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Signal grid resitors values & output tube distortion???

7/16/2000 11:30 PM
T.B.	Signal grid resitors values & output tube distortion??? Well, I'll try this one more time. AOL has been a real pisser tonight. I wrote a nice long post asking about the values of signal grid resistors and effects on output tube distortion and tube life. Just as I was reviewing my post, I got knocked off. About ready to go to cable or another server real soon! Anyway, if I can remember and clear my now clouded brain, ha, ha. What I've read in recent months is to lower the "typical" 220K resistor to a 100K or even lower to 68K and I've even seen them as low as 48K (silverface twin?) Anyway, that's like installing a "set" master volume as I see it. So the idea with this lower resistor is that we'll be kinder to the output tubes and we will also have more preamp and PI distortion. Is any of this correct? My understanding is that it is. Question now is wouldn't it be easier to drive the output tube and consequently the output transformer by allowing more signal to hit the grids? If that is the goal, then why lower the resistors in the first place? I personally do not crank my amps to the max but on a rarest of occasions so can't see that I'd be burning up tubes any too fast. Also, does not the bias setting control the current and what about the screen grid voltage in relation to the plate, etc.? Bottom line is if a person desires to get more ass from their output tubes wouldn't it make sense to hit them with more juice up front by keeping the grid resistor up to the max? How does this affect P-P and single ended? I'd love to hear from some real pros on this subject, also anyone else who has opinions, ideas, etc.

7/16/2000 11:30 PM

T.B.

Signal grid resitors values & output tube distortion???

Well, I'll try this one more time. AOL has been a real pisser tonight. I wrote a nice long post asking about the values of signal grid resistors and effects on output tube distortion and tube life. Just as I was reviewing my post, I got knocked off. About ready to go to cable or another server real soon!

Anyway, if I can remember and clear my now clouded brain, ha, ha. What I've read in recent months is to lower the "typical" 220K resistor to a 100K or even lower to 68K and I've even seen them as low as 48K (silverface twin?) Anyway, that's like installing a "set" master volume as I see it.

So the idea with this lower resistor is that we'll be kinder to the output tubes and we will also have more preamp and PI distortion. Is any of this correct? My understanding is that it is. Question now is wouldn't it be easier to drive the output tube and consequently the output transformer by allowing more signal to hit the grids? If that is the goal, then why lower the resistors in the first place? I personally do not crank my amps to the max but on a rarest of occasions so can't see that I'd be burning up tubes any too fast.

Also, does not the bias setting control the current and what about the screen grid voltage in relation to the plate, etc.?

Bottom line is if a person desires to get more ass from their output tubes wouldn't it make sense to hit them with more juice up front by keeping the grid resistor up to the max? How does this affect P-P and single ended?

I'd love to hear from some real pros on this subject, also anyone else who has opinions, ideas, etc.

7/17/2000 5:49 AM
JohnC	If the previous stage supplied the signal from its plate then lowering the grid resistor will lower the gain of that stage. This will reduce output tube distortion. Some amps have lower values for grid resistors because they use low gain tubes (like 12au7) that are better for low voltage gain but higher current.

7/17/2000 12:44 PM
T.B.	Hi John, Is it better to have more voltage into the grid of your output tube or more current or is that subjective? I usually think of gain in terms of volts and to this date don't really get the current thing for driving another tube. I understand current of each individual tube and how that can affect tone, etc., but not as a driving force into the next stage. Can you or anyone explain that a little? Thanks, T.B.

7/17/2000 12:44 PM

T.B.

Hi John,

Is it better to have more voltage into the grid of your output tube or more current or is that subjective? I usually think of gain in terms of volts and to this date don't really get the current thing for driving another tube.

I understand current of each individual tube and how that can affect tone, etc., but not as a driving force into the next stage. Can you or anyone explain that a little? Thanks, T.B.

7/17/2000 8:20 PM
Aaron V.	Hi TB, Under idle conditions the grid of the output tube is sitting at a negative voltage - the bias voltage - and lets suppose it is -30V. When the grid is hit with a large signal, say a 40V peak signal, the voltage on the grid will be positive for a period of time. When this happens, some of the negatively charged electrons coming from the cathode will be attracted to the positve charge on the grid. In terms of conventional current, there will then be current flowing from the grid to the cathode, much like the current flowing from the plate or screen. One of the problems here is that the capacitive coupling from the plate of the previous stage cannot feed the grid the current it wants to draw. That is what the cathode follower is for. Try to think of it as still being voltage driven, but we just have to find a way to give the grid the current it wants. IOW, the current isn't doing any driving, it is just a side-effect of hitting the grid with such a big voltage. Look on GEO under the Mosfet Follies section for a more detailed description of the negative things that happen when driving the grids positive via capacitive coupling. I think RA might have a description on his site too. HTH, Aaron

7/17/2000 8:20 PM

Aaron V.

Hi TB,

Under idle conditions the grid of the output tube is sitting at a negative voltage - the bias voltage - and lets suppose it is -30V. When the grid is hit with a large signal, say a 40V peak signal, the voltage on the grid will be positive for a period of time. When this happens, some of the negatively charged electrons coming from the cathode will be attracted to the positve charge on the grid. In terms of conventional current, there will then be current flowing from the grid to the cathode, much like the current flowing from the plate or screen. One of the problems here is that the capacitive coupling from the plate of the previous stage cannot feed the grid the current it wants to draw. That is what the cathode follower is for.

Try to think of it as still being voltage driven, but we just have to find a way to give the grid the current it wants. IOW, the current isn't doing any driving, it is just a side-effect of hitting the grid with such a big voltage.

Look on GEO under the Mosfet Follies section for a more detailed description of the negative things that happen when driving the grids positive via capacitive coupling. I think RA might have a description on his site too.

HTH,

Aaron

7/19/2000 7:27 PM
JohnC	I'm not really an expert on these things so I dont want to overstate what I know. The job of the signal amplifying stages (the non output stages) is merely to amplify voltage. To get the maximum output of the output stage you want to provide more voltage swing onto the grids of the output tubes. In a perfect world the grids would be and infinite impedance to any other circuit and would swing the voltage without drawing any current. In the real world there are leakage currents and interelectrode capacitances which will draw current as a voltage changes on them. Therefore in reality the driver stage needs to supply some current as well as voltage. If the valve in the driver stage does not pass alot of current (eg a 12ax7) then the voltage amplification of that stage will sag. Its output is loaded down by the low impedance of the stage after it. For high outputs or paralleled output valves it is better to look at valves that pass more current such as 12at7s and 12au7s. These have lower gain but can supply the current needed by high output stages without sagging. I hope this is of some help John

7/19/2000 7:27 PM

JohnC

I'm not really an expert on these things so I dont want to overstate what I know.

The job of the signal amplifying stages (the non output stages) is merely to amplify voltage. To get the maximum output of the output stage you want to provide more voltage swing onto the grids of the output tubes.

In a perfect world the grids would be and infinite impedance to any other circuit and would swing the voltage without drawing any current. In the real world there are leakage currents and interelectrode capacitances which will draw current as a voltage changes on them. Therefore in reality the driver stage needs to supply some current as well as voltage.

If the valve in the driver stage does not pass alot of current (eg a 12ax7) then the voltage amplification of that stage will sag. Its output is loaded down by the low impedance of the stage after it.

For high outputs or paralleled output valves it is better to look at valves that pass more current such as 12at7s and 12au7s. These have lower gain but can supply the current needed by high output stages without sagging.

I hope this is of some help

John

7/17/2000 11:03 PM
Stephen Conner	Actually the value of that grid resistor is a compromise between two things. For maximum drive to the output tube grid, it should be as high as possible. But, to soak up grid current and prevent thermal runaway, which can destroy the power tubes, it should be as low as possible. This is especially important if you're driving the tubes hard and causing them to heat more. That's why you're advised to swap the 220K for a 100K. With tubes like EL34s, you will still have enough signal left to push them right into hard clipping. Steve C.

7/17/2000 11:03 PM

Stephen Conner

Actually the value of that grid resistor is a compromise between two things. For maximum drive to the output tube grid, it should be as high as possible.

But, to soak up grid current and prevent thermal runaway, which can destroy the power tubes, it should be as low as possible. This is especially important if you're driving the tubes hard and causing them to heat more. That's why you're advised to swap the 220K for a 100K. With tubes like EL34s, you will still have enough signal left to push them right into hard clipping.

Steve C.

7/19/2000 5:15 PM
SpeedRacer	*Re: Anyone with a real understanding* of power tube distortion?** Hey Steve - I've had the oppotunity to spend some quality time with Dr. Z over the years and one thing he advocates is what he terms "gain scheduling". In a nutshell, what parts distort when. There are 3 basic topologies if you view the amp as 2 objects - pre-amp and power amp joined typically at the input to the phase inverter... You have: 1. Pre distorts first (clean Power) 2. Power distorts first (clean pre) 3. Pre and Power both clip at the same signal level Which of these your circuit matches will determine a lot about how it sounds and feels in overdrive. It not only requires tweaking but a good look at the rail voltages, gain stages etc.. too many folks don't really think it all out to this level IMHO. (maybe they do?) Anyhow, I think it is critical to how the whole amp goes into overdrive and how it feels to the player as it does. Is it dynamic and punchy or is it squishy, etc. Also, in amps like 4-pronger Marshalls, I think the pre-amp breaking up first is key since the waveform "excites" the output tubes into doing what they do best. Send the EL34's a clean signal and they don't act the same. (or you could say that the outputs "process" the dirty pre-amp signal based on their own operating parameters) I don't have the calculus to get into it, but my pet theory is that set of typical harmonics you get from the pre-amp excites the output circuit into producing a complementary set of harmonics via intermodulation distortion etc. One without the other is "ok" but not inspiring. Both together in a well set up amp is nearly a religious experience. It's not "Even" or "Odd" but rather the dynamic mix of them you get as a result of pre-amp and power-amp working off each other.

7/19/2000 5:15 PM

SpeedRacer

Re: Anyone with a *real understanding* of power tube distortion?

Hey Steve -
I've had the oppotunity to spend some quality time with Dr. Z over the years and one thing he advocates is what he terms "gain scheduling". In a nutshell, what parts distort *when*. There are 3 basic topologies if you view the amp as 2 objects - pre-amp and power amp joined typically at the input to the phase inverter...
You have:
1. Pre distorts first (clean Power)
2. Power distorts first (clean pre)
3. Pre and Power both clip at the same signal level

Which of these your circuit matches will determine a lot about how it sounds and feels in overdrive. It not only requires tweaking but a good look at the rail voltages, gain stages etc.. too many folks don't really think it all out to this level IMHO. (maybe they do?) Anyhow, I think it is critical to how the whole amp goes into overdrive and how it *feels* to the player as it does. Is it dynamic and punchy or is it squishy, etc. Also, in amps like 4-pronger Marshalls, I think the pre-amp breaking up first is key since the waveform "excites" the output tubes into doing what they do best. Send the EL34's a clean signal and they don't act the same. (or you could say that the outputs "process" the dirty pre-amp signal based on their own operating parameters) I don't have the calculus to get into it, but my pet theory is that set of typical harmonics you get from the pre-amp excites the output circuit into producing a complementary set of harmonics via intermodulation distortion etc. One without the other is "ok" but not inspiring. Both together in a well set up amp is nearly a religious experience. It's not "Even" or "Odd" but rather the dynamic mix of them you get as a result of pre-amp and power-amp working off each other.

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