Electronics newby seeks advice on circuit please.

1066 · Mar 16, 2011

Hi
I have put together a simple working circuit and have drawn up a schematic with a view to getting some PCB's made.

So - Could some of you guys who actually know what you are doing (I really don't have much of a clue) cast your eyes over the drawing and point out any glaring errors.

The object of the circuit is that a momentuary press of SW1 will energise D2 infra-red emmiter (and start the small DC motor running) - Photo transistor D1 will then energise the relay and so latch the circuit on and the motor will continue to run until optical switch (D1 D2) is broken.

The reason I layed the schematic out with the photo-switch out on a limb is partly so I could get my head round it and also I want to extend them away from the board by a 3 conductor cable.

I'm sure there are better ways of doing this but cost, size and simplicity are all important.

I have the relays at about 40p each and the photo diodes/transistors at arounf 12p each.

Any feedback would be appreciated please.

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(*steve*) · Mar 16, 2011

When you say "working" -- is it?

Have you breadboarded this?

You should breadboard your circuits.

The relay should really be a collector rather than an emitter load.

A photodiode is perhaps not the best element to use -- I'd probably go an LDR but you will probably have problems unless you have some means of adjusting the sensitivity. Ambient light may be a problem.

The photodiode, the way around it currently is, is not going to work. You would probably need a lot more gain with a photodiode anyway.

1066 · Mar 16, 2011

Thanks for the reply Steve - Yes the circuit works - I build them on a 2x2 inch vero board.

This is the photo-transistor I use, a Lumex 2 pin NPN.
http://www.dynamictoys.com/Products/071228/st-8l.pdf

Instead of the infra-red emmiter I have been using a 3mm "grain of wheat" bulb but have tried the IR Led in the interest of cutting battery consumption.

The interrupter gap is around 20mm with the photo-transistor housed in a black tube and lined up directly with the light source. I opted for a photo-transistor rather than an LDR as I need a quick and consistant shut off with varying ambient light conditions.

When you say the relay should be a collector load, is this as simple as just swapping the relay the other side of the transistor and what are the advantages of this?

(*steve*) · Mar 17, 2011

If it works, then I'd leave it how it is.

Your symbol for the phototransistor is wrong, and that threw me.

The advantage of having the relay as a collector load rather than an emitter load is increased sensitivity. Again, if it's currently working, I wouldn't change it as it may upset other operating parameters.

1066 · Mar 17, 2011

Thats good news Steve, and I can see what your are saying about the photo-transistor symbol.

Now another couple of questions - I have run this circuit sucessfully on 6v with 4AA batteries and while running draws around 160ma (using "GoW" bulb). By changing the motor and relay the circuit runs reliably on 3v using 2AAA batteries and substituting the IR led for the bulb helps with reducing current.

So - The little motor runs happier on about 2 - 2.5 volts, is it exceptable practice to just plonk a 1N4000 type diode (or 2) in series with the motor to drop the volts down a little?

How do I go about getting a few prototype PBC's made up and about what sort of cost would be involved for say 10 pcb's about 60 x 40 mm. (cost is of the essence here) Or, would it be more practical to do it myself for a few prototypes? (I made a few small boards around 30 years ago but they didn't look very professional, I guess its easier now with computers and printers)

Any help appreciated

Allan