Driving a LED

[Hero999]

If you want simple youd could always use a jugfet to give you a constant current source of about 10mA.

 

[audioguru]

The 22 ohm resistor in the two-transistors constant current sink circuit determines the current. The trimpot adjusts so that the upper transistor has enough base current at the minimum battery voltage.

With only two LEDs in series for 6V, then nearly half of a 12V battery is wasted. Use 3 LEDs in series with a 12V battery.

Use a 9V battery then two LEDs in series will remain bright until the battery voltage drops to 7V.

Darlington transistors aren't needed in the simple circuit.

 

[Cabwood]

Without a switching current regulator, you are forced to limit current through a diode with an additional load in series with it. Even with PWM to provide large current pulses, the linear limiter will dissipate the same fraction of power as it does with continuous current. There is no efficiency improvement obtained from using PWM.

The trick to maximising efficiency of linear current limiting is to maximise the ratio of voltage across the useful load to voltage across the current limiting load.

Of course you can't control the voltage that a diode drops, and so the only way to do this with LEDs is to string as many of them in series as you can, until their combined voltage drop is a little less (a couple of volts or so) than the power supply voltage.

This way, the remaining voltage (dropped across the limiting load) is kept small, and the ratio of power dissipated in the LEDs to power dissipated in the limiter is high.

Due to the variations in LED forward voltage, it is impossible to predict with any accuracy what that remaining voltage will be. Since it is so small, and so variable, calculating a simple limiting resistance in order to control LED current is not a good idea. Even if you could measure the combined LED drop, any reduction in battery voltage will quickly dim the LEDs.

It is possible though, with a few easy to obtain and cheap components to create a simple current regulator which will supply a precisely controlled LED current, even as the battery fades. That's what the diagram is.

Know these things:
a) Vbatt: Battery voltage. I'll assume 9V
b) Vled: LED forward voltage. I'll assume 2.0V
c) Iled: LED operating current. I'll assume 20mA

The two regular silicon diodes in series provide a stable 1.4V at the base of the transistor. The voltage at the transistor's emitter is about 0.7V less than this - about 0.7V. With ohm's law calculate Re to provide 20mA:
Re = V / I = 0.7 / 0.02 = 35. I'll use 33 Ohms, which is close enough.

If we used 4 LEDs, 8V is lost across the diodes, and 0.7V across Re. For a fully charged 9V battery that leaves only 0.3V of freedom for the transistor. If we use only 3 LEDs, that leaves the transistor with a headroom of 9V - (2V x 3) - 0.7V = 2.3V, which is enough. It means also that the battery voltage can drop down to 7V without any loss of current regulation.

The attached diagram is based on these values. Rref is chosen so that Iled is large compared to Iref (25 x Iref in this case), but keep the ratio Iled/Iref significantly smaller than the transistor's beta.

The current drawn from the battery is about 22mA, for about 200mW of power. 130mW of power is delivered to the LEDs, so the overall efficiency is 67%. This efficiency increases as the battery voltage drops.

The power dissipated in the TR1 is 46mW, and in Re 16mW, so no need for hefty components. Even with 200mA through the LEDs Re can be a 1/4W type, but TR1 will need to be chunkier - like a BFY51. For D1 and D2, any silicon signal diode will do, such as 1N4148.

A fully charged 200mAh PP3 battery should last about 8 hours powering this circuit.