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Unformatted text preview: Handout 2 ECE 315, Cornell University 1 PN Junctions and Ideal Diode IV What you will learn: Electrostatics of PN junction diodes Ideal diode IV formulation Practical diode IV: reverse bias, reversebias with injection, weak forwardbias, resistive. Handout 3 Handout 2 ECE 315, Cornell University 2 3.1 The Diode Electrostatics and IV Diode electrostatics: charge, electric field and electrostatic potential. Current components across a PN junction under bias. The diode IV in linear and log scales The subthreshold slope in small forward bias Carrier injection into regions with large reverse bias Sedra and Smith, 3.7 3.9 Handout 2 ECE 315, Cornell University 3 Ideal Diodes from Circuit Viewpoints Diode circuit representation Ideal diode IV Open circuit reverse bias Short circuit forward bias Ideal diodes can be used in restricting signal directions and waveform clamping . Fig. 3.1 i v + cathode anode v < 0 i = 0 i > 0 v = 0 Handout 2 ECE 315, Cornell University 4 Doping Inside a PN Junction Diode ptype, N A =10 17 cm3 , N D =0 ntype, N D =10 17 cm3 , N A =0 N A (Ptype doping) N D (Ntype doping) p n N D + = n p = n i 2 /n p n + N D + =0 N A = p n = n i 2 /p p N A n=0 Holes: majority carrier Electrons: minority carrier Holes: minority carrier Electrons: majority carrier Handout 2 ECE 315, Cornell University 5 Formation of the Depletion Region ptype, N A =10 17 cm3 , N D =0 ntype, N D =10 17 cm3 , N A =0 Metallurgical Junction N A (Ptype doping) N D (Ntype doping) p n + + + + + Close to the interface, or the metallurgical junction, in equilibrium the holes will diffuse to the Ntype side to leave some N A exposed, and the electrons will diffuse to the Ptype side to leave some N D + exposed. How large will be that charge region, or socalled depletion region ( depletion of majority carriers )? This will be set up by the exact balance between drift and diffusion.!! Handout 2 ECE 315, Cornell University 6 DriftDiffusion at the Junction Depletion region approximation : We will assume that there is net charge density ONLY in the transition region (the depletion region), where the only dominant term in linear scale is N A to the left or N D + to the right. The question basically is: what is the dipole charge necessary to support the potential difference in the ntype and ptype materials. Yes, it is the solution from the Poisson equation!! ptype, N A =10 17 cm3 , N D =0 ntype, N D =10 17 cm3 , N A =0 N A = p n = n i 2 /p pN An+N D + =0 pn junction in equil. N D + = n p = n i 2 /n pN An+N D + =0 n p J ndiff J pdiff J pdrift J ndrift Handout 2 ECE 315, Cornell University 7 Builtin Potential in PN Junction Diode Independent of the transition region, the Pside and Nside have a potential difference in equilibrium with respect to the same reference potential to the intrinsic silicon!!!...
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 Spring '07
 SPENCER
 Microelectronics

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