I initially thought that using cathode bias meant things were handled 'automatically' and, for example, a PP amp would run Class A because of this.
Now I read Valvewizard's pages on setting the correct cathode resistor for the bias you want.
If I look at the specs for (say) a 6L6, it tells me that a bias of -16v is normal for Class A, and -22v for AB.
If I set the cathode resistor for -16v bias, can I expect it to run in Class A, or will it hop into AB as and when appropriate?
Now I read Valvewizard's pages on setting the correct cathode resistor for the bias you want.
If I look at the specs for (say) a 6L6, it tells me that a bias of -16v is normal for Class A, and -22v for AB.
If I set the cathode resistor for -16v bias, can I expect it to run in Class A, or will it hop into AB as and when appropriate?
In class A, you can or not decouple the cathode resistor. In class AB it is a must, in both of them supposing common cathode resistors for both tubes.
The resistor is simple to compute, let the bias necessary, and the current drained by the TWO tubes, and a simple quotient will give the resistor needed. Use the nearest value you can get. And, multiplying both values (current and voltage drop), you will get the MINIMUM power dissipated, use almost a two times this value resistor.
The resistor is simple to compute, let the bias necessary, and the current drained by the TWO tubes, and a simple quotient will give the resistor needed. Use the nearest value you can get. And, multiplying both values (current and voltage drop), you will get the MINIMUM power dissipated, use almost a two times this value resistor.
Using cathode resistor bias for Class AB is one of those things which in simple theory should not work but in practice (and full correct theory) can work reasonably well provided that the music is real music with a significant dynamic range so that most of the time the amp is putting out very little power.
If -16V puts the valve in Class A then it will stay in Class A, apart from a small bias shift due to second-order distortion. If you try to overdrive it to get AB then you will probably just get grid current which will add distortion and possibly blocking too. Going for -22V will give you AB and then you can (for a short time) put in a bigger grid signal. A bigger signal for a long time will shift the bias to 'cooler' conditions and may give crossover distortion.
If -16V puts the valve in Class A then it will stay in Class A, apart from a small bias shift due to second-order distortion. If you try to overdrive it to get AB then you will probably just get grid current which will add distortion and possibly blocking too. Going for -22V will give you AB and then you can (for a short time) put in a bigger grid signal. A bigger signal for a long time will shift the bias to 'cooler' conditions and may give crossover distortion.
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You also have to consider the plate voltage vis-a-vis the bias voltage to set the class you want to run. -16v is not the same operating point with a 300v B+ compared to a 400v B+.
There's a lot I don't understand. That's why I'm here!
From what I gather, there is waveform distortion when amplifying through a tube. Using push-pull adds opposite distortions together, thus correcting them. This will only happen if you have both sides of the signal, i.e. with Class A, or Class AB with an appropriate bias in relation to how hard the amp will be driven.
I'm only slowly putting the pieces together in my head, and working out what is relevant from a practical perspective and what isn't.
This cancelling affect is mostly any powersupply noise and the idle currents, since the PP OPT has the idle current and B+ flowing in opposite directions in each half. Hum gets cancelled because it's from a common source, the B+ input CT. The actual music signals are of the same polarity as they pass through the tranny because one side is inverted, first. Otherwise the music would also cancel. So any tube created distortion is actually passed into the tranny and to the speakers because the signal is flowing in the same direction, adding, to create music power.
Unfortunately I don't want to post a picture from Valvewizard's site (it helps) on here as it might be copyright, but here's a link to the page: http://valvewizard.co.uk/Common_Gain_Stage.pdf, Figure 1.16 in section 1.11 about harmonic distortion.
As I read it, during amplification one side of the waveform gets stretched and the other compressed a little. If the other out-of-phase waveform goes through the same process, the part that is stretched in one tube will get compressed in the other. After the inverting of one side in the tranny they are added back together, so the stretched form is added to the compressed form which gives us the correct undistorted waveform back (save for any non-linearities).
Please correct me if I am wrong...
By nature, a push-pull amplifier with transformer output is a class AB or B arrangement. But, ignoring that misunderstanding, cathode bias is a generic way to operate a tube without a separate grid bias supply. In other words, the cathode bias method works fine for low level voltage amplifier stages that do not have to supply power, or be operated near their maximum allowable voltages. I would describe this method as quick and easy when absolute linearity is not at issue.
For any type of class A power amplifier, a separate negative grid bias supply is a better approach, and allows easy adjustment for lowest distortion. Since the bias supply does not have to supply current, it can usually consist of a simple rectifier and filter capacitor. It may also happen that a cathode resistor is used in addition to a bias supply, but in that case it would be intended to provide local negative feedback.
RA
'Even harmonic' distortion produced in the 'push' and 'pull' output tubes will be cancelled, while 'odd harmonic' distortion will not be cancelled.
There is only significant cancellation when the 2 signals are @ equal. But when one tube is going negative and approaching cutoff while the other tube is going full positive and at the highest output, the distortion products are not an equal mirror of each other and one tube will be throwing all the distortion through the tranny while the other tube is not conducting and not producing anything.
Push-pull (which can be Class A, but isn't the best use of costly tubes) will cancel all even-order, even if driven to AB or B. (Assuming perfect balance.)
Some of the "sweetest" amps I have known were single-ended. Even-order distortion is not horrible.
OTOH you can design a perfectly balanced push-pull class A amp, and over-drive it to horrible fuzz.
Some of the "sweetest" amps I have known were single-ended. Even-order distortion is not horrible.
OTOH you can design a perfectly balanced push-pull class A amp, and over-drive it to horrible fuzz.
Not necessarily. Constant current bias enforces Class A. Cathode resistors do not.
Now I read Valvewizard's pages on setting the correct cathode resistor for the bias you want.
If I look at the specs for (say) a 6L6, it tells me that a bias of -16v is normal for Class A, and -22v for AB.
If I set the cathode resistor for -16v bias, can I expect it to run in Class A, or will it hop into AB as and when appropriate?
When using cathode bias, you have to pay attention to the screen voltage. Making the cathodes +16V above ground reduces V2K by 16V, and that can drop your Q-point plate current.
As for whether the Class A finals will go a bit into Class *2 depends on the grid drive and whether it can supply the grid current or not, and if the grids are capacitor coupled. If they are, then turning on the grid-cathode parasitic diode will add negative bias that can shift the operating point into less linear territory. (How RF amps derive some -- or all -- of their grid bias.)
Half true, half false. Even order distortion is cancelled in a push-pull OPT so never gets to the speaker. Odd order distortion is not cancelled and so gets through.20to20 said:
No. PP can be Class A, AB or B (or even Class C - but not for audio!).RestAssured said:
You seem to imply that fixed bias is particularly useful for Class A. The opposite is true. Fixed bias is more or less essential for Class B, desirable for AB and offers only a small advantage for Class A
No. You may be 'double-counting' the distortion.20to20 said:
As is often the case, some simple and sensible questions from a newbie have brought out lots of misinformation which then needs to be corrected.
OK, I understand the word no, in this case, means you disagree. Then, you say I might be double-counting the distortion... That's your counter enlightenment? No other explanation of what is in the PP OPT when one tube is in cutoff, or approaching?
If one tube cuts off at a certain point in the positive going part of the cycle, the other tube must cut off at the same point in the negative going part of the cycle (assuming exact balance between the tubes). This creates symmetrical distortion of the waveform, which is by definition odd-harmonic distortion.
Similarly, if one tube reaches grid-limited clipping, the other tube must do the same - again this is odd-harmonic distortion.
Similarly, if one tube reaches grid-limited clipping, the other tube must do the same - again this is odd-harmonic distortion.
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More precisely: assume that the ‘push’ tube has a nonlinear transfer function g(x). Assuming exact tube matching the ‘pull’ side must also have a transfer function g(x). But if the PI is supplying ‘x’ to the push side it will be supplying ‘-x’ to the pull side. The push-pull connection then means that the combined transfer function is
g(x) - g(-x).
This satisfies the definition of an ‘odd’ function ( i.e. f(x) = -f(-x) ), as follows
f(x) = g(x) - g(-x)
-f(-x) = -g(-x) + g(x)
Hence only odd harmonics can be produced.
g(x) - g(-x).
This satisfies the definition of an ‘odd’ function ( i.e. f(x) = -f(-x) ), as follows
f(x) = g(x) - g(-x)
-f(-x) = -g(-x) + g(x)
Hence only odd harmonics can be produced.
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