Anderton Tube Sound Fuzz [inverters?]

Started by Leftrights, December 24, 2003, 09:48:41 PM

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Leftrights

Can anyone give me a quick run down on how the inverters in the tube sound fuzz distort the signal?  Does it have anything to do with the inverted signal cancelling itself out?  (just a guess probably a weak one).

The reason I ask is I am wondering how important the type of IC is.  I have a bunch of nand gates.  Can I tie together the leads and use these as the inverters.  Or if I find some other than the ones specified can I sub in?

Any info would be great,

And Happy Holidays everyone.  I hope everyones significant others figured out how to order from Mouser.

gez

Quote from: LeftrightsThe reason I ask is I am wondering how important the type of IC is.  I have a bunch of nand gates.  Can I tie together the leads and use these as the inverters.  Or if I find some other than the ones specified can I sub in?

You CAN tie the inputs of NAND/NOR gates together to give you an inverter and bias them as a 'linear amplifier', but most chips are buffered and aren't suitable for distortion (gain is too high - will oscillate).  

Unbuffered NAND/NOR chips are really hard to find (well, over here they are), and will bias off-centre (will give you some asymettry).

As to how they work, there's a good explanation in "the art of electronics" if your library has a copy.  This question came up a few weeks/months ago.  Do a search (RG gave a good reply).  I'd offer to explain myself, but after all it is Xmas :D
"They always say there's nothing new under the sun.  I think that that's a big copout..."  Wayne Shorter

gez

OK, 'My Fair Lady' is on the box, so I have plenty of time now!

An inverter is just a p-channel MOSFET stacked on top of a n-channel MOSFET (many data sheets will have a schematic) joined at the drains and with their gates tied together.  When both gates are pulled to ground the n-channel is completely off and the p-channel is fully on so the output is within a smidgen of 9V (assuming a 9V supply is used).  When the gates are tied to +9V the opposite occurs and the output is within a smidgen of 0V.  

Tie the gates to the wiper of a trimpot whose outer lugs are connected across the rails, and at some point in the travel of the pot’s wiper both devices will be conducting.  A current will flow (‘class A’ current).  Both devices are matched in a chip so the point where this current will flow will be somewhere in the middle of the supply.  By connecting a resistor from output to input the device will self bias and will act as an inverting amplifier.  Distortion results from overdriving the amplifier until the signal is clipped.

Here’s a little more from the archives:

“CMOS distortions work like any saturating amplifier.

They can only swing a certain amount of their power supply range linearly. In the case of CMOS logic chips, the linear range is very small compared to the power supply voltage, but the onset of distortion is gradual. This makes for distortion that increases gradually over a wide range of signals, instead of suddenly like other technologies.

This gradual onset of distortion is linked to the way CMOS transistors work: they are best looked at as voltage variable resistors. When two of them, a P and an N channel are hooked in series and their gates connected like they are in CMOS inverters, the combination of one turning on and one turning off results in the output signal approaching the power supplies only gradually, not abruptly, in both directions.

In more techie terms, the transfer function of a simple CMOS inverter is linear over only a small amount of its range, and has a gradual curve to it that causes relatively gentle limiting as long as feedback does not cause a lot of linearization.

There are transfer function curves in the RCA CMOS databook and in National Semiconductor's data.

R.G.”
"They always say there's nothing new under the sun.  I think that that's a big copout..."  Wayne Shorter

donald stringer

Thats really interesting. I am going through my 4049 kick right now and have a question.If you take a p channel and n channel moss and wire them as you noted[ I have seen that which you are refrring to on the data sheets but wasnt aware that they were p and n. With the right component values ,in theory you would have a perfect low parts count miniature tube sound fuzz just by duplicating one inverter of the 4049...this could lead to a whole other series of revolutionary stompboxes. O k the coffe is wearing off , I am crashing now, need more coffee , so difficult, cant concentrate now, but must hang on...........
troublerat

gez

Quote from: donald stringer.If you take a p channel and n channel moss and wire them as you noted[ I have seen that which you are refrring to on the data sheets but wasnt aware that they were p and n. With the right component values ,in theory you would have a perfect low parts count miniature tube sound fuzz just by duplicating one inverter of the 4049...this could lead to a whole other series of revolutionary stompboxes

Inverter chips are designed so that they don't draw huge amounts of current within the specified supply range - not so with many discrete MOSFETS.  If you went down this route you'd need current limiting resistors.

Inverter chips are low gain when used as amps, and this is a lot of the reason they sound so good.  Discrete devices have a lot higher transconductance figures so there's no guarantee that clipping will be soft.  

The devices in a chip are matched in terms of threshold so they bias up at around half the supply - you'd have to match a p-channel and a n-channel to do this with discrete devices and that's extra work.  Alternatively use a trim pot as part of the bias network.

Would one stage be enough?  Probably not.  Tot up the extra parts count and extra work and it's easier to stick with inverter chips.
"They always say there's nothing new under the sun.  I think that that's a big copout..."  Wayne Shorter