3V IR transmitter problem!

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AFAIK it will simply kill the LED if you drive it too hard - as it's only a low duty cycle you can pulse the LED considerably higher than it's specifications, but usually about 1A is used - probably as a good compromise between range and power consumption?.

If you don't mind killing an LED or two, feel free to experiment :lol:

Let us know what happens!.
 
I am confused!

I bought a BC639 and a BD139 today. I have only tested the BD now but that one has the same voltage drop that the darlington had. And the person i bought them from said that there allways is a voltagedrop about 0.7V when using a BC transistor. So how do you mean that i could get just a 0.2 V drop?

I am using the first scematic that I posted, but with just one led.

Like this;
+3V--------[IR_LED]--------[0.6 ohm]--------[BD139]--------GND
 

Yes, a transistor turned fully on in that configuration will have a low voltage drop across it. Using the BD139 probably isn't a good idea, it won't have very high gain so probably won't turn fully on with the existing resistor - use a higher gain transistor, that's why I used a small BC337 rather than a power transistor - it has much higher gain than a power transistor.
 
I don't know what i am doing wrong. I tested with a BC547 (500mA) now. But that one drop 1.2 V!
Only 0.2V over the 1.2 ohm resistance! What am i doing wrong.

IR-Circuit
3V----[IR-LED]----[1.2ohm]----[BC547]----GND


The base is driven by a PIC via a 158ohm.
 
BC547 is too wimpy for 1 - 1.5 amps. A bipolar transistor that can saturate at 1.5 amps of collector current with 25ma of base current is going to be difficult to find.
Do as Sebi suggested - use a power MOSFET with Rds(on) less than 0.1 ohm at 3 volts Vgs, such as Fairchild FDN327N. This is a surface mount part, which might be a problem for you.
Keep in mind that the MOSFET must be specified at Vgs=3.0v or less.
 
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