Bottlehead Forum
General Category => Technical topics => Topic started by: anthony on October 21, 2009, 07:30:17 AM
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Hello again!
Same topic, third thread. Previously, on this thread:
The thread scrolled off the screen so I'm bumping it up to a new one.
I said "...This creates a voltage below ground, which biases the grid of the output triode to a negative voltage." and you asked "Is this because the high negative voltage below Rg yields a positive voltage at the Cc-Rg-Rstop node which then gives you a negative bias at the second tube's grid because of Rstop?"
The grid is connected to the "-bias" node through Rstop and Rg (diagram in the manual). Since the grid draws no current, and that resistor chain is isolated from the driver by Cc, the grid DC voltage is the same as "-bias". Now you have to look at the power supply at the same time; you will see that all the power supply current goes through Rbias. Most of it is the output triode current, about 27mA if I recall correctly. That current, through that resistor places "ground" about 18 volts more positive than "-bias". Thus the grid is about 18 volts below the cathode which is grounded.
"...does Cpf stand for parafeed capacitor?" Yes, I tried to label the components with descriptive names.
Generally "cathode bias" means the cathode is raised above the grid which is grounded, so my use of the term is not precise. Another meaning is that the cathode current generates the bias voltage (through a resistor usually; sometimes we use an LED for another example) - this is the way I used it. Sorry for the confusion! Anyhow, if the cathode current is too high, the resistor Rbias will drop more voltage, placing a more negative voltage on the grid and counteracting the excess current. That is the negative feedback action. Notice that Rbias is shunted with a large capacitance; that makes the negative feedback ineffective at audio frequencies; it only works at DC.
Hope that helps!
@Paul Joppa: I've had a chance to ponder and review that last post in the now-archived forum. Thanks again. I get what you're saying about setting the -bias at the output triode's grid now and the Cc blocking the current and isolating the two resistors. Very cool idea haha. While I was trying to understand the bit about Rbias, which I get now, I came across another question, how did you decide on the value for Rbleed in the supply?
The feedback operation makes a lot more sense now too, thanks, but how can the cathode current get too high, in other words, why is this necessary?
thanks as always,
-Anthony
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Rbleed is easy - I just chose a value that would dissipate less than 1/3 watt so we could use inexpensive 1-watt resistors. Bleeders should never fail since they are a safety part to discharge the power supply capacitors. And resistors at full rated power get extremely hot, another safety factor. So I de-rate resistors by a factor of 3 most of the time.
Tubes will vary in their parameters from one tube to another, and will change as they age. The variation is typically 30%, much better than transistors which will vary by a factor of 5, but still enough that the tube's current for a given bias may be quite different from what is shown on the published curves. So tubes either use negative feedback (also called "automatic bias" in the old days) or they have an adjustable bias voltage and a means to monitor the current periodically.
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Excellent, this all makes sense. I forgot about the fact that the tubes "warp" over time. Man, going back to the impetus of this thread, there is so much to consider when designing an amp, with or without tubes. I'd love to get to the point where I can design something like the S.E.X. amp that is effective and eloquent, but the road seems difficult and time-consuming. Do you have any final advice for me? I think that building this kit and talking to you was extremely helpful regardless, so thanks for everything!
-Anthony
PS- I hope you would be as willing to answer any further questions I may come up with as you have been so far (crosses fingers)!
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Hello,
I hope that I have not worn out my welcome with questions, but I have a few more. One, that I posted in the wrong spot, I will repost here, and this is, why is there no DC-blocking cap at the input? Is it just not necessary?
Another question I have is: I thought that the grid of the first stage would need to be negatively biased to operate in a linear way, yet looking at the schematic, that doesn't seem to be the case. Am I missing something obvious? (here I was, thinking I understood the input stage. sigh.).
Lastly, does it matter that the input stage triode in the 6dn7 is an oscillator, according to the datasheet? Is that just what they intend it for?
If anyone can find time and patience for these questions, I would be very appreciative.
thanks,
-Anthony
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Describing the first half as an oscillator gives the reader of the data sheet an idea of the triode's ideal application. If you look at various data sheets for the 12AT7, you will also see it referred to as an oscillator.
Grid bias for the two halves of the 6SN7 is accomplished in two different ways in the circuit. The amplifier triode gets a negative bias applied to its grid, and the driver half pulls current through its cathode resistor to bias the cathode above the grid.
If we wanted to also apply a negative voltage to the grid of the driver triode, we would indeed want a capacitor or a transformer at the input of the amplifier. With cathode bias, however, the driver triode operates as long as the grid voltage is close to zero.
In a solid state amp, there is not nearly the amount of squeeze room that a tube amp has. A couple of volts at the input of an op amp might cause a minor explosion, while accidentally biasing the driver grid on a 6DN7 will give you distortion and a blown fuse.
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@2wo: Got it, so there is no input cap needed since there is no DC bias there? And you are right on about the cathode resistor... I should have been able figured that out myself. Thanks though!
@caucasian blackplate: nice handle. looking at the schematic, i can see exactly what you are talking about in the first part of your response, it goes along with 2wo's response. I am slightly confused though at the second part. If we were using negative grid bias instead of cathode bias, why then do you want a cap/transformer? Is it to protect the audio source from the negative voltage? Also, and this is probably just a semantics thing, but why is the first tube the "driver" tube and the second one the "amplifier" tube?
@Paul Joppa: you said earlier " Anyhow, if the cathode current is too high, the resistor Rbias will drop more voltage, placing a more negative voltage on the grid and counteracting the excess current." I thought a more negative grid voltage would increase cathode current, not decrease it. I understand now though, how the cathode current generates the bias.
Thanks for your responses, you are all helping a young engineering hopeful understand analog concepts and vacuum tube technology and I am very grateful!
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@caucasian blackplate: nice handle. looking at the schematic, i can see exactly what you are talking about in the first part of your response, it goes along with 2wo's response. I am slightly confused though at the second part. If we were using negative grid bias instead of cathode bias, why then do you want a cap/transformer? Is it to protect the audio source from the negative voltage? Also, and this is probably just a semantics thing, but why is the first tube the "driver" tube and the second one the "amplifier" tube?
The cap/transformer at the input with negative grid bias will protect the source components from seeing that voltage. As for the driver/amp nomenclature, we generally refer to the amplifier half as the half with lower plate resistance and less mu. A driver stage typically has more gain but less ability to driver lower impedance loads (high Rp, low plate current).
Some tubes blur the line (10Y, 6C45, etc), so this doesn't always work. Also, you might use a 300B as a driver for an 845...
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One thing that probably hasn't been stressed enough here is that we tend to ignore reference when we refer to things like the Voltage at thus-and-such node in a circuit, probably as a form of verbal shorthand because everyone "knows" what the reference point is by custom when discussing the various parts of a circuit.
For instance, when discussing negative bias on the grid of the tube, it is more accurate to say that the Voltage developed on the grid is negative with respect to the cathode. While "ground" is often the reference point, in the case of the input stage to the SEX amp (and many, many other common tube amplifier stages), the Voltage on the grid is only negative with respect to the cathode. As others have explained, this is accomplished in practice by making the cathode positive with respect to ground while the grid is held essentially at ground potential.
You can, of course, verify this with your meter: Measure the Voltage from grid to ground, then cathode to ground. You will find little if any Voltage at the grid with respect to ground. At the cathode, you will find positive Voltage with respect to ground. Then, measure between the actual grid pin and the actual cathode pin. The difference you will see in the potential between these two points is the bias. If you use the cathode pin as the reference point (black lead on your DMM), with the red lead on the grid, you will see "negative" Voltage.
It is important to note, again, that the "polarity" of the Voltage is also relative depending on which meter lead you apply to which pin. With a DMM, you can switch the red and black leads back and forth, and the only difference will be whether there is a + or a - before the reading. Back in the day when mechanical meters often only went in the positive direction from zero, it was rather important to make a good guess as to which direction the Voltage was going before you attached the leads of the meter to the circuit! Most meters would not react well to a "negative" Voltage applied for a long period of time, and the higher the Voltage the more severe the reaction might be!
All of this is to point out that a Voltage reading taken with a meter is literally only the difference in potential between the two leads of the meter. It is only pertinent information in the context of that very exclusive little world.
In the larger sense, a Voltage is always nothing more than the potential difference in electro-motive force between two specified points; it is always relative to that context and essentially meaningless otherwise.
Another good example in your tube data sheet is the "Maximum Plate Voltage" specification; although the reference point may just be assumed, it means "Maximum Plate Voltage with respect to the Cathode". IOW, if the potential on the Cathode is 100 VoltsDC with respect to ground, and the potential on the Plate is 300 VoltsDC with respect to ground, then the potential difference between the Plate and the Cathode is 200 VoltsDC. If the Cathode is the reference point, then the Plate will be said to be 200 Volts positive with respect to the Cathode.
I apologize if this seems a bit pedantic, but, trust me, as you gain more experience with circuit analysis or even just troubleshooting, these fine points will take on more importance.
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@caucasian: thank you much for clearing this up for me.
@JC: While this hasn't been a big problem for me, i know exactly what you mean. At first i had no idea that grid voltage was usually referenced to cathode voltage. Most of the tube-amp hobbyists and professionals who take time to go into online forums have an advanced knowledge of these topics, so a lot of the simpler concepts are left out. For someone like me who is a beginner, this can be very confusing. The people here seem to understand that i'm a newbie (i wonder how they ever figured that out! haha) and are considerate enough to explain things to me that would be taken for granted by them and their peers. So I thank you for raising this point.
A great example of what I mean is, why does positive voltage on a cathode even work? My understanding of vacuum tubes (or a triode, at least) is that the cathode filament is heated and this releases electrons into the vacuum which are then attracted to the positive anode and this connection generates a current through the device. So if you apply a positive bias at the cathode, why don't the negative electrons get attracted to it and not even bother with the anode? Im sure there is a valid explanation for this, maybe cause the typical anode potential is much stronger than a typical cathode bias, but you can see why it is confusing.
Now I'm not trying to say that this is something newbies like us cannot look up on our own (and i intend to after this post). I completely agree that someone constantly asking questions that are easily answered by wikipedia or a google search gets annoying and wastes everyone's time. All I am saying is that when i ask a specific question that i can't look up, I really appreciate when people don't assume that I know as much as they do.
Getting back on topic, i feel like i'm rapidly gaining a greater understanding of vacuum tube amplifiers based on this thread and the previous one, on the old forum. Thanks again all.
(though i'm still curious about how more negative voltage at the grid of the second stage decreases cathode current, if anyone feels like explaining that :D )
-Anthony
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Well, at least on this forum, an intelligent question is always welcome. The only questions that tend to get curt answers are ones on which there is ample discussion already available in the archives, and even the short answers are generally polite.
With regard to the triode amplifier with cathode bias, one needs only go back to the previous discussion: Voltages are measurements of electro-motive force and are always relative to the potential difference between two points. In the case of the cathode biased triode amplifier, let's say that you have such a circuit with +100Vdc on the plate with respect to ground, and the Voltage drop across the cathode resistor is +1Vdc with respect to ground. That puts the potential on the plate at +99Vdc with respect to the cathode. So, as far as the electrons being emitted by the heated cathode are concerned, they see the two other electrodes as follows: The grid, which since it is held at ground potential, is 1Vdc negative with respect to the cathode and therefor tends to repel the electrons back toward the cathode. And, the plate which is at +99Vdc with respect to the cathode and therefor tends to attract the electrons toward itself.
Think of it in terms of a scale from 0 to 100; 1 is higher than 0 by one unit and 100 is higher than 1 by 99 units. Since the electrons are being emitted by the element that is at 1, they are naturally attracted to the element that is 99 units more positive.
The important point to recognize is that the electrons react according to the potential difference between the cathode and the other elements in their own universe and the relative potential differences between those elements. IOW, as far as the electrons are concerned, they themselves are the baseline and they don't know from that Volt on the cathode because that is where they came from; they only react to the potential differences on the other elements in their universe with respect to themselves.
Another thing to recognize is that relative physical proximity also has an effect on the influence these elements have on the electron flow. The grid, for instance, is much closer to the cathode than the plate, and therefor its potential has a much more pronounced effect on the electrons at a given potential than the plate can have. That is why the 1 Volt of bias on the grid can set the idle (no signal) current flow from cathode to plate: That one Volt of potential has more influence due to its proximity relative to the influence of the potential on the farther-away plate.
It may be worth your while to hunt up a copy of "The Beginner's Guide to Tube Audio Design" by Bruce Rozenblit. It is a great addition to all the info you can find on line, IMO, when it comes to the basics of how tube audio amps work.
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JC: Thank you for this. here's a potentially intelligent question:
It seems to me that the resistance between the grid and ground is 220 ohms plus whatever is left on the pot (between 0 and 100k). Doesn't this create a high pass filter with the coupling cap after the first stage? with a .1uF for the Cc, and let's say 20k from grid to ground (relatively low volume), wouldn't the lower roll-off freq be about 80Hz (assuming fc@ Xc=R) and get worse (higher) as the volume (grid to ground resistance) decreases? Why not increase the 220-ohm resistor? Is there something I'm missing, or maybe I'm just wrong about the High-Pass effect?
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I am afraid I don't understand your question. Presumably, you are referring to the resistances of the volume pot and the 220 Ohm grid-stopper resistor at the input to the grid of the tube. If so, I can only say that these interact chiefly with the tiny inter-electrode capacitances (grid-to-cathode and grid-to-plate), but the "active" tube is in between the input resistors and the output coupling cap, so I don't believe there can be much interaction there.
What the coupling cap does interact with, of course, is the dynamic plate resistance of the tube, so that must be considered when deciding on the value of the coupling cap.
Again, I probably don't understand your question.
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I see, so what I thought occurred was that a filter would be generated by the grid resistance to ground and the coupling capacitor at the anode. If there is no interaction, then I'm worried about nothing. This is where is got the idea: (I'm reading from the 3rd edition of Morgan Jones' "Valve Amplifiers") "The traditional value of 1 Megohm grid leak [grid to ground] and 0.1uF [output coupling capacitor] forms a filter whose -3dB point is 1.6Hz." And he is describing the same common cathode amp stage as the driver stage of the S.E.X., so I don't know.
One thing I have always wondered though, and I'm totally switching gears again, is how does the power supply get the 350 and 366 volts for the B+'s? I see 160V at the secondary on the schematic, and I understand that the supply is supposedly a voltage "doubler" but that's still only 320 volts. I did measure those high voltages, so I know i'm missing something here... those sneaky extra volts are coming from somewhere.
well, again, thanks for answering me, I really appreciate it!
-Anthony
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The cap/resistor combo does form a high pass filter, but it is the cap and the resistor that follows. (.1uf and 249k in the SEX amp)
The doubler can be simulated in PSUDII or other software to get the output voltage. A lot depends on other transformer parameters besides the AC output voltage. For example, the extra steps taken to make the Eros transformer uber quiet also cause it to produce lower B+ through the doubler than the SEX amp transformer.
If you measure the DC resistance of the winding and the AC voltage with no load, that tends to provide enough information to get a good value from PSUD.
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Just a WAG, but I expect this refers to the grid resistor on the following stage, as CB points out.
I note that Morgan Jones identifies this as the "grid leak" resistor, which is a common phrase from some time ago; it is not much in current usage, and neither are grid resistors of that large a value from what I can tell. But, the use of the term "grid leak" is evocative of an interesting design point: Search up some info on "grid leak bias", which is yet another method by which the negative bias on the grid can be established without the use of a cathode resistor or a negative bias supply.
This method is also not much in use in current designs, but you may possibly run into it in older gear. I used to see it on the input stages of old guitar amps all the time, and it was a real puzzler until I learned more about it!
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Oh, and excuse me, I neglected to respond to your power supply question. CB offered good advice, and I will only add that it is important to remember that the secondary Voltages specified for the transformer are generally given as RMS Voltages. In the power supply, after rectification, the caps charge to the peak Voltage of the AC, which is, of course, much higher. A good rule-of-thumb calculation for converting RMS to Peak is to multiply by a factor of 1.414.
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@caucasian blackplate: I see. that makes a lot more sense than the way I was looking at it (the filtering effect)
as for the power supply, i guess i'm just ill-equipped to do my own measurements. I brought my multimeter to work and it was stolen :( also I do not own an o-scope and i can't run useful software since my PC won't turn on (i'm on my macbook, which i typically use for audio recording and editing). So again, I apologize for my inability to do some of this work myself. I was going to steal someone else's meter at work, but I decided that would be uncool, so i just gotta suck it up and buy a new one i guess, and not bring it to work!
What you and JC said, however is very helpful. Sometimes I don't know when to use RMS and p-to-p, i guess, but thanks for pointing that out, JC. One thing JC, what is a WAG? hahaha
I have one more question that I've been meaning to ask, that I haven't been able to find the answer to elsewhere. In the filament heater circuit, why is the resistor only 0.1 ohms? What is the purpose of it?
Again, I can't thank you guys enough for all your help. The reason I'm so curious here is cause after I built the kit, I wanted to try to design an amplifier myself. I started (back when I was in college and had the necessary equipment) and soon realized that I have a significant amount still to learn before I can build something that will work as intended haha. I think another kit build will be a good idea (more $ for bottlehead haha), so i've been saving up for that and looking forward to my next project.
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Wild Ass Guess!
I can only imagine that the .1 resistor is in there to drop a tiny bit of voltage to keep the result at the tube pins at or somewhat below the 6.3 Volts specified for the 6DN7 tubes.
As you can see, after rectification and filtering, the DC ends up closer to the peak of the waveform coming out of the secondary of the transformer. When the load (tube filaments) is applied, it no doubt lowers the voltage due to the fact that the secondary Voltage on the transformer drops according to how much load is applied to it and how well it maintains its Voltage under load.
I suspect in this case that even the load was not quite enough to lower the Voltage to the safe level, so a small resistance was needed to lower it a bit further.
This will be a good exercise in the use of Ohm's Law. I am away from home, so I don't have the data readily available, but if you do you can use the heater amperage specification x 2 for the current flowing through the resistor. That multiplied by the .1 value of the resistor will tell you how much Voltage it is dropping and, therefor, about how much the Voltage would probably be too high if it were not in there.
Of course, .1 Ohms doesn't seem like much resistance, but since the amount of Voltage it drops is also a factor of how much current is flowing through it, it can be significant.
I have often had to use fractional resistors like this in old guitar amps to lower the heater supplies, mostly AC ones. See, the amps were designed and built at a time when the typical AC Line Voltage in the US was 117 VAC or less. So, the transformers were specified to operate at that input Voltage, not the 120 - 125 VAC typical today. That increase on the input translates to a corresponding increase on the secondaries of the transformer, so often an older amp was applying heater Voltages that were higher than +10% of nominal. This not only shortened tube life, as you can imagine, but the extra heat also caused the cathodes to emit more which could alter the operating points enough for audible differences.