Voltage Divider Circuits with Potentiometers - Page 1
Forum section: electronics / beginners

author: Legion
author_rank: Frequent Contributor
timestamp: February 22, 2014, 10:26:24 pm
content: I've been trying to solve these voltage divider circuit problems with pots but so far I've only managed to figure out how to find the wrong answer many different ways.The goal is to find the output voltage Vout without using Thevenin or current division. Just using the voltages and resistance values.My strategy has been:1.) Use the wiper position values to separate the pots into discrete resistors. So the first 5k would turn into a 4K followed by a 1K and so on.Looks like this:2.) Calculate the total resistance in the circuit.3.) Find the voltage at point A using this: Va = Vin * (Rtotal - 4k) / Rtotal4.) Find the output voltage using this: Vout = Va * (40k + 4k5) / (60k + 40k + 4k5)The answer I get is always wrong. I think you're supposed to use something like this:However, they never explain how these formulas are derived so I don't understand how to use them. As soon as the circuit presented is slightly different from the four examples I'm lost. If I could understand how they derived those formulas, maybe I could solve these types of questions.

author: IanB
author_rank: Super Contributor
timestamp: February 22, 2014, 10:37:41 pm
content: Your approach is generally correct, but you have neglected to mark a point "B" at the bottom of the 40 k resistor. Once you do this, you can calculate the voltage at point "B" similarly to the voltage at point "A". In the final step the 40 k and 60 k resistors will divide the voltage difference between points "A" and "B" and you will have the answer you seek.

author: IanB
author_rank: Super Contributor
timestamp: February 22, 2014, 10:40:58 pm
content: By the way, you can do much to avoid confusing yourself if you avoid the use of "ground" symbols anywhere on your diagrams. Draw all your circuits as closed loops with lines that show clearly where the currents flow. You will have a better understanding of how circuits behave if you do that.

author: Legion
author_rank: Frequent Contributor
timestamp: February 22, 2014, 10:49:22 pm
content: Quote from: IanB on February 22, 2014, 10:37:41 pmThank you so much! After spending 12 hours trying to solve that, it's nice to finally get the right answer.Could you explain why it was necessary to find Vb instead of the method I was using? What was I not accounting for that made my method wrong?Is it because I was treating half of a parallel circuit as being in series with the last resistor?

author: IanB
author_rank: Super Contributor
timestamp: February 22, 2014, 11:30:51 pm
content: If we draw the circuit as shown below, does it make things clearer?The circuit is symmetrical. Given the symmetry, why would it be necessary to find a voltage at point "A" on one side, but neglect the voltage at point "B" on the other?Once you forget any concept of "ground" and think of current flowing in loops (which is after all the meaning of the word "circuit"), then the need to treat all parts of the circuit with the same rules becomes more evident.From this diagram it should be possible to see that the required voltage is equal to VR6+ VR3. The voltage at point B is equal to VR6, and then you have to add on VR3to get the output voltage.

author: IanB
author_rank: Super Contributor
timestamp: February 22, 2014, 11:45:27 pm
content: Also, remember the basic circuit rules:1. The sum of the currents at any junction is equal to zero.2. The sum of the voltage steps around any loop is equal to zero.Therefore, in the diagram above, we must have:IR1+ IR2+ IR4= 0IR3+ IR5+ IR6= 025 + VR1+ VR4+ VR5+ VR6= 0VR2+ VR3+ VR5+ VR4= 025 + VR1+ VR2+ VR3+ VR6= 0etc.(In all the above equations, make sure to assign the appropriate positive or negative sign to the value according to the direction of measurement.)The last equation is the one you could use to solve your given problem.

author: Rerouter
author_rank: Super Contributor
timestamp: February 23, 2014, 12:00:37 am
content: I'm a bit old-school, but my approach is to find the equivalent resistance, which comes to 9977.8 ohmsfrom there you have a current of 2.506mA flowing through the series elements,thus you can trunicate them by removing there voltage drop,So A = 25-4000*0.002506 = 14.98Vand B = 4500*0.002506 = 11.27Vwe then take the division across the final pot which is 40% of this,vout = 0.4* (A - B) + B = 12.76V

author: Legion
author_rank: Frequent Contributor
timestamp: February 23, 2014, 12:01:14 am
content: Quote from: IanB on February 22, 2014, 11:30:51 pmIt's amazing how drawing the same circuit a different way can make it so much clearer. Thanks again.

author: aroby
author_rank: Regular Contributor
timestamp: February 26, 2014, 01:42:09 pm
content: Quote from: IanB on February 22, 2014, 11:30:51 pmI was able to figure out the answer from the original diagram, but this redraw got my completely confused! I can't figure out how to calculate VR3or VR6from this.Your point about drawing the complete circuit is a great tip - it especially helped me when I couldn't figure out why my initial answer was so low and then I realized that that value should not be from a 0v base, but from the voltage at B.Anthony

author: IanB
author_rank: Super Contributor
timestamp: February 26, 2014, 03:28:06 pm
content: Quote from: aroby on February 26, 2014, 01:42:09 pmJust apply the voltage divider formula in each case.First, you can replace the block of resistors between points A and B with a single resistor value--call it RAB.Now you know that:VR6/ R6 = 25 / ( R1 + RAB + R6)From which you can determine VR6.Similarly, find the voltage VABbetween points A and B.In step 2 you know that:VR3/ R3 = VAB/ (R2 + R3)So this gives you VR3.Lastly you compute:Vout = VR6+ VR3

author: aroby
author_rank: Regular Contributor
timestamp: February 26, 2014, 08:18:33 pm
content: Got it. Thanks for the explanation.Anthony