IR LED is 5V and i think shouldn't be, because i wanted it to be 1.6V \$\endgroup\$ Jul 24, 2012 at 14:38
  • \$\begingroup\$ Okay - what is the voltage measured across the resistor (R5)? \$\endgroup\$
    – Oli Glaser
    Jul 24, 2012 at 14:43
  • \$\begingroup\$ 0V across it. I think that it is possible that the LED is not receiving enough current to saturate no? \$\endgroup\$ Jul 24, 2012 at 14:46
  • \$\begingroup\$ Aha, thought so - it looks like you have blown your LED (it is open circuit) \$\endgroup\$
    – Oli Glaser
    Jul 24, 2012 at 14:47

1 Answer 1


If you are measuring 0V across your resistor, then it looks like your LED is blown (open circuit)

V / R = I, so 0V * 220 = 0mA. This means there must be a high impedance in series with the resistor (i.e. open circuit LED)


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