Get the Best Price at Indiamart: Power Supply Systems: dir.indiamart.com/impcat/power-supply-systems.html?YTI& DC Power System: dir.indiamart.com/impcat/dc-power-system.html?YTI& High Voltage Power Supply: dir.indiamart.com/impcat/high-voltage-power-supply.html?YTI&
I love this video and the explanation keep going . It’s very organised and clear . I am not indian and with the help of the subtitles I understood thank you for your time to add English subtitles
Well, the maximum input voltage of the 7805 is specified to 35 V. You operate the device across its break-though voltage, which is obviously around 46V. Only the input capacitor and the internal break through resistance of the 7805 limits the undefined current (here considerably smaller than 12 mA) across the device. I am surprized that the 7805 is still working if it's break-through voltage is exceeded. Nevertheless, this operation can harm the 7804 after a certain time and there will be no more 5V limiting behaviour after a while and the diode can be damaged too. My suggestion is to eliminate the 7804 and to use a 5 V zener diode behind the smoothing capacitor to get always a well defined voltage across the LED and its resistor. It will also draw down the rectified voltage to well defined 5 V. So you can use a smaller electrolytic capacitor with a voltage rating of 10 V (how did you decide your voltage rating or did you use a 400 V specified device). The current across the zener diode and the LED is well defined. Even when the LED gets disconnected, the circuit will not fail.
Zener diodes dissipate power all the time. I prefer using regular ics but the input side must be protected with higher voltage zener diode (close to 35v) . At higher voltage, zener diode's power dissipation should be lower.
@@δωμάτιομελέτης no, the power dissipation of a Zener diode at a higher voltage is higher. In fact, at voltages higher than 8 V Zener diodes doesn't work with the Zener effect but with avalanche breakdown. Considering the data sheet for the BZX55-series in all cases you need to set the operating current for the Zener diode to at least to 5 mA. So for 5.1 V you only dissipate 25 mW. At 30 V you dissipate 150 mW. Furthermore you dissipate power in the linear regulator. For 12 mA and a voltage drop from 30 V to 5 V you dissipate 300 mW. So in total you dissipate 450 mW. This is much more than 25 mW. The most important rule for a capacitor voltage supply is to develop the active power at the lowest useful voltage. Otherwise you not only increase the power dissipation but also the reactive power at voltages considerably lower than the mains voltage. The reactive power for the zener diode regulator is 2 W. The your solution it is 40 W!
Please check the aritmethic, the calculated capacitor should be 125 nF not 1.25 uF, besides your are not considering the bridge and 220 ohm resistance voltage drop,about 3-4 volts, the circuit works because of the 5v regulator
You are right the capacitor calculator is wrong. 125 means 1.25 uf. But output works because of low current output and regulator IC7805. Naturally, if the capacitor value is more then current too will be more 👍
It's Not a Good Explanation. As The 2 Bias Points ie; Cathode Points Of a Diode With Another One One The Same Line In Series. Will Not Allow Current Flow - Result. You Will Create Back EMF And At The Junction. Where Both Cathodes Meet & Are In Series - Very, Very Serious Heat Damage Will Occur! R H Fraser.
Thanks for the lesson. very useful and practical. I have a doubt. The calculation started assuming 7V at input of regulator. But after building the circuit the measurement shows 46V. Why ? How do you design for a specific current draw.
@@inderdasssetia5243 46V is far higher than the rated max input voltage of the regulator. The high drop across it can overheat the device. How do we design it such that the input voltage is not more than 5v from output.
Watch... the voltage on the regulator input measures 46V. The absolute maximum input voltage for the LM7805 is 35V. Also the filter cap looks like max. 25V, so Boom expected on these two parts. Better maybe, but never perfect, is to place the 220Ω resistor in the plus line to the regulator, and use a large zener, e.g. 20V 1W (at least, but measure to be sure) to ground between resistor and regulator to limit the voltage there, so that the input at the regulator an filter cap does not exceed 20V. Simple thought, have not tested.
This video is for dropping capacitor calculation.. For exact value of filter capacitor you can refer to the another video on the channel capacitor value calculation formula.
Hi. I think you have made a mistake in your circuit. You calculated the circuit for 12mA. But you have a voltage drop of 15V above the 220 Ohm resistor. That means a total current of 68mA. The calculation for the voltage drop capacitor is wrong. 1 ------------------------------- = 125nF not 1.25 µF 2x3.14 x 50 x 25333 I think the voltage regulator is close to end in a gammaburst ;-) Nearly 50 Volts is very much for a 7805. That could be the reason therefore that the output voltage is only 4.8 Volts. The regulator is running critical. Normaly the output voltage of a 7805 is very close to 5 Volts. But regardless of that it is a nice video. Thanks.
Need the explanation for led 9, 12, 15, 18, 20, 25, 35, 40 Watt bulbs and 9,12 Watt rechargeable 3.2 bettery and 3.2*2bettery circuits can you explain this
Please don't do this. The first problem is that absolute maximum rating for a 7805 is 35V. Your circuit applies 46V to it. Of course, it's possible to get away with this some of the time, but there's no guarantee that you won't simply destroy the 7805. The second problem is that you're assuming 12 mA is flowing through the circuit. Actually, using a 470 ohm resistor with your red LED (which drops around 2 V when conducting) will give a current of around 6.4 mA (i.e. (5V - 2V)/470R). That's one reason why the rectified DC voltage is not the 7V you expected. The third problem is that if anything disrupts the current through the LED, such as one of the wires coming loose), you'll immediately get full mains voltage available to the rectifiers and thence to the filter capacitor, which then needs to be rated for rectified mains, which it clearly is not.
Yes, he has not shown the DC voltage measurements before feeding to 7805 ic. This worked only for low current and instantly, Don't know if it could stand for 1 hr. But one has to be kept distance and the circuit into a closed wooden box, in fear of capacitor explosion 💥 and is dangerous. 👍
12.46, There is a calculation error, If you divide 1 by 7944k (you missed k there) you will get 0.125uF capacitance(not 1.25uF). 13.55, The capacitor you are holding is 125J denoting 1.2uF, not 1.25uF (J is for 5% tolerance). Still I don't get it how you got the results.
Sir, your calculation for Capacitive Reactance = 25.3K ohms (25333 ohms) is correct, but your further calculation of Capacitance C (1.25 micro F) is incorrect. It works out to only 0.125 micro Farad. Also, how to know how much voltage drop an X-rated capacitor will give, kindly clarify.
Sir, I was searching this type electronic circuit with proper mathematical and practical experimentle which giving correct way of getting OUTPUT VALUES. many many THANKS
Sir, your way of explaining is having fully touch of an Engineer. You explain the things with theory and problem calculation and with reasoning which helps to understand the subject without any doubt. Thanks for your nice explanation.
Very clear explanation of transformer less circuit designing video you have made. This is the one video wherein I could gather the information on taking the RMS voltage for designing the correct PS. Kudos.
One of the best explanations of transformerless circuits based on explanation, calculations and diagrams and you have been honest at least you have said that these kind of circuits can only work on low Amps unlike other videos of these kind, so thanks in advance. The video was in Hindi but at least the subtitles help 👍
Super explained. Explained strightly no time waste. Subtitles manner is good .Some times subtitles came across(Sometimes I could not see the circuits ). Please put explaining article little bit above.
Sir आपके वीडियो बहुत ही उम्दा होते है और आपके समझाने का तरीका बहुत बढ़िया है क्यों की वो प्रैक्टिकल के साथ होता है आपसे रिक्वेस्ट है की एक वीडियो । पैनेल में लगने वाले डिजिटल वोल्ट मीटर और एम्पीयर के सर्किट डायग्राम पे भी एक वीडियो बनाए की डिजिटल वोल्ट मीटर और एम्पीयर मीटर कैसे काम करता है ये बोल्टमीटर और एम्पीयर मीटर जो आजकल सबमरसेबिल पंप के स्टार्टर पैनेल में लगा होता है
You have probably explained this already, but as I am not able to understand the language in the video, but can read the comments, it seems that most folk are regarding the capacitor as a voltage dropper. While this is the effect of the capacitor, it should be remembered that the primary effect of the capacitive reactance it to limit the Current. This then allows us to pass that current through our load, the load actually defining the voltage across it.
voltage regulator heat up and burn its self when you connect it in 46 v in higher load. Best regulator voltage is 7 to 12 v DC for (LM7805) Over all best explanation i didn't know any thing about it Thanks
please describe all components details and circuits diagram,ur circuit explanation is very nice and ur use very simple languages. i hope ur continue to ur circuitry video upload in you tub. Thanks
It's a perfect and ideal presentation... Your all vedios are decently planned and clear... They are all most useful to the learners and hobbyists and lovers of electronics.👍
Get the Best Price at Indiamart:
Power Supply Systems: dir.indiamart.com/impcat/power-supply-systems.html?YTI&
DC Power System: dir.indiamart.com/impcat/dc-power-system.html?YTI&
High Voltage Power Supply: dir.indiamart.com/impcat/high-voltage-power-supply.html?YTI&
Very clear good teaching ability. Keep it up. I am from Sri lanka
I love this video and the explanation keep going . It’s very organised and clear . I am not indian and with the help of the subtitles I understood thank you for your time to add English subtitles
Thanks 😊
9
@@TEKNOISTIX Bhai can you explain with 110V AC input how thsi can be done to get 12v DC output
@@TEKNOISTIX how to contact or connect sir
बहोत ही अच्छे से समझाया आपने ऐसे समझाने वाला वीडियो पूरे यूट्यूब में नही मिला धन्यवाद
Amazing video! Never have I seen such an easy-to-understand lecture!! Thanks!
Great to hear!
Your explanation is very clear. And u born for the teaching. Good job keep it up. I am from Sri lanka
Well, the maximum input voltage of the 7805 is specified to 35 V. You operate the device across its break-though voltage, which is obviously around 46V. Only the input capacitor and the internal break through resistance of the 7805 limits the undefined current (here considerably smaller than 12 mA) across the device. I am surprized that the 7805 is still working if it's break-through voltage is exceeded. Nevertheless, this operation can harm the 7804 after a certain time and there will be no more 5V limiting behaviour after a while and the diode can be damaged too.
My suggestion is to eliminate the 7804 and to use a 5 V zener diode behind the smoothing capacitor to get always a well defined voltage across the LED and its resistor. It will also draw down the rectified voltage to well defined 5 V. So you can use a smaller electrolytic capacitor with a voltage rating of 10 V (how did you decide your voltage rating or did you use a 400 V specified device). The current across the zener diode and the LED is well defined. Even when the LED gets disconnected, the circuit will not fail.
Thanks
Zener diodes dissipate power all the time. I prefer using regular ics but the input side must be protected with higher voltage zener diode (close to 35v) . At higher voltage, zener diode's power dissipation should be lower.
@@δωμάτιομελέτης no, the power dissipation of a Zener diode at a higher voltage is higher. In fact, at voltages higher than 8 V Zener diodes doesn't work with the Zener effect but with avalanche breakdown.
Considering the data sheet for the BZX55-series in all cases you need to set the operating current for the Zener diode to at least to 5 mA. So for 5.1 V you only dissipate 25 mW. At 30 V you dissipate 150 mW.
Furthermore you dissipate power in the linear regulator. For 12 mA and a voltage drop from 30 V to 5 V you dissipate 300 mW. So in total you dissipate 450 mW. This is much more than 25 mW.
The most important rule for a capacitor voltage supply is to develop the active power at the lowest useful voltage. Otherwise you not only increase the power dissipation but also the reactive power at voltages considerably lower than the mains voltage.
The reactive power for the zener diode regulator is 2 W. The your solution it is 40 W!
Well said
@@δωμάτιομελέτης omg VR IC also consume power all the time
Best electronics video I have come across
Please check the aritmethic, the calculated capacitor should be 125 nF not 1.25 uF, besides your are not considering the bridge and 220 ohm resistance voltage drop,about 3-4 volts, the circuit works because of the 5v regulator
Agree with you
Right
I was also calculating and finding the Capacitor value was coming to 0.125Uf or 125nF. I'm glad you also found this mis-calculation.
You are right the capacitor calculator is wrong. 125 means 1.25 uf. But output works because of low current output and regulator IC7805. Naturally, if the capacitor value is more then current too will be more 👍
Yeah, he miscalculated the capacitor, so I'm can't trust the rest of the video.
I like the way you present your videos it's clear and precise,I have trouble with math formulas you make it easier to grasp it better. Thank you.
Sir I am in 9th grade and I am very very thankful to you for providing free high quality content to us.
I am very passionate in electronics
Thanks.. Keep supporting
Sir I'll always support you 👍
👍
Well done Phaji...great .....theory with practical in one video...thanks
Good explanation,keep it up,make video's for variable power supply with explanation for different loads.thanks in advance.
Thanks. I will make video on variable supply soon.
Keep supporting
धन्यवाद सर वैरी नाईस।
It's Not a Good Explanation. As The 2 Bias Points ie; Cathode Points Of a Diode With Another One One The Same Line In Series. Will Not Allow Current Flow - Result. You Will Create Back EMF And At The Junction. Where Both Cathodes Meet & Are In Series - Very, Very Serious Heat Damage Will Occur! R H Fraser.
Dear Sir,
Your are the best Instructor in the world.
Great Explained, I'm new Subscriber from Today.
Thanks
Thanks for the lesson. very useful and practical. I have a doubt. The calculation started assuming 7V at input of regulator. But after building the circuit the measurement shows 46V. Why ? How do you design for a specific current draw.
I have the same question
Sir told that "7805 Data Sheet" says that it needs minimum 7v to give regu 5v output.
As 46v is more than 7v, the circuit has to work.
@@inderdasssetia5243 46V is far higher than the rated max input voltage of the regulator. The high drop across it can overheat the device. How do we design it such that the input voltage is not more than 5v from output.
Thanks for the video, Its really helpfull not just the theory but the practical testing good too.
Watch... the voltage on the regulator input measures 46V. The absolute maximum input voltage for the LM7805 is 35V. Also the filter cap looks like max. 25V, so Boom expected on these two parts. Better maybe, but never perfect, is to place the 220Ω resistor in the plus line to the regulator, and use a large zener, e.g. 20V 1W (at least, but measure to be sure) to ground between resistor and regulator to limit the voltage there, so that the input at the regulator an filter cap does not exceed 20V.
Simple thought, have not tested.
This video is for dropping capacitor calculation.. For exact value of filter capacitor you can refer to the another video on the channel capacitor value calculation formula.
वाहे गुरू की जय हो जय श्री गुरु गोबिंद सिंह जी की जय best बताते हैं धन्यवाद आपका दिन मंगलमय हो i like sabcrise your चैनल सर thanku welcome
Thanks 😊
Hi. I think you have made a mistake in your circuit. You calculated the circuit for 12mA. But you have a voltage drop of 15V above the 220 Ohm resistor. That means a total current of 68mA. The calculation for the voltage drop capacitor is wrong.
1
------------------------------- = 125nF not 1.25 µF
2x3.14 x 50 x 25333
I think the voltage regulator is close to end in a gammaburst ;-)
Nearly 50 Volts is very much for a 7805. That could be the reason therefore that the output voltage is only 4.8 Volts. The regulator is running critical. Normaly the output voltage of a 7805 is very close to 5 Volts.
But regardless of that it is a nice video. Thanks.
i know hindi but full video understand. i like you and your video. Bangladesh
Dear sir
Please Kindly provide information on 12v 1, 2, 3 amp transformer less power supplies.
For higher current i ll recommend to use transformer. This is for educational purpose.
Excellent presentation with proper clarifications makes a perfect project. Thank you very much.
Need the explanation for led 9, 12, 15, 18, 20, 25, 35, 40 Watt bulbs and 9,12 Watt rechargeable 3.2 bettery and 3.2*2bettery circuits can you explain this
sir huge respect and love from pakistan .sir your way of teaching is very good.
Thanks.. Do share these videos and channel with your frends if you think it can help someone.
Please don't do this.
The first problem is that absolute maximum rating for a 7805 is 35V. Your circuit applies 46V to it. Of course, it's possible to get away with this some of the time, but there's no guarantee that you won't simply destroy the 7805.
The second problem is that you're assuming 12 mA is flowing through the circuit. Actually, using a 470 ohm resistor with your red LED (which drops around 2 V when conducting) will give a current of around 6.4 mA (i.e. (5V - 2V)/470R). That's one reason why the rectified DC voltage is not the 7V you expected.
The third problem is that if anything disrupts the current through the LED, such as one of the wires coming loose), you'll immediately get full mains voltage available to the rectifiers and thence to the filter capacitor, which then needs to be rated for rectified mains, which it clearly is not.
Yes, he has not shown the DC voltage measurements before feeding to 7805 ic. This worked only for low current and instantly, Don't know if it could stand for 1 hr. But one has to be kept distance and the circuit into a closed wooden box, in fear of capacitor explosion 💥 and is dangerous. 👍
Perfectly described, no one shows like you. Always keep doing great.
Thanks
12.46, There is a calculation error, If you divide 1 by 7944k (you missed k there) you will get 0.125uF capacitance(not 1.25uF). 13.55, The capacitor you are holding is 125J denoting 1.2uF, not 1.25uF (J is for 5% tolerance). Still I don't get it how you got the results.
but sir even if k is missed(just as an example) doesn't 1/7944 = 0.000125... which equals 125 microfarad? please help and not 1.25 microfarads?
@@zohaibshaikh5755 Yeah, on both sides he is wrong but the practical part shows he is getting correct output. Donno how😕
i've read on an article,, 1uf capacitor dropper in the transformerless circuit will give 50ma in the output current..
@@jokolewung9093 around 160 mA for 105 j(1 microfarad )
Use MOSFET to amplify less current .....
explanation and calculations are splendid. Look forward to many more simple projects.
thanking you . K. subramanya Banbalore.
Thanks
Sir, your calculation for Capacitive Reactance = 25.3K ohms (25333 ohms) is correct, but your further calculation of Capacitance C (1.25 micro F) is incorrect. It works out to only 0.125 micro Farad.
Also, how to know how much voltage drop an X-rated capacitor will give, kindly clarify.
You are absolutely correct and that is the reason 46 Volts(much higher) is being shown after rectification.
I was also calculating and finding the Capacitor value was coming to 0.125Uf or 125nF. I'm glad you also found this mis-calculation.
your captions are very useful bro and easily understood.... bcoz I m not a hindi.... i m tamil.
Thanks
Sir,
I was searching this type electronic circuit with proper mathematical and practical experimentle which giving correct way of getting OUTPUT VALUES.
many many THANKS
Do share with others if you really like
It will also help me to reach more interested people
Tgis channel is new so not much people know about it
But jandar, shandar video hai. You are most intelligent person. Thanks
Thanks
Sir, Explanation is amazing 👍
Keep it up
बढ़िया वीडियो। ट्रांसफॉर्मर नही होने से थोड़ा रिस्क भी है इस मे। वो भी आप को बताना चाहिए।
great video and thanks for the English subtitles
Excellent solution, thank you best of luck.
I don't know hindi but Amazing class all doubt cleared thanks 🙏
From kerala
Method of explanation is best. God bless you
Thanks
Sir you deserve 1000000000000000000000000000 subscribers 👍 because you are helping us too much. Thank you sir for these amazing videos. Keep it up 👍
Thanks 😊
Your welcome Sir
Do share these videos and channel with your friends
Sure sir why not 🙏🏻
👍
Your explanation is superb. Very good vidios.
Thanks 😊
It was great idea to check the voltages across diffrent sections .That gave a clear picture. Thanks .Stay 🙌
Thanks
Smjhaan da treeka kya baat bhaji👍🙏
Sir, your way of explaining is having fully touch of an Engineer. You explain the things with theory and problem calculation and with reasoning which helps to understand the subject without any doubt. Thanks for your nice explanation.
Very clear explanation of transformer less circuit designing video you have made. This is the one video wherein I could gather the information on taking the RMS voltage for designing the correct PS. Kudos.
Thanks 😊
Thanks for your clear explanation
One of the best explanations of transformerless circuits based on explanation, calculations and diagrams and you have been honest at least you have said that these kind of circuits can only work on low Amps unlike other videos of these kind, so thanks in advance. The video was in Hindi but at least the subtitles help 👍
Thanks 😊
Good sardar ji vadiya samjhaya
Thanks 😊
Great video .very good information sir
Very useful description. Thank you.
your teaching method is excellent, thank you. 😀👍🌼
Thanks
Super explained. Explained strightly no time waste. Subtitles manner is good .Some times subtitles came across(Sometimes I could not see the circuits ). Please put explaining article little bit above.
You can turn off subtitles on your phone or pc just clik the cc button below the video
Superb explained 👍👌👌❤️❤️
U are my ideal teacher. Thanks
Thanks
Mantap thank you from indonesia
Thank you for the video!
Professionally well done sir....
I like ur method of teaching
Tha ks
I love electronics and i am your fan
Thanks
I can't stop dancing since I stumbled upon this rather new but great channel. Hope to learn a lot. I can expect what's coming.
Thanks.. Keep supporting
Please post videos of your dance
@@daljeetsinghsokhey605 i already have a few videos of my own. Can you check them out?
Aap bahot badiya samjhate ho sir
Excellent
Regards
🙏🏼
Excellent video teaching. Kudos
I loved it. I never knew how to calculate values . Thanks to u
Thanks a lot boss... 👍👍👍👍👍👍
Simple calculations explained nicely. Thanks
Thank you for imparting this knowledge .
Super fully help FULL video 😍😍😍
Thanks
Theroitically and practically excellent sir....
Thanks for your videos...
मै २५ साल से ईलेक्ट्रॉनिक्स विषय मे काम कर रहा हु । ये मैने २००५ मे बनाया था । एल ईडी बल्बवर प्रयोग करतांना। 👍
जय महाकाल
Thanks for your knowledgeful video .
Sir
आपके वीडियो बहुत ही उम्दा होते है और आपके समझाने का तरीका बहुत बढ़िया है क्यों की वो प्रैक्टिकल के साथ होता है
आपसे रिक्वेस्ट है की
एक वीडियो ।
पैनेल में लगने वाले डिजिटल वोल्ट मीटर और एम्पीयर के सर्किट डायग्राम पे भी एक वीडियो बनाए की डिजिटल वोल्ट मीटर और एम्पीयर मीटर कैसे काम करता है ये बोल्टमीटर और एम्पीयर मीटर जो आजकल सबमरसेबिल पंप के स्टार्टर पैनेल में लगा होता है
Volt current meter already bhut sari videos me use kiya hai waha explanation hai
@@TEKNOISTIX
Link dijiye
🙏🙏🙏 thank you very much sir 👍💅💅💅
You have probably explained this already, but as I am not able to understand the language in the video, but can read the comments, it seems that most folk are regarding the capacitor as a voltage dropper. While this is the effect of the capacitor, it should be remembered that the primary effect of the capacitive reactance it to limit the Current.
This then allows us to pass that current through our load, the load actually defining the voltage across it.
Orignl formullla. Power supply very nice. God bless you.. Thank.u. srdarr gi. Naeem Anjam.
Thanks 😊
Sat Sri Akal, Veer bhot vadiya Sikhate Si tussi great Video God bless you and keep the good work going 👍🏻👍🏻🙏🏻🙏🏻❤️
Thanks 😊
Very good teacher God bless you
बहुत पसंद theory plus practical, what more does one need.Thanks.
Nice explanation. Keep up the good work.👍
Thanks
Super Explanation about the stepping down voltage, Rectifying the voltage, filters, regulating voltage and getting required output voltage.
Tnq.
Explanation was very clear keep it up
Thanks
Do share videos with your frends
If you thik it can help someone
@@TEKNOISTIX sure i will and keep making such videos....
Thanks
I ll do my best
Nice video and very good explain sir 👍
Bahut acchs laga vedio. SMPS par ek vedio bane to accha hoga. Iski details hame milegi.
Jaldi hi bana du ga
thanks for the clear explanation
Excellent tutorial and practical experience analise video
Wow, very descriptive and knowledgeable video
Thanks
btw thanks for showing the calculations!!! helpful video!
👍🏻
voltage regulator heat up and burn its self when you connect it in 46 v in higher load. Best regulator voltage is 7 to 12 v DC for (LM7805)
Over all best explanation i didn't know any thing about it Thanks
7805can take upto 35v input at 500mA current
Here current is very small about 20 mA so it is working
please describe all components details and circuits diagram,ur circuit explanation is very nice and ur use very simple languages.
i hope ur continue to ur circuitry video upload in you tub. Thanks
Sirji apki vedio bahut achi aur details mai hai sab same vedios banaye sir hame seekneko milege
Very very good presentation thanks
You are most welcome
Very nicely explained🙏
Thanks
very easy and clear superb vedio for understanding nice pl.put the vedio for auto cut power emergency led light with circuit
Thanks
Well done sir je, what an explaination.
Thanks and welcome
Very useful video thanks
👍🏻
Bahot khoob sir
Thanks
Very nice...n use full for education purpose also....in next ..how to replace...regulator by zenner diode....
Ok
Thanks
It's a perfect and ideal presentation... Your all vedios are decently planned and clear... They are all most useful to the learners and hobbyists and lovers of electronics.👍
Thanks
Excellent presentation
Thanks, Good information.