Series Diode Configurations

[@subbusiva6014]

please change the polarities of diode equalent model for example-2

[@sauravdas280]

Grateful to you Master (Neso Academy) and thank you so much, Mr. Subbu Siva :)

[@94D33M]

@@sauravdas280 i never get annotations anymore....i heard youtube removed that feature ?

[@darshanhollam4401]

was useful ... cleared my doubt thank u

[@tyanne]

sorry, which example do you mean?

[@Vishal___Yadav.01BR]

@@tyanne 2nd last question ...i mean before homework question

[@dheerajkumarraghuvanshi532]

Answer to the homework problem is 1 A. Explaination- since one of the diodes in the middle branch is reverse biased it should be considered as open circuit. Hence no current passion through any of the diodes. Further I= 10 V / 10 Ohm that is 1 A.

[@tech4toil621]

Thanks for explain

[@mvenkatesh6490]

Yes right me also get same answer

[@thermodynamics_724]

-1A

[@shafiaafin8945]

you teach so well!I wish I had a faculty like you.

[@pujithch4078]

same feeling....

[@nimitgarg2531]

Same feeling

[@AmanSharma-ii7uq]

Sir ,kasam se apse jaise logo ki wajha se ache engineers h warna agar college faculty k upar depend ho toh pass bhi ni ho sakte , i promise that i will donate alot after my job.

[@AmanSharma-ii7uq]

Sir communication engineering k bhi lectures upload kr di jiye please

[@programming_world187]

job lgi ?

[@crazygirl-cg8iv]

​@@programming_world187😂😂😂

[@el-drago9688]

@@programming_world187 hn sahi baat hai😅

[@muhammadsamiullah6614]

Bta bhaii job kaha lagiii

[@elugotiramyasree8403]

sir depth and quality education u r providing sir..really proud of u sir even average students also can understand sir..

[@bungbang-e9n]

the middle line is open circuit. so I = 1 amp. right?

[@malothnaveenkumar8808]

s

[@104payalmittal5]

Yup

[@adityaagrawal3443]

Yes

[@surajkantsharma6427]

Sir in network theory you have said when we go from - to + polarity then we use + sign but here you used -0. 7v instead of +o.7v.

[@susantamondal3035]

That is my doubt also.

[@biswajeetnitrourkela2383]

Actually it is + to - so wrongly interpreted

[@31nehayaduvanshi92]

Sir has wrongly replaced the diode .It must be from + to - as current flows from +to - .

[@hm2715]

@@31nehayaduvanshi92 Yaa

[@swagatika2206]

hey at 7.37 time the diode polarity should b reversed right???

[@swagatika2206]

tq....I also corrected in my note,,😃

[@Jagan-ko1vr]

Yes

[@aliyarc.a150]

Yes you're correct

[@Shikshasila]

Rt

[@Kanekikenn999]

I thought i was wrong for considering polarity reversed 😅 thnx🤜🤛

[@wisdomjamesmurombo1292]

The diodes are representing an open circuit hence we have only one big loop giving us a current of 1A (10-10I=0).

[@atulsharma-zy6nq]

Sir the polarities of diode replaced was mistakenly taken in reverse it must be the other way.

[@104payalmittal5]

U r awesome sir I wish u can have session on how to teach with our teachers

[@prajjwalsrivastava9674]

sir i learn here everything very clearly and nicely more than i do in class

[@ananda_saha_shorts]

Current =1Amp..As in the middle one diode is in reverse biased so become open ckted ..Hence current will go through 10ohm only..Thus ans is 1 Amp

[@subhashmandal6027]

Sir, please check the polarity of the battery which replaced the diode, At the time of KCL u have mentioned right but, in the equivalent diagram, battery polarity got reversed.

[@zubaerahammed]

Exactly, you are right

[@vishakbhat3032]

in the problem one while replacing the diode with the cell the polaririty is reversed ....rest the lecture was amazing thanks a lot neso academy

[@chayandas8204]

yes u r totally correct

[@zohebahmad9633]

@7:40, why aren't we supposed to include the voltage rise from a current travelling in the mesh on the right side (given that we are using KVL)? Do we ignore the right mesh because of the Earthing between the two meshes?

[@changerequired7851]

Same doubt

[@pradeepkumarmohanty5412]

In example the polarity of 0.7v should be interchanged.equation written is correct.I through diode is also correct i.e 0.965A

[@ROHITKUMAR-lh3ps]

I want to thank from bottom of my 💓.. Really awesome job sir... I wanna salute uuu..,

[@ROHITKUMAR-lh3ps]

Sir I think you should upload emft also because it is very important for gate. It's our request.... There are lots of KZbinrs which does not give concept like u... So we need... Please help the students..

[@sandipanpaul4272]

@@ROHITKUMAR-lh3ps exactly emft is hard to find on youtube...but the way neso acedemy teaches ..wish they would hav started the subject..

[@kundurupavan1429]

7:37 polarity is reversed

[@sahilgupta5472]

Example 2 (already acknowledged by tutor :)) A little mistake , the diode equivalent polarity should be reversed so current is: i=20.7/20=1.035A Like this comment so as to help others:)

[@meditationtales5866]

yeah i am thinking the same .. thanks for clarification dude

[@sahilgupta5472]

@@meditationtales5866 :)🤝

[@udattam]

No the voltage that replaced diode gives a voltage drop of 0.7 V. So the answer given by Sir is correct but when he replaced it the polarity of the diode was incorrect which you can rectify. Check pinned comment for validation

[@puneetray4045]

@@udattam if polarity is reversed then result will also change . So current is 1.035 A .

[@puneetray4045]

Same doubt and I came with same answer.

[@sankhamitra3862]

Ans of the h.w. problem is I=1 amp because 2nd diode will act as an O.C. AT 7:37 the diode equivalent voltage source polarity should be reversed as you mentioned in the lecture.. BTW great explanation sir with great concept 👍👌.

[@brahmingamer2374]

in first example the polarity of the equivalent voltage source is changed as the p side of the diode is at higher potential but you have shown n side on higher potential

[@nainsigupta6784]

y do you take oly one loop equation for calculating the current in the example problem ,there are two loops right?

[@adityaanand7185]

Absolutely right.

[@ShAmS894]

Will there be no current in left loop?

[@potatoguy8390]

Do you perhaps use the book of boylestad?

[@aashimagaba27]

I think, you applied wrong KVL. It should be 20- 20i + 0.7 = 0

[@vishalmahla6762]

The direction of diode equivalent circuit is wrong ...

[@xtremecuber9644]

​@@vishalmahla6762 yes

[@ashish3212]

Yes

[@k_____l2502]

Yes

[@Sirra-p7n]

Yes

[@dwinovianto1250]

1 A i already simulate it using proteus &, the value was same, 1A

[@shreeyashrasal8574]

Sir I have a doubt in 1st prblm where polarity of diode in equivalent ckt is shown along the current but it should be reversed bcoz of the barrier potential will flow in opposite direction to the direction of current shown. Can I get the explanation?

[@tshibulal]

U r ryt.. That was a mistake.

[@suprabhatchoudhary6402]

Sir, I don't understand the polarity marking of equivalent circuit for barrier potential.

[@saikirannaramwar2886]

suprabhat choudhary I too don't understand

[@sandeeptiwari3655]

Saikiran Naramwar brother...this thing will be clear when you saw the construction of pn junction diode ......when we add impurities in a pure semiconductor whatever it is silicone or germanium....you saw there is an action taking place.....electrons from the majority n side goes to p side and recombined with holes....and there is an loss of electron in n side so it will become + charge and due to excess of negative charge in p side...it will become - charge....after reading this I hope you got some idea about it...... after all your comment is 2 years old...but now I studied this so I answered you....you are our senior....good day sir

[@Hari-lg3fi]

7:59 The eqn should be 20V - 20 I + 0.7V = 0

[@yashprajapati7699]

NO THE EQUATION THAT SIR HAS WRITTEN IS CORRECT BUT THE POLARITY OF DIAGRAM IS WRONG, THE POLARITY SHOULD BE REVERSED

[@harshsinghal2350]

follow one sign convention as you wish.there it is mixed. In case of diode you are moving from -ve to +ve and it is taken -ve voltage drop and in case of voltage source again you are moving from -ve to +ve but this time the voltage drop is +ve in KVL equation.

[@powerstarfanikkada1320]

y do you take oly one loop equation for calculating the current in the example problem ,there are two loops right? can anyone explain this

[@teetanrobotics5363]

He is the best

[@nagendrarachamallu6201]

Sir , how u will gave a + or _ in resistance ????

[@abhishekbawa5713]

Answer to homework problem is 1A because if current has any direction whether up or down, one diode will be on and other will be off. means in both cases any one diode will be in off state so open circuit.

[@snehaarya9979]

8:21 Krichoff law ...20V -20I +0.7 hoga

[@aneetakushwaha9058]

Thanku so much Sir its really helpful.

[@kashishshukla23]

1 amp as the diode in reverse biased is making the line in the middle of the circuit open so no current will flow through it & all the current will fllow through the 10 ohm resistor.

[@abeshbiswas7007]

Plz give me answer I am so confused 😭

[@hamizanbinmuhamedganasan3886]

cn help with homwork prblm? how to solve if 2 diodes are in opposite directions?

[@Kanekikenn999]

Sir homework problem me....jo middle ka connection remove hoga na?...

[@deepikadeepika8595]

in example 1, wouldn't the equation be -20v-20i+0.7v ?

[@anupkumaryadav1562]

Sir ne equations shi likhi hai but diode se replace voltage ka diagram galat bnaya hai

[@makeitsimple247]

Bhaiyaa ye smj ni a rha ki jb diode ko uske equivalent se replace krte hai to polarity kis tarah rakhni hai equivalent battery ki,aapne pehle example me alg rakhi or dusre me alag ?

[@jonasyamba8671]

loving this.

[@basappabasappa4685]

How barrier voltage is taken negative?

[@charankumar1323]

in the example the diode repalce by battery written reverse once clarify it

[@hitenpatel9794]

solution for the homework problem is 1 amp because there are 2 diode face opposite direction so one or the other will act as an open circuit so no current will pass through the doide and the circuit will become a simple circuit with a battery and a resistor and we can easily find the ans by the help of ohms law

[@kirtichandrajal4633]

As there is one reverse biased condition which will act as open circuit.so the current can't flow in this circuit. Hence I=0A

[@mahmoudeid7107]

thank you sir you are a great help

[@amishibansal]

How did u define the polarities of the equivalent battery used?... That is not explained plz help

[@நக்மாசெல்]

First time anode is plus sign. Secobd time anode is minus sign. Totally confused and i have to go back to kirchhoff again.

[@DeepakKumar-kk5qv]

how we have to choose the direction of Vb in equivalent circuit

[@siddarthningthoujam5181]

won't the current through the resistor be 0amp as the diodes are opposite the voltage cancel out and there is 0 voltage drop which means it is short circuit so all the current flows through it and no current through the resistor.... plis correct me if I am wrong

[@hadiaamin1466]

siddarth ningthoujam the diodes cannot cancel each other because one of them has Vb=0.7 V and the other has Vb=infinity (coz its open ckt). Hope you understood

[@apurbamandal6535]

Is 1 Amp. is the correct answer? Please comment.

[@gitakumari-2098]

In example ,the polarity applied should be in the direction of arrow naaa...but u have applied it in the opposite direction .

[@onkarrandad1043]

what happened to resistance of diode in first example

[@SamriddhaDhali]

in your hometask problem the answer will be 1 amp. there will be no current in diode section due to open circuit for second diode..

[@satyajitsarkar8810]

@NesoAcademy Respected Sir, my answer for the current I for the configuration in Example Problem 1 is 2.89 A, as there is a second resistor of 10 ohms parallel to the first one. So, I have taken the equivalent resistance that they offer to the circuit which stands at 6.66 ohms, which we must do first, and then applied the KVL to obtain the forementioned result. Sir, am I correct or wrong?

[@SS-me1ss]

You are wrong because the I shown in th figure is for 20ohm, so you cannot consider that I for both the resistors because both the resistors are in parallel and so current flowing in them will be different

[@sele.7358]

@@SS-me1ss but why he did not do mesh analysis and find currents for each mesh then find I , I=1.035 A

[@hanochhans9601]

bro i got same doubt but i then i remembered that equivalent resistance is the total resistance not the single loop or connection which are in parallel so it is correct.

[@aashishsaini6037]

sir in the question, you have taken the wrong direction of diode as a batter of 0.7 volts. Barrier potential always oppose the external battery source if it is in forward bias state.

[@saraswatisharma7861]

1A is wrt ans??

[@CodePinaka]

yes

[@nageswarrao2848]

Sir in constant voltage drop model rd should be 0 but in this resistance is there .How is it constant voltage model.please anyone tell immediately.

[@mebratumekonnen3636]

it was ver nice but 20i-0.7-20=0 i=20.7/20=1.035 mA

[@architgoyal2934]

Very very thanks sir

[@dawsh8658]

why do we replace diode by 0.7 voltage source ? it is known that only zener diode is replaced by a voltage source!

[@churchilljovs6861]

Sir please add led,solar cell , photo diode , varactor diode in this playlist . It will be grateful for our preparation.

[@whatcomesafterinfinity3291]

10-0.7+0.7 = 10 Net V and then 10I = 10 so I=1; elsewise the middle path is open ckt too for they are having opposite polarity

[@saurabhgupta1595]

Thanks you sir

[@naveenkumarandukuri3490]

Kindly confirm the diode polarities in equivalent circuit.

[@saivamsi1480]

The ans. of HW problem is I =1A

[@pranavs4445]

ans for ex. 2 is 1A

[@ashutoshjaiswal5426]

mistake at 7:37. it should be reverse biased

[@alterguy4327]

Homework Problem Current is 1A : )

[@lakshmy18791]

way you calculated the current for the problem is not correct. i think we have to do mesh analysis here because current is mentioned inside a loop

[@saket_saurav]

Lakshmi Asokan applying KVL in both loops results I=1.035A

[@venkataapparao2699]

Yeahhhh I too taught the same

[@118johnm]

1 Amp as two diodes in the middle cancel each other out and act as a short circuit

[@jahansaid6382]

As the current goes through the first Diode it is, but the second diode makes it off, which is equal to pen circuit. I= v/r=-10/1= -1 amp

[@aryanpatel368]

what the answer of the diode problem

[@yashagarwal8476]

sir wont the answer be 0 because the middle network will act as short circuit

[@hadiaamin1466]

Yash Agarwal It will act as open ckt and there is no path for the current to flow. Short is the path from which whole of the current flows

[@ashishkarn7688]

sir what about for 2nd loop in your's question. battery will dissipate 3mA ,and current through diode and r (20 ohm) will be ,1 A.

[@nikhilsahu9039]

I =1amp because second diode act as open ckt

[@rajibbarman5194]

Required answer for the H/W problem is I=1.07A

[@piyushsahu1746]

how

[@shubhnamdev5265]

added 0.7 barrier pot. i dont get it,.. is diode is a power source in your terms.. pls clarify

[@pic1989able]

with these polarities of diodes i think correct answer is 1 A.

[@AnotherGamer21]

1 Amp.

[@bhanuprakashreddy]

Current =1A.the second diode acts as open

[@Motivational_hub_98]

i=0 A

[@puretiajay3840]

take two loops of individual currents and do mesh analysis. Answer is 0.965A

[@shivajisanap9308]

Ans is 1A as one is opposite circuit will be open there

[@mohankrishna2093]

Diode equivalent was taken with reverse polarities...

[@abeshbiswas7007]

Sir please give me answer with explanation of example 2 I can't understand plz help me 🥺

[@vedantukarde7473]

1 A is the answer for H.W. Que

[@muskangpt19]

I=1 A bcz every time one of the diode is opened.

[@rockeykumar3348]

Middle diode is o.c so it conduct 0 current so current I will be 1 ampere.

[@brahmingamer2374]

so uss hisaab se 1st example mein current ki value 1.035A aani chahiye

[@ABDULREHMAN-mk8he]

Vin = 10V ,D1 = 9.3 drop across it and D2 is open so 9.3/10 = 0.93Ampere

[@matthewsmalatji5994]

Lol its 1amp.. by inspection you know that voltage across the resistor will be same as the source because of the parallel configuration therefore by applying ohms law you should get the answer: 10/10=1A

[@raviSHANKAR-zy2yj]

In the homework problem, the two opposite diodes result in 0 V through the middle arm, resulting in a short circuit, and hence all the current passes through the middle arm, thereby leaving I = 0A. Many of u have said, the middle arm acts as an open circuit, but an open has certain voltage but 0 current.

[@Motivational_hub_98]

yes...

[@PJ-jn6gg]

Please upload the new lecture in place of it it can make doubt to new learning students

[@ghanasyamasahu2250]

The equivalent circuit used for diode reversed.

[@NazmulHaque-iw2zk]

using kvl, -10+i*10=0 so, i=1A

[@jitendrapratapsingh9189]

sir , ig you take wrong kvl , here +0.7 hoga

[@saikrishna-kw2gd]

I think that problem u solved polarity for the diode should be reversed

[@jankipatel2392]

1 amp is the answer

[@sandeeptiwari3655]

Janki Patel can you please tell me how to make equivalent circuit for diode....I mean it is confusing me .... please help me.....all commentors are old.... only you are new... please help me in this

[@ghanasyamasahu2250]

0.93 amp???

[@tulasirambandi8102]

No bro see the diode which is opposite to the direction of current so the diode is open so no current is flowing in the diodes. There fore current is flowing in the 10 ohm resistance with 10v so current =1amp