Help understanding voltage dividers - Page 1
Forum section: electronics / beginners

author: Frankentronics
author_rank: Regular Contributor
timestamp: April 28, 2020, 02:42:23 am
content: I've read several online articles and watched a few YouTube videos about voltage dividers. Although I fully understand all that is explained in those resources there are some unanswered questions I can't find online.If one of the two resistors is of a known value it is easy to calculate the other. But I can't understand what criteria is used to determine the value of the first resistor.One resource explained that you can take a random value for the first resistor and then calculate the other. Sounds reasonable, but I ran into some issues on the breadboard.To make things really easy I set up to take a 12V input and make it come out 6V at the other end to power up a circuit. So, I took a couple of 1K resistors and it all worked until the point when I hooked up a load. The load I chose (just what I had on hand) was a VC288 panel volt meter, which should work with 6V. But it didn't. The voltage which was 6V without any load dropped to 4V once I hooked up my load.Then I used two 10K resistors to see what happens. Again, I got 6V without a load, but as soon as I hooked up my load the voltage dropped to under 2V.So, then I took a couple of 10 Ohm resistors and this time I didn't even have time to measure the voltage because the resistors started smoking up, so I quickly yanked out the power from by breadboard.Clearly I am missing some critical information and none of the resources I looked into explain about that. Which the exception of the Khan Academy video which mentions that if you expect the current to be a bit higher then you should account for that in your calculations. End of video, no further explanation.If anyone could shed some light or point me to some specific resources that explain about that, I would appreciate it.Thank you...

author: bob91343
author_rank: Super Contributor
timestamp: April 28, 2020, 03:05:44 am
content: Hit the books. For the first part of your message, learn Thevenin's theorem. That will explain it.Basically, the divider has an impedance that causes the meter to measure low when it's connected. A high impedance meter will not cause the voltage to drop as much as a lower resistance unit.The power consumed by the circuit depends on the voltage and resistance, which is also explained in the books.You can't learn this stuff by the seat of your pants.

author: WattsThat
author_rank: Frequent Contributor
timestamp: April 28, 2020, 03:33:33 am
content: It’s classic rookie mistake, attempting to use a voltage divider to power something, that’s not the purpose of a divider. The current taken from a divider must be kept to a minimum.If you’re trying to power something at a lower than the supply voltage you have to know what the load is and how it behaves. If it’s current consumption changes, that change in current determines the resulting change in voltage, simple ohms law will give you that range. Ultimately, you may not be able to use resistance dropping to get to the desired voltage. If the circuit is okay with the range of voltage change, one resistor is all you need to drop the voltage, using two in a voltage divider configuration doesn’t change anything (in a practical sense).If you don’t have an answer to your question by now, consider your question faulty. Just any way of saying a better description of what you’re trying to do will get you better answer.

author: Cubdriver
author_rank: Unknown
timestamp: April 28, 2020, 07:56:21 am
content: The load that you're adding is a parallel resistance to whichever resistor you place it across. This parallel resistance lowers the total resistance of that section of the circuit, changing the overall balance of your voltage divider and increasingly the total current in the circuit. You then wind up dropping more voltage across the lone resistor and less across your 'new' parallel circuit.As you discovered by changing the resistor values in your divider circuit, this imbalance is greater when higher value resistors are used for the divider. This is because the load your adding has a greater influence on the higher value resistor - a parallel circuit's resistance will always be lower than the lowest value resistor in the circuit. Say for example the meter you were connecting had a resistance of 100k ohms. If you make your divider with 1k resistors, putting that 100k meter in parallel with one will result in the combination being the same as a 990 ohm resistor - a relatively small change of about 1%.Use 10k resistors in your divider and the meter will make it look like a 9.09k resistor - now the resistance of that side of your divider has decreased by just over 9% - a bigger change.For the last example, say you use 100k resistors for your divider. Now your meter will have a huge impact - when put in parallel with the 100k divider resistor, the 100k meter will halve the value of that section of the divider - it will now have an effective resistance of 50k - 50% of what you want it to be.Hope that makes sense to you.-Pat

author: Mp3
author_rank: Frequent Contributor
timestamp: April 28, 2020, 09:49:13 am
content: I made the mistake of trying to power stuff directly off a voltage divider. The way I see them implemented correctly in commercially made circuits, is they go through several other components before likely ending up at a regulator along the way.I would personally strongly suggest using a DC-DC buck/boost device. They are inexpensive (you should be able to get a handful for $10), readily available, very useful and very safe. I've used them in many circuits where I am adding something that needs voltages not found in the original circuit.MY first project where I attmepted to power other components directly off the voltage divider, ended fairly quickly with magic smoke.

author: tunk
author_rank: Super Contributor
timestamp: April 28, 2020, 10:11:43 am
content: Smoke: Not surprising, you're trying to dissipate7.2W through two (most likely) 0.25W resistors.Edit: I guess many of us have done things like this,and this one is quite easy to overlook.

author: Frankentronics
author_rank: Regular Contributor
timestamp: April 28, 2020, 12:09:14 pm
content: Thank you all... It looks like I just made the classic rookie mistake, as WattsThat pointed out. Good I yanked out that power wire from the breadboard, quickly.Quote from: Cubdriver on April 28, 2020, 07:56:21 amWhat you explained makes 100% sense. I think that's one piece of information that is missing from the tutorials I've studied. They should probably explain that at the very beginning.Clarification: I wasn't actually trying to build a circuit to power up my device. I simply used the first device I could grab to check how the circuit works. What I was in fact trying to do was to try and figure out what difference it would make to use two 10K resistors, instead of two 100K (or two 10 Ohm) to build a voltage divider. I still don't understand if it makes a difference and what would make one decide to use lower value vs. higher value resistors to get the same end result. Or is it really the same end result, and if not, what's the difference?But since we touched on the subject of powering up devices I now have additional questions.So, if I had a 12V source and I needed 6V for some device I could use a buck converter, as Mp3 pointed out. I guess I could also use a 7806, but I guess the voltage regulator is wasting energy into heat, correct? So, the efficiency is (I guess) why a buck converter is a better option, especially if my circuit is battery powered. Is my understanding correct on this issue?Also, let's imagine that I have a reason to use a 7806 instead of a buck converter and let's imagine that my source voltage is somewhat higher than the maximum recommended input voltage for the 7806 (which I believe is 25V). Could I use a voltage converter in that scenario, to lower the voltage coming into the 7806?And my related question would be, if I had say 20V coming into the 7806 and I wanted to have less energy wasted into heat, could I use a voltage divider to feed, say 9V, into my 7806, so that there's less heat? Or would that just make the resistors heat up?Thank you all...

author: Cubdriver
author_rank: Unknown
timestamp: April 28, 2020, 03:29:13 pm
content: Quote from: Adinol on April 28, 2020, 12:09:14 pmI'm glad to hear that my explanation helped. This is the sort of thing that's much easier to get across with lots of sketches and hand waving, as opposed to a wall-o-text, so it's good that you made sense of it.Quote from: AdinolIt does make a difference as I noted above in my earlier reply - the 'balance' of the higher value resistors will be affected much more by whatever you load the output with - that extra parallel path changes the drop across the 'output' resistor because it in effect makes it a smaller value. To get a better handle on it, sketch up some voltage dividers with various value resistors, and do the calculations. Then pick a resistor value to be your 'load' (to simulate your meter, or whatever else you might connect). Now add that resistor in parallel to the output of your several voltage divider sketches, and recalculate the currents and voltages with the new, lower effective parallel value of the 'output' resistor - this should clearly show how it changes the balance of the voltage drops in the circuit.In short, high value resistors are more efficient in that they don't draw as much current, but are more load-value sensitive. Low value resistors will change their output much less for a given load, but will draw far more power from the voltage source. Simple voltage dividers are not optimal for 'powering' much other than a known, fixed load value, and in that case that load value will be included in the calculations to determine the output resistor value (if there even is one - more likely the output device would in effect BE the output resistor, as is the case for instance with an LED and a series resistor which effectively for a voltage divider circuit). If the load is something that will vary, then the voltages will swing up and down with current draw changes and all bets are off.Quote from: AdinolYes. The gist is that a linear regulator acts like the first resistor in a voltage divider, with your load circuit acting as the second one. It monitors the output voltage and actively changes its resistance to maintain that voltage under varying load and supply conditions (within the limits it can handle). And yes, a linear regulator dissipates the excess energy as heat. A switching regulator is more efficient from a wasted power point of view.Quote from: AdinolYes, a buck converter would work there, and be more efficient - the closer to the output voltage plus some dropout margin that the input voltage to a linear is, the less energy it will need to dissipate as heat.Quote from: AdinolThe resistors would just be the things dissipating the heat, instead of the regulator - it would still need to go as heat. Another thing to keep in mind in that sort of circuit is that if the load goes away, then the current through the series resistor for all intents and purposes does too (other than the regulator's quiescent current draw, which is very low), and it will no longer be dropping voltage - practically the full supply could then appear at the input of your linear regulator. This condition could make the regulator unhappy.-Pat

author: John B
author_rank: Frequent Contributor
timestamp: April 28, 2020, 10:40:10 pm
content: Have a think about ohms law. The concept of a voltage divider, ie, the ratio of voltage drops is the same as the ratio of resistances, holds true because the current is the same through each resistor. So mathematically, a common term can be eliminated.When you load a voltage divider, ie "powering" something with it from some point in the divider, some current takes another path and so the current flowing through each resistor in the divider is no longer equal, therefore the ratio of voltage drops change.This is important to note because we are ALWAYS "powering" something with the divider. The important thing to consider is the ratio of current flowing through the divider vs current flowing into your connected load. The greater the divider current vs the load current, the less the error will be in the divider.For example if you were setting a bias voltage for the input of an op amp or comparator, the divider current may be in the order of mA, while the "load" current would be in the order of nA or pA. Therefore the error is inconsequential.However, as you found, if the divider current would normally be in the range of mA, and the load current would be in the range of mA (assuming if powered by a power supply of correct voltage), the voltage divider method wont work because of the large error.I guess you CAN make the divider current in the range of amps, but you also found that things get smoky

author: Frankentronics
author_rank: Regular Contributor
timestamp: April 28, 2020, 10:47:21 pm
content: Quote from: Cubdriver on April 28, 2020, 03:29:13 pmNow that you explain this detail it's one of those things that I can say, in retrospect, that it should have been obvious.Quote from: Cubdriver on April 28, 2020, 03:29:13 pmWhat you called a "switching regulator" is in fact a buck converter, correct?Am I correct to assume that a buck converter will generate noise and which would make it less suitable to be used for any audio applications?Thank you for taking the time to explain everything so well. It was really helpful.And I will definitely do those breadboard experiments. Nothing beats putting things into practice.Thank you very much...

author: Frankentronics
author_rank: Regular Contributor
timestamp: April 28, 2020, 10:55:46 pm
content: Quote from: John B on April 28, 2020, 10:40:10 pmThis is actually very nicely explained, from yet another perspective. It really is helpful to think of the same thing from two different perspectives.Quote from: John B on April 28, 2020, 10:40:10 pmI think that if things don't get smoky then we're missing out on the excitement of experimenting.Things got smoky for me 2 days ago when I soldered one of the two electrolytic caps backwards in a +/-20V rectifier. It was the one on the negative output, as I soldered the negative lead to the 0V trace. I didn't make the mistake on the breadboard but then I was not careful when I was populating the PCB. When I powered it up I first heard a noise that was increasing in volume and I thought it was coming from the fan. Then the cap erupted and I immediately realized what I had done.I won't do that mistake again.Thanks for your thoughts...

author: Cubdriver
author_rank: Unknown
timestamp: April 28, 2020, 11:17:24 pm
content: Quote from: Adinol on April 28, 2020, 10:47:21 pmIt's funny how many things are like that once the light bulb in your head goes off, but until that happens they're confusing as hell.Quote from: AdinolYes - a buck converter is a form of switching regulator, as is a boost converter (steps up the input voltage). There are also circuits that can do both.Quote from: AdinolA switcher will generate more electrical noise than a linear will, yes. Whether this will make it less suitable for any particular application is dependent on the regulator circuit and what it's powering. Whre I work we've had issues in the past with switching regulators in some RF circuits in the past, but have also been using them successfully in others - the answer (like so many answers in engineering) isit depends. Off the top of my head most switchers operate at supersonic frequencies, so they're probably less likely to cause issues in audio circuits.Quote from: AdinolGlad to help!-Pat

author: Jwillis
author_rank: Super Contributor
timestamp: April 29, 2020, 03:28:45 am
content: Voltage dividers also act as current limiters. To get the available current you need to divide the supplied voltage by R1 + R2 . So with 2 10K resistors in you divider the available current will only be 0.6 mA . If more amps are being drawn than what can be supplied the voltage will droop. Also keep an eye on the power rating as well . As you decrease the value of R1 and R2 the required power rating of the resistors goes up .

author: Frankentronics
author_rank: Regular Contributor
timestamp: April 29, 2020, 04:34:34 am
content: Thank you guys.I did some more Googling to understand how a voltage divider would be used in a transistor amplifier circuit.I found this articlehttps://electricalacademia.com/electronics/transistor-amplifier-working-circuit-npn-transistor-amplifier/.Specifically referring to this image in the article......it says, "The series resistance combination of RF and RB must be large enough so that the current drain from the supply battery is small. This ensures a long battery life."This clearly provides further clarification to my initial question about how to determine the value of the resistors in a voltage divider.I want to learn more about these calculations but I have to read up about it so that I can engage in discussions on a meaningful level.I think this bulletin board should have a category forum just to discuss math and calculations. Although I'm not sure how much people still calculate things in the old fashion way, in electronic engineering, as I'm sure it's mostly done by software. I have a mechanical engineering background and I always enjoyed calculating. But my mind understands mechanics better because it's something I can feel, touch and interact with. When it comes to electronics I understand tubes much better than semiconductors. Just the way my mind works, I guess.At the end of the day what I really want to learn is how to put together my own audio amplifiers (mostly guitar amps) as well as power supplies for those amps. And I'm also interested in developing battery powered guitar amps. Copying schematics that I find on the internet doesn't quite satisfy me because I don't fully understand what I'm copying.Hope this makes sense.Thanks...

author: Jwillis
author_rank: Super Contributor
timestamp: April 29, 2020, 06:18:29 am
content: You could think of a Triode much the same way as a NPN transistor . Think of the grid as the base , The plate as the collector and the kathode as the emitter. As you know when a negative voltage is applied to the grid electron flow from the cathode to the plate stops. As a small positive voltage is applied to the grid the voltage from the Kathode to the plate increases proportional to the tube's gain (amplification) factor or mu . Much the same way a transistor works. The only thing is a transistor is dependent on a small current at the base proportional to the gain or hfe of the transistor for a high current across the emitter and collector . Where as the Triode is voltage controlled . This of course is a very basic explanation but it helped me understand as well.

author: Frankentronics
author_rank: Regular Contributor
timestamp: April 29, 2020, 03:52:49 pm
content: Oh, yes, I am already familiar with the triode comparison. One thing that my mind is having trouble with is the fact that my mind thinks in terms of electron flow (i.e. what is actually happening) and that makes it easy to understand how tubes work. It is closer to a mechanical explanation imagining little particles flowing. In terms of transistors that emitter arrow is pointing the other way and we have to think of hole current (and I do understand that this is to eliminate all the negative signs from calculations). When I look at a transistor my mind tries to think of the electrons and that makes things confusing.I think it's time for me to read up some more on transistors and start thinking more in terms of hole current to make things easier for myself.

author: Jwillis
author_rank: Super Contributor
timestamp: April 29, 2020, 07:02:42 pm
content: Quote from: Adinol on April 29, 2020, 03:52:49 pmI know what you mean . That caught me in the beginning as well and then again when I started using MOSFETs . But half the fun is blowing some things up to. I like to take things to the limits to see what I can get away with . Having things go wrong has just as useful information as when things go right.I'm trying to put together a tube amplifier but stalled because of this quarantine . I can't get the materials from a local machine shop for the chassis . It's annoying . But I'd like to pick your brain now and again when I get started if you don't mind.

author: Frankentronics
author_rank: Regular Contributor
timestamp: April 30, 2020, 12:57:37 am
content: Quote from: Jwillis on April 29, 2020, 07:02:42 pmIf there's any way I can help or contribute I'm happy to do it.How do you think I can be of help with your project?