When first time warming/powering up a tube the anode voltage of the lower triode slowly charge from 0 to 137v. The connected output capacitor act if the loading is AC signaal and passes it thrue to the output. Ones heated up there is a solid 137v DC and that is blokked by the output capacitor. The bigger the capacity of output capacitor the more this phenomen occours.
Just use output short relay with 10 to 20sec delay.
Ronny
Not open, shorted to ground (short the 120k resistor until the tubes are hot).
The simplest solution that does not involve more components and still works is to reduce the output cap and increase the loading resistor outside the cap. That way the cap may be charged slow enough while the tube rectifier starts conducting 1uF / 12k is a reasonable suggestion. This will allow bass to any home speakers, and allow the output cap to be a folie cap. 0.47 would also be acceptable with small speakers.
10uF/12k is more reasonable, and only costs 1 cent to implement since the caps are there already.
Remember this resistor is in parallel with the amplifier Zin (10k?) and this will make the LF roll off even worse as stated below:
1uF/12k = LF -3db cut off is 13Hz, 0db is 130Hz
1uF/10k||12k = LF -3db cut off is 28Hz, 0db is 280Hz.
10uF/12k = LF -3db cut off is 1.3Hz, 0db is 13Hz.
10uF/10k||12k=LF -3db cut off is 2.8Hz, 0db is 28Hz.
Unless you want to castrate the bass response, I'd either use a relay to short it, or change the resistor for 10k from 120k. You will always get a voltage from this cap charging if you don't short it because the bottom and top tubes rarely heat and conduct exactly the same. One hogs the voltage until the other conducts in my experience. This appears as a signal through the cap.
Remember this resistor is in parallel with the amplifier Zin (10k?) and this will make the LF roll off even worse as stated below:
1uF/12k = LF -3db cut off is 13Hz, 0db is 130Hz
1uF/10k||12k = LF -3db cut off is 28Hz, 0db is 280Hz.
10uF/12k = LF -3db cut off is 1.3Hz, 0db is 13Hz.
10uF/10k||12k=LF -3db cut off is 2.8Hz, 0db is 28Hz.
Unless you want to castrate the bass response, I'd either use a relay to short it, or change the resistor for 10k from 120k. You will always get a voltage from this cap charging if you don't short it because the bottom and top tubes rarely heat and conduct exactly the same. One hogs the voltage until the other conducts in my experience. This appears as a signal through the cap.
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This is highly plausible, it might also oscillate on shut down. To verify, measure AC voltage across 150k plate resistance while start up .
That's a perfectly acceptable but requires a relay, a power supply for the relay - do you really want mains on the coil of a really that has signal on it's contacts? And it's more complex.
I like SIMPLE (like me)
I suggested a switch, but a relay would work, too. Could get the power from the filament winding.
The filaments are all elevated.
(120k X 275VDC)/(120k + 270k) = 84.6VDC on all filaments.
If the 12AU7 cathode follower has a bad filament to cathode leakage, it will elevate the cathode to 84.6VDC before the bottom 12AU7 warms up.
One old manufacturer's 12AU7 specified Spiral Filaments. I have seen 12AU7 tubes that did Not have Spiral Filaments.
I use JJ ECC82 - 12AU7 tubes. JJ ECC82 Do have spiral filaments; you might try one of those.
The Uk/f is rated for 180V. I get my JJ tubes from Eurotubes.com, they are re-tested at Eurotubes. They ship worldwide.
Note: the grid of the 12AU7 cathode follower is at 275VDC until the 12AX7 that drives it is warmed up.
(120k X 275VDC)/(120k + 270k) = 84.6VDC on all filaments.
If the 12AU7 cathode follower has a bad filament to cathode leakage, it will elevate the cathode to 84.6VDC before the bottom 12AU7 warms up.
One old manufacturer's 12AU7 specified Spiral Filaments. I have seen 12AU7 tubes that did Not have Spiral Filaments.
I use JJ ECC82 - 12AU7 tubes. JJ ECC82 Do have spiral filaments; you might try one of those.
The Uk/f is rated for 180V. I get my JJ tubes from Eurotubes.com, they are re-tested at Eurotubes. They ship worldwide.
Note: the grid of the 12AU7 cathode follower is at 275VDC until the 12AX7 that drives it is warmed up.
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The comment about the grid slipped by my eyes until it was mentioned here. This design will damage the top tube as the grid will have high positive voltage relative the cathode. A diode between pin 2 ( grid) and catode (3) will prevent damage. When tubes conduct the diode will be isolating and out of the circuit.
A benefit of this is that the output cap will charge much slower, thus have a better chance to charge without excessive voltage on the output. A 1n4148 will do just fine here !
A benefit of this is that the output cap will charge much slower, thus have a better chance to charge without excessive voltage on the output. A 1n4148 will do just fine here !
With the 1N4148 diode across the grid to cathode of the cathode follower, as soon as B+ comes up, but before the other tubes warm up, you have: 275V B+ driving 150k, driving a 1N4148, driving a 10uF, driving a 120k to ground. (275V x 120k) / (150k + 120K) = 122V peak transient.
What you get depends on the order of the various tubes warm up times: rectifiers, 12AX7s, 12AU7s.
And when the warmed up preamp is turned off, there is 137V across the 10uF cap, when the voltage at the cathode follower drops, the output will drop too by an equal amount, until the 10uF cap discharges all of the 137V charge that it has.
Oh . . . And you also have 137V across the other 10uF cap, it drives the 24k and 1k negative feedback network. When the power is turned off, that 137V across the 10uF cap will discharge, according to the other 10uF cap, the voltage change at the cathode follower voltage which is connected to the plate of the 12AU7 below it.
How about connecting the negative feedback 10uF cap to the Output of the other 10uF cap. This might reduce the size of the voltage transients. I have to think about what happens during the power up, and the power down sequences. Anybody want to take it from here? my brain is needing a re-charge.
What you get depends on the order of the various tubes warm up times: rectifiers, 12AX7s, 12AU7s.
And when the warmed up preamp is turned off, there is 137V across the 10uF cap, when the voltage at the cathode follower drops, the output will drop too by an equal amount, until the 10uF cap discharges all of the 137V charge that it has.
Oh . . . And you also have 137V across the other 10uF cap, it drives the 24k and 1k negative feedback network. When the power is turned off, that 137V across the 10uF cap will discharge, according to the other 10uF cap, the voltage change at the cathode follower voltage which is connected to the plate of the 12AU7 below it.
How about connecting the negative feedback 10uF cap to the Output of the other 10uF cap. This might reduce the size of the voltage transients. I have to think about what happens during the power up, and the power down sequences. Anybody want to take it from here? my brain is needing a re-charge.
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