[sdiy] 4049/4069 linear mode

[Scott Nordlund]

As far as I know, the 4049 can source/sink more current, so it's suitable for driving things like LEDs.

Forget looking at the schematics that use the 4049 or 4069, look at the datasheet (for the 4069UB, not the buffered 4069B, since it's got additional stages that ruin the fun).  An input voltage of Vdd will turn the N MOSFET on (Vgs = Vdd) and the P MOSFET off (Vgs = 0), thus giving a low output voltage.  Conversely, an input voltage near Vss will turn the N MOSFET off (Vgs = 0) and the P MOSFET on (Vgs = -Vdd), producing a high output.  So it works as a digital inverter.  An intermediate voltage will result in both MOSFETS being partially turned on and operating in a more or less linear region, though they weren't designed with this in mind.  This isn't always a good thing, as it can allow a rather high current to flow from Vdd to Vss (though these chips seem to be able to handle it).  If you've got it biased at 1/2 Vdd, a small signal superimposed on top of this will be amplified and inverted.  Thus it's also an amplifier, though not a very good one.  It uses a lot of power and isn't particularly linear, especially for large input amplitudes, but sometimes this is what you want.  The very high input impedance is an added bonus.

Negative feedback reduces gain just as it does in an op amp circuit.

----------------------------------------
> Date: Sun, 31 May 2009 03:51:53 +0200
> From: derek at umatic.nl
> To: synth-diy at dropmix.xs4all.nl
> Subject: [sdiy] 4049/4069 linear mode
>
> I've been trying to understand how the 4049 and 4069 inverters work in
> linear mode. Studying the Wasp Filter, Craig Anderton's 4049 distortion
> (http://www.tonepad.com/project.asp?id=65) and the schematics which
> Nicholas Collins included in his circuit-bending book, I'm still not so
> clear on why they work.
>
> First off, are these essentially the same chip (aside from pin outs), or
> does the 4049 give additional gain while the 4069 only reaches unity?
>
> Second, why does this happen? For gain less than unity, I can see that
> adding the inverted signal to the original signal via a feedback
> resistor could create attenuation, but how does this circuit produce
> positive gain??
>
> Thirdly, is the amplified or attenuated signal resulting from using the
> inverter in this way inverted or not?
>
> Sorry if these are super basic questions. I'm coming at this from an
> artist's rather than an engineer's background.
>
> Best!
> Derek
>
> --
> ::: derek holzer ::: http://blog.myspace.com/macumbista :::
> http://www.vimeo.com/macumbista :::
> ---Oblique Strategy # 5:
> "Abandon normal instructions"
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