In zener diode, since NA and ND is large, there are no enough minority carriers to participate in drift under the influence of E during breakdown, so E has to do an additional work of creating these minority carriers by knocking them off from their atoms and then they can drift. But in avalanche breakdown since NA and ND are small, minority carriers are already available to drift during breakdown.
Actually I think one of the reasons your videos work so well is the black background with the different colors of red, blue yellow and green to signify different things. I think it's something to do with the human brain and primary colors which makes watching the videos enjoyable just off color scheme alone. Did you plan it that way on purpose?
I liked those basic colors, hence used them at the time of making the videos! Thank you for noticing, taking time and expressing your views! Thank you!!
Hi Sreekanth, In an intrinsic semiconductor 4N states vacant in CB, 4N states filled in VB. Even in ntype extrinsic semiconductor, due to doping very small fraction of CB gets filled - lots of vacant states in CB. For silicon N is 5x10^22, doping is in 10^15 to 10^19. Try this problem to get feel:ua-cam.com/video/o4C2jq2dhuQ/v-deo.html
souldn't the equation say q*(Vbi+V) as we are in reverse biase so the E field resulted by the reverse biasing voltage points to the same direction as the E field in the depletion would in an unbiased state?
techgurukula just asked ‘cause in your video that discussed the forward and reverse biase if the pn junctiom the equation was q*(Vbi+V) in the reverse biase. That made sense as both the np junctions own E field and the reverse biasing voltage’s E field points into the same direction which is only possible if the voltages have the same polarity
@@techgurukula I you look at your video ua-cam.com/video/eUQeWB7Psbo/v-deo.html (Electronic Devices: pn junction - forward and reverse bias) the equation you write up at 11:55-ish. That says q*(Vbi+Vr). I believe this one is the correct one, right?
Sir, if depletion width is small how can we conclude that electric field is very high in the depletion region. I am confused coz in the formula of electric field the depletion width comes in the numerator i.e Emax = -(qNdWn)/(epsilon).
In avalanche effect, how atoms will exist in the depletion region which is made up -ve and +ve ions of Na and Nd. i didnt understand about impact ionization.
i loved how you explained their breakdown effect using energy band daigram ..great
exquisitely explained! so much better than other explanation on youtube. thank you so much!
splendidly explained and well understood. Thank s so much sir. Keep it up
Incredible lecture. Great job!
Excellent explanation
In zener diode, since NA and ND is large, there are no enough minority carriers to participate in drift under the influence of E during breakdown, so E has to do an additional work of creating these minority carriers by knocking them off from their atoms and then they can drift. But in avalanche breakdown since NA and ND are small, minority carriers are already available to drift during breakdown.
very nice and useful make more video on imp topics
Such a good explanation
You are really blessed.
Actually I think one of the reasons your videos work so well is the black background with the different colors of red, blue yellow and green to signify different things. I think it's something to do with the human brain and primary colors which makes watching the videos enjoyable just off color scheme alone. Did you plan it that way on purpose?
I liked those basic colors, hence used them at the time of making the videos! Thank you for noticing, taking time and expressing your views! Thank you!!
How come vacant states present in CB of n+ side in Zener Effect? I think vacant states will only be created in VB right...Isn't it that way?
Hi Sreekanth,
In an intrinsic semiconductor 4N states vacant in CB, 4N states filled in VB. Even in ntype extrinsic semiconductor, due to doping very small fraction of CB gets filled - lots of vacant states in CB.
For silicon N is 5x10^22, doping is in 10^15 to 10^19. Try this problem to get feel:ua-cam.com/video/o4C2jq2dhuQ/v-deo.html
@@techgurukula Thanks :)
Sir... In RB condition barrier potential isn't q(Vbi+Vd)???
souldn't the equation say q*(Vbi+V) as we are in reverse biase so the E field resulted by the reverse biasing voltage points to the same direction as the E field in the depletion would in an unbiased state?
q*(Vbi-V), in forward bias V value is +ve and in reverse bias V value is -ve.
techgurukula just asked ‘cause in your video that discussed the forward and reverse biase if the pn junctiom the equation was q*(Vbi+V) in the reverse biase. That made sense as both the np junctions own E field and the reverse biasing voltage’s E field points into the same direction which is only possible if the voltages have the same polarity
@@techgurukula I you look at your video ua-cam.com/video/eUQeWB7Psbo/v-deo.html (Electronic Devices: pn junction - forward and reverse bias) the equation you write up at 11:55-ish. That says q*(Vbi+Vr). I believe this one is the correct one, right?
@@marcellmurgas7823 V=Vf and V=-Vr, substitute in the equation q*(Vbi-V).
The energy difference between Ec of p side and Ec of n side will be q(Vbi+Vd)...
Well explanation, thank you sir
Sir, if depletion width is small how can we conclude that electric field is very high in the depletion region. I am confused coz in the formula of electric field the depletion width comes in the numerator i.e Emax = -(qNdWn)/(epsilon).
aishwarya nair, E=U/d, agree? d is smaller, U is the same, E - bigger :)
in this formula Wn is width of N side semi conductor not the total depletion width (W=Wn+Wp); Emax= (2*Vbi)/W from this we conclude...... 😊😊😊
@@mayanksisodiya8567
But depletion region also decreased right??
Wn means width excluding DR in N side??
In R.B Electric field increases compared to F.B, then for Breakdown effect voltage should be high,then Electric field increases.
May I know the name of the faculty and where is he from??? He is incredible and master in this subject..Respect🙏🙏
beautifuly explained..
in which effect electrons will be more accelerated, zener or avalanche, please tell......great video man .
why there are so many vacant states in conduction band of n side? ????
sir, which software you used to record these lessons
In avalanche effect, how atoms will exist in the depletion region which is made up -ve and +ve ions of Na and Nd. i didnt understand about impact ionization.
i think this can be better understandable in bond model.
rajeswararao macherla, do you think there is no Si atoms in the doped Si? :)
I think there are si atoms exists in depletion also...,those atoms ionized and accelerated...
It was very useful...
thanks
alot
best ever
(y)