ECE-305: Fall 2013 PN Junctions V: IV characteristics Professor Mark Lundstrom Electrical and Computer Engineering Purdue University, West Lafayette, IN USA [email protected] 10/17/14 outline 1) Electrostatics under bias 2) Ideal diode equation 2 equilibrium e-band diagram E qVbi EC EF EF I=0 VA = 0 EV W x !xn xp 3 effect of an applied bias V =0 VA > 0 Where does the voltage drop? N-side? P-side? junction? 4 3 resistors in series. e-band diagram under forward bias E EC Vbi ! VA V = 0 EF VA > 0 EV W x !xn xp The applied voltage drops across the junction, but… 5 QFL’s split E EC VA > 0 Vbi ! VA V = 0 Fn Fn > Fp qVA Fp EV W x !xn 6 xp e-band diagram under reverse bias E Vbi EC V = 0 EF VA < 0 EV W x !xn xp 7 e-band diagram under reverse bias Vbi ! VA = Vbi + VR E Fp EC VA < 0 V =0 Fn Fn < Fp EV W x 8 !xn xp electrostatics under bias N P N D = 1016 N A = 1016 W !xn xp 0 depletion region 1/2 Vbi = k BT ! N D N A $ ln # q " ni2 &% ) 2K ! " N + N D % , W =+ S 0$ A (Vbi ( VA ). ' q N N # D A & * - E (0) = 2 (Vbi ! VA ) W xn = NA W NA + ND xp = ND W NA + ND 9 key points Vbi ! k BT " N D N A % ln $ (difference in Fermi levels / q) q # ni2 '& 1/2 # 2K ! & W = % S 0 (Vbi " VA ) ( $ qN A ' E (0) = 10 W ) Vbi " VA W) 1 NA 2 (Vbi ! VA ) E ( 0 ) " Vbi ! VA E ( 0 ) " N A W outline 1) Electrostatics under bias 2) Ideal diode equation 11 ECE 255 I ( mA ) ! VA 0.6 ! 0.7 V 12 VA + NP junction in equilibrium n0 P ! ni2 N A n0 N ! N D Vbi = k BT N A N D ln q ni2 qVbi EF EF n0 P ni2 N A ! = e" qVbi n0 N ND P 0 = e! qVbi k BT = P0 p0 P ! N A p0 N ! ni2 N D k BT 13 NP junction under bias P 0 = e! qVbi k BT nP >> n0 P "? P = e! q(Vbi !VA ) kBT q (Vbi ! VA ) P >> P 0 P = P0 e Fn qVA kBT nP = n0 P eqVA 14 kBT = ni2 qVA e NA Law of the Junction kBT Fp p0 P ! N A (This assumes nearequilibrium, J ≈ 0, conditions.) excess electrons on the p-side of junction n0 N ! N D nP >> n0 P q (Vbi ! VA ) Fn Fp p0 P ! N A p0 N ! ni2 N D x 0 15 What is Δn(x) on the p-side? Ans. Solve the MCDE. Δn(x) on p-side of junction !n ( x ) !n ( 0 ) = nP " n p0 !n ( 0 ) = ni2 qVA (e NA kBT " 1) !n ( x ) = !n ( 0 ) e" x/Ln !n ( x ) " 0 W p >> Ln 0 16 Wp x What is Δn(x) on the p-side? Ans. Solve the MDE. total recombination on p-side of junction !n ( 0 ) = !n ( x ) ni2 qVA (e NA kBT " 1) How many electrons recombine per second? One answer: Wp !n ( x ) = !n ( 0 ) e " x/Ln RTOT = A " !n ( x ) dx xp #n = A!n ( 0 ) Ln #n x 0 17 total recombination on p-side of junction !Dn !n ( x ) d"n dx 0 How many electrons recombine per second? Another answer: !n ( x ) = !n ( 0 ) e" x/Ln # d"n & D RTOT = A % !Dn = A n "n ( 0 ) ( dx 0 ' Ln $ x 0 18 recombination and current minority carriers injected across junction FP qVA Fn ID ! VA + 19 Every time a minority electron recombines on the p-side, one electron flows in the external current. current !n ( 0 ) = !n ( x ) ni2 qVA (e NA kBT " 1) !n ( x ) = !n ( 0 ) e W p >> Ln I D = qRTOT = qA I D = qA " x/Ln Dn ni2 qVA (e Ln N A !n ( x ) " 0 x 0 20 Dn !n ( 0 ) Ln kBT ! 1) diode current Jn = VA > 0 V =0 Jp = " D n2 % q!n ( 0 ) Dn = q $ n i ' ( eqVA /kBT ( 1) Ln # Ln N A & " D n2 % q!p ( 0 ) D p = q $ p i ' ( eqVA /kBT ( 1) Lp # Lp N D & J D (VA ) = J p (VA ) + J n (VA ) 21 diode current ideal diode equation ID Shockley diode equation VA J = J 0 ( eqVA /kBT ! 1) ! D n2 D n2 $ J0 = q # n i + p i & " Ln N A L p N D % 22 diode current I D ( pA ) I D ( mA ) VA VA 0.6 ! 0.7 V 23 ( ) I D = I 0 eqVA /kBT ! 1 diode current: qualitative picture I D ( pA ) J = J 0 ( eqVA /kBT ! 1) VA 24 NP junction in equilibrium P0 = e ! qVbi kBT J1 J0 qVbi EF EF p0 P ! N A J1 = J 0 25 NP junction in forward bias P = e! q(Vbi !VA ) kBT = eqVA J1 = J 0 eqVA /kBT J0 J = J1 ! J 0 = J 0 ( eqVA /kBT ! 1) 26 P0 kBT J1 J0 q (Vbi ! VA ) Fp p0 P ! N A NP junction in reverse bias (VR=-VA) P = e! q(Vbi +VR ) kBT = e! qVR P0 kBT P << P 0 J1 J0 q (Vbi + VR ) Fp J1 << J 0 Fn J = J1 ! J 0 " !J 0 27 diode current: qualitative picture I D ( pA ) J = J 0 ( eqVA /kBT ! 1) VA 28 narrow p-side We have assumed that WP >> Ln (long diode). What happens if WP << Ln (short diode)? 29 narrow p-side !n ( 0 ) = !n ( x ) ni2 qVA (e NA kBT " 1) # x & !n ( x ) = !n ( 0 ) % 1" $ WP (' J n = qDn Jn = q Dn !n ( 0 ) Wp W p << Ln x 0 30 Wp d!n ( x ) dx x=0 diode current ! D n2 $ J n = q # n i & ( eqVA /kBT ' 1) " Wp N A % VA > 0 V =0 ! D n2 $ J p = q # p i & ( eqVA /kBT ' 1) " Wn N D % 31 Ideal diode equation ID ( ) I D = I 0 eqVA /kBT ! 1 VA long ! D n 2 D p ni2 $ I 0 = qA # n i + & " Ln N A L p N D % 32 short ! D n 2 D p ni2 $ I 0 = qA # n i + " WP N A Wn N D &%
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