chap3 - Diode Introduction Carrier Diffusion A diode is...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Diode Introduction • A diode is formed by interfacing an n -type semiconductor with a p -type semiconductor. •A pn junction is the interface between n and p regions. Diode symbol Carrier Diffusion Carriers diffuse from high concentration to low concentration. Drift Currents Diffusion currents lead to localized charge density ( ρ ) variations near the pn junction. Gauss’ law predicts an electric field due to the charge distribution: Assuming constant permittivity ( ε S ), Resulting electric field gives rise to a drift current. With no external circuit connections, drift and diffusion currents cancel. There is no actual current, since this would imply power dissipation, rather the electric field cancels the diffusion current ‘tendency.’ ∇⋅ E = ρ c ε s E ( x ) = 1 s ( x ) dx (Change in electric field is proportional to net charge per unit volume.) Space Charge Region Formation at the pn Junction Potential across the Junction Charge Density Electric Field (define E =0 in neutral region) Potential (define φ =0 at junction) φ j =− E ( x ) dx = V T ln N A N D n i 2 , V T = kT q Width of Depletion Region w d 0 = ( x n + x p ) = 2 s q 1 N A + 1 N D j Combining the previous expressions, we can form an expression for the width of the space-charge region, or depletion region. It is called the depletion region since the excess holes and electrons are depleted from the dopant atoms on either side of the junction. The 0 subscript indicates that no voltage is applied to diode terminals.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Width of Depletion Region (Example) Problem: Find built-in potential and depletion-region width for given diode Given data :On p -type side: N A =10 17 /cm 3 on n -type side: N D =10 20 /cm 3 Assumptions : Room-temperature operation with V T =0.025 V Analysis: m 113 . 0 1 1 2 0 µ φ ε = + = j D N A N q s d w () V 979 . 0 /cm 10 /cm 10 /cm 10 V)ln (0.025 ln 6 20 3 20 3 17 2 = = = i D A T j n N N V Diode Electric Field (Example) Problem: Find electric field and size of individual depletion layers on either side of pn junction for given diode Given data :On p -type side: N A =10 17 /cm 3 on n -type side: N D =10 20 /cm 3 from earlier example, Assumptions : Room-temperature operation Analysis: V 979 . 0 = j m 113 . 0 0 = d w m 4 - 10 13 . 1 1 0 1 1 0 × = + = + = + = + = A N D N d w n x D N A N p x A N D N n x p x n x d w kV/cm 173 m 113 . 0 ) V 979 . 0 ( 2 0 j 2 = = = d w MAX E m 113 . 0 1 0 + = D N A N d w p x Internal Diode Currents j n T = q n nE + qD n n x =0 j p T = q p pE qD p p x From Chapter 2: in a diode with no external connections, the total currents are equal to zero.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/11/2011 for the course ECE 322 taught by Professor Staff during the Spring '08 term at Boise State.

Page1 / 10

chap3 - Diode Introduction Carrier Diffusion A diode is...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online