1. Win a Broadcaster or one of 3 Teles! The annual Supporting Member Giveaway is on. To enter Click Here. To see all the prizes and full details Click Here. To view the thread about the giveaway Click Here.

Help with Power Supply, Dropping Resistors

Discussion in 'Shock Brother's DIY Amps' started by Jeru, Oct 7, 2015.

  1. Jeru

    Jeru Tele-Holic

    Posts:
    723
    Joined:
    Nov 17, 2006
    Location:
    Chicago
    This is a corollary to my 'Bogen PA --> Bassman' thread. I've tried to figure this out, and it's time to ask my forum friends here.

    I'm not sure how to determine the current portion of the calculation for resistance value and wattage rating of needed dropping
    resistors along the power rail. I absolutely need to be able to do this to get the right plate voltages to the right places. Varying articles
    (on the Online) say to add up the current draw for tubes 'upstream' of the resistor. Common assumption seems to be 1.1 Ma for 12aX7 triodes.

    For power tubes, though, "current draw" for calculation purposes is unclear to me. If it's "plate current" on the data sheet then I'm confused.
    Looking at the GE 6L6 data sheet for two tubes in AB1, idle plate current is 88ma at 360v and 116ma at 450v, and
    maximum-signal plate current is 140ma at 360v and 210ma at 450v.

    Which current number to use..? And, if my amp's B+ is closer to 500v (which it might be), then what..?

    Even better, can you run through an example for me -- say, using an AA864 schematic (link). For purposes of calculating the values
    (resistance/needed wattage rating) of each of the 1K/1W and 4.7K/1W dropping resistors (if we didn't know them or, say, needed to adjust
    for high/low B+), what is the 'current' in the R=V/I equation?

    Thanks for any insight you'd care to offer.
     
    Last edited: Oct 7, 2015
  2. jtcnj

    jtcnj Tele-Holic

    Age:
    56
    Posts:
    596
    Joined:
    Feb 2, 2015
    Location:
    N.J. USA
    http://www.valvewizard.co.uk/psu.html

    Wizard says average:
    "The amplfifier circuit (or whatever happens to be draining current from the reservoir capacitor) can be represented as a single load resistance, Rl. This can be estimated simply as:
    Rl = Vdc / Iload
    Where:
    Vdc = DC supply voltage after rectification (i.e., the voltage across the reservoir capacitor).
    Iload = average DC current demanded by the amplifier circuit. "


    Not sure how you determine that average value.

    I would use peak to determine the power resistor wattages and add something like 20-30% for tolerance. W=I^2R.
     
  3. clintj

    clintj Friend of Leo's

    Posts:
    4,957
    Joined:
    Apr 4, 2015
    Location:
    Idaho
    For an accurate current number, you'd need to establish load lines and bias points for all tubes fed by a particular power supply node. For example, a 12AX7 can dissipate about 2ma per triode, but is usually roughly center biased hence the 1.1ma assumption.

    For the power tubes, assume a desired bias percentage, say 65% for class AB for example, and do the math as you would to figure bias at idle.
     
  4. robrob

    robrob Poster Extraordinaire Ad Free Member

    Posts:
    8,143
    Joined:
    Dec 29, 2012
    Location:
    United States
    This online calculator will get you in the ballpark for total amp B+ current. It's good for calculating dropping resistors before the first filter cap.

    I would use idle current and voltages because that's what everyone uses and what's on the schematics. When someone asks what's your B+3 voltage they mean idle.
     
  5. Jeru

    Jeru Tele-Holic

    Posts:
    723
    Joined:
    Nov 17, 2006
    Location:
    Chicago
    Thanks all. Rob, I've come across that calculator in my online search for answers on this. Very helpful -- I'm getting my head around this.

    Thanks again.
     
  6. tubeswell

    tubeswell Friend of Leo's

    Posts:
    2,429
    Joined:
    Jul 1, 2008
    Location:
    NZ
    In a guitar amp, its close enough to approximate the 'average value' in a signal circuit as being equal to the summed idle current draw. (Signal current goes 'up and down' around an idle point). Under big signal loads, the current through the dropping supply rises a bit because of the added screen current and plate current for the output stage increasing somewhat. But its not worth getting hung up about. Its geetar amps after all.

    Therefore calculate the DC load line for each stage, and then sum the idle tube current for all eh stages that hang off any given supply node. Total current through each dropping resistor supply any given node is the sum of all currents that come from, and after, that node - like the picture in Merlin's webpage suggests.
     
IMPORTANT: Treat everyone here with respect, no matter how difficult!
No sex, drug, political, religion or hate discussion permitted here.