The timestamps for the different topics covered in the video: 1:02 Graphical Method (Using the Load Line) 3:19 Diode Approximations 4:29 How to Solve a circuit problem using diode approximation 7:46 Example 1 ( Series connection of Diode) 9:54 Example 2 12:11 Example 3 (Parallel Connection of Diode) 13:41 Example 4 (Parallel Connection of Diode with different diodes (Si and Ge)) 16:11 Example 5 (Parallel connection of diode with different voltages) For more Solved Questions on the Diode Circuits, check out this playlist: ruclips.net/video/f0xMLVpu0zo/видео.html
I study in one of the top university in Bangladesh having one of the best physics teachers in South Asia. I admit you make it clearer than those over learned professors. Thank you so much sir.
@@Honest-King you probably are virgin because nobody around you looks like Kim Kardashian but you know with that face of yours you won't be able to impress bar dancers 😂
Thanks a lot, sir, my electronics exam is just 3 days away, and despite covering the syllabus, I was just blanked out of how to solve problems. Thanks a lot for making my concept clearer & helpin me on solving problems ❤️🙏
Thank you very much for the tutorial. Yet, I wish you had emphasized the importance of the directions and the signs. What could happen if your assumption were wrong and how can you check whether it is or not? If I am not mistaken, I have figured out this issue as follows: We always assign positive voltage to the entering port and negative to the leaving port of a diode. If we assume that the diode is ON, then after solving the circuit(maybe by the virtue of node analysis and substituting a corresponding voltage source whose potential difference is V0(threshold voltage), we should look at the direction of the current coming into the diode. If it is consistent(with the assigned voltage direction), we're right, if not the diode was actually OFF. If at first, we assume that the diode is OFF, it requires the potential difference on the diode is smaller than V0. After solving the eq. by assuming that the diode acts as an open circuit, the potential difference condition( must be smaller than V0) must be checked to see whether the assumption is correct or wrong.
You are life saver!!!!👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍
The video is excellent with the clear explication ! Congratulation ! For me, I just have one question in the final example. In effect, what is the value of V0 if the first source is 3.5 V in stead of 2 V ? And the voltage of the first diode is 0.2 V in this case ? Thanks very much for your contribution !🎉🎉🎉
i think there would be a 1.3V at the 4V source diode which results in the diode conducting because 4V is bigger than 1.3V , im not confident with my guess tho lol
Excellent...but please answer to it...as soon as.. While drawing equivalent ckt of diode we are taking opposite to voltage supply (but it is actually inside diode electric field is from higher potential+ to lower -) so same direction of supply...
Sir, in example 5 you said that upper diode will off but for diode to be off potential drop across diode should be less than diode voltage but when you say that at right terminal of upper diode potential is 3.3volt then for upper diode potential drop is 1.3 volt which is more than that zener voltage then how it can be off ? please tell sir
For example 5, can you not also make the assumption that the top diode is ON and the bottom diode is OFF? This assumption would still be correct, but v0 and a different value of 1.3 V instead?
At the example 5, 18:0, i think Vout should be (Vo= -1+Vr) because we have a -1V source at the bottom, therefore, it should be 4.4V right? pls i need explanation, someone helps me
An ac voltage of peak value 30V is connected in series with a germanium diode and a load resistor of 400 ohm. the forward resistance of the diode is 20 ohm and the barrier voltage is 0.3V. Find the peak current through the diode and peak voltage across the load,
Perfect explanation 👍🏻....but because of that subtitles the stuff u were writing on that part was not visible.....btw u don't need subtitles...ur English can be understood by everyone 👍🏻🔥
In example 5, why is it that the diode with the 2-volt source will be the one to stay off instead of the one with the 4-volt source? is it only based on assumption?
Let's say, for example, the diode which is connected with a 2V voltage source is ON. in that case, the voltage at the cathode of the diode is 2- 0.7 = 1.3V. The same voltage will also appear at the cathode of the second diode. (The connected with 4V source). With that assumption, the voltage at the anode of the second diode is 4V, and the cathode is 1.3V. That means the second diode should also conduct. And in that case, the voltage drop across the second diode is more than 0.7V (which is not possible) On the other end, if second diode is conduciton alone, then voltage at the cathode both diode is 4 - 0.7 = 3.3 V. And in that case, the first diode will remain off. (also there is no violation of KVL) I hope it will clear youir doubt, why first diode is OFF.
I exactly don't know the theory...but.. suppose the -5v wasn't there...v0 would be v2 (I think that's clear)...now...we r adding this -5 v...adding v2 and -5v we get the result
I didnt understood the 5th example.....can someone please help me with this like what if the second diode remains OFF then too the voltage will remain same at both the nodes (1.3V)
it really confuses me that at 07:00 you say it is -5 V using Thevenin's voltage. that's fine but let's assume that the diode is connected normally, by which i mean it's forward biased. in that case there would be no change in the Thevenin's voltage except for the sign of it since we have anode - cathode. this means that if you put a forward biased diode instead of reverse biased, it will not let the current flow. why? because, as we said, left side of that forward biased diode will be 5V and right side of it will be 0V so that the potential difference can be 5 V. we also know that + part of voltage source is 5V and - part of it is 0V. now, as you see, the resistor will have no potential difference across its two ends since both of them are 0V, and that means no current flows. i would be incredibly thankful if someone could explain this to me.
Please check the playlist of diode and diode applications. You will get more such questions. In case, if you are not able to find it on the playlist page then let me know here, I will send you a link.
Because they are connected in the series. when in doubt, just draw the supply voltages with + and - signs, (In this case, just draw the 10V and -5V voltage sources) and draw the complete circuit. You will get it.
The timestamps for the different topics covered in the video:
1:02 Graphical Method (Using the Load Line)
3:19 Diode Approximations
4:29 How to Solve a circuit problem using diode approximation
7:46 Example 1 ( Series connection of Diode)
9:54 Example 2
12:11 Example 3 (Parallel Connection of Diode)
13:41 Example 4 (Parallel Connection of Diode with different diodes (Si and Ge))
16:11 Example 5 (Parallel connection of diode with different voltages)
For more Solved Questions on the Diode Circuits, check out this playlist:
ruclips.net/video/f0xMLVpu0zo/видео.html
In ex 3..why did u only subtract 0.7 ..there was 2 Si diode..so u should have to subtract 1.42??
@@baljeerai3789 because only 0.7volt drop between common node will be there of two diodes suppose them as single element with 0.7 volts
In example 5, how do you choose which diode is conducting?
In example 4 there are two diodes , what happens to the current , does it pass equally through two diodes?
Last one answer should be 4.3V
Thank you indian guy, you're saving a student for doing this tutorial video
For every course I cannot understand, I can rest assured that there's and indian guy with a dedicated youtube channel for it somewhere.
Not all heroes wear cape, some make educational diode circuits solving problems video
Yeah they are everywhere ...
Yeah literally man.
@@CarsonScholz being Indian I also know there'll some Indian to help me out 😂
@@GStar1 lmao same
I study in one of the top university in Bangladesh having one of the best physics teachers in South Asia. I admit you make it clearer than those over learned professors. Thank you so much sir.
This is the most perfect video ever. Clear and concise with perfect explanations. Thank you!
The accent though ...
man not really. I can still understand him pretty well with the accent
yeah u r right
@@Honest-King you probably are virgin because nobody around you looks like Kim Kardashian but you know with that face of yours you won't be able to impress bar dancers 😂
@@covertmediocrat bro itna offend kyu ho gya?😂
I was so confused with this diodes stuff but you saved my llife. Thank you, you are the best
Before that video I was very much confused about diode solving but now I got confidence in this topic .Thanks sir for this amazing tutorial ✌️👍👍
9:49
Current = 4 mA
Voltage = 4 volt
9:49
I=4 ma
V=4 volt
Right 👍
@MD TAIYAB 21PKBEA500 it should be 4.3V I guess the other terminal is at -1 hence total difference is of 4.3V
This is a great channel for studying circuits and electronics
Thanks a lot, sir, my electronics exam is just 3 days away, and despite covering the syllabus, I was just blanked out of how to solve problems. Thanks a lot for making my concept clearer & helpin me on solving problems ❤️🙏
I am here before 30 min lol
I can see an immediate use for the last circuit. Thanks. Excellent demonstration!
Again you proved better than my professor.
Thank's Sir. It's the best tutorial. After watching this tutorial, nobody else has any problem, I think.
the professor might be iitian or overlearned
the thing which matters is "THE TEACHING STYLE" in which this guy is overlearned
hatsoff !!!
The voltage across the 1k ohm resistor is 4 V and the current for the circuit is 4 mA for the example with one silicon diode and one germanium diode.
I came here for concept used at 15:45
And it is cleared thank you 😊
thank you sir, i'm studying about this and transistor at my school, and this video helps me a TON!!!
best explanation I have ever seen.... Thank you Sir for making video like this....
Tamam... Your explanations are always Crystal Clear...
Thanks have a test on diodes your way made everything more clear.
you just saved my life sir.. thank you so much for this.
Perfect video. So clear and direct.
I wouldn't say "clear"
Im 2022 neet aspirant...thank u for this vdo n u r perfect ❤️
Thanks man ! Love from Pakistan 🇵🇰
Problem in 9:50 :-
current = 4 mA
voltage across Si diode = 0.7 V
voltage across Ge diode = 0.3 V
voltage across Load resistance = 4 V
Nice
This video is very helpful guys.Thankyou ower of this channel
Thank you for the quick and simple explanation.
For item 9:38 the value of I= 4mA and Vo= 4V
How is that ??
@@KhalidAli-tr4nw 5V - (0.7V for Si + 0.3V for Ge)
Hell clear now. Thank you sir!
Indian are the best.
From morocco ❤
Excellent video. Very clear and helpful.
Thank you very much for the tutorial. Yet, I wish you had emphasized the importance of the directions and the signs. What could happen if your assumption were wrong and how can you check whether it is or not? If I am not mistaken, I have figured out this issue as follows:
We always assign positive voltage to the entering port and negative to the leaving port of a diode. If we assume that the diode is ON, then after solving the circuit(maybe by the virtue of node analysis and substituting a corresponding voltage source whose potential difference is V0(threshold voltage), we should look at the direction of the current coming into the diode. If it is consistent(with the assigned voltage direction), we're right, if not the diode was actually OFF.
If at first, we assume that the diode is OFF, it requires the potential difference on the diode is smaller than V0. After solving the eq. by assuming that the diode acts as an open circuit, the potential difference condition( must be smaller than V0) must be checked to see whether the assumption is correct or wrong.
You are life saver!!!!👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍
I was in a big trouble.This is a nice video.love it!Thank you very much
ya same here thankyu sir
For the first time i understood electronics ❤❤
Your examples selection makes your teaching amazing sir👍👍👍
very nicely Explained
massive W for this whole series
RUclips University > Any University
The video is excellent with the clear explication ! Congratulation ! For me, I just have one question in the final example. In effect, what is the value of V0 if the first source is 3.5 V in stead of 2 V ? And the voltage of the first diode is 0.2 V in this case ? Thanks very much for your contribution !🎉🎉🎉
Easy to understand the concept. Nice explanation.
Thank you brother. Nice illustration and presentation.
Mast Sir.. I like it...this is recall my past study's
In example 5, how do you choose which diode is conducting?
i think there would be a 1.3V at the 4V source diode which results in the diode conducting because 4V is bigger than 1.3V , im not confident with my guess tho lol
in the last example how about the -1V source and 1k resistor ? do they have any effect on the output ?
I too have that doubt
11:00 the -(-5) part is a little confusing.
Mera life save kar dhiya
Thanks bhai
Voltage across 1 kilo ohm resistance is equals to 4 volt
Excellent...but please answer to it...as soon as..
While drawing equivalent ckt of diode we are taking opposite to voltage supply (but it is actually inside diode electric field is from higher potential+ to lower -) so same direction of supply...
Sir, in example 5 you said that upper diode will off but for diode to be off potential drop across diode should be less than diode voltage but when you say that at right terminal of upper diode potential is 3.3volt then for upper diode potential drop is 1.3 volt which is more than that zener voltage then how it can be off ? please tell sir
great explanation !!! take love :)
Curious why Vo is calculated different in example 2 and example 5? Why not care about the voltage drop across the 1k ohm resistor?
Yes the answer should be 3.3+1 = 4.3V
@@orbhaisabkaisahoothik14 So current will be I= v/R which is 3.3-(-1)/1=4.3mA....so Vo will be -1+(I*R) = -1+(4.3*1) that is 3.3v 😅
In example 2 there is a resistor before the diode so there will be a voltage drop
For example 5, can you not also make the assumption that the top diode is ON and the bottom diode is OFF? This assumption would still be correct, but v0 and a different value of 1.3 V instead?
same ques
Same doubt
Very well done!!!
At the example 5, 18:0, i think Vout should be (Vo= -1+Vr) because we have a -1V source at the bottom, therefore, it should be 4.4V right? pls i need explanation, someone helps me
thankyou brother for guiding us🇵🇰
Why In 2 example , u didn't consider V1?
12:13 for output voltage why we minimize it from V2 not v1
An ac voltage of peak value 30V is connected in series with a germanium diode and a load resistor of 400 ohm. the forward resistance of the diode is 20 ohm and the barrier voltage is 0.3V. Find the peak current through the diode and peak voltage across the load,
Please solve this Question........
Thnx for this video. It really helped a lot.
This really helped. Thanks!
very nice explaination and tricky sums thankyou sir.
Perfect explanation 👍🏻....but because of that subtitles the stuff u were writing on that part was not visible.....btw u don't need subtitles...ur English can be understood by everyone 👍🏻🔥
If required, you can manually turn off the subtitles in the video settings.
Bro please use spot light while explaining the questions...🙏
Thank u for this amazing video...💖💖
Nice Narration
Quite helpful for 12th boards
Very much appreciated! Thanks.
Super explanation brother 👌. Keep it up!!
good teaching, thank you.
very clear explaination..thank you...
You are amazing man! Thank you
In Eg. 2 for calculating V0 why not consider V1
4mA current across each diode nd voltage across 1 kili ohm is 4
Because it is Seri es conection both
In example 5, why is it that the diode with the 2-volt source will be the one to stay off instead of the one with the 4-volt source? is it only based on assumption?
Let's say, for example, the diode which is connected with a 2V voltage source is ON. in that case, the voltage at the cathode of the diode is 2- 0.7 = 1.3V.
The same voltage will also appear at the cathode of the second diode. (The connected with 4V source).
With that assumption, the voltage at the anode of the second diode is 4V, and the cathode is 1.3V. That means the second diode should also conduct. And in that case, the voltage drop across the second diode is more than 0.7V (which is not possible)
On the other end, if second diode is conduciton alone, then voltage at the cathode both diode is 4 - 0.7 = 3.3 V.
And in that case, the first diode will remain off. (also there is no violation of KVL)
I hope it will clear youir doubt, why first diode is OFF.
why in example 5 the diode with 4V input will be considered as on while with 2v input will be considered as off?
Answer I=4mA VO=4v
Well explained,thank you❤️😁
In ex.5 you calculated that V0=4-0.7. but there is a negative voltage of -1v also. Then can i say v0=(4-0.7-0.1)volt
sir , in last case why didnt we consider the -1v acting at that node ...pls sir explain??
Very good explanation. Thank yy
very fine explanation. Thank you
4mA is answer for the series circuit with Si and Ge diode
Wow, so helpful!
I do not understand what I have to know in theory to calculate vo = v2 - 5v, what If I assume vo = 10v-v1-0.7? help! for 11:59
I exactly don't know the theory...but.. suppose the -5v wasn't there...v0 would be v2 (I think that's clear)...now...we r adding this -5 v...adding v2 and -5v we get the result
So helpful thank you Sir and keep going🖒
I didnt understood the 5th example.....can someone please help me with this like what if the second diode remains OFF then too the voltage will remain same at both the nodes (1.3V)
Worthful video
Nice video, helps me a lot
it really confuses me that at 07:00 you say it is -5 V using Thevenin's voltage. that's fine but let's assume that the diode is connected normally, by which i mean it's forward biased. in that case there would be no change in the Thevenin's voltage except for the sign of it since we have anode - cathode. this means that if you put a forward biased diode instead of reverse biased, it will not let the current flow. why? because, as we said, left side of that forward biased diode will be 5V and right side of it will be 0V so that the potential difference can be 5 V. we also know that + part of voltage source is 5V and - part of it is 0V. now, as you see, the resistor will have no potential difference across its two ends since both of them are 0V, and that means no current flows.
i would be incredibly thankful if someone could explain this to me.
Nice explanation...Thanks Sir.
💓💓💓💓💓💓💓💓
Need more questions on the same,I hav and managed to get the concept
Please check the playlist of diode and diode applications. You will get more such questions.
In case, if you are not able to find it on the playlist page then let me know here, I will send you a link.
Aapna muja acha SA samja diya thanks,💖💖💗💗💗
Sooo nice video sir...sooo nice.it is physics with beautyy.thanks for your good explanation.
In the last example why you consider that first diode is off and 2nd is on but reverse can also be true that 1st is on and 2nd is off ?
EX2 : why did we assume the current through v1 and v2 are same they are in parallel right?
Because they are connected in the series. when in doubt, just draw the supply voltages with + and - signs, (In this case, just draw the 10V and -5V voltage sources) and draw the complete circuit. You will get it.
thanks. when the courant is negative and when it is positive???
thankYou for the video. :)
its very helpful.
love from agra
9:55 Answer is I = 4mA and V = 4 V
Ans: 4 V (si&germanium) series ckt
Yes, that's correct.