0
\$\begingroup\$

I'm using an FODM8061 Optoisoltor with and open collector Logic Output. I'm referencOptoisolator Logic Output Question

Here's my current circuit: Circuit

LED 1(on the left) connects to an open-collector gear-tooth sensor. The led is used to help set the location of the sensor. It's intended to light up when the sensor is correctly positioned. On the right side, the signal coming out of R6 goes to the MCU.

My diagnosis: The LED will light up when I bring cathode to ground, but I get no output on optoisolator. When I short the LED(remove the led from the circuit) and bring it to ground, The optoisolator works as intended.

Why would the LED cause this issue? Can I provide any other information?

\$\endgroup\$
  • \$\begingroup\$ Have you done the math yet? \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 26 '16 at 17:11
  • \$\begingroup\$ @IgnacioVazquez-Abrams no, I haven't. Can you explain what calculations should be done? \$\endgroup\$ – GisMofx Jul 26 '16 at 17:13
  • 2
    \$\begingroup\$ Current calculations for the input side. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 26 '16 at 17:13
  • \$\begingroup\$ @GisMofx: What's the purpose of D2? \$\endgroup\$ – Curd Jul 27 '16 at 7:30
  • \$\begingroup\$ @Curd see here: electronics.stackexchange.com/questions/178961/… \$\endgroup\$ – GisMofx Jul 28 '16 at 21:31
3
\$\begingroup\$

You should not have the LED set up like this. With the 3.3V source, two 330 ohm resistors, an LED (presumably a voltage drop of around 1.8V) then that only leaves 1.5V for the LED in the opto-isolator. The typical forward voltage of that LED (according to the opto datasheet) is 1.45 (max 1.8V) so you are already pushing your voltage limits. Not to mention you have two 330 ohm resistors in series, which drops even more voltage and reduces the current to the point where it's completely useless. I'm actually surprised LED1 lights up at all.

You cannot have LED1 in series with the opto-isolator. If you want to do that you'll have to increase the voltage to ensure both diodes "switch on" (light up) and to make sure enough current can flow through them to produce enough light.

You should have done the math first, you would have seen this was a very clear problem.

\$\endgroup\$
  • 1
    \$\begingroup\$ Visible LEDs light up with a very small current, but the FODM8061 has a controlled threshold current. \$\endgroup\$ – CL. Jul 26 '16 at 21:33
3
\$\begingroup\$

3.3V is not enough to operate two LEDs in series.

Instead, you need to put them in parallel, each with its own current-limiting resistor:

schematic

simulate this circuit – Schematic created using CircuitLab


EDIT: After taking a closer look at the datasheet, the worst-case minimum threshold current for the optoisolator is 7.5 mA, so let's put 10 mA through it to give us some design margin. With a supply voltage of 3.3 V and a worst-case Vf of 1.8 V, we need a total resistance of (3.3V - 1.8V)/10 mA = 150 Ω. If you want to split that across 2 resistors, then they need to be 75 Ω each.

\$\endgroup\$
  • \$\begingroup\$ If the OP does this he will need to reduce the resistor values as well. The Datasheet suggests a minimum Vf of 1.05V and an If of 50mA. The TOTAL series resistance would need to be around 45 ohms in order for the optocoupler to work correctly. As it sits, the OP will have a best-case current of only 3.4mA. \$\endgroup\$ – DerStrom8 Jul 26 '16 at 17:39
  • \$\begingroup\$ @DerStrom8: I'm just taking the OP at his word that the opto works correctly without the extra series LED. \$\endgroup\$ – Dave Tweed Jul 26 '16 at 18:02
  • \$\begingroup\$ @DerStrom8 Aargh! Never use an absolute maximum rating as a design goal! \$\endgroup\$ – CL. Jul 26 '16 at 21:34
  • \$\begingroup\$ @CL. I'm choosing values that would contribute to the worst case. You should always plan for the worst case! EDIT: Woops, I did misread the datasheet (DOH!). Even so, the design provided still won't work for the OP's needs as not enough current is allowed to flow through the opto \$\endgroup\$ – DerStrom8 Jul 27 '16 at 1:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.