My previous post, about coupling cap & grid current. Hint: why audiophooles hate coupling caps?
Not so, though it may appear so to the untrained eye.
The bias determines the operating point of the tube(s). The bias relationship to the signal peak determines the Class.
What has Class B got to do with it? B1 is no different to AB1 wrt grid current. (Of course, there is always B2 . . .)
I see this a lot -- don't confuse letters with numbers!
Last edited:
This is a new one on me. Infinite-improbability drive, yes.
Class A - conduction during 360 degrees of a sinewave cycle.
Class AB - conduction during greater than 180 degrees of a sinewave cycle and less than 360 degrees.
Class AB1 - grid is not driven positive WRT cathode, no grid current.
Class AB2 - grid is driven positive WRT cathode, hence grid current flows
Class B - conduction during 180 degrees of a sinewave cycle
Class C - conduction during less than 180 degrees of a sinewave cycle
w
Funny, one would think that there should be class B1 and B2 to denote grid current or lack thereof.
However, looking over a lot of tube datasheets, it becomes apparent that, at least once upon a time, grid current was commonly implied in "class B" operation.
Whether the bias is determined relative to the peak grid drive, or conversely the grid drive is determined relative to the bias, seems to be a sort of chicken-or-egg question. Clearly they are both determined together based on a load line analysis.
Anatoliy speaks of unintentional grid current causing bias shift by charging the coupling cap?
Happy Friday!
Michael
However, looking over a lot of tube datasheets, it becomes apparent that, at least once upon a time, grid current was commonly implied in "class B" operation.
Whether the bias is determined relative to the peak grid drive, or conversely the grid drive is determined relative to the bias, seems to be a sort of chicken-or-egg question. Clearly they are both determined together based on a load line analysis.
Anatoliy speaks of unintentional grid current causing bias shift by charging the coupling cap?
Happy Friday!
Michael
Exactly!
You never know when some peaks jump above zero volts, if you listen to the real music instead of sinewave from signal generator.
I know exactly when the peak jumps over zero. The amp farts and I know to turn it down.
If I want it that loud... the distortion and compression is my problem.
Music does not follow a perfect gaussian distribution. Real noise doesn't even follow a perfect gaussian distribution. A good noise source might do around four sigma, which means 1V RMS might have peaks to 4V if you wait long enough to see them.
Tim
Brain fart or amplifier fart?
Hello,
Brain fart or amplifier fart?
I think we are experiencing cognitive blocking distortion. The coupling capacitor, drain resistor time constant defines the sampling window. Even if music or for that matter noise falls on the bell curve the finite time constant will cause not so periodic blocking.
Now the other side of the coin. Now with our new and improved compressed and limited digital sources. We know exactly what the peaks will be. We should be able design interstage coupling that will never block.
The brain farts remain, unfortunatlly.
DT
All Just for fun!
Hello,
Brain fart or amplifier fart?
I think we are experiencing cognitive blocking distortion. The coupling capacitor, drain resistor time constant defines the sampling window. Even if music or for that matter noise falls on the bell curve the finite time constant will cause not so periodic blocking.
Now the other side of the coin. Now with our new and improved compressed and limited digital sources. We know exactly what the peaks will be. We should be able design interstage coupling that will never block.
The brain farts remain, unfortunatlly.
DT
All Just for fun!
If you're talking about Class *1, the input impedance will be the grid DC return resistor in parallel with Cgk + Cmiller + Cstray. For an 807 based design:
Cgk= 12pF
Crt= 0.2pF
Given the operating conditions, the voltage gain was 13.1:
Cmiller= 0.2(1 + 13.1)= 2.8pF
Guesstimate Cstray= 10pF
Ci= 10 + 2.8 + 12= 24.8pF
Bc= wC (where: w= 2pi X f)
Bc= 4.68E-6 (Set f= 30KHz)
I= (4.68E-6)(22.5)= 0.11mA
You can borrow the "Rule of Five" from solid state design, and set the grid drivers to 0.55mA of Q-Point plate current. Class A* pents aren't a real tough load.
For practical designs, you'd probably want a bit more drive current capability since the transient overdrive behaviour will be much improved if the grids can slip transparently a bit into Class *2 operation. It's also better to use fixed bias, and DC coupling to the final control grids to prevent that problem of blocking distortion that can result from capacitor coupling.
Another consideration that needs attention is the cutoff frequency that results from driver impedance. Even though it looks like something like a 12AX7 or 6SL7 can drive pentode finals, these types also have very large plate resistance, and typically high Zo. You could be losing high frequencies that way.
Or Class C1 and Class C2. The reason they never made that distinction is that Class B was the lotsawatts class where the main consideration was efficiency, and not sonic quality (i.e. PA systems, AM plate modulators, or RF power amps). Therefore, it was always assumed that the finals would always be driven into grid current to maximize power output.
Let's take such simple example that everybody must understand what determines the operating class of the pp-output stage.
We have a 6L6G PP-stage biased with -30 V. In addition we have a driver stage with low output impedance connected to our pp-stage
When the driving voltage swing is set to 58 Vpp ( from -1 V to -59 V), there is no grid current flowing in the output tubes.
What is the operating class now ? Obviously it is AB1.
Next we increase the driving voltage swing to 88 Vpp (from + 14 V to - 74 V). Now grid current starts to flow during that part of drive swing when the grid is having positive voltage.
What is the operating class now ? Obviously it is AB2 since grid current exists.
What have we done above when the amplifier was "transformed" from class AB1 to class AB2 amplifier ?
We increased the driving voltage of the output tubes.
What we did not do ?
We did not change the bias voltage at all.
Conclusion: The magnitude of the driving voltage determines the operating class, not the bias voltage.
For those who still have doubts about what was said above, please look at the data sheet at the link:
http://www.mif.pg.gda.pl/homepages/frank/sheets/127/6/6L6G.pdf
There is at the end of page 3 data given for both AB1 and AB2 operating class. There you can see that in both operating classes are as follows:
- equal bias -22,5 V
- equal quiescent current 44 mA / tube
- .....all other conditions are equal, but
- driving voltage for AB1 = 45 Vpeak
- driving voltage for AB2 = 72 Vpeak
- output power for AB1 = 18 W
- output power for AB2 = 47 W
Once again: What makes this difference, bias or driving voltage ?
You should already know.
Silly question. The difference between AB1 and AB2 is set by the difference between bias and peak signal voltage, not by either on their own.
The question as you posed it is like asking "What is the difference between being solvent or bankrupt? Is it due to income or spending?" Then people argue about it. The obvious simple answer is that it is the difference between income and spending which matters.
The question as you posed it is like asking "What is the difference between being solvent or bankrupt? Is it due to income or spending?" Then people argue about it. The obvious simple answer is that it is the difference between income and spending which matters.
Back to issues:
To use cathode follower to avoid this phenomenom makes sense.
In the case of overloading, the output stage will be driven to grid current due to low output impedance of the cathode followers and the output signal is not distorted and the blocking effect avoided.
In my own designs I have usually used grid series resistors up to 10 ...22 kohms ( at output tubes ) to avoid the blocking effect, but ofcourse these resistors can not avoid the clipping of the transient.
This must be tested in the future.
To use cathode follower to avoid this phenomenom makes sense.
In the case of overloading, the output stage will be driven to grid current due to low output impedance of the cathode followers and the output signal is not distorted and the blocking effect avoided.
In my own designs I have usually used grid series resistors up to 10 ...22 kohms ( at output tubes ) to avoid the blocking effect, but ofcourse these resistors can not avoid the clipping of the transient.
This must be tested in the future.
I am not sure I can be more technical than pointing out the necessity to use the subtraction operation to find the difference between peak signal and bias, rather than trying to use one on its own. Simple arithmetic may seem very basic and non-technical, but in the right place it can provide the right answer. I even gave a financial analogy to help those struggling with elementary electronics.
- Status
- This old topic is closed. If you want to reopen this topic, contact a moderator using the "Report Post" button.