Huh ? If you're referring to B+ subsiding after power-off this is a matter of energy storage, not rectification method. Capacitors store charge and until this charge is depleted B+ is non-zero.
You can deplete it sooner by adding load across the capacitors (to "bleed" the charge off) but too large a capacitor will affect filtering ability of the capaictors adversely, not to mention dissipation on bleeder resistor. A sane value will be in the range of 10+K or even 100+K, possibly with an indicator (LED) in series.
This however has nothing to do with either rectification method (once you cut the power there is no further source of energy ahead of the rectifier to feed the B+ rail from).
You can deplete it sooner by adding load across the capacitors (to "bleed" the charge off) but too large a capacitor will affect filtering ability of the capaictors adversely, not to mention dissipation on bleeder resistor. A sane value will be in the range of 10+K or even 100+K, possibly with an indicator (LED) in series.
This however has nothing to do with either rectification method (once you cut the power there is no further source of energy ahead of the rectifier to feed the B+ rail from).
There is no perfect diode as far as I know and tubes cool and semiconductors remain in the same state.
Perhaps one for wavebourne...LOL
I must admit I have never checked because the loading of circuits will change the time!
I suppose the only way to know would be to charge a Cap with both and time the B+ volt drop at power off!
Just to make it more interesting if the heaters cool before B+ is Low do we not have the same as B+ on with no heaters?
Regards
M. Gregg
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I once changed a leaky B+ cap in my one and only valve rectified amp (EZ81) with a new modern cap. It still lost it's charge after switch off. After much scratching of the wooden stuff I decided that valves just do that and until now never looked back.
Perhaps I need to go looking for another leaky cap
Perhaps I need to go looking for another leaky cap
It's not anything inherent about valve amplifiers. It may indicate the designer who made the circuit included a "bleeder" resistor from supply to ground - a safety device intended to discharge the B+ supply after the amp is turned off. It might be around 150K ohm. It will only draw a few milliamps while the amp is running, but it will ensure the caps don't stay charged up when the amp is off.
Leaving capacitors charged with hundreds of volts is dangerous. Someone might turn the amp off (and maybe even unplug it) and think it is safe to go troubleshooting inside.
Two things:
Discharge through any reverse leakage in the rectifiers is a looooooong process. Bleeders totally dominate.
If you use a temporary bleeder for discharge, beware the DA voltage. You can have 20V or more come back on an electrolytic after you thought you discharged it. That can be a major hazard if you brush those terminals accidentally with something conductive (screwdriver, wedding ring, scope ground clip...).
Discharge through any reverse leakage in the rectifiers is a looooooong process. Bleeders totally dominate.
If you use a temporary bleeder for discharge, beware the DA voltage. You can have 20V or more come back on an electrolytic after you thought you discharged it. That can be a major hazard if you brush those terminals accidentally with something conductive (screwdriver, wedding ring, scope ground clip...).
That's a jolly good idea - one can check the resistor is working when the set is on - as the led will be lit!!
The spookiest live cap I found was on a Pioneer sx838 (transistor) that leaves the capacitors charged at 40V per rail for weeks, just a strange effect leaving no drain path through the two connected amplifiers when off!
If it went open circuit how would you know?
Just like to quote "If it can go wrong it will go wrong".
The switch in idea would work with an LED i guess. It would come on, at switch off and go dimmer until off.
The same would be true of a relay in the heater supply then if the heaters failed the HT and the amp would fail safe. so it could be put in the closed side of the HT delay on.
Regards
M. Gregg
Just like to quote "If it can go wrong it will go wrong".
The switch in idea would work with an LED i guess. It would come on, at switch off and go dimmer until off.
The same would be true of a relay in the heater supply then if the heaters failed the HT and the amp would fail safe. so it could be put in the closed side of the HT delay on.
Regards
M. Gregg
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You really don't want 10 mA wasted through bleeder in a tube (pre)amplifier. LED will light up with 1-2 mA through it, you'll waste less power.
@piano3: Shunting lower resistance across the capacitor will work, but people tend to mix up to many switches. A better solution would be a 230V AC coil relay which shunts the bleeder with significantly lower resistance (10x lower for instance) once the power is turned off. Since tubes need to cool down a bit before you can start rummaging throuh amplifier's internals a time constant of 5 seconds or something along these lines is just fine.
When you switch off a valve amp, much of the remaining HT charge goes through the valves as the cathodes retain some ability to emit electrons for quite a few seconds. This is assuming you haven't fiddled with "stand-by" switches etc.
Most of the rest should disappear via bleeder resistors, which will be present in any good design. These also hold down any tendency for dielectric absorption to make charge reappear in an apparently discharged capacitor.
Almost no charge will go back through the rectifier, whatever type it is.
Most of the rest should disappear via bleeder resistors, which will be present in any good design. These also hold down any tendency for dielectric absorption to make charge reappear in an apparently discharged capacitor.
Almost no charge will go back through the rectifier, whatever type it is.
Piano3,
It depends on your HT voltage.
Arnulf says 1-2ma is OK for the LED to light so its your HT Voltage minus 2V (LED volt drop) divided by .001mA LED current. This will give you your LED series resistor.
Your HT voltage across the bleeder divided by the bleeder resistance will give you current through the bleeder.
If you multiply this current by the voltage across the resistor it will give you the bleeder Wattage. So if you use 10K bleeder you have your HT voltage divided by 10000 gives you current through the bleeder. Multiply this by voltage across the bleeder (HT voltage) will give you the beeder wattage.
You have to accept that you are going to loose a lot of energy as heat as you discharge the HT caps! so try the calc with different values of bleeder resistance until you get a wattage that is OK. 100k across 450V is just over 2W.
See what happens with 10K ...20W
Hope this helps!
It depends on your HT voltage.
Arnulf says 1-2ma is OK for the LED to light so its your HT Voltage minus 2V (LED volt drop) divided by .001mA LED current. This will give you your LED series resistor.
Your HT voltage across the bleeder divided by the bleeder resistance will give you current through the bleeder.
If you multiply this current by the voltage across the resistor it will give you the bleeder Wattage. So if you use 10K bleeder you have your HT voltage divided by 10000 gives you current through the bleeder. Multiply this by voltage across the bleeder (HT voltage) will give you the beeder wattage.
You have to accept that you are going to loose a lot of energy as heat as you discharge the HT caps! so try the calc with different values of bleeder resistance until you get a wattage that is OK. 100k across 450V is just over 2W.
See what happens with 10K ...20W
Hope this helps!
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