How Can I Create an Adjustable Optical Switch Circuit for an Inductive Load?

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The discussion focuses on creating an adjustable optical switch circuit for controlling an inductive load, specifically a relay. The circuit requires a 5V DC supply for the optical components and a regulated 3-7V DC for the relay, with a need to keep the relay activated for a short duration after light is blocked. Suggestions include using an LM555 timer for debouncing and setting a minimum on-time, while also considering a simpler design using a Darlington transistor for switching. The idea of incorporating a capacitor on the transistor's base to control the timing is proposed, allowing for adjustments with a potentiometer. Overall, the participants emphasize experimentation with components and circuit design to achieve the desired functionality.
ashankan
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i'm trying to make an optical switch circuit.
i'm no expert on electronics, so i was hoping you people could give me a helping hand:)
in advance, thank you very much:)

the purpose of the circuit is to switch an inductive load such as a relay.
when the light is blocked i need the relay to be switched on.
the optical circuit needs a 5vdc supply, and the the relay will be powered by a regulated psu 3-7vdc.
the relay will have a resistance about 3-4ohms.
i also need to keep the switch that turns on the relay open some micro seconds(about 0-0,03seconds) after the light has been cut.
i was thinking that i could hold the switching transistor open with a capacitator and a potentiometer
to regulate how long it would stay open after the light have been cut.

i have looked at this circuit:
http://www.circuit-finder.com/categories/sensor/light-sensor/331/simple-optical-switch

i could use that one, just need to have a adjustable circuit that holds the switching transistor open some micro seconds.
it says 9volt, but the lm555 operates fine with 5volt, so the circuit should work fine with that?
can anybody give me a tipp to how i could make the adjustable switch?
would it be possible to just drop the lm555 chip and hock the switching transistor directly on to the opto switch?
if i could keep the circuit as simpel as possible it would be the best.

so i was thinking that i could just have an opto switch which turns the switching transistor on\off,
and a regulation on the transistor switch which holds it open for some micro seconds.

this is the parts i was thinking of using for the opto switch:
optical transistor: vishay BPW17N, http://docs-europe.electrocomponents.com/webdocs/0de2/0900766b80de2cc1.pdf
IR-diode: vishay CQY37N, http://docs-europe.electrocomponents.com/webdocs/0e22/0900766b80e22c70.pdf

i'm not sure which transistor i should use for the relay switch.

i'm not asking you people to do the work for me.
if you could just give me some advice on what i should check out to figure this circuit out, that would be fine.
then i will draw the circuit and post it. then you could say if it's fine, or point out faults or improvements.
 
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Yes, you could probably dispense with the 555 and just run your switching transistor from the opto-diode, assuming the light is bright enough to keep the opto-diode's voltage below the switching threshold of your power transistor. You can help that out by using a "darlington" NPN transistor as the switch, something like the TIP 111 or pretty much anything that will handle the power you need.

The 555 has the advantage to "debouncing" so your switch doesn't go on'n'off a bunch of times as the light is interrupted, and it allows you to set a minimum time for how long it is on. You could use a capacitor across the opto-diode for similar effect.

I'd say get some general-purpose-whatever-you-can-find components and breadboard the thing to see what you get.
 
thank you.

how would i go about setting the minimum time?
i'm going to buy some parts soon. just need to find out exactly what i need.

the diode and opto-transistor will be completely covered. there won't be any leakage of IR-light. do you still think it would go on'n'off?

but maybe i should go for the lm555 circuit, then i can try other circuits later.

the thing i was wondering about was if i could put a capacitor on the base of the transistor-switch, so that when the power to the base was cut, the discharge from the capacitor would hold the switch open about 0,03 seconds.
and if i put a potentiometer between the capacitor, base, and ground i could adjust the leakage to base, and ground.
could that work?
 
Try out the capacitor idea... bigger capacitor, longer decay time.
 
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