Voltage Dividers - Electronics Basics 12
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- Опубликовано: 19 авг 2024
- What are Voltage Dividers, and how do they work? Watch this video to find out: Try the circuit: goo.gl/a81Zpb
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Every little helps :)
This is by far the simplest explanation of voltage dividers on RUclips. A natural teacher giving only facts that are useful, no BS on this channel.
I agree w/ you 100%. I watched three other videos before this one, and seeing this explanation it all made sense-it was intuitive.
This animated way of teaching is great. I can actually understand what's being taught as it's being basically shown in action. Awesome stuff. Thank you.
That is literally the best explanation of voltage dividers I've run across, I've struggled to figure out how to apply Kirchhoff's law when figuring the resistance of two base-bias resistors in an NPN transistor, I can figure out one simple enough, but making a voltage divider for the base would make it more stable and thus I couldn't wrap my head around the math for figuring both resistors out. So thank you so much for that, count me subbed :)
What he said.
Best circuit tutorials on the internet. Thanks, mate.
Excellent explanation so people with no electronic can understand. Thank you for making this video! 👏👍
nice job you make electronics more simpler and easy to understand
You are a wonderful teacher. Thank you very much for your time you expended to make these videos. Great job.
thank you, this is the only video I could find that would actually explain what everything was and why we were doing this.
Really awesome explanation and formula. I just found out your channel and I think you are doing a good job. Thanks again!
Or, you can use the voltage divider formula...The value of the resistor in question, divided by the total resistance, multiplied by the total voltage!
Excellent explanation. I get it, but, remembering it is my problem.. I have to keep my Firefox bookmarks like a library of formulas.. It works for me..
A quick story : I went to school in 60's an 70's.. Learning was much more difficult cuz all we had was books and "a" teacher.. Both, teachers and books explain things in the way they think are best.. But if it wasn't right for you, You had few options.. Finding another book that explained it "Your way" was time consuming and frustrating.. They were only a couple of Teachers, at the most, and they were usually very busy.. Now-a-days, the Internet has lots of different teachers and text, to explain it the way you understand best. It makes learning stuff much easier. You can find what you need to know, in any way you need it.. Learning is fun again !
From the same generation, and totally totally right. Started electronics in 6th grade with Radio Shack Fairchild Heathkits. The only fast way to understand it was by figuring it out yourself, and reading books from city library. We definitely didn't have RUclips!
@@thurmanwatson9693 From the same generation, and what really kills me is that so many things are availabe for todays Generaion as learning material that is incredible yet they have great difficulty learning,,, I wish I had all those possibilities,,, I guess our generation wanted to make it easier to the next one, but with huge amount of information I guess the students are getting lost.
Yep kids get it easy these days
@@W3TFART Kids nowadays always complain,,, there don't understand how difficult live was back then. I had a house strait outof colege and a stable job and worked hard. If I had this information back then, I wouldve been much better of now,,, Cant change it I guess.
You had me at "Thank God for equations." :-) Good video. I learned something today.
Another excellent video. I am going work through your series. Maybe I will finally learn electronics!
Thank you very much. That was very helpful. Well explained, well animated, precise and short.
Thank you, I was reading a guide on esp8266, and they showed the formula, and I did quick calculation, and realized that im getting two answers, if I swapped the resistor value, and your video discusses why its like that.
this is so well explained... thank you so much!
Thanks for sharing this tutorial. I now have EveryCircuit and am enjoying it very much.
Kudos for your beautiful presentation!
Please give a link to the circuit engineering program you're using this tutorial.
Everycircuit.com
Thanks. I was having a hard time getting why.
so basically the drop to 0V is evenly shared by all resistor units (Ohm). The drop across a single resistor device is the proportional drop per unit (Ohm) times the units in that resistor. :)
Thank you for this video sir. This is very helpful. May I ask, what circuit simulator did you use?
thanks i guess we are left with the current divider...and the 4 major circuit analysis...i believe in u
Thanks man you just gave the clear solution of my problem ❣️❣️
Long doubt of only V & R circuit cleared , TNX 👍.
You say that it's hard to find a resistor with 7.14 Ohm's resistance. Wouldn't it be possible to tune a potentiometer to that resistance instead of using the closest one available? Or is there any economical/technical issue with that? Thanks for the videos!
Well, potentiometers are expensive compared to resisters, and if you start to put a load and suck more amps through your voltage divider, it will change the characteristics and your volts might drop
N maybe u could make parallel resistor to get 7.14 ohm's.
Great video. Although when you use that voltage it will change the value of resistance and the voltage will change. Am I right?
Thank you.
You are inspiring a beginner.,👍
Learning electronics becomes nice
it was awesome explaination. i really understand the voltage divider rule.thank you!
I love this series and it is quite helpful, but you lost me on this video. I get the math, but I just don't get the usefulness so it is hard for me to assimilate the information as useful. all of the previous videos seemed to me to do a better job. I am certain it will be clear as I move through and learn more, but wanted you to know about the experience I had. Thanks, and keep up the good work.
Curious to know if you have found a purpose for resistor dividers?
Like all the electronics videos on RUclips It goes zero to algebra in 3 seconds
This was great I totally get it. A little hard to remember that second equation though.
understandable , and interesting means of presentation . many thanks S.E !
Is that a schematics simulation application? If so could you share it with us? I would like to play around it.
Thanks for the playlist and especially this video.
simple explanation and neat representation.
sorry if this is a silly question but if you wanted 5 volts for example. why wouldn't you just use a larger single resistor to drop from say 12 v to 5 then connect your load?
It depends on what your load is, voltage dividers are generally for very low current applications such as a reference for an IC, where current will be minimal and quite stable.
This video is poorly motivated and very superficial. A voltage divider is not very efficient.
You may be correct but can you demonstrate it as clearly as Simlpe Electronics demos the principles of voltage division via resistors?
Omission with a lack of qualifications is not clarity.
If you can explain with clarity please do, otherwise why do you criticize ?
GREAT video
what software you use to simulate those schemes ?
everycircuit. it is $15
@@cachepage6261 no it's free to use. You can pay for more features if you want
@@j5892000 no. it's 24 hour TRIAL. Or you make new accounts every day.
@@GameBacardi no that's false
@@j5892000
everycircuit.com/
has 24 hour TRIAL, after trial you can't do nothing.
Period.
excellent sir....very clear explanation......
now i finally understand ! thanks for the awesome tutorial.
Simply Excellent - Thank You ☘️
THX So much. great link also
Can’t seem to pull it up and make it work. I downloaded the latest version of Chrome but still cannot get it to work is this application? Thanks. Excellent videos I think I put everyone I’ve been my favorites. Totally helpful thank you so much
I am loving your videos! Thank you sir!
Best explanation
i ve just subscribed to ur channel u r amazing sir...... tnx for the such valuable info
Good explanation. However why we need chips like 7805 to decrease voltage if we can do it simply by using a relevant resistor pair?
Brother, your explanations are neat and precise.. But as of now i can see your videos are bit long..and are less in number... You can add few other concepts like; Wheatstone bridge, working of switches, embedded system, complex analysis, star delta conversion, battery, short circuit, soldering, earthing much more.. And if you are an engineer then analogue, microprocessors, micro controlers would be highly appreciated🙂 hope for the best!!!!
Great video
this guy should teach, if you want my opinion ;)
May I ask that´s the ground needed for, if we anyway already have a negative-power-source-pole? And what did all the fuzz bring us after all? I mean, wasting my electricity over 1, 2 or 3 resistors won´t matter for me (and the exact mode/proportions of this wasting), as long as all I cared about, was to waste electricity. Where can one connect something on this circuit, in order to make it have some utility?
Well understood🎉
nice tutorial.very helpful.thanks
Or say your water is providing a bridge at 7 mega ohms of resistance, and you want to light an LED whenever the water makes the connection.. how do you light an LED, with 7 mega ohms of resistance?
I did a video about that. Thanks to excellent teachers such as yourself.
Hi, When current passes through a resister, Isn't it the current which should change since we are providing a constant voltage source? Sorry if it is a dumb question.
Hi, thank you for your great videos, What i know about resister is it is resist current not volt, please correct me if i am wrong, and What is the name of the simulator you are using
3:18 "forumula" it just means the voltage is divided proportionally over the resistors
i think the best way to think about it is that the current thru each component is constant. V_total/R_total= V/R = I (current) :)
@@andrescoca983 why?
@@andrescoca983 nice, that's elegant!
@@userou-ig1ze because that is the assumption this circuit and formula makes. In the example circuits, the current IS the same through R1 and R2. It cannot be different, in these example circuits.
If you use V2 (the portion of Vtotal, measured across R2) as the input to a Hi-Z (or high impedence) component (like an OpAmp), there will be "no" current flow, and so The total current through the divider network (Itotal) will not change, which is where this formula and such circuits are useful. If you parallel circuitry with R2, and it draws current, then Itotal changes, and this change occurs through R1, so I1 will change, and the whole calculation changes. Then you have to mess around calculating the effective resistance of R2 and the circuit part which is drawing current, using the total current through that part of the circuit (R2+parallel circuit) and pump those numbers back into the equation with R1 to get the actual voltage drop across R2.
Or just use a Hi-Z component, and be happy.
I hope you will do more videos on EveryCircuit.
thank you, can you make a video on amplifiers? it's really confusing
With regard to wanting a particular voltage from a voltage divider (3:48 in the video, trying to get 5 volts across R2), rather than calculating the value of R2 relative to a given R1 of 10 ohms (resulting in a very hard-to-find 7.14 ohm resistor needed for R2), couldn't you just replace the 10 ohm resistor (R1) with a 7 ohm resistor and then change R2 to 5 ohms?
this concept is cleared after this video
Maaaaaaan. Thank you for this!!!
great explanation
Correct me if I'm wrong, but that power rating would blow up those resistors at 5:11 ... I think?
If P = V x I you would get a power rating 4.9 watts on the first resistor.
there are many 100W or even 1000W resistors, extremely huge
I want to modulate AC and DC voltage by using voltage divider, means two sources AC and DC. Output should be about 0.6A. How can I select the Resistances ?
Use Ohms law. You know the V and the I, so calculate the R. Once you know the R needed at the point of measurement, you can then also calculate the 2nd R value to make up the divider circuit you need.
Fantastic, a channel that helps me, a software engineer, get into electronics! Q: why can't I just use this, rather than a buck stepdown converter? Is this less efficient? If so, is it significant if I'm just using it to step from 5 (arduino) to 3.3v (esp8266)?
Consider a voltage divider as "pass by value" to a function, as opposed to "pass by reference".
You're supplying a voltage level to your circuit without changing the current flowing through the R1+R2 divider. Say if you use V2 as the input to an OpAmp.
If you connect a circuit or component that DOES draw current, parallel to R2, then that WILL change current through the R1+R2, through R1, and the circuit changes. This is like a "pass by reference", which causes a change in the scope of the "calling" portion.
What computer program are you using. It’s great
Nice video ☺️ Thanks sir
That was good explanation, known as BIDMAS.
How much easier would it be to remember:
Vdrop = (R/Rtotal) x Vtotal
Also, if you want a specific ratio (say, divide by 10), then R1 and R2 must be in a ratio of 9:1. Because 9x(some ohms) + 1x(some ohms) is 10x(some ohms) [where (some ohms) is the same for each part]
So 9 ohms and 1ohm, or 900ohms and 100ohms, or 90kohms and 10kohms... there are 10 "units" and one side has 9 of those units (or 9/10ths of the units) and drops 9/10s of the voltage, while the other side has 1 of the units (or 1/10 of the units) and drops 1/10 of the voltage.
Divide by 2: is a 1:1 ratio, eg 1ohm + 1ohm
So 2 ohm total in the divider, 1 ohm is half of that total, so you get half of the total voltage.
Divide by 3: is a 2:1 ratio, eg 2ohm + 1ohm. (2/3rds and 1/3rd. If R1 is 2ohm and R2 is 1ohm, then R1 drops 2/3rds of the voltage and R2 drops 1/3rd)
Of course, it can be reversed to be 1ohm + 2ohm for divide by 3, or 1ohm + 9ohm for divide by 10 and so on. You decide where you put the resistor value which gives you the portion of Vtotal you want at that point.
Perhaps you want 90% (9/10ths) of Vtotal, then you put the 9ohm resistor where you want that voltage, and the 1ohm takes the other position.
The key to remember is for a given ratio, the parts must add up to the desired divisor. So for divide by 10 ratio must be 9:1 (because 9+1=10). And because you want 1/10th of Vtota, you would would use R2 = 1 unit = V2 or 1/10th of Vtotal (where "unit" is an order of magnitude ohms... like kiloohms or megaohms or milliohms or just ohms)
Also, for values like 10%, 20%, 30% etc, think of them as 10ths in the ratio:
10% is 1/10th so 100% = 90% + 10% = 9:1
20% is 2/10th so 100% = 80% + 20% = 8:2
30% is 3/10th so 100% = 70% + 30% = 7:3
and so on. To get 30% of the voltage, you might use 300ohm and 700ohm, using the 300ohm where you want to measure that 30%
what program is this? is it in the description? good video!
thank you
Thanks mate, great video ☺
Excellent Sir 👍💪👍👌👌
What program are you using?
Pls let me know that electrons flow from negative polarisation to positive or positive to negative ?
Electron flow is from negative to positive. Conventional is from + to -.
Current flows opposite to electron flow.
I feel "flow" is all about "charge". Conventional flow is the flow of positive charge. Electron flow, is clearly the flow of negative charge.
So, conventional flow is the opposite to electron flow.
As such, electrons are physically moving. They increase negative charge at their destination, and their absence increases positive charge at their source. So negative charge flows with electrons, and positive charge "flows" into the absence of negative charge.
@ 2:30 The equation is practical, but not very intuitive. I rather work out the current (1 Amp) in the circuit first and then calculate the voltage-drop over resistor-1: V = R . I = 2 Ohm x 1 Amp = 2 Volt. Now I understand what's going on. Nice video though.
So does this interactive doohickey only work if you pay the $15? I add the voltage meter across a resistor in the circuit, but then I can't get it to play. Or is there a button I don't see to get the current flowing again?
This app he's using?
Good second formula!
Good day. I have a problem you may have the answer to. i have a 120w amplifier and a VU meter that is rated at 60w. This VU meter connects to the output of the amp. I was wondering if i could use a resister to cut my wattage in half to safely run the VU meter.
A VU meter is made with a resistor in parallel to the wire wound "motor" of the meter. It basically measures the voltage across the resistor inside the meter. That resistor has a capacity, and running a larger supply through it would break it.
So you want to put an extra resistor in parallel with the meter, which will take 1/2 of the Power away. Then you're left with a VU meter which will read 1/2 of the true Power. Your scale would be wrong. The VU meter would work safely, but the reading would be inaccurate because the scale is 1/2 what it should be.
Check out some videos on shunt resistors to see how voltage is calculated according to current flow through a resistor, and you will understand how the VU meter works. Apply that knowledge to be able to calculate the necessary resistor to use in parallel with the VU meter, and then try it out. If you get it to work, don't forget to somehow change the scale printed on the meter.
Maybe you can find a data sheet for the particular meter which will tell you what the internal resistor value is, and just use the same value resistor in parallel with the meter.
At the 4:10 mark of the video resistor 1 is 2 ohms and resistor 2 is 10 ohms. Then when he goes to work out the equation, resistor 1is 10 ohms and resistor 2 is 2 ohms. Does that make a difference? Thanx ahead of time. Good video
The 2 ohm resistor will always have 2v and the 10 ohm resistor will always have 10v, using the example circuit, regardless of the order of the resistors.
i am useing hybrid 24/12volts on my m37 army truck via center tap 2 12 volt batt. dist is 24v also starter all else 12v for economy alter. is 28 v dc one wire conn no diodes or res used no prob 53 dodge just keep it simple
How much current can I take if my vout is paralleled to one of the resistors?
thank you. regards keep doing more videos please.
Great tutorial. What software did you use for the presentation? Thanks.
Very easy to understand, but how to get the simulation software?
Very helpfull, just one question. Say you have resistors of 1 - 10 ohm, only rounded numbers. Is there a method to calculate if there is any way to get the desired voltage with round numbers, treating both resistors as a variable, but rounded. And if there is, how do you do it? I know in practice this will probally never be a big problem, but I can imagine if you have really fragile equipment it might be.
I'm not quite sure I understand the question correctly. But, if you want to know the resistance of both resistors to get a desired voltage, there isn't an equation that will tell you what both resistors should be. However, you can work out the ratio between the two. First you need to know, supply voltage (Vin), and the desired voltage across a resistor(Vout).
Ratio = Desired Vout/(Vin - Vout).
If my supply voltage is 12V and desired voltage is 5V.
Ratio = 5 / (12 - 5)
Ratio = 1.4
I now have the ratio, but I still need to chose a random value resistor. Let's say I choose 10ohms for R2. Because the ratio should be 1.4, R1 should be 1.4 times the resistance of R2. In this case, R1 = 10 x 1.4. R2 = 14ohms.
If we now work out the voltage across R2...
V = (12v/24ohms) x R2
R2 Voltage = 5v
If you need rounded numbers and you get decimal answers, just round the result up or down to the nearest whole number. Your voltage won't be truly accurate, but good enough.
There are different voltage dividers such as accurate potentiometers that can provide much better accuracy and variable resistance.
Sorry if i didn't adequately answer your question.
play.google.com/store/apps/details?id=it.android.demi.elettronica&hl=en this tool has a calculator that will do just that
Thx fot the responses. When I wrote the question I was wondering if somebody would understand it, but it happens to be I got my answers. Thx!
no problem M8
Seems weird that in order to get more volts around resistor 2 you have to INCREASE the resistance 😅 they should change the size or opacity of the green dots to differentiate current and voltage
How will you implement a voltage divider circuit?
what was the name of the simulator software that is used by this particular page ? i want to know the name ! we use Labview
romel tilak the app is "every circuit"
when do we use voltage dividers in parallel or series Circuits ???
all parallel things have the same voltage
@@iwantitpaintedblack That's a risky generalisation. Risky in that someone may read too much into your statement.
Aya, I've left a few replies to other comments in this video. If you still need an answer, maybe one of my other replies can help?
@@CollinBaillie *Have the same voltage applied to them but may have a different voltage drop across them
corrected? :D
What is the practical application to this, what would it be used to do?
To reduce the voltage output of the circuit, incase the voltage output previously was too high for the load
0:12, if you put an LED after the 2 resistors (near the negative terminal), will it not work? Since the voltage is 0?
The circuit changes (obviously) so it will no longer be 0v. The 0v will be on the negative terminal of the LED.
Thank You
Smashing!
Thanks so much
What is the software used in this video please, and is there a Linux version?
He said it’s in the description. Hope it helps
Does anyone know if Everycircuit simulates the effect of meter loading?
Will it follow the same rules under ac?