Thanks for the explanation (and calculation). I admit that I am a bit math-phobic. I know that they are very simple equations, but I still get confused.Dr d wrote:Hi Brewdude, By my calculations at 430v B+ and a cathode voltage of 30v, cathode current = 30/470R =0.063 amps. This is shared current so 0.063/2=0.031amps. 430-30 = 400v plate voltage. Therefore plate dissapation in watts equals 400v x 0.031amps =12.8watts. Max plate dissapation for JJ 6V6 =@14-16watts. This means they should be running at about 80% which some might argue is a touch on the hot side for a Dumble design. Hope this is accurate!!!
edit: I was just looking at a 5E3 schematic. I guess that the 5E3 must be running at much lower voltages since it uses a 250R cathode resistor shared between 2 6V6's. I wanted to try the math out on that schematic, but I didn't have any voltages to calculate from.