Pn x 2 aewhere lp be d p x e lxp lp b dp x lp a nd

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Unformatted text preview: VA kT e 1 ND ni2 qVA kT e 1 ND 0 and A pn ( x' 0) N n x ' / LP kT e 1e for x' 0 N D ECE 3080 - Dr. Alan DoolittleE 332 Spring 2013 E2 qV A ni x ' / LP ECE 3080 - Dr. Alan Doolittle kT 2 i qV A p ( x' ) e 1e Quantitative p-n Diode Solution 0 Hole Diffusion Current 0 n-side E at Excess Spatial Region Depletion Gradient E x’=0 0 0 of Excess Minority Carriers n-type E0 x’=0 pn ( x ' ) ni2 qV A kT e2 1e qV ND ni kT pn ( x ' ) Jp ND e A x ' / LP 1e for x' 0 x ' / LP for x' 0 d pn qD p dx p d pn Jp qD dx Jp A D p ni2 DqVn 2kT qV q J q e p i e 1 kT x1/ Le e 'P p Lp N D Lp N D A for for x0 0 x' ' x ' / LP EE 332 Spring 2013 p-type Depletio Excess ElectronQuantitative0CurrentSolution-side E Diffusion Current at P E p-n Diode at P-side Excess Electron Diffusion p-type E Depletion Region 0 E0 x’’=0 Similarly for electrons on the p-side… x’= x’’=0 Similarly for electrons''/ on the p-side… ni2 qV kT n p ( x' ' ) e 1e x L for x' ' 0 2 qVN ni x '' / L kTA A n p ( x' ' ) Jn NA e J n d qD n np qD n A n for x' ' 0 d np dx dx 2 Jn 1e n qV A qV Dn ni2 DnAni q J n q e kT 1 e xkT/ Ln e '' Ln N A Ln N A '' for 1 e x x/'L'n 0 for x' ' 0 EE 332 Spring 2013 Total Diode Current Total Diode Current p-n Diode Solution Quantitative p-type E 0 Depletion Region E 0 x’’=0 n-type E 0 x’=0 Total on current is constant throughout the device. Thus, we can characterize the current flow components as… J Jp Jn Jn Jp J Jn Jn J Jp e x '' / Ln Jp e x '/ L p Recombination Recombination EE 332 Spring 2013 I-V Characteristics of P-N Diode I-V Characteristics of P-N Diode EE 332 Spring 2013 hus, evaluating the current components at the depletion Thus, evaluating the current egion edges, we have… components at the depletion Ideal Diode Equation region edges, we have… Evaluating the current components at the depletion region edges we have: J J n (x' ' 0) J p (x' 0) J n (x' ' 0) J p (x' ' 0) J n (x' 0) J p (x' 0) J J n (x' ' 0) J p (x' 0) J n (x' ' 0) J p (x' ' 0) 2 n i2 Dn JJ q Dn ni qLN Ln N A n A D p n2i2 D p ni Lp N D Lp N D e e qV A qV A kT kT J n (x' 0) J p (x' 0) for all 1 for all x x 1 or or qV A qV I I II o e oe A kT kT 1 1 where o where II o 2 2 Dnini2 DD p2ni p ni qA Dn n qA Lp N LLNN A L p N D D nn A IIoo is the " reverse saturation current" " is the reverse saturation current EE 332 Spring 2013 R=1000 ohms R=1000 ohms p-n Diode Solution Quantitative Examples: Diode in a circuit Example: Diode in a Circuit I V1=IR V VA A R=1000 ohms 9V I(1000) VA V=9V, 5V, I(1000) -9V V 2V, VA I 1eV A 12 e 12 e 0.0259V or A 0.0259V 1 I 1 9V I(1000) VA I IqV Ae o Io e qV A kT 1 kT 1 VA where I o where I o Solutions qV A Solutions V I I e VA 1 1 pA 1 pA In forward I 1e 12 e 1 or VA V 9V VA Solutions mA In bias (VA>0) e 12 e 0.0259V 1 (1000) VA 0.59V I 8.4 forward VA or the VA is VA I V 0.0259V ias In forward 1 (1000) VA V 0.0259V 9V 5V 0.59V 8.4 4.4 mA bias~(V(VA>0) mA constant 0.58V VA b 9V 1e 12 e 1 (1000) VA 9V 0.59V 8.4 mA theforAlarge VA>0) is the VA is 1e 9 e 0.0259V 1 VA ~ constant 5V0.55V4.4 mA 4.4 mA 5V 2V 0.58V 0.58V 1.5 mA ~constant V VA differences 0.0259V for 9V 1e 9 e 1 VA 0.0259V forlarge large 2V 1.5 e 1 VA in current differences -9V 0.55V 0.55V mAmA -9.0V 1.5 -1 pA 2V differences in current -9V -9.0V -1 pA In reverse bias (VA<0) the current current in In reverse bias (VA<0) the -9V -9.0V -1CE 3080- -Dr. Alancurrent pA is ~constant (=saturation current) current) E ECE 3080 Dr. Alan Doolittle is ~constant (=saturation Georgia Tech Doolittle In reverse bias (VA<0) the current EE 332 Spring 2013 is ~constant (=saturation current) ECE 3080 - Dr. Alan Doolittle VA A A 0.0259V o kT where I o 1 pA I...
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This document was uploaded on 02/18/2014.

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