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Zener Diodes to drop Heater Voltage ?

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  • #16
    Originally posted by Helmholtz View Post
    The winding only cares about total current drawn. The 5V6s might have somewhat higher heater current than the 6V6s though.

    Edit: Just checked 6V6: 0.45A, 5V6 ; 0.6A. So some increased stress on the PT/heater winding.
    Right. My point was any added device dropping volts will also be drawing current from the winding. In your example above with a 2.7 ohm resistor dropping 1.6V the resistor would seem to be drawing about .6A of current. If two tubes are used and your proposal of a resistor for each tube is employed that should mean an additional 1.2A demand on the winding. No? Add the .15A per tube for the 5V6 additional requirement and now we have an extra 1.5A of current demand on the filament winding. 1.5A extra may easily be enough to run the winding over spec. And then, even if the winding isn't beyond max current there's still the complication of more voltage sag from the winding itself at the higher current demand. Not that it's likely to be a big deal, but maybe get a couple of 2.2 ohm resistors just in case the circuit needs to be fine tuned.

    EDIT: Oh, wait... That 2.7 ohm resistance would have applied to the current through BOTH tubes, right? So using a resistor for each tube would still only total another .6A and the numbers get a lot better.
    "Take two placebos, works twice as well." Enzo

    "Now get off my lawn with your silicooties and boom-chucka speakers and computers masquerading as amplifiers" Justin Thomas

    "Being born on third base and thinking that you must have hit a triple is pure delusion!" Steve A

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    • #17
      My point was any added device dropping volts will also be drawing current from the winding. In your example above with a 2.7 ohm resistor dropping 1.6V the resistor would seem to be drawing about .6A of current. If two tubes are used and your proposal of a resistor for each tube is employed that should mean an additional 1.2A demand on the winding. No? Add the .15A per tube for the 5V6 additional requirement and now we have an extra 1.5A of current demand on the filament winding. 1.5A extra may easily be enough to run the winding over spec. And then, even if the winding isn't beyond max current there's still the complication of more voltage sag from the winding itself at the higher current demand. Not that it's likely to be a big deal, but maybe get a couple of 2.2 ohm resistors just in case the circuit needs to be fine tuned.
      No, series wired resistors don't draw additional current (and they don't waste more power than diodes if voltage drop is the same). The heater current difference between two 5V6s and two 6V6s is 0.3A. This is what the heater winding sees.
      There's only one current in a simple series circuit so each component gets the same current.

      Each tube needs an individual 2.7 Ohm resistor.
      Last edited by Helmholtz; 10-26-2019, 06:34 PM.
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      • #18
        Originally posted by Chuck H View Post
        Something not yet discussed would be stress on the winding. Wouldn't any form of reducing voltage in line basically be "loading down" the winding to reach the lower voltage? Is the winding up to it?
        I had thought about the .3A additional current as well, but I (hope) the transformer is up to it. Currently with the stock tubes it runs pretty cool even after an hour or more, only around 118 degrees Fahrenheit peak, but I realize that I am not measuring the 6.3vac winding specifically. The power transformer on this amp seems to be no smaller than the ones used in my Valco amps with Tremolo circuits in addition to the same other tubes, so the addition of .3A more heater current would be about the same... Provided it was the size that mattered. I looked up a Stancor transformer about the same size from the same period, and it's able to handle 3A for the 6.3vac draw. I have in the amp : ( 6j5 .3A + 12ax7 .3A + Two 5V6 = 1.2A = Total 6.3vac draw of 1.8A ) so it should be good, and if I was wrong and it was a smaller 2A version, that would pass as well.

        I think that is what you were referring to ?

        Thanks for keeping me honest Chuck, I will keep tabs on it for sure, and so far the power or output transformer don't seem to be stressed in the least on this "New" Zenith amp I am referring to. I wish I could find the exact specs on this old 1958 transformer, but so far no luck.
        Last edited by HaroldBrooks; 10-26-2019, 08:11 PM.
        " Things change, not always for the better. " - Leo_Gnardo

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        • #19
          Originally posted by Helmholtz View Post
          No, series wired resistors don't draw additional current (and they don't waste more power than diodes if voltage drop is the same). The heater current difference between two 5V6s and two 6V6s is 0.3A. This is what the heater winding sees.
          There's only one current in a simple series circuit so each component gets the same current.

          Each tube needs an individual 2.7 Ohm resistor.
          I'll have to take your word for it. My simple mind says that if your loading the voltage down and something is heating the resistor then current is being used. But I'll guess that it has something to do with the net figure between the 5V filament and the resistor in series that is the same to the winding as the original 6.3V filament. Other than any additional draw in the tube specs I mean.
          "Take two placebos, works twice as well." Enzo

          "Now get off my lawn with your silicooties and boom-chucka speakers and computers masquerading as amplifiers" Justin Thomas

          "Being born on third base and thinking that you must have hit a triple is pure delusion!" Steve A

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          • #20
            if your loading the voltage down
            The additional resistor doesn't load the heater winding's terminal voltage down. The master principle is the law of conservation of energy. As energy is power times time, power must be conserved as well.
            Let's see if it works. Assuming a simple heater circuit with just one 5V6, a dropper resistor of 2.7 Ohm and 6.3V heater winding voltage. The sum of resistor dissipation IČ*R and heater power I*4.7V should equal the power delivered by the heater winding. The latter is 6.3V*0.6A= 3.78W. Resistor dissipation is 0.97W and heater power is 2.8W. Does it work?
            Last edited by Helmholtz; 10-26-2019, 09:07 PM.
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            • #21
              Well I've never been formula strong, but that's easy enough to see. Laymans terminology, the voltage is split (unequally relative to resistance) between the resistor and the filament. The current remains equal to it's share of the voltage. Got it.
              "Take two placebos, works twice as well." Enzo

              "Now get off my lawn with your silicooties and boom-chucka speakers and computers masquerading as amplifiers" Justin Thomas

              "Being born on third base and thinking that you must have hit a triple is pure delusion!" Steve A

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              • #22
                Originally posted by Helmholtz View Post
                No, series wired resistors don't draw additional current (and they don't waste more power than diodes if voltage drop is the same). The heater current difference between two 5V6s and two 6V6s is 0.3A. This is what the heater winding sees.
                There's only one current in a simple series circuit so each component gets the same current.

                Each tube needs an individual 2.7 Ohm resistor.
                The difference is quite small for this particular case, but in point or fact, due to the waveform distortion the version with the diodes will use about 2 -3 % more power than the purely resistive version.
                Experience is something you get, just after you really needed it.

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                • #23
                  Each tube needs an individual 2.7 Ohm resistor
                  There you go, that solves the open heater problem.
                  Education is what you're left with after you have forgotten what you have learned.

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                  • #24
                    Originally posted by nickb View Post
                    The difference is quite small for this particular case, but in point or fact, due to the waveform distortion the version with the diodes will use about 2 -3 % more power than the purely resistive version.
                    Interesting. I wonder where the additional power goes. Need to think about it.
                    Last edited by Helmholtz; 10-26-2019, 11:33 PM.
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                    • #25
                      Originally posted by Enzo View Post
                      There you go, that solves the open heater problem.
                      Yes, this is what I had in mind since post #2.
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                      • #26
                        Originally posted by Chuck H View Post
                        I'll have to take your word for it. My simple mind says that if your loading the voltage down and something is heating the resistor then current is being used.
                        A series resistor doesn't load the voltage down. Imagine a 6V transformer winding driving a 5 ohm resistor. That's 1.2A from the winding. Now add a 1 ohm series resistor to make the load on the winding 6 ohms. The current is now 1A so the load on the transformer is lower with the series resistor not higher.
                        Last edited by Dave H; 10-27-2019, 11:39 AM.

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                        • #27
                          Originally posted by Dave H View Post
                          A series resistor doesn't load the voltage down. Imagine a 6V transformer winding driving a 5 ohm resistor. That's 1.2A from the winding. Now add a 1 ohm series resistor to make the load on the winding 6 ohms. The current is now 1A so the load on the transformer is lower with the series resistor not higher.
                          Yep. Helmholtz taught that class in post #20
                          "Take two placebos, works twice as well." Enzo

                          "Now get off my lawn with your silicooties and boom-chucka speakers and computers masquerading as amplifiers" Justin Thomas

                          "Being born on third base and thinking that you must have hit a triple is pure delusion!" Steve A

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                          • #28
                            Yep. Helmholtz taught that class in post #20
                            The extra point mentioned by Dave H is that the series resistor actually reduces the circuit current. Without the resistor the heater current would higher (too high!). Also a series resistor always takes/steels voltage and power from the actual load (heater). And that's what is needed here: Without the resistor heater voltage and heater power would be too high.

                            Summarizing: The series resistor lowers heater current, voltage and power.
                            Last edited by Helmholtz; 10-27-2019, 01:48 PM.
                            - Own Opinions Only -

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                            • #29
                              Right guys! My mechanical mind (mechanics rather than pure physics) was interpreting only an addition of current loading the voltage. Which I now understand is not the case. The voltage and current sort of see-saw with a redistribution of the current. It makes sense to me now. Thank you.
                              "Take two placebos, works twice as well." Enzo

                              "Now get off my lawn with your silicooties and boom-chucka speakers and computers masquerading as amplifiers" Justin Thomas

                              "Being born on third base and thinking that you must have hit a triple is pure delusion!" Steve A

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                              • #30
                                Originally posted by Helmholtz View Post
                                Interesting. I wonder where the additional power goes. Need to think about it.
                                Indeed. I'm starting to think the result is not real.

                                I went back to my simulation and carefully tweaked things for better precision. There is now a small 0.2mW difference in power dissipated in the load (truncation errors) and 10.8mW in the input power, so about 0.2% rather than 2 to 3 %.

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                                Experience is something you get, just after you really needed it.

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