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HW 4: Plotting carrier concentrations in forward biased diode and deriving circuit quantities
For this exercise, we will use a one-sided p-n diode, with p-side Na=2×1018cm-3, and n-side Nd=2×1016cm3
. For the carrier lifetimes, use 1ns, and the included electron mobility (1000cm2/Vs) and the included hole
mobility (400cm2/Vs).
1.
2.
3.
4.
5.
6.
7.
8.
9.
Calculate the EQUILIBRIUM minority carrier concentrations on the 2 sides, i.e. pn0 and np0.
Calculate the fermi level positions on the 2 sides in the quasi-neutral regions
Draw the full band diagram under EQUILIBRIUM
Assuming the quasi-neutral regions on the 2 sides are 10x the diffusion length, estimate the
series resistances in Ωcm2 on the 2 sides.
Draw the band diagram under 0.2V forward bias. Where does this voltage drop? Make sure to
indicate the quasi-fermi levels, i.e. how they split under application of voltage.
Where is the peak electric field?
Plot the carrier distributions and current densities on the 2 sides using the attached Scilab code.
At V=-.2V, calculate the DEPLETION region width, and the associated capacitance in F/cm2.
Diffusion capacitance extraction)
a. Now, calculate the total excess charge in the diffusion induced minority carriers at 0.2V
in #/cm2 and C/cm2. Remember to do this for both pn and np.
b. Calculate the total excess charge in the diffusion induced minority carriers at 0.26V
forward bias in #/cm2 and C/cm2. Remember to do this independently on the 2 sides.
c. From the expressions (in the notes)
−𝒊𝒏𝒇
∆𝒏𝒑 (𝒙)𝒅𝒙]/𝒅𝑽
𝑪𝒅𝒊𝒇𝒇𝒖𝒔𝒊𝒐𝒏,𝒑−𝒔𝒊𝒅𝒆 = 𝒒𝒅[∫
−𝒙𝒑
+𝒊𝒏𝒇
𝑪𝒅𝒊𝒇𝒇𝒖𝒔𝒊𝒐𝒏,𝒏−𝒔𝒊𝒅𝒆 = 𝒒𝒅[∫
∆𝒑𝒏 (𝒙)𝒅𝒙]/𝒅𝑽
+𝒙𝒏
Calculate the diffusion capacitances on the 2 sides in F/cm2.
d. Which side has larger diffusion capacitance?
e. How does this diffusion capacitance compare with your depletion capacitance?
10. Sketch the equivalent circuit diagram (remember from the notes)? Include Rn, Rp¸Cdepletion,
Cdiffusion,n-side and Cdiffusion,pside.
11. Now, perform the integration analytically using the following equations for the carrier
distributions we derived in class:
𝑛𝑖2
𝑉𝑎𝑝𝑝𝑙𝑖𝑒𝑑
𝑥 − 𝑥𝑛
∆𝑝𝑛 (𝑥) = ( ) (exp (
) − 1) exp(−
)
𝑁𝐷
𝑉𝑡ℎ
𝐿𝑛
𝑥 + 𝑥𝑝
𝑛𝑖2
𝑉𝑎𝑝𝑝𝑙𝑖𝑒𝑑
∆𝑛𝑝 (𝑥) = ( ) (exp (
) − 1) exp(
)
𝑁𝐴
𝑉𝑡ℎ
𝐿𝑛
This is actually not so hard. Integral of exponential is exponential, and when exp(-x)0 as
xinf.
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