The micro voltate regulator

SARCASM
Voltage regulators are needed all the time. But which one should you take? A simple zener diode? A zener diode and a pass transistor? They have both poo regulation. Don't they? Well. What should you do? You could use one of those hard to get and strange ICs called 78xx or 79xx. But do we really (i mean really) know what they are doing. Nooooo. ICs are bad. What you need is a completely discrete solution. No ICs, no hard to get parts, simply adaptdable. And here it is: The micro voltage regulator.
/SARCASM

Schematic

Board Layout


The board size is about 15mmx19mm. The input is marked with I, the output with O and the ground with a ground symbol.

Test Results

I built a prototype with VZD=6.8V, R5=330Ohm, R4=10k, R3=3.3k which leads to an output voltage of 9V. The small signal transistors are 3 BC546B. Th Load regulation from zero load to 1A is within about 120mV which is more than twice as much as in a simulation, but remember that in simulation all Transistors have exactly the same parameters which is not true in reality. (Load regulation test was done with 13V input.) The minimum voltage drop depends on the current amplification of the transistors and the output current, but is at least UBEQ4 + UBEQ3. The ripple rejection was done by turning the knob of the Lab power supply because I don't have any well suited AC source to deliver this power. This caused the output voltage to float around some millivolts (approx 10mV) as the transistors heat and cool. The input ripple rejection test was done at 0.55A load.
Output versus input voltage at ILoad=0.55A
Input Voltage [V]Output Voltage [V]
128.9
138.95
149.02
159.066
169.117
179.14
As you can see, the input/output voltage relation within these limits is approximately linear. And i calculated a output voltage change of:
(9.14-8.9)/(17V-12V)=48mV/V
This a ripple rejection of about 26dB.

Of course a 4 transistor solution can't be as good as a fully fledged integrated voltage regulator, but it does its job quite well. Q1 and Q2 should be coupled thermally as close as possible, e.g. put heat sink compound between them and wrap a bit of wire around them to hold them together (I didn't do that, but it should be better). Note that the input compliance range of the differential amplifier doesn't go down to 0V by far. The worst case is when Q1 is open, and Q2 conducts completely. In this case you can derive the minimum output voltage quite easily. A zener diode of 4.7V should provide enough room. R3 and R4 cound be replaced by a Pot to make a variable output voltage supply, but remember that you'll never be able to go down to ground. Additionally you'll get problems with the zener-current which would vary a lot if the output voltage range is large.
The best would be to replace R1, R2 and R5 with constant current sources, but this increases the parts count. It would be better to buy a regulator IC in this case.

Minumum output voltage

To calculate the minimum output voltage, we assume that the base of Q1 is on ground. This will lead to the minimum output voltage. Further we assume that no load is connected to the output and thus we can neglect the base currents. (If we don't neglect the base currents, the complexity of the calculation would increase quite a lot, because the unknown variable would increase from 2 to 5. The two IB's and the collector current of the pass transistor would have to be found too.) Consider this simplified schematic: 
   ^        ^ Ub
   |        |
  .-.       |
  | | Rc    |
  | |       V ic2
  '-'       |
   |      |/
   *------|  T2
   |      |\
   V ic1    |
   |        |
    \|      |
  T1 |------*      |
    /|      |      |
   |        |      |
   |        |      |
  .-.      .-.     | Ua
  | | Re   | | Rl  |
  | |      | |     |
  '-'      '-'     |
   |        |      V
  ---      ---
(Note that in the "real schematic T1 is called Q2 and T2 is the combination of Q3 and Q3. The base-emitter voltaged Ube1 and Ube2 are not drawn here.)
You'll see that we need neither ic2 nor Rl for our calculation, because of the simplifiction mentioned above. Using Kirchoff's laws it is simple to find the following equations: 
I:  Ua - ic1*Re - Ube1 = 0
II: ic1*Rc + Ube2 + Ua = Ub
Now we solve I for ic1 and get: 
ic1 = (Ua - Ube1)/Re
We can put ic1 into II, and solve for Ua: 
Ua = (Ub*Re + Ube1*Rc - Ube2*Re)/(Rc+Re)
If we insert our component values, using 0.7V for Ube1 and 1.4V for Ube2 (because it's a darlington) and a assumed input voltage of Ub=20V we get 1.84V. That's the minimal output voltage at no load, T1 is already as open as it can in this condition.