p-n junction under bias

[@ganiulrajjak4180]

Sometimes I am compelled to think how these lectures are free......thank you sir for another awesome conceptual video...God bless you

[@animeshtiwari1005]

Preparing for GATE?

[@renatoberaldo2335]

Sedra Chapter 3 + this full course = you rock!!!

[@IPLHava]

i have a doubt, how to apply all these concepts in creating some practical applications.

[@naveensagar1101]

29:53 It should be exp(+(x+wp)/ln) With x < -wp As electrons on p-side decay along -ve x direction

[@saicharanmarrivada5077]

Yes,correct bro

[@saisagardasari227]

No bro..I think minus should be included ..since it is a decaying fn.

[@vamsimohan5369]

@@saisagardasari227 hey try copying this 2\cdot e^{\left(x+2 ight)} on www.desmos.com/calculator you can see that + sign is correct for the curve needed

[@louerleseigneur4532]

Awesome Thank you very much sir

[@MMTRINATHSOLASA]

Thnq sir 🙏🏻

[@Virus-ke8xj]

hyperphysics.phy-astr.gsu.edu/hbase/Solids/pnjun2.html For more details☝🏻

[@freeandreliablejeeprep820]

thank you sir

[@praneshkandasamy3175]

In 25:30 delta p(x) = delta p(Wn)*exp((Wn - x)/Lp) is substituted from what we learnt from continuity equation, but there we derived delta n proportional to exp(-x/Ln) after assuming that the semiconductor is not experiencing any field but that is not the case here. Then how can apply that result here. Please can anybody explain

[@CHAUHANANAND-hn5mu]

There we assumed that somehow excess carriers are generated (remember shining light) and then we derived an equation for those excess carriers profile with respect to x. Here instead of shining light we are using electric field for excess carriers generation. but at the end the equation tell the excess carrier profile. Hope it helps. Correct me if I am wrong

[@farihminan2512]

maybe it is because E=0 in neutral regions, the applied field drops the potential barrier in the depletion region. we derived the concentration in a region where E = 0. Correct me If I am wrong.