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In Reply to: RE: No. posted by Chip647 on July 26, 2014 at 08:28:18
The resistor value is 270R, so the cap value....??
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F = 1,000,000/R x C x 2PI
where
F is the Frequency of cutoff (-3 db point) in Hz.
R is the resistor in ohms
C is the cap in *microfarads*
2PI is 6.28
Normally the formula is shown with a 1 instead of the 1 million but this way solves for actual values so its a little easier to use. I would look at the timing constants in the rest of the amp- the cathode bypass should be about the same or a little lower then the timing constant in the succeeding stage of gain. If you are dealing with an output transformer I would set it at the same value as the cutoff spec of the output transformer (IOW if specc'ed do 5 Hz set the cutoff at 5Hz).
In a push-pull output section, I would not use a bypass cap- the differential effect of the cathode resistor will help with the push-pull operation.
Ralph
Sorry it is not that simple ...
You need to take into account the impedance
of the tube section that is in parallel
with the cathode resistor.
From the linked website -
It can be shown that the resistance seen looking into the cathode (Rk unbypassed) is:
Rk' = (Rp+ra)/(mu+1)
Therefore, the total cathode resistance, or the output impedance if you take the signal off the cathode, is the parallel combination of the cathode resistance, Rk', and the cathode resistor, Rk, as below:
R = Rk' || Rk
DanL
-
"Sorry it is not that simple ...
You need to take into account the impedance
of the tube section that is in parallel
with the cathode resistor."Dan, you're right!
The formula is, bypass cap value in uf = 1000000/ 6.28 * F3 * r
r = (RK * rk) / (RK + rk) This is what is being bypassed, not just the cathode resistor value. "r" represents the impedance of the cathode itself in parallel with the value of the cathode resistor.
rk = (RL + RP) / (mu + 1)
RK = value of cathode resistor
RP = plate resistance of tube
RL = value of plate resistor (or load that the plate is seeing)
Mu = in circuit Mu of tube
F3 = the 3db down point of the filter
Tre'
Have Fun and Enjoy the Music
"Still Working the Problem"
Edits: 07/28/14
The value of the cathode bypass cap is determined by more than just frequency response. Things like the class of operation, topology, and even the designer's philosophy regarding issues such as overload recovery must be considered. That aside, typical values to bypass a 270 ohm cathode resistor in a AB1 PP amplifier would be 47-100uF. The exact value isn't critical, as you will see if you research schematics of commercial equipment online. Note that while the cathode should have sufficient bypassing so as not to create significant degeneration or phase shift within the audio passband, frequency response is usually determined during the design phase primarily at the grid.
--------------------------
Buy Chinese. Bury freedom.
Edits: 07/26/14
A class AB1 push-pull amplifier should not have a cathode resistor at all beyond something for current sensing.
The mention of overload recovery above deserves some attention. In the context of guitar amps that are frequently driven into distortion, a too-large cathode bypass will result in very unpleasant sounds during the time it takes for the output stage (for example) to return to normal operating points following a momentary overload. It has to do with the time constant of the resistor and capacitor. If the cathode bypass is smaller, the recovery may be barely noticeable.
Of course, we don't want HiFi amps to be driven into distortion, but a very large cathode bypass can make the amp more fussy about transient overloads. Some bandwidth limiting at the lower end could also make an amp less susceptible to subsonic signals like record warps.
It's usually not that difficult to experiment with different cathode bypass values to determine which one sounds best. Use your ears :-)
TK
Andy has a input 6GK5 CCS loaded
with a 270R cathode resistor.
That makes a 120uFd for 5Hz F3.
DanL
~!
The Mind has No Firewall~ U.S. Army War College.
Without checking Chip's calculated values, it would be around 250uF, for the 5 Hz cut-off he chose. The simplest way to look at it is that the values of Rk and Ck multiplied together should be a constant for a given desired cut-off frequency. Note that 50uF*1000R is roughly equal to 500uF*150R (the numbers Chip chose; obviously you could use a 450uF or 400uF cap with the 150R resistor for a result closer to that which you get with Rk = 1000R, or you could use a 100R resistor with 500uF).
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