A Maybe Not That Complicated Either Headphone Amp (Part 4)

Now we're on to the power circuit.  The result is disappointingly boring, but I wanted to look at the circuit I tried first.  Let's look at how we're getting power.

The phantom power (+48V) is applied to both the "HOT" and "COLD" lines, which are what carry the audio.  Whatever we do, it needs to affect both of those lines equally.  Anything we do to one signal and not to the other will be amplified by the input of the mixer.  And we don't want the actual microphone signal either, just the DC that it's riding on.

So here's my initial layout:

After drawing this up I felt pretty proud of my somewhat involved circuit.  Here's how it breaks down: it's just a zener voltage regulator.  If we just look at R1, D2, and Q1 we see that zener regulator.  Ideally, the base of Q1 will always be 5.1V.  Subtract $V_{BE}$ and we see the emitter (and effectively VCC if we establish R15 and R16 as small) as 4.5V.  Q1 will draw as much current as needed to maintain that emitter voltage.  So when that kick drum hits, it'll turn on more to keep Vcc from sagging.  Now I said before we have to affect both audio lines equally.  Well, we can just add R2 and Q2 as a current mirror.  With the bases tied together, and R1=R2, both transistors should pull half the required current needed to power the amplifier.
In reality though, Q1 and Q2 will always vary slightly, and you need ballast resistors (R15 & R16) to balance them out.  Their gains and actual $V_{BE}$ values will differ, and you can get thermal runaway.  This is a phenomenon where if one transistor pulls a bit more current than the other, it will get warmer and start conducting even more of the shared load ($V_{BE}$ will decrease as temperature increases).  The ballast resistors try to counteract that.  As one of the transistors conductors more current, the resistor tied to it will have a greater voltage drop (simple V=I*R).  This prevents thermal runaway from... well running away.
The last little bit, being R3, R4, and D1 are soley there because I only had 2N3904's on hand.  The max $V_{CE}$ of these transistors is spec'd at 40V.  If the amp is drawing little or almost no current, then Q1 and Q2 will seem like large resistances, dropping the majority of the phantom P\power across them.  Worst case being $V_{Phantom}-V_{Vcc}=V_{CE}$ or $48V - 4.5V = 43.5V$.  Normally, as the amplifier draws power, the "HOT" and "COLD" line DC voltages will be pulled lower because the 6.8K$\Omega$ resistors in the mixer drop some of the voltage.  So as the voltages on those lines start to get close to 40V, D1 starts to conduct, keeping them around 38V max.  Even though D1 is rated at 27V, any current it draws will drop a voltage over R3 and R4, which are large enough not to load down the audio, but small enough to let D1 operate.

CMRR, What CMRR?

So how did this design fare?  Not so awesome actually.  Unfortunately, even a few millivolts difference across Q1 and Q2 shows up in the mixer.  While playing music through the amplifier, you could actually hear the songs thru the mic input.  It was lower than the microphone level, but still audible, and distorted to boot.  R15 and R16 can only get so big before they start causing problems, and they couldn't get big enough.  So I ended up going to a simpler, proven design:

Many less parts, and in the end, it does work.  It draws a more or less constant current, either thru the zener diode or the load.  Even so,  there is still enough juice for a powered microphone.  After much testing, there was no observable bleed from the amplifier into the mic signal.  Nor was there bleed of the mic into the amp, if R1 and R2 are equal, any audio cancels itself out at the Vcc node (which is the intent of phantom power).  Not nearly as interesting as my first circuit, but still the best solution.  We can wrap this project up in the next post.