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stunner5000pt
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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?
[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!
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!