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Thread: London Power JCM 800

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    London Power JCM 800

    Hello,

    Im trying to build a London Power JCM 800 (for the fourth or fifth time... I like to learn the hard way I guess), and I'm having some trouble. Has anyone tried building this circuit, or verified that it works?

    Here are some voltages that Im getting:

    Va: 433V
    -Vb:-28.1 (at the zener and about a volt less at pin 5 on the tube)

    The latter seems rather low (this is with bias pots turned all the way low, as negative as possible).

    Im thinking that my zener may be whacked. Or I'm dropping too much across the 15k before the zener. If you dont have the schem, I can try to explain better. Thanks for your time,
    -Pete
    Last edited by peteko; 08-19-2006 at 07:46 PM.

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    Va Fluctuations

    I meaured Va again today and it has changed to 450V. Does anyone have any idea why this would change 20 volts?? I haven't made any changes to the circuit.

    Thanks,
    Pete

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    Just a thought - could it be the mains voltage has changed? Around where I live it can vary by 5% or so depending on customer load, and that could account for the increase you have seen.

    Mark

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    Mains

    Hi Mark,

    Thanks for the reply. I just checked the mains again, and since yesterday its down about 7 volts, so that explains the Va change. I have been playing around a bit with the bias pot values, etc, and found that if I increase their values to about 50k, I can get a more negative voltage at the control grid. Can you recommend a (negative) voltage range to shoot for?

    Thanks again,
    Pete

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    Lifetime Member Ray Ivers's Avatar
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    Pete,

    Do you have a book and page reference for the circuit?

    Ray

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    Yes, the book is TUT3 and the schematic is on pages 8-8 through 8-9 (The800 Chapter). Thanks,
    Pete

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    Lifetime Member Ray Ivers's Avatar
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    Pete,

    OK, thanks. I'd first pull your output tubes and disconnect the -Vb output to the bias pots, then measure your DC voltage to ground on each side of the 15K resistor and see what you get - then lift the Zener and measure again.

    You're following the schematic religiously, or just using parts you have on hand?

    Ray

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    Ray,

    I am using the same values, but higher power ratings on some components (i.e., the 15k is a 3 watt resistor instead of a 1 watt. And the Zener is a 100v 5 watt diode). Also, I'm using switching jacks instead of a rotary for the OT.

    Ok, so I disconnected the bias supply from the pots, and got some better readings. On the 1n4007 side of the 15k, i am reading 126.5v and on the Zener side (actually the drop on the Zener) is 100.5 v.

    I have played around with the bias pot values a bit, and got more negative voltage on the grid with higher pot values. 100k gave too much and when I tried to bias, I couldn't get near the -30mV reading I was going for (it was too negative). 50k pots worked a bit better, but then I got to thinking that something may be wrong with a components somewhere because I don't think that there would be a mistake in the book. Of course there will be tolerance differences, so some tweaks might have to be made, but I'm still not sure about making the pot change. What do you think?

    Thanks for your help,
    Pete

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    Lifetime Member Ray Ivers's Avatar
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    Pete,

    I am using the same values, but higher power ratings on some components (i.e., the 15k is a 3 watt resistor instead of a 1 watt. And the Zener is a 100v 5 watt diode).
    FWIW, I'd do the same. OTOH, I've seen switching jacks go intermittent sometimes - not usually a big deal in an input jack, but a speaker jack... well, YMMV.

    Ok, so I disconnected the bias supply from the pots, and got some better readings. On the 1n4007 side of the 15k, i am reading 126.5v and on the Zener side (actually the drop on the Zener) is 100.5 v.
    Exactly what I thought would happen - it appears to me your bias supply (up to & including the Zener) is "fine" - but please keep reading. .

    I have played around with the bias pot values a bit... but I'm still not sure about making the pot change. What do you think?
    If you change the pot values, you'll have to change the other resistor values as well, to keep the pot adjustment range roughly the same. If it were me I would put back in the 25K pots, use the resistor values on the schematic, and make the change outlined below.

    I just did the math, and Kevin's supply (using the specified 100V, 1W Zener, and assuming -126V pre-resistor -Vb) will allow only 1.7mA of Zener current, meaning the Zener could only supply a bit less than this, maybe 1.6mA or a little more... but the bias circuit pots & associated components draw at least 8.33mA of current just by themselves, assuming no grid-conduction current or component/AC-line variations.

    Using a 5W Zener (as you are, and which is what I would do) this is no longer a problem, so... assuming @ -126V raw bias under load, try replacing the 15K 3W with a 1.5K 2W or 3W resistor; this will allow about 17mA Zener current, which should be more than enough for practically any contingency short of a failure mode, using the original 25K bias pots. 1.5K Rdiss will be @ .5W, and 100V 5W Zener dissipation about 1.73W.

    Ray

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    Hi Ray,

    I will try the replacement. When you calculate the current draw of the bias pot and resistors, are you just counting the 25k pots, the 220k resistors from the " o " lug to the wiper (on the bias pot), and the 6k8 resistor to ground from the bias pot " x " lug? Without the tubes, the grid stop resistor doesn't come into play and the 22n caps block dc.

    So at one extreme of the bias pots we have 220k in parallel with 25k and at the other end, we have just 25k because the 220k is shorted, right? Thanks for your time,

    Pete

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    Lifetime Member Ray Ivers's Avatar
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    Pete,

    I used 12K for my approximate Zener load; four 25K pots in parallel, in series with the 6.8K resistor to ground; I see now that the actual total load is @ 13K... so the bias-adjustment-circuit current draw will be about 7.7mA, not the 8.33mA I posted - not a massive difference IMO.

    I didn't include the 220K 'pot-failure safety resistors', they will increase overall bias-circuit current draw slightly (from 7.7mA to @ 8mA) if all were set to the minimum-bias-voltage setting, which I figured was unlikely - and my suggestions more than cover this possibility anyway.

    My original recommendations should still work fine. The idea is to build a bias supply that can supply more current than needed, so that you don't have to worry about 500uA here, a couple of mA there, etc.

    Ray

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    Hi Ray,

    With the 1.5k I get -31.1mV at the grid, much better (I'm shooting for about 27mA on the grid). However here's the next problem, when I plug in the el34's, the current begins to rise through the grid, and keeps rising past 27mA, and I'm turning the amp off when it gets to 50mA. Do you think this is due to faulty tubes? (I may have fried these in previous attempts).

    Pete

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    Lifetime Member Ray Ivers's Avatar
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    Pete,

    With the 1.5k I get -31.1mV at the grid, much better (I'm shooting for about 27mA on the grid).
    I assume here you mean 'cathode' instead of 'grid' - TP1 thru TP4 on page 8-9 of TUT 3. What negative voltage do you read at pin 5 of each output tube socket?

    when I plug in the el34's, the current begins to rise... and keeps rising past 27mA, and I'm turning the amp off when it gets to 50mA. Do you think this is due to faulty tubes? (I may have fried these in previous attempts).
    If you're reading a ballpark-correct negative bias voltage at each output socket pin 5 (@-34V to -45V, but YMMV), I would leave your meter connected to one pin 5 and plug in one output tube at a time in that socket - you may find one or more of them start(s) to 'run away' and pull the bias voltage down, allowing the current to rise, etc., and may have to only replace that one (although if you think they're all damaged, I would replace them all - it's hard enough to troubleshoot a new build without having tube problems to contend with).

    Ray

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    Hi Ray,

    Sorry for the confusion, I got my terms mixed up. I am reading -31.1V at the grid (pin 5), in retrospect still a bit low.

    The cathode current is running away (test points). It's here that I want the 27mA, but the current keeps rising way past 27mA (up to 100mA before I could turn it off).

    Pete

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    Lifetime Member Ray Ivers's Avatar
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    Pete,

    I am reading -31.1V at the grid (pin 5), in retrospect still a bit low.
    You should be able to adjust the voltage at pin 5 from a bit less negative than -100V to a bit less negative than -30V.

    The cathode current is running away.
    What does the bias voltage at pin 5 do during runaway?

    Ray

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    Hi Ray,

    The most negative voltage is -31.1 and I can only go up to -17v at pin 5, on all the sockets.

    When the cathode takes off, the bias voltage stays put. I tried all four tubes, like you suggested and all four of them run away.

    Pete

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    Pete,

    You'll have to reduce the value of both 100K, 1W resistors from the PT secondary to the 1N4007 diodes, most probably to something like 15K to 22K, 5W.

    Just as a troubleshooting hint... if you're getting -100V across the Zener, which then connects directly to another point where you're getting only -31V or so, this means that the -100V Zener voltage is dropping to -31V, a fault condition. To remedy this fault, the Zener voltage must be brought back to -100V with the bias adjustment circuit connected.

    This thing just keeps getting better and better. Are you starting to understand why you've tried to build it so many times without success?

    Ray

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    Ha Ha, yeah, I guess not completely understanding electronics will do that. The other reasons being sketchy construction methods. I finally went to an eyelet board to make replacing components a bit easier.

    Anyway, I'll give the resistors a try. Will they draw so much more current that they should be stepped up to 5W?

    Thanks,
    Pete

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    Lifetime Member Ray Ivers's Avatar
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    Pete,

    The power dissipation will be about 3W per resistor using 15K's, and 2W each using 22K's. You'll be stealing about 18mA from the B+ line, but I don't think it will make much difference. It's quite a bit of current for a bias circuit, but to draw less you'll have to redesign the whole thing, change the pots, etc. so I would just 'bite the bullet' and make this one last change. Hopefully the Zener will pop right into conduction, you'll have a rock-steady -100V max bias voltage, and your output tube currents can then be set wherever you want them to be.

    Ray

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    Hi Ray,

    I just got my resistors today (20k was the closest I could find, 5W) but now im only getting a sweep of -100 to -54 volts at the grid. When I try to bias my tube, it only gets up to 3.3 mV ( so 3.3 mA at the test point). I guess the next step would be to increase the resistors a bit to get a les negative voltage at the grid right? Maybe 30k?

    I'm not exactly sure how you are calculating the power rating of the resistors. The bigger the resistor value, the higher the power dissipated (P = I^2 * R), or am I way off?

    Thanks for your help,
    Pete

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    Senior Member Enzo's Avatar
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    If current stayed the same as resistance went up, yes power would increase. But the higher the resistance, the less current will flow in most circuits. And current would be dropping by the square while resistance grows linearly.

    Power is simply the voltage dropped across the resistor times the current through it. 100 volts end to end and 100ma flowing means 100x.1 = 10 watts. (You can figure what the resistance of that one must be.)

    Note that the current through the tube cathode to plate is NOT the current through the bias circuit.

    If you have a voltage divider circuit across 100 volts, you can figure out what the voltage is across each resistor. That let's you calculate the current as well by Ohms law. Power is then VxI or just use the voltage and resistance for P = E^2/R. I^2xR works too, but you need to know the current flowing.

    If a resistor has to dissipate 2 watts, you need a 5 watt resistor, not a 2 watt.

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    Lifetime Member Ray Ivers's Avatar
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    Pete,

    im only getting a sweep of -100 to -54 volts at the grid.
    Oh man... I should have just worked you up a bias circuit from scratch. Replace the (incorrect-value) 6.8K common bias shunt resistor with a 3.3K (.5W is fine) and your bias range should improve quite a bit; this will draw a little more Zener current (see now why I added some in reserve?) but should still be fine. You may have to drop its value further to get the full range you'd like.

    FWIW, this may be a two-output-tube bias supply doubled up for a four-tube output stage; doing this is no problem, but you have to change at leaast one resistor if it's to work the same way in both circuits.

    I'm not exactly sure how you are calculating the power rating of the resistors.
    Using Ohm's Power Law: Esquared/R if the voltage is known, and !squared x R if the current is known. I realize you know this already, but I just wanted you to know I wasn't using anything more sophisticated than this.

    The bigger the resistor value, the higher the power dissipated (P = I^2 * R), or am I way off?
    If the current stayed the same through the resistor, you're absolutely correct - but the current never stays the same in a straight-ahead resistive-voltage-divider circuit like this, so each resistor change has to be accompanied by a new power calculation at the new current/voltage levels created by the change.

    Ray
    Last edited by Ray Ivers; 08-25-2006 at 02:24 PM.

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    fwiw, i do not like to have a regulated bias supply in an amp that does not have a regulated b+ supply: either regulate both, or neither.

    jm2c
    ken

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    Ray,

    I see how lowering the 6.8k will give more zener current, I'll give that a shot.

    Quote Originally Posted by Ray Ivers View Post
    FWIW, this may be a two-output-tube bias supply doubled up for a four-tube output stage; doing this is no problem, but you have to change at leaast one resistor if it's to work the same way in both circuits.
    Ray
    Do you mean lowering the bias shut resistor (6.8k) even more to get more current from the zener for the two extra tubes?

    About the voltage/current changes, how do you calculate the voltage/current changes? Am Iím still looking at 18mA draw from the B+ supply? Iím not sure what numbers to use for these calculations.

    Ken,

    Thanks for the reply. Do you think that the same method, by zener diode, would work to regulate the B+ supply?

    Thanks to all of you for your time (and patience) in helping me learn about all of this.

    -Pete.

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    Lifetime Member Ray Ivers's Avatar
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    Pete,

    I see how lowering the 6.8k will give more zener current, I'll give that a shot.
    Well, it will raise the Zener current, but we're not doing it for that reason, but rather to reduce the minimum negative bias voltage available by adjustment; the increase in Zener current is just an unfortunate side effect.

    About the voltage/current changes, how do you calculate the voltage/current changes?
    OK - suppose you have 100V, and you need to supply 10mA at 10V; you need to drop 90V across the resistor, so by Ohm's Law:

    R=E/I; R=90/.01; R=9,000 ohms

    Rdiss = Esquared/R = (90 x 90)/9000 = 900mW

    Now you decide you need 20mA:

    R=E/I; R=90/.02; R=4,500 ohms

    Rdiss = Esquared/R = (90 x 90)/4500 = 1.8W

    Am Iím still looking at 18mA draw from the B+ supply?
    Yes,


    Ray

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    Quote Originally Posted by peteko View Post
    Ken,

    Thanks for the reply. Do you think that the same method, by zener diode, would work to regulate the B+ supply?
    pete,

    if i'm following the thread correctly, you're using the zener as a shunt regulator. a shunt regulator is quite effective (and lots of folks say is the least detrimental to sound), but a pretty wasteful way of regulating voltages.

    while certainly possible, a shunt reg large enough for the B+ would be blowing off a lot of power all the time and would not be a very efficient solution to the problem.

    hth
    ken

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    Hi Ken,

    Yeah, I am using a zener as a shunt regulator for the bias supply. Can you suggest other ways of regulating the b+ supply, or what would you recommend?

    Thanks for your help,
    Pete

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    Hi Ray,

    Thanks for your help with the bias supply. I can finally bias my tubes at the correct current. And thank you for explaining the power calculations to me.

    (The saga continues however as now the amp is still not working right. When I plug in, with all controls down, I get sound through the speaker and the only control that works is the 'preamp' which seems to adjust the volume. I am going to check out my wiring, but please let me know if you have any ideas.)

    Sorry, I misdiagnosed the problem, whats really going on is that on the high input, with all controls down, I can hear sound through the speaker. I think I also had a connection to ground in my eq, because I fixed a shield ground, and now the rest of the controls seem ok.

    Does this seem like a problem (that there is sound with all controls down)?

    Thanks again,
    Pete
    Last edited by peteko; 08-26-2006 at 02:07 AM.

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    Senior Member Enzo's Avatar
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    Sound with the controls down is called crosstalk. The circuits AFTER the volume control have enough gain or sensitivity to pick up the signal from the parts of the circuit BEFORE the control. They are only inches apart after all. The two halves of the circuit are talking to each other across some gap.

    Is it really a problem? How often do you find yourself needing to play through an amp and have NOTHING come out?

    Crosstalk is a serious problem if it occurs in a mixer or tape deck. We wouldn't want the snare drum leaking into the vocal track or something.

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    Thanks Enzo,

    I just wasn't sure if this was a problem that was going to fry the circuit or anything like that.

    Thanks for your help,
    Pete

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    Hello again,

    I changed the Bias shunt resistor from 6.8k to 3.9k (the closest I could find to Ray's suggestion) and things were good for a bit. My eq died again and I am trying to figure out whats going on with that (only the preamp control works and increases the volume).

    However, I noticed that the (now) 20k resistors from the PT secondary to the 1N4007 diodes get really really hot. They are dropping about 92v so 4.6 mA. Also, the zener diode is only dropping 89v not 100.

    The 3.9k is dropping 34.5v so the bias circuitry after the supply is pulling 8.8mA, still about half of the supply (17mA) as mentioned in Ray's previous post.

    So, any ideas why the resistors are overheating and why the diode isn't dropping the full 100v?

    Thanks,
    Pete

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    Lifetime Member Ray Ivers's Avatar
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    Pete,

    I changed the Bias shunt resistor from 6.8k to 3.9k (the closest I could find to Ray's suggestion) and things were good for a bit... However, I noticed that the (now) 20k resistors from the PT secondary to the 1N4007 diodes get really really hot. They are dropping about 92v so 4.6 mA.
    This is less than .5 watt - a 20K, 5 watt resistor should get a little warm, but not 'really really hot'. You may want to actually check the current through these with your meter just in case things are not what they seem by Ohm's Law analysis.

    Also, the zener diode is only dropping 89v not 100.
    I would check the DC voltage on the 1.5K resistor, on the 'upstream' side away from the Zener, and see what you get. It should be at least -126V.


    The 3.9k is dropping 34.5v so the bias circuitry after the supply is pulling 8.8mA, still about half of the supply (17mA) as mentioned in Ray's previous post.
    This is correct - the Zener will be conducting the remaining 8mA or so of the 17mA total.

    So, any ideas why the resistors are overheating and why the diode isn't dropping the full 100v?
    Assuming all resistor values/ratings are as I specified, w/power tubes removed, and short of a component failure, no ideas at all, sorry to say. You may want to lift one end of the Zener and check current through it, and also lift the bias circuit to remove the load and see if the Zener pops back up to 100V; if it does, connect your meter (in current-measurement mode) to the bias circuits and see what current they're trying to pull (it should be 8.8mA, but you never really know till you read it on the meter).

    Until the bias circuit is 1000% squared away and running reliably, I wouldn't even think about the wonders in store for you in the preamp section. Did anyone notice the bias diodes are drawn backwards?

    If I had the amp in front of me I could go through it and do each of the dozen or so checks I would need to eliminate each and every possible cause, but in "long-distance troubleshooting" inevitably I'll leave one out - usually the critical one, too.

    Ray

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    Hi Ray,

    Thanks for the reply. I looked at a few other marshall schematics and noticed that the diode going to the bias supply had its cathode facing the pt. You mentioned that the bias diodes are drawn backwards, so the anode should be connected to the pt through the 20k resistors?

    Thanks,
    Pete

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    Lifetime Member Ray Ivers's Avatar
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    Pete,

    The drawing I'm looking at - from page 8-8 of TUT 3, the same one I've been referring to throughout this thread - shows the the 2 x 1N4007 bias-supply diodes with their anodes connected to the 2 x 100K resistors, which are connected to the PT HV secondary. This is the same way the 4 x 1N4007 B+ rectifier diodes are connected (minus the resistors, of course).

    For a negative output voltage, the rectifier diodes' cathode should face the AC source (i.e., transformer), and the anode should connect to the filter and/or load. Yours are obviously connected correctly, just wanted to point this out for anyone else who might try to build this circuit.

    Ray

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    Hi Ray,

    That's strange, I guess I have a different edition. Im my book, the bias diodes are drawn as you describe they should be drawn.

    Ok, something wierd is going on. With the bias set circuit (pots, etc...) disconnected, the zener drops about 103V. However, on the 1n4007 side of the 1.5k, im reading 319v!!!

    Could this be attributed to diode failure. I checked the diodes and compared them with the other rectifier diodes, and they seem to check out, but thats an awefully high voltage.

    Thanks,
    Pete

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