That's right, you will confuse people who are just learning this stuff. You should repost a correction. When I first started electronics, one little mistake in the numbers and it was so hard to learn. Still good info.
I just discovered your channel and it's exactly everything I need to know. No explanations on why it works, or its history. Just how to use it, I mean I'm not studying for collage, I'm just an 11 year old with some free time and access to materials.
As a life long design engineer, I think this is a good down to earth channel for all newbies, Just watch the calculations and formulas your advising on.
I really enjoyed this video. It is basic but helpful. It is a good refresh for me. Thanks for the help!!! I have been watching your videos and so far they have been very clear to understand.
Correct me if I'm wrong, but at 4'04" the current flowing in the circuit surely is 10 Volts divided by 110,000 Ohms, which is 0.00009 Amps which you were rounding up to 0.0001. It's therefore 0.1mA, not 0.01mA, and the power consumed/wasted by the resistors is 1mW, not 100uW. Not a criticism, your videos are very informative, just thought I'd point it out in case others were confused. I have liked and subscribed! :)
Fantastic demonstration! I like how You explained “Conventional Flow” and “Electron current Flow” in your AO 19 video. In this one you said the electrons move from pos to neg...oops! As an instructor, if I don’t quickly explain it both ways there is always polar arguments or confusion, unless I mention conventional or positive flow. The best way is both ways, because each is very important. Thank you for taking the time and extra effort to produce some of the best videos out there. 😊
You should have told that the resistance between Vout and GND (lets call it R3) should be at least 10 time bigger than the resistance of R2. This holds the voltagedrop at R3 as low as possible to provide the desired voltage. This is very critical if the wanted tolerance is small. But ok, the arduino analog inputs have a pretty high impedance so that about 200kohms in total should be ok.
A great video. An important subject simply explained. Please don't use buzzer sounds. I spilled coffee on my breadboard. Now my circuits are totally amped up.
Could anybody please explain to me why Vout is the voltage drop across the second resistor and not across the first one? I mean I'm trying really hard to visualise electricity using the water analogy and to me it makes more sense that the first resistance (R1) puts some sort of barrier in the direction of flow and therefore reduces pressure, so it kinda "eats up" a couple of volts from the source and what's left comes out the other side and gets channeled towards the Vout wire. Sorry I am really new to this.
That is exactly what is happening. Well “exactly” in the context of a crude analogy. Vout is the voltage drop after R1 drops some voltage. Kirchhoff’s Law. Volt drops in series add up to the voltage source. Voltage is the same in parallel branches, which is why Vout is the same as the drop across R2.
@@AddOhms omfg thanks I totally forgot about parallel voltage drop. It's been like 7 or 8 years since I've come into contact with anything remotely approaching circuits in highschool so I've been trying to relearn everything :)) Thanks a lot and keep making vids!
Bonjour monsieur, Merci pour vos video. Ma question est la suivante. Étant donné que l'on peut abaissé la tension avec un pont diviseur de tension, pourquoi utilise t-on les régulateur linéaire pour abaisser des tensions. Merci
Very nice tutorial (as always!), but, u forgot to mention the voltage drop man! In real life, it's so annoying to know that the voltage divider isn't reliable for more than 10 mA. Otherwise, u need a huge power resistors!
Yes and no. Datasheets and what is important varies by component type and by the application. So it is not possible to summerize how to read "any" data sheet. Something more general on a few component types is in my plan, but far down the road.
Great video, you mentioned wasting energy when lower value resistors are used. I often get into a quandary as to when to use a voltage divider or a voltage regulator when it comes to reducing waste. I would love to see a video around efficiency of a circuit and the pros and cons of a divider vs regulator.
Hey there, nice video! I have one question. Assuming a 10V battery, wouldn't be more simple to just put a 20ohm resistor to get a 5V output, instead of all this? Thanks!
Hi Sir, What is the Voltage at Point 1 (Before R1), Point 2 (After R1 & Before R2) and Point 3 (After R3). Vout was calculated at 1.13v, i thought it just passed R1, voltage should be at 7.86V, correct me if i am wrong.
You're confusing voltage drop of the individual resistor to the voltage relative to ground. The voltage across R1 is 7.86 volts. Put your voltage meter across it and that's what you'd measure. That means the voltage across R2 is 9 volts - 7.86 volts = 1.13 volts. Which is why Vout (or your Point 2) is 1.13 volts. R2 is connect to ground and Vout is relative to ground. Voltage drops in a series circuit add up.
In your initial example of 2 10Ω resistors, why is Vout not = to 0 volts? It passes through a voltage drop of 5v twice which means you are effectively reducing your input voltage to zero, or at least, that is what I would assume. According to the math at the end though, Vout = 2.5v
Voltage is pressure. So you have to look at the difference between two points. In the case of Vout, the assumption is "Vout compared to Ground." Since Vout is in parallel with the 2nd resistor, it's voltage drop is the same as the resistor. So it is 5V.
I think the answer you are looking for is that the Vout is the voltage that can be supplied from BETWEEN the two resistors, not the voltage after both resistors. The resistors take up all of the voltage combined, meaning you are looking to grab the drop after the first resistor, which can not be accomplished without more than one resistor. The voltage after the last resistor and before the negative terminal of the power source will always be basically nothing.
AWESOME! so here's a dumb question i'm sure, but here goes.......If i need 5v at the load, why i cant i simple replace R2 with the 5v load, then to ground?
Voltage drop is the difference between two points. Output voltage could called the "voltage drop" from a device's Output to Ground. (but I don't usually hear that.)
hmm interesting if say my voltage was 16v -12v at 12mA all I have to do is use a 3.3k and a 1k .Basically all I did was Vout = vin × r2/r1 + r2 .Is this correct?
How about a video on how you make your videos? I love your animations. If it's not something you wish to keep secret could you tell us what tool(s) you use to make them? You've put a lot of work into these and it's greatly appreciated!
I use a combination of Illustrator, After Effects, Motion, and Photoshop. The most important tool though is a very good script. I usually spend more time writing than I do animating.
Hey there! Great video and fun to watch. Just wondering how voltage is effected by adding a load in parallel to R2. It will obviously change a small bit as the R total will be less that the original R2? Maybe you have a video on this?
Neil Hayes If the load is constant, like a LED and current limiting resistor, then you just calculate that as part of R2. However, if the load is variable like a variable speed motor, it'll be difficult to determine what R2 will be at any given time. This is especially true if the load is an IC, like a microcontroller. The VCC to the microcontroller will be very (very) unstable.
Thanks for the reply! The solution to the problem that I had popped into my head just as I was going to bed last night lol. Your video helped a lot! Thanks again!
Voltage dividers only work well in cases where the next stage is high impedance, or draws very little current. Which is why they work well for references into something like a comparator or op amp.
The answer depends on the application. The larger the resistors, the less current, which means the divider is more susceptible to noise or interference. It's the same issue with high impedance inputs into chips and MCUs. If you're creating a divider that goes into an analog-to-digital converter, you have to consider its input impedance. For example, on an Arduino (aka ATmega328p) analog input, you want to keep the resistors in the 10k-100k range. If you used 1M range resistors, there isn't enough current for the ADC to operate correctly.
I noticed this only now... I don't know anyone let u know too... but the part where you calculated Voltage across R1 is 0.91V but the the brackets you wrote 91mV but that's 910mV ..... lol I'm seeing this in 2k15 😂
Nobody born knowin! You're okay. New stuff, you read and you do something else or rest for a while and you do that enough and words and terms get familiar. It's because your brain has to grow connections and that takes time. It's the same being body, hand, eye coordinated, nerves grow to do that. Start with literacy, the terms. Any term you don't know, you look it up, and the terms in that until you get to what you DO know and make the connection. If what you know is right then it will start making sense..
Can you please explain what is the role of the second resistor. I mean only one resistor can drop or divide the voltage so why use two resistors ? or what is the role of the second resistor. Btw nice editing and good way of teaching. ,,👏👏👏👏
So the v-out goes via the load to what? You're probably gonna say ground, but what about devices that arent connected to ground, like a cellphone or mp3 player?
Everything is connected to ground. Ground is THE reference point for the entire circuit. If the divider's output is connected to a circuit/load/whatever that doesn't share the same ground, then obviously, the divider won't work.
So, does that mean that all the lines in a schematic, that end with the ground symbol can be seen as connected to the same point, AKA each other? I'm really trying to understand this but my brain doesn't speak electricity.
Thank you. I just found found your channel like an hour ago, and i like the way you describe things. My father was an extremely talented electrician/electronics guy but he was never good at explaining things. Now, a bunch of years after his passing i suddenly got interested in how it all works. Subscribed.
But you cant use this technique as 9 volt battery to create 5 volts, that supplies good ammount of current that you can use? is there no way to do that?
BeucaN YOLO They're used to change the level of a signal (and not power). So like the example in the video, I am measuring the voltage of a battery, but not powering anything from the divider. Another example would be say you want a 5V device to talk to a 3.3V device. You could use a voltage divider from the 5v I/O to the 3.3v I/O, to drop the voltage down. Lastly, say you want to set a voltage for the Arduino (or any ADC) Analog Reference to something other than 1.1v, 3.3v, or 5v. You could use a divider to be whatever reference you need.
AddOhms well i wanted to create circuit to light an LED when you put "wet" finger with resistance around 0.5MegaOhms between two wires, i wanted to make a voltage divider to turn on an Bipolar transistor, but it doesnt work because it cant supply any current. Hoiw can I solve this without MOSFETs (i dont have any) or without darlington (it would make any finger not just wet! conduct )
BeucaN YOLO It CAN supply current but if you draw current from a voltage divider you are essentialy adding another resistor in parralel to the R2 !!! That increases the voltage drop across R1 and you get less voltage than expected. To solve this problem try this method: Make sure that the voltage divider will run at at least 10 times more current that you want to draw from it. That way you will be making the error significantly lower.
The LED creates a 3rd voltage drop. That drop depends on the amount of current, which is mostly determined by the resistor values. Location (or order) doesn’t matter in a series circuit.
Think that if you really want to make it easy to understand you should say 0.5 amps rather than 500mA. To a newer would be engineer, understanding how 10x500=5 isn’t that obvious
I have a question. On voltage dividers such as these is R2 always on the ground or negative side of the supply? Also to use a voltage divider for negative voltage, say -1.13v would you just need to swap the resistor values?
Quick Fix is needed at 3:10 because 0.91V isn't equal to 91mV, it is 910mV
Good eye. Damn.
I even scripted it as "910", but recorded it saying 91. Then mindlessly did the animation wrong.
Thanks
That's right, you will confuse people who are just learning this stuff. You should repost a correction. When I first started electronics, one little mistake in the numbers and it was so hard to learn. Still good info.
I noticed that too!!😁
Of the 1000’s of tutorial videos out there, I find your teaching style works really well for me.. Thank you
I'm starting to learn about voltage dividers and MOSFETS and found this channel explains things really well.
Please never stop making videos. Tour are an excelent teacher!
I just discovered your channel and it's exactly everything I need to know. No explanations on why it works, or its history. Just how to use it, I mean I'm not studying for collage, I'm just an 11 year old with some free time and access to materials.
What does "voltage drops" mean to you? That term was in the video.
As a life long design engineer, I think this is a good down to earth channel for all newbies, Just watch the calculations and formulas your advising on.
I love watching your videos they’re so clear and easy to understand..... keep up the good work 😃
Found this channel yesterday and nothing new for a year. Please. Well done. Do more when you can!!. THANKS!!!!!
+BladeforgerKLX Um. 7 videos since this one, and the last one was like a month ago.
Ack! My bad. Watching on the Kindle they don't sort the same. In that case, THANKS FOR ALL THE VIDEOS!!! I'm subscribed and intend to watch the ALL.
No problem. #20 is on its way. Script is done and already started the shooting / animating.
I really enjoyed this video. It is basic but helpful. It is a good refresh for me. Thanks for the help!!! I have been watching your videos and so far they have been very clear to understand.
best electronics channel so far
Wow, awesome video!!!! Now I understand why voltage dividers work, really really amazing how good you're at teaching!!
Correct me if I'm wrong, but at 4'04" the current flowing in the circuit surely is 10 Volts divided by 110,000 Ohms, which is 0.00009 Amps which you were rounding up to 0.0001. It's therefore 0.1mA, not 0.01mA, and the power consumed/wasted by the resistors is 1mW, not 100uW. Not a criticism, your videos are very informative, just thought I'd point it out in case others were confused. I have liked and subscribed! :)
+Spoolz07 Yup. I missed the zero.
also 1W is not less than 100uW (typo if someone wonders)
James tutorials are better than the other guy's.
Fantastic demonstration! I like how You explained “Conventional Flow” and “Electron current Flow” in your AO 19 video. In this one you said the electrons move from pos to neg...oops! As an instructor, if I don’t quickly explain it both ways there is always polar arguments or confusion, unless I mention conventional or positive flow. The best way is both ways, because each is very important. Thank you for taking the time and extra effort to produce some of the best videos out there. 😊
Excellent video, but when you mentioned voltage around the loop was conserved, I imagined a loud "ka-ching!" and "Kirchhoff" flashing on the screen.
You should have told that the resistance between Vout and GND (lets call it R3) should be at least 10 time bigger than the resistance of R2. This holds the voltagedrop at R3 as low as possible to provide the desired voltage.
This is very critical if the wanted tolerance is small.
But ok, the arduino analog inputs have a pretty high impedance so that about 200kohms in total should be ok.
It is awsooome!!!!!! Thank you!!
A great video. An important subject simply explained. Please don't use buzzer sounds. I spilled coffee on my breadboard. Now my circuits are totally amped up.
I think your coffee is making you jumpy
That reverb cracked me up, haha. xD
Thanks. :)
gracias, excelente forma de explicar...entendi en español sin traducir....
+anibal zuñiga Estoy feliz de escuchar eso.
Could anybody please explain to me why Vout is the voltage drop across the second resistor and not across the first one? I mean I'm trying really hard to visualise electricity using the water analogy and to me it makes more sense that the first resistance (R1) puts some sort of barrier in the direction of flow and therefore reduces pressure, so it kinda "eats up" a couple of volts from the source and what's left comes out the other side and gets channeled towards the Vout wire. Sorry I am really new to this.
That is exactly what is happening. Well “exactly” in the context of a crude analogy. Vout is the voltage drop after R1 drops some voltage.
Kirchhoff’s Law. Volt drops in series add up to the voltage source. Voltage is the same in parallel branches, which is why Vout is the same as the drop across R2.
@@AddOhms omfg thanks I totally forgot about parallel voltage drop. It's been like 7 or 8 years since I've come into contact with anything remotely approaching circuits in highschool so I've been trying to relearn everything :)) Thanks a lot and keep making vids!
Good video, thx
Bonjour monsieur,
Merci pour vos video. Ma question est la suivante. Étant donné que l'on peut abaissé la tension avec un pont diviseur de tension, pourquoi utilise t-on les régulateur linéaire pour abaisser des tensions. Merci
really like your videos. well done!
could you turn the sound effects down a bit on your next videos? they feel too loud...
Great!! is very interesting!!
GREAT videos, I am bald too. I went bald when I was 12:)
Very nice tutorial (as always!), but, u forgot to mention the voltage drop man!
In real life, it's so annoying to know that the voltage divider isn't reliable for more than 10 mA. Otherwise, u need a huge power resistors!
Good video thnx!
If we use three resistors as r1,r2,r3 to divide the voltage to get two Vout,
What will be the formula...?
You're videos Are AWESOME...!
+Vishnu Murugesan Just combine the two resistors for V1 and V2
yeahh,,,,
Thanks Man.....!
Can you make a video on "How to read any electonics component datasheet ? What are the paramters must have to care about ? "
Yes and no. Datasheets and what is important varies by component type and by the application. So it is not possible to summerize how to read "any" data sheet. Something more general on a few component types is in my plan, but far down the road.
AddOhms No problem sir. I will wait for your video. Keep spreading knowledge.
Good video... however, electrons do not flow from the +ve terminal to the -ve terminal. I think you meant 'conventional current'.
Aaqil Khan Yes, all of the AddOhms videos show current flow as conventional and not electron.
When do you use a voltage regulator rather than the divider system?
You almost never use a voltage divider to power a device. Dividers are meant for "signals" and references. Not power.
Great video, you mentioned wasting energy when lower value resistors are used. I often get into a quandary as to when to use a voltage divider or a voltage regulator when it comes to reducing waste. I would love to see a video around efficiency of a circuit and the pros and cons of a divider vs regulator.
You should never use a divider as a regulator: th-cam.com/video/-kEh0TYjYYE/w-d-xo.html
Hey there, nice video! I have one question. Assuming a 10V battery, wouldn't be more simple to just put a 20ohm resistor to get a 5V output, instead of all this? Thanks!
Leon Brasero No, that wouldn't work. You would still have the full 10V dropped across the single 20ohm resistor.
Hi Sir,
What is the Voltage at Point 1 (Before R1), Point 2 (After R1 & Before R2) and Point 3 (After R3).
Vout was calculated at 1.13v, i thought it just passed R1, voltage should be at 7.86V, correct me if i am wrong.
You're confusing voltage drop of the individual resistor to the voltage relative to ground. The voltage across R1 is 7.86 volts. Put your voltage meter across it and that's what you'd measure. That means the voltage across R2 is 9 volts - 7.86 volts = 1.13 volts. Which is why Vout (or your Point 2) is 1.13 volts. R2 is connect to ground and Vout is relative to ground.
Voltage drops in a series circuit add up.
In your initial example of 2 10Ω resistors, why is Vout not = to 0 volts? It passes through a voltage drop of 5v twice which means you are effectively reducing your input voltage to zero, or at least, that is what I would assume. According to the math at the end though, Vout = 2.5v
Voltage is pressure. So you have to look at the difference between two points. In the case of Vout, the assumption is "Vout compared to Ground." Since Vout is in parallel with the 2nd resistor, it's voltage drop is the same as the resistor. So it is 5V.
I think the answer you are looking for is that the Vout is the voltage that can be supplied from BETWEEN the two resistors, not the voltage after both resistors. The resistors take up all of the voltage combined, meaning you are looking to grab the drop after the first resistor, which can not be accomplished without more than one resistor. The voltage after the last resistor and before the negative terminal of the power source will always be basically nothing.
+Mr_Doublebutt yes that is what I was trying to understand. Thank you.
AWESOME! so here's a dumb question i'm sure, but here goes.......If i need 5v at the load, why i cant i simple replace R2 with the 5v load, then to ground?
I explain why you cannot in this video: Voltage Dividers as Regulators ?! th-cam.com/video/-kEh0TYjYYE/w-d-xo.html
How do you calculate the voltage drop of a resistor connected to a load (e.g. LED)?
th-cam.com/video/81zNcctopBI/w-d-xo.html
So voltage drop is the same as output voltage?
Voltage drop is the difference between two points. Output voltage could called the "voltage drop" from a device's Output to Ground. (but I don't usually hear that.)
ah ok I just watched your load video great to know diode acts as a load .What if we wanted to to add bf494 bias transistor to it ???
I don’t understand your question. Add a bias transistor to what?
@@AddOhms to attach to the voltage divider to make a oscillator or amp stage
Nice
Great series bald engineer. Just reporting another typo at 00:04:03 1W < 100uW
Thank u so much ... u r genious...
Can I use this circuit as voltage regulator ?
Nope. I explain why here: www.baldengineer.com/regulator-basics.html.
hmm interesting if say my voltage was 16v -12v at 12mA all I have to do is use a 3.3k and a 1k .Basically all I did was Vout = vin × r2/r1 + r2 .Is this correct?
What do you mean "at 12 mA?" Voltage dividers are not suitable for driving loads. th-cam.com/video/-kEh0TYjYYE/w-d-xo.html
How about a video on how you make your videos? I love your animations. If it's not something you wish to keep secret could you tell us what tool(s) you use to make them? You've put a lot of work into these and it's greatly appreciated!
I use a combination of Illustrator, After Effects, Motion, and Photoshop. The most important tool though is a very good script. I usually spend more time writing than I do animating.
kaltenbaugh If you look at this post, I go into more detail about how the videos are made: www.baldengineer.com/learning-a-new-tool.html
AddOhms Thank you so much! This is very helpful.
3:24 10X !
Hey there! Great video and fun to watch. Just wondering how voltage is effected by adding a load in parallel to R2. It will obviously change a small bit as the R total will be less that the original R2? Maybe you have a video on this?
Neil Hayes If the load is constant, like a LED and current limiting resistor, then you just calculate that as part of R2.
However, if the load is variable like a variable speed motor, it'll be difficult to determine what R2 will be at any given time.
This is especially true if the load is an IC, like a microcontroller. The VCC to the microcontroller will be very (very) unstable.
Thanks for the reply! The solution to the problem that I had popped into my head just as I was going to bed last night lol. Your video helped a lot! Thanks again!
SO Vout is that the voltage across R1 or the voltage across R2?
+Philip Pepperell R2
Thanks for the info.
But the truth is not the info it's the presentation skills that are cool
Why would the ardunio blow? Due to it running on 5v rather than 10v?
+Simon betts The absolute maximum rating for I/O pins is 6 volts. So 6 volts and higher damages the pin.
+AddOhms thank you AddOhms. By the way I find your show very interesting. :)
What is v out connected to? Where does that current go
Voltage dividers only work well in cases where the next stage is high impedance, or draws very little current. Which is why they work well for references into something like a comparator or op amp.
so bigger resistors mean less power loss. Is there a limit to how big they should go? what is the ideal amount of "wasted power"?
The answer depends on the application. The larger the resistors, the less current, which means the divider is more susceptible to noise or interference. It's the same issue with high impedance inputs into chips and MCUs. If you're creating a divider that goes into an analog-to-digital converter, you have to consider its input impedance. For example, on an Arduino (aka ATmega328p) analog input, you want to keep the resistors in the 10k-100k range. If you used 1M range resistors, there isn't enough current for the ADC to operate correctly.
@@AddOhms thank you
I noticed this only now... I don't know anyone let u know too...
but the part where you calculated Voltage across R1 is 0.91V but the the brackets you wrote 91mV but that's 910mV .....
lol I'm seeing this in 2k15 😂
+Ramya Narendran Yup, I missed a zero. I thought I had put an annotation in the video, better double check. Thanks.
+AddOhms 👍😃
AddOhms still no remark visible (at least in the ChromeCast, dunno every/any annotation would be shown)
It's definitely there.
What if you have three resistors lined up?
Depending on which point you make Vout, two of the resistors would be in series. Treat them as one.
Great channel sir. At 3:11 you say 0.91v == 91mv ... but it equals 910mv (0.91 x 1000 = 910)
In the formula to find Vout, you used R2/(R1+R2)*Vin, why NOT R1/(R1+R2)*Vin?
+RAKESH V Because you solve for the voltage across R2.
As much as I want to understand this stuff my brain won't let me learn it.... #deerinheadlights
Nobody born knowin! You're okay. New stuff, you read and you do something else or rest for a while and you do that enough and words and terms get familiar. It's because your brain has to grow connections and that takes time. It's the same being body, hand, eye coordinated, nerves grow to do that.
Start with literacy, the terms. Any term you don't know, you look it up, and the terms in that until you get to what you DO know and make the connection. If what you know is right then it will start making sense..
until now what is Ground??? how it work???
Amazing video, thank you so much! One thing, at 3:08, I think 0.91V = 910mV not 91mV?
buck, boost and buck-boost operation please thank you in advance
Covered here: th-cam.com/video/CPvA-uS0dNY/w-d-xo.html
Switching Regulator (Buck, Boost) Introduction | AO #18
Now... If you could change the value of one of those resistors... ;)
Quelle sont les inconvénients d'une alimentation à base d'un pont diviseur de tension
Why you cannot use voltage dividers as regulators: th-cam.com/video/-kEh0TYjYYE/w-d-xo.html
I like the way you explain the equations but the numbers don't add up.
Can you please explain what is the role of the second resistor. I mean only one resistor can drop or divide the voltage so why use two resistors ? or what is the role of the second resistor. Btw nice editing and good way of teaching. ,,👏👏👏👏
1 resistor drops the entire voltage. 2 resistors divide it.
@@AddOhms thanks. Btw
your fast response for my question and this video made me to subscribe.
So the v-out goes via the load to what? You're probably gonna say ground, but what about devices that arent connected to ground, like a cellphone or mp3 player?
Everything is connected to ground. Ground is THE reference point for the entire circuit. If the divider's output is connected to a circuit/load/whatever that doesn't share the same ground, then obviously, the divider won't work.
So, does that mean that all the lines in a schematic, that end with the ground symbol can be seen as connected to the same point, AKA each other?
I'm really trying to understand this but my brain doesn't speak electricity.
Yes.
Thank you. I just found found your channel like an hour ago, and i like the way you describe things.
My father was an extremely talented electrician/electronics guy but he was never good at explaining things.
Now, a bunch of years after his passing i suddenly got interested in how it all works.
Subscribed.
Say for instance I wanted to use 16v and at output would be 12v .Same idea with your 9v battery if I 9v ÷ 2 = 4.5v .4.5 volts would be the output
You can’t use a voltage divider as a source. You need a regulator.
@@AddOhms ok I understand now thanks
3:25 TEN TIMES BIGGER
Why not just use one resistor to drop voltage?
Explained in this video: Voltage Dividers as Regulators ?! | AO #22 th-cam.com/video/-kEh0TYjYYE/w-d-xo.html
Can you make tutorial on BJT ASAP
th-cam.com/video/sRVvUkK0U80/w-d-xo.html
@@AddOhms Can you make more tutorials on it
like basics of BJT, BJT as power, Current, Voltage amplifiers and so on
No
Why do we need dividers?
I can't believe not a single voltage divider video explains the effect of adding a load and how it affects the voltage and current ratio.
th-cam.com/video/-kEh0TYjYYE/w-d-xo.html
But you cant use this technique as 9 volt battery to create 5 volts, that supplies good ammount of current that you can use? is there no way to do that?
No, you cannot. Voltage dividers can't supply power to a device. You need a regulator for that.
why would voltage by voltage divider be useful then except arduino input pin or like mosfet?
BeucaN YOLO They're used to change the level of a signal (and not power). So like the example in the video, I am measuring the voltage of a battery, but not powering anything from the divider.
Another example would be say you want a 5V device to talk to a 3.3V device. You could use a voltage divider from the 5v I/O to the 3.3v I/O, to drop the voltage down.
Lastly, say you want to set a voltage for the Arduino (or any ADC) Analog Reference to something other than 1.1v, 3.3v, or 5v. You could use a divider to be whatever reference you need.
AddOhms well i wanted to create circuit to light an LED when you put "wet" finger with resistance around 0.5MegaOhms between two wires, i wanted to make a voltage divider to turn on an Bipolar transistor, but it doesnt work because it cant supply any current. Hoiw can I solve this without MOSFETs (i dont have any) or without darlington (it would make any finger not just wet! conduct )
BeucaN YOLO It CAN supply current but if you draw current from a voltage divider you are essentialy adding another resistor in parralel to the R2 !!! That increases the voltage drop across R1 and you get less voltage than expected. To solve this problem try this method: Make sure that the voltage divider will run at at least 10 times more current that you want to draw from it. That way you will be making the error significantly lower.
If your name is not Adam, that is one of the most wasted puns on Earth.
Well, it says my name in the beginning. The name of the channel has nothing to do with it.
So if you drop 5v twice in a 10v Circuit, the voltage past the second resistor would be 0, right? So if you put an LED on there; you’d have no light
The LED creates a 3rd voltage drop. That drop depends on the amount of current, which is mostly determined by the resistor values. Location (or order) doesn’t matter in a series circuit.
Think that if you really want to make it easy to understand you should say 0.5 amps rather than 500mA. To a newer would be engineer, understanding how 10x500=5 isn’t that obvious
I have a question. On voltage dividers such as these is R2 always on the ground or negative side of the supply? Also to use a voltage divider for negative voltage, say -1.13v would you just need to swap the resistor values?
A voltage divider won't provide a negative voltage.
How will you divide 17v DC into 12 & 5v DC?
You don’t use a divider. With a 12V regulator and a 5V regulator.