"Contour" control

[Funkalicousgroove]

It's not just a treble bleed, it's a tone control that actually does get "Brighter" one way and "darker" the other.

[greiswig]

Well, I tried the contour control. As I implemented it, it does NOT work as TR states: instead, it just rolls of highs. Bass never increases (hint: it's a passive control, so I didn't expect it too) except in a relative sense as you roll off the highs.

I settled on a .0033 cap as my favorite, but even then I didn't care for the tone with the presence control dimed. I liked it much better with the presence set fairly about halfway, which is more than I usually have it...by about halfway.

It definitely allows for a greater range of adjustment than the presence control I have does: you can go from fairly dark sounding all the way up to nails-on-chalkboard screech if you have the presence high enough.

I could see it as a potentially useful control, but I am beginning to think that there is something off with my presence circuit, which is based on the Bluesmaster. Maybe I should just go back to the modified D'Lite presence and try that again.

I think the Brown Deluxe tone control can be found at http://www.schematicheaven.com/fenderam ... _schem.pdf

[Fischerman]

A Resonance circuit like Soldano etc. will just increase bass and gain at lower frequencies...just like the Presence does for high frequencies. So unless you want more (and 'looser) bass then you likely wouldn't like a Resonance control.

I've tried 'Cut' controls in several amps and I generally don't like them. I think I'd rather use a dual-pot and have a post-PI dual/parallel treble bleed than a Cut control (I've tried this and liked it much better). But this is probably very amp dependant since the Cut control is a push-pull cancellation-thing.

I almost always like a much bigger Presence cap than the typical .1uF you see in so many amps. Usually .47uF or .68uF works for me. That makes the Presence control more of a 'boost everything except the bass' control.

[greiswig]

I almost always like a much bigger Presence cap than the typical .1uF you see in so many amps. Usually .47uF or .68uF works for me. That makes the Presence control more of a 'boost everything except the bass' control.

Stock D'Lite presence: 4ohm tap->4k7->390ohm/2kPot->1uF cap
Bluesmaster presence: 4ohm tap->100k->4.75kohm/15kPot->.1uF cap

These are very different circuits. I'd like to understand better how they actually behave differently in the circuit.

The BM presence seems to have a lot less feedback overall, and the lower cap value at the end seems like it would raise the knee of the affected frequency of the presence pot?

[ayan]

I almost always like a much bigger Presence cap than the typical .1uF you see in so many amps. Usually .47uF or .68uF works for me. That makes the Presence control more of a 'boost everything except the bass' control.

Stock D'Lite presence: 4ohm tap->4k7->390ohm/2kPot->1uF cap
Bluesmaster presence: 4ohm tap->100k->4.75kohm/15kPot->.1uF cap

These are very different circuits. I'd like to understand better how they actually behave differently in the circuit.

The BM presence seems to have a lot less feedback overall, and the lower cap value at the end seems like it would raise the knee of the affected frequency of the presence pot?

Assuming both amps are 100W, the BM does has less overall damping (4.75/104.75 x 100 = 4.5% VS 0.39/5.09 x 100 = 7.7% for the "classic" Dumble feedback loop), so the sound will be looser all around -- Note: if one amp is 50W and the other 100W, some scaling needs to occur and in a nutshell, all other things being equal, the feeback resistor on the 50W amp needs to ~70% of the value it would be in a 100W amp to keep the negative feedback effect the same.

However, once you get to the frenquency selective part of the circuit, the "classic" Dumble pair is 390ohm//1uF... and what really matters is the product RC (1/RC to be more accurate). If you scale the cap down to .1uF instead, you'd have to have a resistance ten times larger to achive the same effect, which would be 3.9K. The next standard value up from there is 4.7K, so we're not talking about a huge difference here. But, in the end the BM has less negative feedback and its presence control works down to lower frequencies, so the sound will be rawrer due to both effects. Not my cup of tea for the Dumble sound.

Cheers,

Gil

[greiswig]

Thanks, Gil! Between this and reading Aiken's web site, I've got a better grasp now. Very, very helpful...that's what this forum is great for.

I just like understanding the theory behind what I'm hearing. Knowing that helps me make educated guesses as to how to change something when I need to.

[greiswig]

Assuming both amps are 100W, the BM does has less overall damping (4.75/104.75 x 100 = 4.5% VS 0.39/5.09 x 100 = 7.7% for the "classic" Dumble feedback loop), so the sound will be looser all around -- Note: if one amp is 50W and the other 100W, some scaling needs to occur and in a nutshell, all other things being equal, the feeback resistor on the 50W amp needs to ~70% of the value it would be in a 100W amp to keep the negative feedback effect the same.

However, once you get to the frenquency selective part of the circuit, the "classic" Dumble pair is 390ohm//1uF... and what really matters is the product RC (1/RC to be more accurate). If you scale the cap down to .1uF instead, you'd have to have a resistance ten times larger to achive the same effect, which would be 3.9K. The next standard value up from there is 4.7K, so we're not talking about a huge difference here. But, in the end the BM has less negative feedback and its presence control works down to lower frequencies, so the sound will be rawrer due to both effects. Not my cup of tea for the Dumble sound.

Cheers,

Gil

Okay, just as soon as I say I understand better, more questions come to mind that tell me I don't. Gil, in my case I'm talking about a 2x6L6 amp.

Quoting Aiken: "The actual resistor values used in the feedback attenuator aren't that important, as their ratio determines the amount of feedback. The shunt resistor value is usually fixed by the phase inverter design requirements, and the series resistor is then sized according to the desired amount of feedback, given the voltage available at the output."

Can the "frequency selective" (shunt) part of the circuit be treated somewhat independently of the series resistance part? Aiken talks about the ratio of series R to shunt R being important, but I'm not getting it.

Aiken also talks about the shunt resistance (is that both the pot and the R to ground in parallel, or just the R to ground, or...) being dependent on the PI. So does this mean that having a Bluesmaster PI negates the possibility of changing the presence circuit to more standard Dumble values? What would be the consequences?

Basically, I'm trying to make my amp so the presence control is somewhat useful. As it is, the amp and speaker I'm using are bright enough that the presence sounds best at 0.

[dogears]

Try this.

First off, the presence pot is a 25K in the Bluesmaster. I had one here and have pics. If you use 15K then your presence is 1/3 on all the time.

Secondly, try either using the 8 ohm tap, or changing the feedback resistor to a 68K or 75K. (assuming this is a 50 watter) That will tame those highs and smooth the mids alot!! Combined with the 25K pot and you are in biz!!

I almost always like a much bigger Presence cap than the typical .1uF you see in so many amps. Usually .47uF or .68uF works for me. That makes the Presence control more of a 'boost everything except the bass' control.

Stock D'Lite presence: 4ohm tap->4k7->390ohm/2kPot->1uF cap
Bluesmaster presence: 4ohm tap->100k->4.75kohm/15kPot->.1uF cap

These are very different circuits. I'd like to understand better how they actually behave differently in the circuit.

The BM presence seems to have a lot less feedback overall, and the lower cap value at the end seems like it would raise the knee of the affected frequency of the presence pot?

[greiswig]

Thanks, Scott...I'll give those changes a try. I appreciate the positive feedback.

[ayan]

Assuming both amps are 100W, the BM does has less overall damping (4.75/104.75 x 100 = 4.5% VS 0.39/5.09 x 100 = 7.7% for the "classic" Dumble feedback loop), so the sound will be looser all around -- Note: if one amp is 50W and the other 100W, some scaling needs to occur and in a nutshell, all other things being equal, the feeback resistor on the 50W amp needs to ~70% of the value it would be in a 100W amp to keep the negative feedback effect the same.

However, once you get to the frenquency selective part of the circuit, the "classic" Dumble pair is 390ohm//1uF... and what really matters is the product RC (1/RC to be more accurate). If you scale the cap down to .1uF instead, you'd have to have a resistance ten times larger to achive the same effect, which would be 3.9K. The next standard value up from there is 4.7K, so we're not talking about a huge difference here. But, in the end the BM has less negative feedback and its presence control works down to lower frequencies, so the sound will be rawrer due to both effects. Not my cup of tea for the Dumble sound.

Cheers,

Gil

Okay, just as soon as I say I understand better, more questions come to mind that tell me I don't. Gil, in my case I'm talking about a 2x6L6 amp.

Quoting Aiken: "The actual resistor values used in the feedback attenuator aren't that important, as their ratio determines the amount of feedback. The shunt resistor value is usually fixed by the phase inverter design requirements, and the series resistor is then sized according to the desired amount of feedback, given the voltage available at the output."

Can the "frequency selective" (shunt) part of the circuit be treated somewhat independently of the series resistance part? Aiken talks about the ratio of series R to shunt R being important, but I'm not getting it.

Aiken also talks about the shunt resistance (is that both the pot and the R to ground in parallel, or just the R to ground, or...) being dependent on the PI. So does this mean that having a Bluesmaster PI negates the possibility of changing the presence circuit to more standard Dumble values? What would be the consequences?

Basically, I'm trying to make my amp so the presence control is somewhat useful. As it is, the amp and speaker I'm using are bright enough that the presence sounds best at 0.

OK, here I go: I am not as elloquent as Randall about any of this stuff and I'm sure his treratment of the subject in his website is top notch - like any other piece of information ever to come from Randall. Let's first forget about the cap and presence pot, for a moment...

The deal is this: the (negative) feedback loop in the power section is what's commonly known as a voltage divider. The "college" model for this, for a basic EE course, would be to show a voltage source (in this case the output generated across the windings of the secondary of the OT), a rersistor in series with that (our feeback resistor Rf) and a load resistor, Rl. All three are in series and form a nice little loop; google "voltage divider" and you will see a little diagram that is bound to help.

Now, if you apply Ohm's law to that, you can see that the voltage that develops across the load resistor, Rl, is equal to the voltage of the source, Vs, mupltiplied by th following "voltage divide" ratio: Rl/(Rf+Rl). In turn -- although completely inconsequent here - the voltage across Rf is Vs multiplied by Rf/(Rf+Rl). To double check the math, you can verify that the voltage drop across Rf plus the voltage drop across Rl add up to the source voltage, Vs.

Applying the above to the case in question here, Rf would be the feeback resistor (either 4.7K or 100K, depending upon the design) PLUS the small resistance across the OT secondary terminals. That is 4 ohms (in this case) and is totally non existent compared to the feedback resistor, so we leave that one out for the sake of simplicity of calculations.

The voltage that is applied (negatively) to the PI at its tail (so yes, this is fixed by the PI design!) is therefore the voltage at the output of the OT multiplied, in the case of the Dumble original design: 390/(390 + 4,700) = 0.077. So, the circuit basically feeds back 7.7% of the output at the 4-ohm jack to the "subtracting" half of the PI. If you want to make a further approximation, because the numbers work that way, since 4,700 ohms is much greater than 390 ohms, one can say that the feedback voltage is approximately given by the ratio of Rl/Rf, which in this case is 390/4700 = 0.083, or 8.3%. Close, maybe a wee bit on the crude side, and perhaps this is the ratio that Randall talks about. In either case, if you consider the PI design fixed, since changing that has a number of other implications, the way to control the feedback applied is to: (1) Change the feedback resistor; (2) Derive the feedback voltage, Vs, from another tap in the OT, if available.

I hope the above makes sense. Assuming it does, what happens when you add a cap in series with a presence pot, all of that in parallel with Rl? As you increase the setting on the presence control, the cap ends up bypassing Rl more and more, which allow high frequencies in the loop to go to ground. If they go to ground, they will not participate in the "negative feedback" being injected into the PI. Thus, not subtracting those highs means that they will appear in the final output, and that is how turning up a presence control adds highs to the sound. Choosing the value of the bypass cap strategically you can "tune" the frequency response of the loop. Make the cap smaller and only the highst frequencies will be decoupled. Make the cap larger and larger, and all of a sudden lower and lower frequencies will be decoupled from the loop. At some point, the presence control starts working on the midrange, rather than the high end, and if you use a large enough cap, it will approach "bandpass" (i.e., it will decouple all frequencies).

You can do the "analogous" thing to the loop with Rf and put a cap in series with it. That will inhibit low frequencies in the loop and therefore end up boosting the low end; if you shunt that cap with a variable resistor, you can either bring it into the loop or out of the loop. That is what's commonly known as a "resonance" control. Similar to the presence control, but it works on the low end. You can definitely have both controls coexist very happily in the same amp.

I hope this helps.

Cheers,

Gil

[greiswig]

Great explanation, Gil. I've got nothing but gratitude to Aiken for what he's contributed on his web site, but the technical expertise and patient explanation present here is amazing as well. The voltage divider piece is what I was missing about the ratio being important. Light bulb on.

[Fischerman]

Great post Gil.

Ever notice how older Fenders (which only had one output impedance...not selectable or anything) generally had a 100 ohm resistor as the 'to ground' resistor for a 4 ohm amp and a 47 ohm there for an 8 ohm amp? They all had an 820 ohm NFB resistor but the OTs were usually either 4 ohm or 8 ohm so to compensate they changed the resistor to ground instead of the NFB resistor (and didn't seem to take the power output into account as much). I always wondered why they did it there instead of at the NFB resistor and also tailor it to the power output as well. The obvious exception is the Super Reverb which had a 2 ohm OT @ ~40W but they still used the 100 ohm resistor so it has less NFB...but SRs are known for being one of the better/best BF Fenders cranked up. But a Twin Reverb had a 4 ohm OT @ 88W and the same 100 ohm resistor...more NFB and is known as the clean machine.

I should have mentioned in my above post that I was referring to the BM PI...since I'm kind of a 'Marshall guy' that PI is what I think of first.

[jelle]

Gil,

That is the best NFB explanation I have ever read!

[greiswig]

Note: if one amp is 50W and the other 100W, some scaling needs to occur and in a nutshell, all other things being equal, the feeback resistor on the 50W amp needs to ~70% of the value it would be in a 100W amp to keep the negative feedback effect the same.

Gil, I'm puzzled by something. If I look at the Hybrid A schematic, it has values for both 100W and 50W circuits. Rf for 100W is listed as 4k7, and for 50W it lists it at 8k. Rl for 100W is listed as 390, and for 50W is 1k. These seem to be scaled counter to what you suggest.

In other words, I'd expect Rf, at least, to be 70% of 4k7, or 3k3 or so. So either I'm misunderstanding somebody, or what you say above and what the Hybrid A schematic seems to say are contradictory.

In my build, which is a 50W D'Lite with Bluesmaster PI, I started with the values listed on the 100W bluesmaster schematic I could find: 100k Rf, 4k7Rl, and a .1uF cap. If I scale these values according to the Hybrid A schematic, I'd be moving to a 200k Rf, a 9k Rl, and a .05uF cap.

Or does all this only really need to scale if the OT isn't changed when shifting from a 100W to 50W circuit?

[ayan]

Note: if one amp is 50W and the other 100W, some scaling needs to occur and in a nutshell, all other things being equal, the feeback resistor on the 50W amp needs to ~70% of the value it would be in a 100W amp to keep the negative feedback effect the same.

Gil, I'm puzzled by something. If I look at the Hybrid A schematic, it has values for both 100W and 50W circuits. Rf for 100W is listed as 4k7, and for 50W it lists it at 8k. Rl for 100W is listed as 390, and for 50W is 1k. These seem to be scaled counter to what you suggest.

In other words, I'd expect Rf, at least, to be 70% of 4k7, or 3k3 or so. So either I'm misunderstanding somebody, or what you say above and what the Hybrid A schematic seems to say are contradictory.

In my build, which is a 50W D'Lite with Bluesmaster PI, I started with the values listed on the 100W bluesmaster schematic I could find: 100k Rf, 4k7Rl, and a .1uF cap. If I scale these values according to the Hybrid A schematic, I'd be moving to a 200k Rf, a 9k Rl, and a .05uF cap.

Or does all this only really need to scale if the OT isn't changed when shifting from a 100W to 50W circuit?

The math doesn't lie: I said that in a 50W amp Rf should be decreased to 70% of the 100W value, ALL OTHER THINGS REMAINING EQUAL. But if you will be changing the TAIL resistor on the PI as well, you need to modify my statement to read: In a 50W amp, the voltage divide ratio (= Rt/(Rt+Rf))needs to be 1.4 times (reciprocal of 0.7) greater than in a 100W amp to have the same NFB loop.

So, for the 100W amp the voltage divide ratio is: 390/(390+4700) = 0.077

For the 50W amp the voltage divide ratio is: 1000/(1000+8000) = 0.11

The 50 to 100W ratio therefore is: 0.11/.077 = 1.4

Looks good to me.

Gil