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Thread: How does the B+ go high?

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    How does the B+ go high?

    I have a Mission Amps Scholar amp (SE 6V6) and it ran at 390-400v on the B+ for years, solid state diodes. An "event " happened whereby a metal standoff got loose and contacted something, frying a resistor in the B+ chain. I removed the standoff, measured no continuity from the power supply plus to ground, replaced the 470 ohm, five watt and started it up and phssst again.

    I measured the B+ and it was 513v. I desoldered the power xfmr B+ leads and got 750 ac across them. I wondered if the primary had shorted some turns, causing the whole secondary to be higher, but he heater voltage was exactly the same, 6.3v, under load.

    I know I have a crap power xfmr and have to replace it, but I'm at a loss to think of a model of what's happening.

    Ideas?

    Thanks

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    Does "phssst again" mean it burnt again?
    Without the loading from the rest of the circuit (mostly the power tube), the B+ will rise.
    Chances are the burning resistor is due to a faulty power tube.

    Power transformer is probably ok.

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    Lifetime Member Enzo's Avatar
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    And shorted transformer turns do not make voltages rise, it loads voltages down and usually blows fuses.

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    A transformer with 750ac across a ct high voltage output makes 375vac on each half secondary to be rectified. The UNLOADED voltage when rectified is 530Vdc peak, which agrees well with your 513 loaded.

    The question I have is how it was 400V before. Enzo is, as always, correct about shorted turns in a transformer making the output very low if there is a shorted turn. If the secondary was open, there would be no voltage at all, at least no voltage on the half of the HT secondary that is open. Your measurements don't say it's open.

    I suppose that it may have been dramatically low because one side of the original HT winding was not being rectified, so there would have been a lot of losses, letting 500+ volts to drop to 400.

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    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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    I couldn't find a Mission Scholar schematic online. does "SE" mean single ended? Maybe can you post a schematic and few gut photos?

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    Quote Originally Posted by R.G. View Post
    The UNLOADED voltage when rectified is 530Vdc peak, which agrees well with your 513 loaded.
    I took this 513V to be unloaded, as the supply resistor was burnt.

    A schematic would be very helpful here, although I'm guessing it's a simple circuit like a single-ended Princeton.

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    A few clear gut shots would probably do the trick.

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    Quote Originally Posted by nickb View Post
    A few clear gut shots would probably do the trick.
    Thanks to all for your time.

    Enzo, I said a shorted turn(s) on the primary would cause a higher secondary. However, that doesn't go along with same VA on the heater winding. Hence the post.

    The 513 vdc with one 22u/450v cap and no further load, and 750 AC unloaded, go together, that was my point: that the high voltage is there with out any weirdness in the circuit.

    When normal, the power supply string and the power tube idle load did not sag the power supply more than 15v, if that.

    I have a paper schematic but no way to upload it, but the power supply is bone simple: two s.s. diodes to 22u/450v cap, then through a 510 ohm/2w R, to a 22u/450vcap, which feeds to the top of the SE transformer and to the preamp RC's. It was 400V prior to the event.

    Thanks for your thoughts.

    Dan

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    Were you measuring your 400V at the first 22uF/450 or the second one?

    I guess, put another way, what are the current (sorry, meaning now) voltages on both caps.

    If the event opened the OT primary or made the tube not pull current through the 510R/2W, it would have no voltage drop so the voltage would be the same on both sides. Ohm's Law again.

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    Amazing!! Who would ever have guessed that someone who villified the evil rich people would begin happily accepting their millions in speaking fees!

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    Enzo, I said a shorted turn(s) on the primary would cause a higher secondary
    yes, I understood, but that isn't true. Any shorted turns on any winding will load the whole transformer down.

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    Quote Originally Posted by Enzo View Post
    yes, I understood, but that isn't true. Any shorted turns on any winding will load the whole transformer down.
    Enzo , you've seen the results of a a whole lot more stuff going down than I have, but it seems to me the hypothetical power transformer primary, with nothing else wrong but with a 20%, say, reduction of primary windings, will be no challenge for the power company and the effective new higher reflected load ratio might make the current rise not so high. Could you be more specific about "load the whole transformer down" ?

    R.G., The previous measurements of 390-400vdc were at the diode. Bruce's schematic had it at 387vdc. That was ballpark. On the retry, I didn't want to blow out my 450v caps with the 513v (without downstream load), so I shut it down as soon as the 510 ohm R phssshed. I understand that with out the load it will read higher, but 20v higher, not 100v higher, IME.

    Thanks

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    Lifetime Member Enzo's Avatar
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    But a shorted primary turn is NOT just a reduction in turns. If you simply unwound a few turns on the primary, then yes the secondaries would rise. But a shorted turn is NOT just taking off a few turns. The new path might be a few turns shorter, but that leaves the now isolated wire turns around the core with its ends shorted together. Now think about any secondary, and we short the ends together. If we do, then huge currents flow. Fuses blow. Why would your shorted turns be any different? Those cut off turns become a secondary, simply because they are wrapped around the same core.

    "with nothing else wrong with it" Well, a shorted turn IS something else wrong with it.


    In fact, in many modern amps a toroid power transformer is used, helps fit an amp into a lower profile chassis. Usually a large bolt sticks up through the bottom of the chassis through the center hole of the toroid. They are very careful that that bolt is short enough that it will not contact the top lid panel. If we replace that bolt with a longer one that hits the top panel, then that forms a one turn "winding" around the toroid and will shunt it out, usually blowing fuses. The winding is the path through the bolt, across the top panel, down the side and across the bottom panel, and back up through the bolt. It may not look like one, but all that metal is a winding as surely as it can conduct electricity.

    This can be maddening, you work on an amp open, and it works fine. Screw the lid on it, and BAM, fuses blow. Yet there is nothing wrong with the amp, just the bolt is too long.

    I used to work on a lot of arcade video CRT monitors. Most power supplies in a CRT monitor derive from the horizontal output transformer, or "flyback". This included the voltage for the CRT heater. The ferrite core of this transformer was often exposed down one side. I recall one brand of monitor that wrapped a piece of wire twice around that exposed core. It generated the heater current for the CRT.

    It doesn't take much to be a winding.

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    ^^^^^^^^^ thatīs how it works.

    To add a little, tranformers work a little more complex than what we "see".

    Everybody knows about different windings, turns ratio, etc, but not too many go down to the fine detail

    HOW does electricity, energy, pass from one winding to another without touching it, in fact being fully insulated?

    And how do we get higher voltages? ... isnīt it like seeing water run uphill?

    Or current out higher than current in?

    Fact is there is a huge transformation inside ... literally.

    A winding creates a strong alternating magnetic field in a magnetic core, period, nothing else.

    And any other winding "captures" that magnetic flux and turns it into "X" volts, feeding as many Amperes as load demands, depending on its own resistance.

    Now a when shorted turn is wound around that same magnetic core, will try to develop some voltage ,and feed a short

    How much current does a voltage, however low, develop into a short?

    I = V/R

    If R is 0 , then I is infinite
    But ... but ... you canīt pull infinite Amperes from a transformer!!

    Of course not, but try all available current.

    The transformer, all of it is shorted, because all of the magnetic flux is shorted.
    Even healthy windings, because they are all being fed from the same sick (shorted) magnetic field.

    The fact that not all transformers explode instantly when having shorted turns is because there is a little resistance between frayed ends, usually burnt carbonized enamel , insulating tape, burnt immersion varnish, which offers a fraction of an ohm.

    In any case, transformer is doomed.

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    "But a shorted primary turn is NOT just a reduction in turns. If you simply unwound a few turns on the primary, then yes the secondaries would rise. But a shorted turn is NOT just taking off a few turns. The new path might be a few turns shorter, but that leaves the now isolated wire turns around the core with its ends shorted together."

    Thank you for the education, Enzio. Thanks JM.

    While your at it, how come the shorted copper strap soldered around the cores doesn't suffer the same fate?

    Thanks

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    Lifetime Member Enzo's Avatar
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    It is soldered around the whole thing, not around the central core.

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    Quote Originally Posted by dcoyle View Post
    Thanks to all for your time.

    Enzo, I said a shorted turn(s) on the primary would cause a higher secondary. However, that doesn't go along with same VA on the heater winding. Hence the post.

    The 513 vdc with one 22u/450v cap and no further load, and 750 AC unloaded, go together, that was my point: that the high voltage is there with out any weirdness in the circuit.

    When normal, the power supply string and the power tube idle load did not sag the power supply more than 15v, if that.

    I have a paper schematic but no way to upload it, but the power supply is bone simple: two s.s. diodes to 22u/450v cap, then through a 510 ohm/2w R, to a 22u/450vcap, which feeds to the top of the SE transformer and to the preamp RC's. It was 400V prior to the event.

    Thanks for your thoughts.

    Dan
    You didn't mention the all important 470 ohm resistor that keeps burning up. What is it connected to?

    No scanner? Digital camera? Smart phone? Camera on computer?

    There are lots of online schematic capture tools where you could draw it in half an hour and then export e.g. https://www.digikey.co.uk/schemeit/project/

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    "You didn't mention the all important 470 ohm resistor that keeps burning up. What is it connected to?"

    The 470 ohm resistor I mentioned is the 510 in the above quote. Look at a fender 5c1 champ.

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    a 2W/500R resistor is designed to fail when passing more than about 60-ish ma. If the OT is after the 2W resistor, then at that point there is no load at all on the PSU, allowing it to raise to what you see. Swapping a 5W resistor in will allow 100ma at design max, long enough to take some voltage readings assuming all OK with the tube.

    If we assume 10% regulation for the PT, then the 500v (513vdc) at the diodes will drop to around 450v, and a 500R resistor will drop the remaining 50 volts with 100ma current. Not that I'd expect 100ma, closer to 40ma at idle, even with SE class A at 100% dissipation. That would allow for the 2W resistor, in a correctly-working amp (no failures).

    My guess is a problem 6V6 tube. Did you power up with tubes in or out?

    edit: oh, snap. my hypothesis was mentioned in post #2

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    Quote Originally Posted by dcoyle View Post
    While your at it, how come the shorted copper strap soldered around the cores doesn't suffer the same fate?
    To add to Enzo's reply, the external strap doesn't see (thus doesn't affect) the flux that stays in the core itself, but it does see the flux that "gets away." For that flux, it is exactly a shorted turn. A current is generated in it that cancels the escaping flux. You can think of it as loading down the escaping flux in the same way that a shorted turn inside the transformer would load down the primary current.

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    Quote Originally Posted by eschertron View Post
    If we assume 10% regulation for the PT, then the 500v (513vdc) at the diodes will drop to around 450v, and a 500R resistor will drop the remaining 50 volts with 100ma current. Not that I'd expect 100ma, closer to 40ma at idle, even with SE class A at 100% dissipation. That would allow for the 2W resistor, in a correctly-working amp (no failures).

    My guess is a problem 6V6 tube. Did you power up with tubes in or out?

    edit: oh, snap. my hypothesis was mentioned in post #2
    Well it needed to be reiterated anyway.

    Quote Originally Posted by dcoyle View Post
    The 513 vdc with one 22u/450v cap and no further load, and 750 AC unloaded, go together, that was my point: that the high voltage is there with out any weirdness in the circuit.

    When normal, the power supply string and the power tube idle load did not sag the power supply more than 15v, if that.
    It seems to me the high voltage is there without any circuit at all (aside from the diodes and cap).
    You are comparing to a time when all you disconnected was the power tube. With the resistor removed, you have everything disconnected, and the voltage will rise much more.

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    Quote Originally Posted by dcoyle View Post
    "You didn't mention the all important 470 ohm resistor that keeps burning up. What is it connected to?"

    The 470 ohm resistor I mentioned is the 510 in the above quote. Look at a fender 5c1 champ.
    I'm almost speechless...

    OK so the problem is after the 470/510 resistor. The 6V6 has been mentioned several times. It could be that, it could be it's cathode bypass cap or...other things.

    There is almost certainly nothing wrong the PT so pull the 6V6 and retest for that resistor overheating. If OK check the DC resistance from pin 8 of the 6V6 socket to ground. Take a look to see what the cathode resistor value is and that is what you should measure. If wrong remove the bypass cap and retest. If it's now right fit a new cap. If the DC resistance was OK then try a known good 6V6. Report back with test results and we'll go from there.

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    If the screen supply resistor is burning up, could be a shorted screen in the 6v6, or a shorted smoothing cap (screen supply node filter cap) - assuming the ‘470R’ is the screen supply dropping resistor. Without a schematic, we might as well be whispering in the dark.

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