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Marshall 20W Clone Build Redplating?

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  • #16
    Originally posted by GregS View Post
    Assuming 0.136A current regardless of what resistor value is used here (I'm doubting this is correct) and I want to drop the voltage 45V that would mean 45V/0.136A = 330R. If I use a 330R resistor it will have to be able to handle 45V x 0.136A = 6.12W. Does this make sense or am I way off here? I'd rather not find out the hard way that my logic is wrong.
    Your logic is fine. But incomplete. The larger resistor will have higher current across it. I can't say exactly why, it's probably something to do with the plate and OT primary resistance being in a parallel circuit to the B+ rail. So maybe the closer you get to that resistance (the plate and OT primary resistance) with the dropping resistor, the nearer the two current levels match. At any rate, though I can't say exactly why, I can say that a larger resistor will have more current across it. But by the same token that means you don't need as high value a resistor to drop the volts. Even though I can't pin down the exact cause, I do own a working example of this. In my cathode biased 2xel84 amp I have 398 volts at the rectifier. I'm using a 150 ohm dropping resistor and it drops about 43 volts from the supply with the tubes set to idle at near 100% dissapation. So...

    If my 150 ohm resistor (operating under very similar circumstances to yours) is dropping 43 volts. 43/150*43=12.33 watts

    That's at idle. In use the current across that resistor will be on the order of 150% of 12.33 watts. 12.33*150%=18.495 watts.

    With safety margin that's a 25 watt resistor. But wait... There's more...

    Your amp has a higher voltage PT secondary than mine. So how come we both have 390 +/- volts at the rectifier??? It's because your amp is drawing excessive current right now and this is loading the HV rail. When you re-bias your power tubes (with a bigger cathode resistor) to draw 100% (or preferably a little less) of their rated max dissapation, your voltage is going to go up and you'll be back to square one. So you'll need a bigger dropping resistor than I am using, but probably not 330 ohms. I guessed at 180 ohms and I think that's a good start. I bet it'll be 180 to 220 ohms. So...

    The higher value resistor again will require a slightly higher rating. Now the 25 watt won't offer much safety margin and live a hot, hard life close to it's max rating much of the time. Since most resistor ratings jump from 25 watts straight to 50 watts, I say go with 50 watts. You could use two 15 watt or three 10 watt resistors in series or parallel too. But since one resistor looks tidier and is easier to wire in, I say just go with the 50 watt. And as I said before, do otherwise at your own peril.

    Chuck
    "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
      Your logic is fine. But incomplete. The larger resistor will have higher current across it.
      I suspected that something wasn't right in my assumptions. I'll try out around 180 Ohms as you suggest for a start and see where that gets. Also, it's good to know what is and isn't a safe power rating to use.

      Thanks again,

      Greg

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      • #18
        Another choice you have is to use an EZ81 tube rectifier. That will bring the B+ down considerably. If it has to be SS rectification, as I stated before, stringing 10v 5w zeners works fine to lower the B+ (It also works for tube rectified B+). The best thing to do is mount the zeners on a separate piece of perf board in a zig zag arrangement mounted away from the chassis where they can breath. They get tend to get rather hot.

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        • #19
          The good thing about stringing 5watt diodes is that they come in the standard DO-XXX case type so there's no chassis mounting. Mouser has 5watt 14V DO-201 zeners for 52 cents. But try to get some other stuff you need while your at it (gun tips, scope probes, shrink tube, lead wire, etc.) because their minimum shipping charge is about $10.00

          Chuck
          "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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          • #20
            Here's a link with some good information on Zener diodes....

            https://taweber.powweb.com/store/vdump.htm

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            • #21
              Great. Thanks for the info from all.

              Greg

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              • #22
                I ordered the VDump kit from the Weber site. I chose a 40V Zener rated for 25W (only this power rating was available).

                After installation all was great at first. The voltages were dropped exactly to spec, there was no more redplating, the amp sounded great. When trying it out again today an unfortunate incident happened that damaged the zener I believe. While I was was inserting the DMM probes into the jacks I have to test the cathode current of the power tubes (effectively measuring the voltage drop across 1 Ohm resistors before the main cathode bias resistor) there was a clicking in the speaker, it seemed the lights dimmed in the room slightly, and power was gone to the amp. It turns out that the AC cord wasn't plugged into its recepatacle properly and was probably dislodged when I was trying to measure the cathode current. Once put back in properly the amp came on again, however, now one of the tubes was redplating again. When re-checking the voltages they were all high again although not quite as high as before installing the zener.

                The information on the Weber site concerning the zener mentions that if your amp has a solid state rectifier and a very high value of filter capacitance you can get a large current spike when the amp is turned on. He mentions putting a NTC in-rush thermistor in series with the diode to protect it. I didn't get this thermistor as I wasn't sure what it was, couldn't find it on the site, and thought it probably wasn't necessary for a 20W amp. The amp does have a solid state rectifier and uses 50uF filter caps between ps nodes.

                I suspect what happened was the power was turned on and off rapidly by the power cord not being in properly which caused a large current spike that has damaged my zener so that it is no longer dropping as much voltage as it should. I did have a 500 mA slo blo fuse in the HV CT that did not blow so I'm not totally certain of this.

                Anyway, the point of this post is does my theory of what has happened make sense or could the problem be that something else is damaged??? If what I think happened makes total sense then I'll have to go ahead and order another zener and this time get this thermistor thing as well to prevent this happening again.

                Thanks for any input,

                Greg

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                • #23
                  Yes, from your description that sounds like you had a voltage spike that killed the zener. Weber has the NTC inrush current limiter thermistor listed on the bottom of the page with diodes.

                  https://taweber.powweb.com/store/diodeord.htm

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                  • #24
                    Thanks for the information on where to find that current limiting device Diablo. I'll be sure to get it along with the replacement zener diode.

                    I'm glad this happened during my test while I was watching things closely. Otherwise a lot more damage could have been done.

                    Greg

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