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Yesterday, 04:27 AM  #11 
diyAudio Member
Join Date: Apr 2004
Location: Monroe Township, NJ

Apply Ohm's Law to calculate the cathode resistor value. You need a 600 part.
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Eli D. 
Yesterday, 04:29 AM  #12 
diyAudio Member
Join Date: Jun 2016

mr2racer,
For your triode with a CCS as the plate load: Set the CCS to 10 mA (many CCS use a single resistor to set the current, calculate for 10 mA, and use that resistance). Connect the grid through its grid return resistor to ground (0V) Calculate the cathode resistor, 6V/0.010A = 600 Ohms. The self bias resistor is 600 Ohms, connect the cathode through a 600 Ohm resistor to ground. Turn the filaments on, and the B+ on. The voltage on the cathode Will Be 6V (600 Ohms x 0.010A) Done. You may or may not want to use a bypass capacitor across the 600 Ohm self bias resistor. The cathode impedance is as high as it ever can be, because the plate is loaded by an extremely high impedance (CCS). The grid resistor of the next stage is quite often lower than the impedance of a good CCS plate load. Example gain of the circuit: Suppose the circuit has the plate and CCS plate load, driving a coupling cap, driving the next stage grid resistor: Suppose the CCS is not real good, so it only has an impedance of 100k Ohms. Suppose the grid resistor of the next stage is 100k Ohms. The plate load is effectively 50k Ohms. Suppose the triode has a Mu (u) of 20. Suppose the triode has a transconductance, Gm, of 1000 micro mhos. That is 1000 Ohms cathode impedance. With a u = 20, and a transconductance of 1000uMhos, rp = 20,000 Ohms (u/Gm = rp) But the 1000 Ohm cathode impedance is raised by the 'plate load/u' 50k/20 = 2500 Ohms. The total cathode impedance is 1000 Ohms + 2500 Ohms = 3500 Ohms. 3500 Ohm cathode impedance is in parallel with 600 Ohms = 512 Ohms. If you use a capacitor that has 51 Ohms of capacitive reactance at 10 Hz, the low frequency will be no more than 3dB at 10 Hz, and only 1dB at 20 Hz (single pole roll off). 51 Ohms = 2 x pi x f x C C = 1 / (51 x 2 x pi x 10Hz) C = 312 uF With the bypass cap, the gain will be 20 x (50k/(20k + 50k)) = 14.3 If the CCS impedance is 400k, and the next stage grid resistor is 400k, the plate load is 200k. Now the gain is 20 x (200k/(20k + 200k)) = 18, much closer to u. And do not forget that the impedance at the cathode is 600 Ohms in parallel with the cathode impedance (1000 Ohms in series with 200k/20) = 1,000 + 10,000 = 11,000 Ohms. You can see that 11,000 Ohms is much much greater than 600 Ohms, so the bypass capacitor is Not needed in this case (and some people are very happy when they do not need to use a bypass capacitor). 
Today, 06:28 AM  #13 
diyAudio Member
Join Date: Jun 2003
Location: Maine USA

You have picked a current, and a tube.
You must also pick a Plate voltage. Usually near midpoint of your B+. The platecathode voltage will be somewhat larger than Mu times the gridcathode voltage. It *really* helps to work this out on the tube curves. 
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