Led current = V/R = 4V/330 ohms = 12mA. Maximum surge current will be V/R = 230V x 1.4 / 56 ohms = 5.75A which is good (1N4007 surge current is 30A). The 47uF capacitor will protect the zener diode and the led from surge current. The 47uF capacitor, the zener diode and the 330 ohms resistor will limit the led current, so the surge current will not reach the led. 47uF/0.2uF = 235, so the maximum inrush voltage on the zener diode will be around 230V x 1.4 / 235 = 1.3V, which is good, under 6.2V and the zener diode will be protected. (the 2 capacitors will react like a resistive divider).
Sir I'm from Sri Lanka that i watched your channel regularly it is very important topics for me also you started to explain in English it is Very useful for none Urdu people like me continue jazakallh khirah
Its been a while since I did any electronic projects but this brought me joy going through the design of this and seeing I haven't forgot everything I taught my self .
Excellent video, as always. Here if I don't use the bridge rectifier, only the zener diode, the led will work only half of the cicle, and the average led current will be 10mA = half of 20mA :)
if we don't use rectifier, then LED reverse voltage are only 5 volt in datasheet but in practical it can sustain only 20V reverse. although the current is limited by dropper, but it may lead to drive in reverse breakdown, in case that the line voltage are increased than the rated values for that reason we must use at least 1 diode to clip off the reverse peak. so that LED must be protected
You are right. In the absence of the bridge rectifier we can count on the zener diode to reduce the reverse led voltage to the 0.7V zener direct drop voltage.
Salam alaekun engr Haseeb. I am and Rofiu from Nigeria, I really love ur blog and I have learnt alot especially, something I have been fine difficult in power electronics, I.e SMPS. May God Almighty Allah continue to increase your knowledge. Pls. I would like to ask you two questions: (1) in your CAPACITOR VOLTAGE DROPPER CIRCUIT (transformerless power supply circuit), you posted few week ago. How did you get "0.636" with ripples when you are calculating filter capacitor?. (2) pls, I have 12V/200AH sealed AGM Lead Acid batteries two, and I need to charge these batteries with at least, 30% of 400AH (60A). I don't want to use iron core transformer, I need SMPS, using ferrite core transformer. Pls I need your guidance on which topology should I used? with high current. Pls, giude me with schematic design diagrams and few calculation. (3) lastly, I have gone through all your tutorial about SMPS, I want to know how to design choke for input line filter and what type of toroid should be used with few calculation. Thanks God bless you.
Supply voltage is 230 Need voltage is 4V Need current is 0.1A Need power is 4v*0.1A = 0.4W Then capacitor voltage is 230-4 = 226V capacitor heating power is 226V / 0.1A = 22.6W But it doesn't get hot like 22.6W
Thank for sharing your knowledge..! sir, please give complete calculations with timing too..on bleeder resistor as there is no complete info on dicharging capacitor on youtube channel..
i think you forget to take into consideration that when calculating the capacitance(C), you took Xc (apparent resistance of capacitor) as the whole resistance of the dropper , but in reality the total resistance is Xc + R(of surge resistor), this is way you got different output voltage than you calculated
Hello, thank you very much for this excellent video, I ask you, why did you use the peak voltage (317 V) instead of using only the input rms voltage of 250V for the calculation of the capacitor? pro security issues???Thank you, greetings.
Warning this circuit could potentially kill unless double insulated. In the event of a faulty component there is the potential to electrocute by direct connection to mains potential. The circuit is fine & safe if it is impossible to accidentally touch live parts or any metal work which the unit could be connected. Recommend also the addition of a fuse, and earth connection to provide protection in the event of a fault. An ELCB or RCCB may also provide protection.
Salam M Haseeb. I am having hard time to understand some aspects of this technical video. Why not putting the surge resistor before the capacitor and what is the function of the zener ? Thank you.
i confuse..about calculate math n fomula of err electronic.calculate capasitor....dificult for me..😅😅..but it good video ...learn new today..26 december 22..
Some design uses drum core choke to limit inrush current and noises. Sir plz make a video on drum core choke design for differencial mode, common mode, buck or boost converter choke.
Nice video Sir I have 24v 750mamp transformer and i need 12v but 750mamp and if I drop voltage by using resistors, will i get amps also reduced by this process?
If I use a resistor R to drop 12V, R=V/I = 12V/0.75A = 16 ohms, P=V x I = 12V x 0.75A = 9W, so, I need a 16 ohm resistor / minim 9W. The resistor will heat, and need good airflow. The amps will not be reduced. If the load resistance is constant, this circuit will work, but if the load don't use 0.75A continuously, then the voltage on the load will be unstable, and can reach values greater than 12V. If the load use DC voltage, then we can use a voltage regulator as L7812 on a heatsink to provide a constant 12V on the output.
Led current = V/R = 4V/330 ohms = 12mA. Maximum surge current will be V/R = 230V x 1.4 / 56 ohms = 5.75A which is good (1N4007 surge current is 30A). The 47uF capacitor will protect the zener diode and the led from surge current. The 47uF capacitor, the zener diode and the 330 ohms resistor will limit the led current, so the surge current will not reach the led. 47uF/0.2uF = 235, so the maximum inrush voltage on the zener diode will be around 230V x 1.4 / 235 = 1.3V, which is good, under 6.2V and the zener diode will be protected. (the 2 capacitors will react like a resistive divider).
perfect analysis
special thanks and best regards
Thank you. I searched for months to find someone who could and would explain the design of a electronic circuit as you have.
it is my pleasure
Sir I'm from Sri Lanka that i watched your channel regularly it is very important topics for me also you started to explain in English it is Very useful for none Urdu people like me continue jazakallh khirah
මේ විඩියෝ එකේ විනාඩි 16.02 තියන එකේ 6.2 v / ×0.636 කියෙනේ මොක්කද ඒ ඒකේ ආදේසය පැහැදිලි නෑනෙ. 6.2v ඒක පැහැදිලි අවුට්පුට් Dc වෝල්ට් එක. ප්රස්නේ තියෙන්නේ x 0.636 කියලා තියෙන්නේ ඒමොකේ ආදේසයද .
උඩ ආදේසය C= i 🔼 T / 🔼 v
ඒ සූත්රය සන්කේත මොක්කද
C = කැපසිස්ටය කියලා දන්නව
I = අමිපියර් කිය්න්න දන්නව
ඒ සූතයේ 🔼 ත්රීකෝනේ වගේ තියන එක මොක්කද කියන්න දන්නෑ. පුලුවන් නම් කියලා දෙන්න
Wonderful video dear friend, have a happy Sunday 😊
Thanks again! The bleeder and surge resistor calculations were great. I am building my project to provide a LED light to a microwave light.
Its been a while since I did any electronic projects but this brought me joy going through the design of this and seeing I haven't forgot everything I taught my self .
Salaam Karne Ke Liya Hazir Ho Geya Hun Bhai Saab
wa alaikum salaam dear brother
Excellent job done keep sharing
Excellent video, as always. Here if I don't use the bridge rectifier, only the zener diode, the led will work only half of the cicle, and the average led current will be 10mA = half of 20mA :)
if we don't use rectifier, then LED reverse voltage are only 5 volt in datasheet but in practical it can sustain only 20V reverse. although the current is limited by dropper, but it may lead to drive in reverse breakdown, in case that the line voltage are increased than the rated values
for that reason we must use at least 1 diode to clip off the reverse peak. so that LED must be protected
You are right. In the absence of the bridge rectifier we can count on the zener diode to reduce the reverse led voltage to the 0.7V zener direct drop voltage.
Buht umdah 👍👍👍
Very excellent information! Thank you, Haseeb!
My pleasure
Salam alaekun engr Haseeb. I am and Rofiu from Nigeria, I really love ur blog and I have learnt alot especially, something I have been fine difficult in power electronics, I.e SMPS. May God Almighty Allah continue to increase your knowledge. Pls. I would like to ask you two questions: (1) in your CAPACITOR VOLTAGE DROPPER CIRCUIT (transformerless power supply circuit), you posted few week ago. How did you get "0.636" with ripples when you are calculating filter capacitor?. (2) pls, I have 12V/200AH sealed AGM Lead Acid batteries two, and I need to charge these batteries with at least, 30% of 400AH (60A). I don't want to use iron core transformer, I need SMPS, using ferrite core transformer. Pls I need your guidance on which topology should I used? with high current. Pls, giude me with schematic design diagrams and few calculation. (3) lastly, I have gone through all your tutorial about SMPS, I want to know how to design choke for input line filter and what type of toroid should be used with few calculation. Thanks God bless you.
Supply voltage is 230
Need voltage is 4V
Need current is 0.1A
Need power is 4v*0.1A = 0.4W
Then capacitor voltage is 230-4 = 226V
capacitor heating power is 226V / 0.1A = 22.6W
But it doesn't get hot like 22.6W
thank you sir, you are a good teacher.
Great analysis sir, love from India
Thank You Sir
Thank for sharing your knowledge..! sir, please give complete calculations with timing too..on bleeder resistor as there is no complete info on dicharging capacitor on youtube channel..
Very useful information sir thanks
it is my pleasure
👍👍👍👍👍
Harika bilgiler için teşekkürler kardeşim. Allah cc seni ve sevdiklerini korusun .
Thanks! Your explanations are amazing!
Informative video bro
i think you forget to take into consideration that when calculating the capacitance(C), you took Xc (apparent resistance of capacitor) as the whole resistance of the dropper , but in reality the total resistance is Xc + R(of surge resistor),
this is way you got different output voltage than you calculated
Thank you so much; As always awsome!
Pretty video. How do I modify the circuit to charge small 12v lead accumulator battery?. I am aware charging current should be way high.
woud you please inform me can this circuit use for rechargeable torch 'and can use 105 j 400v one instead of two sir
Hello, thank you very much for this excellent video, I ask you, why did you use the peak voltage (317 V) instead of using only the input rms voltage of 250V for the calculation of the capacitor? pro security issues???Thank you, greetings.
Good job thanks for sharing
best regards
great video
How long can it be used, can we apply as a continuously....
if it is designed properly, it works long.
Warning this circuit could potentially kill unless double insulated. In the event of a faulty component there is the potential to electrocute by direct connection to mains potential. The circuit is fine & safe if it is impossible to accidentally touch live parts or any metal work which the unit could be connected. Recommend also the addition of a fuse, and earth connection to provide protection in the event of a fault. An ELCB or RCCB may also provide protection.
Excellent
Thank you so much 😀
Nice job dear brother
Thank you Sir 🙏
Salam M Haseeb. I am having hard time to understand some aspects of this technical video. Why not putting the surge resistor before the capacitor and what is the function of the zener ? Thank you.
Great video. Where is the value "0.636" coming from (in your calculation 16 minutes into the video)? Thanks
Looks like it's an average value of a sine wave after diode bridge over its period Pi that is equal to 2/Pi=0.6366
i confuse..about calculate math n fomula of err electronic.calculate capasitor....dificult for me..😅😅..but it good video ...learn new today..26 december 22..
👍🙏🏻👍👋good
Some design uses drum core choke to limit inrush current and noises.
Sir plz make a video on drum core choke design for differencial mode, common mode, buck or boost converter choke.
great
thanks a lot dear brother
Nice video Sir
I have 24v 750mamp transformer and i need 12v but 750mamp and if I drop voltage by using resistors, will i get amps also reduced by this process?
If I use a resistor R to drop 12V, R=V/I = 12V/0.75A = 16 ohms, P=V x I = 12V x 0.75A = 9W, so, I need a 16 ohm resistor / minim 9W. The resistor will heat, and need good airflow. The amps will not be reduced. If the load resistance is constant, this circuit will work, but if the load don't use 0.75A continuously, then the voltage on the load will be unstable, and can reach values greater than 12V. If the load use DC voltage, then we can use a voltage regulator as L7812 on a heatsink to provide a constant 12V on the output.
@@N0G0MAIL ok ji
Thanku so much
Cảm ơn bạn!
Thanks a lot for your valuable support 💕
Sir can you please make a video on budget Multimeters.
request is for a 6kv solar circuit and load 1.5 tonn ac ( using igbt - steup transformer -battery 100 ah X 12 v and panel 8no
Aslam o alikum sir
Can i make this type supply for 12v 500mA.
👌👌👌👍
thanks a lot
salut 🙂
salutes dear friend
🙏👍
so nice of you
Sir hindi wala chanel pe vido bana plise sir
Excellent explains