Charles Hansen said:
Freeman Dyson was janneman, not me. The new one, well...
In any case, my question was quite sincere, I really don't understand the distinction you're making- even in a follower, there still seems to be a relative delay between input and correction signal. I guess that's why they call it a follower rather than a walk-alongsider.
And I'm a much handsomer fellow than Steve.
I'm confused (again!). I thought Janneman used Richard Feynman for a brief time. Who was the guy that looked kind of like Picard in Star Trek NG. And who was the robot? And who is the young woman? Ah, the mysteries of life...
I'm sorry, I was unclear in my original posting. What I intended to say was that there was no delay in the action of the feedback. Upon re-reading it, I can see that it would be easy to interpret it as saying there was no delay in the signal at all. Of course that is not the case, and I would presumably explain why Steve Eddy was making snide comments about a mythical device.
Let's look at your example of a single cathode follower and think about what's going on. For the moment, let's tie the grid to the cathode but leave the heater and power supply connected. The tube is now acting like a diode (we can pretty much ignore the grid for now).
There is a stream of electrons heading from the cathode to the plate, and the current flow is limited by the combination of the cathode resistor and what is called the plate resistance. There current is expressed in amperes, which is expressed in Coulombs per second (a Coulomb is 6.022 x10^23, or Avagadro's number of electrons), so even with a very small current there is a large number of electrons flowing in the vacuum.
There is even a way to calculate the velocity of the electrons. Their kinetic energy is expressed in electron-volts, and is a function of the voltage impressed upon the tube. I'm not a tube guy and I'm too lazy to look up the proper constants, but suffice to say that both the effective potential (voltage) and electron velocity will vary as a function of the distance between cathode and plate, and that there is a finite transit time for the electrons' flight.
Into this nice little picture we now apply a voltage to the grid that is negative relative to the cathode. So some electrons go marching up the wires to end up on the grid itself. Relatively speaking, it doesn't take a whole lot of them, as all we have to do is charge up a small capacitor. Using the formula Q = CV (where Q is again expressed in Coulombs), let's assume 1 volt applied to the grid and an interelectrode capacitance of 10 pF (about the right order of magnitude for a small signal tube). Then we will stuff about six-quadrillion electrons up onto the grid.
This doesn't happen instantaneously, of course, but instead as the exponential charging of the capacitance. Since this is a tube circuit, let's assume this circuit is being driven by a source impedance of 500 ohms along with another 500 ohms of grid stopper resistance for a total of 1000 ohms. (We know the grid stopper makes the circuit sound better, although we're not exactly sure why....) This gives a time constant of 10 nS, which means it will charge to 1/e = 63% of the final voltage (1 volt) in that time. After three time constants, it will have charged up to 95% of the final voltage. So right off the bat we have created a low-pass filter with a -3 dB point of around 17 MHz.
Back inside the tube, now. As the electrons move onto the grid they radiate an EM field that propagates in all directions at the speed of light. It starts to get a little messy now, as we not only consider the action of this field on the electrons leaving the cathode, but also the electrons already in transit. The electrons already past the grid will experience a slight acceleration and those not yet to the grid will experience a decceleration. This causes some interesting things to happen and is the basis for the operation of things like klystrons and traveling-wave tubes.
If we ignore this mess and focus only on the steady state (not a bad approximation for an audio amplifier), we will find that the beam current has changed. This change in beam current creates a voltage drop across the cathode resistor, and hence the cathode voltage also changes. Now the cathode voltage is changed by number of electrons present there, and these electrons have to charge up any capacitance between the cathode and other electrodes, including the grid, plate, and heater. So this change doesn't happen instantaneously either, but instead as another exponential function as those capacitances are charged through the cathode resistor.
At this point we have at least 3 distinct mechanisms contributing to the delay between the input signal and the output signal:
- Charging time of the grid capacitance
- Propagation delay of the field from the grid
- Charging time of the cathode capacitance
But please note that as the cathode (output node) itself changes potential ("following" the grid), that the beam current changes essentially instantaneously, with only propagation delay of the field from the cathode having any real effect. The speed of light in a vacuum (ain't tubes great?!) is 3 x 10^8 m/S, so assuming a cathode-grid spacing of 1 mm (about the right order of magnitude for a small signal tube) the delay is around 3 pS, which corresponds to a frequency of around 300 GHz. Now, maybe I'm being sloppy, but for audio I'd call that "no delay", hence the statement in my posting.
Please contrast this to the simplest two-tube circuit in which a feedback loop is added to create a similar action. We would have a common-cathode stage driving another common-cathode stage. The output (plate) of the second stage is coupled back to the cathode of the first stage. As they say in the physics texts, the solution to this problem is left as an exercise for the reader. Suffice it to say that I think one will find that the results are somewhat different for a two-tube circuit compared to the cathode follower.
I'm sorry, I was unclear in my original posting. What I intended to say was that there was no delay in the action of the feedback. Upon re-reading it, I can see that it would be easy to interpret it as saying there was no delay in the signal at all. Of course that is not the case, and I would presumably explain why Steve Eddy was making snide comments about a mythical device.
Let's look at your example of a single cathode follower and think about what's going on. For the moment, let's tie the grid to the cathode but leave the heater and power supply connected. The tube is now acting like a diode (we can pretty much ignore the grid for now).
There is a stream of electrons heading from the cathode to the plate, and the current flow is limited by the combination of the cathode resistor and what is called the plate resistance. There current is expressed in amperes, which is expressed in Coulombs per second (a Coulomb is 6.022 x10^23, or Avagadro's number of electrons), so even with a very small current there is a large number of electrons flowing in the vacuum.
There is even a way to calculate the velocity of the electrons. Their kinetic energy is expressed in electron-volts, and is a function of the voltage impressed upon the tube. I'm not a tube guy and I'm too lazy to look up the proper constants, but suffice to say that both the effective potential (voltage) and electron velocity will vary as a function of the distance between cathode and plate, and that there is a finite transit time for the electrons' flight.
Into this nice little picture we now apply a voltage to the grid that is negative relative to the cathode. So some electrons go marching up the wires to end up on the grid itself. Relatively speaking, it doesn't take a whole lot of them, as all we have to do is charge up a small capacitor. Using the formula Q = CV (where Q is again expressed in Coulombs), let's assume 1 volt applied to the grid and an interelectrode capacitance of 10 pF (about the right order of magnitude for a small signal tube). Then we will stuff about six-quadrillion electrons up onto the grid.
This doesn't happen instantaneously, of course, but instead as the exponential charging of the capacitance. Since this is a tube circuit, let's assume this circuit is being driven by a source impedance of 500 ohms along with another 500 ohms of grid stopper resistance for a total of 1000 ohms. (We know the grid stopper makes the circuit sound better, although we're not exactly sure why....) This gives a time constant of 10 nS, which means it will charge to 1/e = 63% of the final voltage (1 volt) in that time. After three time constants, it will have charged up to 95% of the final voltage. So right off the bat we have created a low-pass filter with a -3 dB point of around 17 MHz.
Back inside the tube, now. As the electrons move onto the grid they radiate an EM field that propagates in all directions at the speed of light. It starts to get a little messy now, as we not only consider the action of this field on the electrons leaving the cathode, but also the electrons already in transit. The electrons already past the grid will experience a slight acceleration and those not yet to the grid will experience a decceleration. This causes some interesting things to happen and is the basis for the operation of things like klystrons and traveling-wave tubes.
If we ignore this mess and focus only on the steady state (not a bad approximation for an audio amplifier), we will find that the beam current has changed. This change in beam current creates a voltage drop across the cathode resistor, and hence the cathode voltage also changes. Now the cathode voltage is changed by number of electrons present there, and these electrons have to charge up any capacitance between the cathode and other electrodes, including the grid, plate, and heater. So this change doesn't happen instantaneously either, but instead as another exponential function as those capacitances are charged through the cathode resistor.
At this point we have at least 3 distinct mechanisms contributing to the delay between the input signal and the output signal:
- Charging time of the grid capacitance
- Propagation delay of the field from the grid
- Charging time of the cathode capacitance
But please note that as the cathode (output node) itself changes potential ("following" the grid), that the beam current changes essentially instantaneously, with only propagation delay of the field from the cathode having any real effect. The speed of light in a vacuum (ain't tubes great?!) is 3 x 10^8 m/S, so assuming a cathode-grid spacing of 1 mm (about the right order of magnitude for a small signal tube) the delay is around 3 pS, which corresponds to a frequency of around 300 GHz. Now, maybe I'm being sloppy, but for audio I'd call that "no delay", hence the statement in my posting.
Please contrast this to the simplest two-tube circuit in which a feedback loop is added to create a similar action. We would have a common-cathode stage driving another common-cathode stage. The output (plate) of the second stage is coupled back to the cathode of the first stage. As they say in the physics texts, the solution to this problem is left as an exercise for the reader. Suffice it to say that I think one will find that the results are somewhat different for a two-tube circuit compared to the cathode follower.
Charles, sorry, I didn't mean this to be such a long essay question. Appreciate your effort! At what point do you think delay time is significant to make a feedback circuit "audibly" poor compared to open loop (all things being equal, of course, which they never are)?
Re: avatars, Jan used Feynman, then switched to Dyson. I had changed the dope-smoking monkey to Pim Fortuyn (the bald guy), then to someone else who Jan is still guessing about. The new young lady here is someone quite special, IMO. I'm pushing my friends in Nederland, CO to send me a picture of The Dead Guy to use for my next change-ola.
Re: avatars, Jan used Feynman, then switched to Dyson. I had changed the dope-smoking monkey to Pim Fortuyn (the bald guy), then to someone else who Jan is still guessing about. The new young lady here is someone quite special, IMO. I'm pushing my friends in Nederland, CO to send me a picture of The Dead Guy to use for my next change-ola.
The long essay was made for a reason. In it we found step-by-step how much delay there will be between the change in potential at the output of the cathode follower and the subsequent action on itself. The answer is around 3 pS, which corresponds to 300 GHz.
I urge readers to spend the time to perform a similar analysis with a simple two-tube circuit. (All of the information on how to do so was given in the long essay.) Somebody, please let us know how long of a delay there is between the time the output changes (plate of the second tube) and the time that the output changes again based on the feedback loop.
SY, when you have the answer to that question, then you will also have the answer to the question you posed, since in my experience the second circuit will be audibly poorer than the first (all things being equal, of course, which they never are).
Thanks for the avatar info. Please feel free to e-mail me the answers privately. I'm too lazy to try and figure out who they are, and I'll probably miss the thread where the answers are revealed. The Dead Guy is something of a local legend here. Takes all kinds to make the world go 'round....
I urge readers to spend the time to perform a similar analysis with a simple two-tube circuit. (All of the information on how to do so was given in the long essay.) Somebody, please let us know how long of a delay there is between the time the output changes (plate of the second tube) and the time that the output changes again based on the feedback loop.
SY, when you have the answer to that question, then you will also have the answer to the question you posed, since in my experience the second circuit will be audibly poorer than the first (all things being equal, of course, which they never are).
Thanks for the avatar info. Please feel free to e-mail me the answers privately. I'm too lazy to try and figure out who they are, and I'll probably miss the thread where the answers are revealed. The Dead Guy is something of a local legend here. Takes all kinds to make the world go 'round....
For me, yes, although this is a separate issue from feedback. Each active device creates a copy of its input signal. Obviously (all things being equal, of course, which they never are) the fewer stages the better. Like Einstein said, "All things should be a simple as possible, but no simpler."
Pavel, no apologies necessary. I know for a fact that your English is at least 10^6 times better than my Czech, but sometimes I can't even understand my own writing... 
I don't know too much about the Czech republic, but I know a Czech man that knows the family that escaped in the homemade hot-air balloon about 25 years ago. That family lives in the next town from Boulder (at least they used to). That was a brave trip!
I don't know too much about the Czech republic, but I know a Czech man that knows the family that escaped in the homemade hot-air balloon about 25 years ago. That family lives in the next town from Boulder (at least they used to). That was a brave trip!
till said:
The schematic attached to the following post:
http://www.diyaudio.com/forums/showthread.php?postid=334963#post334963
Has 32 dB of gain. This is enough to run a CD through a volume control straight into the amplifier. This design has only 3 FETs between input and speaker.
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