JCM800 2204 Cameron HG Jose build thread

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FourT6and2 said:
I think you might be confused about asymmetrical vs symmetrical clipping. If you have two zeners arranged for diode clipping and they are both 20v, that's symmetrical. If one is 20v and the other is 16v, that's asymmetrical. If you have two 2N5401 or 2N5400 transistors, that's symmetrical. Two MPSA06 are still symmetrical. Mix and match two different transistors and wire them up to clip like zeners, and that's asymmetrical.

The symmetry or asymmetry of a voltage clamping circuit is determined by the positive or negative clipping thresholds. If you clip the signal at +15 volts and -15 volts, that's symmetrical. If you clip the signal at +15 volts and -5 volts, that's asymmetrical.

It has nothing to do with the polarity of the diode or transistor.

Make sense?
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My basic understanding with zeners is the striped end is negative side of the zener. If you orient the negative side facing outward from each other that is assymmetrical clipping. If you orient then positive sides out then that would be symmetrical clipping.

I'm always open to hear another viewpoint.:yes:
 

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SpiderWars said:
I see them now, the older pics were straight to pin 5 so that’s why I asked.
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I'm using my old camera so the pics aren't as crisp as they should be, the damn thing sometimes doesn't auto focus the best. I really need another digital camera, I'm just used to it and it still works............:LOL::rolleyes:
 
harddriver said:
My basic understanding with zeners is the striped end is negative side of the zener. If you orient the negative side facing outward from each other that is assymmetrical clipping. If you orient then positive sides out then that would be symmetrical clipping.

I'm always open to hear another viewpoint.:yes:
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That would still be incorrect. @FourT6and2 is correct and explained it well - it’s not a viewpoint it’s just factual.

What you’re describing above is still considered symmetrical clipping if the Zener diode clamp voltages are identical. Their orientation for AC signals doesn’t matter - it’s “symmetrical” regardless which node has the highest voltage potential hence the word symmetrical clipping.
 
harddriver said:
My basic understanding with zeners is the striped end is negative side of the zener. If you orient the negative side facing outward from each other that is assymmetrical clipping. If you orient then positive sides out then that would be symmetrical clipping.

I'm always open to hear another viewpoint.:yes:
Click to expand...

No, that is incorrect. The polarity of the diodes has nothing to do with asymmetrical or symmetrical clipping. It's the clamping voltage that determines symmetry. Think of a sine wave. +20v / -20v = symmetrical. And +20v / - 15v = asymmetrical. How you physically orient the zeners (anode to anode or cathode to cathode) has nothing to do with it.

Diode clipping is a clamping circuit.
You are literally cutting (clipping) the top and bottom off the audio signal.
This manifests through the speaker as distortion.

The voltage rating of the zeners determines how much of an audio signal gets clamped/cut/clipped.

For example, a 20v zener will clip/cut/clamp any part of a signal that is greater than 20v. A 12v zener will clip/cut/clamp any part of a signal that is greater than 12v. This means MORE of the signal gets cut/clipped with a lower-value zener. This equals more of a volume drop, more clipping, and more distortion.

If you cut both the top (+) and bottom (-) of your original AC audio signal equally, using equal value zeners, that gives you a symmetrical signal. If you cut the top of the signal more than the bottom of your signal, using unequal value zeners, that vies you an asymmetrical signal.
 
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From my testing, a pair of MPSA06 have the same volume/compression/loudness as a pair of 12v zeners (or was it 6v?).
However they sound different (better to me), so I would deduce they cant be clipping exactly the same as diodes do, maybe something to do with their (non)linearity around the conduction point
 
Talking about the orientation of Zeners or transistors for clipping, How do you take multiple pairs to increase clipping. I know it can be done, but not sure how. For example, let’s say I have some transistors that I like, but they are 20v, how can I use multiple 20v transistors to get clipping down to 10v?
 
psychodave said:
Talking about the orientation of Zeners or transistors for clipping, How do you take multiple pairs to increase clipping. I know it can be done, but not sure how. For example, let’s say I have some transistors that I like, but they are 20v, how can I use multiple 20v transistors to get clipping down to 10v?
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You can go the other way - two 20V oriented in the same way to get 40V, but it’s not possible to go the other direction and take two 20V and make 10V with orientation alone. By design the 20V aren’t going to clamp until 20V is reached.
 
WarHed said:
From my testing, a pair of MPSA06 have the same volume/compression/loudness as a pair of 12v zeners (or was it 6v?).
However they sound different (better to me), so I would deduce they cant be clipping exactly the same as diodes do, maybe something to do with their (non)linearity around the conduction point
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They’re more like 9-10v when comparing to a Zener. I’ve tested 12v, MPSA06 and 8v on a rotary switch. The volume drop is fairly linear between them , with the transistor being a bit closer to the 8v Zener than the 12v. MPSA06 is quieter than 12v and louder than the 8v.

I’ve also tested this on my scope about a million times and it’s the same result.
 
glpg80 said:
You can go the other way - two 20V oriented in the same way to get 40V, but it’s not possible to go the other direction and take two 20V and make 10V with orientation alone. By design the 20V aren’t going to clamp until 20V is reached.
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Thx. I knew something could be done, just was hoping for more clipping. 🤣

How would you wire up two pairs of 10v zeners/ transistors to get 20v?
 
psychodave said:
Thx. I knew something could be done, just was hoping for more clipping. 🤣

How would you wire up two pairs of 10v zeners/ transistors to get 20v?
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Put them in series oriented the same way. So two in series cathode facing up and two in series cathode facing down. Like a regular Jose style but two up top and two on bottom.
 
FourT6and2 said:
No, that is incorrect. The polarity of the diodes has nothing to do with asymmetrical or symmetrical clipping. It's the clamping voltage that determines symmetry. Think of a sine wave. +20v / -20v = symmetrical. And +20v / - 15v = asymmetrical. How you physically orient the zeners (anode to anode or cathode to cathode) has nothing to do with it.

Diode clipping is a clamping circuit.
You are literally cutting (clipping) the top and bottom off the audio signal.
This manifests through the speaker as distortion.

The voltage rating of the zeners determines how much of an audio signal gets clamped/cut/clipped.

For example, a 20v zener will clip/cut/clamp any part of a signal that is greater than 20v. A 12v zener will clip/cut/clamp any part of a signal that is greater than 12v. This means MORE of the signal gets cut/clipped with a lower-value zener. This equals more of a volume drop, more clipping, and more distortion.

If you cut both the top (+) and bottom (-) of your original AC audio signal equally, using equal value zeners, that gives you a symmetrical signal. If you cut the top of the signal more than the bottom of your signal, using unequal value zeners, that vies you an asymmetrical signal.
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Thank you for that clear and concise explanation correcting my flawed understanding of how asymmetrical and symmetrical clipping applies to zener diodes in an audio circuit. :yes::2thumbsup: Essentially the diode clamp is taking a full sine wave and cutting the top and bottom off the sine wave in to a square wave that I fully understand.

Since diodes in an AC circuit only allows current to flow in one direction only conducts one half of the AC cycle and that's how they rectify AC into DC.
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So how does orienting the different polarities of the anode and cathode of the diode affect the clipping characteristics in a circuit then? There must be some effect on the distortion characteristics when applied in a clipping circuit such as a Jose?

If I ran a 20V/20V with the negative/cathode stripes facing out this is symmetrical clipping and then I ran the same diode set with the negative/cathode stripes facing inward towards one another are the clipping characteristic effects the same besides the fact that they are both symmetrical clipping??????:unsure:

With the cathodes facing out is the clipping harder and less saturated/compressed?
 
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Jason does a dive into the zener clipping and the Friedman SAT switch both vids are worth the watch. Jason orients his 20V diodes with the anodes facing out and cathodes facing in, the reverse of how I have always seen the orientation in the original Jose amp schematic with the cathodes facing out.

 
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harddriver said:
So how does orienting the different polarities of the anode and cathode of the diode affect the clipping characteristics in a circuit then? There must be some effect on the distortion characteristics when applied in a clipping circuit such as a Jose?

If I ran a 20V/20V with the negative/cathode stripes facing out this is symmetrical clipping and then I ran the same diode set with the negative/cathode stripes facing inward towards one another are the clipping characteristic effects the same besides the fact that they are both symmetrical clipping??????:unsure:
Click to expand...

Try it and see!
But I haven't noticed any difference.
 
Alrighty.....all wiring is complete.:2thumbsup: I made a couple of component changes but the amp is ready for initial startup and voltage checks and inspection prior to installing tubes. That is going to have to wait until my schedule permits. I have taken my time with everything else so why rush now at the end.

I kept all the wiring stock Marshall colors for reference sake and authenticity and yes... I kept the factory stock untwisted heater wires but I figured with the elevated heaters the lack of twists should be a non-issue....we'll have to wait and see. I will probably have to make an adjustment to the voltage divider on the elevated heater circuit once I know exactly where the B+ from the screens will be running, I shot for about 50 DCV with the high end of 467 on the B+ with a 470K and 56K on the voltage divider, I didn't want things above 60 V for initial start up.

I am going to do a full wiring check prior to initial startup just to be 100% sure everything is where it is supposed to be before any voltage hits the circuit board. I'll update this thread when I get to putting some voltage to it and tubes in the amp.

I also know I put labels over the JCM800 silkscreening but absolutely want to know where my NFB will be set at....hence the labeling. I ended up going with 3000pf on the depth mod pot lugs and another 2000pf in parallel on the depth pot switch for a total of 5000pf.

I've had fun putting this thing together!
 

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