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: th-cam.com/video/f0xMLVpu0zo/w-d-xo.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.
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 !🎉🎉🎉
You are life saver!!!!👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍
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 👍🏻🔥
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,
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...
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
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
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
Because if Vo = 1.3V then the second diode is already conducting. The anode is at 4V and the cathode would be at 1.3V. So, if we consider the drop of 0.7V across the diode, then Vo =3.3V. Right, so output can't be 1.3V and 3.3V simultaneously. (KVL will be violated) With 3.3Vas output, the first diode will remain off. As the anode of the first diode is at 2V. I hope it will clear your doubt.
Just apply the KVL in that output branch. The voltage at the upper node is Vo.The other end of 3k resistor is not at ground potential but, it is at -5v. (Vo is the voltage between that Upper node and the ground.) Therefore, if you apply the KVL, in the output branch then Vo is V2 - 5. I hope, it will clear your doubt. If you still have any doubt then you can simulate the same circuit and check.
The Calculations For Example #2 Are Incorrect. I = (10V - (-5V) - 0.7V) / (4K + 3K) = 614.29 uA The drop across R1 is 2.46 volts and R2 is 1.84 volts. Vo is therefore 6.9 volts.
I would recommend you to check your calculation once. (10 - (-5) - 0.7) / 7k is 2.04 mA Instead of adding 5V, you have subtracted it. And that is why you are getting incorrect result.
I too checked it and I am getting around 2.04 mA. Please check the polarity of -5V. I think, you must have put +5V instead of -5V. That's the only case, where you will get 614 uA current.
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.
![BOWKYLION - วิงวอน (ex-change) [Official MV]](http://i.ytimg.com/vi/cLdnZWDaO-w/mqdefault.jpg)
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:
th-cam.com/video/f0xMLVpu0zo/w-d-xo.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
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
Thank you indian guy, you're saving a student for doing this tutorial video
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?😂
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 ✌️👍👍
I was so confused with this diodes stuff but you saved my llife. Thank you, you are the best
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
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
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.
This is a great channel for studying circuits and electronics
I can see an immediate use for the last circuit. Thanks. Excellent demonstration!
the professor might be iitian or overlearned
the thing which matters is "THE TEACHING STYLE" in which this guy is overlearned
hatsoff !!!
thank you sir, i'm studying about this and transistor at my school, and this video helps me a TON!!!
Thank's Sir. It's the best tutorial. After watching this tutorial, nobody else has any problem, I think.
Again you proved better than my professor.
I came here for concept used at 15:45
And it is cleared thank you 😊
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.
best explanation I have ever seen.... Thank you Sir for making video like this....
Im 2022 neet aspirant...thank u for this vdo n u r perfect ❤️
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 !🎉🎉🎉
Tamam... Your explanations are always Crystal Clear...
you just saved my life sir.. thank you so much for this.
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
Thanks have a test on diodes your way made everything more clear.
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)
This video is very helpful guys.Thankyou ower of this channel
Indian are the best.
From morocco ❤
Thanks man ! Love from Pakistan 🇵🇰
For the first time i understood electronics ❤❤
Perfect video. So clear and direct.
I wouldn't say "clear"
Your examples selection makes your teaching amazing sir👍👍👍
TH-cam University > Any University
I was in a big trouble.This is a nice video.love it!Thank you very much
ya same here thankyu sir
Voltage across 1 kilo ohm resistance is equals to 4 volt
Thank you for the quick and simple explanation.
massive W for this whole series
very nicely Explained
You are life saver!!!!👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍👍
Excellent video. Very clear and helpful.
Easy to understand the concept. Nice explanation.
Mast Sir.. I like it...this is recall my past study's
Answer I=4mA VO=4v
Mera life save kar dhiya
Thanks bhai
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.
Hell clear now. Thank you sir!
Nice Narration
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........
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...
Thank you brother. Nice illustration and presentation.
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
Bro please use spot light while explaining the questions...🙏
Thank u for this amazing video...💖💖
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
11:00 the -(-5) part is a little confusing.
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
thankyou brother for guiding us🇵🇰
Quite helpful for 12th boards
Super explanation brother 👌. Keep it up!!
9:55 Answer is I = 4mA and V = 4 V
Thnx for this video. It really helped a lot.
4mA is answer for the series circuit with Si and Ge diode
Ans: 4 V (si&germanium) series ckt
Yes, that's correct.
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
love from agra
very nice explaination and tricky sums thankyou sir.
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
Why In 2 example , u didn't consider V1?
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 :)
CURRENT FLOWING IN CIRCUIT IS 4ma & voltage across the diode is 4v in given ciruit
Nice explanation
Brother, in last question why answer can't be 1.3 volt.
Nice explanation.
Because if Vo = 1.3V then the second diode is already conducting. The anode is at 4V and the cathode would be at 1.3V.
So, if we consider the drop of 0.7V across the diode, then Vo =3.3V.
Right, so output can't be 1.3V and 3.3V simultaneously. (KVL will be violated)
With 3.3Vas output, the first diode will remain off. As the anode of the first diode is at 2V.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS Thank you, this explanation is what I needed!
good teaching, thank you.
Aapna muja acha SA samja diya thanks,💖💖💗💗💗
4mA current across each diode nd voltage across 1 kili ohm is 4
Because it is Seri es conection both
Sooo nice video sir...sooo nice.it is physics with beautyy.thanks for your good explanation.
very clear explaination..thank you...
Very much appreciated! Thanks.
Well explained,thank you❤️😁
12:13 for output voltage why we minimize it from V2 not v1
Id=4mA,Vo=4v
Very well done!!!
Answer : current I flowing is equal to 4 milli ampere and voltage across1k ohm resistor is 4 volt
So helpful thank you Sir and keep going🖒
very fine explanation. Thank you
Very good explanation. Thank yy
You are amazing man! Thank you
This really helped. Thanks!
Nice explanation sir...
Nice video, helps me a lot
Sir I am unable to understanding exam 2 that how V2 - 5 gives Vo but all the other explanation was genius
Just apply the KVL in that output branch. The voltage at the upper node is Vo.The other end of 3k resistor is not at ground potential but, it is at -5v. (Vo is the voltage between that Upper node and the ground.) Therefore, if you apply the KVL, in the output branch then Vo is V2 - 5. I hope, it will clear your doubt. If you still have any doubt then you can simulate the same circuit and check.
@@ALLABOUTELECTRONICS thank you very much sir for explaining the part very clearly 😇😇
Nice explanation...Thanks Sir.
💓💓💓💓💓💓💓💓
Current=4mA
Output voltage=4V
Nice sir which is this app you use
Worthful video
current 4ma voltage = 4 v
awesome explanation
solution:
5-0.7-0.3-1000I=0
=4mA
The Calculations For Example #2 Are Incorrect.
I = (10V - (-5V) - 0.7V) / (4K + 3K) = 614.29 uA
The drop across R1 is 2.46 volts and R2 is 1.84 volts. Vo is therefore 6.9 volts.
I would recommend you to check your calculation once. (10 - (-5) - 0.7) / 7k is 2.04 mA
Instead of adding 5V, you have subtracted it. And that is why you are getting incorrect result.
@@ALLABOUTELECTRONICS Run a Spice sim and you get my result!
The current is 633 uA.
I too checked it and I am getting around 2.04 mA. Please check the polarity of -5V. I think, you must have put +5V instead of -5V. That's the only case, where you will get 614 uA current.
@@johnpautler4872 yours is wrong
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.