I took the power supply section from Figure 7 of this app note, also copied here, as part of a phantom-powered project, but I can't figure out what the author was thinking for R3 and R14:


simulate this circuit – Schematic created using CircuitLab

I've studied it a while and simulated it in my head, and I think I understand how it works:

  • R1,R2 provide the base current for Q1,Q2 to operate. Other than that job, they're supposed to stay out of the way to avoid loading the AC signal.
  • R3 completes a voltage divider with R1||R2 to set the base voltage around 10.7V or so, which is stabilized by C1.
  • Q1,Q2 are used as emitter-followers, providing a stable 10V or so, which is stabilized/decoupled/bypassed/[your term here] by C2,C5 and used as the supply for the rest of the circuit.
  • As seen from the SIGnal lines, Q1,Q2 are high-impedance current sinks and therefore don't affect the differential-mode signal very much while drawing common-mode power from them.

What I can't figure out is the author's comment of:

R14 (marked **) is described as 'S.O.T', or select on test. This resistor needs to be chosen so that the DC is about 10V with a normal phantom supply of 48V. A zener may be used instead, which will give a little more voltage range - depending on the opamp used.

I thought the better place for a zener would be in place of R3, not R14. Then R14 could be removed entirely. Is this also a high-impedance current source into the ~10V point, so as to behave like a classic zener regulator as the author suggested?

  • \$\begingroup\$ I agree, R3 is where I would put a zener not R14. \$\endgroup\$
    – Andy aka
    Aug 18 '16 at 9:27

Your analysis is good so far.

Understand that the phantom power source is effectively 48V with a source impedance of 3.4 kilohms, so the voltage on Q1,Q2 collectors is strongly dependent on the load current.

This circuit will work as intended provided the load current is somewhere around 10mA (dependent on the gains of the transistors, since the base currents are of the same order as current in R3).

Coincidentally, 10mA is quite a good current for a low noise JFET (2N5459 for example) somewhere off on the left, to operate at its lowest noise figure - though I have my doubts about the TL072's noise contribution.

You and Andy are correct that a Zener is an alternate way to provide voltage stabilisation, though when you come to measure the resulting circuit's performance, you may find (as I did in similar circumstances) that its dual role as a noise generator is difficult to overlook. Note that zeners around the 10-12V range (and higher) are technically avalanche diodes, and quite noisy

IMO this circuit is quite cunning in overcoming that problem and providing "just good enough" voltage regulation without that unnecessary issue.

  • 1
    \$\begingroup\$ Hmm, I didn't know about zener noise, but it makes sense now why a resistor would be preferable given a (mostly) constant load. I've even seen intentional noise generators based on the base-emitter junction of a BJT (collector not connected), but never made the logical connection to zeners. (or diodes in general, according to a google search) \$\endgroup\$
    – AaronD
    Aug 18 '16 at 20:00
  • \$\begingroup\$ Coming back to this question, and realizing that I'd have to re-spin the board now to do this beyond a prototype: Would two low-voltage zeners in series be quieter than a single high-voltage zener? \$\endgroup\$
    – AaronD
    Sep 1 '16 at 0:30
  • \$\begingroup\$ Might be better to use a zener at fairly low current (probably not lower than 1 mA), followed by an R-C filter (big C, time constant around 1 second for audio) to the base of an emitter follower (which is also decoupled). That's roughly what I did. \$\endgroup\$ Sep 1 '16 at 8:37
  • \$\begingroup\$ Now there's an idea! I think I'll have to try that. This project does have a pretty big budget for startup time, so a long time constant for noise rejection should be okay. It's going to be a little while before I get to it though - other projects keep getting in the way. \$\endgroup\$
    – AaronD
    Sep 1 '16 at 18:35

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