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Thread: 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 !

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    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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    Quote 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

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    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.

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    Last edited by Chuck H; 10-26-2019 at 09:40 PM. Reason: bad math
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    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.

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    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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    It was implicit in my approach to have cross coupled diodes, but you are right, I should have stated it that way.

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    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.

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    Last edited by Helmholtz; 10-26-2019 at 02:08 PM.
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    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

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    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.

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    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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    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?

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    Quote 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.

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    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.

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    Bent Member Chuck H's Avatar
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    Quote 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.

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    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.

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    Member HaroldBrooks's Avatar
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    Quote 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.

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    Quote 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.

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    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?

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    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.

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    Quote 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.

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    Lifetime Member Enzo's Avatar
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    Each tube needs an individual 2.7 Ohm resistor
    There you go, that solves the open heater problem.

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    Quote 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.

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    Quote 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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    Quote 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.

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    Quote 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

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    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.

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    Bent Member Chuck H's Avatar
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    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.

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    Quote 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 %.

    Click image for larger version. 

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    Quote Originally Posted by nickb View Post
    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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    Thanks.
    Now it seems that the diodes actually take a little less power than R1. But the difference too small to justify deeper analysis.

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    Quote Originally Posted by Chuck H View Post
    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.
    There are scenarios where loading is parallel (as 'ballast') rather than series. Sometimes for very minor tweaks (whether intentional or not), like changing the value of the 'virtual C.T.' resistors, or sometimes big power resistors from supply rail to ground to lower the supply node ("The Twin").

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