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Zener Diodes to drop Heater Voltage ?

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  • Zener Diodes to drop Heater Voltage ?

    A while back I picked up two absolutely perfect 5V6 tubes in a second hand store. I realize they are just like 6V6's with the exception of the heater voltage (5vac) and the heater warm up time I believe. I've used them briefly in one of my amp, and they sounded awesome, and this may be do in part to the too high heater voltage spiking things a bit, but the tubes are also pristine, and look like they have never been used for even a minute.

    My question is, how can I safely drop heater voltage in a 6.3vac amp to (around) 5vac ? I realize there is 10% or so leeway. Can I use Zener Diodes ? or do I need to do it with Resistors ? What size resistor ?

    I also do not want to change the other heater voltages for the PI and preamp tubes.

    Thanks for your help, again !
    " Things change, not always for the better. " - Leo_Gnardo

  • #2
    Acc. to 5V6 datasheet the tube draws 0.6 A at 4.7V nominal heater voltage. So each tube needs a 2.7 Ohm series resistor to drop 1.6V. Resistor power rating > 1W. Better use something like 5W resistors as current is higher during warm-up and resistor surge power varies.
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    • #3
      Originally posted by Helmholtz View Post
      Acc. to 5V6 datasheet the tube draws 0.6 A at 4.7V nominal heater voltage. So each tube needs a 2.7 Ohm series resistor to drop 1.6V. Resistor power rating > 1W. Better use something like 5W resistors as current is higher during warm-up and resistor surge power varies.
      Thanks, I had thought to use resistors, and I have 3ohm 10 watt cement ones. I measured the current voltage of the heaters and it is 6.9vac, so the 3 ohm resistors would work I believe.

      Any thought on using Zener diodes ? I saw someone mention this on another site, but he didn't go into detail regarding the orientation of the band, wattage, etc. I currently have 12v 5 watt versions (Motorola 1N5349B). I was experimenting with using these to drop B+ voltage, and they worked rather well... But not sure If the same would apply for A/C volts on a heater.

      Here's a small schematic I just found from another post on this site :


      https://music-electronics-forum.com/...0&d=1169606859
      " Things change, not always for the better. " - Leo_Gnardo

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      • #4
        http://www.mh-audio.nl/Calculators/D...gResistor.html

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        • #5
          I don't see any advantage to Zeners vs. plain ol resistors. There's not much happening in the amp to DE regulate the actual voltage of the circuit. And there's no heat advantage. Either component needs to dissipate the unused current. Zeners are an overkill consideration IMHO. Higher current Zeners are expensive and often require heat sinking. Maybe plain diodes.?. Two high current diodes in series should drop about the right amount due to their forward voltage drop and are pretty cheap. I've used resistors and haven't had any problems.
          Last edited by Chuck H; 10-26-2019, 08:40 PM. Reason: bad math
          "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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          • #6
            How were you planning to use zeners anyway? You need to lose 1.6v. Gonna use 1.6v zeners? There is no changing load, so simple resistors would work. No one regulates heaters. If I needed to reliably drop a volt or two, simple diodes have a voltage drop. Two diodes in series ought to about do it. But I'd stick with resistors.
            Education is what you're left with after you have forgotten what you have learned.

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            • #7
              If you just use two diodes in series it will only have half the waveform. It needs two more diodes in reverse parallel with the first two so now you have four diodes when one resistor would do.

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              • #8
                It was implicit in my approach to have cross coupled diodes, but you are right, I should have stated it that way.
                Education is what you're left with after you have forgotten what you have learned.

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                • #9
                  Zener diodes are used to drop or stabilize a DC voltage. In reverse direction they drop the zener voltage, in forward direction they drop around 0.7V just like standard diodes. In principle 2 zeners with appropriate and equal zener voltages wired in anti-series configuration can be used to drop an AC voltage.

                  Using a common dropper resistor for several tubes is no good idea, as voltage drop changes with current/number of operating tubes.

                  The version with 4 rectifier diodes has the advantage that the total voltage drop of around 1.4V depends very little on current/load. So if more than one tube is connected and one of the tubes fails or is pulled, heater voltage for the remaing tubes won't rise like with a common resistor.

                  But using individual dropper resistors is just as safe. And remember that the failure rate of well chosen resistors is much lower than that of any semiconductor (diode etc.). And total failure rate multiplies with the number of diodes used.
                  Last edited by Helmholtz; 10-26-2019, 01:08 PM.
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                  • #10
                    A couple of things bothered me.

                    1) The diode drop at 1 Amp is closer to 1V than it is to 0.7V.
                    2) The diodes distort the waveform with a dead zone around zero Volts.
                    3) The peak of 6.3VAC is over 8.8V. (How much does the peak matter ?)

                    So I did an experiment. 4.7V at 0.6A is 7.83 Ohms. I used 8.3 Ohms in parallel with 150 Ohms, measured 7.904 Ohms. I used a 6.3V at 4 Amp transformer connected to a variac. I used a Agilent True RMS meter and for comparison, a 3-1/2 digit Fluke (not true RMS). I used four 1N4003 diodes.

                    Raw xfmr Volts: TRMS 6.336 Fluke 6.30
                    Across resistor: TRMS 4.775 Fluke 4.52
                    WARNING! Musical Instrument amplifiers contain lethal voltages and can retain them even when unplugged. Refer service to qualified personnel.
                    REMEMBER: Everybody knows that smokin' ain't allowed in school !

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                    • #11
                      Using a common dropper resistor for several tubes is no good idea, as voltage drop changes with current/number of operating tubes.
                      Yeah but the number of tubes is not going to change, the load is not going to change. The guy is building an amp using the odd tube. Once it is wired up it will stay that way.

                      2) The diodes distort the waveform with a dead zone around zero Volts.
                      yes, it does, but since all it is doing it heating the power tube, does that matter? Heaters can be run on most any current, if it were a input stage, I might worry a little about some noise pickup, but not really in the power tube heater.

                      3) The peak of 6.3VAC is over 8.8V. (How much does the peak matter ?)
                      I shouldn't think it matters unless our diodes or resistors were at some low rating right on the edge.
                      Education is what you're left with after you have forgotten what you have learned.

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                      • #12
                        Yeah but the number of tubes is not going to change, the load is not going to change. The guy is building an amp using the odd tube. Once it is wired up it will stay that way.
                        As said above, if one of the tubes' heaters fails, the other one will see too much heater voltage. Might not be critical, but I prefer the individual resistors (or the 4 diodes) over a common dropper resistor.
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                        • #13
                          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?
                          "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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                          • #14
                            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?
                            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.
                            Last edited by Helmholtz; 10-26-2019, 04:45 PM.
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                            • #15
                              1) The diode drop at 1 Amp is closer to 1V than it is to 0.7V.
                              Voltage drop at high current depends on diode type. Larger chips have lower drop. A 1N4003 is not suitable for 1.2A anyway.
                              For the heaters only RMS voltage and current matter as their product gives real power which relates to heater temperature.

                              BTW, there is only one correct RMS value. So RMS and true RMS means exactly the same. Simple AC meters often just register peak value and multiply by 0.707 which works only with pure sine signals. The term "True RMS" is a marketing concept to distinguish correct RMS fom false RMS. But also True RMS meters have their limits given by the max crest factor of the measured signal.
                              Last edited by Helmholtz; 10-26-2019, 05:57 PM.
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