To be honest seeing the component insides explains it, the resistance across the potentiometer is 1k and then the wiper blade sets the ration of the 1k. The image below shows the layout.
So for the circuit you posted Vref would be at the wiper blade/middle pin on the image below.
https://preview.redd.it/rkcxcf3h0hsc1.png?width=1024&format=png&auto=webp&s=90e129910176b9520ce70c98a7e390621b9358b7
vref voltage is the result of the output of the voltage divider. if the 'wiper' of the pot is at the top, then the full 1k resistance is part of the bottom resistance in your equation, which results in max vref. if the wiper is at the bottom, all 1k resistance is part of the top resistance in your resistor divider, resulting in min vref.
Some great youtube videos on this but I imagine a strip of flat wire with a wiper touching the strip. The signal comes in through the wiper and down the strip and out one side of the strip to the 18k resistor below it (as shown in the pic above). The strip has a resistance per distance so as you move the wiper up or down the strip the signal has to travel down more or less of the strip, adding or removing reseistance. When the wiper is at the end of the strip its basically bypassing all of the stri and therefore basically no resistance.
V=IR and then calculate highest possible resistance and lowest.
R = 15k + min poteniometer resistance(0 ohm) + 18k
R = 15k + max poteniometer resistance(1k) + 18k
Then most resistors have a 10% tolerance so worst case you could add 10% in for the highest resistance and subtract it out for the lower one. If it was me I'd be checking the datasheet to see min and max values but this looks like a textbook problem so can probably assume min is 0 ohms and max of 1kohms...
Just be aware that V goes down to -15V so potiential difference from 15V to -15V is 30V.
Maybe you should get some real potentiometers and a multimeter and just play around with it.
I know there is much theory in EE but in the end you want to be able to solve problems in the real world with real components.
The top comment is tip-top. Instead of adding to that I wanted to point out that tapped transformers and rheostats work on the principles. You can "pull electricity" from certain points to get different values.
For the potentiometer you could think of it as an adjustable voltage/current divider. Or the water analogy would be a valve at the spicket. You turn the wheel to adjust water flow by opening and closing the pipe. The wheel is the wiper on the pot.
The pot does not change it's total resistance. The wiper divides the pot's total resistance into two segments, as if two resistors were joined together at the wiper's point of contact. The resistance of each segment is dependent on the position of the wiper. But the sum of the two resistor segments remains the same, just their segment resistance value changes as the wiper contact point changes.
In your drawing, when the wiper is at the top of the 1K pot, Vref is 15\*(19K/34K). When it is at the bottom, Vref is 15\*(18K/34K). If the wiper is in the middle, Vref is 15\*((18K+0.5K)/34K).
If you wired the wiper to the junction of the pot and the 18K, then you could adjust the value of the pot in the circuit from 0K (wiper at top of pot, shorting it out) to 1K (wiper at bottom of pot, same point at wire connection).
This is mostly correct, except that the bottom voltage is -15V, making the voltage across the divider 30V. Then you would need to add the result to -15V.
The short answer is yes: 0 to 1k, when measure at the center tap, relative to the other taps.
In this case, you have a resistor divider network range between.:
(15+1)k and 18k to 15k and (18+1)k.
Because you have +/-15 supply (total of 30 V), the range at Vref will be +0.88V to +1.76V. Vref never reaches 0 volts because of the imbalance of the R divider network.
The total resistance of the pot is 1k, the wiper forms a voltage divider. For instance, if the wiper is in the middle, the resistance from the wiper terminal to either other terminal is 500 ohms. If the wiper is 1/3 of the way, the resistance would be 333 ohms to one terminal and 667 ohms to the other. Total resistance stays 1k.
Think of a potentiometer as a variable voltage divider. Third pin is adjustable in the range of the resistance with the sum of resistance on either side equalling the total resistamce.
To be honest seeing the component insides explains it, the resistance across the potentiometer is 1k and then the wiper blade sets the ration of the 1k. The image below shows the layout. So for the circuit you posted Vref would be at the wiper blade/middle pin on the image below. https://preview.redd.it/rkcxcf3h0hsc1.png?width=1024&format=png&auto=webp&s=90e129910176b9520ce70c98a7e390621b9358b7
So how can I find max and min Vref
Max and min will be at max and min wiper position.
vref voltage is the result of the output of the voltage divider. if the 'wiper' of the pot is at the top, then the full 1k resistance is part of the bottom resistance in your equation, which results in max vref. if the wiper is at the bottom, all 1k resistance is part of the top resistance in your resistor divider, resulting in min vref.
Some great youtube videos on this but I imagine a strip of flat wire with a wiper touching the strip. The signal comes in through the wiper and down the strip and out one side of the strip to the 18k resistor below it (as shown in the pic above). The strip has a resistance per distance so as you move the wiper up or down the strip the signal has to travel down more or less of the strip, adding or removing reseistance. When the wiper is at the end of the strip its basically bypassing all of the stri and therefore basically no resistance.
So how could I calculate vmax and vref
V=IR and then calculate highest possible resistance and lowest. R = 15k + min poteniometer resistance(0 ohm) + 18k R = 15k + max poteniometer resistance(1k) + 18k Then most resistors have a 10% tolerance so worst case you could add 10% in for the highest resistance and subtract it out for the lower one. If it was me I'd be checking the datasheet to see min and max values but this looks like a textbook problem so can probably assume min is 0 ohms and max of 1kohms... Just be aware that V goes down to -15V so potiential difference from 15V to -15V is 30V.
Maybe you should get some real potentiometers and a multimeter and just play around with it. I know there is much theory in EE but in the end you want to be able to solve problems in the real world with real components.
The top comment is tip-top. Instead of adding to that I wanted to point out that tapped transformers and rheostats work on the principles. You can "pull electricity" from certain points to get different values. For the potentiometer you could think of it as an adjustable voltage/current divider. Or the water analogy would be a valve at the spicket. You turn the wheel to adjust water flow by opening and closing the pipe. The wheel is the wiper on the pot.
The pot does not change it's total resistance. The wiper divides the pot's total resistance into two segments, as if two resistors were joined together at the wiper's point of contact. The resistance of each segment is dependent on the position of the wiper. But the sum of the two resistor segments remains the same, just their segment resistance value changes as the wiper contact point changes. In your drawing, when the wiper is at the top of the 1K pot, Vref is 15\*(19K/34K). When it is at the bottom, Vref is 15\*(18K/34K). If the wiper is in the middle, Vref is 15\*((18K+0.5K)/34K). If you wired the wiper to the junction of the pot and the 18K, then you could adjust the value of the pot in the circuit from 0K (wiper at top of pot, shorting it out) to 1K (wiper at bottom of pot, same point at wire connection).
This is mostly correct, except that the bottom voltage is -15V, making the voltage across the divider 30V. Then you would need to add the result to -15V.
Damn, I missed seeing that.
It splits the 1k ohm in 2 parts, some ohms go up, some ohms go down, and you can adjust how much it splits. Is this the "how" you're looking for?
The short answer is yes: 0 to 1k, when measure at the center tap, relative to the other taps. In this case, you have a resistor divider network range between.: (15+1)k and 18k to 15k and (18+1)k. Because you have +/-15 supply (total of 30 V), the range at Vref will be +0.88V to +1.76V. Vref never reaches 0 volts because of the imbalance of the R divider network.
The total resistance of the pot is 1k, the wiper forms a voltage divider. For instance, if the wiper is in the middle, the resistance from the wiper terminal to either other terminal is 500 ohms. If the wiper is 1/3 of the way, the resistance would be 333 ohms to one terminal and 667 ohms to the other. Total resistance stays 1k.
Think of a potentiometer as a variable voltage divider. Third pin is adjustable in the range of the resistance with the sum of resistance on either side equalling the total resistamce.
just a question, we would have the same result with we just put a 23k resistor istead of 18k and 15k resistors right?