Another voltage divider problem - Page 1
Forum section: electronics / beginners

author: Weez
author_rank: Contributor
timestamp: February 23, 2014, 03:14:02 am
content: After reading through Legion's voltage divider using potentiometers thread, I was reminded of a problem in the textbook I'm working through:Develop a variable voltage divider to provide output voltages ranging from a minimum of 10V to a maximum of 100V using a 120V source. The maximum voltage must be at the maximum resistance setting of the potentiometer. The minimum voltage must be at the minimum resistance (zero ohms) setting. The maximum current is to be 10 mA.I was able to find the correct answer, but I think I took the long way around in finding it.I'd like to see how some of you would go about solving this...probably something most of you could do in your sleep.

author: Galaxyrise
author_rank: Frequent Contributor
timestamp: February 23, 2014, 04:24:35 am
content: The simplest is probably a configuration where 120V is applied across an unchanging total resistance, and a pot is changing where the output voltage is tapped from. That means the "max current" is "the" current, so ohms law tells us the total resistance is 12k. And since the current is unchanging, Ohms law also tells us that 10ma generates 10V at 1k and 100V at 10k, supplying the resistance from ground to the pot wiper at the two settings. That leads me to:Code:Gnd------[9k pot]------120V^|VOutUsing the pot as a variable resistor instead, you could get something like this:Code:Gnd------^-+----120V|VOutthen you've got two ratios,(R1+0)/(R1+R2) = 10/120(R1+VR)/(R1+VR+R2) = 100/120The constraint of 10mA max current sets the resistances in the first equation; R1+R2 must be = 12k. Solving R1/(12k) = 10/120 yields R1 = 1k, and thus R2 = 11k.Plug that in to the second equation and solve for VR, then VR is 54k at the max setting.This second solution meets the criteria, but has a very functional difference. Do you see it?

author: Weez
author_rank: Contributor
timestamp: February 23, 2014, 05:37:20 am
content: Thanks Galaxyrise, I think I get the ratios you explained, but that's not how I solved it.Here's what the textbook showed as the solution, and what I eventually came up with:The 11k - 1k split was easy, what gave me trouble was the pot. What I did was simply start plugging in values and increasing them incrementally until I arrived at 54k. I knew there was a formula to use, but for the life of me I couldn't figure it out. I believe I was just over-thinking it and making it more difficult on myself.I'm sure if I was in a formal classroom setting I'd be absorbing this stuff differently, but I don't have that luxury right now.

author: IanB
author_rank: Super Contributor
timestamp: February 23, 2014, 06:08:42 am
content: Quote from: Weez on February 23, 2014, 05:37:20 amI doubt it. In my experience, the need to maintain schedule, the rapid pace of delivery and the need to accommodate all the students at once makes the formal classroom setting the worst place to absorb complex information.You absorb new things away from the stress of the classroom, doing tutorial examples and exercises by yourself or with a study partner, where you have time to reflect on the problem and pick it apart. Solving problems like this one until you understand them is how you learn.In short, an internet forum like this with helpful people in it is as good as it gets where learning is concerned.

author: IanB
author_rank: Super Contributor
timestamp: February 23, 2014, 06:28:56 am
content: Quote from: Weez on February 23, 2014, 03:14:02 amLooking at the wording, I guess the key phrase here is "variable" voltage divider. So let's see how to calculate that.For the minimum output voltage, we need a voltage divider with two resistances, such that:RB / (RA + RB) = 10 / 120For the maximum voltage we need to add more resistance to the divider, such that this time:(RB + RC) / (RA + RB + RC) = 100 / 120In the first case, can rearrange so that:120 RB = 10 (RA + RB)or110 RB = 10 RAthusRA = 11 RBNow we plug this result into the second equation:(RB + RC) / (11 RB + RB + RC) = 100 / 120or(RB + RC) / (12 RB + RC) = 100 / 120therefore120 (RB + RC) = 100 (12 RB + RC)Rearranging:20 RC = (1200 - 120) RBor RC = 54 RBLastly we need to find the value of RB. If the maximum current is to be 10 mA, the minimum resistance should be:Rmin = 120 / (10/1000) = 12 000 ohms = 12 kWe know thatRmin = RA + RBso that12 k = RA + RB = 11 RB + RB = 12 RBhenceRB = 1kRA = 11 RB = 11 kRC = 54 RB = 54 kwhich is the answer given.

author: Rigby
author_rank: Super Contributor
timestamp: February 23, 2014, 06:49:17 am
content: you made that look harder than it actually is, somehow.

author: IanB
author_rank: Super Contributor
timestamp: February 23, 2014, 07:10:19 am
content: Quote from: Rigby on February 23, 2014, 06:49:17 amMy teachers always taught me to work every problem in full, without missing out any steps.The time for taking shortcuts is later, when you are confident in what you are doing.

author: C
author_rank: Super Contributor
timestamp: February 23, 2014, 09:13:40 am
content: Are Both answers here a FAIL?Unless I am wrong both require a custom pot!Would the proper answer be a voltage divider that uses a standard pot?C

author: Weez
author_rank: Contributor
timestamp: February 23, 2014, 04:42:25 pm
content: @IanBThanks for the explanation, makes more sense to me now.@CIt is an odd value for a pot, but I don't believe that using standard values is expected ( I don't have the instructor copy of the text). As I work my way through the book, it does seem that some amount of thinking outside the box is expected/required from time to time. I appreciate that, even though it can be frustrating.I was obviously making it harder on myself, as I should have been able to rearrange already-known formulas to come up with the answer.Thanks for the repliesRichard

author: C
author_rank: Super Contributor
timestamp: February 23, 2014, 05:36:41 pm
content: Note that IanB's circuitThe current from the 120v changes with the setting of the potWhere Galaxyrise's circuitThe current from the 120v is constant with changes with the setting of the potand If you want a standard pot resultFor Galaxyrise's circuitChange the 9K pot to a 10k standard pot and parallel the pot with a 90k resistorC

author: IanB
author_rank: Super Contributor
timestamp: February 23, 2014, 06:05:39 pm
content: Quote from: C on February 23, 2014, 05:36:41 pmGalaxyrise's circuit is the simple and obvious configuration, but apparently it is not what the textbook was looking for.Perhaps the book was just presenting a contrived example (hence the non-standard pot value) as calculation practice?Or perhaps the book example has some advantage over the simple potentiometer arrangement?

author: uwezi
author_rank: Unknown
timestamp: February 23, 2014, 06:23:04 pm
content: I wonder what textbook this was...This could be a typical question which I give to my students, and I for sure would not expect the quoted textbook solution, but rather a potentiometer with two fixed resistors as in Galaxyrise's answer.Non-standard valuefor the potentiometer?normal off-the-shelf carbon potentiometers are most often ±5%, not rarely ±10%. Therefore a 10k potentiometer can in fact as easily be found to be 9k as 11k.Since potentiometers most often are used as variable voltage dividers, the exact value over the whole range is not really important.the quoted original question stated: "The maximum current is to be 10 mA."By changing Galaxyrise's solution to an accurate 10k potentiometer, we need to scale the upper resistor to 2.2k (E3+) and the lower resistor to 1.1k (E24+) - no problem here and the current woulr drop to 9 mA which is lower than 10 mA and hence a legitimate answer.

author: Weez
author_rank: Contributor
timestamp: February 24, 2014, 01:01:35 am
content: Quote from: uwezi on February 23, 2014, 06:23:04 pmSorry, I neglected to mention that. It's Electronics Fundamentals - Circuits, Devices, and Applications, 4th Ed. by Floyd

author: Weez
author_rank: Contributor
timestamp: February 24, 2014, 01:37:32 am
content: Quote from: uwezi link=topic=27274.msg393575#msg393575 date=1393179784[bWell, what I meant by that was, most pots come standard in whole numbers like 10k, 25k, 250k, 500k, and so on, and 54 is an odd value that is probably only available as a custom order. I don't know this for sure, it's just that I've not seen one like that. I do understand that the rating has a tolerance range, but I just assumed the book was looking for a precise value.Quote from: uwezi on February 23, 2014, 06:23:04 pmWell, I don't think so, because if the two resistors were 2.2k and 1.1k, then the minimum voltage and current (at the pot's "off" or zero ohms position - I do understand it's never truly "0") would be:120/3300 = 36mA current (too high).036 x 3300 = 118.8V (again, too high)We need a minimum of 12k when the pot is "off" to get the required 10V, as well as to keep the current within spec.Unless I'm misunderstanding something. Please correct me if I'm wrong, I'm here to learn.

author: IanB
author_rank: Super Contributor
timestamp: February 24, 2014, 02:03:02 am
content: Quote from: Weez on February 24, 2014, 01:37:32 amIn Galaxyrise's solution (which is the sensible, rational solution, by the way), the pot is never "off". The load resistance is always constant and is equal to 2.2 k + 10 k + 1.1 k = 13.3 k. Hence the load current is 120 / 13300 = 9 mA.In Galaxyrise's solution the pot is being used as a potentiometer, to tap off a variable voltage.In the apparent textbook solution the pot is being used as a rheostat, or variable resistor, to adjust a series resistance.The textbook solution is odd, and we don't understand why it is giving that as the answer. Can you give more context from the textbook to explain why it proposes that particular solution instead of a more conventional solution?

author: Weez
author_rank: Contributor
timestamp: February 24, 2014, 02:44:29 am
content: Quote from: IanB on February 24, 2014, 02:03:02 amAs in, tapping the voltagedropacross the pot?Quote from: IanB on February 24, 2014, 02:03:02 amI really don't know. The only thing the book has is a schematic like the one I drew. No explanation, nothing else.If I had to guess, it would be simply in keeping with the examples given in the voltage divider section(s). They always show Vout as being a "tap" in between resistors, not from pots.

author: IanB
author_rank: Super Contributor
timestamp: February 24, 2014, 02:56:44 am
content: The book seems to be misleading.This is a potentiometer:And this is a variable resistor (rheostat):A potentiometer literally means "variable voltage device". A potentiometer is nearly always used in a configuration similar to the first picture when variable voltages are needed (such as for the input attenuator or volume control on a pre-amplifier).

author: Weez
author_rank: Contributor
timestamp: February 24, 2014, 04:31:54 am
content: Fair enough, but, although there are a number of different types of pots, they are all just adjustable resistors, yes?So, calling it a rheostat or potentiometer is dependent upon its particular implementation in a circuit?In my (admittedly limited) experience, I've seen them drawn both ways in schematics, yet they're always just called "pots".Sorry to keep asking these dumb questions, but I'm a little confused now...maybe I should get a different book

author: IanB
author_rank: Super Contributor
timestamp: February 24, 2014, 07:56:56 am
content: Sure, it's just the same device wired up in different ways.But when you want to tap off a variable voltage, the arrangement in the first picture is what should immediately spring to mind.It's usually called a pot when it is a small device used in electronics for adjusting voltages.It's usually called a rheostat when it is a big, wirewound device used to adjust the brightness of a lamp or other such thing (rarely these days with the advent of electronic dimmers).

author: Galaxyrise
author_rank: Frequent Contributor
timestamp: February 25, 2014, 12:38:26 am
content: Quote from: Galaxyrise on February 23, 2014, 04:24:35 amQuote from: C on February 23, 2014, 05:36:41 pmI want to highlight this difference. One of the reasons my first solution is the "standard" configuration is because the output is linear (if using a linear potentiometer.) We've only been solving for the endpoints, but check out the graphs for voltage out with respect to wiper setting:This is using the potentiometer-based circuit:This is using the variable resistor-based circuit:Quote from: Weez on February 24, 2014, 04:31:54 amA variable resistor is just one varying resistance. A potentiometer is a variableratioof resistances. When using a potentiometer as a variable resistor, you're only using half of it.

author: rstor22
author_rank: Regular Contributor
timestamp: April 18, 2020, 06:15:51 pm
content: I also came across this question in Electronics Fundamentals A Systems Approach by Floyd and Buchla. The question and answer was similar to the one provided in Electronics Fundamentals Circuits Devices and Applications Eight Edition Floyd and Buchla which has the pot wired as a rheostat (see attached)In Principles of Electric Circuits Eighth Edition Floyd a similar question is found however the answer has the pot wired as a potentiometer (see attached)Interesting to note that in both questions it mentions the currentis to be10mA.

author: Wimberleytech
author_rank: Super Contributor
timestamp: April 18, 2020, 08:24:13 pm
content: Quote from: Weez on February 24, 2014, 04:31:54 amA resistor has two terminals.A potentiometer has three terminals.So it is not accurate to call a potentiometer a variable resistor.