[sdiy] headphone driver for synths
brianw at audiobanshee.com
Sun Aug 29 10:04:42 CEST 2021
On Aug 29, 2021, at 00:21, Roman Sowa wrote:
> It's the last transistor at opamp's output that drains the most current from supply pin, so in that sense the output IS controlling the current draw from supply rails in big part. In case of low Iq opamps it may be vast majority, or should I say basicaly all current flowing via supply pins is controlled by last transistor inside opamp's output stage.
That makes sense.
> I can't see how supply current depends on difference between in and out voltage. It depends only on the current going to the load. There is no difference between input and ouotput, as it is unity gain follower.
Instantaneous differences, not constant differences. If the input voltage rises, there will only be a voltage difference between input and output until enough current flows to raise the output voltage to match. Same for falling voltages.
> W dniu 2021-08-29 o 08:30, Brian Willoughby pisze:
>> Thanks for the explanation, and sorry for the question, but is the opamp *output* controlling everything? ... or is it actually the "amount of current draw on each supply pin" that is controlling each transistor?
>> Yes, the opamp will work (pump or sink current) until the output voltage matches the input voltage (due to unity negative feedback), but it still seems like the "control" is the supply current. Everything else seems to be a reaction to that.
>> Technically, the supply current depends on the difference between the input voltage and the output voltage, so the input voltage is controlling everything. The supply current is controlling the transistors. The load is controlled by a combination of the transistor pair and the op-amp feedback.
>> I still can't remember all the active modes of the bipolar transistor, so I'm probably only thinking about the typical mode.
>> On Aug 28, 2021, at 22:49, Roman Sowa <modular at go2.pl> wrote:
>>> 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 everything.
>>> 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 capacity.
>>> 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.
>>>> 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.
>>>> 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.
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