This circuit provides a constant current to the LED. The LED can be replaced by any other component and the current through it will depend on the value of R2. Suppose R2 is 560R. When 1mA flows through R2, 0.56v will develop across this resistor and begin to turn on the BC547. This will rob the base of BD 679 with turn-on voltage and the transistor turns off slightly. If the supply voltage increases, this will try to increase the current through the circuit. If the current tries to increase, the voltage across R2 increases and the BD 679 turns off more and the additional voltage appears across the BD 679.
If R2 is 56R, the current through the circuit will be 10mA. If R2 is 5R6, the current through the circuit will be 100mA - although you cannot pass 100mA through a LED without damaging it.
3 comments:
What does this circuit do to the voltage? Will the output stay the same as the input?
Assuming the load to be straight resistive, the output voltage will be a function of current times load resistance (Ohms law).
However, LED's are very different (non linear) as the forward voltage stays pretty much the same once you reach the threshold, and only increases slightly (as the diode gets brighter) up to the point where you burn out the LED. This is why you want the constant current in the first place. By varying the current (and modern LED's such as some of the the Cree XP and XT 3 watt series, can take up to 1.2 amps), you vary the brightness. To determine the voltage you need on the output, you look at the LED's Vf (forward voltage) and the input voltage must be somewhat higher (my electronics is a little rusty, so I cannot say exactly, but it isn't a critical value once you are above a certain point). So, if you have a typical 700mA LED with a Vf of, say, 3.2 volts, you could run 12 LED's in series at 700mA from a constant current source capable of outputting some 38 volts at 700mA. Or you could use two parallel chains of 12 LED's each from a 1400mA source (though fuses and balancing resistors are a good idea here.)
By changing R2 for a potentiometer, you can adjust the output current and hence the brightness although if you go too low, the diode will simply turn off.
Oh, I should have added to the last post that the circuit given probably needs beefing up to drive high power Cree LED's and the like.
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