# Trouble understanding pots/voltage dividers...

Discussion in 'Homework Help' started by aBarrett, Apr 17, 2012.

1. ### aBarrett Thread Starter New Member

Apr 17, 2012
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0
In the past month or two I've been teaching myself about electronics from the ground up with various textbooks (mainly Grob's Basic Electronics). But I'm stumped when it comes to potentiometers/voltage dividers. The concept of reducing voltage, especially through resistors, doesn't make any sense to me. I was under the impression that resistance only affects current, not voltage. But apparently pots are the exception to this rule... I just don't understand why.

EDIT: I should mention that I understand the mathematics. My question pertains to how a pot actually works.

Last edited: Apr 17, 2012
2. ### Wendy Moderator

Mar 24, 2008
21,604
2,978
How about two resistors? That is all a pot is, and circuits use 2 resistors all the time to create voltages.

If you have two resistors, the total controls current to a set level. Then Ohm's Law (if you haven't learned it yet you need to) will tell you what the voltage is on one of the resistors based on the current.

3. ### panic mode Senior Member

Oct 10, 2011
1,664
474
assume that pot is in some position (maybe about half the turn, does not really matter).
suppose top portion is R1 (portion from one end to wiper).
then the remainder is R2 (portion from wiper to other end).

when you turn pot backwards, wiper gets closer to end2 so R2 is smaller.
at the same time wiper is further from end1 so R1 is larger.
but R1+R2=R=const.

also since we are dealing with voltage (circuit is called "voltage divider"), then any 'disturbance' such as current drawn by load shall be negligible.

then we can say that only (noteworthy) current is through R1 and R2 (good approximation). and since there is no output current (towards the load), current through R1 and R2 must be same. (this point is known as KCL).

I=Vin/R=Vin/(R1+R2)

consider for simplicity that the input voltage Vin is constant. then the current is constant too because total resistance of potentiometer (end1 to end2 is R1+R2=R) is fixed. voltage drop across R1 or R2 is proportional to this current as well as resistance of particular segment (R1 or R2).

V1=I*R1 voltage drop across R1
V2=I*R2 voltage drop across R2

note:
Vin=V1+V2 (this is known as KVL)

note that output voltage V2 is

V2=I*R2=R2*I

V2=R2*Vin/(R1+R2)

V2=Vin* R2/(R1+R2)

note that R2 is 0-100% portion of R (where R is 100%).
so by adjusting position of potentiometer, we can cet 0-100% of Vin on output (V2).
this is valid regardless if Vin is fixed or changing.

4. ### aBarrett Thread Starter New Member

Apr 17, 2012
2
0
Aha! I was listening to Yes while drawing some diagrams off the top of my head, and suddenly it hit me like a rock. A potentiometer is a gate for a separate pathway. The "main" pathway is independent of the alternate pot pathway, but the pot pathway is dependent on the main pathway's current.

If a primary voltage source is providing 100V, and there's a pot with a resistance of 10 ohms, the current is therefore 10 amps. When the wiper is placed directly in the middle of the resistor, it creates a potential difference of 50V between pot's first and second terminal, because the current stays the same!

After really thinking about it and actually working with it on paper, it's fairly obvious. The current can't change in this context, and there's a variable amount of resistance. The voltage of the electrical source is irrelevant, which is what I was having trouble grasping.

5. ### WBahn Moderator

Mar 31, 2012
23,980
7,427
You've begun to understand it, but you're not quite there yet. In the case you discribed, the resistance (seen by the 100V source), isn't changing at all, which is why the current doesn't change. But the 100V is distributed across the two resistors differently depending on where the wiper is located. For instance, if the wiper is positioned so that there is 1 ohm between it and ground, then it will have 10V across it, while the other 9 ohms will have the remainding 100V across it. Notice that, in both cases, the current flowing through each portion is still 10A and that the voltage, current, resistance relationship imposed by Ohm's Law is satisfied for both (as well as the whole thing).

Now, this is all provided the wiper is not connected to anything. But let's say that you connect the wire to one end of the pot. Then the current flowing will change significantly as you move it because you have created a rheostat (a variable resistor). Similarly if you connect the wiper to some other point in the circuit, say to a resistor that goes to ground. As you move the wiper, the current in the pot will change.

As I think you have already discovered, when analyzing a circuit with a pot, replace it with two resistor having values $\alpha R\text{ and } (1-\alpha R)$ with the wiper connected to the junction between them and the other two ends representing the fixed ends of the pot. The parameter $\alpha$ can vary between 0 and 1 with 0.5 representing the wiper being right in the middle.

Last edited: Apr 18, 2012