A lower valued plate resistor value will give smaller voltage swings, but a higher current swing; a higher plate resistor value, lower current swings, but a wider voltage swings. In the end, do all plate loads equal in results? No. If you do the math, you will see that Ra=2rp ratio gives the greatest power delivery into the plate load resistor (which is why this ratio is used so often in the design of SE amplifiers).
     What if we do not wish to abide to Ra=2rp ratio? What if we want to use a plate resistor = 4rp?
    To get the largest, symmetrical voltage swings with a plate resistor of our own arbitrary choosing, we must pick the optimal idle current for a given plate resistor value. Once again we need to find the range that defines the usable area of operation. With this new information, setting the optimal idle current becomes trivial. First we must find Imax, the maximum current draw the amplifier can under go without leading to positive grid current:

Imax = Vb / (Ra + rp)

where Vb is the B+ voltage. Next we need to set the idle current:

Iq = Imax / 2

    These two formulas ensure the widest possible voltage swing with any plate resistor value, but not the greatest power delivery into the load resistance, for that, the famous "2rp" formula re-enters the picture.

Tube CAD does the hard math for you.


Windows 95/98/Me/NT/2000/Xp

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