[sdiy] headphone driver for synths
Michael E Caloroso
mec.forumreader at gmail.com
Sat Aug 28 19:05:06 CEST 2021
> p.s. Is there a type or brand of op-amp where this circuit would not work as designed?
My bigger concern is the tiny wires between the substrate and the
pins. Too much current on the opamp output and they blow open like a
And with today's shrinking design rules for substrate, they can't
dissipate much power.
On 8/27/21, Brian Willoughby <brianw at audiobanshee.com> wrote:
> 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.
> 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
> 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.
> p.s. Is there a type or brand of op-amp where this circuit would not work as
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