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Drop DC heater voltage by 1 volt, what resistor

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  • #16
    Just kidding

    You need to know your filament winding rating. If it's high enough to handle the extra current a resistor could be placed in series with one of the filament leads. But...

    Is it really something you want to do? In your other post you mention that your amp is sounding as good as it ever has. And as noted by LT in post #13 lowering the filament voltage will reduce the gain of your cascaded stages with unknown results. But I don't suppose there's any harm in trying it.

    The rating and value given by Diablo in post #2 would apply and as noted by R.G. in post #4 your filament winding needs to be rated for at least an extra 1A of current than the tube filaments are using now or you will overheat the winding.

    The answer to your question was right here in the thread all along.
    "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

    "If you're not interested in opinions and the experience of others, why even start a thread?
    You can't just expect consent." Helmholtz

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    • #17
      If the resistor is in series, there should be no extra current required from the winding. A watt dissipated in the resistor is removed from the tube heater.
      If you are putting resistors in parallel across the winding to drop voltage, then this would increase current required of the winding.
      Originally posted by Enzo
      I have a sign in my shop that says, "Never think up reasons not to check something."


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      • #18
        I'm confused. Then where is the power being dissipated by the resistor coming from? Is it possible then to run a series filament string from a 0VA rated winding?!?
        Last edited by Chuck H; 12-27-2013, 08:00 PM.
        "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

        "If you're not interested in opinions and the experience of others, why even start a thread?
        You can't just expect consent." Helmholtz

        Comment


        • #19
          Ok, well lets say a series R is not a issue as far as current. What value might i use ? I assume it's a very small value on the order of a ohm or a few? I think i have a few 5 watt sand type that are under 1 ohm i could string together if necassary. If for no other reason that to see what it does to gain. I gave one I built like this for a friend a few years back and i could never figure out why it had more gain than mine. I never thought to check the heater V, but i tried everything else imaginable so maybe thats it. Same PT but they may vary some i would think.

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          • #20
            Originally posted by Chuck H View Post
            I'm confused. Then where is the power being dissipated by the resistor coming from? Is it possible then to run a series filament string from a 0VA rated winding?!?
            Chuck, the power dissipated by the resistor is the power we are "robbing" from the heater. We are decreasing the voltage across the heater by 1 volt. The heater current is basically constant.

            Originally posted by daz View Post
            Ok, well lets say a series R is not a issue as far as current. What value might i use ? I assume it's a very small value on the order of a ohm or a few?
            It depends on your total heater current and how many volts you want to drop across the resistor. In post #2, Diablo solved it for the OP where he needed to drop 1V and had total heater current of 1A.
            Originally posted by Enzo
            I have a sign in my shop that says, "Never think up reasons not to check something."


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            • #21
              Originally posted by g-one View Post
              Chuck, the power dissipated by the resistor is the power we are "robbing" from the heater. We are decreasing the voltage across the heater by 1 volt. The heater current is basically constant.
              The edit indicated on my post was me answering that question for myself! I figured that out easily enough. But then I removed it because the second observation still applies. Can you run a series filament circuit on a 0VA winding? I just don't have my head around it completely yet.

              EDIT: OK, obviously if there is a filament between the winding ends there is a load. And more filaments (also in series) increases resistance. So... The more filaments there are on a series circuit the lower the load. I assume that to run two 12V filaments in series we need a 24V winding. Ergo, current is halved and voltage is doubled. So a series resistor to reduce voltage is just doing the same thing, Got it.
              Last edited by Chuck H; 12-28-2013, 12:16 AM.
              "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

              "If you're not interested in opinions and the experience of others, why even start a thread?
              You can't just expect consent." Helmholtz

              Comment


              • #22
                Well, like i have said in the past, my math skill are literally grade school level and worse i don't understand formulas as written above. My PT's heater winding is 8A, and i don't know what to use so i started with a 1R, then a 3R. neither brought it down more than .1v.


                edit: Eh, never mind. I tried a 7R too and it went UP. Splain that Lucy. I guess it has to do with the posts above being in respect to DC heaters? Well, the heater wires are too short and the connection is slop so i finally got it back on and screw it, i'm leaving it. Too much trouble for what will likely amount to nothing.
                Last edited by daz; 12-27-2013, 09:05 PM.

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                • #23
                  Originally posted by daz View Post
                  Well, like i have said in the past, my math skill are literally grade school level and worse i don't understand formulas as written above. My PT's heater winding is 8A, and i don't know what to use so i started with a 1R, then a 3R. neither brought it down more than .1v.


                  edit: Eh, never mind. I tried a 7R too and it went UP. Splain that Lucy. I guess it has to do with the posts above being in respect to DC heaters? Well, the heater wires are too short and the connection is slop so i finally got it back on and screw it, i'm leaving it. Too much trouble for what will likely amount to nothing.
                  Was that with the tubes in or out?
                  You have to load the circuit.
                  T
                  "If Hitler invaded Hell, I would make at least a favourable reference of the Devil in the House of Commons." Winston Churchill
                  Terry

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                  • #24
                    in

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                    • #25
                      Here is a formula that you can use. R = Vactual - Vdesired /2 x Ih

                      Where Ih is the total heater current.
                      I got it here: The Tube CAD Journal, Heater Concerns

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                      • #26
                        Originally posted by daz View Post
                        ...I tried a 7R too and it went UP. Splain that Lucy...
                        My guess is that you were measuring the voltage across the transformer winding. The voltage will go up at that part of the circuit because the transformer isn't loaded down as much. (Yep - Additional series resistance is LESS loading in electronics speak) Measure the heater voltage after the resistor (at one of the tube sockets) and you will find that it goes down as you increase the series resistance.
                        Cheers,
                        Tom

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                        • #27
                          Originally posted by Tom Phillips View Post
                          My guess is that you were measuring the voltage across the transformer winding. The voltage will go up at that part of the circuit because the transformer isn't loaded down as much. (Yep - Additional series resistance is LESS loading in electronics speak) Measure the heater voltage after the resistor (at one of the tube sockets) and you will find that it goes down as you increase the series resistance.
                          Cheers,
                          Tom
                          No, at a preamp tube. I'm dumb, but not THAT dumb.

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                          • #28
                            According to the Valve Wizard you can drop heater voltage .7v by adding a pair of 60S1 Diodes.
                            The Valve Wizard
                            T
                            "If Hitler invaded Hell, I would make at least a favourable reference of the Devil in the House of Commons." Winston Churchill
                            Terry

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                            • #29
                              Originally posted by daz View Post
                              No, at a preamp tube. I'm dumb, but not THAT dumb.
                              Well...then we must consider some form of measurement or setup error or some other parameter that changed at the same time. Otherwise, a couple of participants in this thread seem to be on the verge of creating free energy.

                              Comment


                              • #30
                                Hey now! I corrected my misunderstanding above with an edit. Of course NO ONE stepped up to explain so I just had to roll it around in my head until it made sense.

                                I AM on the verge of creating free energy though! I stole the idea from Gary (mooreamps)
                                "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

                                "If you're not interested in opinions and the experience of others, why even start a thread?
                                You can't just expect consent." Helmholtz

                                Comment

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