0
\$\begingroup\$

One of the projects I am working on requires a number of photo-interrupters to be used.

I was wondering if someone could explain how to use them correctly? I was thinking something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The problem, is I don't know how to pick a photo-interrupter and transistor for this to work.

Edit:

I just realized that I probably didn't give enough information. Please see the amended schematic:

schematic

simulate this circuit

Edit: I just thought of a better phrasing of my question. What should I look for in the specs that tells me what the values for these:

schematic

simulate this circuit

Edit:

So, would this next schematic work? enter image description here

Or even something like this one if I got the correct current? enter image description here

Edit: I believe I found the photo-interrupter I will use, but I am not sure if the above circuit can power it. (Each of the current limiters 'CVCCs' are adjustable

IR Diode Forward Current 50mA IR Diode Reverse Voltage 5V Transistor Collector Current 20mA Photo Transistor Collector-emitter Voltage 30V Photo Transistor Emitter-collector Voltage 5V

What I get from that is:

The LED needs 50mA

The phototransistor needs 20mA

and I have no idea about the voltage, as I can't make sense of what they said.

Any help unraveling this puzzle appreciated.

Edit:

As I said, I think my problem is that I don't know what voltage these are running at. I got two values for the voltage of the photo-interrupter, and I don't know which one to use.

Photo Transistor Collector-emitter Voltage 30V

Photo Transistor Emitter-collector Voltage 5V

I have no idea what Collector-emitter voltage is.

\$\endgroup\$
6
  • \$\begingroup\$ What have you found about photo-interrupters so far? \$\endgroup\$ Aug 18, 2014 at 23:49
  • \$\begingroup\$ Well, they will burn out if you pass too much current through them, the photo-transistor will not conduct if there is light, but will conduct in darkness. I know the transistor on the load has a threshold voltage or current. I don't know what the current after the photo-interrupter is like, and I don't know where to look on the spec sheets to find the maximum current the photo-interrupter can handle. \$\endgroup\$
    – CoilKid
    Aug 18, 2014 at 23:53
  • \$\begingroup\$ Where did the requirement to run the photo-interrupter from 1.5V come from? Why do I ask? A photo interrupter is very unlikely to work at 1.5V. \$\endgroup\$
    – gbulmer
    Aug 19, 2014 at 1:05
  • \$\begingroup\$ Well, its not required to operate at 1.5V, but I figured a common AA battery would be easy to use. (If it could meet the requirements for the photo-interrupter). I think A 9V alkaline battery would work well too, but I assumed I would be dropping the current to the mA range. A smaller resistor means less wast heat. At least, that was my reasoning. \$\endgroup\$
    – CoilKid
    Aug 19, 2014 at 1:12
  • \$\begingroup\$ @Coil: No, you have it backwards. Light causes the phototransistor to conduct, as if it were being fed some base current. Therefore the output transistor will conduct when nothing is in the slot, and be off when something is in the slot that breaks the beam. \$\endgroup\$ Aug 19, 2014 at 2:17

2 Answers 2

2
\$\begingroup\$

The easiest approach is to use logic-level output photointerrupter like this one. There is no need for two power supplies, connect the transmitter side through a dropper resistor to set IR LED current (usually 50mA at 1.2V).

Output of this particular series is open collector, so if your load draws less than 50 mA,you can connect it directly between output and power source. For greater currents, add a PNP power transistor like BD136 on high side.

The output transistor has to be PNP because the photointerrupter output can only sink current (although it has a weak pull-up built in).

schematic

Edit: new schematic for split supplies

schematic2

No way this could work reliably with single AA which can be anywhere between 1.6 and 0.9 Volts - there would be no margin for current regulation.

The pull-up resistor from your schematic is not needed since you need to amplify current and not to sense voltage.

\$\endgroup\$
3
  • \$\begingroup\$ Well, the load is running a current of 1A at 9.6V, while I wanted the photo-interrupter to run off a 1.5V(AA-battery). \$\endgroup\$
    – CoilKid
    Aug 19, 2014 at 0:27
  • \$\begingroup\$ And I don't suppose it would work with 2 AAs in series... What about a standard 9V alkaline battery? I suppose I could get another current limiter... \$\endgroup\$
    – CoilKid
    Aug 19, 2014 at 1:29
  • \$\begingroup\$ Schematic images are gone, do you mind updating them? \$\endgroup\$
    – bkarpuz
    Mar 3, 2021 at 15:32
1
\$\begingroup\$

A circuit using a phot-interrupter would look something like this:

schematic

simulate this circuit – Schematic created using CircuitLab

D1 and Q1 are inside the photo-interrupter. There will be a gap in the package between them so something can interrupt the light path between the LED and photo-transistor.

If light can pass from the LED (D1) to to Q1, then Q1 will conduct, pulling the output towards ground.

If the light path between D1 and Q1 is blocked, Q1 will not conduct, and R2 will pull the output towards +9V.

The datasheet should list the typical forward voltage and maximum current for the LED, and probably a "current transfer ratio (CTR)" for the whole device. If the CTR is 50% (or 0.5) the transistor current will be about half the LED current (if the load resistor (R2) will allow that much currrent).

\$\endgroup\$
1
  • \$\begingroup\$ How can we plut an LED to output part? Can you provide a sample schematic? Thank you. \$\endgroup\$
    – bkarpuz
    Mar 3, 2021 at 15:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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