Junction Diode as a Rectifier Video Lecture | Physics for JEE Main & Advanced

c'mon agent physics help me solve this
mystery I have connected a sine wave at
the input so this is what the input
looks like but at the output half the
wave has disappeared I only get one half
of the wave any guesses what could be
there inside this box over here yes you
got it right
it's a diode
today we are going to talk about the
working of a junction diode and let's
see why does it allow only the positive
half cycles to go through and why does
it stop the negative half cycles this is
the circuit symbol for a diode
it's basically an arrow and this side is
the P side and this side is the N side a
diode is a PN Junction so let's first of
all draw a PN Junction
the majority carriers on the inside are
these electrons and the positive charge
that you can see are the lattice sites
on the p-side the majority charges are
the whole you can see that the negative
charges have diffused across the
junction and combined with the holes
present over here for instance say these
are the electrons and these are the
holes now when the electrons diffuse to
the other side they occupy these
vacancies so once the electron occupies
this vacancy over here then the hole no
longer exists now this region in which
there are no free charge carriers is the
depletion region and due to this layer
of positive charge on the inside and
negative charge on the P side there is a
potential difference and that potential
difference can be shown on the graph you
have a positive potential on the N side
and a negative potential on the P side
now this potential difference creates a
potential barrier that stops further
diffusion of electrons from taking place
because when the electrons move to this
side they will be repelled back by the
negative charges which are present on
this side
now comes the most important part we
join a battery with its positive
terminal connected to the P side and the
negative terminal connected to the N
side when we connect the diode in this
manner we find let's say we have a
galvanometer over here in the circuit
then the current in the galvanometer
increases when the battery is connected
in this manner the PN Junction is said
to be forward biased and in this case
the current increases see actually if
you increase this biasing voltage then
the current will further increase we can
show this on a graph let me just draw
the graph we have V over here this is I
is the current and O so as you increase
V I increases alone Oh a now why do you
think the current increases when we
increased the forward bias see due to
the forward bias this is the positive
side and this is the negative side so
the holes are being pushed away from the
end towards the center towards the
junction and if you look at the
electrons since this end is negative
then the electrons are also pushed
towards the junction so you have more
electrons close to the junction and more
holes close to the junction due to this
the width of the depletion region
decreases in addition to that the
potential barrier also reduces so the
opposition to the flow of current is
less now this means in a forward bias
all conditions are such that it helps
the current to increase because the
depletion region has decreased the
potential barrier has decreased so the
electrons can cross over to the other
side
easily now
now let us consider a reverse bias which
means we connect the negative terminal
of the battery to this end and the
positive terminal to the other end so to
the p-side we connect the negative
terminal and to the inside we connect
the positive terminal this is called the
reverse bias now as you probably guessed
the current will reduce in case of a
reverse bias the current the current
decreases in case of a reverse bias
let's just draw this on the graph now
this graph
oby shows a tiny amount of current that
flows you to reverse bias and this small
current is due to the minority carriers
C on the N side there are a few holes
that are present and on the P side there
are a few electrons which are present so
see this layer of positive charge on the
N side which is present at the junction
attracts these electrons so these
electrons cross over to the other side
and combine with these holes but the
concentration of these minority carriers
is very low so this minority current is
very less now let's take a look at the
majority carriers see this negative
potential at this end will pull the
holes towards this side of the junction
so the holes are getting attracted to
this side of the junction and they are
moving away from the junction
similarly the electrons are attracted by
the positive terminal of the battery so
the electrons will tend to move towards
the end of the junction and away from
the junction this leads to an increase
in the depletion region because the
mobile charge carriers tend to move away
from the junction when the depletion
region increases then the potential
barrier also increases
so if the potential barrier increases it
becomes more difficult for the electrons
to cross over and combine with the holes
on the other side so this leads to a
decrease in current and you know that
the tiny amount of current which is
present is due to the minority carriers
now coming back to the mystery of the
disappearing half-wave I am sure you can
answer this see for the positive half
cycle the diode is conducting so since
the diode is conducting we get an output
for the first positive half cycle now
during the negative half cycle the diode
is reverse-biased
because this end will be negative so in
that case the diode will not conduct so
since there is practically no current
flowing we say that the diode is non
conducting and we get a zero output
again during the next positive half
cycle the diode is conducting so we have
an output during the next negative half
cycle once again it will be reverse
biased and we have zero output so the
output is only in one direction it is
only in the positive direction and there
is no negative half cycle present the
output is now said to be rectified
because it is only throwing in one
direction though you can still see that
this is pulsating which means it is
varying it is not constant but still it
is a DC output because it doesn't
reverse direction
to test your understanding try these
questions the answers are given a short
while later
and here are the answers if you've got
all of them right don't forget to give
yourself a pat on your back and if you
haven't snowboarders go bad go to the
lesson once again and try the collection
again