[sdiy] headphone driver for synths

Roman Sowa modular at go2.pl
Sun Aug 29 07:49:06 CEST 2021

Yes, these are common emitter circuits and supply current is the key here.
On the contrary what you wrote, the opamp's ouptut is controlling 

When an opamp wants to drive something with its output, something low 
impedance like, say a speaker or motor, it sources some current to it 
(let's talk about positive swing for now). But that current must come 
from somewhere - the supply pins. So the current at positive supply pin 
is now sinking more current from the supply. That current is now flowing 
via base-emitter of PNP upper transistor, because it's so big that we 
have reached 0.6V on that resistor. So the PNP adds pretty thick 
collector current to the current that poor opamp is already sourcing. 
And by the magic of negative feedback the opamp is driven just as much 
to tell: yeah, the PNP gave us a helping hand, no need to drive our tiny 
guts more than that. And obviously negative swing works exactly the same.

As others pointed out, it was popular in 70's. I haven't tried it yet, 
that's why my question was, have any of you did. But it's so simple and 
effective, not to abuse word "clever" that it's simply beautiful. Which 
opamp would work best? Anyone with low Iq. So when there is no signal, 
or very small amplitude, low Iq will not cause enoough drop on the 
resistors and opamp is on its own. But when output load rises, 
transistors start to chip in. Well, even with an opamp with high Iq, 
it's only more critical to select proper resistor values to turn 
transistors at best usable level, say half of opamp's output current 
Such a thing is just asking for crossover distortion, but probably not 
at level that I could hear.


W dniu 2021-08-28 o 00:39, Brian Willoughby pisze:
> On Aug 25, 2021, at 00:42, Roman Sowa <modular at go2.pl> wrote:
>> Has anyone ever tried to explore this not-so-new idea of using opamp's supply currents to drive boosting transistors?
>> Schematics attached.
>> Roman
> Hi Roman,
> Are these common Emitter transistor circuits? I need to crack open my college textbooks this weekend...
> The circuit seems to depend on the op-amp supply current to boot-strap the biasing resistors.
> Basically, what it looks like to me is that the transistors run on ±12 V supply rails, but the output range is slightly less due to the Base-Emitter diode drop. I think this design avoids reverse-biasing the transistors by running the op-amp on the reduced voltage of ±11 V or ±10 V, whatever it happens to be.
> So long as current is flowing through the Resistors, the Base nodes of the circuit are each one diode drop away from the corresponding voltage rail, at least when the transistors are on. I don't know whether it's important, but the op-amp never sees the full ±12 V unless there's no current flowing anywhere.
> What's bizarre about this topology is that the "output" of the op-amp is not exactly controlling anything. But, due to the feedback, any current draw on the op-amp rails should quickly result in a multiple of that current flowing through the headphones until the op-amp output matches the input. The transistors supply the bulk of the current, compared to the op-amp, by a factor of the transistor beta.
> Brian
> p.s. Is there a type or brand of op-amp where this circuit would not work as designed?

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