[sdiy] headphone driver for synths

Brian Willoughby brianw at audiobanshee.com
Sat Aug 28 22:13:09 CEST 2021


It looks to me like the transistors would carry the bulk of the current required by the load. The schematic doesn't show a transistor part number, so the beta is unknown, but assuming the typical beta of 100, the op-amp would be carrying only about 1% of the total headphone current. That seems safe.

Seems like the driver coils of the headphones would overheat before the op-amp.

Brian


On Aug 28, 2021, at 10:05, Michael E Caloroso <mec.forumreader at gmail.com> wrote:
>> 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
> fuse.
> 
> And with today's shrinking design rules for substrate, they can't
> dissipate much power.
> 
> MC
> 
> 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.
>>> 
>>> 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




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