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20th September 2004, 01:29 PM  #1 
diyAudio Member
Join Date: Jun 2004
Location: bs.as.

capacitor reactance
Looking for info about like calculates capacitor value to bypass cathode in bias automatico; I find that:
when I now the value of cathode Resistor in ohms. it formulates: Reactance=(1/50).Ohms ....to the Low Frecuency operate the circuit (for example at 20Hz.) now, as the reactance of a capacitor for a given frequency calculates? 
20th September 2004, 04:46 PM  #2  
diyAudio Member
Join Date: Sep 2002
Location: Sacramento, CA

Re: capacitor reactance
Quote:
Where f is frequency in Hertz and C is capacitance in Farads. se 

20th September 2004, 09:15 PM  #3 
diyAudio Member
Join Date: Jun 2004
Location: bs.as.

Thanks steve.
I confused because: In one PushPull EL84/6BQ5 use for comon cathode resistor arround 140 Ohms I want Bypass this resistor, and calculate: Reactance=(1/50)*140 Reactance=2.8 Now find capacitor for 2.8 Ohms reactance at 20Hz 1/(2*pi*f*c)= 1/(2*3.141516*20* "test value capacitor in farads")= for 2800uf 1/(2*3.141516*20*0.0028)= 2.84 Ohms Ok. now;.... for correct bypass cathode resistor 140 Ohms at 20hz need 2800uF capacitor ????? Please help !!!!! 
20th September 2004, 10:49 PM  #4 
diyAudio Senior Member
Join Date: Aug 2002
Location: Belgium

Hi,
Actually the formula you should use is: f = 1/(2*PI*R*C) or: C = 1/(2*PI*F*R) f = 3dB point. R= Rk'//Rk Rk' = (Rp+ra)/(mu+1) (Commoncathode stage) Rp = Plate resistance ra = Ri or internal resistance of the tube. Rk = cathode bias resistor. Note that C should always be much larger than Rk but not much larger than necessary for the lowest 3dB point wanted: t = R*C Cheers,
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Frank 
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