Is G always equal to R in equilibrium, as seems to be assumed in this video? Because technically you could have dq/dt = 0 even though R isn't G (you could generate a net amount of charges that are then carried out through a net current), but I'm guessing that this wouldn't work forever since we'd run out of electrons? Is it always reasonable to assume R = G at equilibrium?
Excellent question. This is a good assumption under equilibrium. The reason is this - imagine you did manage to extract some charge from the semiconductor (*and* that you successfully extracted only the electrons without the holes). The semiconductor would now have a net positive charge, and it becomes increasingly difficult to rip additional electrons out of it (the net positive charge is attracting back the electrons you try to take away). This is by definition not equilibrium. Eventually you won’t be able to get out any more electrons because it will just be too hard, and your semiconductor will be in equilibrium, at which point R=G. This is equilibrium.
How could electrons diffuse from the p region to the n region should not they diffuse from the high concentration region (the n region) to the low concentration region (the p side)?
a percentage of electrons that are generated within 1 diffusion length of the space charge region on the p side will randomly drift around into the SCR region and then get swept across to the other side by the electric field. You are thinking of diffusion which is what happens to the majority carriers, drift current is what happens with the minority carriers (a force is applied due to an electric field as opposed to a concentration gradient) and is a weak function of the applied field
@@jacobcorr337 yes so minority carries in the n side which simply defined to be the holes get drifted to the p side and same for the minority charge carries on the other side I got it thanks brother
@@JordanEdmundsEECS in the prior video, you used the equation of Vbi developed in equilibrium condition, in a PN junction under forward bias which is not equilibrium. Could you please explain it. Thank you
@@user-ge8hj9br6w my teacher said that the model breaks because the expression for the depletion zone width becomes the square root of a negative number.
Great series of videos really gives a good overview!
Thank you so much, this is so clear and well explained
at moment 03:19, why does the "NA" pass through the deplition to "n" area??
Is G always equal to R in equilibrium, as seems to be assumed in this video? Because technically you could have dq/dt = 0 even though R isn't G (you could generate a net amount of charges that are then carried out through a net current), but I'm guessing that this wouldn't work forever since we'd run out of electrons? Is it always reasonable to assume R = G at equilibrium?
Excellent question. This is a good assumption under equilibrium. The reason is this - imagine you did manage to extract some charge from the semiconductor (*and* that you successfully extracted only the electrons without the holes). The semiconductor would now have a net positive charge, and it becomes increasingly difficult to rip additional electrons out of it (the net positive charge is attracting back the electrons you try to take away). This is by definition not equilibrium. Eventually you won’t be able to get out any more electrons because it will just be too hard, and your semiconductor will be in equilibrium, at which point R=G. This is equilibrium.
@@JordanEdmundsEECS I think I get it now. Thank you so much! These answers you're giving are incredibly helpful!
sir, is the mass action law valid throughout the pn diode? pls answer.
How could electrons diffuse from the p region to the n region should not they diffuse from the high concentration region (the n region) to the low concentration region (the p side)?
a percentage of electrons that are generated within 1 diffusion length of the space charge region on the p side will randomly drift around into the SCR region and then get swept across to the other side by the electric field.
You are thinking of diffusion which is what happens to the majority carriers, drift current is what happens with the minority carriers (a force is applied due to an electric field as opposed to a concentration gradient) and is a weak function of the applied field
diffusion length being the average distance travelled before recombinating with a majority carrier
@@jacobcorr337 yes so minority carries in the n side which simply defined to be the holes get drifted to the p side and same for the minority charge carries on the other side
I got it thanks brother
What is equilibrium condition
‘Equilibrium’ just means you aren’t actively putting energy into the system. For example, when you aren’t applying voltage to the diode.
Even temperature is 0
@@JordanEdmundsEECS in the prior video, you used the equation of Vbi developed in equilibrium condition, in a PN junction under forward bias which is not equilibrium. Could you please explain it. Thank you
@@JordanEdmundsEECS even if we apply VD we still assume that in infiite time there's no net charge movement and no accumulation of charge?
is Vd always less than Vbi?
Nope, it can be larger, but then the model we are using starts to fail.
@@JordanEdmundsEECS won't the internal resistance of the voltage source take most of the drop with it?
@@user-ge8hj9br6w my teacher said that the model breaks because the expression for the depletion zone width becomes the square root of a negative number.
I would recommend the 'tao of pooh' by Benjamin Hoff, it is an incredible introduction to Western teachings of Taoism
.