Diode Clipping

Discussion in 'Shock Brother's DIY Amps' started by KaldtFjell92, Dec 4, 2019.

  1. KaldtFjell92

    KaldtFjell92 TDPRI Member

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    Alright! Experimenting with LED's-to-ground off a pot wired as a variable resister. Firstly, only ONE LED seems to light, and it remains lit at any level of any control on this amp, although particularly brighter with the 500k pot it is sent from turned LOW. The sound seems good but I don't notice much difference across the pot range really. Would a capacitor where the LED's tie together to ground help, and if so, what value?

    Other ideas I am considering are to place a capacitor in parallel with the diodes as I have seen in many schematics, or to use FOUR LED's with a rectifier diode as a bridge the way we see in JCM900 schematics...

    Please observe the schematic and video below, and thanks for any help in advance!

    Frank.
     

    Attached Files:

  2. DADGAD

    DADGAD Friend of Leo's

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    One LED is on all the time because you have DC across them both. You need a blocking capacitor in seres with the LED diodes. Anything will do, depending on the sound you want. Start with a 10mfd electrolytic.
     
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  3. KaldtFjell92

    KaldtFjell92 TDPRI Member

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    Hey DADGAD. I put in a 25uf electrolytic where the diodes connect to ground. I tested the LED's with a multimeter and either will light depending on the polarity of my multimeter leads. While on, one LED still remains lit, and when I turn the amp off, the other lights and dims. Overall, nothing is happening with this circuit. When I crank the "Drive" knob (500k pot) that runs signal to the LED's, the sound cuts out. the tone also gets thin right before this point as well.

    I'm wondering if running them off a cathode follower is part of my issue but I can't yet see why. I'm also considering running the diodes plate-to-grid with a 2m pot, but I'm not there yet.
     
  4. Ten Over

    Ten Over Tele-Meister

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    The red LED breaks over around 1.6Vp while the cathode follower is probably swinging 100Vp. You would need to string like 25 LED's in series in both directions to put the clipping point in the range of the signal coming off the cathode follower. Or use Zener's.

    Diode clip KaldtFjell92.png
    I haven't tried this, but this is the ballpark for the values. An on-off switch is better because the 50K pot is where most of the adjustment happens but it would take a much larger pot to get rid of the clip altogether.
     
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  5. KaldtFjell92

    KaldtFjell92 TDPRI Member

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    Awesome schematic and info, Ten Over! I injected the clipping circuit all over the preamp before landing back in that same spot after the cathode follower, and did change the pot to 100k and added a 1k5 resister from the diodes to ground at one point because, as you said, things happen very quickly with the diode clipping knob turned up.

    Side note; can you reason to me the voltage divider you added? I'm interested in that, and also the Zenners you added.

    Demo of amp currently

     
  6. Ten Over

    Ten Over Tele-Meister

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    If you mean the 15K/4M7 combination, that is such a lop-sided voltage divider that it can be ignored. The 4M7 resistor is there to prevent the voltage from rising up to a couple hundred volts when the switch is open, hopefully eliminating a pop when the switch is initially closed.

    The 15K resistor flattens out the positive side of the signal as it gets bigger by forming a voltage divider with the zener circuit. The cathode follower can drive a pretty heavy load on the positive side with just some rounding. The voltage divider flattens the signal when the zener breaks over. Of course both sides of the signal will be hard-clipped by the zeners when the pot is completely shorted out.

    Before the pot is completely shorted out, the negative side of the signal hard-clips when the cathode follower bottoms out. As the resistance of the pot becomes less, the negative side of the signal clips sooner.
     
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