How to replace a power resistor with a Zener diode string?

Discussion in 'The Projects Forum' started by phoenixjazz, May 9, 2010.

  1. phoenixjazz

    Thread Starter New Member

    May 9, 2010
    3
    0
    Hi,
    I'm working on an old Tube amp and I have a problem I need help with.

    This amp uses a 20w 8k power resistor as a bleeder for the screens on the output tubes. The design has this resistor dissipating 18watts.

    I want to replace it as it is severely under rated, I also would like to eliminate the heat associated with this device

    I would like to use a zener string to provide the necessary voltage drop but I'm having trouble getting my mind around the problem.

    The screens are at +380Vdc
    The current is at .046amps
    The dissipation is 17.48 watts

    Thanks for any comments or direction on how to work this out!

    Chris
     
  2. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    I don't see your schematic posted.

    Without that, it'll be hard to make much progress.

    380V times 46mA is 17.48 Watts, no matter how you calculate it. A Zener clamp would dissipate the same amount of power.

    Maybe the thing to do is to use a Zener string and a HV transistor to regulate the voltage on the screen, but without a schematic, we're kind of dead in the water.
     
  3. Bychon

    Member

    Mar 12, 2010
    469
    41
    Here's another guess based on no drawing...the screen grid does not live at zero volts. If you are dropping 368 volts in the 8 k resistor and the screens are at 380, you must have started with 748 volts of B+. Is that true?

    and ps, changing to zeners will do absolutely nothing to reduce waste heat.
     
    Last edited: May 9, 2010
  4. phoenixjazz

    Thread Starter New Member

    May 9, 2010
    3
    0
    Sorry bout that! I was in such a hurry to post the question I forgot the schematic.
    Here it is.
    The power resistor is R30.
    The voltage is marked as 375 on the schematic but measures 380 in a meter.
     
  5. Bychon

    Member

    Mar 12, 2010
    469
    41
    Typing fast so the D*** website doesn't invalidate my login (again) b4 I get done.

    Electrons are always coming out of the cathode and leaving through the screen grid (G2). You have to provide that path. The original designer wasted a lot of power to make sure G2 is at 80% of B+. There must be a reason, and I think it's about distortion.

    You could leave out R30 and increase R29 until the voltage and current on G2 is right but it will change the distortion.

    Replacing R29 with a 100V zener is a terrible idea because a zener's job is to have low impedance. G2 could dump disasterous current through a zener to B+ in addition to changing the distortion characteristics.

    G2 runs with about 10 to 40 ma. 6 watts max at idle and 10 watts peak during operation. Use those numbers if you want to mess with R29. Put it back the way you found it if you want the distortion quality it was born with.
     
  6. retched

    AAC Fanatic!

    Dec 5, 2009
    5,201
    312
    Are you clicking the remember me box when you log in bychon?
    Sorry, off topic for a sec, back to you regularly scheduled programming.
     
  7. Bychon

    Member

    Mar 12, 2010
    469
    41
    no. my browser wipes everything when i close it, so i didn't think "remember me" would accomplish anything. thanks 4 the idea.
     
  8. retched

    AAC Fanatic!

    Dec 5, 2009
    5,201
    312
    I would give it a try. That way it wont forget you while your in the middle of a long reply.
     
  9. phoenixjazz

    Thread Starter New Member

    May 9, 2010
    3
    0
    Thanks Bychon,
    I'll look at this tonight when I'm home. The distortion is an issue so perhaps my solution will be to replace the single 20w resistor with multiple units to spread the load and a decent heat sink.

    Chris
     
  10. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Four 2k 10w resistors spaced out for air circulation would be an improvement. Non-inductive resistors would be good if you can find them.

    Using several wirewound resistors in a row will add inductance to the circuit, which will change the characteristics. The amount of inductance will be difficult to determine without having a sample.

    Found these at Mouser:
    http://www.mouser.com/ProductDetail...=sGAEpiMZZMvhlCB8CTbT5OrDT5vv518R36C1a/0H7dI=
    However, I can't determine from the datasheet whether they are wirewound or metal oxide; and if metal oxide if they are non-inductive.

    These are thick-film 2k 10W resistors:
    http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=TA810PW2K00JE-ND
    A good bit more expensive, but they would have nowhere near the inductance of a wirewound.
     
Loading...