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DC Coupled Soundcard DAC Mod

Started by Ken Morrow October 13, 2015
All you need is a depletion mode FET, a current source for the source of said FET and a transistor in common collector with the load resistor of your choice. What's the problem ? Need more DC ? I doubt it but if you do, the depletion mode FET is the only real way to do it. And you need the emitter follower on the tail of it because that is the nature of FETs.
Incidentally, where I figured this out was when someone wanted to INPUT to a soundcard and needed � Vcc as a setpoint. You want to output, you have many other options.
On Wed, 14 Oct 2015 06:01:28 -0700 (PDT), George Herold
<gherold@teachspin.com> wrote:


>As others said that's weird, with a 10k load how can it draw too much current. >But maybe there is some weird feedback path. >What about if you reduce the amplitude below 3.3 (3.0) V or so? >
As stated in my OP, what makes it run hot is when the input signal, generated in PC software, is deliberately given a positive DC offset. By design, this would normally be filtered out by the soundcard's output cap, but when you short it with a resistor, the chip is then forced to pass more DC which apparently causes it to heat irrespective of the load. The more the DC shift, the hotter it gets. As others have suggested, a buffer is one possible solution. The problem with that is that if I leave the soundcard output cap intact I lose the desired DC shift want. Looks like a heatsink might be the most practical solution. or perhaps to decrease the generated signal's amplitude and restore it after the soundcard. Ken Morrow
On 10/15/2015 4:39 AM, Ken Morrow wrote:
> On Wed, 14 Oct 2015 06:01:28 -0700 (PDT), George Herold > <gherold@teachspin.com> wrote: > > >> As others said that's weird, with a 10k load how can it draw too much current. >> But maybe there is some weird feedback path. >> What about if you reduce the amplitude below 3.3 (3.0) V or so? >> > > As stated in my OP, what makes it run hot is when the input signal, > generated in PC software, is deliberately given a positive DC offset. > > By design, this would normally be filtered out by the soundcard's > output cap, but when you short it with a resistor, the chip is then > forced to pass more DC which apparently causes it to heat irrespective > of the load. The more the DC shift, the hotter it gets.
But you said you were only "shorting" the output with a 1M or so scope probe and it gets hot.
On Thursday, October 15, 2015 at 5:40:12 AM UTC-4, Ken Morrow wrote:
> On Wed, 14 Oct 2015 06:01:28 -0700 (PDT), George Herold > <gherold@teachspin.com> wrote: > > > >As others said that's weird, with a 10k load how can it draw too much current. > >But maybe there is some weird feedback path. > >What about if you reduce the amplitude below 3.3 (3.0) V or so? > > > > As stated in my OP, what makes it run hot is when the input signal, > generated in PC software, is deliberately given a positive DC offset. > > By design, this would normally be filtered out by the soundcard's > output cap, but when you short it with a resistor, the chip is then > forced to pass more DC which apparently causes it to heat irrespective > of the load. The more the DC shift, the hotter it gets.
OK, but that can not be from current in the output circuit that you have drawn. It's less than mA's of current... that can't generate much heat. So more current is flow somewhere else... a buffer on the output, (where no extra current is flowing) is not going to help. (Unless there is something you are not telling us.) How about a different sound card? George H.
> > As others have suggested, a buffer is one possible solution. The > problem with that is that if I leave the soundcard output cap intact I > lose the desired DC shift want. > > Looks like a heatsink might be the most practical solution. or perhaps > to decrease the generated signal's amplitude and restore it after the > soundcard. > > Ken Morrow
On 15/10/2015 10:39, Ken Morrow wrote:
> On Wed, 14 Oct 2015 06:01:28 -0700 (PDT), George Herold > <gherold@teachspin.com> wrote: > > >> As others said that's weird, with a 10k load how can it draw too much current. >> But maybe there is some weird feedback path. >> What about if you reduce the amplitude below 3.3 (3.0) V or so? >> > > As stated in my OP, what makes it run hot is when the input signal, > generated in PC software, is deliberately given a positive DC offset. > > By design, this would normally be filtered out by the soundcard's > output cap, but when you short it with a resistor, the chip is then > forced to pass more DC which apparently causes it to heat irrespective > of the load. The more the DC shift, the hotter it gets.
The chip does exactly the same whether the decoupling capacitor is shorted out by 100R or not. The difference is that its output drivers now see any external load directly and its dissipation will scale with the square of the voltage being output as V^2/R. I suspect the thing would also get warm if driving a maximum amplitude square wave into the same resistive load. The other possibility is that whatever you are connecting it to has a different idea about the potential of mains earth.
> > As others have suggested, a buffer is one possible solution. The > problem with that is that if I leave the soundcard output cap intact I > lose the desired DC shift want.
You put the unity gain buffer between the DC coupled output and the resistive load. The thing should not run hot if all you hang on the output is a scope probe. Something else is wrong.
> Looks like a heatsink might be the most practical solution. or perhaps > to decrease the generated signal's amplitude and restore it after the > soundcard. > > Ken Morrow >
-- Regards, Martin Brown
On Thu, 15 Oct 2015 14:38:58 +0100, Martin Brown
<|||newspam|||@nezumi.demon.co.uk> wrote:


>You put the unity gain buffer between the DC coupled output and the >resistive load. The thing should not run hot if all you hang on the >output is a scope probe. Something else is wrong. >
Yes, heating up just by changing the waveform with no significant load is a mystery. It appears to be something happening within the chip itself. The circuit diagram here appears to be of a similar device. http://www.electronics-diy.com/PCM2706_USB_Soundcard.php It would appear the if C9 was shorted, and a DC offset was present, excess current would flow into R8. Perhaps the solution is to remove that resistor. Ken Morrow
On 10/15/2015 4:52 PM, Ken Morrow wrote:
> On Thu, 15 Oct 2015 14:38:58 +0100, Martin Brown > <|||newspam|||@nezumi.demon.co.uk> wrote: > > >> You put the unity gain buffer between the DC coupled output and the >> resistive load. The thing should not run hot if all you hang on the >> output is a scope probe. Something else is wrong. >> > > Yes, heating up just by changing the waveform with no significant > load is a mystery. It appears to be something happening within the > chip itself. > > The circuit diagram here appears to be of a similar device. > > http://www.electronics-diy.com/PCM2706_USB_Soundcard.php > > It would appear the if C9 was shorted, and a DC offset was present, > excess current would flow into R8. > > Perhaps the solution is to remove that resistor.
The current required would be minimal, so that current can't be what is heating the chip... directly. The issue may be that the chip does not expect to see *any* DC load. Drawing DC current might be upsetting internal biasing leading to higher currents purely within the device. But I wonder what "heating up" means to the OP. The fact that he can feel it get warmer is not significant. Unless it is hot enough to affect the performance of the chip or its useful life, why bother? -- Rick
On Thu, 15 Oct 2015 17:52:25 -0400, rickman <gnuarm@gmail.com> wrote:

>> http://www.electronics-diy.com/PCM2706_USB_Soundcard.php >> >> It would appear the if C9 was shorted, and a DC offset was present, >> excess current would flow into R8. >> >> Perhaps the solution is to remove that resistor. > >The current required would be minimal, so that current can't be what is >heating the chip... directly. The issue may be that the chip does not >expect to see *any* DC load. Drawing DC current might be upsetting >internal biasing leading to higher currents purely within the device. >But I wonder what "heating up" means to the OP. The fact that he can >feel it get warmer is not significant. Unless it is hot enough to >affect the performance of the chip or its useful life, why bother? >
I take your point, but the fact than it was getting hot_TER with the DC shift and no load was a cause for concern. Not too hot to touch or smoke though. I'll add a heatsink for peace-of mind. Thanks to everyone who responded. Ken Morrow
rickman <gnuarm@gmail.com> wrote:
> On 10/15/2015 4:52 PM, Ken Morrow wrote: >> On Thu, 15 Oct 2015 14:38:58 +0100, Martin Brown >> <|||newspam|||@nezumi.demon.co.uk> wrote: >> >> >>> You put the unity gain buffer between the DC coupled output and the >>> resistive load. The thing should not run hot if all you hang on the >>> output is a scope probe. Something else is wrong. >>> >> >> Yes, heating up just by changing the waveform with no significant >> load is a mystery. It appears to be something happening within the >> chip itself. >> >> The circuit diagram here appears to be of a similar device. >> >> http://www.electronics-diy.com/PCM2706_USB_Soundcard.php >> >> It would appear the if C9 was shorted, and a DC offset was present, >> excess current would flow into R8. >> >> Perhaps the solution is to remove that resistor. > > The current required would be minimal, so that current can't be what is > heating the chip... directly. The issue may be that the chip does not > expect to see *any* DC load. Drawing DC current might be upsetting > internal biasing leading to higher currents purely within the device. > But I wonder what "heating up" means to the OP. The fact that he can > feel it get warmer is not significant. Unless it is hot enough to > affect the performance of the chip or its useful life, why bother? >
What's the scope grounded to and what's the PC grounded to?