Using a L14F1 IR phototransistor in a switch

univeruser

As aforementioned, I have a L14F1 phototransistor. I am using this in a DC circuit.

A stupid question it is. Please take it seriously however. Read carefully; I repeat: how do I make a switch in a DC circuit using a L14F1 (3-pin) phototransistor?

And coming to why I don't use a 2-pin phototransistor rather than 3-pin, the cause for it is that there is nowhere I find 2-pin phototransistors.

Please post a diagram if possible, illustrating the circuits.

Note: Please don't tell me any locations or websites to find a 2-pin phototransistor

Adjuster

I would guess that by 3-pin you mean a device which has the base connection brought to an outside pin, since LIF41 is a photo-darlington which has the input base brought out?

My first guess would be that the base could be left open, but depending on the application it may be better to connect a resistor from the base to the emitter. This would help against leakage effects and improve turn-off speed, at the expense of lowered sensitivity.

Could you please post a diagram and fuller details of your application?

univeruser

All right, here it goes, the diagram you asked for.

You may be knowing that the current passes from n to p, as shown in this circuit layout. Basically, I am simply using it as a switch to make the motor spin whenever beamed at with an IR LED.

I am not really well versed with transistor circuits, so I believe that you may have to check out from this site below from where I am going to purchase the transistor, whether it is n-p-n or p-n-p.

http://cgi.ebay.in/L14F1-IR-Photo-Tr...item2a031b352f

P.S. please post me the circuit diagram using the L14F1 phototransistor quickly, I have less time on hand.

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Adjuster

Well Univeruser, I think I'm in a different time zone to you. My last post was sent during my lunch break. It's evening now, so time for another look at the computer. It probably isn't a good idea to expect express replies from people on a forum like this though.

Frankly, I don't think the simple arrangement shown in your diagram is likely to work very well. I'm going to give you a few tips about that, but you are going to have to get your requirements better defined before going any further. Here goes:

1. To design this you first need to specify the motor power requirements. What current will the motor take?
   
2. You have a 330 ohm series resistor in the circuit. This would limit the current to 27mA. Does the motor really need so little current, or was this copied from, say, an LED circuit?
   
3. A data-sheet for L14F1, http://www.datasheetcatalog.org/data...hild/L14F1.pdf specifies minimum 7.5mA collector current at .125mW/cm² illumination - you might do better with a powerful LED right next to it, but I suspect that this is telling us that some sort of driver would be required to provide more current.
   
4. Finally, unless you are planning to have all other sources of light excluded, things like room light are likely to interfere. If what you want is really some sort of remote control, I'm afraid that a more complex modulated system would be needed.

berkeman

As Adjuster says, an IR LED at basically any distance away will not give you enough current gain to turn the motor. Can you use a laser pointer instead? If you hit the base junction of the phototransistor with a laser pointer from not too far away, that might give you enough photocurrent out.

univeruser

All right, I won't mind using a laser pointer. But however, could someone quickly post the diagram so that I can decide if I should buy this L14F1 phototransistor or not, before the end of 13^th January, 2010 (in accordance with GMT +5½, Kolkata, India)?

As to explain why I am in such a hurry, the cause for which is that I have a gift voucher for Rs. 200 to be used up before or on 13^th January, 2010, and my parents are not willing to pay any money for it by not claiming the voucher.

Please make your post with the circuit diagram as soon as possible. And yes, that circuit diagram is partly copied. And talking about the resistor, I shall make it 100 Ω; in fact, I'm not specific, because I'm not well-versed in electronics (of course, being just 12 and in VII^th class).

As for the motor, I shall use a brushed or brushless motor, except that brushed ones are slower. If 330 Ω limits the power that flows to the motor and making in work at a lower revolution rate per minute, then I shall take a brushless inrunner (7700 rpm/volt) for the purpose. I am in fact trying to make as wirelesly controlled toy (say, a car) with it.

I hope I get my requirement without further questions by the next post or any sooner. Consider the circuit using a brushed DC motor, not a BLDC. Lastly, could I leave one pin unconnected? I suppose so, but which one?

Adjuster

I am sorry to note that the difference in our time zones may mean that this message comes too late to help with your decision. In any case , I think that it might be very difficult to control a model car in the way you suggest with such an insensitive circuit. It would be quite difficult to keep a laser pointer accurately aligned with a detector on a moving model.

It would of course be possible to design some more sensitive arrangement with more components, but this might be a bit too ambitious given what you have said about your age and limited knowledge of electronics. Depending on how you were planning to build this (soldering?) there may also be safety issues, so on the whole I think it's better to leave it there.

Sorry if that's a bit disappointing, but we have to be realistic about things.

univeruser

I believe soldering shouldn't be much of a safety concern as I have used them before and have gone for a robotics class wherein we had to solder the parts provided onto the provided PCBs (in a way, they didn't explain how they work, but just doing the soldering). Let us trash the safety issues as of now.

Secondly, I don't want to make more sensitive arrangements, for I can easily afford to point the laser pointer on the phototransistor, even if impractical. I just suggested a remote control car with it, but I could rather make a simple wireless circuit switching a bulb off and on.

Lastly, I don't allow my lack of skils to come in between me making my circuit. This was exactly why I asked a circuit diagram. Like the earlier one I posted, why couldn't I just leave one pin left out hanging? But then, which one?

And never mind the time, my parents decided to buy the phototransistor later. So again, I hope the diagram will be posted without further questioning.

Adjuster

Hello univeruser

Since you have said that you don't want to make any more sensitive arrangements, here is a simple diagram adapted from what you originally uploaded, showing the photodarlington transistor driving a LED - in this case, the 330 ohm resistor is needed. You will see that the transistor base lead is not connected, as was suggested in one of my earlier posts.

In another posting I also sent you a link to a data sheet for the L14F1. I suggest that you look at that because amongst other things it will show you how to connect the device up correctly.

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univeruser

Thanks. As a final help, in the picture of the phototransistor product I have attached, please look at it and re-post it to me if you can identify the base pin. I believe it is the pin that juts out a little bit, which should be the base pin (I have marked my guess in the photo attached). If not, I don't mind, a little bit of trial-and error can be done.

By the way, could you tell why the resistor is used in the circuit? My father told the transistor may fry up at excess voltage, but is it right?

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Adjuster

Hello Univeruser - here are the answers to your questions:

The base connection is the MIDDLE pin. This is shown on Page1 of the datasheet - is there a problem with the link which I sent earlier? It would be better to avoide trial-and-error, as you might damage the device.

The resistor is required so that, no matter how powerfully the transistor is illuminated, the current cannot become too high. Your father was correct in that without a resistor, the transistor could overheat, as could the LED. The resistor should therefore not be left out.