DC Coupled Soundcard DAC Mod

I want to generate a signal in my computer that has a DC offset and output them to the real world.

Unfortunately, the DC offset does not pass through my soundcard.

So I found this "solution" online.

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What the article does not state is that the large chip on the circuit board gets very hot trying to pump out DC instead of the bipolar signal it was desgned to produce.

The author suggests a 100R in paralle with the output cap. So I then tried upping this to 10K which provided less but still acceptable shift, but still caused heating.

I then tried a 1000Uf cap in parallel, which again caused heating.

However, these are cheap and handy ltittle devices.

My question is, apart from clamping a large heatsink on the overheating chip, and accepting as little DC shift as possible, is there any way I could safely use this particular USB soundcard to output a DC offset signal?

BTW my application requires me to use the computer to produce the signal, not a function generator, etc.

Ken Morrow

Reply to
Ken Morrow
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That chip outputs a unipolar signal, always positive.

Maybe it always runs hot.

I can't see why the mod would make the chip get hot.

What is your load?

Reply to
John Larkin

OK. It must be the DC offset that is causing the problem then. IOW the chip is being asked to output more current than it is designed for.

No it only happens when I make these mods to the board. The more the DC offset, the hotter it runs.

See my guess above.

Just a scope probe at this point.

Ken Morrow

Reply to
Ken Morrow

If you're driving a scope probe, adding 100 ohms, or 10K, across an output coupling cap does nothing to the chip.

Makes no sense to me.

Reply to
John Larkin

When you say "hot", how hot? Does the chip burn your finger if you hold it on the chip? If so, how long does it take? If you can keep your finger on the chip for a couple of seconds, it is likely not too hot for the chip.

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Rick
Reply to
rickman

The chip outputs the same voltage as it ever did it is just that the poor thing is now seeing whatever external load you are offering it.

Either that or you have splashed solder across a land somewhere and it is now overloaded.

I expect the chip would also run hot (although about half as hot) if you forced it to output a maximum amplitude square wave into the same load. You need a buffer or unity gain voltage follower to protect the chip from abusive external loads.

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Regards, 
Martin Brown
Reply to
Martin Brown

One way is to use a second channel output (or other half of stereo) to generate a PWM signal which when averaged provides your variable offset.

See for example:

piglet

Reply to
piglet

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?

George H.

Reply to
George Herold

You are not supposed to pump out large currents from such a chip. When you have a low resistance DC load, use a buffer amplifier.

Reply to
Rob

All you need is a depletion mode FET, a current source for the source of sa id 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 d epletion mode FET is the only real way to do it. And you need the emitter f ollower on the tail of it because that is the nature of FETs.

Reply to
jurb6006

Incidentally, where I figured this out was when someone wanted to INPUT to

many other options.

Reply to
jurb6006

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

Reply to
Ken Morrow

But you said you were only "shorting" the output with a 1M or so scope probe and it gets hot.

Reply to
John S

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.

Reply to
George Herold

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.

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.

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Regards, 
Martin Brown
Reply to
Martin Brown

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.

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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

Reply to
Ken Morrow

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?

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Rick
Reply to
rickman

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

Reply to
Ken Morrow

What's the scope grounded to and what's the PC grounded to?

Reply to
Ralph Barone

The DC offset is already happily flowing down R5 & R6 which are across the chips outputs to ground. It would double the current and quadruple the power dissipation at a given output voltage but that is all.

This chip can certainly drive a DC resistive load of 3k to ground - it is already doing so. I am a bit amazed that it will drive a pair of 8R headphones when AC coupled. It might be a better choice for you.

The resistor on yours looked like 47k. That cannot be causing excess power dissipation sufficient to get the chip warm.

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Regards, 
Martin Brown
Reply to
Martin Brown

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