pot /w series parallel resistors not working properly...  - Page 1
Forum section: electronics / beginners

author: bigjoncoop
author_rank: Regular Contributor
timestamp: August 04, 2024, 03:45:45 pm
content: Im trying to modify a potentiometer to go from 80 ohms to 115 ohms by adding series / parallel resistors.I have a assortment of pots:1K 2K 5K 10K 20K 50K 100K 250K 500K 1M(Screenshots Below) ppIn the simulator if I have a 1k pot. I connected a 35Ω resistor between pins 1 & 2 and a 10Ω resistor between pins 2 & 3. Then I have one side of a 80Ω resistor conneced to pin 1.If I take a reading between the 80Ω resistor and pin 2, the pot does exactly what I need it to! It goes from 80Ω to 115Ω...Unfortunately after building it for real it doesn't work like that at all! The pot for some reason measures 115ohm turned either direction... I triple checked all my connections and made sure that the wiper in the simulator is the same as the pot I'm using. The wiper is in the middle.Below are some screenshots and a short 5 second video of the pot working in the simulator.If some can help me I would greatly appreciate it! Maybe point out what I'm doing wrong or suggest a simpler method to achieve this... THANKS!

author: tooki
author_rank: Super Contributor
timestamp: August 04, 2024, 04:34:02 pm
content: You sure you’re measuring the real-world pot at the wiper, and not the other end? Not all pots put the wiper in the middle pin (particularly multiturn pots and some trimmers).Have you made sure the pot works properly to begin with, without the parallel resistors installed?FYI, uploading animated gifs as attachments doesn’t work, the animation is lost entirely.

author: SteveThackery
author_rank: Super Contributor
timestamp: August 04, 2024, 04:52:04 pm
content: I've just worked it out on pen and paper, and I get a resistance range from 80 ohms to 113.8 ohms. So, I think you have made a mistake when you did the physical wiring up.Also, the 10 ohm resistor does nothing and you can replace it with a link.

author: ledtester
author_rank: Super Contributor
timestamp: August 04, 2024, 04:55:18 pm
content: Quote from: SteveThackery on August 04, 2024, 04:52:04 pmIn fact, it is shunted by the wiper and can be removed entirely.Update:Here is a Falstad simulation -- with a current source used to measure the resistance.(Link to Falstad Simulation)You can adjust the pot using the slider on the right panel. Note that no current flows the 10R resistor.The total resistance is indeed highly non-linear -- at the pot midpoint the resistance is already 33 Ohms.

author: SteveThackery
author_rank: Super Contributor
timestamp: August 04, 2024, 05:07:02 pm
content: Here is the equivalent circuit, without the 10 ohm resistor.Unfortunately the resistance varies in a highly non-linear manner. Graph to follow.

author: SteveThackery
author_rank: Super Contributor
timestamp: August 04, 2024, 05:37:05 pm
content: Sorry about the terrible quality! But as you can see, the action all happens in the first few degrees of rotation of the pot.

author: DavidAlfa
author_rank: Super Contributor
timestamp: August 04, 2024, 05:38:19 pm
content: Of course, 1K in parallel with 35ohms makes barely any difference.Only when the pot reaches 100omhs it's start to make some work, which is just the last 10% of the trim.

author: perieanuo
author_rank: Frequent Contributor
timestamp: August 05, 2024, 05:31:03 am
content: simple, make a digital one

author: pcprogrammer
author_rank: Super Contributor
timestamp: August 05, 2024, 05:44:00 am
content: Quote from: SteveThackery on August 04, 2024, 05:37:05 pmThis is true if the pot is linear.The OP did not state the type of potentiometer.When it is logarithmic it will be a different story.

author: SteveThackery
author_rank: Super Contributor
timestamp: August 05, 2024, 09:14:56 am
content: Quote from: pcprogrammer on August 05, 2024, 05:44:00 amYes, obviously!But we have to work with what we've been told. Plus, I wanted to illustrate to the OP the underlying principle that his approach will almost invariably lead to serious non-linearity. The curve I sketched is neither logarithmic or anti-logarithmic, so the response curve will be "interesting" whatever the case.Perhaps you would like to plot or sketch the response curve with a log pot, and then the OP will have even more useful information to work with.

author: tooki
author_rank: Super Contributor
timestamp: August 05, 2024, 09:29:05 am
content: Don’t hold your breath for an update. OP has a history of starting threads and never responding to them even once.

author: MrAl
author_rank: Super Contributor
timestamp: August 05, 2024, 10:55:38 am
content: Quote from: bigjoncoop on August 04, 2024, 03:45:45 pmHello,It does not look like the high end resistance would be 80+35=115 because there would be 1k in parallel with that 35 Ohms which would reduce the top end slightly. To get that 115 you would have to make that 35 Ohm resistor about 36.3 Ohms.You can also remove that 10 Ohm resistor.However, there is a bigger problem.In theory, all that above works just fine. In practice potentiometers do not usually go to zero Ohms at either end, there is usually some resistance that remains between the wiper and either end terminal. That means that when you adjust it fully counterclockwise (which would normally short out that 35 Ohm resistor) you could see a higher value than 80 Ohms. How high it is depends on the quality of the potentiometer.The pot is effectively being used as a rheostat where the resistance is being altered from 1k max to 35 Ohms max. That's a really big change. It would be better to try this with a 100 Ohm pot although it looks like you do not have one of those unfortunately.Maybe what you meant to do is connect the external leads to the two END terminals of the pot. Make that 80 Ohm resistor 70 Ohms and that 35 Ohm resistor around 45 Ohms, then that 10 Ohm resistor would actually do something useful. You'd still have to check the range of resistance though.What you should do is test the potentiometer separately without any other resistors and see what the end resistances are, and even if the pot is working right at all. If the pot is faulty, it will never work.