Calculate Current and Current Density of PN Junction

In summary, 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. The current is 2 x 10^-4 C and the current density is 1.50 x 10^4 C/m^2. For a copper wire and an iron wire of the same length and potential difference, the ratio of their radii must be equal to the square
  • #1
stunner5000pt
1,461
2
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?
[tex] i = \frac{A \Delta V}{\rho L} = \frac{\pi r^2 \Delta V}{\rho L} [/tex]
equating for both the copper(C) and the iron(I)
[tex] \frac{r_{I}}{r_{C}} = \sqrt{\frac{\rho_{I}}{\rho_{C}} [/tex]
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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  • #2
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
 
  • #3
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?
 
  • #4
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
 
  • #5
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?
[tex] i = \frac{A \Delta V}{\rho L}[/tex] ...
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.
 
  • #6
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!
 
  • #7
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.
 

1. What is a PN junction?

A PN junction is a type of semiconductor junction formed by joining a p-type (positively charged) and an n-type (negatively charged) material. This junction creates a barrier that allows for the flow of electric current in one direction only, making it a crucial component in electronic devices.

2. How is current calculated in a PN junction?

The current flowing through a PN junction can be calculated using Ohm's Law, which states that current (I) is equal to the voltage (V) across the junction divided by the total resistance (R) of the junction. The equation is I = V/R.

3. What is the formula for current density in a PN junction?

The formula for current density (J) in a PN junction is J = I/A, where I is the current and A is the cross-sectional area of the junction. This formula takes into account the size of the junction and is often used to determine the efficiency of the junction.

4. How does temperature affect the current and current density in a PN junction?

Temperature can affect the current and current density in a PN junction in several ways. As the temperature increases, the resistance of the junction decreases, allowing for a higher current to flow. However, at very high temperatures, the junction may become damaged and its efficiency may decrease. Additionally, temperature can also affect the recombination rate of charge carriers, which can impact the current density.

5. What factors can affect the current and current density of a PN junction?

Several factors can affect the current and current density of a PN junction, including temperature, voltage, and the properties of the materials used to create the junction. The doping levels, thickness, and purity of the p-type and n-type materials can also impact the current and current density. Additionally, external factors such as light intensity and magnetic fields can also influence the behavior of the junction.

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