- #1
hogrampage
- 108
- 1
Homework Statement
A 1-cm cube of p-type silicon (ρ = 0.1Ω-cm) acquires a linear electron distribution in the x-direction, such that n = 1014/cm3 at one side and n = 105/cm3 at the opposite side.
Wires are attached to the sides of the cube via ohmic contacts, and a 0.1mV voltage source is applied. Find the values of the electron, hole, and total currents that flow in the external circuit.
Homework Equations
ρ = R[itex]\frac{A}{L}[/itex] = [itex]\frac{E}{J}[/itex]
E = [itex]\frac{V}{L}[/itex]
Je = qnμeE + qDe[itex]\frac{dn}{dx}[/itex]
Jh = qpμhE + qDh[itex]\frac{dp}{dx}[/itex]
i = [itex]\frac{V}{R}[/itex]
n0p0 = n2i
The Attempt at a Solution
I started by finding R:
R = ρ[itex]\frac{L}{A}[/itex] = (0.1Ω-cm)[itex]\frac{(1cm)}{(1cm^{2})}[/itex] = 0.1Ω
Then, the total current is:
I = [itex]\frac{V}{R}[/itex] = [itex]\frac{(0.1mV)}{(0.1Ω)}[/itex] = 1mA
p0 = [itex]\frac{n^{2}_{i}}{n_{0}}[/itex] = [itex]\frac{(1.5x10^{10})^{2}}{10^{14}}[/itex] = 225x104/cm3
p1 = [itex]\frac{n^{2}_{i}}{n_{1}}[/itex] = [itex]\frac{(1.5x10^{10})^{2}}{10^{5}}[/itex] = 225x1013/cm3
Now, [itex]\frac{dp}{dx}[/itex] = 225x1013/cm4 and [itex]\frac{dn}{dx}[/itex] = -1x1014/cm4, since dx = 1cm.
Which values would I use for n and p in the following equations, assuming the above steps are correct?
Je = qnμeE + qDe[itex]\frac{dn}{dx}[/itex]
Jh = qpμhE + qDh[itex]\frac{dp}{dx}[/itex]
Last edited: