[Amps] Tank circuit Q

[Peter Chadwick]

Gary asked:

>If I then change the coil to one with more loss (higher resistance) but the
same amount of inductance as before and keep the same plate tune C then does
the overall Q still remain the same even though the coil may have much more
loss? <
No, but.........
Look on the coil as having a parallel shunt loss resistance, rather than a series loss resistance. (At any one frequency, this is perfectly valid)
Say the plate load resistance is 2000 ohms, and working Q = 10, and for the sale of the example, it's a parallel tuned circuit with a link couplig to the load.
The parallel tuned circuit looks like transformer (ignoring leakage reacatance) and there's a step.up from 50 ohms to 2000 ohms. We want a Q of 10, so the shunt XC (and for the matter, XL to a first approximation) has 200 ohms of reactance. Now if the coil Q unloaded is 100, that's approximately an effective shunt resistance of  20,000 ohms (I'm approximating here XL/Rseries = Rparallel/XL) So the net effect is pretty negligible, and the coil loss is the RMS plate voltage squared divided by 20kohms.
Now use a coil with an unloaded Q of 20, and R parallel is about 4000 ohms. Tibe load is now 4000 ohms in parallel with the 2000 ohms transformed up, so the effective plate load is 1333 ohms, and the coil losses go up. 
So the answer is that the Q does change, but how much depends on the actual Q values of the coils. Obviously, changing from a coil Q=500 to a coil Q=250 when the working Q is 10 won't make a lot of odds, but changing from Q=200 to Q=20 will.
Hope this helps
73
Peter G3RZP