Voltage Dividers - Electronics Basics 12
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- čas přidán 3. 08. 2016
- 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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This is by far the simplest explanation of voltage dividers on CZcams. 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.
Best circuit tutorials on the internet. Thanks, mate.
You are a wonderful teacher. Thank you very much for your time you expended to make these videos. Great job.
nice job you make electronics more simpler and easy to understand
Really awesome explanation and formula. I just found out your channel and I think you are doing a good job. Thanks again!
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.
Excellent explanation so people with no electronic can understand. Thank you for making this video! 👏👍
this is so well explained... thank you so much!
Another excellent video. I am going work through your series. Maybe I will finally learn electronics!
thank you, this is the only video I could find that would actually explain what everything was and why we were doing this.
Thanks for sharing this tutorial. I now have EveryCircuit and am enjoying it very much.
Thank you very much. That was very helpful. Well explained, well animated, precise and short.
Thanks for the playlist and especially this video.
I am loving your videos! Thank you sir!
now i finally understand ! thanks for the awesome tutorial.
I've never seen that version of the voltage divider equation before. Its so simple! Thank you.
You had me at "Thank God for equations." :-) Good video. I learned something today.
Kudos for your beautiful presentation!
Thanks man you just gave the clear solution of my problem ❣️❣️
simple explanation and neat representation.
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.
Great video
it was awesome explaination. i really understand the voltage divider rule.thank you!
Thank you for this video sir. This is very helpful. May I ask, what circuit simulator did you use?
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
excellent sir....very clear explanation......
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!
Thanks mate, great video ☺
Simply Excellent - Thank You ☘️
nice tutorial.very helpful.thanks
understandable , and interesting means of presentation . many thanks S.E !
Liked the explanation. Keep it up.
Thank you.
You are inspiring a beginner.,👍
i ve just subscribed to ur channel u r amazing sir...... tnx for the such valuable info
Long doubt of only V & R circuit cleared , TNX 👍.
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 CZcams!
@@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.
thanks i guess we are left with the current divider...and the 4 major circuit analysis...i believe in u
great explanation
thank you. regards keep doing more videos please.
Great tutorial. What software did you use for the presentation? Thanks.
Please give a link to the circuit engineering program you're using this tutorial.
Everycircuit.com
THX So much. great link also
I hope you will do more videos on EveryCircuit.
GREAT video
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.
Maaaaaaan. Thank you for this!!!
Nice video ☺️ Thanks sir
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.
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. :)
This was great I totally get it. A little hard to remember that second equation though.
Best explanation
Is that a schematics simulation application? If so could you share it with us? I would like to play around it.
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.
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
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
Excellent Sir 👍💪👍👌👌
Good explanation. However why we need chips like 7805 to decrease voltage if we can do it simply by using a relevant resistor pair?
Very easy to understand, but how to get the simulation software?
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.
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 ?
What computer program are you using. It’s great
great video, could you explain please what happens in this circuit after I connect LED in series after second resistor, lets say i change power supply to 2V, resistors to 1 Ω each, in between R2 and ground i add a LED ( 2V, 20mA ) , now my voltage is not divided equally, is it because LED drains these 2V?
Yes. If you consider that a 2v LED using 20mA, there must be a resistance across it which can be calculated.
V=IR
2=0.020 X R
R = 2/0.02
R = 100
So it acts like a 100 ohm resistor in the circuit (for calculations at least)
So you have 102 ohms total in series. The LED get 100/102 parts of that, and the two resistors get 2/100th or 1/50th of the total.
The LED gets 1.96v and the resistors get 0.04v total.
If you have 0.04v across both resistors, then one would get 0.02v across it, and the other would get 0.02v across it (because they're equal, they will get 1/2 each)
Well understood🎉
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?
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?
How much current can I take if my vout is paralleled to one of the resistors?
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"
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?
Does anyone know if Everycircuit simulates the effect of meter loading?
Thanks so much
thank you!
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!!!!
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.
what program is this? is it in the description? good 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.
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.
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?
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
this concept is cleared after this video
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.
Please tell me the name of the application you designed the circuit
That was good explanation, known as BIDMAS.
How will you implement a voltage divider circuit?
Learning electronics becomes nice
Will it follow the same rules under ac?
Thank You
thank you, can you make a video on amplifiers? it's really confusing
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.
Good second formula!
Come back dude!
Is this in IGCSE?
Can u compare Linear Voltage Regulators and this method? Which is better?
A voltage divider is all about getting a specific voltage level to part of your circuit. A Linear Voltage Regulator supplies both Voltage and Current to the circuit.
You use a voltage divider to set a voltage level to supply a component which "draws no current" like the input of an OpAmp.
What is the software used in this video please, and is there a Linux version?
what is the simulation software you are using.
What program are you using?
What happens when you have no load on your selected voltage? Doesn't this simply make an almost short causing the resistors to heat up to destructive temps?
But, the resistors are the load, in these simple examples.
At 1 or 2 ohms, the current flow would be significant, yes.
12v with 2ohms (total) resistance would give:
V=IR
V/R=I
I=12/2
I=6A
So, the circuit has 6A running through it. And we can work out the power given:
P=IV
P=6x12
P=72W
The resistors would have to be 36W each to handle that.
If you have large magnitude resistors (1k and 1k) the divider circuit gives the same result, but the total current in the circuit would be 1000 times lower, or 1000th of the calculated voltage for the 2ohm example, or 6mA. The Power will also then be 1000th, or 0.072w or 72mW.
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
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%
I am wondering if it's really as simple as that? Why wouldn't we only use one resistor? I have an old 12V 2A power supply and as it seems that 5V is very useful for many electronics applications so I only need a voltage divider as in the video to provide that 5V? Why have I seen some circuits that use some kind of IC would I also need to use one of those to make sure that the voltage from my old 12V is regular?
Answering your question about using only one resistor, you can think of the second resistance (in this video 10 ohm) as a separated circuit in which you need the potential difference of the two nodes which connect to the outer circuit (the whole circuit in this video) to be 10v. Because this inner circuit cannot handle 12v in those two nodes, you need a voltage drop that allows the potential difference to be 10v, and that's why you should add a resistance that will make a 2v drop across the circuit. The idea of a second resistance is an abstraction, you only care about the potential difference between the 2 nodes.
A voltage divider generally ISN'T used to supply POWER (that is, Voltage AND current).
A voltage divider is generally use to feed a voltage to a high impedance component or circuit which is considered to draw no current. Like an ADC or the input pin of a microcontroller, or an OpAmp. This is more like a reference voltage, not a supply to power a device, component or circuit.