EMRFD Message Archive 4027

4027 2010-01-19 06:13:16 Nick Kennedy Class C tube amp load resistance Message Date From Subject I'm working on my first-ever tube TX design. Getting to the output network (pi-net), I need to know the resistance it should transform my 50 ohm load to. From ancient history I recall load lines and published values of rp, but I don't think they're applicable for class C. Right? Does the simplified formula often used for transistor amps work, that being Z = Vcc ^2 / 2*Po ? In my case, figuring 250 volts B+ and 6 watts out, I get 5,200 ohms. Taking another approach, my 1969 ARRL Handbook (pg 158) gives charts for looking up pi-net reactances of C1, C2 and L using the ratio of plate voltage to plate milliamps and a chosen value of Q (10, 15, or 20). I figured plate current by assuming 70% efficiency and using the 6 watt RF output value, giving me 34 mA. Now, taking the long way around, there's a formula in QST Technical Correspondence January 1984 for computing Rin based on C1, C2 and Rout. (Whew!) With this I reverse-engineered the values from the Handbook's charts to give a plate resistance value of 3,360 ohms. The two numbers are in the same ballpark I guess, but I wonder if there's a more direct way. I realize that a pi-net with variable C1 and C2 will allow me to adjust loading, so my question is more academic than anything. 73-Nick, WA5BDU Nick, With 250 VDC and 34 Ma I make the plate load resistance at 7352. Lou, W7JI ----- Original Message ----- I believe that you are off by a factor of two. The plate load resistance for a pi - net is the plate voltage divided by the pate current x 2 That would be 250 /(.034 x 2 ) = 3676 ohms The modulation resistance would be 250/.034 = 7352 The old handbooks use the formula for plate load resistance = Plate volts x 500/plate mA That gives the same results as mine 250 x 500 /34 = 3676 ohms Pat N4LTA ----- Original Message ----- You are correct......guess it's been tooooo long since I built anything with a tube hi hi. 73, Lou ----- Original Message ----- I would think this would not be any different than designing with transistors such as MOSFET (which is very tube like). Once you have determined the desired output power, Power = V*V/2R. If you want 6 watts, 6 = 250*250/2*R, or R = 5200 ohms. As far as the DC power, if the amplifier is 70% efficient, the DC power will be 6/0.7 or 8.6w Power = V*I, 8.6 = 250* I, I = 0.032A or 32 mA. - Dan, N7VE ________________________________ It is very easy to get mixed up with Modulation impedance and Plate impedance - I just finished doing it for a dual 1625 am transmitter - so it was fresh in my mind. Pat N4LTA ----- Original Message -----