In transformer - coupled output stage with output triode we must do a graphical load - line analysis to find load resistance Ra (primary resistance, or reflected secondary load). Ra is max. AC voltage "swing" divided with max. current "swing" through the load, orRa = Ua pp/ Ia pp = Ua p / Ia p = Ua rms / Ia rms
Power at the primary: Pa=Ua rms^2 / Ra = Ia rms^2 * Ra = Ua rms*Ia rms
Note that Ia p = Ia dc, Ia pp = 2*Ia dc, Ia rms = Ia dc / 1,4142
The "goal" is to avoid graphic analysis, and find the simple formulae, "good enough" for "everyday use". ![](http://www.audioroundtable.com/emoticons/smiley.gif)
Our triode output tube with its "bias" Ugk, can have max. peak AC input voltage in class A1 equal to Ugk, or Ugk rms = Ugk/1,4142.
With very high load Ra, AC voltage at the load Ra is:
Ua = mu * Ugk
But, our "real" load Ra form voltage divider with tube plate resistance rp, and voltage at the load Ra is actually lower:
Ua = (mu * Ugk) / (1 + rp/Ra)
And from Ra = Ua / Ia, we have Ua = Ra * Ia
If we put together these equations:
Ra*Ia = (mu*Ugk)/(1+rp/Ra), and Ra*Ia = (mu*Ugk)/((Ra+rp)/Ra), and
Ra*Ia = (Ra*mu*Ugk)/(Ra+rp), and Ia = (mu*Ugk)/(Ra+rp), and
Ra+rp = mu*Ugk/Ia, and finally:
Ra = ((mu * Ugk) / Ia) - rp FORMULAE FOR Ra
What does this mean in practice? If we have some DC operating point for our output triode, say 300B - Uak=400V, Ia=80mA, Ugk=-85V and we know (about) rp and mu from tube manuals (simplification, assumed that rp and mu are constant, but error is minimal and negligible). Say, rp = 700 Ohms and mu = 3,9. Then:
Ra = ((3,9 * 85) / 0,08) - 700 = 3443,75 Ohms ~3,5 kOhms
In agreement with load - line analysis and tube manual data! ![](http://www.audioroundtable.com/emoticons/smiley.gif)