> to power some 24V preamps.
Does it have to be 24V exact?
Is your "48V" known to be a steady 48V, or is it unragulated?
What value is R1?
Why is R2 variable to zero? Do you ever want zero V out?
Is there some objection to a 25V Zener?
The way I see it:
The "48V" probably is regulated.
"some 24V preamps" is -probably- under 100mA. Therefore over 240 ohms.
The emitter impedance should be much-much less than 240 ohms. Pencil 24 ohms.
The emitter impedance is the impedance at the base divided by current gain.
Assume Darlington gain is over 1,000. Pencil 24K base resistance.
The approximate voltage ratio is 2:1, so two equal resistors of 48K each will work fairly well.
These resistors draw 24V/48K or 0.5mA, dissipate 0.012 Watts. There is no objection to taking more power. This will also reduce output impedance. Pencil two 10K resistors.
> R1 ... 10K
OK, you are ahead of me.
Effective base impedance is 5K, effective emiter impedance is <5 ohms, which is reasonably small for ~~240 ohm loads. A 0.1A load will cause <0.5V sag, which is good for a simple buffered divider.
The actual output voltage is 1.2V less due to Darlington Vbe drop. Two diodes would give excellent compensation. Or observing that 1.2V is 5% of 24V, you could select R2 10% higher than R1, plus a bit for current sag: here is where you trim. R1=10K, R2=10K fix plus 2K trimmer. That gives you roughly 25V-23V range, assuming perfect 48V.
Now short the output. (Don't tell me this never happens.) What current flows? While C2 holds, current is limited ONLY by Q1 self resistance, possibly 50 AMPS, and near-certain destruction. If it gets past that, the <5 ohm output impedance suggests >10A current, at 48V, is 500 WATTS dissipation, which will distress Q1 to death.
There are a bazillion ways to limit current.
But you also have (48V-24V)*0.1A = several Watts heat in Q1. You can heat-sink it easily.
But imagine a 200 ohm resistor in series with Q1 collector. 0.1A will force 20V across the resistor. Now there is 4V across Q1, it dissipates 0.4W, and needs no sink.
The resistor is dissipating 2 Watts. A 10W part is inexpensive, and perhaps easier to mount than a heatsink.
AND the maximum current to a short is 48V/200= 0.24A, well within Q1's rating. And while that is happening, Q1 voltage is nearly zero, dissipation is maybe 0.3W, still smoke-proof. The resistor is 11 Watts... a "10W" part will sit there and stink for many minutes. No instant-death, and if you ignore it, it will burn-open in minutes or days.
If you -know- your load, figure this resistor to sop-up most of the excess voltage. If uncertain, figure to sop-up some of the excess, to allow some leeway, yet limit "infinite" disasters to some slow-smoke situation.
> use a switch, followed by an LDO.
Good for your company. Possibly cool and clean. But "some 24V preamps" implies an era when power was pretty crude, yet sound was good. Maybe they sound better when powered better. Or maybe the onboard caps do their job: make power quality moot. And he's not concerned with cleanliness nor efficiency. I'd use 1970 power techniques with 1970 classic-amps.
Anyway, doesn't TI still make/market/license the "TIP" parts? That's a penny for TI.
> good ole' 2N3055, it will run cooler.
How? It still dissipates V*I. TIP122 is a reasonably large part, not as big as 2N3055, but plenty big for this chore. (Anyway I already eliminated the heatsink.)
And the '3055's much lower current gain will demand much more current in the base feed. The 10K penciled above, as a simple divider, has to be more like 500 ohms. A Zener won't need huge current for low impedance, true. But it still wants over 2mA divider current to ensure a '3055 can dump 100mA without starving the base.