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Hot Rod Deluxe CR13 Diode replacement Or sub

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  • Hot Rod Deluxe CR13 Diode replacement Or sub

    Hi all Ive got a Fender HRD, older one on the bench I want to do the resistor fix as the board is scorched the diode is looking bad to (but still working) for now. He brought it in with the power tranny fried, I confirmed this, ordered a new Classic tone. He wants me to do a little preventative maintenece as I recomended. My issue is the only replacement for the cr13 diode is a 1n5353b instead of a 1n5353v as the schemo calls for. Does anyone know what would be a suitable replacement to get this up off the board a bit. also where would be a good source for it, as I need several parts for this amp like new input jack new 250ka for the master vol and the 2, 470 ohm resistors and this diode which I cant seem to find to easy. I see newoldsounds has most of what I need. Will any rated 16v 5 watt zener due ?..... thanks guys for any input..

  • #2
    I do not know of anything special about the zener diode.
    16 volt/ 5 watt. DO-201 case.
    I did notice that there are two operating temperature ratings available.
    150C or 200C.
    Find 1N5353 Stock and Compare Prices Across the Most Reputable Distributors in the Industry.

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    • #3
      The schematic doesn;t say 1N5353V, it says 1N5353B. The parts list says 1N5353V and is clearly a typo.


      In any case look at the circuit. This is a plain old vanilla white bread 16v 5W zener diode. Nothing remotely special about it.

      You want to heat sink it? OK, use its own wires. Mount the thing up in the air a ways, not close to the board. The more exposed wire, the more the wire can radiate heat. And underneath, bend the wire over and solder as much as possible to copper traces. DOn;lt jsut solder it into the hole and trim the excess. ANy extra wire lead you can lay onto copper and solder down is even more heat sinking. It makes a difference, honest.

      And the 470 ohm resistors? Go ahead and get anal. Grab a couple in the aluminum finned package like Dale makes, run wire to them and mount them on a chassis wall. Don;t worry about the wires, they are carrying DC - nothing to couple.
      Education is what you're left with after you have forgotten what you have learned.

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      • #4
        AH Thanks for the info Enzo & Jazz. Also for the link this will come in handy..

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        • #5
          I mainly use large suppliers like Mouser, Digikey, Allied. There are good others, I just don;t happen to use them. For commondity parts like these zener diodes, I usually find the prices not too spread out, and I order from wherever i am ordering other things. And I keep in mind that even if I find them for 40 cents less somewhere else, if it means setting up a separate order, then any parts cost savings will be instantly wiped out by the additional shipping cost of the second order.
          Education is what you're left with after you have forgotten what you have learned.

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          • #6
            The zener diode low voltage supplies on the Fender De-Whatevers are failures waiting to happen. It's a testament to luck that more of them don't go postal.

            The problem is that it's a cheap design, and works... mostly. The use of a 5W resistor and a 5W zener there theoretically works if nothing goes wrong. However, if you pull too large a load on the +16 or -16V supplies, the resistor goes into the thermal destruction we all know about these amps.

            I did some modelling work on the circuit. I thought the failure syndrome was zero load on the zener making the zener eat too much current, die, then kill the series resistor. I now believe it's just too much loading on the +/-16V supplies running the zener out of current, which fries the resistor. For a 470 ohm resistor on a 48Vdc supply, 5W implies only a little over 100ma. So if something goes a little funny out on the +16V or -16V line, like leaky power filter caps, soft/cracked solder joints, even a modestly failing part, this can put 5W in the 5W power resistors long term.

            Problem is, they are not 5W resistors at any temperature other than 25C. It's hot inside that amp. And as the resistor heats up, it gets hotter, and the resistor's ability to dissipate heat goes down too. The old rule of thumb we had was that resistors had a surface temperature of 200C at their rated power. That's close to the melting point of tin-lead solder, which we can't use any more because it... well, never mind. It's close to the melting point of solder, and from experience, some of these resistors get there because they melt away the solder holding them in the holes.

            There are only two good ways to get around this and make the thing more reliable. One is to put in a much larger power rated resistor. 10W might do it. I'd go for more than that, or heat sink the ones that are there with some kind of home-made resistor heat sink to get the resistors' ability to get rid of heat up. The other way is a series current limit.

            If you do the calculations, this circuit should not be capable of putting out more than 68ma [that's (48-16)/470], which would all be good. What kills it is if something else dies and pulls more than that. Only another 30ma on the +16 or -16V supplies will get the resistors into China-syndrome territory.

            I modelled a series current limiter of a MOSFET, 8.8 ohm resistor, 2N5551 NPN, 47K resistor and 1K resistor set up to have the MOSFET not let through more than 100ma; also, I changed the 470 to 330 ohms. The result was that the resistor could only get up to 3.3W, the zener was unaffected, and the MOSFET and 8.8 ohm resistor stayed at low powers - 0.725W on the MOSFET and 88mW on the 8.8 ohm sense resistor.

            It's clear why it was done this way. A 470 ohm/5W resistor and a 1N5353B both can be had for under $0.15 in the USA in 1000-5000 piece lots. It's cheap. The better performance answer, a three terminal regulator for +15V and -15V can be had for $0.15 also in the same quantities, but it has a crucial flaw it can't withstand input voltages over 35Vdc. The resistor/zener solves the issue, costs twice as much, and ...mostly... works.

            I have some ideas about how to do better, but for now anyone fixing these should replace the two 5W resistors with either 470 ohm/10W or four 5W/1K resistors, two in parallel each place. Or do some creative heat sinking.
            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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            • #7
              I always assumed that the reason for all of these problems was the fact that they tried to use the same power transformer secondary for the power amp bias supply to power the chips. There are thousands of Zener based designs that work just fine to supply whole preamps when the input voltage is a realistic value.

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              • #8
                That's another way to look at it. Designing shunt regulators is just as complex as designing series regulators, and perhaps more so since the failure modes are not obvious.

                Strictly speaking, shunt regulators work best from the highest possible raw DC supply and the biggest possible resistance in series with the shunt element (zener in this case) because that results in the smallest possible change in the current from no load (i.e. all current in the shunt/zener) to a semi-short or dead-short. But for this to be practical, the series resistance has to be able to withstand the short-circuit current forever. That is what is marginal in this design. A change from the design-max current of 68ma (which runs through the resistor all the time because of the zener shunt) to 105ma which happens if something shorts the zener is what pushes the power of the resistor up to where Bad Things happen.

                In some ways, it would have been better to run this with a much higher DC input voltage and a correspondingly higher resistance to limit current to 68ma, or whatever. The change in power on the dropping resistor from no load to short circuit gets much smaller as a percentage.

                One can sidestep the issues by either using a much higher power resistance to go ahead and dissipate the short circuit current, or by clamping the current so it is OK for normal operation, but stops before the resistor melts down. My thumbnail current clamp was set for 75ma.

                This particular design is difficult to work with, other than putting in bigger resistors. It would cost a couple of bucks in components to fix both the + and - sides of the regulator, and you'd have to find places to put the parts on the PCB.

                My approach would be to get two 1K/5W wirewound resistors and mount them between two plates of aluminum, running bolts between them to clamp the aluminum to the sides and putting a little heat sink goo on each. Then I'd bend the leads of one down to the other and solder, then solder the original into the PCB. Some additional mechanical stabilizing would be nice, perhaps some of that automotive 500F RTV stuff. Using two 1Ks in parallel ought to get 500 ohms @ 10W, and the aluminum plates ought to about double that ability. The available current from this goes down a bit, to (48-16)/500 = 64ma, but that's probably OK. If the output is shorted, the power in the resistors goes to 48*48/500 = 4.6W, and a 10W, let alone a 20+ watt setup would dissipate that at a temperature low enough not to melt the solder or scorch the PCB.

                I suppose one *could* just rig a couple of aluminum plates clamped to the existing resistors. That would be better than nothing.

                As a practical matter, shunt regulators are very inefficient. They waste a lot of power as heat in order to do regulation. Series regulators are much more efficient if you can run them from a voltage nearer their output voltage, and they only let through what current is needed, not 100% of the possible current all the time.
                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.

                Comment


                • #9
                  My approach would be to get two 1K/5W wirewound resistors and mount them between two plates of aluminum, running bolts between them to clamp the aluminum to the sides and putting a little heat sink goo on each.
                  That is why I like these:

                  Click image for larger version

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                  Education is what you're left with after you have forgotten what you have learned.

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