Can I convert a potentiometer to 2 resistors?

[gneill]

Yes!

Just to check,


I'm asked if there's a linear connection between how you set "P" in precent, to Vout

Now, I presume that when Vout is -8 volts it's considered a higher voltage than -5 and even 0, since "voltage" by definition is all about potential difference?

So, in my opinion there is a linear connection since Vout increases the lower the potentiometer is!

Linear implies a straight line relationship, so it doesn't matter if the values are positive or negative

You should be in a position to write an expression for Vout as a function of a to see if the relationship is linear or not (form y = mx + b).

 

[Femme_physics]

Ahh...there...

What I wrote is

"no , the relationship is not linear because according to the data P and Vout don't change at a constant rate"

I also proved it with math and graph

http://img811.imageshack.us/img811/5623/saifgimel.jpg

 

[gneill]

Ahh...there...

What I wrote is

"no , the relationship is not linear because according to the data P and Vout don't change at a constant rate"

I also proved it with math and graph

http://img811.imageshack.us/img811/5623/saifgimel.jpg

My apologies, but I suffered a brain fart when I agreed with your pronouncement that Vout is zero when a = 100% If a is 100% and so Va = 4V, the input current is zero since there's no potential difference across R1. That means that there's also no potential drop across R2, making Vout the same as Va: Vout = 4V.

For your graph, I see you've put Vout = 0V when a = 0%, yet your table has it at -8V.

I think you'll find that the relationship of Vout vs a is in fact a linear one...

Write an expression for Va given a in percent. Write an expression for Vout given Va. Substitute the first expression into the second and collect on "a".

 

[Femme_physics]

My apologies, but I suffered a brain fart when I agreed with your pronouncement that Vout is zero when a = 100% If a is 100% and so Va = 4V, the input current is zero since there's no potential difference across R1. That means that there's also no potential drop across R2, making Vout the same as Va: Vout = 4V.

How come? If Vb = 4[v] and we have the Earth point defined...current will always flow from high to low! And Vout must go through R2 and R1, which therefor experience voltage drop...

http://img339.imageshack.us/img339/8015/4voltsdifference.jpg

You're right that it's not 0... but if I use KVL I get


Sigma V = 0; Vout -4 -4 = 0

Vout = 8 [V]

 

[gneill]

How come? If Vb = 4[v] and we have the Earth point defined...current will always flow from high to low! And Vout must go through R2 and R1, which therefor experience voltage drop...

http://img339.imageshack.us/img339/8015/4voltsdifference.jpg

You're right that it's not 0... but if I use KVL I get


Sigma V = 0; Vout -4 -4 = 0

Vout = 8 [V]

Where's Vin in your diagram? It's a 4V source. It'll be happy to supply the current through the potentiometer to ground.

If Va is at the same potential as Vin, no current can flow through the input resistor since there will be no potential difference.

 

[Femme_physics]

You're right, come to think of it. So no current flows through R1 and R2. My original position was correct in that respect. I'm unfamiliar with the rule that says that if Va and Vb are 4 volts and no current flows through R2 and R1 then Vout must also equal 4 volts...although it does make sense...in this case there's no amplification, though, so the op-amp is kinda useless there

 

[gneill]

You're right, come to think of it. So no current flows through R1 and R2. My original position was correct in that respect. I'm unfamiliar with the rule that says that if Va and Vb are 4 volts and no current flows through R2 and R1 then Vout must also equal 4 volts...although it does make sense...in this case there's no amplification, though, so the op-amp is kinda useless there

It's just the circumstance for one particular setting of the potentiometer. The user can twist the knob to set a desired output, and 4V happens to be one choice. The op-amp is just doing its job to set Vout to a value that makes Va equal to Vb.

 

[Femme_physics]

Fair enough, but just to clarify:

For general cases of an op-amp, when Va = Vb and no current passes through R1 and R2, then Va = Vb = Vout

 

[gneill]

Fair enough, but just to clarify:

For general cases of an op-amp, when Va = Vb and no current passes through R1 and R2, then Va = Vb = Vout

It's certainly true for the given configuration where there's a single input resistance and feedback resistance. Essentially, when there's no current through a resistor the potential must be the same at either end of it. This is true whatever the circumstances. If that resistor happens to be the feedback resistor, then the potentials at the output and the op-amp inputs must be the same.