• WARNING: Tube/Valve amplifiers use potentially LETHAL HIGH VOLTAGES.
    Building, troubleshooting and testing of these amplifiers should only be
    performed by someone who is thoroughly familiar with
    the safety precautions around high voltages.

Understanding Cathode Decoupling

Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.
The 5670 is unusual in having a large square-law region in its curves. However, by my eye, if you are biased into the centre of the straightish section of the curve then the curve is slightly downwards so degeneration will make it worse. Degeneration would help in the low current/low gm region where the curve goes the other way, but this would be the wrong thing to do as overall linearity is low so you could get lots of higher order terms. There might be a hint of upward curve on the 300V one, but this is 300V at the anode not 300V supply so probably not relevant to many preamps.

Most valves would have more curvature, and in the wrong direction.
 
No, but used with care. The feedback reduces total distortion in roughly the same proportion as gain, but there will be a shift in balance with the third/second ratio increased.

For example, I once worked out that for an ideal 3/2 triode with an unbypassed cathode resistor equal to 1/gm, the ratio between the 3rd-order and 2nd-order coefficients in the transfer function was increased by 2.5, although both were smaller than before.

As with other forms of feedback, if you are going to use it at all you need to use enough of it. The one exception is that for valves with unusual curvature a little degeneration can cancel 3rd - although it could still increase other higher-order ones.
 
Thanks all for the interesting discussion so far. I'm reading Jones's Valve Amplifiers as well, and I've got a kind of nuts and bolts question:

In Chapter 2 (p. 78 of 3rd Ed.) he talks about calculating the size of the cathode decoupling capacitor. He states we need to know the resistance the capacitor sees from it's positive terminal to ground. So, he calculates the resistance looking into the cathode, then finds the total resistance by putting it in parallel with the cathode bias resistor.

In Jones's example, the resistance looking into the cathode is 2.38k, the bias resistor is 1.56k, and so the total resistance is 946R.

However, when he finds the value of the cap, he uses 2380 as the value for rk, not 946. Is this an error? I skipped ahead to his plans for the scrapbox se amp, and it looks like there he does use the parallel resistance.
 
However, when he finds the value of the cap, he uses 2380 as the value for rk, not 946. Is this an error? I skipped ahead to his plans for the scrapbox se amp, and it looks like there he does use the parallel resistance.

It isn't an error. The r(k) of the tube is inherent to the device. Even though it acts like a cathode resistor, there is no physical component associated with it, therefore no way to install a capacitor to bypass it.

If you use the parallel resistance, this will give a "high side" error, and it's usually best to have a cathode bypass that's too large than one that's too small.
 
Yes, that must be what happened. I have the 2003 edition from the library, and while he gets 170 uF, he uses 2380 in the equation: the two sides of the equation don't equal. He also makes reference to the 2380 value in the very next sentence when he says: "Note that we had to convert the 2.38 k cathode bias resistor back into ohms..."

Thanks for the help! I was pretty sure that 946 was correct; I'm slowly learning. The best part about this book is that it's possible to work along using the datasheets.

In response to Miles, since the two resistances are in parallel, to bypass one is to bypass both, no? The capacitor is a third route to ground, and is therefore in parallel with both, and bypasses both.
 
Last edited:
This curvature issue of gm versus Vg is interesting. Seems that constant Mu triode tubes have the positive gm curvature (see 6CS7 section 1). High 2nd harmonic triode tubes like 12AT7 and 12BR7 have wildly varying Mu and a negative gm curvature. And then some in-betweens (see 6CS7 section 2) have a flat (no curvature, still tilted) gm curve.

Taking this further, Kiebert and Mallory present a cathode degeneration calculation (single tail resistor for both triodes) for a P-P triode stage to cancel 3rd harmonic. (2nd and 3rd page of linked paper). They also mention that pentodes have the wrong coefficients for this to work. They never mention any specific triodes, but say triodes have a plus coefficient for vg^3 while pentodes have a negative coefficient for vg^3. Seems to me this scheme would only work for the negative gm curvature triodes. ???????

On a slightly different track, many datasheets give gm versus plate current curves, which almost always curve positively in that format. But some are closer to flat, like the 12AT7/12BR7/6LQ8/6LB8 triodes. I would guess that these should work well in P-P with a constant current source on the tail. Current now being the constraint, the two flat ramped gm curves in class A should add up to a near constant gm for low distortion of any order. While the usual constant Mu triode type gm curves (positive curvature) would add up to a curved gm total, giving 3rd harmonic. Same thing for the pentodes in P-P. (although the 6LQ8 pentode gm versus Vg curve is oddly negatively curved too, unlike most pentodes. 6LY8 pentode listed below for a normal pentode gm versus Vg curve. Maybe 6LQ8 just always operates in the lower section of a normal pentode gm curve)

http://www.clarisonus.com/Archives/...System Design Factors for Audio Ampifiers.pdf

http://scottbecker.net/tube/sheets/137/6/6CS7.pdf
http://scottbecker.net/tube/sheets/137/1/12BR7.pdf
http://scottbecker.net/tube/sheets/049/6/6LQ8.pdf
http://scottbecker.net/tube/sheets/135/6/6LY8.pdf
 
Last edited:
Seems that the Kiebert paper was discussing a 5687 dual triode diffl. pair with a 1.2K tail resistor for 3rd harmonic cancellation. This is a near constant Mu triode tube, most likely with positive curvature to its gm versus Vg curve. (can only find gm versus Ip curves unfortunately, they are of course the usual positive curvature in this format, and not flattish ramps like the oddball tubes 12AT7, 6KR8=6LQ8 triode, 6S4, 12BR7 ... when shown in gm versus Ip format)

http://scottbecker.net/tube/sheets/135/6/6KR8A.pdf
http://scottbecker.net/tube/sheets/127/5/5687.pdf

Puzzling that this would then work to reduce 3rd harmonic here (P-P), when it doesn't work in SE mode. Although I can see that the tail resistance would be increasing total current, and hence combined gms, at +Vg extremes, it does the same thing in SE mode too. ??
 
Last edited:
That "third harmonic compensation" is actually extension of transfer function by driving control grid positively, so it is bent on higher output power than without proper grid current drive. In case of SE it could be called "2'nd harmonic compensation", while it is not.

Also, diodes on fig 16 are drawn backward...
 
Last edited:
In response to Miles, since the two resistances are in parallel, to bypass one is to bypass both, no? The capacitor is a third route to ground, and is therefore in parallel with both, and bypasses both.

No. Consider what happens when you use fixed bias, and connect the cathode directly to ground. If you had: Rk || 0= 0R, then Gm= 1/0= INF. You would have infinite Gm, and infinite gain. That doesn't happen.

A bypassed cathode resistor has a shelving function. The DC gain is limited by the cathode resistor since the bypass capacitor is ineffective at DC. As the frequency is increased, the gain increases at the usual 6.0db(v) / octave rate. Once the bypass is fully effective, the gain shelves.
 
Status
This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.