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Re: Throwing Away Gain


 :
7/22/2005 8:33 PM
BK Re: Throwing Away Gain
So why not two identically biased stages around -0.5V or -1.0V in series. Take your sine wave around 1-2V p-p. Through one stage the top gets lopped off. Signal gets inverted. Passes through the next stage the other side gets lopped. (Assuming appropriate attenuation).  
 
Can this aproximate symmetrical clipping? Ray, any insight from your study?  
 
BK
 
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7/22/2005 9:13 PM
Ray Ivers

BK,  
 
This is pretty much what happens in most cascaded-SE-gain-stage preamps; if you look at stage #3's output while turning up the gain control, you'll see very symmetrical clipping behavior. The thing is... it's all cutoff clipping, which IMO is the harshest-sounding tube distortion. It's possible to minimize muddiness and the really atonal IM stuff through careful choice of R/C values and filter out the high-order harmonics with post-distortion low-pass filter techniques - you can end up with a pretty good heavy-distortion sound if you do it all right - but starting off with a distortion that doesn't sound at all harsh or muddy (and thus doesn't need 'repair') would be way cooler to my way of thinking, and would also be applicable to more musical styles.  
 
Ray
 
7/22/2005 9:01 PM
Ray Ivers

Shea,  
 
Hey, thanks for the kind words... I guess I am kinda passionate about that, come to think of it. And if it wasn't for that damned oscilloscope, I never would have known... :D  
 
I'd say that pretty much every guitar-amp guy with a 'scope has done it more than once - turned up the gain control while monitoring gain stage #2's output, and seen the waveform immediately flat-top while the negative swing just keeps going down, down, down, just as you said. This is the voltage-divider grid-circuit action, and it's definitely a crude form of compression - but since it's only on half the waveform and combined with the hard cutoff clipping, it doesn't sound particularly compressed to me at all, especially after the next gain stage has converted it to a hard cutoff clip.  
 
I really wish sometimes that tubes had some kind of soft-clipping mechanism that activated at 0Vg, but as you know, they don't - hence the classes of operation that end in -2 (A2, AB2, B2, etc.) which generally use steeper load lines that intersect with characteristic curves not often found in common trace families, the +1Vg, +2Vg, +5Vg, etc. curves. To get plate-output clipping on the negative-going waveform swing, you have to drive the grid positive until the plate current stops increasing; this is one of the very smoothest of clips IMO, but also one of the hardest to obtain as well. NOTE: By 'smooth' I don't necessarily mean 'mellow'; you can make both a killer blues amp and a fizz-free metal monster using this type of operation. I know you're familiar with all this already, Shea, I'm just throwing it out in case it might be helpful in general.  
 
I would like to have direct-coupled, but the 12AX7 grid apparently draws only about 40uA, in bursts - so I think most any good cap can supply that kind of current. With a low-impedance driver you can always oversize the cap, and then it won't even notice the 12AX7 grid conducting.  
 
El Passionato
 
7/23/2005 10:14 AM
BK
Ray,  
 
I'm a little confused now. I thought when the grid potential was at 0V relative to the cathode, the plate was passing maximum current, hence saturation. So, if you biased your stage hot, the forward swing of your signal would reach saturation, but the negative swing would have enough headroom not to bottom out (assuming 2V p-p) biased at -0.5V. The characteristic curves don't seem to reach cutoff until the grid potential goes below -3 or -4V. (considering 250-300V on a plate resistor of 100k).  
 
BK
 
7/23/2005 11:25 AM
Ray Ivers

BK,  
 
quote:
"I thought when the grid potential was at 0V relative to the cathode, the plate was passing maximum current, hence saturation."
 
 
If a 12A_7's control grid is pushed past 0 in the positive direction, the plate current will continue to increase. Since practically all driving stages found in these circuits are incapable of doing this, it's something that almost never happens - so your assumption was very understandable.  
 
There are some output tubes designed for Class B operation that have zero (or close to it) plate current at 0Vg (e.g., the original Svetlana SV572-160) - many power-grid-drive RF power tubes also have similar characteristics - both to simplify design and to supply built-in protection against bias-supply failure (no bias supply, no problem!) which would not be pretty in a megawatt-range AM transmitter. Enhancement-mode tubes? ;)  
 
quote:
"The characteristic curves don't seem to reach cutoff until the grid potential goes below -3 or -4V."
 
 
Despite the fact that most 12A_7 characteristic-curve families assume fixed-bias operation which is almost unheard of in guitar preamps, this is still pretty close; my circuit cuts off at about -3.4Vg.  
 
Ray
 
7/24/2005 9:54 PM
Shea
Question, Ray . . .
I guess I'm a little slow to catch on, but I haven't figured out how this circuit that you designed is clipping both halves of the waveform. Are you cascading 2 gain stages that each have this particular combination of plate load resistor, cathode resistor, and B+ voltage? Or is it one gain stage followed by a cathode follower? Or does just one of these gain stages clip symmetrically all by itself, as long as you have that low-impedance grid driver?  
 
Shea
 
7/25/2005 6:20 AM
Ray Ivers

Shea,  
 
Here's the clipping mechanism on each plate-output waveform half, as I see it:  
 
Positive swing: The grid is driven in a negative direction by the input waveform from the preceding stage, and at some point the tube cuts off. During this cutoff interval you're basically listening to your B+ line, and the tube could be removed from its socket without affecting tone.  
 
Negative swing: The grid is driven more and more positive, and at @ 0Vg begins to pull positive grid current through the grid circuit. The tube conducts harder and harder, and at a point determined by the cathode/plate resistances, tube type, and B+ voltage, the tube will cease to conduct more current with further increases in grid voltage. Unlike the nuance-free "shutoff" of cutoff clipping, this is a lot more like push-pull power amp clipping in that the 'clip zone' is wider and less rigidly defined. Combined with the cutoff clipping - which is still present on the other waveform half - you should end up with a hybrid of "preamp" and "power-amp" distortions, so to speak.  
 
quote:
"does just one of these gain stages clip symmetrically all by itself, as long as you have that low-impedance grid driver?"
 
 
This is pretty much it. Since you have to provide that low-Z drive which may entail redesigning or replacing the preceding stage, retrofitting this thing in an existing amp may well involve 2 stages - drive and overdrive. ;) I think for best results, a high-gain, low-impedance output 1st gain stage set up for best symmetry (tube/MOSFET SF combo, JFET, small-signal N-depletion-mode MOSFET, etc.) would be followed by the symmetrical-clip stage, which might then be MOSFET-buffered and overdrive the following symmetrical-clip stage, etc.  
 
Ray
 

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