PN Junction and MOS structure

PN Junction and MOS structure

Basic electrostatic equations We will use simple one-dimensional electrostatic equations to develop insight and basic understanding of how semiconductor devices operate Gauss's Law Potential Equation Poisson's Equation It puts together Gauss's Law and the potential equation

Gauss's Law E = electric field [V/m] de dx charge density [C/m 3 ] permittivity [F/m] charge per area in the interval from x a to x [C/m 2 ] x x a x d E x x E x a E x a x a xdx S x The possibility of a change in permittivity due to a material interface has been accounted for by keeping the permittivity together with the field

Potential Equation reference point x x x R E xdx x R E x d x dx

Poisson's Equation It directly links the potential with the charge distribution (there is no need to go through the field) de dx E x d dx d 2 x dx 2 de x dx

Boundary Conditions Electronic devices are made of layers of different materials We need conditions for and E at the boundary between two materials

Potential at a boundary An abrupt jump of ( along x ) would lead to an infinite electric field at the boundary E x d dx Infinite electric fields are not possible (they would tear the material apart) Therefore (x) must be continuous: 0 0 Where the boundary is located at x = 0

Electric Field at a boundary The electric field usually jumps at a boundary d Ex 2 E x 1 E x By letting 0 : x dx0 x dx 2 E x0 1 E x0 0 2 E x0 1 E x0 S x dx S There can be a sheet of charge at the boundary A sheet of charge is an Infinite amount of charge all distributed on the boundary surface (that is a Dirac function)

Boundary between materials

Boundary conditions x0 x0 continuity of potential at a boundary x dx0 E x0 1 2 E x0 electric field jump for charge free boundary x dx S E x0 1 2 E x0 S 2 electric field jump for charged boundary

Oxide-Silicon interface Example of very common interface in microelectronic devices Permittivity of vacuum si 11.7 0 ox 3.9 0 0 8.8510 12 F m E si 0 ox E ox 0 E ox0 si 3.0

In many IC processes there are two or more levels of metal separated by silicon oxide Metal Metal Capacitor

Metal Metal Capacitor

Metal Metal Capacitor Since there is no charge present in the oxide the electric field in the oxide is constant Since the electric field in the oxide is constant and the voltage drop across the gap is V de dx = 0 t d t d 0E ox dx 0 V 0t d it follows that: E ox V t d E ox metal ox E metal S ox V t d S ox S d S dv ox t d No electric field inside metals Capacitance per unit area [F/m 2 ]

Example Find potential, electric field and charge distribution for a metal-metal capacitor with t d =1 μm and an applied voltage of 1V

M-O-S Capacitor

M-O-S charge distribution 0 q N A Q B charge per area in the substrate (C/cm 2 ) B 0 X d G G ox V ox t ox V X d oxv t ox 0 ox V t ox qn A

M-O-S charge density profile

M-O-S Electric Field The electric field is confined in the region t ox < x < X d NOTE: the total excess charge in the region t ox < x < X d is zero M O S In the metal the electric field is 0 In the oxide ( t ox < x < 0) the charge density is zero ((x)=0), therefore the electric field is constant fort ox x0: de x dx x 0 0 E xe ox ox

M-O-S Electric Field NOTE: The electric field jumps only at the interface between two different materials M O S Outside the charged region of the silicon (x > X d ) the electric field is 0 In the charged region of the silicon (0 < x < X d ) the charge density is constant ( 0 ) therefore the electric field is a linear function of x de x 0 X d 0 dx s E x d s E 0 0 dx s 0 E 0 0 X d s

M-O-S Electric Field M O S The boundary condition at the oxide/silicon interface is: ox E ox s E 0 E ox s ox E 0 3 E ox 0 X d ox

Potential plot through M-O-S Potential is continuos at boundaries: x = t ox metal/oxide boundary x = 0 oxide/silicon boundary potential at the metal gate : m t ox surface potential : s 0 substrate potential : sub X d M O S Poisson ' s Equation: d 2 x x dx 2 V m sub V ox V B drop across the oxide drop across the charged region of the silicon

Surface potential fort ox x0: t ox m s E ox dxe oxt oxv ox 0 s m E ox t ox m 0 X d =Q G =Q B =1/C ox ox t ox x is negative in the region considered Linear s m B ox and V ox B ox G ox The drop across the oxide is proportional to the charge stored on each side of the oxide t ox m x E oxdx = x = E ox t ox x V ox t ox t ox x

Potential drop across the substrate for 0xX d : d 2 x dx 2 0 s >0 charged region of silicon The potential is concave up in the charged region V B 0 X d 1 2 0 s X 2 0 X d d 2 X d B s 2 s X d Quadratic

PN Junction in Thermal Equilibrium If no external stimulus is applied (zero applied voltage, no external light source, etc) the device will eventually reach a steady state status of thermal equilibrium In this state ( open circuit and steady state condition) the current density must be zero: J tot,0 J p, 0 J n,0 0 Eventually, the populations of electrons and holes are each in equilibrium and therefore must have zero current densities J p,0 0 J n,0 0

Diffusion Mechanism The charge on the two sides of the junction must be equal (charge neutrality) Under open circuit and steady state conditions the built in electric field opposes the diffusion of free carriers until there is no net charge movement N A > N D We assumed the n-side is the more lightly doped

Diffusion Current It s a manifestation of thermal random motion of particles (statistical phenomenon) In a material where the concentration of particles is uniform the random motion balances out and no net movement result (drunk sail-man walk Brownian walks) If there is a difference (gradient) in concentration between two parts of a material, statistically there will be more particles crossing from the side of higher concentration to the side of lower concentration than in the reverse direction. Then we expect a net flux of particles

Diffusion Equations I n Aq e dn dx charge in the cross section The more non uniform the concentration the larger the current Proportionality Constants dn I n D n Aq e dx D Aqdn n dx I p D p Aq h dn dx D p Aq dn dx Assuming the charge concentration decreases with increasing x It means that dn/dx and dp/dx are negative, so to conform with conventions we must put a sign in front of the equations. J n D n q dn dx J p D p q dp dx

L Drift Current h W v n n E v p p E cross section area I n v n Whnq e Mobility (proportionality constant) I p v p Whpq h Eventually v saturates too many collisions effective electrons' mass increases charge per unit of volume volume travelled per unit of time J n n Enq e n Enq J p p Epq h p Epq

Drift and Diffusion currents J n, drift n Enq e n Enq J p, drift p Epq h p Epq dn J n, diff D n q e dx D q dn n dx dp J p, diff D p q h dx D p q dp dx qq h q e 1.610 19 C

Built in Voltage At equilibrium: (drift and diffusion balance out) J n J n, drift J n, diff n EnqD n q dn dx 0 J p J p, drift J p, diff p EpqD p q dp dx 0 Let's consider the second equation: x p x n p EpqD p q dp dx p p x p dp p dv D p p V x p V x n p x n dv dx pd p dp dx V x p V x n D p p pdvd p p dp ln px p px n

10 17 10 4 Built in Voltage x n x p 10 18 10 3 N D e.g.10 17 n 2 i N D N A e.g.10 18 n 2 i N A Since both μ and D are manifestations of thermal random motion (i.e. statistical thermodynamics phenomena) they are not independent Einstein ' s Relation : D p D n KT p n q V T MassAction Law : npn i 2 If n then p A larger number of electrons causes the recombination rate of electrons with holes to increase 0 V x n V x p D p ln p x p p p x n 0 KT q ln N A N D 2 n i

Applying KVL to the PN junction in equilibrium I 0 R We cannot have current. Something is wrong! P x p x n N + + + + + + 0 R

Built in Voltage E KVL at equilibrium: pm + mn + 0 = 0 P x p + + + x n N Metal-semiconductor contact potentials + + + + pm 0 mn If a free electron in the P region or a hole in the N region somehow reach the edge of the depletion region get swept by the electric field ( drift)

Neutrality of charge There are 4 charged particles in silicon, two mobiles (holes and electrons) and two fixed (ionized donors and ionized acceptors) The total positive change density and the total negative charge must be equal positive ions concentration N D pn A n holes concentration negative ions concentration electrons concentration

Depletion region in equilibrium The doping concentrations N A on the p side and N D on the n side are assumed constants xq pnn D N A the depletion regionis freeof electrons and holes N D N A n p x q N D N A on the P side N D 0: xqn A on the N side N A 0: xqn D +q N D x p (x) q N A E(x) x n x x Positive and negative excess charge in the depletion region must balance out (neutrality of charge): q N D x n = q N A x p Gauss ' Law: de x x dx s E(0)=E max =qn A x p / s =qn D x n / s

Depletion region in equilibrium 0 x p E(x) x n E max =qn A x p / s =qn D x n / s (x) x x 1 2 E maxx n 1 2 1 2 E maxx p 1 2 N D x n N A x p 2 qn D x n Area Triangle0 E max x n s qn A x p 2 s Area Triangle0 E max x p N x n A x N p D x p N D N A x n Potential Equation : d E x dx 0 1 2 qn D s x 2 n 1 2 qn A s 2 x p

Depletion region in equilibrium x n N A N D x p x p N D N A x n X dep x n x p 1 N A N D x p 1 N D N A x n x p X dep N D N A N and x n X D dep N A N A N D 0 1 2 qn D s x 2 n 1 2 qn A s x 2 p 1 2 1 q X dep 2 s N A N D 2 qn D s 2 X dep N A N A N D 2 1 2 N D N 2 A N A N 2 D qx 2 dep 2 s qn A s 2 X dep N A N D N A N D N D N A N D 2

Width and max field of the depletion region in equilibrium X dep 2 s q N A N D N 0 D N A with : 0 KT q ln N A N D 2 n i E max qn A s x p qn A N X D dep q s N A N D s N A N D N A N D X dep E max 2q s N A N D N A N D 0

Biased PN Junction P N v pm mn v j 0 0 pm v j mn v + - v j 0 v E(x) v>0 v=0 v<0 x

0 KT q ln N A N D 2 n i Biased PN Junction X dep 2 s q N A N D N A N D 0 v BV v E max 2 q s N A N D N A N D 0 v 0 i if we keep decreasing the voltage eventually we'll break the material. For silicon the breakdown point is reached for an electric field of approx. 10 7 V/m. NOTE: the depletion region can't get bigger than the length of the bar! if we keep increasing the voltage the depletion region will disappear (v= 0 ). As v becomes comparable with 0 the PN junction behave like a sort of resistor (the current is determined by the ohmic contacts and the resistance of the semiconductor)

Reverse Biased PN Junction E 0 Under reverse bias the depletion region becomes wider Then, it gets harder for the majority carriers to cross (diffuse through) the junction and easier for the minority carrier to be swept (drifted) across the junction E Since there are only a FEW minority carriers, the current carried under reverse bias is negligible

Reverse Biased PN Junction E 0 E NOTE: As soon as a minority carrier, let's say an electron on the P side, is swept across the junction, on the N side it becomes a majority carrier. Every time a minority carrier on the P side is swept toward the N side it leaves one less minority carrier on the edge of the depletion region in the P region. The same is true for holes swept from the N side to the P side. Under reverse bias the minority carrier concentration at the edges of the depletion regions is depleted below their equilibrium value. Since the number of minority carriers is small anyway this won't be a major difference

Forward Biased PN Junction Under forward bias the depletion region shrinks Then, it gets easier for the majority carriers to cross (diffuse through) the junction and harder for the minority carrier to be swept (drifted) across the junction. Since there are a LOT of majority carriers we expect the current to be considerable

Forward Biased PN Junction

I/V characteristic of PN Junction I qad n diode L n 2 n i qad p N A L p 2 n i N e D V diode V diode V T V 1I s e T 1 I s with : Across section area of the diode 2 n i holes' concentration in the N region (minority carriers) N D 2 n i electrons' concentration in the P region (minority carriers) N A D p diffusion constant for the holes in the N region L p diffusion lenght for the holes in the N region D p p p average time it takes for a hole into the N region to recombine with a majority electron...

I/V characteristic of PN Junction Since the only region where we have net charge is between -x p and x n such region (a.k.a. space charge region) is the only one where there is electric field. The regions from A to -x p and from x n to K are quasi-neutral (it is like they were perfect conductors and in perfect conductors there is no electric field inside)

I/V characteristic of PN Junction The situation is similar to the one at equilibrium but now the built in voltage is 0 V diode instead of 0 p p n n gnd V diode n p p n A X -X p 0 X n K

I/V characteristic of PN Junction Under forward bias close to the depletion edges we have: a greater hole concentration than normal on the N side (minority carriers) a greater electrons concentration than normal on the P side (minority carriers) P In the P region we have a lot of holes that will diffuse toward the N region N In the N region we have a lot of electrons that will diffuse toward the P region

I/V characteristic of PN Junction Extending the result derived at equilibrium we can write the voltage across the space charge region (between -x p and x n ) as: V j V x n V x p 0 V diode V T ln px p px n V T ln n x n nx p 0 V diode V T 0 V diode V T ln px p px n ln nx n nx p px n px p px p e e 0 V diode V T nx p n x n e 0 V diode V T e 0 V T n x n e 0 V T V diode V T V diode V e T

I/V characteristic of PN Junction And noting that: at the boundary of the quasi neutral P region at -Xp the hole density (majority carriers) is approximately equal at equilibrium as well as under bias, and the same is true for the electron density (majority carriers) at the boundary of the quasi neutral N region (at Xn) px p 0 p p0 0 p n0 n i V e T V e T n x n 0 n n0 0 n p0 n 2 i N V e T V A e T 2 N D the concentration of the majority carriers in the quasi neutral regions is approximately the same as the concentration at equilibrium

I/V characteristic of PN Junction Thus: px n px p px p e e 0 V diode V T nx p n x n e 0 V diode V T e 0 V T n x n e 0 V T V diode V T V diode V e T p x n n 2 V diode i V e T N D nx p n 2 V diode i V e T N A px p 0 p p0 0 p n0 n i V e T V e T n x n 0 n n0 0 n p0 n 2 i N V e T V A e T 2 N D

I/V characteristic of PN Junction I diode I n I p I n,drift I n,diff I p,drift I p,diff If we consider the quasi neutral regions, since in the quasi neutral regions there is no field there will be no drift I n I n, diff I p I p, diff neutral regionw N neutral region W P Then, the most suitable traverse sections for the evaluation of the total current I diode are those at the boundary of the depletion layer (x= x p or x=x n ) I diode I n x p I p x p I n.diff x p I p,diff x p

I/V characteristic of PN Junction If we make the simplifying assumption that the flow of the carriers in the depletion region is approximately constant (in other words we assume the recombination in the depletion region is negligible) I p,diff x p I p,diff x n I diode I n.diff x p I p,diff x p I n.diff x p I p,diff x n The currents due to diffusing carriers moving away from the junction are given by the well know diffusion equations: I n, diff xqad n dn p x dx I p,diff xqad p dp n x dx

I/V characteristic of PN Junction If we assume that the carriers distribution is linear (SHORT DIODE) dn p x dx n px p n p A n p0 e x p W p V diode V T n p0 W p n p0e V diode V T 1 W p dp n x dx p nx n p n K p n0e x n W n V diode V T p n0 p n0e W n V diode V T 1 W n

PN Junction: I/V characteristic Recalling that: n p0 n 2 i N A (electrons are minority carriers in P) p n0 n 2 i N D (holes are minority carriers in N) dn p x p dx n p0e V diode V T W p 1 n 2 V diode i V e T 1 N A dp n x n dx V diode p e V T n0 W n 1 n 2 V diode i V e T 1 N D I n, diff x p qad n dn p x p dx qan i 2 dp I p, diff x n qad n x n 2 p qan dx i V D diode n V e T 1 N A W p V D diode p V e T 1 N D W n

PN Junction: I/V characteristic And finally: 2 D I n, diff x p qan n i e N A W p V diode V T 1 V 2 D diode I p,diff x n qan p V i e T 1 N D W n I diode I n.diff x p I p,diff x p qan i 2 D n N N A W D W n V diode e V T 1 p In the case of a LONG DIODE the minority carriers will recombine before reaching the diode terminals I diode I n.diff x p I p, diff x p qan i 2 I s D n N A L n D p N D L e p V diode V T 1

PN Junction: I/V characteristic SHORT DIODE: I diode I n.diff x p I p, diff x p qan i 2 D n N A W p D p N D W e n V diode V T 1 LONG DIODE: I s I diode I n.diff x p I p, diff x p qan i 2 D n N A L n D p N D L e p V diode V T 1 Where L n is a constant known as the diffusion length of electrons in the P side and L p is a constant known as the diffusion length for holes in the N side. The constants L n and L p are dependent on the doping concentrations N A and N D respectively. I s

Diode Capacitances Depletion Capacitance (= Junction Capacitance) Diffusion Capacitance C j C d Reverse Biased Diode Depletion Capacitance Forward Biased Diode Diffusion Capacitance + Depletion Capacitance

Depletion Charge The depletion region stores an immobile charge of equal amount on each side of the junction ( it forms a capacitance!!) q J q P qn A x p A q N qn D x n A P side has ions N x p X dep D N A N and x n X D dep N side has + ions N A N A N D q J q N A N D N A N D AX dep X 2 s dep q N A N D N A N D 0 v D The decision to take q J negative is totally arbitrary. (But it turns out to be a good one if we prefer to work with positive capacitances) q J v D A 2q s N A N D N A N D 0 v D The charge of the depletion region is a function of the voltage v D applied to the diode

Depletion Capacitance Since the depletion charge does not change linearly with the applied voltage the resulting capacitor is non linear!! q J 0 v D q J v D A 2q s N A N D N A N D 0 v D An important physical consideration: we are dealing with a capacitor that no matter where I put the + of the applied voltage it always accumulate the positive charge on the N side of the junction, and the negative charge on the P side of the junction.

Small Signal Depletion Capacitance For small changes of the applied voltage about a specified DC voltage V D we can derive an equivalent linear capacitor approximation V D D = V D + d q J 0 D since q J vs. D relationship is non linear the capacitor is non linear q J v D slope d q J = Q J + q j Q J q J v D q J V D dq J v q j dq dv D V D V D D J slope v dv D V d D Common conventions: total quantities (applied) v D V D v d V d V D v d small quantities DC quantities

Small Signal Depletion Capacitance C j C j V D dq J dv D V D d dv D A 2 q s N A N D N A N D 0 v D 12 V D A 2 q s N A N D N A N D 12 d dv D 0 v D 12 V D A 2q s N A N D N A N D 12 2 1 2 v 0 D 1 V D A q s 2 N A N D N A N D A 2 N A N D q s N A N 0 V D D 12 0 V D 1 2 A 2 q s N A N D N A N D 0 V D 12 A 2 N A N D q s N A N D = C j0 0 1 V D 0

Small Signal Depletion Capacitance Zero Bias Capacitance = junction capacitance in thermal equilibrium (V D =0) A C j0 C j V D 0 2 N A N D q s N A N 0 D A s X dep, 0 X 2 s dep,0 q N A N D N A N D 0 s 2 q s N A N D N A N D 0 C j C j V D q j v d dq J dv D V D C j0 1V D 0 X dep, 0 A s A s 1V X D dep 0

Small Signal Depletion Capacitance: Physical Interpretation C j q j v d A s X dep Capacitance of a parallel plate capacitor with its plates separated by the depletion width X dep (V D ) at the particular DC voltage V D. The charges separated by X dep are the small signal charge layers ±q j. For vd 0, the small signal charges become sheets that are separated by a gap width of exactly Xdep

Physical Interpretation C j q j v d A s X dep Capacitance of a parallel plate capacitor with its plates separated by the depletion width X dep (V D ) at the particular DC voltage V D. The charges separated by X dep are the small signal charge layers ±q j. For vd 0, the small signal charges become sheets that are separated by a gap width of Xdep

Small Signal Depletion Capacitance In the practice the depletion capacitance is usually provided per cross-section area: j0 j V D 0 j j V D dq J dv D s X dep, 0 V D s X dep 1 2 N A N D q s N A N 0 D s 0 X 1V D dep, 0 j0 1V D 0 NOTE: when the diode is forward biased with v D = 0 the equation for C j blows up (i.e., is equal to infinity). As v D approaches 0 the assumption that the depletion region is free of charged carriers is no longer true.

Graded Junctions All the equations derived for the depletion capacitance are based on the assumption that the doping concentration change abruptly at the junction. Although this is a good approximation for many integrated circuits is not always true. More in general: j j0 1V D 0 M j Mj is a constant called grading coefficient and its value ranges from 1/3 to depending on the way the concentration changes from the P to the N side of the junction

Large Signal Depletion Capacitance The equations for the depletion capacitance given before are valid only for small changes in the applied voltage It is extremely difficult and time consuming to accurately take this non linear capacitance into account when calculating the time to charge or discharge a junction over a large voltage change A commonly used approximation is to calculate the charge stored in the junction for the two extreme values of applied voltage, and then through the use of Q = CV, calculate the average capacitance accordingly jav V 2 V 1 V 2 V 1 The approximation is pessimistic

Large Signal Depletion Capacitance V 2 q s N A N D N A N D 0 V 2 q s N A N D 0 N A N 1 V D 2 0 1 V j0 0 0 j0 q s 2 0 N A N D N A N D q s N A N D N A N D j0 2 0 jav V 2V 1 V 2 V 1 2 0 j0 1 V 2 1V 1 0 0 V 2 V 1

Example Find a rough approximation for the junction capacitance to be used to estimate the charging time of a reverse biased junction from 0V to 5V (or vice versa). Assume 0 0.9 V jav 2 0 j0 1 V 2 1V 1 0 V 2 V 1 0 j0 1 5 0.9 1 20.9 50 j0 2 5V V D 0V