Voltage divider question
- Thread starter Locrian99
- Start date
owlexifry
Well-known member
Yes. My understanding is that you will end up with the same voltage.
However, there is some other nuance to the selection of those resistor values, but I'm not smart to explain how / why it matters, and to what degree... (it's minimal)
the difference between 47K/47K and 56K/56K is not likely to matter that much.
Locrian99
Well-known member
I think it affects impedance or something in some way. I was 99.99% sure it wouldn’t matter just want to double checking.
iamjackslackof
Well-known member
The voltage will be the same, but the current drawn will differ. I think it's negligible in this case since 47K and 56K are so close, but if you were to use a pair of 1M resistors it would affect the current draw much more.
Big Monk
Well-known member
Winner winner, chicken dinner.
This is the right answer. Voltage doesn’t care what resistors you use but current does.
Brett
Well-known member
Not quite.
Edited because much more concise answers were given below.
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Locrian99
Well-known member
Interesting. My way of thinking about it it would be the same as you have the same as a voltage divider which is basically second how much signal to send to ground and how much to let through. I could be thinking about this completely wrong though too which is why I ask. But it’s still just a resistance of how much signal is passing and how much is going to ground fully ccw we’ve got max resistance and everything is being allowed to go to ground. Fully cw we’ve got no resistance and max resistance going to ground. I suppose I could understand how a higher resistance to ground there could allow more volume.
Brett
Well-known member
Surely, someone wiser will respond with a bit more detail but I think the output pot also plays an important role in impedance.
Locrian99
Well-known member
I’m not by any means saying you’re wrong, heck I’m the one asking the question. Just like to wrap my head around this stuff. And your comment makes some sense.
BuddytheReow
Breadboard Baker
You can use nearly any value pot on an output control and it will work just fine. The problem is what comes after that effect. That's where the impedance part comes in. Standards are 50k or 100k to keep the impedance up there on output.
Big Monk
Well-known member
IF there is an output cap directly couple to the output volume/level control, value will matter because changing the resistance to ground changes the frequency content of the output signal.
An example would be the Fuzz Face where the output cap is kept constant 10nF but you tweak the Volume control from 100k up to 500k.
A 100kA pot all the way up will let more highs through to the output than a 500kA.
If you wanted to use a different pot but keep the frequency response the same, you’d tweak the cap along with the volume pot.
An example would be the Fuzz Face where the output cap is kept constant 10nF but you tweak the Volume control from 100k up to 500k.
A 100kA pot all the way up will let more highs through to the output than a 500kA.
If you wanted to use a different pot but keep the frequency response the same, you’d tweak the cap along with the volume pot.
Prolly a stupid question, but I gotta ask! Is it different than in a guitar volume pot? Where a larger value volume pot let’s more highs thru? I.e. Fender uses 250k to quell some of the highs, and Gibson, at least on the reissues, uses 500k pots to let more highs thru to get rid of the ‘wooly’ sounding neck pickups.
PunchySunshine
Well-known member
With the guitar, my understanding is that the source (pickups) are different so the pots need to match. I'm sure someone much more wise can answer with assuredness.
Big Monk
Well-known member
It is different.
In a guitar volume pot, the tone pot filtering is like a separate addendum to the signal. Signal comes into the pot and the filtering is separately tapped off the signal.
In the nominal output stage of a pedal, the cap AND volume pot are part of a filter in the signal path.
That makes sense! Thank you for replying.
giovanni
Well-known member
If you’re talking about the output pot at the end of a gain stage, I don’t think that’s completely accurate. The output impedance of a gain stage is determined by the circuit topology. I believe a fuzz face has a common emitter at the end which has Rc as output impedance. As long as Rc is smaller than the volume pot, it is the dominating output impedance and the frequency response will be the same. I would be surprised if Rc were not much smaller than 100K, in which case 100K vs 500K won't make an audible difference (the frequency response will be slightly different but probably not to our ears).
giovanni
Well-known member
I am not sure I follow what you mean here? I think the volume and tone pot are actually more interactive in a guitar wiring than in a pedal because there is no gain stage preceding them: the output impedance of the pickup determines the frequency response and "loads" the pots. This impedance is usually much higher than most gain stages, and higher for humbuckers, which is why we usually use higher pots with humbuckers.
Edit: the guitar tone pot is usually (non 50s wiring) one side of the pot in series with the cap, both in parallel with the volume pot. At very low volume the interaction between the three components and the pickup impedance has a substantial effect on high frequency content, which is why when you lower the volume on the guitar you lose some high frequencies (unless you use a treble bleed).
