In Reply to: DC couple a 5687 to 2A3 ? posted by Aria Seven on September 30, 2002 at 22:56:33:
A 5687 are most probably more suited than a 12AX7/ECC83 (no comment on your values though).
But why not try the E88CC/6922 (6DJ9-ECC189 which is simmilar but unlinear, ie. a remote cut of 6DJ8/ECC88) you already have figures for?
Can't really see how the gain from a E88CC could be enough but if the scheeme says so it may.
I already have the 5687 tubes ... will give it a shot & measure some voltages ...
Yes now you have the hang of the technique you can try various valves.
I agree with PAR. Buty everything depends on what goes before. I design in expectation of the need for a maxcimum input voltage of 1v rms which makes 1.414v peak. That is quite an average requirement, not enough gain for every situation but a satisfactory compromise.
Sush requirements make two stage amps a little difficult. We have to use such as a 6sl7 or 12at7 or indeed 12ax7. The gain of the 6dj8 is not in my view sufficient alone.
The 5687 is a really nice sounding but hard running valve. The gain of which is not sufficient alone. Were this not to matter an octal choice in the same ball park is the 6SN7.
Both these would sound great and really improve your amp, but you would not achieve full power. You have reached a compromise. Additional stage or valves with more gain but less favourable sonically. Not insignificant people have gone the route of better sounding valves in more stages, the two stage amp too has it's followers.
There is nothing wrong with the method you wish to adopt of using the right valves in two stages, but only if you are happy with your volume levels.
Thanks for your post Paul, I'm learning slowly ...
Now that I have learnt how to read the data sheets, and see what DC current my tube will draw given a plate voltage & bias, how could I determine what sort of output I can expect ?
You mentioned I am compromising my output level by using the 5687, is it a dramatic reduction ? Just by way of comparison, the 'Darling' I built has an output power of 1.5W, and thats enough for me, this original 12AX7/2A3 schematic I'm following had an output of 3.5W.
If the explanation of calculating my output power is too lengthy, are you able to instead 'ball park' estimate how much lower my output might be ... e.g. half ?
Thanks for your assistance in solving my original cathode bias problem and taking to time to explain the workings to me.
In Britain there is a plethora of surplus Marconi 10 watt output power meters that are sold at every Radio Rally. That's a good starting point.
To determine graphically takes a little longer.
Let's look in theory at your 5687 and assume you have a 1v rms source. I like to work in peak values so we are talking 1.414 volts as the input voltage. This is multiplied by the 5687. No valve evr multiplies by the mu of the valve, and not many valves depict the same mu as that in the specifications. With a mu of 17 but a very efficient load that you have in the 5687 you may achieve 80% of the mu in voltage amplification so you would have a peak voltage on the grid of the 2a3 of 19 volts. This means that the grid wil swing from -45 to -26 and to -64, for full power you would expect a swing of from 0 to + 90.
For the next step I need to look at the published curves for the 2a3 and see what output voltage and current swing I would expect from a movement along the 2.5k loadline from grid line -26 to grid line -64. Before I look I predict we can expect less than a watt of output power.
Firstly we must draw the 2.5k loadline. Start with the quiescent current, that is the current at idle, in your case this is about 60ma or 0.06 for the formula. Using a load impedance of 2500 ohms. To determine the point on the x axis that the maximum theoretical voltage is reached we multiply the quiescent current by the impedance and add the sum to the quiescent voltage (250v in your case) to arrive at 400v. To determine the maximum theoretical current at which there is no voltage we divide this 400v by the impedance to arrive at 0.16 or 160 mA this is the point on the Y axis. Connect the two and the resultant straight linbe intersects the quiescent operating point of 250v and 60mA with the bias curve -45v (in theory, in practice you could expect a 20% shift in any parameter).
Frank Philips has published a set of curves in which this 2.5k loadline is drawn for us http://frank.nostalgiaair.org/sheets/021/2/2A3.pdf
Look at the intersection of the -26v grid line with this loadline and read off the current and voltage from the Y and X axis, make a note. I would guess by eye that we are dealing with 82 mA or 0.082amps and 190 volts. Now we need to look at the -64 volt grid line and we can guess at 0.034 amps and 310 volts. To be perfect we should blow up the drawing and measure with a ruler but I just want to illustrate the point. We need to know change in voltage and change in current. Change in voltage is 310 - 190 = 120. Change in current is .082 - .034 = .048. Now we have the necessary tools. There are various formulae for graphical determination of power, but the one I remember without looking it up is ((change in volts) times (change in amps)) over 8. so that is (120 X 0.048)/8 which is 0.72 watts.
Paul, thanks for your reply ...
You are right, an output power meter is a good start !
I am reading your explanation and it's slowly coming to me, am I right in deducing this ...
My peak input source, say from a CD player, can be taken as 1.414 volts, this will then be amplified by the driver valve's amplification factor less say 20% for inefficiencies, then this will drive the grid of my 2A3 ...
So in a perfect world, and working backwards, my 2A3 is biased at -45V, so I am looking for an amplification factor of 40 for my driver valve ...
-45V(bias of 2A3) divided by 1.414 (input voltage) divided by 80% (efficiency loss) = ~40 (amplification factor)
You are on the right lines. This is the laughable thing about some designs out there. The driver could not either handle the voltage swing on it's output nor amplify the voltage swing on it's input sufficiently. As in the case of the 6dj8 option in your chosen design. A mu of 40 would be a minimum for such a circuit, but there would be no comfortable margin.
The 12ax7 is probably a little too sensitive, grossly over priced and not of any particular sonic benefit. But more headroom in the driver is sometimes aimed at, as that stage can be operated within a very linear region, leave the clipping to the output valve. The 12at7 (ECC81) or ECC85 (don't know the US equivalent) would be a better choice, but over all those high gain valves I would prefer the 6SL7, that's just the way I am.
As I said already, if you only need 0.7 watts carry on with the 5687. If you need more power and must remain Novel based try a 12at7, if you can change bases go 6sl7. The 6sn7 would sound better though less powerful again.
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