{664} How to Make LED Indicator FOR 220 Volt AC, Calculation
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- Опубліковано 6 вер 2024
- How to Make LED Indicator FOR 220 Volt AC.
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#ledindicator #resistance #calculation
i demonstrated how to calculate a proper resistor for LED to drop voltage and control the LED current to make led indicator lights to connect directly to ac line voltage 220 / 230 / 110 VAC, it is normally used as led indicator to show the ac presence. for that purpose we need to calculate a series resistor, i shared the calculation formula to calculate resistor ohmic value and power rating for the resistor.
using this method i demonstrated, how to make ac 220v led indicator, and how to connect led light to 220v ac or 110v ac,
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Very good brother always the best brother my request is about calculation gate resistor for MOSFETs or IGBTs if we replace IGBTs or MOSFETs with totally defrent value so how to calculate gate resistor or gate to source resistor or the gate diode ?
Thank you very much and best regards from Philippines 🇵🇭
best regards and special thanks. yes bhai, i have already your 9 comments on my both channels pending for the same topic, i am arranging for that. very soon we will study it
wish you best of Luck
your brother
Muhammad Ashraf
@@HaseebElectronics than you 10×°10000 time's
I think mains LED indicator lights should not be done with resistors. They will always bake the circuit board and turn everything brown around the resistor. Much better to use capacitive dropper. In USA 120V I always use 0.1uF 400v film capacitor with 1M bleeder resistor, 1K inrush resistor , 1n4007 full bridge rectifier then DC to a red 3mm LED. Takes a couple minutes to twist together the components, solder and shrink wrap. Very reliable and produces no heat..
Exactly, I did the same thing to turn a Led on with mains. The capacitor will have an impedance in the circuit and drop the voltage just like a resistor but this drop will not generate any heat since the the wasted power is in reactive power. I didn’t use a full bridge for the circuit, just half bridge (one diode). Technically the led flashes with a rate of 20ms which is okay to me and is not noticeable, pretty solid.
@@abolfazlabbasi4854 👍
LED in reverse parallel with the single diode correct? if diode is in series the capacitor no longer sees AC and can't pass current, no?
@@GannDolph ideally the led is a kind of diode and there is no need for another diode, but practically, led will be fine with positive half cycle of the AC and will turn on without any issue ( with proper capacitor and resistor selected ). In the negative half cycle of the AC, the led will have a very small current leakage in reverse bias ( around 100uA) which is enough to close the circuit, in this case the capacitor will drop a very small portion of the voltage ( V= I * X) and the rest of the voltage will drop on the led which can easily blow up the led.
To protect the led you can have a diode in reverse parallel to the led. The diode will conduct in negative half cycle of the AC and protect the led.
Led only conducts in half cycles. If you have another diode in series, basically will not affect anything except adding more protection which is still not enough to save the led ( I haven’t tested it yet, not sure though).
All in all, you definitely need a reverse parallel diode to save your led if you are using a half bridge configuration
@@abolfazlabbasi4854 Makes sense. When i tried it once with a series diode but no reverse parallel diode, with a blue LED, the LED did not light up. Adding a reverse parallel diode fixed that. I think full bridge is the way to go - diodes are so cheap, easy to salvage from dead CFL ballasts and other junk electronics - basically free if you have any access at all to e-waste.. full bridge won't have any failures , and little to no flicker.
Thanks for your suggestion. I agree with you but..............................
Dear, if proper calculations are done there is no question of bake.....
As it is already said by Mr. Haseeb sir.
Excellent video, and very good explanation. I do the calculation in this way: At 220V RMS, when we use 220Kohms, the power will be P=VxI=VxV/R=220mW, and because we use only half of the wave, the power on the resistor will be 110mW. The same formula we apply to an instant water heater of 15 ohms, and we will find P=VxV/R = 3,2KW. (we can calculate an integral to reach the same result). The maximum led current will be V/R = 220 x 1,4 /220 mA = 1,4mA, but the average value will be V/2R = 220V/(2x220k) = 0.5mA.
Thank you haseeb sir for answering my question with a nice explainer video. As i read the comments below from @Gann Dolf and @Abolfazl Abbasi there are many useful suggestions like 1) diode connected in reverse parallel to LED 2) using 0.1uF 400v film capacitor with 1M bleeder resistor, 1K inrush resistor , 1n4007 full bridge rectifier then DC to a red 3mm LED and many other feedback. May be it can be another video with title "Ideal/ perfect way to make LED light indicators for AC mains switches"
You are most welcome
Assalaamulaikum,
Great video
I had learned a lot from this.
I love the way you explain with clarifying small small points.
Thanks Sir.
Jazak Allahu Khaira
Excellent video dear friend, have a happy new week and August 😊
Конденсатор небольшой надо ставить паралельно светодиоду до 1 мкф (от 0,01, 0,1 до 1 мкф), что бы не действовало обратное напряжение на него, иначе диод может выдержать, а светодиод пробьёт. Стабилитрон ещё можно.
0.25W is ok, the current is only happening during only a portion of the cycle, so on average it will be acceptable as the failure mode of the resistor is in this case thermal.
Crystal clear 👍🏻 Thx for educational video!
Thanks for your efforts
I suggest if you can make every think about converting frige with inverter compressor into one ac compressor .
Because so many people made videos on UA-cam and said that's working fine but I saw many ambiguities because there is no scientific explanation
Only what I see people practice a hit and try
Very nicely and simply explained.
Thank you very much Haseeb. I understand from your explanation that if we have any ac circuit containing a rectifier diode, we must consider the peak voltage in our calculations not rms voltage whether the load is led or not. mustn't we?
if we didn't put a rectifier diode in the circuit in this tutorial the peak voltage will be equal to 327V like your calculation in the video or the peak voltage value will differ?
On calculating power, one should divide by 2 since only get voltage every other cycle. Also I believe it would be correct to use rms value of current and reduce power again by factor of 0.707.
in this circuit we must consider that the LED is being used in unidirectional current. so we must use LED forward current to calculate everything related to it.
either it is serie resistance, either it is power dissipation,
we must not consider RMS values because we are allowing diode to drive at it rated current at 1.8 volt. if we will use RMS values, it will dedicate more voltage to LED, which will cause to damage the LED
@@HaseebElectronics won't the resistor power dissipation be equal to (V- 1.8) ² / (2R) = 218.2 ² / (2 * 220000) ≈ 0.11 W ?
🙏💓very nice teach sirji pranam 💓🙏
Mashallah very good information upld keep it stay connected
Hi shouldn't you use the Irms for the power calculation in the resistance, as in P = VRMS * IRMS = (IRMS)^2 * R ? Because the 20mA is the peak current in this case.
here we must consider LED continues forward current.
that is DC current
so that's why we need to convert rms voltage into peak voltage.
and we must deal this circuit in DC sense. not in rms
because we need to allow the forward current at allowed voltage rating. because LED is unidirectional conductor. so we have to deal in unidirectional calculation
moreover we are allowing the led to conduct in specific portion out of a 360 degree signal. that is peak voltage
@@HaseebElectronics Oh ok i see thank you for the answer...
@@7GIGEO7 it is my pleasure
Good presentation.5mm led will give decent glow from 1ma. So 220k (325/220=1.477ma) is an acceptable value.In place of resistor high voltage 0.01mfd capacitor can be used.No extra diode required. Across Use 3v zener across LED to avoid reverse break down . Average heat generated in the resistor is V²RMS/r=0.24.If there is no zener ,power generated will about half.So 1/4watt resistor can be used. I have been using above circuit for many years.
More informative video
Thank you so much!
👍👍👍👍 very nicely explained thanks, sir.
asslasm alkaum bhi very good information
I've used a .33 uF non polarized polyester capacitor in series with a 1k 1/4 watt resistor to power a white 5mm led to replace a burned out neon indicator light for my freezer and it works great! No wasted power! I also put a 1n4007 diode reverse biased in parallel with the led. 50 hz may require a little more capacitance. What are your thoughts? Nice video! Have a great morning and a great week sir!
Could you kindly explain how smps powered emergency light with charging control works
Naye Saal ki bahut bahut mubarak bad ,,, vdo ke beech me kuchh dangerous sounds aa rahi thi ,,, apna khayal rakhiye ga ,,, jazakallahu khaira ,,,,, 👍
Hello sir
नमस्ते from india
I think that 325 is peak value not DC.
Great job brother 👍
best regards dear sister
@@HaseebElectronics jzakallah Khair
Very nice information
Great work your great man
Good job brother 👍
Very interesting video ! I wanted to ask if it is possible to show how to make a variable (AC not DC) power supply in order not to be scared to work on mains circuits as well as safe test of shorted appliances. Thanks again dear M Haseeb and Salam Aleikum.
nice sir thanks🙏
most welcome
Informative vedeo. Sir i want to make a stabilizer for servo motor three phase 6 kv. Please tell me perfect Diaigrame for same
Thumbs up done nice video sir
LED has maximum reverse voltage rating. In negative cycle of AC, LED will get damaged by high voltage and have a reduced life.....though there will be no current flowing due to Series diode. To protect this, diode should be connected in reverse parallel to LED and not in Series.. This way the maximum reverse voltage would go to a diode drop, that is 0.7 V in negative cycle.
Well done
Nice video sir
Thank you sir
Nise vido hai sir
Good morning sir when someone checking with hand he is not getting shock but in same way i am checking i feelings shock why thanks
Nise vidio sir
it is my pleasure
It will be better to use capacitive reactance for this purpose.
exactly perfect
the capacive voltage dropper is best choice
but it was a question from my viewer to explain the video form Dharmendra experiment channel
Happy Munday Like Video my bro
Hello My Friend
Grazie 👍
Excellent explain thanks for sharing
👌👌👌👍
best regards and special thanks
Lk 78 brother 👍👍👍
Sir vido number 664 hindi me bana sir
yes brother. it is coming tomorrow
👍
Why dont you just measure the voltage across the resistor (both AC and DC) and then calculate the current to check calculations?
Hindi wala chenal pe banai sir
السلام علیکم
Sir yah vido hindi chenal pe bana sir plise
Sir 664 number hindi me bana plise
Plise