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Re: Thanks Paul, how would I calculate output power ?

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.

That may be enough for you, if so all is good.


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