Trying to interpret the LTP.

[Mike_J]

Reading an exciting story by Aiken entitiled "Designing for Global Negative Feedback." Start thinking this is pretty confusing for a retired CPA. Think I took about one days worth of log math instruction in college. Not really a prerequisite for a CPA.


Anyway, I thought better try to make this simple so I can understand it. So I start out thinking I am standing in the backfield which in this case is the junction of both cathodes. I run ahead and smear this little 470 ohm defensive back in front of me but now it is getting serious. See two monster defensive ends, both 1M and another comparatively small 10K defensive back. Realize I don't want to be tackled by the big boys so I hit the defensive back for all I am worth, kind of the path of least resistance. This carries me into the end zone which is ground to complete the circuit.


Then I start thinking that makes no sense. The heater is supposed to heat the cathodes which get excited so when a grid signal comes in they are ready to go. Cathode sends electrons to the grid. A lot of electrons go past the grid and hit the plates of the tube. Sounds like the only way that happens is if we start from ground and work the other way.


Obviously some energy is getting to or maybe it is from the tip of the OT jack then through the 27K resistor and on to the 10K tail resistor where they see the little 470 ohm DB and the huge 1M ohm defensive ends and has a very quick decision to make. Perhaps the defensive ends get a piece of him and some electrons get back to the grids of the LTP tube.


Guessing most of the signal hitting the LTP is coming through the coupling caps to the grids of the LTP tube. Still would need to get their negative power from that ground at the end of the tail resistor I would think.


Please clear up any confusion I may have concerning this matter.


Thanks
Mike

[pdf64]

Provided its heater is powered, the cathode is surrounded by the electron cloud it generates, regardless of whether there’s any (HT) current flowing.
Once we connect up its electrodes and apply HT, then HT current can flow, the valve can assume its operating point and be ready to handle signal.

[Mike_J]

The plates of the LTP are fed their power from two plate resistors filtered from the positive end of a 22uF @ 500VDC ecap. So by extension the ground of the 5K tail resistor pot must be the grounding point for the negative end of the aforementioned ecap. Only question now is does the negative lead of the OT jack take a little visit to the same tail resistor pot ground since the positive side of the OT has made an appearance in the area. I believe Aiken said as much but I don't have time to address it because it is time to go to church.


Thanks
Mike

[Mike_J]

Provided its heater is powered, the cathode is surrounded by the electron cloud it generates, regardless of whether there’s any (HT) current flowing.

Understood. One is only partially related to the other.


Thanks
Mike

[pdf64]

...Only question now is does the negative lead of the OT jack take a little visit to the same tail resistor pot ground...

The principle of the star point is that all 0V returns in that system connect back only to there. The only exception being with 2phase rectification; there the winding’s CT should connect directly to the reservoir cap terminal.

[2deaf]

Reading an exciting story by Aiken entitiled "Designing for Global Negative Feedback." Start thinking this is pretty confusing for a retired CPA. Think I took about one days worth of log math instruction in college. Not really a prerequisite for a CPA.

Anyway, I thought better try to make this simple so I can understand it. So I start out thinking I am standing in the backfield which in this case is the junction of both cathodes. I run ahead and smear this little 470 ohm defensive back in front of me but now it is getting serious. See two monster defensive ends, both 1M and another comparatively small 10K defensive back. Realize I don't want to be tackled by the big boys so I hit the defensive back for all I am worth, kind of the path of least resistance. This carries me into the end zone which is ground to complete the circuit.

Then I start thinking that makes no sense. The heater is supposed to heat the cathodes which get excited so when a grid signal comes in they are ready to go. Cathode sends electrons to the grid. A lot of electrons go past the grid and hit the plates of the tube. Sounds like the only way that happens is if we start from ground and work the other way.

Obviously some energy is getting to or maybe it is from the tip of the OT jack then through the 27K resistor and on to the 10K tail resistor where they see the little 470 ohm DB and the huge 1M ohm defensive ends and has a very quick decision to make. Perhaps the defensive ends get a piece of him and some electrons get back to the grids of the LTP tube.

Guessing most of the signal hitting the LTP is coming through the coupling caps to the grids of the LTP tube. Still would need to get their negative power from that ground at the end of the tail resistor I would think.

Please clear up any confusion I may have concerning this matter.

This sort of reminds me of Roy "Wrong Way" Riegels and Bennie "50/50 Chance" Franklin (what team did he play for?).  It makes a lot more sense if you consider the current to be flowing in the direction that it actually is flowing instead of the way conventional current has it.  I heard the argument from the point of view of the Physics Dept. and the EE Dept.  The Physics Dept. had way far and away the better argument.

Electrons flow from negative to positive so that the overall real current flows from ground to B+.

The grid current under normal operating conditions is very small and it is customary to consider it zero (at least by me).  Nobody gets by those two steroid-enhanced All-Pro 1M defensive ends.

     

[PRR]

The NFB runs through the "Grid 2 Capacitor".

Yes, a teeny amount of NFB flows through the commingled tail. But this is a minor mistake which Leo liked.

[2deaf]

Yes, a teeny amount of NFB flows through the commingled tail. But this is a minor mistake which Leo liked.

The 4.7K commingled tail in the attached diagram creates a bootstrapping effect that makes the constant current resistance appear larger while preserving the headroom.  Figure 5A and 5B have nearly the same idle conditions, gain and headroom, but the difference in output balance is significant due to this bootstrapping effect.

Figure 5B requires a 0.027mA imbalance between the stages in order to get the required 0.68V change across 25.17K of tail resistance.  Figure 5A only needs to raise the tied cathode voltage by 0.19V across 20.47K of resistance because the other 0.54V has already been supplied by the 4.7K NFB resistor.  So the impedance of the tail acts like it is much larger in that it only needs 0.009mA to get the required voltage change at the tied cathodes.  If you wanted to get the same balance with Figure 5B, it would require a 75K tail resistor which would cause an unacceptable reduction in headroom.



   

[Mike_J]

Yes, a teeny amount of NFB flows through the commingled tail. But this is a minor mistake which Leo liked.

The 4.7K commingled tail in the attached diagram creates a bootlegging effect that makes the constant current resistance appear larger while preserving the headroom.  Figure 5A and 5B have nearly the same idle conditions, gain and headroom, but the difference in output balance is significant due to this bootlegging effect.

Figure 5B requires a 0.027mA imbalance between the stages in order to get the required 0.68V change across 25.17K of tail resistance.  Figure 5A only needs to raise the tied cathode voltage by 0.19V across 20.47K of resistance because the other 0.54V has already been supplied by the 4.7K NFB resistor.  So the impedance of the tail acts like it is much larger in that it only needs 0.009mA to get the required voltage change at the tied cathodes.  If you wanted to get the same balance with Figure 5B, it would require a 75K tail resistor which would cause an unacceptable reduction in headroom.



   

Some day I am going to understand what you just said and I hope it is soon. Pretty sure it might have something to do with my quest to balance my LTP.


Thanks
Mike

[pdf64]

The short form paraphrasing of that might be that (within the linear range) compared to its open loop performance, the global NFB loop will act to improve the LTP’s balance.
We can’t rely too heavily on that though, as the downside is that valve guitar amps tend to get pushed beyond their linear range, and when overdriven, the NFB’s corrective abilities are not able to operate.

[Mike_J]

The short form paraphrasing of that might be that (within the linear range) compared to its open loop performance, the global NFB loop will act to improve the LTP’s balance.
We can’t rely too heavily on that though, as the downside is that valve guitar amps tend to get pushed beyond their linear range, and when overdriven, the NFB’s corrective abilities are not able to operate.

Are we talking about more headroom by balancing the pi? If so, probably not an ideal situation since the tweed Bassman is loud as all get out already. Thankfully that is why they made potentiometers. Figure I could replace the first tail resistor with a pot and dial in what I like best. May not be anywhere near balanced. Need some headroom but there is a limit with an amp that is that loud already.


Thanks
Mike

[2deaf]

The short form paraphrasing of that might be that (within the linear range) compared to its open loop performance, the global NFB loop will act to improve the LTP’s balance.

The main upshot of Reply #7 was that Leo's method of inserting NFB with a LTPI, as shown in Figure 5A, has an advantage as compared to the method shown in Figure 5B with regard to LTPI balance.  Any paraphrase of Reply #7 really should include that point. 

Reply #7 didn't compare open loop performance with closed loop performance, but then again you didn't specifically reference Reply #7.  However, there is nothing in the entire thread up to Reply #9 that notes how global NFB improves LTPI balance as compared to open loop performance.  This seems to be a stand-alone observation as opposed to a paraphrase.

We can’t rely too heavily on that though, as the downside is that valve guitar amps tend to get pushed beyond their linear range, and when overdriven, the NFB’s corrective abilities are not able to operate.

And yet when reasonably matched output tubes are used, the overdriven output signal of the amplifier is markedly balanced.   

[Mike_J]

Finally something I can understand. I can purchase and measure power tubes to see that they are within reasonable balance.


Thanks
Mike

[Soulfetish]

The NFB runs through the "Grid 2 Capacitor".

Yes, a teeny amount of NFB flows through the commingled tail. But this is a minor mistake which Leo liked.

Interestingly, it's a teeny amount of positive feedback that gets inserted to the commingled tail. In the Fender/Marshall design, it has a slight bootstrapping effect, but it is largely insignificant with respect to the level of negative feedback.

[Mike_J]

Yes, a teeny amount of NFB flows through the commingled tail. But this is a minor mistake which Leo liked.

The 4.7K commingled tail in the attached diagram creates a bootstrapping effect that makes the constant current resistance appear larger while preserving the headroom.  Figure 5A and 5B have nearly the same idle conditions, gain and headroom, but the difference in output balance is significant due to this bootstrapping effect.

Figure 5B requires a 0.027mA imbalance between the stages in order to get the required 0.68V change across 25.17K of tail resistance.  Figure 5A only needs to raise the tied cathode voltage by 0.19V across 20.47K of resistance because the other 0.54V has already been supplied by the 4.7K NFB resistor.  So the impedance of the tail acts like it is much larger in that it only needs 0.009mA to get the required voltage change at the tied cathodes.  If you wanted to get the same balance with Figure 5B, it would require a 75K tail resistor which would cause an unacceptable reduction in headroom.



   

So today I am waiting for my 250r linear humdinger pots to arrive. What do I get instead are 2.5KL pots. I am thinking this is not ideal until I think based on the PI balance calculation I did not long ago I needed a total of 8K both on the board and the tail resistor that is the presence pot to balance the PI. I am planning to use the 6.8K on the board as the Bassman reissue calls for which has me needing 1.2K if I am going to use this 2.5K pot as my replacement presence pot. Well I know I need something less than 2.5K to get to 1.2K so I grab a 2.2K resistor and run it across the pot from CCW to CW and measure it and it hits 1.2K.


Now I am thinking I better find someone who has some knowledge about whether I can use  a pot that is quite a bit smaller than what Fender called for. The reissue Bassman called for a 25K pot with a 4.7K resistor across it or about 3.8K as I recall. The 5f6a schematic called for a 5K pot with no resistors across it. Any suggestions.? I will definitely be experimenting with this. Can balance the PI with one of these pots with the 2.2K resistor across it, a 6.8K tail resistor on the board and pretty much everything else standard Fender values tweaked a little to dial it in.


Thanks
Mike

[Mike_J]

I was reading Aiken's white page on the LTP this morning. Did a calculation somewhere that the LTP needs to be 8K which includes the presence pot. Still not 100 percent sure it does but it does appear to be part of a voltage divider with the 27K resistor that goes from the OT jack pin to the presence pot cw lug. Apparently if I change the presence pot from a 5K to a 1.2K I would need to change the 27K resistor to a 7.7K resistor to keep everything equal. Changing the .1uF cap from the presence pot wiper to the ccw lug to a 1uF would lower the corner frequency of the boost. Don't know if it is good or bad but it is what it is apparently. Also, what needs to be balanced in the pi tube is the mu between side a and b. That for some reason makes sense to me. Reason to balance the pi in the first place is to get the dBs of the out of phase input to the pi and in-phase input to the pi the same. Would seem that having a different mu on each side would mess up that calculation.


Thanks
Mike

[thetragichero]

having a pot that is "too large" is never an issue, just requires an additional resistor: http://www.geofex.com/article_folders/potsecrets/potscret.htm

[Mike_J]

Had another thought. The original 5f6a Bassman schematic calls for a 5K presence pot and a 10K tail resistor. The Fender reissue amp had a 25K pot with a 4.7K resistor across it which apparently some of the originals had and the tail resistor was changed to 6.8K. I wired my presence pot as a 25K pot with a 4.7K resistor across it and the pot measures about 3.4K from cw to ccw which is the same ratio of presence pot to tail resistor as before (50%). That is 5K presence pot to 10K tail resistor versus 3.4K presence pot to a 6.8K tail resistor. Don't believe the 27K resistor was changed though.

According to Aiken's "this resistor (the 27K resistor) directly affects the amount of negative feedback, and thus, the overall gain of the output section, as well as the linearity, input range, and distortion." It would be easy enough to check with a pot. Dial into 27K and then increase and decrease it the value from there.

Using that logic I could stick with the 2.5K pot and not run a resistor across it. Ideally I would be able to use a 5K resistor for the tail resistor and keep the two to one ratio but I am pretty sure I need a combined resistance of 8K. Probably not the end of the world to make the tail resistor 5.5K and keep the 2.5K pot.

[Mike_J]

If there is any logic at all to my previous reply then I have good news. The 2.5KL pot measures 2.632K. I have ordered some 5.36K precision resistors to get me to the 8K and 2.632K/5.36K equals 49.1% which is close enough to 50% for government work.


Thanks
Mike J

[thetragichero]

there's very little precision necessary in a guitar amp (1 ohm cathode resistor being the only exception i can think of). I'd much rather have 5k1 or 5k6 resistors in my parts bin

[pdf64]

My view is that anything beyond E6 is taking things too far.

[Mike_J]

there's very little precision necessary in a guitar amp (1 ohm cathode resistor being the only exception i can think of). I'd much rather have 5k1 or 5k6 resistors in my parts bin

I understand what you are saying. However the resistor was about $.25 so it won't break the bank and I need it to test my 8K theory.

[Mike_J]

My view is that anything beyond E6 is taking things too far.

Would you explain the meaning of E6. I am sure I am taking things too far. Usually do but it is a good way to learn what to do or mostly what not to do. I am hung up on trying to balance the PI and seeing if it does anything. Until that is disproven as not effective I will keep on the quest.


Thanks
Mike

[pdf64]

See https://www.bourns.com/support/technical-articles/standard-values-used-in-capacitors-inductors-and-resistors

[2deaf]

Here's one with pretty colors.

[Mike_J]

See https://www.bourns.com/support/technical-articles/standard-values-used-in-capacitors-inductors-and-resistors

Can't tell if it is in pF, nF or mF. Took me about twenty times to realize when you were saying 0v you were talking about ground. Every time I saw 0v I thought okay but what about ground. E6 is only 20% according to 2deafs chart. Seems like we can do better than 20%. Now 20% of what I have no idea. Have you seen my sour cream container capacitor switch. Think it may have all of the values contained on your chart but I am not sure because the charts don't mention what those values are measured in.


Thanks
Mike

[Mike_J]

Have attached three pictures which help to explain my obsession with this matter. The first one comes off the original 5f6a schematic. It is the tail resistor of 10K and the presence pot of 5K, combined. The second one is from the Bassman reissue amp. It has a 6.8K tail resistor and a 3.4K presence pot which is the result of using a 25K pot with a 4.7K resistor across it from cw to ccw. The third one is from me using a 5.36K precision resistor for the tail resistor and a pot that measures 2.632K. You will see the dB ratings are the same at that value. Had to adjust the 82K resistor to 81.8K to accomplish it but 8K gets you very close.

[thetragichero]

[2deaf]

Can't tell if it is in pF, nF or mF. Took me about twenty times to realize when you were saying 0v you were talking about ground. Every time I saw 0v I thought okay but what about ground. E6 is only 20% according to 2deafs chart. Seems like we can do better than 20%. Now 20% of what I have no idea. Have you seen my sour cream container capacitor switch. Think it may have all of the values contained on your chart but I am not sure because the charts don't mention what those values are measured in.

Each category has a certain number of choices per decade.  If you put the decimal point after the first digit, then there are a certain number of choices between zero and ten.  E6 has six values between zero and ten.  E24 has 24 choices between zero and ten.  E192 has 192 choices between zero and ten.

The 20% for E6 refers to the tolerance.  Each value in the E6 column can be plus or minus 20% and still be considered the nominal value.  It would make no sense to keep a 20% tolerance when the number of choices increases because plus or minus 20% of the nominal value would overlap the next value up or down.  Therefore, the tolerance percentage decreases as the precision of the values increases.

You can put the decimal point wherever you want to accommodate your needs.  Your units can be Ohms, Farads or Henrys.

Attached is a table of values from an old Mouser catalog.  These are E24 values with the decimal point moved around.
 

[Mike_J]

Can't tell if it is in pF, nF or mF. Took me about twenty times to realize when you were saying 0v you were talking about ground. Every time I saw 0v I thought okay but what about ground. E6 is only 20% according to 2deafs chart. Seems like we can do better than 20%. Now 20% of what I have no idea. Have you seen my sour cream container capacitor switch. Think it may have all of the values contained on your chart but I am not sure because the charts don't mention what those values are measured in.

Each category has a certain number of choices per decade.  If you put the decimal point after the first digit, then there are a certain number of choices between zero and ten.  E6 has six values between zero and ten.  E24 has 24 choices between zero and ten.  E192 has 192 choices between zero and ten.

The 20% for E6 refers to the tolerance.  Each value in the E6 column can be plus or minus 20% and still be considered the nominal value.  It would make no sense to keep a 20% tolerance when the number of choices increases because plus or minus 20% of the nominal value would overlap the next value up or down.  Therefore, the tolerance percentage decreases as the precision of the values increases.

You can put the decimal point wherever you want to accommodate your needs.  Your units can be Ohms, Farads or Henrys.

Attached is a table of values from an old Mouser catalog.  These are E24 values with the decimal point moved around.
 

I understand your point. Tried to use 5% parts as much as possible to try to minimize what you are talking about. The pF tool I have pretty much tracks the values they are supposed to be. Is a concern though that the test tool may contain values that are all over the map.


Thanks
Mike

[Mike_J]

There is an art museum in Houston, Texas that had an exhibit called three shades of white. They were three canvases four feet wide by eight feet tall in a row. No paint on them at all. Artist probably made a fortune for it and didn't do a thing. Called it three shades of white.