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DC offset in fuzz box?

8/17/1996 6:01 PM
Eric M. Wood
DC offset in fuzz box?
Howdy, there.  
I was comparing Craig anderton's Tube Sound Fuzz in his Electronic Projects for Musicians with the MXR Hot Tubes pedal, which I understand is a production model of the same design, when I saw that the latter had it's first two CMOS distortion stages linked to ground at their inputs with 1m5 and 2m2 resistors, respectively. Is this a method of creating a DC offset for asymmetrical distortion? I am not familiar with CMOS ICs, so I do not know if there are pins on a 4049 chip for biasing or not.  
Thanks for any help.  
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8/17/1996 6:04 PM
Anders Westerberg

Yes, itīs probably a method of creating a DC offset for asymmetrical distortion, but there are no pins for biasing a 4049 since itīs a digital (yes!) IC.
8/17/1996 6:13 PM
Erik M. Wood

OK, I think my next quetion is: how does this create a DC offset? If the 4049s are capacitor coupled, causing a 0VDC at the input of each subsequent gain stage, and the ground point should be 0VDC, is it simply the voltage drop across the grounded resistors which causes a negative DC voltage at the input? It's a little confusing to think of negative voltages, but I'm getting it. Though, I still can't figure out whether there is a voltage generated by the input of the IC itself to compare the ground resistor voltages to. Does this matter at all? It does not seem that one can create an electrical potential from a difference of 0 volts. Somewhere, there must be a difference between electrical ground and the input side of the resitors. I know it isn't the previous gain stage because of the coupling caps. Somebody out there know what I'm missing?  
8/17/1996 6:15 PM
Steve Morrison

What you're missing is the feedback resistor, which not only sets the gain but also provides the bias voltage to the input.
4/23/1997 1:48 PM
Jack Orman

The CMOS inverter self-biases to 1/2 Vcc. Since in a no signal condition the both MOSFETs are in a high Z state (megaohms), the resistor across the bottom transistor causes the voltage divider to be unequal and changes the dc levels. (generalized)
4/23/1997 4:43 PM
R.G. Keen

The preceding replies get it right - the input to the inverter is as perfect an open circuit as you'll find. The inverter itself looks like a dc amplifier with a gain of 30 or more, so a single resistor from the output to input causes it to "balance" where neither the input or outpu are causing the amplifier to move around, which occurs at approximately 1/2Vcc where the "approximately" takes up the device differences between N and P channel devices in the chip. It's been within 100mv for all the ones I've seen.  
If you suck some current away from the input, by placing a resistor to some voltage, then the only way the inverter can balance again is to supply enough current through the feedback resistor to get the input back to a point that holds the resistor at that point (isn't iteration fun?). The input winds up being offset by the same amount the input is. If you then put an AC signal into it, the inverter bumps into one power supply before the other one, and you get asymetrical clipping.  
You can take this further. If you tied the asymetry resistor to the dead middle of the power supply, then it would have no current drain through the feedback resistor, and so no offset on the output. If you tied it to a pot strung across the power supply, you could dial in offsets above and below the 1/2Vcc point.  
Even more interesting, if you derive an envelope from the input signal, and connect the asymetry resistor to the envelope voltage, the asymetry offset is determined by the loudness of the signal. Set up properly, you can make it be asymetrical and sweet at low signal levels and tend toward symetrical square waves as you hit it harder. There is an audible shift from softer to harder sounding harmonics as you hit the strings harder.  
So many effects, so little time... I can't lay out circuit boards fast enough.
4/23/1997 5:23 PM

That was beautiful. Read the last one again if  
you don't agree.

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