Electronic – Voltage divider Biasing of transistor

Personally speaking I think R1 is to prevent the voltage regulator from oscillating its little head off.

The output transistor(s) has considerable gain due to its configuration of collector feeding the real output - a lot of voltage regulators feed their output with an NPN emitter - these are easy to design and are more-or-less predictably stable but, they won't work well when Vin is close to Vout i.e. they are not low drop-out regulators. Typical example is the 78xx series - they need input to be about 2V greater than Vout to work properly.

The one in your circuit uses a PNP transistor and this will tolerate Vin getting quite close to Vout before the regulation fails. Probably to within 0.5V - this puts it in the class of an LDO regulator (low drop-out).

So, the PNP has quite high gain (voltage fluctuations on base produce much larger fluctations on the collector. This transistor(s) are fed from an emitter follower (no added voltage gain) and that transistor is fed from an NPN which also has considerable voltage gain - it looks like it's a common base circuit - the base receives a bias from the output voltage via the potential divider - it looks like a 1k1 and a 1k but the diagram has seen better days - this, assuming I'm correct, will put about 6V on the base when the circuit is regulating.

Note also, there is a start up circuit (via the 1N4148 diode feeding the base) - this puts "some" voltage on the base to turn it on whilst the circuit settles down to proper regulation. That start-up voltage is "replaced" by the 1k1 and 1k voltage once output has risen sufficiently. The emitter of this transistor is where the feedback control happens. If the output rises a little too high - the emitter also rises and stops conduction of this transistor thus causing the output transistors to start to switch off.

Like I said earlier there is a lot of loop-gain and this gain is dependant on the undisclosed output load. This is because the series pass transistor(s) feed that load via their collectors. This means the loop gain may cause instability on light loads and go into a situation where the output rockets between minimum and maximum voltage - max voltage could indeed be as high as the input voltage and this would be serious a problem.

In summary, I think R1 is intended to stabilize the output on light loads (maximum loop gain) - the output transistor(s) can turn off fully but there will still be a small feed of current to the output which "dampens" down rogue and extreme changes in those transistors.

That's how I see it anyway. I can't see that R1 offers any other function.

As you said, the LED is build in reversed and so would act like a Zener diode with some volts. But the B-E junction of the transistor forms a diode in forward direction, hence it limits the voltage to about (below) 1V.

When you switch on the circuit, there will be a current into the base, and the transistor is ON (=conductive).

Due to this, a current starts to flow through the primary coil and induces a voltage in the secondary coil. The lower end of the secondary coil becomes negative, pulling down the voltage at the base. The transistor switches OFF, the current in the primary coil stops, you get another voltage peak in the secondary coil in reverse direction, the transistor switches ON and so on...

The diode protects the transistor base from too large negative voltage spikes. I haven't seen it with an LED yet, but with two fast switching diodes like FR207 in series.

From the schematic, it looks as if the base always has the potential of the lower end of the sec. coil. This means it can not affect the B-E voltage of the transistor. However, the sec. coil is coupled to the circuit especially due to the very high fields in this case. So yes, it can have an influence on the B-E voltage. But this is not visible in the schematic.

Another point: I read several functional descriptions of this circuit, some of them were really hilarious. But all of them mention that the polarity of prim. and sec. coil may be wrong. If the circuit doesn't work, one should just swap the terminals of the prim. coil. So, don't think about which polarity is induced in the sec. coil when a certain current flows though the primary.