Question about circuit flow

[Kyle]

Hi forum, I'm fairly new to electronics and I'm trying to get a handle on everything. I'm about to build NYDave's Re..p (just waiting on the Edcor transfomer) and I've got parts for a fuzzface on the way.

A pattern I frequently see in circuits is the center lug on a pot connecting the CW lug. What does that do? Is it the signal coming out of the resistor at that point and folding back on itself, or is it coming in there... I just don't what's happening to the signal there, and I'd appreciate any illumination on the topic.

An example of this is VR2 in the Re..p schematic:
http://www.ionrecords.com/tapeop/ampinterface.jpg

Thank you!!

-Kyle

 

[Svart]

That configuration is considered a "rheostat" or variable resistor. It works like you think with the signal flowing across the resistive element and the wiper just shunts the signal at the point where the wiper it touching.

 

[Kyle]

That much makes sense, but why does it shunt back into itself? Why wouldn't that shunt go to the output?

-Kyle

 

[clintrubber]

[quote author="Kyle"]That much makes sense, but why does it shunt back into itself? Why wouldn't that shunt go to the output?

-Kyle[/quote]

You mean why the third terminal is even used ? You're right, it's in fact redundant, you could as well just use one of the outsides and the wiper.

But since it's there anyway (you already paid for it) why not use it ? If the wiper-contact becomes faulty you still have the safeguard of at least the full resistance between the two circuit-connections.

But indeed, in theory it's of little use. But doing it like this does have a use in real-life.

Bye,

Peter

 

[Kyle]

Ok, starting to make a little more sense.

Let's say the signal is going into the pot, travels halfway through the resistive element and gets shunteohhhHHH!!!

I think I get it!

The shunted signal is discarded (or just cycled back into itself for redundancy), and the rest of the signal travels through the resistor! I was thinking of it the other way around.

Tell me if I have the concept right:

Signal goes into the CW lug, wiper lug shunts half the signal back into itself, and the other half of the signal travels unabated out the CCW lug to the output. If the wiper was all one way, all the signal would be shunted, if it was all the other way, none of the signal would be shunted (or just less of it?). Could the wiper also be shunted to ground?

-Kyle

 

[alk509]

[quote author="Kyle"]wiper lug shunts half the signal back into itself[/quote]

That's not how it works... Remember that current always seeks the path of least resistance. So the signal at the clockwise lug of the pot has a choice to go through the whole 50K resistance of the pot, or skip some of it by going through the wiper (which has close-to-zero resistance). How much of the whole resistance of the pot is skipped then depends on the position of the wiper - fully clockwise forces it to see the whole resistance, fully counterclockwise skips it completely.

 

[Kyle]

AHHHHH, Ok, that makes a lot of sense. So like ClintRubber was saying, the connection to the CW lug only exists as a failsafe against the wiper busting, and the signal is traveling INTO the pot (which make sense, there's even an arrow pointing to it).

Last question, I promise: Is there ANY signal going into the CW lug? Do the electrons just completely ignore that path in the presence of a less resistive route? Would its current measure 0?

Thank you all, this is great!

-Kyle

 

[Svart]

ok lets try to put all this together.

the pot can be used with any of the pins as input or output.

Lets say we have a 10k pot.

if you input the signal to the pin that is connected to the side of the trace with the lowest resistance and then hook up the output to the pin connected to the side of the trace with the highest part of the resistance then you now have a 10k resistor in series with the signal. You can now either hook the wiper pin to the input or output pin. we'll choose the output pin for this discussion. hooking the wiper to the output pin now allows you to choose any position on the resistive material at which to tap into the signal. moving the wiper to center on the pot now allows the signal to leave the trace through the wiper at 5k ohms instead of having to travel through the whole 10k ohms to the output. moving the wiper to the input side will lessen the resistance and moving the wiper to the output side will add more resistance up to the maximum value of the pot.

In this configuration you are dividing the resistance which equates to less or more signal.

i hope that made it a little more clear!

The electrons would for all practical purposes ignore the route of highest resistance. It's likely that you could measure some small amount of current elsewhere but not enough to be of concern.

 

[Kyle]

I think it makes sense now: It doesn't matter if you bind the wiper to the side you're using as an input or the side you're using as an output, either way you're still creating a less resistive path for the signal to flow. Excellent! A million thanks for the clarity.

 

[bcarso]

This reminds me of the Yogi Berra attribution: "When you come to a fork in the road, take it."

This is hard for a person to do---our body doesn't function well divided in pieces---but easy for macroscopic currents :grin:

Current flows proportionally to conductance. Conductance is the inverse of resistance, usually represented by the letter G.

When my current flow comes to a fork in the road like this example, it divides up and flows proportionally according the the conductance of one path vs. the other. A fraction x flows in one path and a fraction 1-x flows in the other. If the two conductances are G1 and G2, x = G1/(G1 + G2), 1-x = G2/(G1 + G2). G1 = 1/R1, G2 = 1/R2. Units of conductance have been dubbed Siemens, abbreviated S (capitalized), but many still prefer the old backwards spelling of ohm, mho(s).

In the case of the pot wiper, the conductance is, relative to the rest of the track, large---that is, relatively low resistance, maybe a tenth of an ohm or so, so G1 (let's call it) = 10 mhos. The remainder of the pot track has 5k based on the example---so G2 = .0002 mho. For whatever current is coming into the "input" end of the pot, fraction x of it flows into the wiper and 1-x flows into the remainder of the track. For this example, x is 0.99998, and 1-x is .0000199996.

Point is---despite the remainder of the track's current being negligible in this example, it is not zero.

At the opposite extreme, if my very bad wiper had 5k of resistance itself the current would divide equally.

Back to your original question---why the wiper is traditionally tied to the one end of the pot---clintrubber gave the answer. In most circuits it is preferable to have some resistance between point A and point B than none at all---and wipers of pots usually fail by ~open-circuiting.

 

[Svart]

In the case of the pot wiper, the conductance is, relative to the rest of the track, large---that is, relatively low resistance, maybe a tenth of an ohm or so, so G1 (let's call it) = 10 mhos. The remainder of the pot track has 5k based on the example---so G2 = .0002 mho. For whatever current is coming into the "input" end of the pot, fraction x of it flows into the wiper and 1-x flows into the remainder of the track. For this example, x is 0.99998, and 1-x is .0000199996.

like i said.. it's too small to worry about.. leave it to Bcarso to bust out the mind numbing equations! :green:

 

[Kyle]

At the opposite extreme, if my very bad wiper had 5k of resistance itself the current would divide equally.

Would that be modeled by throwing a normal 5k resistor between the wiper and the input lug?

 

[bcarso]

[quote author="Kyle"]

At the opposite extreme, if my very bad wiper had 5k of resistance itself the current would divide equally.

Would that be modeled by throwing a normal 5k resistor between the wiper and the input lug?[/quote]

You could do that---between the wiper and one or the other of the track contacts.

In fact some people, using the potentiometer as a voltage divider, like to use linear taper tracks and loading on the wiper to approximate an audio taper pot. If you have a dual or higher multiple of sections on a common shaft, the linear taper + loading R sections can track better than a stack of audio tapers.

 

[Kyle]

In fact some people, using the potentiometer as a voltage divider, like to use linear taper tracks and loading on the wiper to approximate an audio taper pot. If you have a dual or higher multiple of sections on a common shaft, the linear taper + loading R sections can track better than a stack of audio tapers.

How does that work? My understand is that an audio taper pot changes logorithmically as opposed to in a linear (duh) fashion, but I don't see how putting a just putting a load on the wiper would affect that result. Why wouldn't it just be a linear pot with a higher resistance value, like a constant. Would you (or anyone so inclined) be so kind as to elaborate?

 

[aurt]

[quote author="Kyle"]How does that work?[/quote]

Say you have a 50k linear pot. 10VDC at the top, 0V at the bottom. When the wiper is in the middle, it has 5V, right? It tracks proportionally. 3/4ths of the way up, 7.5V, 1/4th the way up 2.5V. Linear, as advertised.
Bypass with a 10kOhm resistor, wiper to ground. Now, when the wiper is at the top, it has 10V on it. The total load is no longer 50k, but just over 8k, 10k in parallel with 50k. Now, turn it 1/5th of the way down, and to be clear, I mean a fifth of the throw, not 72deg. You'll have 10k between wiper and voltage, and 8k between wiper and ground, because now you have 40k in parallel with the 10k, and 4.4V at the wiper. The next 1/5th turn gives you 20k to V+, 10k parallel with 30k, or 7500 Ohms, to ground, and somewhere around 2.7V. The next fifth of a turn you see 30k to V+, and 6.7k to ground, 1.8V. four-fifths down, 40k to V+, 5k to ground, 1.1V.
So, the first fifth of a turn has a large change in voltage. The rest progressively less. Approximates a log curve nicely if you pick the right values. I just used those because the V series monitor pot in my desk drawer uses them. :wink: Well, ok, it's 47k.

 

[Kyle]

I'm still trying to work out the proof myself (I think I've OD'd on math for the evening), but this is great stuff to start with. Thank you so much for your insight. I think I need to go to bed now before I turn into a total ohmosexual and I permanently lose the ability to dance or talk about anything normal.

-Kyle

 

[bcarso]

[quote author="Kyle"]

In fact some people, using the potentiometer as a voltage divider, like to use linear taper tracks and loading on the wiper to approximate an audio taper pot. If you have a dual or higher multiple of sections on a common shaft, the linear taper + loading R sections can track better than a stack of audio tapers.

How does that work? My understand is that an audio taper pot changes logorithmically as opposed to in a linear (duh) fashion, but I don't see how putting a just putting a load on the wiper would affect that result. Why wouldn't it just be a linear pot with a higher resistance value, like a constant. Would you (or anyone so inclined) be so kind as to elaborate?[/quote]

The key word in the first statement is "approximate." You will not get a log function or even a segment of one by loading a wiper of a linear pot. However, as aurt shows the approximation can be pretty useful.

"ohmosexual" :razz: I gotta write that one down!

 

[Greg]

Haha... that's one to remember... LOL