Some “seriously beginner grade” questions regarding resistors

[Gtr0]

There is no grid current and the second stage grid is driven by the signal voltage at the grid side of the mixing resistor. The mixing resistor can be seen as the upper/series part of a voltage divider. THe lower part being composed of the other "channel's"mixing resistor, vol pot and input tube. As the ground path of each mixing resistor runs through the other channel's vol pot, volumes are somewhat interacting.
(Signal and overall peamp gain would be maxed if the other mixing resistor was disconnected.)

Okay, I do know that the "lower" resistor is seen as something of a voltage divider - but I must say, to my mind that would tell me that channel one's signal would increase even more unless channels two's vol pot was turned all the way up. Correct me if I am wrong, but the volume pot (at least ones I have seen) are a voltage divider in and of themselves, so at volume 0 on channel 2, the voltage divider at channel 1 would be roughly 1.5M/470k, and if it were on 10, then 470k/470k. I realize that there is more to it... but is this the gist? To my ear, the sound doesn't really seem to change, but as you said in a later comment a half decibel... probably not noticeable at all when channel 1 volume is already maxed. ??

So, perhaps that was a poor example because I have a grip on dividers... so let me post this little clip I just grabbed from the internets... stage 1 to 2 of an SLO...

after the 500k pot is a 470k resistor to the grid of v2a. I've seen in other high gain schematics that this is often 1 meg - so I suppose because I can't imagine anything other than a signal traveling through the circuit in a linear fashion that I am missing the point here. I would guess that it is working in conjunction with the 500k pot to operate as a grid leak resistor? as well as with capacitance of v2a to form a low pass? and I suppose other things as well... ??

[Enzo]

The 470k and the 500k pot are acting as your grid leak - they provide a DC path to ground for the grid. The 470k at the grid does not diminish the signal, since there is no grid current. Ohm's Law - voltage drop equals current times resistance. And if current is zero, then so will voltage drop be.

In your example drawing, the 470k from the plate of V1b and the 500k pot form a voltage divider, and yes the pot itself is also a voltage divider. A divider in a divider if you like, or just think of the volume control who only sees half the signal in the first place.as a variable element in the overall divider. In either case, the ssignal at V1b plate is cut in half when it hits the top of the pot. Then the wiper of the pot selectively takes signal out of the half level signal.

[Helmholtz]

Okay, I do know that the "lower" resistor is seen as something of a voltage divider - but I must say, to my mind that would tell me that channel one's signal would increase even more unless channels two's vol pot was turned all the way up. Correct me if I am wrong, but the volume pot (at least ones I have seen) are a voltage divider in and of themselves, so at volume 0 on channel 2, the voltage divider at channel 1 would be roughly 1.5M/470k, and if it were on 10, then 470k/470k. I realize that there is more to it... but is this the gist? To my ear, the sound doesn't really seem to change, but as you said in a later comment a half decibel... probably not noticeable at all when channel 1 volume is already maxed. ??

So, perhaps that was a poor example because I have a grip on dividers... so let me post this little clip I just grabbed from the internets... stage 1 to 2 of an SLO...

after the 500k pot is a 470k resistor to the grid of v2a. I've seen in other high gain schematics that this is often 1 meg - so I suppose because I can't imagine anything other than a signal traveling through the circuit in a linear fashion that I am missing the point here. I would guess that it is working in conjunction with the 500k pot to operate as a grid leak resistor? as well as with capacitance of v2a to form a low pass? and I suppose other things as well... ??

Seems you got the voltage divider wrong. A voltage divider has an input and an output. As it's a divider, output voltage is always lower than input voltage. So the divider ratio must be smaller than 1, meaning that a smaller resistance gets divided by a larger one.

In its simplest form the divider consists of 2 stacked (series wired) resistors. The input voltage is applied across both resistors and the output is the voltage across the lower resistor. As its a series wiring the signal current through both resistors is the same and is determined by the sum of the resistances. Let's call the upper resistor Ru and the lower resistor Rl. Then the current is given as I = Vin / (Ru + Rl). The output voltage is the voltage drop across Rl caused by the current: Vout = I*Rl = Vin*Rl/(Ru+Rl). I follows for the divider ratio Vout/Vin = Rl/(Ru+Rl). For higher Rl output increases and for higher Ru output decreases.

When reading a schematic it can be tricky to identify the relevant components of a voltage divider. This requires to understand how the signal current flows, especially how it returns to ground.

The 470k before the V2a grid in the SLO schematic is a grid stopper and carries no current as Enzo said. A resistor that doesn't carry current must have the same voltage at both ends, so the 470k passes signal voltage to the grid without any loss. (Just saw that Enzo already explained.)

[Dave H]

Correct me if I am wrong, but the volume pot (at least ones I have seen) are a voltage divider in and of themselves, so at volume 0 on channel 2, the voltage divider at channel 1 would be roughly 1.5M/470k, and if it were on 10, then 470k/470k. I realize that there is more to it... but is this the gist? To my ear, the sound doesn't really seem to change, but as you said in a later comment a half decibel... probably not noticeable at all when channel 1 volume is already maxed. ??

There's not much volume change because the voltage divider resistors are effectively 470k/470k (Ru/Rl) for both of the above settings. With channel 2 vol on 0 Rl is grounded. With Chan 2 vol on 10 Rl is connected to the tube plate which is low resistance (38k) compared to 470k.

Edit:
There should be a noticeable volume increase when chan 2 vol is set at at its mid position when its max output resistance (500k) is added to Rl.

[Helmholtz]

The voltage divider resistors are effectively 470k/470k (Ru/Rl) for both the above. With channel 2 vol on 0 Rl is grounded. With Chan 2 vol on 10 Rl is connected to the tube plate which is low resistance (38k) compared to 470k.

This means that the voltage divider ratio is 0.5 and that total available preamp gain would double if the other mixing resistor was disconnected.

[gui_tarzan]

You guys continue to amaze me with the theory. I wish I had asked the questions long ago. In light of this discussion, it now makes more sense as to why different amps have different values of resistors and capacitors in similar functions. I'm going through a Harmony H200 right now, bringing it back to life after some 30 years of sitting in a basement. The difference in the circuit compared to other amps is interesting, especially since I have found three different versions of the same schematic including the one inside the case. One has preamp grid resistors, two don't. One has the signal cap in series before the volume pot, two have it after the pot on the preamp grid.

[Helmholtz]

For guitar I would prefer the circuit on the right because of its higher input impedance. Low input impedance (below 500k) noticeably damps PU treble response.

[gui_tarzan]

I'm leaving this one as stock as possible but properly functioning because of the age. I'm not going to keep this one, someone that's actually a good guitarist or harp player needs to use it. The one on the left is the schem from inside the case, I had to take several pics and stitch it together because I left the speaker in place.

[Helmholtz]

...also the input resistors of inputs #1 and #2 in the left circuit attenuate the signal by a factor ~ 4 (if that's actually a 100k pot).

[gui_tarzan]

Yes it is a 100k.

[Enzo]

it now makes more sense as to why different amps have different values of resistors and capacitors in similar functions.

Under the skin, amps are all alike. The differences are the details. It is like meatloaf: meatloaf is ground meat, egg, and filler. beyond that every one is unique. I put onions and diced jalapenos in mine, my mom would never use jalapenos. Some use ground beef, some beef/pork blend, or even lamb. Ketchup on top or not? Etc etc. it is all meatloaf. Just like that old Marshal and that Bassman are so similar. CAthode bypass cap is just a detail.

Sometimes it is useful to hold a schematic at arms length so you can see the circuit without seeing the individual part values and such.

[gui_tarzan]

Yeah, you told me that several years ago when I was searching for a schematic for an old "boutique" amp. I never found one but it doesn't matter anymore, I can identify what's going on a lot more than I could then and I appreciate the coaching.

[gui_tarzan]

And in many cases, food analogies make sense. What is a taco? Meat, tomato sauce, spices and a shell. What is a burrito? Meat, tomato sauce, spices and a shell.

[Enzo]

What is a calzone? Meat, tomato sauce, spices and a shell.

[Mick Bailey]

Here we have Doner Kebabs. Made from connective tissue, fat, rind and sphincter gristle. Like owning an amp, we are in reality merely custodians, but the timescale is a little shorter - for as long as we can keep it down after 6 pints of beer.