I understand the general working of the circuit – R1 biases Q1, Q2 acts as an error amplifier closing Q1 just enough to get 1.4V at it's base. What is the puropse of R2?
simulate this circuit – Schematic created using CircuitLab
Electronic – the purpose of R2 in this discrete voltage regulator circuit
transistorsvoltage-regulator
Related Solutions
First of all, a 9V battery will not output 2Amp. This alkaline battery http://data.energizer.com/PDFs/522.pdf can give you 500mA but even then the rated capacity is halved.
Next, your "Zener circuit" will start conducting above 17V (it starts conducting when Q2 base is 0.7V above ground, so when current through voltage divider reaches 0.7V/4.7kOhm=0.149mA ; this happens when input voltage is 114.7kOhm * 0.149mA = 17V). As-is it is equivalent to the voltage divider alone ; your + input is therefore about 0.2V below 9V. So the op-amp will be fully-on all the time.
Finally, did you use this particular amplifier? That's not an op-amp (gain is 20x) and the output is referenced to Vcc/2 (here to 4.5V). So, when inputs are equal the output voltage for your circuit is 3.8V.
I'd suggest you replace it with a proper op-amp, and choose it so that the output voltage can swing to at least 1V above output (that's 3V below rail) while delivering significant current. Also mind the input voltage range...
Or, if you're not after high precision (as your use of a non-temperature-compensated reference suggests) why not use directly your zener "circuit" and a pass transistor? Something like this :
simulate this circuit – Schematic created using CircuitLab
This works because the zener steals the current from the transistor base as soon as the voltage is above the zener threshold, which is 0.7V above the output (because of the base-emitter voltage drop of the transistor).
Obviously for 2A you also need a beefier pass transistor, something like TIP41 instead of poor little BC547... It will have to dissipate max 2A*4V, that's 8W of power!! Then lower R3 to 47 Ohm.
It's just a bog-standard series-pass linear reg with with a FET instead of a BJT.
As @PlasmaHH notes, it is often used a pre-regulator in more sophisticate setups. It allows you drop some voltage (and thus power/heat) on an external element. Doing this also lets you extend the voltage input range of your more expensive IC regulator. TI has a separate appnote in which they suggest this.
This prereg idea (even with BJTs) not at all an uncommon. Cordless phones do this a lot for instance because they also use the [higher] pre-regulated voltage directly for a few components and they also derive a lower stable voltage for most of the digital parts. So in those setups it serves a double function (more bang for the buck).
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Best Answer
R2 simply makes sure that there's a certain amount of current flowing through Q1 even if no external load is attached to the circuit. This is called a "minimum load".
It also makes sure that a certain amount of current is flowing through D1, which makes its forward voltage drop have a more stable value, improving the regulation of the circuit overall.
Note that as long as R1 = R2, the current through D1 is independent of the actual value of Vout; it only depends on Vin and the resistor value. The current flowing through R1 is (Vin – Vout – Vbe) / R1. The current flowing through R2 is (Vout – Vf) / R2. If you add these together, the Vout terms cancel, leaving (Vin – Vbe – Vf) / R1.
The only current through these two resistors that doesn't flow through the diode is the base current of Q1, which means that the diode current slightly depends on the amount of current being drawn by the load, depending on the actual gain (current transfer ratio) of Q1. For high loads, this would be a good place for a Darlington.