ECE 305 1)
Spring 2015 ECE 305 Homework SOLUTIONS: Week 8 Mark Lundstrom Purdue University The sketch below shows the carrier concentrations in a PN junction at room temperature. Answer the following questions. 1a) Is the diode forward or reverse biased? Explain your answer. Solution: Forward biased because there are excess electrons on the P-­‐side and excess holes on the N-­‐side. 1b) What is the acceptor concentration on the P-­‐side? Solution: N A = 1016 cm -3 1c) What is the donor concentration on the N-­‐side? Solution: N D = 1014 cm -3 1d) What is the intrinsic carrier concentration? Solution: n0 p0 = ni2 On the P-­‐side: n0 p0 = 1016 × 107 = 1023 ni = 1023 = 3.16 × 1011 cm -3 On the N-­‐side: n0 p0 = 1014 × 109 = 1023 ni = 1023 = 3.16 × 1011 cm -3 ni = 3.16 × 1011 cm -3 ECE-­‐305 1 Spring 2015 ECE 305 Spring 2015 HW8 solutions (continued): 1e) Do low level injection conditions apply? Solution: YES. On the P-­‐side: Δn −x p = 1010 << p0 = 1016 ( )
On the N-­‐side: Δp ( xn ) = 1012 << n0 = 1014 1f) What bias (in volts) is applied to the diode? Solution: According the the law of the junction: n2
Δn −x p = i eqVA kBT = n p0 eqVA kBT NA
( )
( )
⎛ 1010 ⎞
k BT Δn −x p
VA =
ln
= 0.026ln ⎜ 7 ⎟ = 0.18 V q
n p0
⎝ 10 ⎠
V A = 0.18 V 1g) Which is longer: the electron diffusion length on the P-­‐side of the junction or the hole diffusion length on the N-­‐side of the junction. Solution: The electron diffusion length on the P-­‐side. (The diffusion length is the distance it takes for the minority carrier concentration to return to its value in the bulk. Question: What would the figure look like for a reverse biased diode? 2)
This problem concerns the energy band diagram shown below for a diode under bias. ECE-­‐305 2 Spring 2015 ECE 305 Spring 2015 HW8 solutions (continued): 2a) Is the diode forward or reverse biased? Solution: Forward biased because Fn > Fp . 2b) What is the value of the applied bias? Solution: qV A = Fn − Fp V A = +0.5 V 2c) What is the bandgap of the semiconductor? Solution: Reading from the graph: EC − EV = 1.25 eV 2d) What is the built-­‐in potential of the junction. Solution: From the plot: V j = Vbi − V A = 0.25 V Since: V A = +0.5 V Vbi = V j + V A = 0.75 V 3)
Vbi = 0.75 V A silicon diode is asymmetrically doped at N D = 1019 cm-­‐3 and N A = 1016 cm-­‐3. (Note that at N D = 1019 the semiconductor is on the edge of degeneracy, but we can assume that non-­‐degenerate carrier statistics are close enough for this problem.) Answer the following questions assuming room temperature. Assume that the minority electron and hole lifetimes are τ n = τ p = 10−6 s. The lengths of the N and P regions are L = 500 µm and L >> x p , xn . ECE-­‐305 3 Spring 2015 ECE 305 Spring 2015 HW8 Solutions (continued): 3a) Find the zero-­‐bias depletion region capacitance per cm2 of diode area. Solution: The junction capacitance per unit area is: Kε
C J = s 0 W (V A )
The depletion region width for a one-­‐sided junction is: 1/2
⎡ 2K ε
⎤
W (VA = 0 ) = ⎢ S 0 Vbi ⎥ ⎣ qN A
⎦
The built-­‐in potential is k T ⎛N N ⎞
Vbi = B ln ⎜ D 2 A ⎟ q
⎝ ni ⎠
Putting in numbers, we find: ⎛ 10191016 ⎞
k T ⎛N N ⎞
Vbi = B ln ⎜ D 2 A ⎟ = 0.026 ln ⎜
= 0.90 V q
⎝ 10 20 ⎟⎠
⎝ ni ⎠
1/2
⎡ 2K ε
⎤
W (VA = 0 ) = ⎢ S 0 Vbi ⎥
⎣ qN A
⎦
1/2
⎡ 2 × 11.8 × 8.854 × 10 −14
⎤
=⎢
× 0.90 ⎥
−19
16
⎣ 1.6 × 10 × 10
⎦
= 3.43 × 10 −5 cm (Note that we used ε 0 in F/cm and N A in cm-­‐3 so that the result would come out in cm not in meters.) K sε 0
11.8 × 8.845 × 10−14
CJ 0 =
=
= 3.05 × 10−8 F/cm 2 2.8 × 10−5
W (V A = 0 )
C J 0 = 3.05 × 10−8 F/cm 2 3b) Find the depletion capacitance at V A = −5 V (reverse biased). Solution: W (V A = 0 )
Kε
K sε 0
Vbi
CJ 0
CJ = s 0 =
×
= CJ 0
=
Vbi − V A
W (V A ) W (V A = 0 )
W (V A )
1− V A Vbi
CJ =
3.05 × 10−8
1+ 5 0.9
= 1.19 × 10−8 F/cm 2 < C J 0 C J (V A = −5 V ) = 1.19 × 10−8 F/cm 2 Reverse bias decreases the junction capacitance. ECE-­‐305 4 Spring 2015 ECE 305 Spring 2015 HW8 Solutions (continued): 3c) Find the depletion capacitance at V A = + 0.5 V (forward biased). Solution: CJ 0
3.73× 10−8
CJ =
=
= 5.6 × 10−8 F/cm 2 > C J 0 1− V A Vbi
1− 0.5 0.9
C J (V A = −5 V ) = 4.58 × 10−8 F/cm 2 Forward bias increases the junction capacitance. 4)
A silicon diode is asymmetrically doped at N D = 1019 cm-­‐3 and N A = 1016 cm-­‐3. (Note that at N D = 1019 the semiconductor is on the edge of degeneracy, but we can assume that non-­‐degenerate carrier statistics are close enough for this problem.) Assume that the minority electron and hole lifetimes are τ n = τ p = 10−6 s. The lengths of the N and P regions are L = 500 µm and L >> x p , xn . 4a) Estimate the applied forward bias at which the P-­‐region enters high-­‐level injection. Solution: High injection will occur first on the lightly doped side, the P-­‐side. The maximum excess electron concentration occurs at the beginning of the P-­‐side. From the law of the junction: n2
Δn ( 0 ) = i eqVA kBT − 1 NA
(
)
Low level injection means: Δn ( 0 ) << N A Let’s estimate the voltage needed to go into high-­‐level injection, this way n2 qV k T
Δn ( 0 ) = i e HL B − 1 = N A (*) NA
(
)
A more conservative estimate would be Δn ( 0 ) = 0.1N A , but (*) is close enough for us now. Using (*) as the criteria: 2
VHL
k T ⎛N ⎞
k T ⎛N ⎞
= B ln ⎜ A ⎟ = 2 B ln ⎜ A ⎟ q
q
⎝ ni ⎠
⎝ ni ⎠
⎛ 1016 ⎞
VHL = 2 × 0.026 × ln ⎜ 10 ⎟ = 0.781 V ⎝ 10 ⎠
VHL = 0.718 V ECE-­‐305 5 Spring 2015 ECE 305 Spring 2015 HW8 Solutions (continued): 4b) Compute the current density at the onset of high-­‐injection. Solution: The current density is given by: J = J 0 eqVA kBT − 1 (
)
From HW8, problem 1): J 0 = 9.1× 10−12 A/cm 2 (
J = J 0 eqVA
k BT
)
(
)
− 1 = 9.1× 10−12 e0.718/0.026 − 1 = 9 A/cm 2 J = 9 A/cm 2 5)
A silicon diode is asymmetrically doped at N D = 1019 cm-­‐3 and N A = 1016 cm-­‐3. (Note that at N D = 1019 the semiconductor is on the edge of degeneracy, but we can assume that non-­‐degenerate carrier statistics are close enough for this problem.) Answer the following questions assuming room temperature. Assume that the minority electron and hole lifetimes are τ n = τ p = 10−6 s. The lengths of the N and P regions are L = 500 µm and L >> x p , xn . What is the reverse breakdown voltage of this diode? Assume a critical field for breakdown of E cr = 3× 105 V/cm. Solution: ⎡ 2qN A (Vbi + VR ) ⎤
2 (Vbi + VR )
2V
E ( 0 ) = bi =
⎥ 1/2 = ⎢
W
K Sε 0
⎡ 2K S ε 0
⎤
⎣
⎦
⎢ qN (Vbi + VR ) ⎥
⎣
⎦
A
1/2
⎡ 2qN A (Vbi + VR ) ⎤
E cr = E ( 0 ) = ⎢
⎥ K Sε 0
⎣
⎦
2
K εE
VBR = S 0 cr − Vbi 2qN A
1/2
Vbi =
⎛ 10191016 ⎞
kBT ⎛ N D N A ⎞
ln ⎜
=
0.026
ln
⎜⎝ 10 20 ⎟⎠ = 0.90 q
⎝ ni2 ⎟⎠
11.8 × 8.854 × 10 −14 × ( 3 × 10 5 )
K S ε 0E cr2
VBR =
− 0.90 =
− 0.90 = 29.4 − 0.90 = 28.5 V 2qN A
2 × 1.6 × 10 −19 × 1016
2
VBR = 28.5 V This assumes a planar junction. If there is junction curvature at the edge of the diode, the breakdown voltage will be lower. ECE-­‐305 6 Spring 2015