EECS 105 Spring 2005, Lecture 27 Lecture 27: PN Junctions Prof. Niknejad Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Diffusion Diffusion occurs when there exists a concentration gradient In the figure below, imagine that we fill the left chamber with a gas at temperate T If we suddenly remove the divider, what happens? The gas will fill the entire volume of the new chamber. How does this occur? Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Diffusion (cont) The net motion of gas molecules to the right chamber was due to the concentration gradient If each particle moves on average left or right then eventually half will be in the right chamber If the molecules were charged (or electrons), then there would be a net current flow The diffusion current flows from high concentration to low concentration: Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Diffusion Equations Assume that the mean free path is λ Find flux of carriers crossing x=0 plane n ( 0) n ( ) n( ) 1 dn dn F vth n(0) n(0) 2 dx dx 1 dn n( )vth 2 F vth dx 1 n( )vth 2 Department of EECS 1 F vth n( ) n( ) 2 0 J qF qvth dn dx University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Einstein Relation The thermal velocity is given by kT 1 2 mn*vth2 12 kT vth c Mean Free Time kT q c vth v kT * mn q mn* 2 th c c kT dn dn J qvth q n dx q dx kT Dn n q Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Total Current and Boundary Conditions When both drift and diffusion are present, the total current is given by the sum: J J drift J diff dn q n nE qDn dx In resistors, the carrier is approximately uniform and the second term is nearly zero For currents flowing uniformly through an interface (no charge accumulation), the field is discontinous J1 (1 ) J 2 ( 2 ) Department of EECS J1 J 2 1E1 2 E2 E1 2 E2 1 University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Carrier Concentration and Potential In thermal equilibrium, there are no external fields and we thus expect the electron and hole current densities to be zero: dno J n 0 qn0 n E0 qDn dx n dno q d no E0 no 0 dx kT dx Dn kT dno dn0 d0 Vth n0 q n0 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Carrier Concentration and Potential (2) We have an equation relating the potential to the carrier concentration kT dno dn d0 Vth 0 n0 q n0 If we integrate the above equation we have n0 ( x) 0 ( x) 0 ( x0 ) Vth ln n0 ( x0 ) We define the potential reference to be intrinsic Si: 0 ( x0 ) 0 Department of EECS n0 ( x0 ) ni University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Carrier Concentration Versus Potential The carrier concentration is thus a function of potential n0 ( x) ni e0 ( x ) / Vth Check that for zero potential, we have intrinsic carrier concentration (reference). If we do a similar calculation for holes, we arrive at a similar equation p0 ( x) ni e 0 ( x ) / Vth Note that the law of mass action is upheld n0 ( x) p0 ( x) ni2 e 0 ( x ) / Vth e0 ( x ) / Vth ni2 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad The Doping Changes Potential Due to the log nature of the potential, the potential changes linearly for exponential increase in doping: n0 ( x) n0 ( x) n0 ( x) 0 ( x) Vth ln 26mV ln 26mV ln 10 log 10 ni ( x0 ) ni ( x0 ) 10 n0 ( x) 0 ( x) 60mV log 10 10 p0 ( x ) 0 ( x) 60mV log 10 10 Quick calculation aid: For a p-type concentration of 1016 cm-3, the potential is -360 mV N-type materials have a positive potential with respect to intrinsic Si Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad PN Junctions: Overview 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 Department of EECS p-type NA −+−+−+−+−+− − −−−−−− V ++++ −+ +−+−+−+−+− + ND n-type University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad PN Junction Currents Consider the PN junction in thermal equilibrium Again, the currents have to be zero, so we have dno J n 0 qn0 n E0 qDn dx dno qn0 n E0 qDn dx E0 dno dx kT 1 dn0 n0 n q n0 dx Dn dpo Dp kT 1 dp0 dx E0 n0 p q p0 dx Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad PN Junction Fields p-type n-type NA ND p0 N a p0 ( x) J diff E0 x p0 ni2 n0 Na xn 0 ni2 p0 Nd n0 N d J diff E0 – –++ Transition Region Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Total Charge in Transition Region To solve for the electric fields, we need to write down the charge density in the transition region: 0 ( x) q( p0 n0 N d N a ) In the p-side of the junction, there are very few electrons and only acceptors: 0 ( x) q( p0 N a ) x p0 x 0 Since the hole concentration is decreasing on the pside, the net charge is negative: N a p0 Department of EECS 0 ( x) 0 University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Charge on N-Side Analogous to the p-side, the charge on the n-side is given by: 0 ( x) q(n0 N d ) 0 x xn 0 The net charge here is positive since: 0 ( x) 0 N d n0 n0 N d ni2 n0 Na J diff E0 – –++ Department of EECS Transition Region University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad “Exact” Solution for Fields Given the above approximations, we now have an expression for the charge density q(ni e 0 ( x ) /Vth N a ) 0 ( x) 0 ( x ) / Vth q ( N n e ) d i x po x 0 0 x xn 0 We also have the following result from electrostatics dE0 d 2 0 ( x) 2 dx dx s Notice that the potential appears on both sides of the equation… difficult problem to solve A much simpler way to solve the problem… Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad 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 0 ( x) qN d 0 x xn 0 The solution for electric field is now easy dE0 0 ( x ) Field zero outside dx s transition region x 0 ( x' ) E0 ( x ) dx' E0 ( x p 0 ) xp0 Department of EECS s University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Depletion Approximation (2) Since charge density is a constant E0 ( x ) x xp0 0 ( x' ) qN a dx' ( x x po ) s s If we start from the n-side we get the following result E0 ( xn 0 ) xn 0 x Field zero outside transition region Department of EECS 0 ( x' ) qN d dx' E0 ( x) ( xn 0 x) E0 ( x) s s E0 ( x) qN d s ( xn 0 x ) University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Plot of Fields In Depletion Region p-type NA ––––– ––––– ––––– ––––– +++++ n-type +++++ +++++ ND +++++ Depletion Region E0 ( x) qN a s ( x x po ) E0 ( x) qN d s ( xn 0 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? Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad 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 EField is continuous What does this imply? E 0 ( x 0) n 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 p-region! Somewhat obvious result. Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad 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 qN a s xp0 ( x' x po )dx' x qN a x' ( x) p x' x po s 2 xp0 2 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Potential Across Junction We arrive at potential on p-side (parabolic) qN a ( x) p ( x x p0 )2 2 s p o 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) qN d 2 qN a 2 n (0) n xn 0 p x p 0 p (0) 2 s 2 s Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Solve for Depletion Lengths We have two equations and two unknowns. We are finally in a position to solve for the depletion depths qN d 2 qN a 2 n xn 0 p x p0 2 s 2 s (1) qN a x po qN d xno (2) 2 sbi N a xno qN d N a N d 2 sbi N d x po qN a N d N a bi n p 0 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad 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 sbi N d 0 qN d N d N a 2 sbi N d 2 sbi qN a N d N a qN a Entire depletion width dropped across p-region Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Total Depletion Width The sum of the depletion widths is the “space charge region” Xd0 2 sbi 1 1 x p 0 xn 0 q Na Nd This region is essentially depleted of all mobile charge Due to high electric field, carriers move across region at velocity saturated speed Xd0 2 sbi 1 15 1μ q 10 Department of EECS E pn 1V V 104 1μ cm University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Have we invented a battery? Can we harness the PN junction and turn it into a battery? ND NA ND N A bi n p Vth ln ln Vth ln 2 n n n i i i ? Numerical example: ND N A 10151015 bi 26mV ln 60mV log 600mV 2 20 ni 10 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad 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: + bi − Department of EECS mn pm 0 bi pm mn n p bi ( pm mn ) University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad 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 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad 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 Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad 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 builtin potential with the effective reverse bias: 2 s (bi VD ) N a xn (VD ) qN d Na Nd V xn 0 1 D bi 2 s (bi VD ) N d x p (VD ) qN a Na Nd VD x p 0 1 bi 2 s (bi VD ) 1 1 X d (VD ) x p (VD ) xn (VD ) q Na Nd VD X d (VD ) X d 0 1 bi Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad 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 charge into two terms QJ (VD vD ) QJ (VD ) q(vD ) Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Derivation of Small Signal Capacitance From last lecture we found dQD QJ (VD vD ) QJ (VD ) dV C j C j (VD ) dQ j dV Cj V VD d dV qN a x p 0 2bi 1 Notice that C j0 qN a x p 0 2bi Department of EECS qN a 2bi VD bi vD VD V qN a x p 0 1 bi V VR C j0 1 2 sbi N d qN a N a N d VD bi q s N a N d 2bi N a N d University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Physical Interpretation of Depletion Cap q s N a N d C j0 2bi N a N d Notice that the expression on the right-hand-side is just the depletion width in thermal equilibrium C j0 s 1 1 2 sbi N a N d q 1 s Xd0 This looks like a parallel plate capacitor! C j (VD ) Department of EECS s X d (VD ) University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad A Variable Capacitor (Varactor) Capacitance varies versus bias: Cj C j0 Application: Radio Tuner Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Part II: Currents in PN Junctions Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Diode under Thermal Equilibrium Minority Carrier Close to Junction Thermal + p-type Generation − - + + - + + − - + + J p ,drift - - ++ + J n ,drift - - ++ ND - + + n-type J n ,diff J p , diff − NA E0 Recombination + qbi − Carrier with energy below barrier height Diffusion small since few carriers have enough energy to penetrate barrier Drift current is small since minority carriers are few and far between: Only minority carriers generated within a diffusion length can contribute current Important Point: Minority drift current independent of barrier! Diffusion current strong (exponential) function of barrier Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Reverse Bias Reverse Bias causes an increases barrier to diffusion Diffusion current is reduced exponentially p-type - ND - - + + + + + + + n-type NA + − q(bi VR ) Drift current does not change Net result: Small reverse current Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Forward Bias Forward bias causes an exponential increase in the number of carriers with sufficient energy to penetrate barrier Diffusion current increases exponentially p-type - ND - - + + + + + + + n-type NA + − q(bi VR ) Drift current does not change Net result: Large forward current Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Diode I-V Curve Id Is d qV kT I d I S e 1 I d (Vd ) I S 1 qVd kT Diode IV relation is an exponential function This exponential is due to the Boltzmann distribution of carriers versus energy For reverse bias the current saturations to the drift current due to minority carriers Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Minority Carriers at Junction Edges Minority carrier concentration at boundaries of depletion region increase as barrier lowers … the function is (minority) hole conc. on n-side of barrier p n ( x xn ) p p ( x x p ) (majority) hole conc. on p-side of barrier e ( Barrier Energy ) / kT p n ( x xn ) e q(B VD ) / kT NA Department of EECS (Boltzmann’s Law) University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad “Law of the Junction” Minority carrier concentrations at the edges of the depletion region are given by: pn ( x xn ) N Ae q ( B VD ) / kT n p ( x x p ) N De q (B VD ) / kT Note 1: NA and ND are the majority carrier concentrations on the other side of the junction Note 2: we can reduce these equations further by substituting VD = 0 V (thermal equilibrium) Note 3: assumption that pn << ND and np << NA Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Minority Carrier Concentration pn 0 e p side qVA kT n side n p 0e qVA kT x qVkTA Lp pn ( x) pn 0 pn 0 e 1 e pn 0 np0 -Wp -xp xn Minority Carrier Diffusion Length Wn The minority carrier concentration in the bulk region for forward bias is a decaying exponential due to recombination Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Steady-State Concentrations Assume that none of the diffusing holes and electrons recombine get straight lines … pn 0 e p side qVA kT n side n p 0e qVA kT pn 0 np0 -Wp -xp xn Wn This also happens if the minority carrier Ln , p Wn , p diffusion lengths are much larger than Wn,p Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Diode Current Densities pn 0 e p side n p 0e qVA kT dn p n side qVA kT dx ( x) n p 0e qVA kT np0 x p (Wp ) pn 0 np0 np0 -Wp -xp xn Wn ni2 Na qVA D diff n kT J n qDn q n p 0 e 1 dx x x Wp p qVA D dp p diff n kT J p qDp q pn 0 1 e dx x xn Wn dn p J Department of EECS diff Dp Dn qn N dWn N aW p 2 i qVA kT e 1 University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Diode Small Signal Model The I-V relation of a diode can be linearized qVd qvd d vd ) q (VkT I D iD I S e 1 I S e kT e kT 2 3 x x ex 1 x 2! 3! q (Vd vd ) I D iD I D 1 kT qvd iD g d vd kT Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Diode Capacitance We have already seen that a reverse biased diode acts like a capacitor since the depletion region grows and shrinks in response to the applied field. the capacitance in forward bias is given by Cj A S X dep 1.4C j 0 But another charge storage mechanism comes into play in forward bias Minority carriers injected into p and n regions “stay” in each region for a while On average additional charge is stored in diode Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Charge Storage pn 0 e p side n p 0e q (Vd vd ) kT n side q (Vd vd ) kT pn 0 np0 -Wp -xp xn Wn Increasing forward bias increases minority charge density By charge neutrality, the source voltage must supply equal and opposite charge 1 qI d A detailed analysis yields: Cd 2 kT Time to cross junction (or minority carrier lifetime) Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Forward Bias Equivalent Circuit Equivalent circuit has three non-linear elements: forward conductance, junction cap, and diffusion cap. Diff cap and conductance proportional to DC current flowing through diode. Junction cap proportional to junction voltage. Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Fabrication of IC Diodes p+ cathode annode p n+ n-well p-type p-type Start with p-type substrate Create n-well to house diode p and n+ diffusion regions are the cathode and annode N-well must be reverse biased from substrate Parasitic resistance due to well resistance Department of EECS University of California, Berkeley EECS 105 Fall 2003, Lecture 27 Prof. A. Niknejad Diode Circuits Rectifier (AC to DC conversion) Average value circuit Peak detector (AM demodulator) DC restorer Voltage doubler / quadrupler /… Department of EECS University of California, Berkeley