I found this quote on another forum and am having trouble visualizing how to orient and place the diodes.
"You can make a unidirectional clipper with a zener diode in series with a regular diode, with the zener's cathode and regular diode's cathode connected together, and this assembly connected across the bias feed resistors, with the zener's anode connected to the coupling cap/output tube grid stopper connection and the regular diode's anode connected to the other end of the bias feed resistor that goes to the bias supply, so you only clip negative-going signals."
This is my understanding so far: the positive ends of two diodes are tied together, the negative end of the standard diode goes to the output side bias feed resistor. Is this correct? I'm also not quite sure where the Zener is connecting. A graphical illustration would be stellar as I am a little fuzzy on the terminology.
PI diode clipping question
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Re: PI diode clipping question
Pretty much anywhere in the signal path to the PI just just throw a pair of diodes to ground and that will do that. Zhhhhhhhh sounds better than a Pod.
At the PI? No idea.
At the PI? No idea.
Signatures have a 255 character limit that I could abuse, but I am not Cecil B. DeMille.
Re: PI diode clipping question
Reading comprehension man...
Quote: "and this assembly connected across the bias feed resistors"
The only thing that it is missing is the remark that for a push-pull amp you need TWO such diode strings. One per each side of the push-pull circuit, across each bias resistor.
By the way. This arrangement merely prevents grid conduction by clipping negative halfwave before its amplitude rises too high to result into grid clipping, bias shifts, crossover distortion and at worst cases blocking distortion when recovering from bias shift. The effect isn't the same kind of reduction in headroom / earlier overdrive as you would achieve by clipping the positive half wave (naturally at a threshold below voltage levels that are enough to overdrive the power tubes themselves).
Quote: "zener's cathode and regular diode's cathode connected together"This is my understanding so far:
the positive ends of two diodes are tied together
Quote: "the regular diode's anode connected to the other end of the bias feed resistor that goes to the bias supply"the negative end of the standard diode goes to the output side bias feed resistor.
Quote: "zener's anode connected to the coupling cap/output tube grid stopper connection"I'm also not quite sure where the Zener is connecting.
Quote: "and this assembly connected across the bias feed resistors"
Took me less than 1 minute to draw. Did you even try? You have ALL the info right there.A graphical illustration would be stellar
The only thing that it is missing is the remark that for a push-pull amp you need TWO such diode strings. One per each side of the push-pull circuit, across each bias resistor.
By the way. This arrangement merely prevents grid conduction by clipping negative halfwave before its amplitude rises too high to result into grid clipping, bias shifts, crossover distortion and at worst cases blocking distortion when recovering from bias shift. The effect isn't the same kind of reduction in headroom / earlier overdrive as you would achieve by clipping the positive half wave (naturally at a threshold below voltage levels that are enough to overdrive the power tubes themselves).
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Re: PI diode clipping question
http://forum.metroamp.com/viewtopic.php?p=199939
Scroll down a bit and there is a schematic. Took me less than one minute to find it ;^)
Scroll down a bit and there is a schematic. Took me less than one minute to find it ;^)
Re: PI diode clipping question
Paul Ruby Zener mod applied to fixed bias.