Posted by Damir [ 213.202.85.246 ] on November 24, 2005 at 12:01:53:
In Reply to: Re: PP 2A3/6B4G anybody? posted by PakProtector on November 24, 2005 at 09:13:45:
Quick "analise" gives Raa ~ 2*(Ugk*µ/Ia - rp) ~ 8kOhms, for OP 300V/50mA/-60V.
Theoretical Pout ~ Paa = Ia^2*Raa = 0,035^2*8000 = 10W, or
Ua=Ugk*µ/(1+2rp/Raa) = 42,4*4/(1+1600/8000)=141,4Vrms, and then
Pout ~ Paa = Uaa^2/Raa = 282,8^2/8000 = 10W
But, realistically, (losses, distortion) - Pout ~ 8W
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