Some “seriously beginner grade” questions regarding resistors

[gui_tarzan]

Ew...

[Gtr0]

So, I didn't mean to fall off of the map there - I appreciate the conversation that is going on here... just been super busy with work and just acquired a new Gibson R6, so in my small free time I have been spending some time to get acquainted :-)

this morning I will be trying to absorb what I can and I most certainly be back with questions no doubt!

Thanks.

[Gtr0]

Yes, the grid voltage causes the modulation. The signal always has the shape of a sine wave but at the plate it also has a DC offset. On the sim below grid V is green, plate V is blue and downstream of the cap is red. You can see that the signal at the plate has a 180V DC offset which is removed by the cap.

Awesome visual and puts in into a much better perspective for me. It was my impression that this is what is going on, but having a visual aid like this really clears it up to a large degree. Thank you!

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.)

Okay, I am clearly lacking in this part here... because I do not work on amps everyday, when I learn something it is easily forgotten. And so when I know that this is the case in other parts of the circuit and I should be applying the same theory in this section of the circuit, I do not, because I am lacking in the "common sense" that a trained individual knows and uses it almost seemingly like a second nature.

IN FACT, while I have noticed it, I did not consciously know that a single resistor in series does not really do anything on its own. I remember having read that two resistors in series will see the input voltage across both, but somehow I STILL stick to thinking that the first resistor drops voltage, then the second resistor sees a lesser voltage then the first, and drops voltage again.

But I should know better, because when I turn an amp on without any tubes to load the circuit, the B+ voltage remains the same down the line even though it has gone through multiple resistors.

I am trying like hell to learn and your replies and breadth of knowledge, as well as that of many of the guys, or gals, here (Dave H, Enzo, nickb, Chuck H, Fahey, and a list of others I can't think of this second), are deeply appreciated!

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.

Going back to my Marshall JTM45 and my Laney LA60BL, I do notice a change depending on the volume setting of the unused channel. I have read that it does make a difference and looking at the respective circuits it certainly makes sense - I just never tried with the purpose of whether or not a change is detected... but I noticed it to be more dramatic on the Laney than the Marshall... for whatever reason.

Getting back to the SLO I attached most recently, if there are no loses across the 470k at grid v2a, what's the point of having it there? Is it to help with grid leak (in conjunction with the vol pot)? Or is it for a grid stopper (radio freq in high gain amps)?

And finally, @gui_tarzan - thanks for your posts - the answers given have given much needed insight!

You guys are awesome - thanks!

[gui_tarzan]

"IN FACT, while I have noticed it, I did not consciously know that a single resistor in series does not really do anything on its own."

I wouldn't say that, there are many places a series resistor works a specific job, the input jack for one, as a current limiter in front of an LED as another just come to mind. Correct me if I'm wrong, guys.

[Helmholtz]

"IN FACT, while I have noticed it, I did not consciously know that a single resistor in series does not really do anything on its own."

I wouldn't say that, there are many places a series resistor works a specific job, the input jack for one, as a current limiter in front of an LED as another just come to mind. Correct me if I'm wrong, guys.

I think he means that a single resistor is no voltage divider. That's true but the resistor could be part of a hidden voltage divider. The other "resistor" of the divider could be a tube, an LED or even a capacitor for AC (signal).

[Enzo]

Or even the ohm meter itself.

[Helmholtz]

Or even the ohm meter itself.

I think you mean voltmeter?
Yes, the internal resistance of a voltmeter can produce a noticeable voltage divider effect in high impedance (especially grid) circuits. Typical result is a too low voltage reading.

An Ohmmeter is different as it actually injects a DC current.

[Dave H]

Getting back to the SLO I attached most recently, if there are no loses across the 470k at grid v2a, what's the point of having it there? Is it to help with grid leak (in conjunction with the vol pot)? Or is it for a grid stopper (radio freq in high gain amps)?

You said it yourself in post #33
"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... ??"

[gui_tarzan]

I think he means that a single resistor is no voltage divider. That's true but the resistor could be part of a hidden voltage divider. The other "resistor" of the divider could be a tube, an LED or even a capacitor for AC (signal).

Ok, someone needs to explain to me how a resistor in series with an LED for current limiting can be a voltage divider. Am I over-thinking the electrical path again?

[Helmholtz]

Ok, someone needs to explain to me how a resistor in series with an LED for current limiting can be a voltage divider. Am I over-thinking the electrical path again?

The supply gets divided into the voltage drop across the resistor and the forward voltage of the LED. Like any forward biased diode an LED has a (strongly) current dependent resistance given by its forward voltage divided by current.
As the DCR of an LED is not constant (reduces with increasing current) and not well specified, it's not used to calculate the series resistor.
Rather one takes advantage of the fact that the forward voltage of an LED is fairly constant (for a given color) and calculates the series resistor by supply voltage minus LED voltage divided by desired current. Same procedure as with zeners.

[Enzo]

Yes, I meant volt meter, hey, I'm old...


When I wrote that, I had in mind the classic situation where you can't directly measure the grid voltage on the long tail pair PI circuits. Not to ground anyway. That happens due to the internal meter resistance.

[Helmholtz]

When I wrote that, I had in mind the classic situation where you can't directly measure the grid voltage on the long tail pair PI circuits. Not to ground anyway. That happens due to the internal meter resistance.

Yeah, that's what I figured. But if the meter resistance is known, the correct voltage often can be calculated using the (inverse) voltage divider formula.

[gui_tarzan]

The supply gets divided into the voltage drop across the resistor and the forward voltage of the LED. Like any forward biased diode an LED has a (strongly) current dependent resistance given by its forward voltage divided by current.
As the DCR of an LED is not constant (reduces with increasing current) and not well specified, it's not used to calculate the series resistor.
Rather one takes advantage of the fact that the forward voltage of an LED is fairly constant (for a given color) and calculates the series resistor by supply voltage minus LED voltage divided by desired current. Same procedure as with zeners.

Math makes my head hurt. Formulas specifically. The problem I have (one of them anyway) is I frequently have this burning desire to know why, but then it takes a while for it to sink in once I get the answers.. After all these years of finding bad components and replacing them, it never dawned on me to really buckle down and ask for the theory. I used to hate theory so I never learned it. I was the same in programming and playing guitar, I'd learn the minimum to get me going and pick things up along the way. I can trace voltages and signals and things like that, but I'm talking about what we've been going over here. Getting down to the DNA of electricity and the components. I never really cared about it in the past, now it's like it's consuming me so I can see WHY designers did the things they did.

[Enzo]

Yeah, that's what I figured. But if the meter resistance is known, the correct voltage often can be calculated using the (inverse) voltage divider formula.

Yep, what I usually suggest in the case of the PI grids is just measure from, grid to cathode, instead of to ground. The difference will be a volt or so and they already measured the cathode voltage, so subtract from that. The meter has very little effect on that.


GT, I tell you, Ohm's Law is not complicated or hard, and I really do use it in a practical sense every day. I mean I have my little pocket calculator next to me, and I am always figuring current or voltage or something. It really is used everywhere. It is so powerful. VOlts, Ohms Amps. If you know two, you get the third.

An example: You have a B+ of 300v on a simple 12AX7 triode stage, And the VERY common 100k plate resistor. And we expect - the schematic might even say it - maybe 200v on that plate. SO 100v dropped across that resistor. Now what if I find 300v on that plate? That means zero volts are dropping across that resistor instead of the expected 100v. Ohm's Law tells me that zero volts across 100k (or any resistor really) means zero current. That tells me the tube is not conducting current. leads me to wonder if the tube heater is working, or if the tube is dead. Ohm's Law.

[gui_tarzan]

See that example makes sense to me. The formula in itself doesn't speak to me like that example does. Had I or my early teachers recognized that I need that type of example that relates to things I'm doing or want to do, I might have been able to pick that up a long time ago. Before today I would have verified the heater was working, then swapped the tube. Until today it wouldn't have made sense logically that the higher than normal voltage reading was caused by a lack of current draw from the tube. Did I notice that in the past? Yes, but I didn't know why. Now that makes sense. No load to draw current to lower the working voltage. My mis-understanding at that time was that the voltage stayed constant, that current flow didn't affect it. Boy was I wrong.

Damn, Enzo... where were you when I was young? I would have loved to have had you as a teacher/mentor.