It should work but be careful, the LED characteristics vary as light falls on them so keep them in the dark if you want a contant voltage. The method of setting the voltage isn't very good because it uses the dynamic resistance of the LEDs which is not a defined parameter. It would be cheaper and more stable if you removed all the LEDs and fitted Zener diode instead. The output voltage will be the Zener voltage + the B-E drop (about 0.65V) in the bottom BC547.
Brian.
It should be wired with the cathode (+) end toward the output point.
Looking closely at the schematic, are you sure the middle transistor is supposed to be a BC547 as there is no way to provide bias to it at the moment. I think either it should be PNP or the 2.2K resistor is in the wrong place.
Brian.
Have you fitted R4 (in the original article), it provides bias for the BC547 but it is missing in your first schematic.
Brian.
You are absolutely right!, I will try it.
Thanks a lot
Try this for a better solution: https://www.ti.com/lit/ds/symlink/lm317.pdf
One of the drawbacks of a three-pin voltage regulator is that the input voltage needs to be 2.5–3 V higher than the output voltage.
This makes these integrated regulators unsuitable for battery power supplies.
If, for instance, the output voltage is 5 V, a 9 V battery could be discharged to 7.5 V or thereabouts only.
On top of this, most of these regulators draw a current of about 2 mA.
Special low-drop versions sometimes offer a solution, but they are not ideal either.
The regulator described here is rather thriftier: it draws a current of only 300 µA and the difference between its input and output is only 100–200 mV In the circuit diagram, T1 is arranged as a series regulator, which means that the difference between input voltage and output voltage is limited to the transistor’s saturation potential.
So the small Vin-Vout difference is not a requirement?
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