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Switchable PS filtering values?

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  • Switchable PS filtering values?

    The new Custom Classic serie Vox AC30s have this: a little switch on the back allows to switch between two levels of filtering, I believe one is the "vintage correct" value, the other is a bit higher therefore tighter (from memory, it's between 22uF and 33uF but I could be wrong).

    I very much liked what it does and I'd like to implement it on my Plexi-ish build since I sometimes do "heavier" gigs and need something tighter.

    I guess I can simply add, say a 50/50uf can, and put it in and out of the circuit with a switch, one half before the EL34 plates and another one at the PI (maybe some experimenting here for the best places and values).

    I'm a bit afraid though: with the switch open, it would see a 400V+ potential across it all the time the amp is on. Sure Standby switches do the same, but those are more often in a closed position, this one could spent hours and hours in the open position. Also, closing the circuit would generate a short current burst while the cap charges, which may or may not be a big deal.

    I thought about maybe putting a big resistor (say 220K or 470K) in serie with the cap and putting the switch accross it. This way, the cap would charge slowly, reducing the voltage accross the switch until it is almost nil. Then closing the switch would bypass the resistor so the capacitor can start doing its work without having to charge first.

    Anyone has thoughts on this? Sounds like a cool idea.

    Thanks!

  • #2
    Have you considered switching the ground of the cap instead of the hot? This would get the high voltage off of the switch while it's in the open position. My other thought is to use a relay for switching the cap, but I don't recall finding high voltage relays when I was searching for them.
    -Mike

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    • #3
      The way I understand it, it doesn't really matter on which side of the cap I put it: if I close the connection on the ground side of the cap, I'm basically elevating it to the full voltage therefore I end up with the same voltage across the switch.

      Maybe I'm wrong though

      Can't be that hard, the switch on the Vox looks anything but heavy duty (... maybe it controls a relay... haven't been able to find a schem)

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      • #4
        A resistor is a good idea but I would go for a lower value like 10K to 47K 2W flameproof.
        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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        • #5
          How about this: TWO switches, one on each end of the cap, put on the top of the chassis for less easy reach. I would put the amp on standby before using them.

          This way, each switche effectively has a zero potential difference accross its terminals until one of them is turned on. By switching while on standby, it's just like turning the amp off and turning it back on with more filtering.

          I would add a bleeding resistor accross its terminal so that it will discharge on its own after the two switches are opened.

          Maybe it's more trouble than required, but it would work and be relatively easy to play with. It's not like I want to change it during a gig, more that I want to adapt the amp to the gig I'm playing.

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          • #6
            I'm basically elevating it to the full voltage therefore I end up with the same voltage across the switch.
            That end of the cap is tied to ground. What you're doing is you're allowing the hot side of the cap to charge when you close the switch, the ground side stays at ground potential. By putting your switch on the ground side it will also be at ground potential.
            -Mike

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            • #7
              Originally posted by Hardtailed View Post
              I thought about maybe putting a big resistor (say 220K or 470K) in serie with the cap and putting the switch accross it. This way, the cap would charge slowly, reducing the voltage accross the switch until it is almost nil. Then closing the switch would bypass the resistor so the capacitor can start doing its work without having to charge first.

              Anyone has thoughts on this? Sounds like a cool idea.
              This is arguably the best way to do it. The switch is almost always at nearly ground voltage, so there is little continuous voltage on the switch when it switches. It does have to withstand high voltage at power-on time though, so it has to be rated to withstand the voltage, but not to switch it. Those are different ratings on switches.
              The way I understand it, it doesn't really matter on which side of the cap I put it: if I close the connection on the ground side of the cap, I'm basically elevating it to the full voltage therefore I end up with the same voltage across the switch. Maybe I'm wrong though
              You are correct. The order of any series string of parts does not matter as long as you look at only the ends of the string, not places in the middle. If you look only at the ends, there is no way to tell the order of the parts in the series string. This works for you in this instance, because you can put the switch at the low voltage end of the string, and if it shorts out the resistor on the low side, that IS changing things in the middle. But the DC level both before and after switching is ground, so you change the operation of the string, and get the advantage of low voltage, the cap always holding off the high voltage. But when the cap and resistor are in series, the performance is the same whether the resistor is at the high voltage end or the low voltage end.
              How about this: TWO switches, one on each end of the cap, put on the top of the chassis for less easy reach. I would put the amp on standby before using them.
              Bad idea. It's more complicated, and now you have one switch floating around at B+. The single switch in the ground side shorting a resistor is enough, and safer.

              This way, each switche effectively has a zero potential difference accross its terminals until one of them is turned on. By switching while on standby, it's just like turning the amp off and turning it back on with more filtering.
              that's a correct way to think, but again it works the same for a single switch across a resistor in the ground side, which is simpler and safer.

              I would add a bleeding resistor accross its terminal so that it will discharge on its own after the two switches are opened.
              Good idea, but adds some complexity.

              The reason that a resistor at the end of the cap works is that the resistance of the capacitor's internal insulation is very, very high - many megohms. The capacitor can be thought of as a "perfect" capacitor, no DC leakage at all, and a parallel resistor which models the DC leakage of the real capacitor. When this is operating, for DC conditions only, the "perfect" cap can be ignored, as only the DC-leakage-imperfections-resistor matters for DC.

              In operation with the switch open, the DC voltage on the ground end of the cap is just the voltage determined by the ratio of the leakage resistance of the cap and the "pulldown" resistor that the switch shorts. If the cap's resistance is 10M and the external resistance is 100K, then the voltage on the bottom end of the cap with the switch open is B+ times 100K/(10M+100K), or 1/100 of B+, or maybe 4Vdc.

              Putting a bleeder resistor across the cap has the effect of raising the DC voltage across the pulldown resistor when the switch is open. If you make this 1M, and use a 100K pulldown, then the voltage across the pulldown resistor is going to go up to about 40Vdc. Still, not bad. A small switch which has 40V across it when it's open and 400V/1M = 400uA through it when it's closed has no great stress on it in either condition.

              Notice also that the pulldown resistor will have in one case 4V across it (for 10M and 100K) and 40V across it (for 1M and 100K) so no great power dissipation results. 40V and 100K dissipates 16 milliwatts, so a 1/4W resistor is a nearly 8X overkill. However, it probably makes sense to decide to either use a 1/4W to act as a fuse if anything fails in the cap, or a 1W or so to make soldering it in and mechanical stability better. Both viewpoints are correct for different situations.

              Notice to the purists who watch my posts: I picked 10M for the cap leakage resistance, 400V for B+, 100K for a pulldown and 1M for a bleeder resistor out of thin air. They are not exact, and operation in an real will vary with conditions in the real amp (duuh...) but they are representative of conditions in real amps, not wildly wrong.
              Amazing!! Who would ever have guessed that someone who villified the evil rich people would begin happily accepting their millions in speaking fees!

              Oh, wait! That sounds familiar, somehow.

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              • #8
                Thanks R.G.!

                For some reason, I was automatically seeing the resistor-bypass-switch on the hot side, but yeah I can do it on the ground side.
                Makes a lot of sense.

                Now remains the question of what resistor value to use: too high and the cap will take an incredibly long time to charge up when the amp is powered up, too low and the cap won't be effectively out of the circuit when the resistor is unbypassed (I want it to be as "vintage correct" as possible when configuring it in classic 60's Plexi mode).

                Your argument seems to favor a higher value to make sure it sees the smallest voltage drop possible. Even if I was to switch it in before the cap had time to charge, it's not any worse than putting the amp out of standby initially. It also means this particular cap will take more time to discharge, but since many commercially sold amps don't even have a bleeder resistor, not that big of a deal (I'll make sure to discharge it when I need to work on the amp!)

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