EE105 Fall 2006 Microelectronic Devices and Circuits Prof Jan M Rabaey jan eecs Lecture 5 Diode Operation Overview Last lecture Capacitance pn Junction This lecture pn Junction cntd Diode operation and models 2 1 Administrativia Labs starting tomorrow 3 PN Junction Summary so far p type n type NA ND p0 N a p0 x J diff E0 x p0 n0 ni2 Na xn 0 J diff n2 p0 i Nd n0 N d E0 Transition Region The most important device is a junction between a p type region and an n type region When the junction is first formed due to the concentration gradient mobile charges transfer near junction Electrons leave n type region and holes leave p type region These mobile carriers become minority carriers in new region can t penetrate far due to recombination Due to charge transfer a voltage difference occurs between regions This creates a field at the junction that causes drift currents to oppose the diffusion current In thermal equilibrium drift current and diffusion must balance 4 Question width of depletion region 2 Depletion Approximation Let s assume that the transition region is completely depleted of free carriers only immobile dopants exist Then the charge density is given by qN a x po x 0 qN d 0 x xn 0 The solution for electric field is now easy 0 x E0 x x xp0 0 x qN dx a x x po s s E0 x qN d s x n 0 x 5 Plot of Fields In Depletion Region p type NA n type ND Depletion Region E0 x qN a s x x po E0 x E Field zero outside of depletion region Note the asymmetrical depletion widths Which region has higher doping Slope of E Field larger in n region Why Peak E Field at junction Why continuous qN d s xn 0 x 6 3 Continuity of E Field Across Junction Recall that E field diverges on charge For a sheet charge at the interface the E field could be discontinuous In our case the depletion region is only populated by a background density of fixed charges so the E Field is continuous What does this imply E 0n x 0 qN a s x po qN d s xno E 0p x 0 qN a x po qN d xno Total fixed charge in n region equals fixed charge in pregion Somewhat obvious result 7 Potential Across Junction From our earlier calculation we know that the potential in the n region is higher than p region The potential has to smoothly transition form high to low in crossing the junction Physically the potential difference is due to the charge transfer that occurs due to the concentration gradient Let s integrate the field to get the potential x x po x xp0 qN a s x x po dx x qN x 2 x p a x x po s 2 xp0 8 4 Potential Across Junction We arrive at potential on p side parabolic op x p qN a x x p0 2 2 s Do integral on n side n x n qN d x xn 0 2 2 s Potential must be continuous at interface field finite at interface n 0 n qN d 2 qN a 2 xn 0 p x p 0 p 0 2 s 2 s 9 Solve for Depletion Lengths We have two equations and two unknowns We are finally in a position to solve for the depletion depths n qN d 2 qN a 2 xn 0 p xp0 2 s 2 s qN a x po qN d xno xno 2 s bi qN d Na Na Nd x po 1 2 2 s bi qN a Nd Nd Na bi n p 0 10 5 Sanity Check Does the above equation make sense Let s say we dope one side very highly Then physically we expect the depletion region width for the heavily doped side to approach zero xn 0 lim N d x p 0 lim N d 2 s bi Nd 0 qN d N d N a 2 s bi qN a Nd Nd Na 2 s bi qN a Entire depletion width dropped across p region 11 Total Depletion Width The sum of the depletion widths is the space charge region X d 0 x p 0 xn 0 2 s bi q 1 1 Na Nd This region is essentially depleted of all mobile charge Due to high electric field carriers move across region at velocity saturated speed X d0 2 s bi 1 1 q 1015 E pn 1V V 10 4 1 cm 12 6 Have we invented a battery Can we harness the PN junction and turn it into a battery bi n p Vth ln ND N N N ln A Vth ln D 2 A ni ni ni Numerical example bi 26mV ln 10151015 NDNA 60 mV log 600mV 10 20 ni2 13 Contact Potential The contact between a PN junction creates a potential difference Likewise the contact between two dissimilar metals creates a potential difference proportional to the difference between the work functions When a metal semiconductor junction is formed a contact potential forms as well If we short a PN junction the sum of the voltages around the loop must be zero 0 bi mn pm bi n p pm mn bi pm mn 14 7 PN Junction Capacitor Under thermal equilibrium the PN junction does not draw any much current But notice that a PN junction stores charge in the space charge region transition region Since the device is storing charge it s acting like a capacitor Positive charge is stored in the n region and negative charge is in the p region qN a x po qN d xno 15 Reverse Biased PN Junction What happens if we reverse bias the PN junction bi VD VD VD 0 Since no current is flowing the entire reverse biased potential is dropped across the transition region To accommodate the extra potential the charge in these regions must increase If no current is flowing the only way for the charge to increase is to grow shrink the depletion regions 16 8 Voltage Dependence of Depletion Width Can redo the math but in the end we realize that the equations are the same except we replace the built in potential with the effective reverse bias xn VD 2 s bi VD N a qN d Na Nd V xn 0 1 D bi x p VD 2 s bi VD N d qN a Na Nd V x p 0 1 D bi 2 s bi VD 1 1 q Na Nd V X d VD X d 0 1 D X d VD x p VD xn VD bi 17 Charge Versus Bias As we increase the reverse bias the depletion region grows to accommodate more charge QJ VD qN a x p VD qN a 1 VD bi Charge is not a linear function of voltage This is a non linear capacitor We can define a small signal capacitance for small signals by breaking up the …
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