Opto Interrupter - Electronic Circuits Projects Diagrams Free

finch661 · 16th October 2007, 05:44 PM

i am trying to use an opto interrupter (IR Source and IR Reciever all enclosed) for a tachometer. The system should detect when the beam is broken. I have tried several circuits, but cannot get an output.

eg http://www.shumatech.com/dro-350/add-ons.htm
any help?

audioguru · 16th October 2007, 06:18 PM

Does the IR LED cause current in the current-limiting resistor?
If it doesn't then it is connected backwards.

Does the photo-transistor conduct when the IR LED has current?
If it doesn't then it is connected backwards.

Does the photo-transistor turn off when there is no IR?

finch661 · 16th October 2007, 06:43 PM

the LED and photo transistor are correctly placed. With the current layout, (5V supply) the output reads 5V regardless of the IR being blocked...

audioguru · 16th October 2007, 08:02 PM

The IR is not supposed to be blocked, it is supposed to be relected.

The output is high because the photo-transistor is not turned on. It is turned on when it sees strong reflected IR from the IR LED directly in front of it.

The IR LED has a 220 ohm resistor for you to measure to see if there is about 3.8V across ir which means that the IR LED has about 17.3mA lighting it.

The shiny reflective surface must be at a 90 degrees angle and only 0.15" away. If the IR LED has 60mA then the photo-transistor will have at least 0.6mA.
0.6mA in the 4.7k collector resistor will cause the collector voltage to be 2.18V.

Your IR LED current is only 43% of the 40mA in the datasheet so your collector voltage will not be higher than 3.8V when it is supposed to be low.

1) Increase the IR LED current to 38mA by replacing the 220 ohm resistor with 100 ohms.
2) Increase the sensistivity of the photo-transistor by replacing the 4.7k ohms resistor with 22k.
3) Make the reflective surface very shiny, at a 90 degrees angle and only 0.15" away.

philba · 16th October 2007, 08:30 PM

What interrupter are you using? I assume you are using an interrupter and not the QRB1114 reflective device.

the first thing I would do is get a digital camera and look at the LED. It is a little hard with an interrupter but it's possible. The LED should show up in the LCD display if it's lit. If you can't see it then the problem is with the LED (backwards, low current or burned out).

If you can see it then you need to determine if the ptrans is working properly. First thing, get a piece of black electrical tape and put it on the ptrans windows. That should cause the PT to stop conducting and you should measure the full voltage across the PT. When you remove the tape, the voltage should drop pretty low - the GP1S53V would have around .4V across it - but not to 0. See the datasheet for your interrupter.

That circuit is pretty simple, not much can go wrong.

finch661 · 30th October 2007, 11:36 AM

i am using a KTIR0611S from www.rapidonline.com it is a not a reflective interuppter! i will have another tinker with it

oh and it seems to be the simple circuits that get me... sigh

audioguru · 30th October 2007, 03:59 PM

Your circuit has 17mA in the IR LED. The datasheet for your KnightBright opto interrupter shows a typical current transfer ratio of only 15% (no minimum is listed) so the output current is 2.6mA which causes the output voltage to be about +0.15V when nothing blocks the IR beam.
Blocking the IR light will cause the output to be +5V.

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16th October 2007, 05:44 PM	(permalink (permalink))
finch661	Opto Interrupter i am trying to use an opto interrupter (IR Source and IR Reciever all enclosed) for a tachometer. The system should detect when the beam is broken. I have tried several circuits, but cannot get an output. eg http://www.shumatech.com/dro-350/add-ons.htm any help?

16th October 2007, 06:18 PM	(permalink (permalink))
audioguru	Does the IR LED cause current in the current-limiting resistor? If it doesn't then it is connected backwards. Does the photo-transistor conduct when the IR LED has current? If it doesn't then it is connected backwards. Does the photo-transistor turn off when there is no IR? __________________ Uncle $crooge

16th October 2007, 06:43 PM	(permalink (permalink))
finch661	the LED and photo transistor are correctly placed. With the current layout, (5V supply) the output reads 5V regardless of the IR being blocked...

16th October 2007, 08:02 PM	(permalink (permalink))
audioguru	The IR is not supposed to be blocked, it is supposed to be relected. The output is high because the photo-transistor is not turned on. It is turned on when it sees strong reflected IR from the IR LED directly in front of it. The IR LED has a 220 ohm resistor for you to measure to see if there is about 3.8V across ir which means that the IR LED has about 17.3mA lighting it. The shiny reflective surface must be at a 90 degrees angle and only 0.15" away. If the IR LED has 60mA then the photo-transistor will have at least 0.6mA. 0.6mA in the 4.7k collector resistor will cause the collector voltage to be 2.18V. Your IR LED current is only 43% of the 40mA in the datasheet so your collector voltage will not be higher than 3.8V when it is supposed to be low. 1) Increase the IR LED current to 38mA by replacing the 220 ohm resistor with 100 ohms. 2) Increase the sensistivity of the photo-transistor by replacing the 4.7k ohms resistor with 22k. 3) Make the reflective surface very shiny, at a 90 degrees angle and only 0.15" away. __________________ Uncle $crooge

16th October 2007, 08:30 PM	(permalink (permalink))
philba	What interrupter are you using? I assume you are using an interrupter and not the QRB1114 reflective device. the first thing I would do is get a digital camera and look at the LED. It is a little hard with an interrupter but it's possible. The LED should show up in the LCD display if it's lit. If you can't see it then the problem is with the LED (backwards, low current or burned out). If you can see it then you need to determine if the ptrans is working properly. First thing, get a piece of black electrical tape and put it on the ptrans windows. That should cause the PT to stop conducting and you should measure the full voltage across the PT. When you remove the tape, the voltage should drop pretty low - the GP1S53V would have around .4V across it - but not to 0. See the datasheet for your interrupter. That circuit is pretty simple, not much can go wrong.

30th October 2007, 11:36 AM	(permalink (permalink))
finch661	i am using a KTIR0611S from www.rapidonline.com it is a not a reflective interuppter! i will have another tinker with it oh and it seems to be the simple circuits that get me... sigh

30th October 2007, 03:59 PM	(permalink (permalink))
audioguru	Your circuit has 17mA in the IR LED. The datasheet for your KnightBright opto interrupter shows a typical current transfer ratio of only 15% (no minimum is listed) so the output current is 2.6mA which causes the output voltage to be about +0.15V when nothing blocks the IR beam. Blocking the IR light will cause the output to be +5V. __________________ Uncle $crooge