Thank you so much for this! As a complete beginner in electronics, I highly appreciate these videos. Hopefully I'll be able to work on some of my projects soon enough!
One of the best videos seen so far . I have seen so many of them in attempt to teach my son , but the way you explained is just awesome . Pls make more videos where kids can be taught to make simple circuits like clap switch or buzzer one . Thanks a lot !!! Great work !!
Ah, thank you so much! I finally powered my first object. I used a 330R and a 100R in series to power the red LED. Your explanation was so simple that even though I didnt have the 470R I still managed to figure it out. Thank you.
Hi Marie Claire. Yes it does. Each LED has a voltage drop across it of about 2v. Therefore, if you use a 6v battery, for example, and you have 3 LEDs, you won't need to use a resistor at all (3 x 2v). With 10 LEDs, you would need a 20v battery, so not really practical. I have 4 10mm white LEDs connected to a 12v battery, and that works ok. Each LED has a voltage drop of about 2.8v, so I'm not using a resistor in the circuit. Cheers, John
Great video! Just one question. Taking into account the ‘voltage drop’ that occurs across the led (9V-2V=7V) and the fact that you used a 470 ohm resistor, will the resulting current be about 14.9 mA in the circuit? (I=7V/470 ohms). Thanks!
I like this simple video. I am looking forward to learning more about electronics. I am a radio enthusiast. I have a radio which I would like to add an LED, to illuminate the dial when it is night time. Do you think this method will work for my project? Thank you for any advice and for helping out, 73's Lee.
Hi Lee, thanks very much for commenting on the video. You could try adding a night light, which will just come on when it is dark. I've added an example here - ruclips.net/video/-yg9h1GxZVw/видео.html. Cheers, John
I am using scrap resistors for this. I have 560Ω resistor and a 9.8 volt battery. I’ve heard that you need 20 to 10 mA for it to light the LED. I used a current calculator. And I got 17.5 mA. Will that still work with an LED?
How come people make circuits from the positive side to the negative? Shouldn't we construct them based on the rule that electrodes (-) are flowing towards the protons (+)?
Thanks for commenting on the video. Yes, I am continuing with the channel, but having a breather at the moment as it took quite a while to write the ebook and produce the videos. Thanks again, John
You could certainly use a 5v supply. You would just need to reduce the resistor value otherwise the LED will be quite dim. I haven't tried a usb power connection, so I'd be interested to know how you get on.
I am not sure how to interpret the flow of current (conventional vs non conventional) as the current flows from negative to positive.....so, is that a bottom rule that the schematics is always the other way around? Ergo, in real time the current flows through the LED from cathode to anode? ...if so, then the resistor could go either end of the LED? Doesn't matter which side ?
I made something like this in middle school. It was an led light and resistor that we connected to two prongs then solidified in a type of epoxy in an ice cube tray. We then could plug it into an outlet as a night light. Would this be the same concept if I replaced the battery in this diagram? Trying to replicate the one I made back then for my kids.
That's great James. It's a good idea to try various resistors to see what the effect is. Of course, if you go too low, the led might blow; too high, and it won't illuminate.
@@quickstartworkbook1532 a few days ago I connected four red LEDs in series with a 50 ohm resistor and they still work... I did however, when i first did it i managed to blow LEDs 😅🤦
really cool. so if you didnt use the resistor you'd blow the LED? I am assuming an option would have been to get a larger capacity LED? really interesting - thank you
Hi Lucy, thanks for commenting on the video. I just used standard jump wires that I bought from an online electronics retailer. Yes, you're correct that as you add More LEDs, the lowt resist value you need. Once you get to about 4 LEDs you don't really need a resistor. Check it out with Ohm's law. Cheers John
Hi great video. I'm new to this can I ask why you kept the current in mA? When I tried this I converted it to Amperes as I thought that's what would be needed to do the calculation in ohms law. Thanks
Hi - Thanks for watching the video and posting a comment. You're absolutely right, when using Ohm's Law you need to use volts, amps, and ohms for the units. So, 20 mA (milli-amps) would need to be converted into 0.020 Amps when it is used in the formula. Cheers, John
I am confused by this, from my understanding electrons flow from negative to positive yet some people state otherwise and put the resistor on the positive side and others the negative side, surely one way is wrong I mean if current is flowing in one direction the resistor should come before the led in order to stop it from frying no? So why are ppls diagrams so different in which direction the current is flowing.
jezmcm It is confusing. Ealier physicists assumed that electrons flow from positive end to the negetive. In fact, they discovered that they flow in the oposite direction from negative to positive. But even today you simply take the „older version“ also called conventional or „technical currentflow“ in german. It is not wrong to use the conventional method, you just realize, that your calculated current is negative for some reason, but why? Its really simple, it just proves that it flows to the other direction. So if you calculated like -20 mA, just switch the direction in which current is flowing and you‘re good to go. Sorry for the poor english, im obviously german haha.
MongoTV is right. You need to make the distinction between Conventional Current and Electron Current. Early pioneers used the Conventional Current thinking electrons were positively charged. Today we still use Conventional Current in our textbooks.
Hi - if I understand correctly, you have 5 LEDs - each having a 2V voltage drop - and 9V battery as the power supply. In this case, the total voltage drop of the combined LEDs (10V) is greater than the supply voltage (9V). The battery won't be sufficient to illuminate the LEDs. Inserting a resistor won't really make any difference. If you had 4 LEDs then you would need to use a very small resistor (about 50 ohms). Resistor value = voltage / current = (9-8) / 0.020 = 1 / 0.020 = 50 ohms Hope this helps Cheers, John
I can make the circuit work with 2 LEDs, but when I try to add more none of them light up. I don't know what I'm doing wrong. I check and recheck to make sure my components are plugged into the right holes, I confirm the LEDs are working. Yet when I add more than 2 I get nothing. So frustrating.
Each LED has a voltage drop of about 2v. So, if you have 3 LEDs, powered by a 6v battery, you don't need to use a resistor. If you have 3 LEDs and a 9v battery then you would need to use a small resistor, say something around 150 ohms. Try reducing the size of the resistor you are using. Keep reducing the size until the LEDs come on. Cheers, John
I have a robotics class decided to go off on my own and decided to make a stoplight and it instantly started to smoke so I need to pay attention to the resistors I guess.
350 would have been absolutely fine. I just used a slightly bigger resistor just to build a bit of additional safety into the circuit so that it would be perfectly safe with any LED. Cheers, John
Current = 20mA ...that seems like an arbitrary number; shouldn’t the Amperage be what the Battery provides? ...not an expert, just a noob wondering where that number actually comes from.
The 20mA is what an LED typically needs in order to light up. If you supply less, eg 10mA, the LED will be a bit dim; too much and you might blow the LED. Be at you know what the battery voltage is, you can select a suitable resistor to give you about 20mA.
Hmm, i must be doing something wrong. When I power my red light-emitting-diode I use a 1.5v battery in series with a switch connected to a red laser diode shining across the room into my solar cell which is connected to the positive input of an LM386 and ground, outputting nine volts direct-current into a 220 ohm resistor in series with my red light-emitting-diode. Yours seems much simpler. ...Lmao
@@quickstartworkbook1532 yea I knew It will go pop but I didn't have resistors AND a big question I powered a white led with 2 batteries it worked and changed to red led it goes pop or does not ilimuninate? What is the reason
I suspect that the white LED can probably cope with a higher current than the red ones, so it doesn't burn out - it maybe very close to doing so though. The red ones that appear to not illuminate might have burned out without it being apparent i.e. no "pop". With the red LEDs, try using one battery instead of two. Or, connect your LEDs in series, which reduces the size of resistor required. As you don't have a resistor, try adding 4 LEDs in series. They will probably look quite dim. Then try it with 3. I have a video that shows how to connect LEDs in series.
The voltage drop across an Led is typically 2 or 3 volts. So, if it's 3v, and you connect 3 LEDs in series, the total voltage drop is 9v. If you then connect a 9v battery as the power supply, you won't need to use any resistors because the total voltage drop = the total supply voltage. Good luck with your circuits.
Thank you so much for this! As a complete beginner in electronics, I highly appreciate these videos.
Hopefully I'll be able to work on some of my projects soon enough!
I've been interested in electronics for a while now, but just made my first LED circuit thanks to your video! Thank you!
Beautifully simple, not over wordy complicating things but all the information needed. Thank you.
Thanks very much for your positive feedback Mark 👍
Very well done. Easy to follow and explained perfectly.
One of the best videos seen so far . I have seen so many of them in attempt to teach my son , but the way you explained is just awesome . Pls make more videos where kids can be taught to make simple circuits like clap switch or buzzer one . Thanks a lot !!! Great work !!
Thanks very much Ankita for your kind comments. I'm glad you and your son found it useful. John
Ah, thank you so much! I finally powered my first object. I used a 330R and a 100R in series to power the red LED.
Your explanation was so simple that even though I didnt have the 470R I still managed to figure it out. Thank you.
Fantastic ! ... thanks very much for commenting on the video - much appreciated.
Clear as day, thank you!
Thank u! I watched a bunch of videos for a simple explanation. Urs worked! Now I get it!
i like how you showed which resistor to use, etc and how to get there, tks for that
Thank you!!!! This was really helpful and I am really happy that you made this video
Thanks for the feedback Jeff 👍
Is it possible for me to do this without the breadboard?
Is it possible to make a sandwhich without bread?!?!?!?!??!
Yes, you can use wire to connect the components.
@@quickstartworkbook1532 Thank you
@Juan Franco Penalba Gonzalez wassup
Tape
Good presentation. Much better than most of the videos demonstrating the same circuit.
Thank you for this video!
If I add 10 LEDs, does that change the resistor?
Hi Marie Claire. Yes it does. Each LED has a voltage drop across it of about 2v. Therefore, if you use a 6v battery, for example, and you have 3 LEDs, you won't need to use a resistor at all (3 x 2v). With 10 LEDs, you would need a 20v battery, so not really practical. I have 4 10mm white LEDs connected to a 12v battery, and that works ok. Each LED has a voltage drop of about 2.8v, so I'm not using a resistor in the circuit. Cheers, John
@@quickstartworkbook1532 Thank you very much for that, I did find another one of your videos that showed me how to do it with multiples!
Absolutely great Tutorial! Really clear an easy to understand. Also great presentation! Many thanks!
Much, much better than those videos that play music and there are no narratives
Really good your tutorial. I watched a lot a videos for beginners others are too complicated. Your tutorial is short but concise. You did a great 👍
Thanks very much for your positive feedback 👍
How good you are! Your step by step guidance were awesome!
Thank alot Very help full for beginners
Simple and easy to understand for kids
Great video! Just one question. Taking into account the ‘voltage drop’ that occurs across the led (9V-2V=7V) and the fact that you used a 470 ohm resistor, will the resulting current be about 14.9 mA in the circuit? (I=7V/470 ohms). Thanks!
Seems like it. A higher resistance gives a lower current.
What kind of LED, have you used ?
So if I had let's say 3 leds I would need 3 seperate resistors on the positives. But all 3 of those could go to a single line to the battery?
Basic but very informative.
Thanks very much for your positive feedback 👍
GOOD JOB, GOOD EXPLANATION
Thanks David
Great video thanks so much for the tutorial!
Nice, thanks for sharing :)
how are you supposed to know what the voltage drop of the LED is and how many mA of current they need?
Sometimes you'll get some specs with the LEDs when you buy them, or you can find generic specs online.
I like this simple video. I am looking forward to learning more about electronics. I am a radio enthusiast. I have a radio which I would like to add an LED, to illuminate the dial when it is night time. Do you think this method will work for my project? Thank you for any advice and for helping out, 73's Lee.
Hi Lee, thanks very much for commenting on the video. You could try adding a night light, which will just come on when it is dark. I've added an example here - ruclips.net/video/-yg9h1GxZVw/видео.html.
Cheers, John
amazing i learn something today ,,thank you
Great - I'm really pleased that you found the video useful
I am using scrap resistors for this. I have 560Ω resistor and a 9.8 volt battery. I’ve heard that you need 20 to 10 mA for it to light the LED. I used a current calculator. And I got 17.5 mA. Will that still work with an LED?
So if I put 5 led's the fifth led won't get power?? ( using a 9v bat )
How come people make circuits from the positive side to the negative?
Shouldn't we construct them based on the rule that electrodes (-) are flowing towards the protons (+)?
Enjoyed the video - just read the ebook - now getting into electronics. Are you continuing with this channel.
Thanks for commenting on the video. Yes, I am continuing with the channel, but having a breather at the moment as it took quite a while to write the ebook and produce the videos. Thanks again, John
Great video mate! Can you show us how to make a chaser light?
sir nice video. is is possible to use a 5v (usb connection) instead of 9V?
You could certainly use a 5v supply. You would just need to reduce the resistor value otherwise the LED will be quite dim. I haven't tried a usb power connection, so I'd be interested to know how you get on.
Thank you very much!
I am not sure how to interpret the flow of current (conventional vs non conventional) as the current flows from negative to positive.....so, is that a bottom rule that the schematics is always the other way around?
Ergo, in real time the current flows through the LED from cathode to anode?
...if so, then the resistor could go either end of the LED? Doesn't matter which side ?
That's correct. In fact, I've seen circuits where that is the case. Give is a go.
Thank you for the video. How would this change using 2 LEDs and would a large gap between LEDs make a difference?
Thanks for your comments. I have created a couple of videos that describe how to add more LEDs, either in series or parallel.
Beginners choice follow this channel .very easy to build circuits
Thanks for the positive feedback 👍
Thank you
Great!!
Thanks !!
I made something like this in middle school. It was an led light and resistor that we connected to two prongs then solidified in a type of epoxy in an ice cube tray. We then could plug it into an outlet as a night light. Would this be the same concept if I replaced the battery in this diagram? Trying to replicate the one I made back then for my kids.
Build a circuit for LED BAR display iski video mil skti hai plzz
i used a 330ohm resistor... worked perfecly
That's great James. It's a good idea to try various resistors to see what the effect is. Of course, if you go too low, the led might blow; too high, and it won't illuminate.
@@quickstartworkbook1532 a few days ago I connected four red LEDs in series with a 50 ohm resistor and they still work...
I did however, when i first did it i managed to blow LEDs 😅🤦
What if I use a pcb board and want my name using LEDs, how will I get the positive and negative from all the LEDs combined? Thanks!
really cool. so if you didnt use the resistor you'd blow the LED? I am assuming an option would have been to get a larger capacity LED? really interesting - thank you
If you use a 1.5V battery as the supply, you don't need to use a resistor.
Does it matter which holes you put the components into, when building the circuit?
Yes. Check out any video on using a breadboard.
Is there some way knowing what is the current is the led?
It's normally about 20mA. Check the specs for your LEDs to see if it is stated there.
How did you figure out which resistors to use so the led won't fry? I mean, the result was 350 and you have chosen 470. Thanks!
you can always use a larger resistor, doesn't matter as far as I understand it
Its better to have greater resistance than less, in that way you will prevent the led from burning though it will light a little dimmer
and if i added more LEDs I'd need less resistance?
Hi Lucy, thanks for commenting on the video. I just used standard jump wires that I bought from an online electronics retailer. Yes, you're correct that as you add More LEDs, the lowt resist value you need. Once you get to about 4 LEDs you don't really need a resistor. Check it out with Ohm's law. Cheers John
so now what if im trying to use 10 LEDS for a model?
thanks for this! what kind of wire do i need to get?
Lucy Paffard just any jumper cable from what I understand
gracias por su enseñanza podria decirme un led efecto flama que resistencia necesita
Should we use an alkaline or lithium battery?
Either should be fine - just check the voltage to make sure you don't burn out the LED
Hi sir, Does this work with the smallest breadboard?
Yes, you can build this circuit on a tiny breadboard if you need to.
Why is the voltage “lost” in the LED
There's a good description of voltage drop on Wikipedia, which is the voltage that is lost in, say, an LED.
Hi great video. I'm new to this can I ask why you kept the current in mA? When I tried this I converted it to Amperes as I thought that's what would be needed to do the calculation in ohms law. Thanks
Hi - Thanks for watching the video and posting a comment. You're absolutely right, when using Ohm's Law you need to use volts, amps, and ohms for the units. So, 20 mA (milli-amps) would need to be converted into 0.020 Amps when it is used in the formula. Cheers, John
That’s just what he did in the demo. 2:05.
I am confused by this, from my understanding electrons flow from negative to positive yet some people state otherwise and put the resistor on the positive side and others the negative side, surely one way is wrong I mean if current is flowing in one direction the resistor should come before the led in order to stop it from frying no? So why are ppls diagrams so different in which direction the current is flowing.
jezmcm It is confusing. Ealier physicists assumed that electrons flow from positive end to the negetive. In fact, they discovered that they flow in the oposite direction from negative to positive. But even today you simply take the „older version“ also called conventional or „technical currentflow“ in german. It is not wrong to use the conventional method, you just realize, that your calculated current is negative for some reason, but why? Its really simple, it just proves that it flows to the other direction. So if you calculated like -20 mA, just switch the direction in which current is flowing and you‘re good to go. Sorry for the poor english, im obviously german haha.
MongoTV is right. You need to make the distinction between Conventional Current and Electron Current. Early pioneers used the Conventional Current thinking electrons were positively charged. Today we still use Conventional Current in our textbooks.
Sir if 5ps 2V led and 9v volt battery then resistor?????
Hi - if I understand correctly, you have 5 LEDs - each having a 2V voltage drop - and 9V battery as the power supply. In this case, the total voltage drop of the combined LEDs (10V) is greater than the supply voltage (9V). The battery won't be sufficient to illuminate the LEDs. Inserting a resistor won't really make any difference.
If you had 4 LEDs then you would need to use a very small resistor (about 50 ohms).
Resistor value = voltage / current
= (9-8) / 0.020
= 1 / 0.020
= 50 ohms
Hope this helps
Cheers, John
Tnx for help sir👍👍👍
You're very welcome. Good luck with your circuit designs !
Cheers, John
YA
I used 9v battery and I connect it in yellow LED and the LED suddenly broke what happen?
Did you use a resistor to protect the LED?
thank bro
I can make the circuit work with 2 LEDs, but when I try to add more none of them light up. I don't know what I'm doing wrong. I check and recheck to make sure my components are plugged into the right holes, I confirm the LEDs are working.
Yet when I add more than 2 I get nothing. So frustrating.
Each LED has a voltage drop of about 2v. So, if you have 3 LEDs, powered by a 6v battery, you don't need to use a resistor. If you have 3 LEDs and a 9v battery then you would need to use a small resistor, say something around 150 ohms. Try reducing the size of the resistor you are using. Keep reducing the size until the LEDs come on. Cheers, John
Why is it that the 20mA equals 0.020 A? I’m not great at math and I’m learning as I go
Hi Kai. 1 "milli" means 1 "thousandth". Look up "Milli" on Wikipedia for an explanation of how it works.
thank you sir!
You are very welcome !
I have a robotics class decided to go off on my own and decided to make a stoplight and it instantly started to smoke so I need to pay attention to the resistors I guess.
Yes you do, you idiot
Kennis van ohm se vergelyking.
You sound like Fomin from the chernobyl HBO series
I'll have to watch it 😊
Why does it have to be 470 ohms not 350 ohms
350 would have been absolutely fine. I just used a slightly bigger resistor just to build a bit of additional safety into the circuit so that it would be perfectly safe with any LED. Cheers, John
Current = 20mA ...that seems like an arbitrary number; shouldn’t the Amperage be what the Battery provides? ...not an expert, just a noob wondering where that number actually comes from.
The 20mA is what an LED typically needs in order to light up. If you supply less, eg 10mA, the LED will be a bit dim; too much and you might blow the LED. Be at you know what the battery voltage is, you can select a suitable resistor to give you about 20mA.
Hmm, i must be doing something wrong. When I power my red light-emitting-diode I use a 1.5v battery in series with a switch connected to a red laser diode shining across the room into my solar cell which is connected to the positive input of an LM386 and ground, outputting nine volts direct-current into a 220 ohm resistor in series with my red light-emitting-diode. Yours seems much simpler.
...Lmao
It does sound as though you've got quite a circuit going on there 😄
@@quickstartworkbook1532 Haha!
YAY I made 1K
I don't have a resistor and Damaged 13 led
You need to use a resistor otherwise your LEDs will go pop.
@@quickstartworkbook1532 yea I knew It will go pop but I didn't have resistors
AND a big question I powered a white led with 2 batteries it worked and changed to red led it goes pop or does not ilimuninate? What is the reason
I suspect that the white LED can probably cope with a higher current than the red ones, so it doesn't burn out - it maybe very close to doing so though. The red ones that appear to not illuminate might have burned out without it being apparent i.e. no "pop". With the red LEDs, try using one battery instead of two. Or, connect your LEDs in series, which reduces the size of resistor required. As you don't have a resistor, try adding 4 LEDs in series. They will probably look quite dim. Then try it with 3. I have a video that shows how to connect LEDs in series.
@@quickstartworkbook1532 series worked but no 1 battery does not work I did 4 battery 2 led white and yellow it worked still thanks for the Info
The voltage drop across an Led is typically 2 or 3 volts. So, if it's 3v, and you connect 3 LEDs in series, the total voltage drop is 9v. If you then connect a 9v battery as the power supply, you won't need to use any resistors because the total voltage drop = the total supply voltage. Good luck with your circuits.
Thank you😂😭
sound like Winnie the poo
Thanks David
Thanks David