Voltage Divider - Page 1
Forum section: electronics / beginners

author: littlebill
author_rank: Regular Contributor
timestamp: June 08, 2015, 12:48:34 am
content: Can some one explain this a bit to a me,why are the resistor not in series, why is one tie up to the voltage being read, and other between the input on the arduino.is it as simple as grounding out the input for the arduino so it does not float?also how did he decide on the resistor values?https://startingelectronics.org/articles/arduino/measuring-voltage-with-arduino/

author: John Coloccia
author_rank: Super Contributor
timestamp: June 08, 2015, 01:09:44 am
content: The resistors ARE in series. The values were decided upon like this:1) he wants a high impedance input so it doesn't load down whatever you're measuring.2) he needs to reduce the voltage going to the Arduino because he wants a useful upper end of 50V. To do that, he figured out that he needs to divide the incoming voltage by about 11, so he makes a voltage divider that divides the voltage by 11.You can figure it all out by looking at the various voltage drops, or you can use the formula 100k/(100k + 1000k) = 1/11...so you have a divider that divides by 11 at the point it's fed into the arduino. If you wanted to divide it in half, for example, you'd use two resistors that were the same (perhaps a couple of 1Mohm). He needs to divide by 11 so that 50V falls within the Arduino's range.Does that make sense?

author: nessatse
author_rank: Regular Contributor
timestamp: June 08, 2015, 11:16:23 am
content: Also note that the author of the article did not consider the input impedance of the atmega328 ADC pin, which is in the order of 10k. Feeding the ADC straight from a 1M divider will affect the accuracy of the readings significantly, unless you are perhaps dealing with very slow changing DC voltages. Its always better to buffer the signal if you are dealing with high impedance sources like this.

author: mikerj
author_rank: Super Contributor
timestamp: June 08, 2015, 11:43:09 am
content: Quote from: nessatse on June 08, 2015, 11:16:23 amThe ADC input impedance is much higher than that. The 10k figure is the recommended minimumsourceimpedance if the datasheet performance of the ADC is to be met.However you are correct that the ~91k impedance of the voltage divider is too high to achieve the best accuracy from the ADC.

author: nessatse
author_rank: Regular Contributor
timestamp: June 08, 2015, 12:35:23 pm
content: Quote from: mikerj on June 08, 2015, 11:43:09 amYou are of course totally correct, I had that 10k number stuck in my memory somewhere, obviously not in quite the correct location.

author: littlebill
author_rank: Regular Contributor
timestamp: June 09, 2015, 12:06:02 am
content: guys dumb it down more please... lol

author: IanB
author_rank: Super Contributor
timestamp: June 09, 2015, 12:19:25 am
content: Quote from: littlebill on June 09, 2015, 12:06:02 amTry an analogy with a long stretchy rope, say 12 ft long. Fix one end of the rope to a post in the ground and pull the other end out until it is straight. Now make a mark on the rope 1 ft away from the post.Let's say you stretch the end of the rope out by 2 ft so it is 14 ft long. How far does the mark in the rope move near the post? If the rope stretches evenly the mark will have moved 2 inches away from the post (one twelfth of the distance).So you have made divider that can divide distance in proportion to length. Replace the rope with resistors and that is how a voltage divider works.

author: mikerj
author_rank: Super Contributor
timestamp: June 09, 2015, 12:46:39 pm
content: Quote from: littlebill on June 09, 2015, 12:06:02 amThings don't get much more simple than a potential divider to be honest. Are you familiar with Ohms law?

author: littlebill
author_rank: Regular Contributor
timestamp: June 09, 2015, 02:37:23 pm
content: i was speaking more about the tangent they were going off about on the input.i just didn't understand why they use 2 resistors and put the input between them, like i said i believe it was so the input wasn't floating.

author: mikerj
author_rank: Super Contributor
timestamp: June 09, 2015, 02:39:58 pm
content: Wikipediahas a reasonable article on voltage dividers which may help.

author: PSR B1257
author_rank: Frequent Contributor
timestamp: June 09, 2015, 02:55:27 pm
content: QuoteThe INPUT to the ADC is the OUTPUT of the voltage divider.

author: John Coloccia
author_rank: Super Contributor
timestamp: June 09, 2015, 03:58:14 pm
content: Quote from: littlebill on June 09, 2015, 02:37:23 pmThe two resistors form a voltage divider. Forget about the Arduino for a second....pretend it doesn't exist. All you have is a voltage source, and two resistors in series. If you put in a voltage V, and measure across BOTH resistors, you will measure V for a voltage. It's basically equivalent to measuring the source with respect to ground...do you see that?Now let's figure out the current across both resistors. For resistors in series, the total resistance is R1 + R2 = 1100k. Let's say you put in 10V.V = IR10V = I * 1100kI = 10V/1100k = 9.09uA. (ok...it's 9.090909090909..... uA, but 9.09 is close enough)So now we know that you will see 9 micro amps of current going through the circuit, for 10V.OK, so now let's figure out the voltage drops across the two resistors. We already know that if you measure across BOTH resistors, you must get 10V, so the voltage drop across the individual resistors must add up to 10V. That's a good way to check you're work.V=IRfor the 1M resistorV= 9.09uA * 1000k = 9.09Vfor the 100k resistorV = 9uA * 100k = .909V9.09V + .909V = 9.999V (would be exactly 10V except for rounding errors)So hookup the Arduino again. It's measuring the voltage across the small resistor...do you see that? Look where it's measuring. Pretend it was a voltmeter instead, and put one probe where the arduino hooks up, and the other one to ground. What's it measuring? It's across the small resistor. What's the voltage? It's .909V, which is 1/11 of the 10V input. See, it's dividing the voltage.... 1/11 gets dropped across the small resistor, and 10/11 gets dropped across the big resistor. If you measure across both resistors, you get the full voltage back (11/11 = 1)And you can use 3 resistors...and 4 resistors...and however many you want. You figure it out exactly the same way. Calculate the current through the entire circuit, and then calculate the drops across the individual resistors. Of course, you could be clever about it and notice that you really end up with the simple ratioR2/(R1+R2)This is what it means to have a voltage divider, and it's a very very very very important concept. From this concept, you not only build up simple dividers like this, just to scale the voltage, but more generally you will use this concept with capacitors to build various kinds of filters.I feel like you need to go back and study how to calculate voltages and currents through series and parallel circuits. There's no way to understand what's going on here without a solid understanding of that.I hope that helps.

author: IanB
author_rank: Super Contributor
timestamp: June 09, 2015, 04:16:52 pm
content: Quote from: littlebill on June 09, 2015, 02:37:23 pmDid you read the analogy of the stretchy rope? One end of the rope was tied to a stake in the ground. If you didn't fix one end of the rope to the ground like that the whole stretchy dividing distance thing wouldn't work. It's the same with the voltage divider. If you don't fix one end of the resistors to a ground reference they won't divide voltage.