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What Plate Dissipation is - something doesn't seem right

Discussion in 'Amp Tech Center' started by peteb, Jul 28, 2017.

  1. Old Tele man

    Old Tele man Tele-Afflicted

    May 10, 2017
    Tucson, AZ
    Huh!

    In the Cathodyne PI there is just ONE current path thru/within the tube.
    In the Long-Tail PI, there are TWO tubes, with differing current paths (when operating), sharing a COMMON "current-source" cathode resistance.

    The LTP is a "differential amplifier" circuit that works like a "teeter-totter" where as one side (tube) goes UP the other side (tube) goes down (and vice-versa), with their diverging PLATE currents basically staying constant (hence its name) thru the shared COMMON cathode resistance.
     
    Last edited: Aug 10, 2017

  2. peteb

    peteb Tele-Afflicted

    Apr 25, 2003
    Cascadia
    Thanks old tele man,

    I see what you mean, that's how they are different,


    But do you see what I mean, they are the same in that they both have feedback between the cathode and the grid?


    Do you have some chips to place?



    To be in the game, you have to ante up.
     

  3. Bendyha

    Bendyha Tele-Afflicted Ad Free Member

    Mar 26, 2014
    Northern Germany
    But then again, you use the word amplify....this has not really the theme here up to now ( has it? )....you want the output to be a larger voltage than the input ? This is not the case with PI's or Cathode followers, although a tube can easily amplify the current, so yes they amplify.

    Up to now it has been a case of whether the input can be, say, 10 to 20V if the bias is only 1V. and the tube can process the input.

    Make up your mind where you want your goal posts.
    If you knew what you were talking about, you might be able to talk about it better.
    That you constantly shift makes it harder to hit you, and facts don't effect you.
    You love to agree with what the last person says, but then you soon twist the comments to present a newer distortion of your misunderstanding.
     
    Last edited: Aug 10, 2017
    robrob likes this.

  4. Old Tele man

    Old Tele man Tele-Afflicted

    May 10, 2017
    Tucson, AZ
    BIAS is fundamentally the DC-voltage across the control-grid to cathode spacing that produces a desired plate current. It is a fundamental function of the tubes grid geometry and spacings, which affect the tubes gm, rp and mu values. It is specifically related to the tubes plate voltage (Vp) and its differential control-grid-to-plate current relationship (gm), which together determine the tube plate current. An example of this relationship is the "projected cutoff" equation:

    Vg(bias) ≈ -Vp/mu

    Also, be aware there are multiple types of control-grid winding characteristics, notably: normal (linear) and remote-cutoff (variable spaced). However, few people use remote-cutoff tubes today.
     
    peteb likes this.

  5. Bendyha

    Bendyha Tele-Afflicted Ad Free Member

    Mar 26, 2014
    Northern Germany
    So you normal = sharp-cutoff I assume.
     

  6. peteb

    peteb Tele-Afflicted

    Apr 25, 2003
    Cascadia

    Come on Bendyha, I take this seriously.






    Isn't this what it still is?


    What changed? I don't understand?



    And please don't talk to me slanderingly.



    Are your chips withdrawn?, or are they still on the table?
     

  7. Old Tele man

    Old Tele man Tele-Afflicted

    May 10, 2017
    Tucson, AZ
    Bendyha: Sharp-cutoff vs. remote-cutoff and variable-cutoff, yes.

    PeteB: Stick to only one PI (Cathodyne or LPT) until you understand--fully--how it operates, and only then begin asking/arguing about the other. At times we cannot discern WHICH PI you're talking about.
     

  8. Bendyha

    Bendyha Tele-Afflicted Ad Free Member

    Mar 26, 2014
    Northern Germany
    You specified "A one volt DC bias is large enough bias for a tube to amplify a signal as large as 10-20 VAC."
    not whether a tube can function correctly with a 1V bias and a 10 to 20V input signal.

    Different things.
     
    peteb likes this.

  9. Old Tele man

    Old Tele man Tele-Afflicted

    May 10, 2017
    Tucson, AZ
    to illustrate Bendyha's above statement: Amplification is a function of control-grid voltage--DC and/or AC--and the tubes transconductance (gm) value and the plate load (Ra) resistance (example with no Rk or fully bypassed Rk):

    A = Vg(AC)*gm/Ra

    The DC-voltage (BIAS) establishes the quiescent plate current resting value about which the amplified ( Vg(AC)*gm ) plate current swings.
     
    Last edited: Aug 10, 2017
    peteb likes this.

  10. jazzguitar

    jazzguitar Tele-Afflicted

    Mar 17, 2003
    Germany
    Just a minor point: it is called a differential amplifier because it has two inputs and amplifies the difference between both inputs (if you change the DC voltage of both inputs in parallel eg both 1 Volt to the positive then the output signal should be 0V).

    It is basically the same circuit as used as input circuit of all silicon IC "op amps".

    The second input of the circuit as used in phase inverters is the grid connected to ground via a cap which removes the DC part of the signal.

    That it has two outputs is optional and only used when operated as phase inverter.

    Yes, all circuits have various kinds of feedback, many of which can be ignored with audio frequencies. All tubes bias via cathode resistor unbypassed have this negative feedback here.
     
    peteb likes this.

  11. Bendyha

    Bendyha Tele-Afflicted Ad Free Member

    Mar 26, 2014
    Northern Germany
    ....and the cathodyne circuit is classified as a voltage amplifier nontheless.

    ...to which I should add that the measuring the output from the cathodynes plate and cathode with a bog standard multimeter will also yealed unbalanced readings when the outputs are balanced, again by the differing nature of the two outputs reaction to loading, or more their effect to the opposing phase during the measurement loading, which debalances the tube, but this can easily be deflected, or at least the disequilibration abated, by measuring both phases simultaneously with matched (preferably infinite impedance) meters.
     
    Last edited: Aug 10, 2017
    peteb likes this.

  12. Old Tele man

    Old Tele man Tele-Afflicted

    May 10, 2017
    Tucson, AZ
    Agreed, it is a differential amplifier and predecessor of IC-opamps, which originally used the two sides of a twin-triode. Burr-Brown and Fairchild made special noval-modules using very high gm and high mu tubes (12AX7 and specially made tubes) and were labelled "differential" amplifiers(*)...a name that carried over to later transistor- and then IC-opamps.

    (*) used a lot in servo systems, where one input is the driving (command) signal and the other is a feedback (sensed) signal, where "null" or zero output occurs only when the two different signals are equal.
     
    Last edited: Aug 11, 2017
    peteb likes this.

  13. peteb

    peteb Tele-Afflicted

    Apr 25, 2003
    Cascadia
    If this doesn't answer the original question for this thread about what plate dissipation is, I don't know what could.




    Page 18, RCA man #16, (1950):


    In a class A amplifier under no-signal conditions, the plate dissipation is equal to the power input, i.e., the product of the dc plate voltage Eo and the zero-signal dc plate current Io.










    I don't think that could be any closer, word for word, with what I wrote in post #1.
     

  14. Old Tele man

    Old Tele man Tele-Afflicted

    May 10, 2017
    Tucson, AZ
    True, at idle, under no-signal, ALL the power supply voltage (Vp) and power supply current (Ip) are being dissipated by the tube. This is the condition where:

    P(tube) = P(in) - P(out)
    P(tube) = P(in) - 0W(out)
    P(tube) = P(in)

    where: P(tube) = tube plate dissipation; P(in) = DC power; P(out) = AC power

    Thus, the tube is dissipating 100% of the plate voltage and current (a DC-power condition) and producing NO AC-power output.

    So, what's the concern?

     
    peteb likes this.

  15. peteb

    peteb Tele-Afflicted

    Apr 25, 2003
    Cascadia

    Thanks old tele man for the reply.


    I don't understand your last comment.
     

  16. peteb

    peteb Tele-Afflicted

    Apr 25, 2003
    Cascadia
    I found in the RCA what I think supports the 1 volt bias on the cathodyne phase inverter:


    On page 29, RCA 16, (1950):


    ...., in this case, the Grid is returned to the junction of the two cathode resistors so that the bias voltage is only the DC voltage drop across the added resistance.






    Is this what you guys are talking about?






    I think it's pretty clear that it says what you guys are saying. The lower, bigger cathode resistor, some how sets a new ground level, that the bias goes up from there across the little 1k (upper) bias resistor.


    I will agree with you that the written evidence supports your side of the argument, the bias on the cathodyne phase inverter could be small like this.




    I still don't see how a 1 volt bias (small for even preamp tube standards) could be big enough for amplifying a much bigger signal, a signal the size of what the power tubes see, I think a larger bias is in order.

    The new thread, bias to grid relationship, will explore that, and in it the RCA has the signal smaller than the bias in AB1 and the signal exceeding the bias in AB2. I was going to post that when I ran across the Catho stuff above
     

  17. Old Tele man

    Old Tele man Tele-Afflicted

    May 10, 2017
    Tucson, AZ
    [
    QUOTE="peteb, post: 7811781, member: 1392"]I still don't see how a 1 volt bias (small for even preamp tube standards) could be big enough for amplifying a much bigger signal, a signal the size of what the power tubes see, I think a larger bias is in order. [/QUOTE]

    You're STILL confusing BIAS voltage with amplification gain:

    BIAS voltage is the DC-voltage that establishes the quiescent (idle) plate current; there is NO amplification taking place (yet).

    It is the AC-signal, Vg(ac), multiplied times the tubes transconductance (gm) which causes the plate current to CHANGE, Ip(ac), and produce a LARGER output voltage change, Vp(ac), across the plate load resistance, Ra. THAT is when/where/how the AMPLIFICATION takes place -- a BIG output change from a little input change:

    Ip(ac) = Vg(ac)*gm
    Vp(ac) = Ip(ac)*Ra


    where amplification (A) gain is Output-voltage change, Vp(ac), divided by Input-voltage change, Vg(ac):

    A = Vp(ac)/Vg(ac)

    Thus, AMPLIFICATION occurs within the output/input AC-voltage relationship, not in the DC-voltage (BIAS) used to establish the quiescent (idle) point.
     
    Last edited: Aug 11, 2017

  18. peteb

    peteb Tele-Afflicted

    Apr 25, 2003
    Cascadia
    No one has ever said the idle plate dissipation, as it gets measurred, is equal to the input power of the tube.


    I said it in post 1, and not one person agreed with me.
     

  19. Old Tele man

    Old Tele man Tele-Afflicted

    May 10, 2017
    Tucson, AZ
    Uh, do you recall reading post #83 that I made earlier--especially the last sentence?:
     
    Last edited: Aug 11, 2017
    peteb likes this.

  20. peteb

    peteb Tele-Afflicted

    Apr 25, 2003
    Cascadia
    That's a good post, I did go back and read it.


    Plate dissipation can't be ALL of those things, that is my point,


    I think the The RCA manual is to blame for the wide spread confusion about what plate dissipation is.



    IN the RCA manual:

    INterpretation of tube data, p. 58:

    Plate dissipation: power dissipated by heat, difference between input power and output power.


    But then as I wrote from RCA in post 193, they say the plate dissipation is equal to the input power.



    Plate dissipation can't be both the input and the difference between input and output at the same time.


    It's still OK, with that disambiguity, but to understand plate dissipation better, ask yourself,







    which are you measuring when you measure the idle plate dissipation in your amp?



    The wasted heat, or the input power. The answer is obvious to me.
     

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