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previous: SteveF Thanks, again. I'm not so interest... -- 1/27/2000 2:59 PM | View Thread |
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1/27/2000 3:27 PM | |
Randall Aiken | Re: Converting from Class AB to Class A Yes, it will continue the "trend" towards warmer sound, and will eliminate crossover distortion. If you are designing from scratch, and want the most output power, you plot various load lines on the characteristic curves until you find the one that puts you in the most linear portion and also touches the max dissipation line. Typically, this will mean that you are running slightly higher than half the normal class AB voltage, and the quiescent current will be maxed out at a value equal to the plate dissipation divided by the plate-to-cathode voltage. Of course, the easy way to do this is to check the tube data sheet for class A SE operating conditions, and use them, except in a push-pull circuit. For instance, an SE EL34 is happy at 250V/100mA, which biases it right at the max dissipation of 25W. If you want to build a PP class A EL34 amp, simply run two of them in a push-pull pair at 250V, with 100mA quiescent bias through each of them, and a plate-to-plate load resistance of twice the recommended SE load. This will mean that your power tranny has to be able to supply 200mA total quiescent current, and the output tranny has to be able to handle 100mA current in each side. The good news is that the max dissipation in class A occurs at idle (when properly biased in the middle of the range), so there is no increase in plate current at full undistorted power (there will usually be a slight increase, or occasionally a slight decrease in average plate/screen current in practice, but not much). This is because the increase in current on one half of the swing is exactly balanced by the decrease in current on the other half of the swing, with a net change of zero. The bad news is that if you intend to run the amplifier fully into clipping for long periods of time, i.e. a square wave output, the current will increase up to twice what is necessary for full power clean operation. This is because the average power of a square wave is twice that of a sine wave of the same peak-to-peak amplitude (the RMS voltage of a sine wave is 1/sqrt(2) times the peak, while the RMS voltage of a square wave is equal to the peak value. When the voltage is squared and divided by the load resistance, the square root goes away in the denominator of the sine wave expression, leaving a 2, which corresponds to dividing by two, so the power of the sine wave is half that of the square wave, or the power of the square wave is twice that of a sine wave). Since music typically isn't a continuous full clipped square wave, you don't need to be able to supply double the max sine wave current, but you do need to be able to supply at least 1.4-1.5 times if you want to be able to run it flat out for long periods of time into an attenuator, for example, without fear of transformer failure. I always design my output and power transformers for twice the rated power output so they can be run at full clipped square wave all day long with no overheating. This is cheap insurance against failure when used with attenuators. Randall Aiken |
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Replies: |
SteveF Thank you, Randall. That's very he... -- 1/27/2000 4:53 PM |