02-11-2016, 07:09 PM
Comments:
1) 50kΩ for the volume control... "why is 1MΩ too high?" Well, the wiper of the control is feeding thye ECC82 grid, which is really hgigh impedance, and virtually capacitive. If the wiper of the control is at mid-position, then the grid would effectively be fed from a resistance of 500kΩ in parallel with 500kΩ (the two halves of the control's resistance), which would give some sort of treble loss when associated with the ECC82's effective input capacitance. But as this is going to be of the order of about 15pF, you can work out using f = 1 / (2πRC) that the frequency roll-off would be 32kHz, well above human hearing. So, while a 50kΩcontrol would extend the upper limit even further, it won't be audible! And as for loading a modern signal source, yes it would drive 50kΩ, but loading with this value rather than 1MΩ, again won't make any noticeable difference.
2) 3µF because "I want to block all frequencies above DC." Why 'all frequencies?' It's actually better to restrict input frequencies - the output transformer is unlikely to work well with much below 50Hz, so putting anything lower won't reach the loudspeaker anyway. But, if your input source has mega-bass boost or whatever, and starts shoving 10-20Hz frequencies (at significant amplitude) into your amplifier, you might find that the input stage is overloaded and distorts, causing clipping of the signal waveform, even though the 10-20Hz then gets stopped right at the output. So, better to use 0.22µF (with 50kΩ volume control), or 0.01µF (with 1MΩ).
I made the mistake, many years ago (teens!) of making a 3-stage amplifier using 0.22µF / 1MΩ coupling between valves, and I couldn't understand why the sound kept disappearing and reappearing every couple of seconds. It wasn't until I started measuring with a voltmeter that I found that the whole amplifier was oscillating at around 0.5Hz and the output stagewas gently going right into and out of cut-off!
3) 12kΩ load for the ECC82. If you have used load lines on characteristic curves, then you've done right, although you are operating the ECC82 at more current than it needs. From knowledge of the '82, I know that the curves are practically as non-curved at lower anode voltages and lower anode currents... 100kΩ would be as good. But that's the only comment. You've got something that works!
4) Load on the EL84. For an 'ideal' pentode, you need to load it with a load equal to: Anode voltage divided by standing anode current. That way, as you wind up the input signal, on one half cycle the thing will just start to distort as the anode voltage reaches zero at the peak of the signal sinewave (EL84 can't turn on any harder) and on the other half-cycle as the current just reaches zero (EL84 can't turn off any harder) at the other peak of the signal sinewave. For a real pentode, the anode-cathode voltage doesn't reach zero before the valve starts getting highly non-linear, so the actual optimum load is more like 0.9 times the 'ideal.' And you also need to allow for cathode bioas, and optput transformer DC winding voltage drop. But hopefully you will get the idea.
Using the above idea, if you want to give your EL84 an easier life by reducing the power, you can! And you can calculkate the optimum load for your reduced current. Obviously the output power will be less, but with the optimum load you can maximise the power available at your chosen standing current.
Have lots of fun - which it seems like you are!
1) 50kΩ for the volume control... "why is 1MΩ too high?" Well, the wiper of the control is feeding thye ECC82 grid, which is really hgigh impedance, and virtually capacitive. If the wiper of the control is at mid-position, then the grid would effectively be fed from a resistance of 500kΩ in parallel with 500kΩ (the two halves of the control's resistance), which would give some sort of treble loss when associated with the ECC82's effective input capacitance. But as this is going to be of the order of about 15pF, you can work out using f = 1 / (2πRC) that the frequency roll-off would be 32kHz, well above human hearing. So, while a 50kΩcontrol would extend the upper limit even further, it won't be audible! And as for loading a modern signal source, yes it would drive 50kΩ, but loading with this value rather than 1MΩ, again won't make any noticeable difference.
2) 3µF because "I want to block all frequencies above DC." Why 'all frequencies?' It's actually better to restrict input frequencies - the output transformer is unlikely to work well with much below 50Hz, so putting anything lower won't reach the loudspeaker anyway. But, if your input source has mega-bass boost or whatever, and starts shoving 10-20Hz frequencies (at significant amplitude) into your amplifier, you might find that the input stage is overloaded and distorts, causing clipping of the signal waveform, even though the 10-20Hz then gets stopped right at the output. So, better to use 0.22µF (with 50kΩ volume control), or 0.01µF (with 1MΩ).
I made the mistake, many years ago (teens!) of making a 3-stage amplifier using 0.22µF / 1MΩ coupling between valves, and I couldn't understand why the sound kept disappearing and reappearing every couple of seconds. It wasn't until I started measuring with a voltmeter that I found that the whole amplifier was oscillating at around 0.5Hz and the output stagewas gently going right into and out of cut-off!
3) 12kΩ load for the ECC82. If you have used load lines on characteristic curves, then you've done right, although you are operating the ECC82 at more current than it needs. From knowledge of the '82, I know that the curves are practically as non-curved at lower anode voltages and lower anode currents... 100kΩ would be as good. But that's the only comment. You've got something that works!
4) Load on the EL84. For an 'ideal' pentode, you need to load it with a load equal to: Anode voltage divided by standing anode current. That way, as you wind up the input signal, on one half cycle the thing will just start to distort as the anode voltage reaches zero at the peak of the signal sinewave (EL84 can't turn on any harder) and on the other half-cycle as the current just reaches zero (EL84 can't turn off any harder) at the other peak of the signal sinewave. For a real pentode, the anode-cathode voltage doesn't reach zero before the valve starts getting highly non-linear, so the actual optimum load is more like 0.9 times the 'ideal.' And you also need to allow for cathode bioas, and optput transformer DC winding voltage drop. But hopefully you will get the idea.
Using the above idea, if you want to give your EL84 an easier life by reducing the power, you can! And you can calculkate the optimum load for your reduced current. Obviously the output power will be less, but with the optimum load you can maximise the power available at your chosen standing current.
Have lots of fun - which it seems like you are!
