Seymour Duncan Convertible 100 Input

[Enzo]

Because V3 is not a gain stage. Note the signal does not flow through it. Instead, it is an electronic volume control. The plate sits on the signal path and by making the tube conduct, we shunt the signal to ground to some varying extent. it does not need to be a gainy tube. Note that while the nice clean drawing shows 12AX7, the same schematic with the wrong 4.7k on it, the cruder drawing does say 12AU7 as well as the 4.7meg. Note the V3 has only 2.2k plate resistors and either grounded cathodes or 100 ohms to ground there. Those are not 12AX7 numbers. The first schematic on the Duncan page is a tif file. It also shows 12AU7 there.

Pull V3 and the amp still plays!!

[bobloblaws]

I own 2 100 watt Convertibles.

Another question for you Graham. Can you (or anybody else) also tell me how the 2 input jacks are supposed to be wired? The schematic only shows one input jack and doesn't offer a clue. I'm looking at how the vintage Fenders do it but it doesn't work because there are no grid stoppers. When plugged into the second jack only, the first jack's tip is grounded by the switch wired to the sleeve. This ends up grounding the second jack's tip as well since there are no grid stoppers (and both tips directly connected to the positive input). Is there some fancy way to this? It just occurred to me, the 170K resistor probably has something to do with it and was not intended as a grid leak as I had assumed. I should have paid attention to how it was hooked up but hindsight is 20/20. The manual describes the behavior thusly:

If you use the Auxiliary Input for a second instrument, the output volume from both instruments will be reduced......Also, the instrument that is plugged into the Auxiliary Input will have less volume than the instrument plugged into the Main Input.

[Enzo]

Just occurred to me, I have a NOS wooden cab for the Convertible 100. Bought new from S-D for a project and never used. Cheap to good home.

[bobloblaws]

Because V3 is not a gain stage. Note the signal does not flow through it. Instead, it is an electronic volume control. The plate sits on the signal path and by making the tube conduct, we shunt the signal to ground to some varying extent. it does not need to be a gainy tube. Note that while the nice clean drawing shows 12AX7, the same schematic with the wrong 4.7k on it, the cruder drawing does say 12AU7 as well as the 4.7meg. Note the V3 has only 2.2k plate resistors and either grounded cathodes or 100 ohms to ground there. Those are not 12AX7 numbers. The first schematic on the Duncan page is a tif file. It also shows 12AU7 there.

Pull V3 and the amp still plays!!

Indeed. The amp came to me with a 12AT7 in V3. The power level pot had very little effect. I swapped the tube for a 12AU7 and the power level feature works as expected, turning it down to 5W results in heavy saturation even at low volume.

Out of curiosity, which one is the phase inverter, V1 or V2, and what is the other one for? I'm pretty green at understanding circuits by looking at schematics but my guess is that V1 (12AU7) is the PI and the top half of V2 (12AX7) looks like an additional gain stage following the effects loop while the bottom half looks to have something do do with the variable damping and/or a negative feedback loop. Am I in the ballpark?

Note: So as not to confuse, I'm looking at the "clean" drawing which has the labels V1 and V2 opposite that of the cruder drawing, presumably following the physical order of the 3 tubes on the amp itself rather the order of appearance on the schematic.

Note the signal does not flow through it.

But the 100K pot is connected to the grids via the 10K resistors. What does this do?

The plate sits on the signal path and by making the tube conduct, we shunt the signal to ground to some varying extent. it does not need to be a gainy tube.

Normally current flows from the cathode to the plate, no? It sounds like you are suggesting in this case the current is flowing (being shunted) from the plate to the grounded cathode. Can you clarify that for me?

[g1]

On the clean drawing, V2 is your PI stage. V1 is a driver, or extra gain stage after PI. Top and bottom halves of V1 are identical.
V3 is what I would call the tube equivalent of a PPIMV. Varying amounts of the signal are shunted to ground depending on the setting of the 100K pot. Look at each side of V3 as a variable resistor shunting the signal to ground.

[bobloblaws]

On the clean drawing, V2 is your PI stage. V1 is a driver, or extra gain stage after PI. Top and bottom halves of V1 are identical.
V3 is what I would call the tube equivalent of a PPIMV. Varying amounts of the signal are shunted to ground depending on the setting of the 100K pot. Look at each side of V3 as a variable resistor shunting the signal to ground.

Ok, good to clear up what V1 and V2 are, but I'm still confused about how V3 actually works. My understanding is that generally signal enters through the grid and exits via the plate with the cathode providing the current that increases amplitude. The cathode is connected directly or indirectly to ground so this is an illustration of electron flow as opposed to conventional flow. In other words, ground/cathode is the source of the electrons that are conducted through the tube and out from the plate. So when I look at the schematic I see that the signal leaves the plates of V1 and in order to be shunted to ground via V3 it would have to travel "backwards" from the V3 plates to the grounded cathodes and/or the ground connections on the other side of the 100K pot. I suspect that I'm missing something to do with the behavior of voltage vs current (or maybe something completely different).

[Enzo]

I say again, V3 is not a signal stage. It is a variable resistor stage. It is not amplifying anything.

Electrons move the same direction through a tube regardless of cathode grounding or not. Ground is always the source of electrons.

V3 plates are on the signal path and the cathodes are grounded. The grids are fed a variable negative voltage from supply F. That controls the conduction through teh tube. That makes the tube a varaible resistance, and that resistance forms a voltage divider with the 2.2k resistors.

[Jazz P Bass]

Best to keep in mind that a vacuum tube is 'thermionic'.

The electrons are literally boiled off of the cathode (which has certain characteristics to achieve this) by the temperature difference induced by the heater (inside the cathode).

The positive high voltage difference of the plate pulls the electrons to it.

The current is then modified by the grid voltage.

[bobloblaws]

I say again, V3 is not a signal stage. It is a variable resistor stage. It is not amplifying anything.

Electrons move the same direction through a tube regardless of cathode grounding or not. Ground is always the source of electrons.

V3 plates are on the signal path and the cathodes are grounded. The grids are fed a variable negative voltage from supply F. That controls the conduction through teh tube. That makes the tube a varaible resistance, and that resistance forms a voltage divider with the 2.2k resistors.

I understand all that for the most part. The V3 tube forms a variable resistance which restricts signal to the power tube grids. Signal does not flow through the V3 tube, as you pointed out. But then how is it that "by making the tube conduct, we shunt the signal to ground to some varying extent"? The path(s) to ground that I see are via the cathode or on the other end of the 100K pot via the 220K resistor. So for part of the signal to reach ground it would have to travel through the tube, no? Can you see where that is confusing to me? Maybe I don't understand what is actually meant by the terminology "shunted to ground"?

[eschertron]

I understand all that for the most part. The V3 tube forms a variable resistance which restricts signal to the power tube grids. Signal does not flow through the V3 tube, as you pointed out. But then how is it that "by making the tube conduct, we shunt the signal to ground to some varying extent"? The path(s) to ground that I see are via the cathode or on the other end of the 100K pot via the 220K resistor. So for part of the signal to reach ground it would have to travel through the tube, no? Can you see where that is confusing to me? Maybe I don't understand what is actually meant by the terminology "shunted to ground"?

I think your confusion is part semantics, and part conceptual. When Enzo says that V3 is not in the signal path, he means it has no direct 'tubey' effect on the sound. however, signal actually does pass through it, on its way to the "sewer" ground - where we never hear it. The shunt effect allows a portion of the signal to be taken away from the signal path, resulting in the reduction of signal strength past that point. Call it 'volume reduction'.

Also don't think of V3 as "restricting" the signal. On the contrary, when V3 conducts it provides additional circuit paths for the signal to go; rather than restricting the flow, it actually reduces the impedance that the signal sees, and allows a portion of the signal through directly to ground (as stated above).

[Enzo]

Fair enough.

The signal path is that abstract concept of the path through the amp from input to output. The signal does not flow THROUGH V3 to get to the output. If you are comfortable saying the signal goes through it to ground, fine, and by that same thinking them the signal finds its way to ground through the tone stack and many other places. But we don't call that the signal path.

It is like a plain old volume control. You can think of it as the signal going to ground with some part taken out the wiper. I myself do not intuit the circuit that way. If I put a resistor to ground from the signal path, I don't think of it as the signal going to ground. To me the signal is a voltage taken with respect to ground. If I tap in half way - like a volume control at 5 - I think of it as a voltage divider.

[eschertron]

Enzo,
I agree. I was just trying to explain how the signal voltage gets attenuated 'across' V3 rather than 'through' V3.
I think the OP may be tripping - conceptually - over the difference between signal path and the signal itself, represented by a signal voltage. You're right, the signal path is not through V3. but some of the signal disappears down through V3, so it has an effect on the signal. What's interesting about this, of course, is that unlike a passive component (in a tone stack, for example) the tube actively modulates how much the signal gets loaded down, and is the reason for its existence. To realize that a tube can be used for more than one kind of function in an amp may be the primary take-away from this thread. I want bobloblaws to have a clear and principled grasp of the circuit under discussion.

[Enzo]

Oh, I am all about alternative explanations. The more ways something can be explained, the better the chance one or more of those ways will reach the listener. I like to read even basic electronics presentations because I can learn other ways to present the material I already present myself.

[bobloblaws]

Thanks guys, it makes more sense to me now.

[GrahamPearson]

How the 2 Inputs are Connected

Another question for you Graham. Can you (or anybody else) also tell me how the 2 input jacks are supposed to be wired? The schematic only shows one input jack and doesn't offer a clue.

The components are the same on both of my amps. I'm confident the 4.7M resistors and the following input wiring scheme have not been tampered with because I bought one of my amps from the original owner and he was definitely not tech savvy.

The upper input: Tip is grounded if no jack is inserted. This ground disables both the upper and lower input if no jack is inserted.
The lower input: Tip is connected to the tip of the upper input via a 100K resistor (brown:black:yellow). Lower input is grounded if no jack is inserted in upper input.

Sorry for my slow reply. I was expecting an email informing me if there was any new thread activity.
Graham