[sdiy] headphone driver for synths

Mike Bryant mbryant at futurehorizons.com
Sat Aug 28 21:42:01 CEST 2021


This is why we have CSP (Chip Scale Packaging) - no wires and far better heat removal.


-----Original Message-----
From: Synth-diy [mailto:synth-diy-bounces at synth-diy.org] On Behalf Of Michael E Caloroso via Synth-diy
Sent: 28 August 2021 18:05
To: Brian Willoughby
Cc: synth-diy at synth-diy.org
Subject: Re: [sdiy] headphone driver for synths

> 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
>
> p.s. Is there a type or brand of op-amp where this circuit would not 
> work as designed?
>
>
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