HOT

4 点赞
1 收藏
0下载

1. Lecture #12 OUTLINE – pn Junctions • narrow-base diode • charge-control model Reading: Finish Chapter 6 Spring 2003 EE130 Lecture 12, Slide 1 Carrier Concentration Profiles Spring 2003 EE130 Lecture 12, Slide 2 1

2. Narrow or Short-Base Diode • We have the following boundary conditions: ∆pn ( xn ) = pno ( e qVA / kT − 1) ∆pn ( x' = xc ' ) → 0 • With the following coordinate system: x’’ 0 0 x’ NEW: 0 x’c • Then, the solution is of the form: x′ / L p − x′ / L p ∆p ( x ) = A1e + A2 e Spring 2003 EE130 Lecture 12, Slide 3 Applying the boundary conditions, we have: ∆pn (0) = A1 + A2 xc' / L p − xc' / L p 0 = A1e + A2 e • So, we have:  e (xc − x ' )/ LP − e − (xc − x ' )/ LP  ' ' ∆p n ( x ' ) = p n 0 ( e qV A / kT − 1) , 0 < x' < xc'  e xc' / LP − e − xc' / LP    • Note: sinh (ξ ) = eξ − e − ξ 2 Spring 2003 EE130 Lecture 12, Slide 4 2

3.• So: ∆pn ( x ' ) = pn 0 ( e qV A / kT − 1) [  sinh (xc' − x ') / LP ], 0 < x' < x ' '  sinh xc / LP [ ]   c • If we write: I = I 0 ' ( e qVA kT − 1) • Then I 0' = qA DLPP ( ni 2 cosh xc' / LP ) ( N D sinh xc' / LP ) where cosh (ξ ) = eξ + e − ξ 2 Spring 2003 EE130 Lecture 12, Slide 5 Note: sinh (ξ ) → ξ as ξ → 0 and cosh(ξ ) → 1 + ξ 2 as ξ → 0 • If xc’ << LP: Spring 2003 EE130 Lecture 12, Slide 6 3

4. Narrow Base Diode: I-V Equation Let WN ≡ width of n - type region WP ≡ width of p - type region and WN′ ≡ WN − xn << LP WP′ ≡ WP − x p << LN 2  DP DN  qVA / kT Then, J = qni  + e −1 ( ) WN′ N D WP′ N A  2  DP DN  qVA / kT I = qAni  +  e ( − 1 = I 0 e qVA / kT − 1) ( ) WN′ N D WP′ N A  Spring 2003 EE130 Lecture 12, Slide 7 Current Flow in a One-Sided pn Junction • Note that the diode current is dominated by the term associated with the more lightly doped side: 2  DP  qAni   long n − side  LP N D  p+n diode: I 0 ≅ I P ( xn ) = 2 DP  qAni   short n − side  WN′ N D  2 D  qAni  N  long p − side pn+ diode: I 0 ≅ I N (− x p ) =  LN N A  2  DN  qAni   short p − side  WP′ N A  i.e. current flowing across junction is dominated by carriers injected from the more heavily doped side Spring 2003 EE130 Lecture 12, Slide 8 4

5.Excess Carrier Profiles: Limiting Cases Long base (xc’ → ∞):  e (xc − x ' )/ LP − e − (xc − x ' )/ LP  ' ' ∆p n ( x ' ) = p n 0 ( e qV A / kT − 1)   e xc' / LP − e − xc' / LP     e x c / L P e − x ' / L p − e − xc / L P e x ' / L p  ' ' = pn 0 (e qV A / kT − 1)   ' ' e xc / LP − e − xc / LP    − x '/ L p ≅ pn 0 (e qV A / kT − 1)e Spring 2003 EE130 Lecture 12, Slide 9 2. Short base (xc’ → 0): ∆p n ( x ' ) = p n 0 ( e qVA / kT − 1) [(  sinh xc' − x' / LP  ) ]   sinh xc ' / LP [ ]   = pn 0 (e qVA / kT − 1) ( ' ) xc − x' / LP   x'   = pn 0 (e qVA / kT − 1)1 − '  '  xc / LP   xc  ∆pn is a linear function of x Æ Jp is constant (no recombination) Spring 2003 EE130 Lecture 12, Slide 10 5

6. Minority-Carrier Charge Storage • When VA>0, excess minority carriers are stored in the quasi-neutral regions of a pn junction: −∞ ∞ QN = − qA∫ ∆n p ( x) dx QP = qA∫ ∆pn ( x)dx −xp xn = − qA∆n p ( − x p ) LN = qA∆pn ( xn ) LP Spring 2003 EE130 Lecture 12, Slide 11 Derivation of Charge Control Model • Consider a forward-biased pn junction. The total excess hole charge in the∞ n quasi-neutral region is: QP = qA ∫ ∆pn ( x, t )dx xn • The minority carrier diffusion equation is (without GL): ∂∆pn ∂ 2 ∆pn ∆pn = DP − ∂t ∂x 2 τp • Since the electric field is very small, ∂∆p n J P = − qDP ∂x • Therefore: ∂ ( q ∆p n ) ∂J =− P − q ∆p n ∂t ∂x τp Spring 2003 EE130 Lecture 12, Slide 12 6

7. (Long Base Diode) • Integrating over the n quasi-neutral region: ∂  1   ∞ J P (∞) ∞  ∫ n  qA ∆p dx = − A ∫ dJ P − qA ∫ ∆pn dx  ∂t  x n  J p ( xn ) τ p  xn  • Furthermore, in a p+n junction: J P (∞) −A ∫ dJ J p ( xn ) P = − AJ P (∞) + AJ P ( xn ) = AJ P ( xn ) ≅ iDIFF dQP Q • So: = iDIFF − P dt τp Spring 2003 EE130 Lecture 12, Slide 13 Charge Control Model We can calculate pn-junction current in 2 ways: 1. From slopes of ∆np(-xp) and ∆pn(xn) 2. From steady-state charges QN, QP stored in each excess-minority-charge distribution: dQP Q = AJ P ( xn ) − P = 0 dt τp QP ⇒ AJ P ( xn ) = I P ( xn ) = τp − QN Similarly, I N (− x p ) = τn Spring 2003 EE130 Lecture 12, Slide 14 7

8.Charge Control Model for Narrow Base • For a narrow-base diode, replace τp and/or τn by the minority-carrier transit time τtr – time required for minority carrier to travel across the quasi-neutral region – For holes on narrow n-side: WN 1 QP = qA∫ ∆pn ( x )dx = qA ∆pn ( xn )WN′ xn 2 d∆pn ∆p ( x ) I P = AJ P = − qADP = qADP n n dx WN′ Q ⇒ τ tr = P = N (W ′ ) 2 IP 2 DP τ = (WP′ )2 – Similarly, for electrons on narrow p-side: tr 2 DN Spring 2003 EE130 Lecture 12, Slide 15 Summary: Narrow (Short) Base Diode • If the width of the quasi-neutral region (e.g. xc’) is not much larger than the minority-carrier diffusion length (e.g. LP), then the solution to the minority-carrier diffusion equation is  x′ − x  sinh  c  L  P  ∆pn ( x' ) = ∆pn ( xn )  x′  sinh  c   LP  • If xc’ < 0.1LP:  x′  ∆pn ( x' ) ≅ ∆pn ( xn )1 −   xc′  Spring 2003 EE130 Lecture 12, Slide 16 8

9.• If LP >> xc’, negligible recombination occurs ⇒ hole current is constant throughout n-type region 2 1 J p = − qDP d∆pn = qDP n i qVA / kT e ( − 1   ) dx ND  xc′  • Compare this to the hole current contribution in the long-base diode: 2  1  n i qV / kT J p = qDP ND (e A ) − 1    LP  Spring 2003 EE130 Lecture 12, Slide 17 9

4 点赞
1 收藏
0下载