Calculate Current and Current Density of PN Junction

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The discussion focuses on calculating the current and current density of a PN junction formed by two cylindrical semiconductors with a radius of 0.065 mm. The current is determined using the formula i = dq/dt, yielding a value of 2 x 10^-4 C based on the flow of 3.50 x 10^15 electrons and 2.25 x 10^15 holes per second. The current density is calculated as J = i/A, resulting in 1.50 x 10^4 C/m². Participants confirm that both electrons and holes contribute positively to the current, clarifying the relationship between charge carriers in the junction.

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A pn junction is formed from two different semiconducting materials in the form of two identical cylinders placed circular face to circular face. The radius of the cylinders is 0.065mm. In one application, 3.50 x 10^15 electrons per second flow across the junction from the n to the p side while 2.25 x 10^15 holes per second flow from the p side to the n side. (A hole acts like a particle with charge +e)
Find a) Current b) Current Density


For the current
i = dq/dt = [(3.50-2.25) x 10^15] (1.6 x 10^ -19) / 1 sec = 2 x 10^-4 C.
For the current density
J = i/A = (2 x 10^-4) / [pi (0.065 x 10^-3)^2 ] = 1.50 x 10^4 C/ m^2
is this correct??

A copper wire and an iron wire of hte same length have the same potential difference applied to them.

a) what must be the ratio of their radii if hte current is to be the same?
i = \frac{A \Delta V}{\rho L} = \frac{\pi r^2 \Delta V}{\rho L}
equating for both the copper(C) and the iron(I)
\frac{r_{I}}{r_{C}} = \sqrt{\frac{\rho_{I}}{\rho_{C}}
This part should be fine

b) Can the current density be made the same by suitable choices of radii?THis is the part I'm having doubts with. SInce J = i/A, would this not be possible? Please do advise!

Thank you for your help!
 
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stunner5000pt said:
A pn junction is formed from two different semiconducting materials in the form of two identical cylinders placed circular face to circular face. The radius of the cylinders is 0.065mm. In one application, 3.50 x 10^15 electrons per second flow across the junction from the n to the p side while 2.25 x 10^15 holes per second flow from the p side to the n side. (A hole acts like a particle with charge +e)
Find a) Current b) Current Density

For the current
i = dq/dt = [(3.50-2.25) x 10^15] (1.6 x 10^ -19) / 1 sec = 2 x 10^-4 C.
You wrote correctly that i=dq/dt. The positively charged holes flow from the p side towards the n side, tending to increase the charge of the n side and their flow determines the direction of current.
What happens with the charge of the n side when negative charges (electrons) flow away from it? As the result of leaving negative charges, the n side tends to become more positive, so the contribution of the electrons to the current is positive.
In principle you get the net current arising from the drift motion of different charged particles as the summ of terms qvn where q is the charge of the particle, v is its drift velocity and n is the concentration of that kind of particle. If both q and v are negative you get a positive contribution.
ehild
 
so in other words i should be adding the numberof holes with the electrons since the holes along iwth the electrons increase the amount of current flowing?
 
Refer to this about holes and conduction -

http://www.hardwareanalysis.com/content/editorials/article/1588.3/

and

http://hyperphysics.phy-astr.gsu.edu/hbase/solids/intrin.html#c4

Holes are formed when an electron leaves an atoms so there is net positive charge. A hole moves when an electron from one atom (lattice site) moves into the hole 'filling it' so that the atom (lattice site) from which the electron originate now has the hole. So holes move as fast as the electrons can fill them.

More generally - http://hyperphysics.phy-astr.gsu.edu/hbase/solids/band.html
 
stunner5000pt said:
A copper wire and an iron wire of hte same length have the same potential difference applied to them.
a) what must be the ratio of their radii if hte current is to be the same?
i = \frac{A \Delta V}{\rho L} ...
b) Can the current density be made the same by suitable choices of radii?THis is the part I'm having doubts with. SInce J = i/A, would this not be possible? Please do advise!
Thank you for your help!
You've written down the expression for the current, i. Now write the expression for the current density, j = i/A. Look at this expression and decide if j can be varied by changing the radius.
 
sooo is the answer to the first one correct or not? Since the electrons flow they create a current and since the holes do they ADD the current??

Please advise

Thnak you so far for the advice!
 
stunner5000pt said:
sooo is the answer to the first one correct or not? Since the electrons flow they create a current and since the holes do they ADD the current??
Please advise
Thnak you so far for the advice!
Yes, they add to the current, as you correctly said in post#3.
 

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