10+PN+junction+I-V - Last Lecture: Electrostatics Summary...

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Last Lecture: Electrostatics Summary N D -N A P N x ρ E C qV bi E i E F E V PN P A N D x N x N = ε ρ = dx d E E x ρ x = 2 ln i A D bi n N N q kT V + = D A D A bi bi N N N N q V V W 2 ) ( -x p x n -x p x n E = dx dV V x V bi -x p x n
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EECS 320 P-N Junction I-V Characteristics
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Current Balanced in Equilibrium In Equilibrium, the Total current balances due to the sum of the individual components
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PN Junction in Forward Bias (V A >0) Current flow is proportional to e (qV A /kT) due to the exponential increase of carriers in the majority carrier bands
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PN Junction in Reverse Bias (V A <0) Reverse current caused by minority carriers being swept away by E , and independent of the size of V A
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Current – Voltage Relationships Exponential increase I = I + -I - -I 0 I V A kT qV A e I / + 0 I I = Independent of V A 0 ) 0 ( = = A V I ) 1 ( / 0 = kT qV A e I I (very small) Constant saturation current Diode = one-way valve
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I-V Relation: Quantitative Determine current flow assuming steady state, low-level injection depletion region AJ I = p n J J J + = E 0 0 0 -x p -x n quasi-neutral p-region dx dn qD nq J n n n + = E µ dx dp qD pq J p p p = E quasi-neutral n-region Need to determine carrier densities, Solve continuity equations Three regions of interest
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Quantitative Studies
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This note was uploaded on 08/09/2011 for the course EE 320 taught by Professor Zhongzhaohui during the Summer '11 term at Shanghai Jiao Tong University.

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10+PN+junction+I-V - Last Lecture: Electrostatics Summary...

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