I don't know what to do!

I have made a IR transmitter that is driven by 2 AA batteries (3V) look at the scematic.

But i have a problem, it is not reaching far enough!

A week ago i made a IR tranmitter with the same scematic but i used 6V instead and used larger resitors and had the led's seriell instead of parallell. That one had ways longer reach.

Why doesn't the 3V version work good?

The transitor i am using is a Darlington switch for currents up to 5A or something. But in my circuit there's just 1 to 1.5 A.


Any ideas what i should do to get it work better? Is the voltage drop in the transitor to big so that the voltage over the leds get to small? What can i do to get aroung this problem? I need to use 3V...

Attached Images

ir_152.jpg (21.1 KB, 776 views)

 

The IR led i am using is this:
http://www.elfa.se/elfa/produkter/en/2021685.htm
nr: 75-225-43

 

I've just done something similar using a 16LF628 (LF for the extended low voltage operation). I used a single LED, with a 1 ohm series resistor fed from a BC337 with a 330 ohm feeding the base - this worked fine.

I would suggest it's probably not a good idea to use a darlington, they often have a fairly high voltage drop from collector to emitter - try using just a normal transistor instead.

If you have a scope try measuring the voltage drop across the darlington!.

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I have meassured the voltage drop with a scope. I meassured between ground and collector, is that right?

There is 0.8 to 1 volt. Is that much? How big is the voltage drop on your BC337 and how much current can it handle?



The 6V version i made could send a data-string perfectly within 20-30 meter in full daylight, but this one can just do 10meters without problem...

 

brodin

Have you measured the current trough the IR diodes? I think that the current is interesting.
You should compare the current to the 6Volts version.

lycka till

Ante :roll:

 

Yes!.

I haven't scoped it, but a fully turned on silicon transistor only drops about 0.1V-0.2V across collector emitter - a darlington drops a great deal more, which you can't afford with a 3V supply.

The BC337 is rated at 1A maximum, but bear in mind it's been pulsed with a low repetition rate (just as the LED's are).

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Use PNP transistor like BC557 to drive your LED. Most I/O pins don't source much current to hard drive any NPN transistor into saturation. Also if you want to drive NPN transitor to saturation, its base voltage needs to be larger than its collector voltage (both junctions should be forward biased) which is many a times not possible because in most cases collector is held at Vcc level and so you require another supply for supplying to base.

It is easy to sink current in an I/O pin. Thus you can easily drive an NPN transistor to saturation by making its base at ground (logic 0) potential while its emitter is connected to Vcc through IR LEDs.

See schematic below.

Attached Images

ir_101.jpg (21.3 KB, 726 views)

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I disagree, with the LED's in the emitter of the transistor you are always losing at least 0.7V across the transistor - with the LED's in the collector it's reduced to 0.1V-0.2V.

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Ok, I have changed the schematic now. Now LEDs should get max. possible power.

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Thanks guys for the good answers!

I will meassure the current through the LED and compare the 6V to the 3V version.

 

I'm just curious, Brodin. What range are you expecting from your IR link?

 

Well, i need it to work perfectly within 15 meters. I am going to use it as an identifications system. I am doing roadracing and i am constructing a laptimer that will stand by the side of the track and notice which bike that's passing and log the laptimes for each bike.
It sends a 6-bit string which is received by the PIC by the side of the road.

I have tested it and it works great with the 6V version. I passed the receiver in 200km/h (120 mph) and that worked just fine. But i need to get the 3V version to work so i need less batteries.


I have meassured the current through the LED which is just about 0.3 A (it is supposed to be 1A!).


I have meassured the voltage drops over the components.

U = 3V

Uled = 1.8 V
Ut = 0.8 V
Ur = 0.4 V

Would it be better if i changed the darlington to a normal transistor? Or can i change the resistor to a 0.4 ohm (0.4/1). Would that work good, not just theoretically?

 

How good did this work? How long does it reach?

 

That’s interesting, because I’ve done a similar project a couple of years ago. I had to bridge a track of 50ft so my experiences may be of help.

I used commercial rx chips and found that with their inbuilt pulse-stretcher they could handle IR pulses of only 10microsecs duration. I was using ‘break-beam’ detection for timing so continuous 10uS pulses were transmitted every millisecond. The detector looked for a break of longer than 2mS. Peak IR emitter power was 800mA, but with a 1:100 duty cycle this meant that the mean IR supply current was only 8mA. Beam range in bright sunlight without elaborate filtering was found to be around 100ft.

I used 3 IR emitters and a single limiting resistor, all in series, switched by a VN10 FET and fed from a 12V battery.

In your case, you have ports to spare on your uC so you could program for a high frequency output at one of them to feed a tripler to get your supply up to around 7V to feed series IR emitters. That would be less wasteful than using the parallel arrangement you have.

If you used my method of ‘break-beam’ detection for the timing, I would suggest you could arrange to send your six-bit word wrapped up with start and stop bits, say, 10mS after the break and use pulse width modulation to differentiate between ‘0’ and ‘1’.

Best of luck.

 

I've never had occasion to measure it (it wasn't a concern for the application). However, I'm a television engineer and I see a LOT! of IR remote controls, I've never seen one use a darlington transistor with a 3V battery - you just can't afford to waste the voltage on 3V.

If required you could decrease the value of the 330 ohm to provide more drive for the transistor to make sure it's hard on (even using more than one I/O pin to get more current) - but there's no way to overcome the extra voltage drop with a darlington.

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