Semiconductor 12 Physics Chapter 14 (L1) | Diodes | EAMCET JEE NEET 2024/25 @vedantutelugu

hello everybody Welcome to vantu telu channel this is your master teacher for physics High Vol High volage series are especially for J [Music] aspirants okay volume some let us close the volume one yes so are we live anyone there watching okay I hope we are live yes josna good evening thank you thank you for responding so we are live welcome josna Chan hello hello everybody so semiconductor devices one of the M importantes when will you start revision oscillation BB M&M oscillation revision so maybe maximum Monday rision start so semiconductor devices semiconductor devices SCD semiconductor devices questions prev questions probably evening Chan good evening Akash everybody good evening high volage they are targeted for J so you can happily watch them previous years questions so semiconductor devices so semiconductor devices okay Jun U sessions January and April last two years highest number of theory questions concept based questions last two three years semiconductors highest number of questions Theory based concept questions hello everybody Ganesh Rd hi hi Ganesh Rd Charlie hi sa Krishna good evening okay first of all so semiconductors non-u El hello I hope everything everything is fine uh I don't know I read something wrong and uh They removed it30 okay daily or 4:30 almost okay message I hope everything is fine God bless them last outermost electr different layers Okay so wait Valance just wait uh ma'am wave Optics wave Optics coming up okay so I hope everything is fine sentence message I just text it so that everything is fine okay so okay yes yes yes yes wait so okay in start Optics yes Valance band condu band conduction BS Optics mechanic properties ofid comp okay so last Valance I mean last energy Bandit Valance band conduction band okay so Valance conduction electr avability units and dimension of electrons electrons Val forbidden energy gap okay forbidden forbidden forbidden energy gap forbidden okay Bond holes okay mains good evening suar thank you thank you so much okay I prepare okay [Music] so okay so uh okay uh [Music] add okay so semiconductors [Music] cond avilability of electrons in the conduction band conduction El okay so and elect so cond insulators semiconductors conductors electron volts conduct forbidden energy gap conductors forbidden energy gap forbidden energy gap forbidden energy El electrons Valu just room temperature cond yes sorry PP okay good evening already only for IP exams ma'am I am in my coaching still IP not started ma'am don't worry greater than elect less elect Medi Gap less than six elect Vols more six elect which is almost very impossible okay so semond correct silicon germanium soic gerum best examples for semiconductors semiconductors best examples enap ger 0.7 Vol silicon 1.1 Vol energy for energy of energy of energy of forbidden energy gap energy of forbidden energy gap forbidden energy silic 1.1 elect ger 0.7 electr Vol main difference conductor conductor key semiconductors M difference conductors semiconductor as the temperature increases conductors temperature increase conduct temper incre resistivity resistivity increases and condu conductivity decreases conductivity decreases electricity row T row T is equal to row into 1 + Alpha into Delta Theta resis so resistivity is linearly proportion to the tempure so right Hara is this enough for J 2024 100% enough next semiconductors as temperature increases semiconduct tempure semiconduct SL temperature increase reverse resistivity resistivity decreases resistivity decreases and conductivity increases conductivity inre semiconductor behaves as a perfect insulator Z k0 absolute temperature because temp sorry absolute at Absolute temperature absolute temperature and absolute temperature 0 Kelvin 0 Kelvin not 0° celius a semiconductor a semiconductor semiconductor material behaves behaves as a perfect insulator perfect ins okay semiconduct question so I hope so in of semiconductors fin in case of semiconductors in case of semiconductors electric current electric current is due to is due to movement of movement of both electrons and holes positive and negative ions movement similarly semiconductor electricity because of the electr and holes movement semiconductors intrinsic semiconductors and extrinsic semiconductors intrinsic semiconductors inin semiconductors and pure form of semiconductors these are pure form of semiconductors pure form of semiconductors pure form of semic semiconduct number of electrons is equal to number of holes number of electrons is equal to number of holes NE and N NE and N electron concentration ne ne is called as NE NH NE electron concentration electron conentration NH and whole concentration electron concentration whole concentration so in case of intrinsic semiconductors current due to electrons is equal to current due to holes so total current is because of current due to electrons and current due to holes semiconductors semond semiconduct total current is due to current due to electrons and current due to holes so intrinsic semiconductors next moving on to extrinsic semiconductors intrinsic semiconductors X transic spelling point extrinsic intrinsic and extrinsic extrinsic semi conductors extrinsic semiconductors and extrinsic semiconductors and these are not pure form not pure form not pure form impure form these are impure form of semiconductors impure form and silicon what is the pra going on in this is doping doping CH what do we call by doping by adding a foreign materi atoron atss extrinsic semiconductors form P type semiconductors and n type semiconductors okay T type semiconductors and N type semiconductors P Tye semiconductors P type semiconductor P semiconduct so P semiconduct doing do so first let us start with P type semi conductor PP semiconductor by doping with a tri valent impurity by doping intrinsic semiconductor pure form of semiconductor trient impur do so doing with trient tri valent or group three element group three element group three element okay only elect elect okay better right semiconduct in boron so mostly Boron do semond so sem majority charge carriers minority charge carriers and okay majority charge carriers majority charge carriers Al minority charge carriers majority charge carriers and minority charge carriers P type semiconductor majority charge carriers holes minority charge carriers electrons okay holes and electrons so majority charge carriers holes minority charge carriers electrons in P type semiconductors okay and energy band Gap Valance band conduction so Val a c c e p t o r acceptor level accept and semicond semond group acceptor impurity important every line can become aent valcy group element dope semiconductor P type semiconductor majority charge carriers holes minority charge carriers electrons P type semiconductor impur because the AC impur energy gap energy level pure semiconductors extrinsic or inic semond band conduction band balance band conduction band FY energy level FY energy level in semond Balance zor FM energy level andaru okay uh five 430 so extrinsic intrinsic semiconductor pure semiconductors per sorry semiconduct enery level Valance band Valance band Valance band conduction band conduction band soyel balance sh and basing on impur AC level next type of extrinsic semiconductor n type semiconductor n type semiconductor n type semiconductor n type semiconductor P type semiconduct semicond by doping with by doping with tri valent impurity n it is made by doping with it is made by doping with pentavalent b a l e n t pentavalent or pentavalent or group Five Element group Five Element pent valent element okay El 1 2 3s El majority charge carriers majority charge carriers minority minority charge carriers minority charge carriers in case of n type semiconductors majority charge carriers elect minity p p majority charge carriers holes minority electr majority charge Carri electr minority Charge car donor type impurity donor type impurity because conduction bandery level it shifts towards the conduction band donor donor energy level H donor energy level dor energy level H Valance semiconduct Valance band shft semond concept clarity about P Type n type semiconductors so intrinsic semiconductors extrinsic semiconductors extr Junction diode junction junction junction P Junction p right so semicond P type majority charge carriers holes n type majority charge carriers electode right so P Junction cross this barrier potential meas let us consider it to be v v barrier potential barrier potential barri acoss the barri potential okay D if D is width of barrier electric field set across barrier formula eal V by D electric field setup across barrier is eal V by D formula important formula formula electric chares Andi electric Rel right intrin semiconductors n i is equal to n sare is equal to NE into NH NE into NH elect concentration n whole concentration n n intrinsic concentration okay intrinsic concentration so n intrinsic concentration whole concentration electron concentration inic concentration relation intr semiconduct Q into mu e v e ne mu e ne e sorry Q is equal to Mu e in color green color color what color mu NE plus mu H nhh mu H nhh Sigma and conductivity Q charge 1.6i moil Mobility electric charges and mual VD by e formula V drift velocity conductors electric charges and Fields current electricity formula Mobility equal VD by E dri velocity e electric field mobility of electrons NE number of electrons mu mobility of holes NH number of holes or whole concentration inic semiconductors WID V across this barri potential v setal v by okay triang problem one week one chapter don't worry about it at all [Music] okay right okay that's [Music] it chaps semiconductors oscillations units and dimensions full revision anyway important problem Techni okay got SE in [Applause] equal Indian Institute of Management one of the unities he got seat so he went to study okay forward bias reverse bi Junction C forward bias condition reverse bias condition forward bias condition reverse bias condition Madam p okay so forward bias condition and reverse bias condition forward bias condition forward bias condition P Type n Tye Junction P should be P should be at higher potential P should be at higher potential the concept p higher potential lower important because problems question a question a okay Qui forward reverse bias + 5 volts minus 3 volts next question B PN Junction diode plus 5 Vol plus 3 volts forward by a b c Vol and - 3 volt D next 0 volts - 2 volts next e 0 Vols minus 2 volts next 2 volts next f F this is at uh -8 volts okay wait this is atus this is at-3 volts and this is at minus 8 volts a b c d tuck tuck tuck tuck which is forward bias which is reverse bias fbrb will you explain all chapters in physics for J Mains boom already 5 n minus 3 high potential low potential so this is in forward bias condition p + 5 n minus 3 high poti low poal forward high potential low potential n is at higher potential p is at lower potential so reverse bias condition 0 Vol grounding symb meaning grounding I cannot read the question so question I don't know what question you have asked I don't know what question has asked so I am unable to answer to your question already one that is the answer to your question ground grounding P at higher potential n at lower potential so forward bias condition - 3 -8- 8-3 forward important next problem very important problem solving techniques problem solving techniques PN Junction diode anaka forward by Junction forward current is due to majority charge carriers majority charge Carri reverse bias condition PN Junction diode characteristic curve characteristic curve important in case of practic semiconductors forward bias condition current due to majority charge carriers majority the goes like this volage knee voltage knee voltage or threshold voltage threshold voltage you can see there is a large amount of current thresh reverse forward bias condition La current due to majority charge carriers reverse bias condition La current due to minority charge carriers right and because majority charge terms of mampers forward bias region current M produce reverse minority Charge car in order of microamp so micro majority Charge car in case of germanium it is only 0.3 volts in case of silicon it is 0.7 Vol ger threshold voltage 0 .3 Vol silicon threshold Vol 0.7 volan problem if the diode if the diode is ideal if the diode is ideal is an ideal condition diode ideal diode ideal diode IDE mean resistance across diode resistance AC resistance of diode resistance meaning resistance of diode diode next find the current in the Circ example find find current in circuit find current in circuit find current through 2 ohms resist Okay so IAL Delta V by r i equal to V by R same formula current electri gerum potential gerum 0 Vol threshold potential 1 2 2 0.33 0 across every device potential drop so across 0.3 Vol so direct I equal V by R V6 r del 6us 0.3 Vol by Resistance 2 same two two ohms two OHS Vol connect vol.3 Vol silic 0.7 vol 01 02 03 03 germanium starts conducting 0.3 Vol already so 6 0.3 by starts thresh okay threshold potential gerum 0.3 silicon key 0.7 next problem solving techniques important problem solving techniques forward bi RSE 10 ohms okay 5 ohms next set up 6 ohms set find the current through the Circ find the through the cward because Junction reverse bi reverse electricity R effective 10 + 5 15 ohms right and diode ideal mention so full resistance IAL V by R 10 by 10 + 5 IDE diode has forward resistance in same problem in same problem diode has diode has forward resistance forward resistance of 5 ohms forward resistance forward resistance effective 5 + 10 + 5 so 20 ohms is the answer hi venkateswar Raju hi bunny hi hello hi anyway IDE fortif next Zena diode uh so next the zena diode zenor diode zenor diode the special application special application of P Junction diode Zen diode P Junction diode of allly doed Highly doped PN Junction diode first of all the very highly doped P it can be operated it can be operated in Reverse bias condition normal P Junction diod reverse bias condition operate Z symb symb normal common CC of Z series resistance because resistance Lo resistance us Zod so Z reg and series resistance Lo right so load resistance RS resistance parall combination parall combination parall combination potential vus V potential across RS v v vus v am so I amount of current I I I I divide I amount of current I I divide according to K's Junction law can I write I equ Al to I + I can I write I equal i+ I problem okay so forward first point par potential potential same second Point next i i z+ i l i z plus I minimum concept of Zen okay sotif half rectifier and full wave rectifier rectifier mean rectifier alternating Curr alternating alady DC output directr so alternating wave rectifier half wave rectifier half half CLE second next positive cycle forward outting outut .46 0.46 into RL by RF plus r r load resistance Lo resistance R load resistance RF forward resistance of diode forward resistance of diu I made second payment for J Mains through internet Center thank you madam for your guidance thank you okay good successful okay pray Kumar successful 40 efficiency almost 50 40 next next full wave rectifier full wave during clei ccle to 0.812 into RL by RF + RL samey half full Bridge full wave Bridge rectifier full wave Bridge re identification based full Bridge diamond full W Bridge rectifier okay next capacitor rectifiers capacitor capacitor is used as filter after using capacitor after using capacitor in case of rectifiers okay so rectifi Capac applications of P Junction phoe photo diode R the LED light emitting diode and third one photo volic cell m let us start with photo diode photo diode photo diode basic it is a sensor it is light sensitive light sensitive and Light automatic on off normal P Junction so it is operated it is operated in Reverse bias condition rever okay firstes phons electrons so no problem AB I think [Music] okay reverse I2 I3 I so I is greater than I3 is greater than I2 is greater than i1 clear nexted lemal light B and this is operated this is operated in forward bias condition operated in forward bias condition previously reverse bias condition operate okay so this is the uh [Music] e equal to HH e equal to H C by Lambda ghani LED blue colored Lambda equal to H C by E Lambda 400 nanom 700 n okay so us photo volic cell photo volic cell photovoltaic cell solar cell meaning photo cell Sol elag waves eles el el short circuit current short circuit current andc completed concept completed problem problem so first problem identify the solar cell is next what is this next in the given circuit the input voltage V in is shown in the figure 10.6 vol. Vol second okay okay okay the cut in voltage cut threshold voltage cut in voltage of PN Junction diode is 0.6 volt di threshold potential 0.6 Vol 0.6 Vol right so cor outut vol. 0 this will be the correct answer because rectifier sorry sorry wait wait this will be the correct answer right so this C is the correct answer am I clear next children I V graph i v graph only the ratio of dynamic resistance corresponding to forward voltage 2 volts and 4 volts 2 volts four volts ratio of resistance resistance Ral Ral Del V Del that's it so R1 by R2 R1 R1 so I can take Delta X by Delta or Delta by Delta X Del V by Delta i v i so Delta v 2.1 minus 2 by Delta I 10- 5 0.1 by 5 R2 con so Del V by Delta I Delta V ENT 4.2 - 4 by Delta I 250 - 200 so 0.2 by 250- 200 is 50 R1 will you find out very good each diode has forward bias resistance of 30 ohms resistance diode is not ideal res 30 e 130 30 Ser so toal 160 ohms 160 ohms 16 16 16 16 160 by 80 80 oh 8 20 OHS are in series effective value 80 + 20 that is equal to 100 ohms okay so 100 ohms question the current current is equal to I equal to V by R so V 200 by R 100 so answer 2 2 Amp is the correct answer next for the Circ shown below Cate the value of IAL I I IUS I I across this resistance 50 vol 50 Vol potential so IAL v l by r l i equal to V by R so I equal to v l by RL I L equal v l 2 1 2 3 by sorry V 50 okay v50 sorry V 50 by 2 1 2 3 so 0 0 cancel 5 1 * 5 40 * so I 1 by 40 already = 100us 50 50 Vol R vol IAL V because V 50 by R okay 20 so 1 by 20 amp I so final i z is equal to i - I I and 1X 20 - 1 by 4 have you understood very standard model standard model very standard model highly standard model Amma cha highly standard model cha highly standard model very good next next question children LED is constructed G G and Gallum arenic phosphide Gallum arenic phosphide us the energy gap of this led is 1.9 electron volts energy gap of this led is 1.9 electron volts calculate the wavelength of light emitted when its color is and its color wavc Lambda lamb equal H C by E okay so Lambda Lambda is equal to H 6.63 into 10 power - 34 into 3 into 10^ 8 by E value e value 1.9 electron volt nanom because so problem 654 nanom 1046 nanomed obvious 1046 nanom RoR 400 to 700 so 700 so 654 red color will be the correcte soed correct red will be the correct answer okay okay so okay Amma next problem next problem [Applause] hello students small problem Studio don't worry listen carefully AB hi vesar hi phys students 2 AB regular student yes tarun yes yes Akash hi K hi yes yes manut hi Harish hi hello hello yes yes minutes study how hi har hi wait two minutes okay okay so uh shall we start continue quick so 5 across x 5 Vol battery connect Vol conect D1 D2 normal silicon find the current supplied by battery if the positive terminal is connected to point x so positive negative find the current supplied IAL V by R okay IAL V by R next if a semiconductor photo diode can detect a photon with a maximum wavelength this is Lambda band Gap c l that's next okay next okay so 10 + 5 + 10 total resistance IAL v r right IAL v r next in the figure potential difference between a and b a potential IAL i1 I2 5 Kil 10 Kil so answer 15 Kil effective okay 15 so problems complete solution okay so the breakdown voltage 6 Vol break Vol next Vol 16 Vol minimum is equal vs by RS so 10 by 2 kilo 10 3 I by RL 6 by 4 Kil 10^ 3 so I I equal to IUS I I equal to IUS I so repeated models okay soete solu next class logic GES

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Id: rKTzRCYvsP0

Length: 112min 47sec (6767 seconds)

Published: Fri Nov 10 2023