Dynamic impedance of LEDs runs from 4-10 ohms. So how much variation is that for a circuit that swings 0.1mA?
The "thermal noise" you speak of is absolutely negligible given the impedance (unless you have some tubes with 5 ohm ENRs, in which case I'd love to hear about them). As a demonstration, look at the noise measurements for my phono preamp (in the articles section of this website), where the LEDs are effectively in series with the inputs. See also Jack Walton's measurements of the ImPasse preamp, posted here and at DIY Test Equipment for Audio and Ham Radio Enthusiasts .
There are numerous posts here showing how to use them in low current circuits- a bit of searching will turn them up and give you a powerful new tool for your designs; one resistor allows them to be used in circuits with minuscule current draws. I'd also strongly suggest looking at the measurements and circuits shown in Morgan Jones's "Valve Amplifiers."
LEDs (and other solid state diodes) are excellent, stable, reliable, low noise voltage references and are a superb way of biasing tubes in a wide variety of circuits without the need for bypass caps. They solve the OP's problem at minimal cost and complication.
The "thermal noise" you speak of is absolutely negligible given the impedance (unless you have some tubes with 5 ohm ENRs, in which case I'd love to hear about them). As a demonstration, look at the noise measurements for my phono preamp (in the articles section of this website), where the LEDs are effectively in series with the inputs. See also Jack Walton's measurements of the ImPasse preamp, posted here and at DIY Test Equipment for Audio and Ham Radio Enthusiasts .
There are numerous posts here showing how to use them in low current circuits- a bit of searching will turn them up and give you a powerful new tool for your designs; one resistor allows them to be used in circuits with minuscule current draws. I'd also strongly suggest looking at the measurements and circuits shown in Morgan Jones's "Valve Amplifiers."
LEDs (and other solid state diodes) are excellent, stable, reliable, low noise voltage references and are a superb way of biasing tubes in a wide variety of circuits without the need for bypass caps. They solve the OP's problem at minimal cost and complication.
Dissertation on Capacitive Reactance simplified for cathodes.
dscott,
My "dissertation" was inspired by the comment you made about one capacitor not producing an audible change from another.
I was pointing out that the important thing about a bypass cap is its impedance relative to the resistance of the cathode resistor(s).
That the crossover frequency, at which the impedance of the cap and resistor were the same, was important to how audible the effect of putting the cap in the circuit would be.
And pointing out that if the new crossover frequency is beyond the effective range of the old one, there would be no noticeable effect.
For example, once you're passing everything above 40 Hz at ten times the current through the cap as could go through the resistor, you're not going to hear anything by making the cap bigger. And this point is determined by the resistor value, which, at 200 ohms in parallel with 2.2k, is about 180 ohms; a cap with an impedance of about 18 ohms at about 40 Hz (220uF) would be beyond improvement. Many a smaller cap would be effectively beyond improvement, too, since we don't really hear everything the book says we should. And given that the attenuation of the signal from a 180-ohm cathode resistor is negligible to begin with, it may be that ANY cap would be hard to notice. (THIS is why the textbook doesn't always have the right answer to the question!)
The formula I described at length were:
(R1 x R2)/(R1 + R2) = RTotal, for resistances in parallel (Calling the Cap "R2." And explaining that it wasn't that simple, but close enough for this purpose.)
and 1/(2 x 3.14159 x f x (uF/1,000,000)) = R for getting the impedance of a capacitor at a given frequency (where this "R" is actually called "Xc")
Once you get that, the impedance at ten times the frequency is 1/10th of the resistance. At one-third the frequency it is three times the resistance. So you don't have to keep using the longer equation for the same cap.
Where 3.14159 is Pi (IIRC)
f = frequency
uF = microfarads
and 1/(1 x 3.14159 x f x R) = C for getting the cap's value if you know the resistance you want it to be.
Where C is Farads and (C x 1,000,000) = uF.
for those of you out there who do not have degrees in physics.
As long as the R of the cap is ten times smaller than the R of the resistor at your lowest frequency of interest, the cathode resistor will not significantly attenuate the signal at any frequency of interest.
At the frequency where the two are equal, the signal will be audibly attenuated by the resistor; it will be less so as the frequency increases from there and more so as it decreases from there.
At ten times the crossover frequency, the attenuation will not generally be audible.
This is, of course, assuming that the resistor is big enough in the first place to produce an audible attenuation of the signal.
dscott,
My "dissertation" was inspired by the comment you made about one capacitor not producing an audible change from another.
I was pointing out that the important thing about a bypass cap is its impedance relative to the resistance of the cathode resistor(s).
That the crossover frequency, at which the impedance of the cap and resistor were the same, was important to how audible the effect of putting the cap in the circuit would be.
And pointing out that if the new crossover frequency is beyond the effective range of the old one, there would be no noticeable effect.
For example, once you're passing everything above 40 Hz at ten times the current through the cap as could go through the resistor, you're not going to hear anything by making the cap bigger. And this point is determined by the resistor value, which, at 200 ohms in parallel with 2.2k, is about 180 ohms; a cap with an impedance of about 18 ohms at about 40 Hz (220uF) would be beyond improvement. Many a smaller cap would be effectively beyond improvement, too, since we don't really hear everything the book says we should. And given that the attenuation of the signal from a 180-ohm cathode resistor is negligible to begin with, it may be that ANY cap would be hard to notice. (THIS is why the textbook doesn't always have the right answer to the question!)
The formula I described at length were:
(R1 x R2)/(R1 + R2) = RTotal, for resistances in parallel (Calling the Cap "R2." And explaining that it wasn't that simple, but close enough for this purpose.)
and 1/(2 x 3.14159 x f x (uF/1,000,000)) = R for getting the impedance of a capacitor at a given frequency (where this "R" is actually called "Xc")
Once you get that, the impedance at ten times the frequency is 1/10th of the resistance. At one-third the frequency it is three times the resistance. So you don't have to keep using the longer equation for the same cap.
Where 3.14159 is Pi (IIRC)
f = frequency
uF = microfarads
and 1/(1 x 3.14159 x f x R) = C for getting the cap's value if you know the resistance you want it to be.
Where C is Farads and (C x 1,000,000) = uF.
for those of you out there who do not have degrees in physics.
As long as the R of the cap is ten times smaller than the R of the resistor at your lowest frequency of interest, the cathode resistor will not significantly attenuate the signal at any frequency of interest.
At the frequency where the two are equal, the signal will be audibly attenuated by the resistor; it will be less so as the frequency increases from there and more so as it decreases from there.
At ten times the crossover frequency, the attenuation will not generally be audible.
This is, of course, assuming that the resistor is big enough in the first place to produce an audible attenuation of the signal.
I'm aware of that. I had to use a resistor in parallel with a certain LED to get it to go on when it was one leg of a low-current voltage divider. That resistor took enough current to drop the needed threshold voltage, and then the LED was willing to conduct and take whatever more was needed to sustain the voltage drop across its own junction. But the LED still requires its own minimum, so the two in parallel will never get below that.
So I'm not convinced that this would work suitably well in a cathode circuit drawing in the tens or hundreds of microamps. YMMV
I see no need to argue this, other than to point out that it only holds true between certain rather narrow current limits.
But I'll pass on the rest of the statement as a matter of personal preference pending further investigation.
No I didn't. I neglected several. And pointed that out. And said that what I had said was a simplification that was good enough for the present case.
I neglected to mention the current lead and voltage lag of an RC circuit, too.
But I don't have time to write an electronics textbook tonight.
And there was no need for it.
I'd agree. That's not the way to get the LED to run more current. Try running a resistor from the cathode-LED junction to B+.
A typical LED will show low impedance and stable voltage between 5-30mA. Some do better at lower currents than others, but there's the resistor trick. For higher currents, they can be paralleled- see the Red Light District amplifier for an example.
Is it, though? The circuit has a 47k plate load. The plate resistance will be something like 80k, mu is 100. So the cathode impedance is (47 + 80)/101 or about 1.25k. That will mean nearly doubling the predicted value of the bypass cap to achieve the same f3.
#1. When the cathode resistor is 180 ohms? Please.
#2/3. Okay, Sy.
#4.f5. Now that you've proven that you're smarter and more knowledgeable than those to whom I was offering a quick and dirty simplification (in case that was sufficient for their needs),
and implied (whether correctly or otherwise) that you're smarter and more knowledgeable than I am,
can we give it a rest, please?
By the way, "Radiotron Designer's Handbook (4th Edition)" by P. Langford Smith contained that same information (in Chapter 12, pp. 484-487) some 42 years before "Valve Amplifiers" copied it onto pages 78-80.
It can be downloaded at P. Millett's website.
At 820R (looking at the schematic in question), yes.
Wouldn't doubt it. It appears in a lot of other references as well, I just happened to have VA nearby.
Have you ever measured the variation in voltage across an LED when its supply resistor/current is varied?
I have.
It can be substantial.
And there is no way to control for frequency-related effects when one is not using a bypass cap.
(Which is why a previous post about niggling little details of the cap equation stuck me as so completely bizarre, coming from an advocate of using diodes as a replacement for resistor and cap.)
Sometimes the cap is a way of controlling rumble, or hum, or motorboating in the low end. When driving a reverb tank, it can help cut off the low end at just above the mechanical shock-induced thumps while preserving all the sound the reverb tank will usefully process. A bass player might find the bypass cap his best friend when the amp he's been unhappy with needs a few tweaks. A guitarist could get more mileage out of a bypass cap than anyone else in this thread yet suggests they know about. Maybe someday I'll show you what I mean by that. But all those things require choosing a cap of a very specific value relative to the resistor.
There are many things one can accomplish with a resistor and a cap that are impossible to do with a diode. Many others that are not too easy to do with a diode. And others that are just downright foolish to do with a diode. The inner workings of a high-end audio preamp have nothing whatever to do with these things, and relatively little to say about what is important to a guitar.
But I don't want to belabor the issue. The OP asked a question. I've seen good answers to it in this thread. And better ones. And also responses I wouldn't exactly characterize quite either of those ways.
But if we ever need to go back to talking about caps and resistors, I have a graph from a tech manual that would be a whole education in a single picture I could post if I ever decided to scan it.
I have.
It can be substantial.
And there is no way to control for frequency-related effects when one is not using a bypass cap.
(Which is why a previous post about niggling little details of the cap equation stuck me as so completely bizarre, coming from an advocate of using diodes as a replacement for resistor and cap.)
Sometimes the cap is a way of controlling rumble, or hum, or motorboating in the low end. When driving a reverb tank, it can help cut off the low end at just above the mechanical shock-induced thumps while preserving all the sound the reverb tank will usefully process. A bass player might find the bypass cap his best friend when the amp he's been unhappy with needs a few tweaks. A guitarist could get more mileage out of a bypass cap than anyone else in this thread yet suggests they know about. Maybe someday I'll show you what I mean by that. But all those things require choosing a cap of a very specific value relative to the resistor.
There are many things one can accomplish with a resistor and a cap that are impossible to do with a diode. Many others that are not too easy to do with a diode. And others that are just downright foolish to do with a diode. The inner workings of a high-end audio preamp have nothing whatever to do with these things, and relatively little to say about what is important to a guitar.
But I don't want to belabor the issue. The OP asked a question. I've seen good answers to it in this thread. And better ones. And also responses I wouldn't exactly characterize quite either of those ways.
But if we ever need to go back to talking about caps and resistors, I have a graph from a tech manual that would be a whole education in a single picture I could post if I ever decided to scan it.
Yes. And the spectrum of that voltage. And the frequency response. There's a reason that many people use them successfully in their designs.
The impedance is flat with frequency from at least 500kHz (my measurement limit) to DC. Adding a bypass cap has the opposite effect of what you intend.
l'Esprit d'Escalier: it may be that using LEDs instead of the RC network could be a detriment in an instrument amp for the same reason that hifi amps/speaker make incredibly boring instrument amps. Too clean and distortion-free. I freely admit that my experience designing guitar amps is painfully limited.
English Lesson
The OP was asking for help with switching bias resistors and caps for precisely the reason that I elaborated upon: the cap allows CHANGING the frequency response so that it is not flat. And changing the bias resistor shifts the point at which overdrive clipping (distortion) occurs. Unlike audiophiles, guitarists like distortion when it happens at the right level of oomph! And that changes from song to song.
When the low E string is honking (with particular amp and pedal settings that are otherwise desirable) so loud that the low A string can't be heard over it, the cure can be to cut the frequency response below 110 Hz. Solution: cap. The tone control can only work with what it gets and operates at a fixed frequency. Bypass cap cutoff can work with a bass cut control to alter the cutoff frequency and also double the rolloff rate.
Diode? No.
When the chords are too bottom heavy for the jangly sound one is seeking: cap.
When one is seeking an especially thin sound with bass and treble rolled off for a nasal midrange bandpass that works better in a mix than a broader frequency range that competes too much with brass and strings: cap.
When a heavy bottom is preferred and the bass control doesn't go high enough: cap.
From one song to the next, the cap may need to be changed.
Guitarists do things that would seem completely irrational to an audiophile-- things like boosting the bass in one part of the signal chain and cutting it in another (because the two circuits act at different frequencies) to color the sound according to what is pleasing to the ear and changes from good to better to bad to worse as other effects are added into the sonic shaping. Every tool is useful at its right time.
That's why the OP wanted to switch the caps.
Diodes no can do. Eliminate pop? Okay, perhaps. But they eliminate the guitar special effect in the process. Ergo: diode is no solution to the OP's problem.
But, . . .
at least you finally caught a scent of the trail that was being followed on this thread and came out of the woods eventually.
By the way, now that you're in the fairway, it might be helpful to notice which forum your in: Instruments (of the musical variety) and Amps.
From here it's a chip shot with a 5 iron and into the cup if you aim it right and put some English on it.
No, please, wrong again. The impedance of a resistor is the same--flat from zero on up. But it attenuates the input signal. The bypass cap puts the cathode at AC ground without interfering with the bias voltage. And yes, it is not flat, but that's the whole idea here. The capacitor has the opposite effect of what you intend. It has exactly the effect the OP intended. It's the diode that has the opposite effect intended in this thread.
The OP was asking for help with switching bias resistors and caps for precisely the reason that I elaborated upon: the cap allows CHANGING the frequency response so that it is not flat. And changing the bias resistor shifts the point at which overdrive clipping (distortion) occurs. Unlike audiophiles, guitarists like distortion when it happens at the right level of oomph! And that changes from song to song.
When the low E string is honking (with particular amp and pedal settings that are otherwise desirable) so loud that the low A string can't be heard over it, the cure can be to cut the frequency response below 110 Hz. Solution: cap. The tone control can only work with what it gets and operates at a fixed frequency. Bypass cap cutoff can work with a bass cut control to alter the cutoff frequency and also double the rolloff rate.
Diode? No.
When the chords are too bottom heavy for the jangly sound one is seeking: cap.
When one is seeking an especially thin sound with bass and treble rolled off for a nasal midrange bandpass that works better in a mix than a broader frequency range that competes too much with brass and strings: cap.
When a heavy bottom is preferred and the bass control doesn't go high enough: cap.
From one song to the next, the cap may need to be changed.
Guitarists do things that would seem completely irrational to an audiophile-- things like boosting the bass in one part of the signal chain and cutting it in another (because the two circuits act at different frequencies) to color the sound according to what is pleasing to the ear and changes from good to better to bad to worse as other effects are added into the sonic shaping. Every tool is useful at its right time.
That's why the OP wanted to switch the caps.
Diodes no can do. Eliminate pop? Okay, perhaps. But they eliminate the guitar special effect in the process. Ergo: diode is no solution to the OP's problem.
This was so obvious it was painful even to see.
But, . . .
at least you finally caught a scent of the trail that was being followed on this thread and came out of the woods eventually.
By the way, now that you're in the fairway, it might be helpful to notice which forum your in: Instruments (of the musical variety) and Amps.
From here it's a chip shot with a 5 iron and into the cup if you aim it right and put some English on it.
I don't see how using LED's instead of resistors would help.
As far as I can see, the problem is this: if you change the voltage at the tube cathode, the tube's plate current will also change in response. That in turn means the voltage drop across the plate resistor changes. And if that change happens fast enough to make it through the coupling capacitor to the next tube in the amp, you will hear a thump.
I don't see how it matter how you achieve the change in cathode voltage. Whether you use different LED's or different resistors, the results will be the same. If you cause, say, an abrupt 0.3 V change in cathode voltage, the tube always will respond with the same size change in its cathode current. Ergo the change at the plate resistor will also be the same. You'll get the same-size thump from resistors, LED's, or any other method of changing the cathode voltage abruptly by the same amount.
The easiest way I can think of to reduce the thump below audibility is to (a) make sure the coupling cap to the next stage of the amp is no bigger than necessary, i.e., move the lower cut off frequency up as high as possible, and (b) make sure the change in cathode voltage happens slowly (that means using a simple mechanical switch is out).
The first of those two changes reduces the amount of low frequency "thump" that makes it through the plate coupling cap to the next tube.
The second change slows down the voltage change at the plate, shifting its frequency components down to a lower frequency, which is less likely to get through the coupling cap.
I think one suggestion given earlier - use a potentiometer to vary the cathode resistance - might be the easiest way to slow down the change in cathode voltage. If a pot is used, the resistance won't change any faster than a human hand can tweak the pot. Even if that happens in 0.1 second, the associated frequency components will mostly be below 10 Hz. Set up the coupling cap to the next tube stage to roll off everything below, say, 100 Hz, and you should be good. (There are only four notes on a guitar with fundamental frequencies below 100 Hz, and you won't hear any change in tone with a 100 Hz cutoff).
I know the OP said he tried a pot and it made scratching noises, but I think a better pot might solve that issue.
If that doesn't work, i.e. pots all sound scratchy in this application, the other idea I have is to use an LDR (light dependent resistor) as the cathode resistor. Light it with an LED, wrap the pair in black tape or something to keep ambient light out, and vary the current into the LED to change the LDR resistance. The trick is to slow down the current change into the LED, and the easiest way to do this is to put a large electrolytic capacitor in parallel with the LED. If the resistance varies over a tenth of a second or longer, I don't think you'll have much trouble with thumping.
-Flieslikeabeagle
As far as I can see, the problem is this: if you change the voltage at the tube cathode, the tube's plate current will also change in response. That in turn means the voltage drop across the plate resistor changes. And if that change happens fast enough to make it through the coupling capacitor to the next tube in the amp, you will hear a thump.
I don't see how it matter how you achieve the change in cathode voltage. Whether you use different LED's or different resistors, the results will be the same. If you cause, say, an abrupt 0.3 V change in cathode voltage, the tube always will respond with the same size change in its cathode current. Ergo the change at the plate resistor will also be the same. You'll get the same-size thump from resistors, LED's, or any other method of changing the cathode voltage abruptly by the same amount.
The easiest way I can think of to reduce the thump below audibility is to (a) make sure the coupling cap to the next stage of the amp is no bigger than necessary, i.e., move the lower cut off frequency up as high as possible, and (b) make sure the change in cathode voltage happens slowly (that means using a simple mechanical switch is out).
The first of those two changes reduces the amount of low frequency "thump" that makes it through the plate coupling cap to the next tube.
The second change slows down the voltage change at the plate, shifting its frequency components down to a lower frequency, which is less likely to get through the coupling cap.
I think one suggestion given earlier - use a potentiometer to vary the cathode resistance - might be the easiest way to slow down the change in cathode voltage. If a pot is used, the resistance won't change any faster than a human hand can tweak the pot. Even if that happens in 0.1 second, the associated frequency components will mostly be below 10 Hz. Set up the coupling cap to the next tube stage to roll off everything below, say, 100 Hz, and you should be good. (There are only four notes on a guitar with fundamental frequencies below 100 Hz, and you won't hear any change in tone with a 100 Hz cutoff).
I know the OP said he tried a pot and it made scratching noises, but I think a better pot might solve that issue.
If that doesn't work, i.e. pots all sound scratchy in this application, the other idea I have is to use an LDR (light dependent resistor) as the cathode resistor. Light it with an LED, wrap the pair in black tape or something to keep ambient light out, and vary the current into the LED to change the LDR resistance. The trick is to slow down the current change into the LED, and the easiest way to do this is to put a large electrolytic capacitor in parallel with the LED. If the resistance varies over a tenth of a second or longer, I don't think you'll have much trouble with thumping.
-Flieslikeabeagle
I don't want any part of the little, erm, manhood contest going on between SY and tubekit, but I'll point out that I can't see any reason why an LED instead of a cathode resistor will reduce distortion (post 73, "distortion-free").
The fact is that LED's are quite nonlinear - any semiconductor diode (including LED's) has an exponential relationship between the current through it and the voltage across it, not a linear one. Putting an unbypassed LED in the cathode circuit means you have added more nonlinearity to the circuit. This can only increase harmonic distortion, not decrease it.
Using a capacitor / resistor instead of the LED does not introduce any harmonic distortion in this way. Resistors and capacitors are both linear devices. The capacitor will introduce changes in frequency response and phase shift, yes, but those two things are far less "ugly" to the ear than the harmonic distortion from a semiconductor diode's exponential I-V characteristic.
That same exponential diode characteristic, by the way, is responsible for the nasty sound of 99% of solid-state distortion/fuzz/overdrive circuits. Usually in these circuits the diode's nonlinear relationship is inverted, i.e. the voltage across the diode is the logarithm of the current through it, and it is this logarithmically distorted voltage that the circuit puts out.
-Flieslikeabeagle
The fact is that LED's are quite nonlinear - any semiconductor diode (including LED's) has an exponential relationship between the current through it and the voltage across it, not a linear one. Putting an unbypassed LED in the cathode circuit means you have added more nonlinearity to the circuit. This can only increase harmonic distortion, not decrease it.
Using a capacitor / resistor instead of the LED does not introduce any harmonic distortion in this way. Resistors and capacitors are both linear devices. The capacitor will introduce changes in frequency response and phase shift, yes, but those two things are far less "ugly" to the ear than the harmonic distortion from a semiconductor diode's exponential I-V characteristic.
That same exponential diode characteristic, by the way, is responsible for the nasty sound of 99% of solid-state distortion/fuzz/overdrive circuits. Usually in these circuits the diode's nonlinear relationship is inverted, i.e. the voltage across the diode is the logarithm of the current through it, and it is this logarithmically distorted voltage that the circuit puts out.
-Flieslikeabeagle
And yet... it does.
The reasons are pretty straightforward, and actual experiments back it up.First, remember that you're forward biased. That means the the exponential characteristic is already far along the curve, that is, you're away from the bottom part and in the part of the curve that only bends a little. The impedance is the tangent of the forward current versus voltage curve and is a small number, like 5 ohms.
Second, the change in current through a tube with signal is fairly low, typically 3-10 times lower than the idle current. For a 12AX7 with a 100k load that is swinging 10V, the current variation is 0.1mA, and the diode impedance isn't changing much.
Third, remember that the AC impedance is low. Typically for a red LED, it's about 5 ohms. That is reflected in the effective plate resistance, which acts as rp + Rk*(mu + 1). For a 12AX7, rp is something like 80k, mu is 100, so the "effective" plate resistance is modified to be 80,500, not much of a change. Now, we start swinging current and the LED's impedance swings as well. I'll be ridiculous and suggest that it swings by 20% (it will actually vary far less than that). So the effective plate resistance will vary from 80,450 to 80,550. Not much of a change. Take a look at the 12AX7 datasheet. How much does rp vary through that swing? WAY more than 100 ohms! That variation in rp is the primary contributor to distortion in a "normal" common cathode voltage amplifier.
As I pointed out earlier, even that minuscule variation can be swamped out by running (say) 5mA though the LED. So as a practical matter, the LED will provide a bias that's as clean or cleaner than a battery. Using different colors allows the amp designer to vary the operating point of the stage, which is what causes the variation in the distortion and overload desired by guitarists (I'm a hifi amp designer, but I've played guitar for 40 years as well).
Why would you want to reduce distortion? - this is a valve amp, it's whole reason for been there is to give distortion.
Sy - sure, the current swing is only a percentage of DC, but the fact remains that a diode is non-linear, and a resistor is linear. The LED can only increase the amount of harmonic distortion compared to using a resistor. It cannot and does not reduce it, as you originally said it would.And yet... it does.
Also, using LED's with different forward voltages will also not help with the problem the OP was having - the problem being the switching transient that occurs at the tubes anode when you switch in a different grid-cathode voltage, by whatever means.
Bringing in the AC impedance of a diode can be somewhat of a red herring when you're looking at nonlinearity. The AC impedance is nothing more than the local tangent to the exponential diode curve. By definition, the tangent line is linear, so this is a linear approximation to the diode curve (first term in the Taylor series expansion). Therefore it's largely useless for any discussion of nonlinearity.
I have reason to think the amount of nonlinearity is not irrelevant. Exactly the same effect plagues any common-emitter transistor circuit with an unbypassed emittor resistor, because the base-emitter junction of the transistor is also a semiconductor diode. Years ago I used a series approximation to estimate the percentage of various orders (harmonics) of distortion as a function of the input signal level. I no longer remember the exact numbers I calculated, but I do recall that you can get several percent harmonic distortion with an input voltage swing of just a few millivolts.
This result is not very surprising if you remember that the current through the diode is proportional to exp(V/Vt), where Vt = kT/e; k is Boltzmann's constant, T the absolute temperature, e the charge on the electron. Vt turns out is only about 0.026 volts at room temperature. Therefore a voltage increase of only 26 mV across a semiconductor diode will nearly triple the current through it, i.e. it will increase diode current by the factor exp(1), or 2.72 times! By the same token, a negative swing of 26 mV will drop the current to nearly one-third of its original value.
I do agree with you that the tubes internal cathode resistance will "linearise" the diode curve somewhat. However, using a diode can still only increase the total harmonic distortion, not decrease it.
-Flieslikeabeagle
Flies, remember that you're operating on the steep part of the exponential which in fact is quite close to linear. Holding the cathode at AC ground (or just a couple ohms away from that) with a low and flat impedance is beneficial for distortion when you consider the alternative- a big, honkin' electrolytic.
You're right about the change in bias- that is indeed a transient. The LED removes the charge/discharge of the electrolytic from the transient and will reduce it but not eliminate it.
edit: The numbers you give are for silicon, not for the semiconductors used in opto. This part, "a voltage increase of only 26 mV across a semiconductor diode will nearly triple the current through it" makes my argument for me if you turn it around- tripling the current only changes the voltage by 26mV!
You're right about the change in bias- that is indeed a transient. The LED removes the charge/discharge of the electrolytic from the transient and will reduce it but not eliminate it.
edit: The numbers you give are for silicon, not for the semiconductors used in opto. This part, "a voltage increase of only 26 mV across a semiconductor diode will nearly triple the current through it" makes my argument for me if you turn it around- tripling the current only changes the voltage by 26mV!
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