CMOS Inverter as a Buffer?

[thehallofshields]

U1F has to be a mistake right?

[Hemmel]

Maybe it's simply showing that you need to short pins 14 and 15?

[Bill Mountain]

Some claim it provides slight compression.  I haven't investigated this enough to know for sure.

[thehallofshields]

Some claim it provides slight compression.  I haven't investigated this enough to know for sure.

How can a shorted component have any effect? It baffles my mind.

Maybe it causes the circuit to consume more current so it 'Sags' a little more?

There should be a 'Summon Mark Hammer' button for threads on old-school/Anderton projects.

[GFR]

Perhaps it works like a diode to Vdd and a diode to ground (altough reverse biased)? Maybe just the extra capacitance to Vdd and ground?

[PRR]

> How can a shorted component have any effect? It baffles my mind.

It's not "shorted". It is connected input to output.

If it were a perfect Follower, it would be an infinite impedance.

If it were a perfect Inverter, it would be a zero impedance.

An unbuffered CMOS is a crappy inverter. It's not zero or infinity, and it isn't the same for all signal levels.

Same for the stage driving it.

What it DOES baffles my mind. But no doubt it does something.

Why ask? You have chips, try it.

[anotherjim]

I've looked at the circuit before but only to study the added FET buffer. Never noticed the "shorted" inverter.
I suppose it will add a power sag influence. The inverters have higher output impedance than op-amps so having them fight each other (U1f will oppose U1a) is an interesting tactic.

[snap]

More info in the accompanying thread: http://www.diystompboxes.com/smfforum/index.php?topic=59106.0

[Keppy]

^There's good info in that thread! I'll try to sum up, mostly as an exercise in getting it straight in my own head.


Connect A to Q on this schematic of an inverter, and what you get is MOSFET-as-diode to both power rails. It's not normal clipping, though, because since the "diodes" run to the rails instead of to a center point, they're both conducting all the time rather than just when a signal exceeds their forward voltage and causes conduction.

I don't know what this accomplishes as far as clipping or compression, since the two "diodes" oppose each other. It's loading the previous stage for sure, though, which seems to make it the opposite of a buffer. It's a load that varies with applied voltage in ways not described on the datasheet, just the kind of messed-up component use we love around here.

Perhaps it works like a diode to Vdd and a diode to ground (altough reverse biased)?

The body diodes are reverse biased, but the MOSFET-as-diode configuration is forward biased.

A couple of people have mentioned sag. I don't think this circuit can sag in the traditional tube amp sense. That type of sag is the result of a the amp drawing more current when signal is present, causing the power supply voltage to drop when a loud note is played. In this circuit, the inverters might draw more current on the positive or negative portion of the input cycle, but will draw less current on the opposite portion of the signal for no overall change to the supply voltage over the duration of the signal. The loaded power supply might cause some asymmetry, but it won't be sag as tube amp guys describe it.

[anotherjim]

I'm still thinking "reactive load".
Draw it differently. Connect pin2 directly to C9. Draw another connection to the side to inverter f pins 14 & 15. Same circuit, just looks different.
Now, look at the inverter f in isolation. What does connecting output to input do with an unbuffered complementary pair? It self biases to Vcc/2. Doesn't matter if the feedback is a wire or a 20Megohm resistor, that's what it'll do.

Because it's an inverter, attempting to change the voltage level on pins 14/15 will cause the mosfets to oppose it. While it's doing the auto bias trick, both mosfets are conducting somewhat, and might have effective channel resistance maybe 300-500 ohm. Even fully on, these mosfets might only get down to 80-100ohm. So it isn't a "hard" correction to signal voltage and enough signal will get through. Other inverter "tube sound" schemes deliberately attenuate the signal on the way out to get back closer to direct guitar level - this one probably doesn't have to.

Anyway ,that's how my own knowledge and logic fathoms it. Could be wrong, but I think the last page of the prior thread was tending the same way (and I've only just read it to avoid bias!).

[thehallofshields]

I've been experimenting with this. Connecting it as shown in the schematic just seems to kill my signal.

However, when I add some series resistance between the shorted Inverter and the Output of the second-stage (1k-50k is the useful range) it drops the Volume, and loads the signal in a way that removes some bassy saturation. It reminds me of switching from 'California' to 'British' on the GT2 if that is familiar to anyone.

[anotherjim]

"Connecting it as shown in the schematic just seems to kill my signal."....
In my theory, that's just what it ought to do, given it was a perfect inverter with zero output impedance -  it should attempt to clamp the signal at Vcc/2. They aren't perfect though so it shouldn't be a complete clamp-up. But, IIRC, some brands of 4000 series have mosfets with lower Ron than others - so could it be a brand difference?
Is the chip getting unusually hot or pulling the supply pin1 voltage really low?

[thehallofshields]

 Yes. This is what it was doing. The signal wasn't completely killed; about 95% attenuation.
Pin 1 was pretty low. 1k series Resistor was was dropping 9.4V (1Spot) to about 5.8V.
I kept checking for Heat. Didn't feel anything.

Adding a 1k-50k Resistor between the Shorted Inverter and the signal creates some nice cleanup. A little less saturated, a little less fizzy, a little more transparent.
But I can't say whether or not the CMOS Inverter creates a unique loading characteristic. Maybe a R-C Load could accomplish the same result more efficiently. - I just don't know.

Is there anything you guys would like me to test while I have this bread-boarded?