Voltage divider - Page 1
Forum section: electronics / beginners

author: iXod
author_rank: Frequent Contributor
timestamp: August 12, 2020, 03:24:48 am
content: In this V divider, if I want to make the point between R5 and R6 adjustable, do I simply replace R5 with a 50k trimmer pot?Thanks.

author: ledtester
author_rank: Super Contributor
timestamp: August 12, 2020, 03:34:56 am
content: You would replace both R5 and R6 with a 50K potentiometer.See especially the section "Potentiometers" on this page:https://learn.sparkfun.com/tutorials/voltage-dividers/all

author: magic
author_rank: Super Contributor
timestamp: August 12, 2020, 06:07:09 am
content: If you want precision adjustment within a narrow range then leave both resistors in place and add a 10k or even smaller pot between them. If you want adjustment between the positive rail and the current value, replace just R5. If you want rail-to-rail adjustment, replace both with one pot. And so on.

author: iXod
author_rank: Frequent Contributor
timestamp: August 12, 2020, 06:19:43 am
content: Let’s say I have two instances of this circuit, unconnected.If I want adjustment between 5v rail and the present value,do I replace R5 with ~33k pot?If I want adjustment between Ov and the present value,do I replace R6 with ~20k pot?In any one circuit I will be using only one of the above choices.Thanks.

author: ledtester
author_rank: Super Contributor
timestamp: August 12, 2020, 06:34:56 am
content: The answer lies in this section of the link I mentioned above:Here R1 is 33K and R2 is 20K.What happens to Vout if R1 varies from 0 to 33K?What happens to Vout if R2 varies from 0 to 20K?

author: iXod
author_rank: Frequent Contributor
timestamp: August 12, 2020, 07:03:49 am
content: Varying R1 varies the voltage at the divider output from supply voltage to 37% of supply voltage (given that R2 is fixed).Varying R2 varies the voltage at the divider output from Ov to 37% of supply voltage (given that R1 is fixed).Next question: do I use just 2 of the pot’s 3 terminals (wiper and one “end” terminal)? Of course solder the unused terminal to an isolated pad?Thanks.

author: Ian.M
author_rank: Super Contributor
timestamp: August 12, 2020, 10:13:33 am
content: What you do with the unused pin of a pot used as a variable resistor depends on the application. The problem is the wiper is a moving contact so is inherently unreliable. You can pay more for a potentiometer with more wiper fingers and better mechanical construction to improve reliability, but there's always a finite chance the wiper will momentarily loose contact with the track. A professional grade design must tolerate that without letting the magic smoke out! The common practice of tying the wiper to the unused track end ensures that it cant go open circuit, but its resistance can and will jump to the maximum value if there is a wiper contact problem. Can your circuit tolerate that?Seehttps://www.eevblog.com/forum/projects/question-about-potentiometer-wiper-failure-in-bench-power-supplies/for an explanation of several ways to get it dangerously wrong, and a few PSU specific mitigation methods.

author: Rick Law
author_rank: Super Contributor
timestamp: August 12, 2020, 08:42:29 pm
content: I am a relatively inexperienced hobbyist, when I was an even less experience (then myself today), I tried to power my 3.3V components with just a 25 turn trimpot to the 5V source. I ran into a few issues. Let me share that with you so you may benefit from that experience.It may seem very attractive to make the voltage divider be adjustable the whole range (0 to 5 Volt in this case) and use just a trimpot alone without fix-resisters, but that may not be the best approach particularly when the circuit doesn't have a "power button" after the voltage divider.These are the issues:1. I don't know where the trimpot is at when I turn on the power. It could be at max (5V) which blows my 3.3V components right the way. So, I needed a "power switch" right there to disconnect the components. With that, I can first disconnect the components and adjust the trimpot down to a survivable voltage (in my case, 4V-ish), then press the power button to connect the component to adjust it to the desirable voltage. That works, but...The problem with such approach is I actually havea self-destruct mechanism build directly into the circuit. I just turn the trimpot the wrong direction too fast and I could get high enough voltage to blow my components. Eventually, I decided to add a fix resistor on the topside so my trimpot at max is just 4V.2. With that added top-side resistor, my adjustment is 0-4 volts, but it occur to me, most of the available adjustment are wasted. My component isn't going to run at below 2.5V-ish, so what's the point of having so many turns on my trimpot. So, I added a resistor at the bottom-side.So these days, I use (top) 1:2:2 (bottom) ratio. So my adjustment is 4V to 2V.- At max, I have top=1+0 (fixed top+trimpot) and bottom=2+2 (trimpot+fixed bottom)That would be a 1 (top) and 4 (bottom) giving me 4 volts.- At min, I have top=1+2 (fixed top+trimpot) and bottom=0+2 (trimpot+fixed bottom)That would be a 3 (top) and 2 (bottom) giving me 2 volts.The bottom part (B) is parallel to the whatever that voltage output is suppose to power, so the bottom part is going to be lower in resistance than just the 2 in the 1:2:2 ratio. As evaluation of parallel resistance go, you havesay"B" being the voltage divider's bottom part, and "C" being the components/parts receiving the power. The actual bottom resistance is going to beactual_B=1/((1/B)+(1/C)). The lower B is as compared to C, the less the actual_B is going to be effected by C and thus the closer it would be to 2V to 4V as the adjustment range.You don't want to use a total voltage divider resistance that is too high - a high total means B is going to be high as well. It may save power, but the divider math is going to increasingly inaccurate unless you factor in the resistance of the load C and evaluate actual_B instead of using just B. It just affects your adjustment range mostly. Once the voltage is adjusted, unless the resistance of C changes significantly during run time, the adjusted voltage is not going to be impacted by much.50K total is very high, but if it is already the recommendation, that means your component isn't going to need a lot of power and C is very high already as compared to B. A 10K top resister above the trimpot, a 20K trimpot, and a 20K bottom resister below the trimpot would work nicely.Personally, I would not use 50K total, 5K total would have 1mA wastage and I sleep better with:(top 1K fixed resister)(2K trimpot, middle is V out)(bottom 2K fixed resister)EDIT: Removed (by strike out) the word "say" which doesn't belong.

author: iXod
author_rank: Frequent Contributor
timestamp: August 12, 2020, 10:06:12 pm
content: Thanks for explaining "Rick's Law".Actually this is not a power supply circuit, it's setting a control input voltage to a fan controller which auto-adjusts the PWM to the fan. This voltage divider simply determines the max fan speed. No critical limits within the Vcc range.All your points are well-taken. Learn by others' mistakes. Some day I'll teach by all my mistakes. (Where's the "escaping Magic Smoke" emoji? I was sure EEVblog had one...).Cheers!

author: Psi
author_rank: Super Contributor
timestamp: August 12, 2020, 10:11:13 pm
content: Just be aware that if you do actually swap R5 and R6 for pots (two pots) then you now have the ability to make a short circuit. If you set both pots to near 0 ohms for example.To avoid that you would add a resistor in series with the R5 pot so that the current through the voltage divider cannot get absurdly high. Maybe 200 ohms. Then max current is 25mA no matter where the pots are set.But the only reason to have two pots, one for R5 and another for R6, would be to experiment around with how voltage dividers work.If you wanted to simply have a pot to adjust the voltage then both R5 and R6 would become the same pot. i.e. a single pot becomes both R5 and R6.This way the value of the pot limits the current and you don't have that short circuit problem as long as you spec a pot value that is over 200 ohms it will only ever draw 25mA.

author: iXod
author_rank: Frequent Contributor
timestamp: August 12, 2020, 10:19:27 pm
content: Yes, as I noted in one of the above posts, "In any one circuit I will be using only one of the above choices," either a pot for R5 or R6, don't need both for my application.Good to be reminded, tho.Cheers.

author: Rick Law
author_rank: Super Contributor
timestamp: August 13, 2020, 12:40:13 am
content: Quote from: iXod on August 12, 2020, 10:06:12 pmHow embarrassing. I missed the part that you were using it as input to a fan controller...Well, smart people learn from their own mistakes. Very smart people learn from other's mistakes. So while other smarter people learn from my mistake, lets hope I am at least smart enough to learn from my own mistake...