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This document discusses the PN junction diode. It begins by explaining how a PN junction forms when P-type and N-type semiconductors are placed side by side. It then discusses the current-voltage characteristics of diodes in different operating regions: equilibrium (open circuit), forward bias, and reverse bias. The document goes on to explain the diffusion and drift currents that occur in the depletion region of a PN junction in equilibrium, and how these currents cancel out, resulting in no net current flow. It also discusses how the built-in voltage arises across the junction in equilibrium due to the diffusion potential.

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Electronics I - PN Junction talarico@gonzaga.edu 1 p++ n NA >> ND A K Chapter 3 Fall 2017
PN Junction (Diode) talarico@gonzaga.edu 2 • When N-type and P-type dopants are introduced side-by- side in a semiconductor, a PN junction or a diode is formed. source: Razavi JUNCTION
Diode I/V characteristics • Our goal is to find out the current-voltage characteristics of the device • Diode operation regions: – Equilibrium = open circuit terminals = no bias – Forward bias – Reverse bias – (Breakdown) talarico@gonzaga.edu 3 D D Equilibrium D D “smoke” ΦB
PN Junction in Equilibrium (no Bias applied) • Because each side of the junction contains an excess of holes or electrons compared to the other side, there exists a large concentration gradient. Therefore, a diffusion current flows across the junction from each side. talarico@gonzaga.edu 4 source: Razavi nn0 pn0 np0 pp0 + ND < NA nn0 ≈ ND pn0 ≈ ni 2/ND pp0 ≈ NA np0 ≈ ni 2/NA
PN junction in Equilibrium (no bias applied) • Diffusion across the junction causes a lot of electrons and holes to recombine leaving a depletion region in proximity of the junction • The fixed ions left behind create an electric field • The electric field opposes the diffusion of holes in the N region and free electrons in the P region • Once the electric field is strong enough to stop the diffusion currents the junction reaches equilibrium – A “barrier” opposing diffusion forms 5 − − − − − − − − − − − − − − − − − − + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + − − − − + + + + Free Electrons Positive Donor Ions Negative Acceptor Ions Free Holes Depletion Region P N ND < NA
Depletion approximation • What happens inside the PN junction in equilibrium is not quite as black and white as depicted talarico@gonzaga.edu 6 ρ(x) = q p0 (x)− n0 (x)+ ND − NA ( ) • On the N-side of the transition region (0 < x < xn0), since there are no acceptors (NA = 0) and the hole concentration is negligible (p0 ≈ 0) • Since n0(x) < ND on the N-side of the transition (and even more so on the P-side of the transition) there is a roll-off in electron concentration ρ(x) = q p0 (x)− n0 (x)+ ND − NA ( )≈ q −n0 (x)+ ND ( ) source: Howe & Sodini
Depletion approximation talarico@gonzaga.edu 7 • The depletion approximation assumes that the mobile carriers (n and p) are small in number compared to the donor and acceptor ion concentrations (ND and NA) in the depletion region (SCR), and the device is charge-neutral elsewhere (QNR) • Depletion approximation – NA >> pp hence ρ ≈ −qNA for −xp ≤ x ≤ 0 – ND >> nn hence ρ ≈ qND for 0 ≤ x ≤ xn – The charge density is negligible in the bulk regions; that is ρ ≈ 0 for x > xn and x < − xp • On the P-side of the transition region (−xp < x < 0), since there are no donors (ND = 0) and the electrons concentration is negligible (n0 ≈ 0) • Since p0(x) < NA on the P-side of the transition region (and even more so on the N-side of the transition) there is a roll-off in hole concentration ρ(x) = q p0 (x)− n0 (x)+ ND − NA ( )≈ q p0 (x)− NA ( ) source: Howe & Sodini That is, there are no free mobile charges in depletion region
One more time … • aa talarico@gonzaga.edu 8 source: Sodini (P-side) (N-side) = 0 = 0 ≈ 0 (minority carriers) ≈ 0 (minority carriers) Na >> p0 (depletion approximation) Nd >> n0 (depletion approximation) P-type N-type NOTE: for the example in this picture NA < ND
Depletion approximation talarico@gonzaga.edu 9 Depletion approximation means we assume the charge density in each transition region to be uniform Charge density x qND −qNA − + source: G.W. Neudeck
PN junction in Equilibrium (no bias applied) • Since the circuit is open no currents exists • Therefore in equilibrium the drift currents resulting from the electric field generated by the fixed ions in the depletion region must exactly cancel the diffusion currents • This must hold for both types of charge carriers: Jn = Jp = 0 • Let’s take one of the two equations and try to solve it: 10 − − − − − − − − − − − − − − − − − − + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + − − − − + + + + Free Electrons Positive Donor Ions Negative Acceptor Ions Free Holes Depletion Region P N ND < NA Jp = 0 = pqµpEx − qDp dp dx Jn = 0 = nqµnEx + qDn dn dx 0 nn0 pn0 pp0 np0 JUNCTION x xp0 −xn0 −pµp dV dx = Dp dp dx pqµpEx = qDp dp dx Ex = − dV dx
Built-in voltage (equilibrium) talarico@gonzaga.edu 11 −pµp dV dx = Dp dp dx − dV V (−xn0) V (xp0) ∫ = Dp µp dp p p(−xn0) p(xp0) ∫ V(−xn0 )−V(xp0 ) =VT ln pp0 pn0 ΦB =VT ln NA ni 2 / ND ! ΦB =VT ln NA ND ni 2 nn0 pn0 pp0 np0 x xp0 −xn0 P N + + − − Ex 0 ΦB + − + + − − Example Find built in voltage of a silicon PN junction at room temperature and with NA ≈ 1018 and ND ≈ 1017 ΦB ≅ 26mV × ln 1018 1017 1020 ≅ 26mV × ln(10)× Log10 1018 1017 1020 ≅ ≅ 60mV ×15 ≅ 900mV depletion region
PN Junction in equilibrium (no bias applied) talarico@gonzaga.edu 12 − − − − − − − − − − − − − − − − − − + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + − − − − + + + + Depletion Region P N + − • Some of the thermally generated holes in the N region (minority carrier) moves toward the junction and reach the edge of the depletion region. There, they experience the strong electric force present in depletion region, and get swept all the way into the P region (where they become majority carriers). • The same happen for the thermally generated minority carriers (free electrons) in the P region. • At equilibrium the flow of carriers swept by the electric field (drift) and the flow of carriers due the gradient in carriers concentration (diffusion) cancel out ID = Idiff – Idrift =0 Ex Idrift Idiff + − ΦB P N ID − − − − − − + + + + + + + + + VD
Width of the Depletion region in equilibrium 13 − − − − − − − − − − − − + + + + + + + + + + + + + + + + + + + + + + + + − − − − + + + + P N −xn0 +xp0 0 E metal contact A K metal contact −xn0 +xp0 x concentration [cm−3] pp0=NA np0≈ni 2/NA nn0=ND pn0≈ni 2/ND W=Xd0 Idrift Idiff −xn0 x +xp0 ρ = charge density [Cb/cm−3] −qNA +qND |Q −|= qNAAxp0 |Q +|= qNDAxn0
Width of the Depletion region in equilibrium talarico@gonzaga.edu 14 −xn0 x +xp0 ρ = charge density [Cb/cm−3] −qNA +qND |Q −|= qNAAxp0 |Q +|= qNDAxn0 Charge conservation: |Q+ |=|Q− | ! QJ 0 " AqND xn0 = AqNAxp0 " ND xn0 = NAxp0 E(x) x −xn0 +xp0 E(0) = −qNAxp0 εsi = −qND xn0 εsi ! Emax PoissonEquation: dE dx = ρ εsi ! dE = ρ εsi dx As long as we know how to compute the area of a rectangle we are fine with the math !!! Permittivity of crystalline silicon: εsi ≈ 11.7×8.85×10−12 F/m εr ε0
• Let’s go from the electric field to the voltage talarico@gonzaga.edu 15 Width of the Depletion region in equilibrium − E(x) x −xp0 +xn0 qNAxp0 εsi = qND xn0 εsi E(x) = − dΦ0 dx ⇔ − E(x)⋅dx = dΦ0 Φ0(x) x Area “red” triangle 1 2 qND xn0 2 εsi Area “blue” triangle = Area “blue” triangle 1 2 qNAxp0 2 εsi Area “red” triangle = 0 As we move in the x direction (from –xp0 to xn) the area under –E(x) keeps growing NOTE: for convenience I flipped the orientation of the p-side and the n-side with respect to the x direction so the name of the depletion region’s edge at the p-side becomes –xp0 and the name of the depletion region’s edge at the n-side becomes +xn0 ΦB=Φ0 (xn0)−Φ0 (−xp0) ΦB = 1 2 qNAxp0 2 εsi + 1 2 qND xn0 2 εsi ΦB + − E − + + parabola N parabola P source: Howe & Sodini −
Metal-Semiconductor Contact Potential • At first glance would seem strange that the potential drops at the contacts would have to add up to the built in potential of the junction. Were this is not the case, KVL would imply that current would flow in thermal equilibrium when a resistor is connected across the metal contacts to the p and n region. This conclusion violate our basic understanding of the meaning of equilibrium: namely the net current must be zero in equilibrium talarico@gonzaga.edu 16 source: Howe & Sodini
talarico@gonzaga.edu 17 Width of the Depletion region in equilibrium W = xp0 + xn0 W = xn0 1+ ND NA ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = xn0 NA + ND NA ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ W = xp0 1+ NA ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = xp0 NA + ND ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ NAxp0 = ND xn0 Charge neutrality: xn0 = W NA NA + ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ xp0 = W ND NA + ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ΦB = qNA 2εsi xp0 2 + qND 2εsi xn0 2 ΦB = qNA 2εsi W 2 ND NA + ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 + qND 2εsi W 2 NA NA + ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 ΦB = q 2εsi W 2 NAND 2 + ND NA 2 NA + ND ( )2 W 2 = 2εsi q ΦB NA + ND ( )2 NAND 2 + ND NA 2 = 2εsi q ΦB NA + ND ( )2 NAND (ND + NA ) W = 2εsi q ΦB NA + ND NAND ! Xdo xn0 = 2εsi q ΦB NA + ND ( ) NAND NA 2 (NA + ND )2 = 2εsi q ΦB NA ND (NA + ND ) xp0 = 2εsi q ΦB ND NA (NA + ND )
Charge on either side of the depletion region (in equilibrium) talarico@gonzaga.edu 18 QJ 0 !|Q+ |=|Q− |= AqND xn0 xn0 x −xp0 ρ = charge density [Cb/cm−3] −qNA +qND |Q −|= qNAAxp0 |Q +|= qNDAxn0 QJ 0 = AqNDW NA NA + ND = Aq NAND NA + ND W QJ 0 = Aq NAND NA + ND 2εsi q ΦB NA + ND NAND = A 2εsiq NAND NA + ND ΦB Junction Charge xn0 = W NA NA + ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ xp0 = W ND NA + ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = xn0 = W=Xd0
Effect of bias voltage on the PN Junction talarico@gonzaga.edu 19 source: Streetman ΦB Φp Φn ΦJ=ΦB−VF ΦJ=ΦB+VR VF=VD VR=−VD + VD − + VD − + VD= 0 − Idiff Idrift Idiff Idrift Idrift Idiff ID = 0 ID > 0 ID < 0 ID = Idiff − Idrift The drift current has the same direction of the electric field (and it is proportional to it) (a) Open circuit (b) Forward Bias (c) Reverse Bias (Equilibrium) Jp,drift = pqµpE Jn,drift = nqµnE ΦB ΦJ ΦJ
PN Junction under Reverse Bias • The external voltage VR is directed from the N side to the P side (same as ΦB), therefore the electric field (ER) due to the bias VR enhances the built in field E. The depletion region W is widened • Since the “barrier” potential rises higher than in equilibrium it gets harder for the majority carriers to cross the junction and easier for the minority carrier to be swept (drifted) across the junction. • The current carried under reverse bias is negligible the minority carriers are too few !! talarico@gonzaga.edu 20 VR = −VD VR = −VD ID − ΦB + − ΦB + source: Millman & Halkias source: Razavi Equilibrium (open circuit) Reverse Bias + ΦB − E E ER + ΦJ − ΦJ = ΦB + VR
PN junction under reverse bias talarico@gonzaga.edu 21 KVL: VD + Φmn + ΦJ + Φmn = 0 ΦJ = −VD − Φmn − Φmp = ΦB−VD = ΦB potential barrier in equilibrium potential barrier in non-equilibrium source: Howe & Sodini
PN junction under reverse bias 22 Example: PN junction under reverse biases VD=−2.4 V and VD=−6.4V Assuming the potential barrier in equilibrium is ΦB = 0.8V As the PN junction gets more reverse biased • the depletion region Xd = xp + xn becomes wider • the potential barrier becomes stronger (a) charge density (b) electric field (c) potential source: Howe & Sodini
Application of Reverse Biased PN Junction: Voltage Dependent capacitor • A reverse biased PN junction can be used as a capacitor. By varying VR, the depletion width changes, and so does the capacitance associated. Therefore a reverse biased PN junction can be seen as a voltage- dependent capacitor. talarico@gonzaga.edu 23 Varactor p n Qualitatively we see that if VR goes up then CJ goes down source: Razavi
Capacitance of Reverse Biased Diode talarico@gonzaga.edu 24 Xd @VD = 2εsi q ΦB −VD ( ) NA + ND NAND = 2εsi q ΦB 1− VD ΦB ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ NA + ND NAND = Xd0 1− VD ΦB CJ @VD = εsi ⋅ A Xd @VD = qεsi 2 NAND NA + ND ( ) 1 ΦB −VD ( ) = qεsi 2 NAND NA + ND ( ) 1 ΦB 1 1−VD / ΦB ( ) = = CJ0 1−VD / ΦB width of depletion region in equilibrium CJ0 = junction capacitance at zero bias (in equilibrium) = CJ0 = εsi A Xdo careful: cap per unit area
Junction capacitance of reverse biased diode talarico@gonzaga.edu 25 CJ0 VR=−VD VD= VF Reverse Voltage Forward Voltage CJ source: Streetman CJ (VD ) = CJ0 1− VD ΦB ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ M = CJ0 1+ VR ΦB ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ M • M=0.5 for the abrupt PN junction we modeled • Typical values for M range from 0.3 to 0.5 • SPICE uses VJ (or PB) to denote the built in voltage in equilibrium (ΦB) The junction capacitance is non linear
talarico@gonzaga.edu 26 QJ @VD = A 2εsiq NAND NA + ND ΦB −VD ( ) = A 2εsiq NAND NA + ND ΦB 1− VD ΦB ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = QJ 0 1− VD ΦB = = QJ 0 1+ VR ΦB charge stored in either side of the depletion region in equilibrium Junction capacitance of reverse biased diode • The relation between charge QJ and voltage (VR=−VD) across the diode is not linear • It is difficult to define a capacitance that changes whenever VR changes source: Sedra & Smith P VP QJ = A 2εsiq NAND NA + ND ΦB +VR ( ) = ς ΦB +VR = ς 2 ΦB 1+ VR ΦB ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = CJO 1+ VR ΦB CJ @VP = dQJ dVR VR =VP = d dVR ς ΦB +VR ( ) @VP = ς 2 ΦB +VR @VP = = ς @VP @VP square root law 𝐶"|@%& = 1 𝐴 𝑑𝑄" 𝑑𝑉- . /01/& /A = 2 3 /A /A
talarico@gonzaga.edu 27 Junction capacitance of reverse biased diode source: Sedra & Smith P VP CJ @VP = dQJ dVR VR =VP ≠ QJ (VR ) VR VR =VP Important result to remember when using non linear devices /A
Application of a reverse biased diode talarico@gonzaga.edu 28 LC fres 1 2 1 p = source: Razavi
PN Junction in Forward Bias talarico@gonzaga.edu 29 ID VD=VF VD=VF − ΦB + − ΦB + source: Millman & Halkias • The electric field (EF) due to the bias (VF) is opposite to the built in field E. The depletion region W is shortened • Since the “barrier” potential becomes lower than in equilibrium it gets easier for the majority carriers to diffuse through the junction and harder for the minority carrier to be swept (drifted) across the junction. • The majority carriers are a lot, so we expect a considerable current flow: ID = Idiff − Idrift Equilibrium (open circuit) + ΦB − + ΦJ − Forward Bias EF E ΦJ = ΦB − VF source: Razavi p-n ID
Diode’s Current/Voltage Characteristic talarico@gonzaga.edu 30 ID = IS (e VD VT −1) ≈ −IS NOTE: Consider |VD/VT| ≈ 2.3 exp(2.3) ≈ 9.97 ≈ 10 exp(−2.3) ≈ 0.1 At room temperature VD = 2.3 × VT ≈ 59.8 mV≈ 60 mV source: Razavi source: Streetman −I(drift) ID VD VD ID ID = I(drift) e qVD KT −1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ very steep !
Diode’s Current/Voltage Characteristic 31 • How can we quantify the I/V behavior of the diode ? • We need to look at how electrons and holes move under the effect of the bias voltage • In equilibrium: • With an external bias voltage applied the situation is similar but now the potential barrier is ΦB – VD instead of only ΦB ΦB = VT ln pp0 pn0 ⇔ pn0 = pp0 ⋅e−ΦB /VT ⇔ pn0 = pp0 eΦB /VT holes at xn0 holes at −xp0 −xp0 and xn0 are the edges of the depletion region in equilibrium pn = pp ⋅ eVD /VT eΦB /VT holes at xn holes at −xp −xp and xn are the edges of the depletion region with non zero bias x −Wp −xp 0 xn Wn WP WN P-side N-side + + − − VD GND A K quasi-neutral region quasi-neutral region Law of the Junction np = nn ⋅ eVD /VT eΦB /VT electrons at −xp electrons at xn The Law of the Junction relates the minority carrier concentration on one side of the junction to the majority carrier concentration on the other side of the junction np0 = nn0 eΦB /VT electrons at −xp0 electrons at xn0
Diode’s Current/Voltage Characteristic talarico@gonzaga.edu 32 • When we apply a forward bias VD the potential barrier is reduced from ΦB to ΦB−VD. This reduces the drift field and upsets the balance between diffusion and drift that exists at zero bias. • Electrons can now diffuse from the N side (where they are majority carriers) to the P side (where they become minority carriers). Similarly, holes diffuse from the P side into the N side. This migration of carriers from one side to the other is called injection. • As a result of injection, more electrons are now present at –xp and more holes appear at xn than when the barrier was higher – There is an excess of minority carriers at the edges of the depletion region • Let’s make some simplifying assumptions to help quantifying pn(xn), pp(−xp), nn(xn) and np(xp) – There is electric field only in the space charge region (that is between –xp and xn). The regions from –Wp to –xp (a.k.a. P bulk region) and from xn to Wn (a.k.a. N bulk region) are quasi-neutral (it is like they were perfect conductors and in perfect conductors there is no electric field inside)
Diode’s Current/Voltage Characteristic talarico@gonzaga.edu 33 • The requirement that the bulk regions on either side remain charge-neutral even under applied bias (remember: E(x) = ρ(x)/ε, so no field means no charge density) implies that any increases in the minority carriers due to injection must be counter balanced by in an increase of majority carriers – On the P side ρ(x=−xp) = 0 implies that: – On the N side ρ(x=xn) = 0 implies that: • Assuming the level of injection is low we can neglect the slight increase in majority carrier concentration due to carrier injection and write: pp (−xp ) = NA + np (−xp ) ≈ pp0 + np (−xp ) nn (xn ) = ND + pn (xn ) ≈ nn0 + pn (xn ) pp (−xp ) ≈ pp0 ≈ NA nn (xn ) ≈ nn0 ≈ ND np (−xp ) << pp0 pn (xn ) << nn0
Minority and Majority carriers concentrations in a forward biased diode • Concentrations Howe & Sodini talarico@gonzaga.edu 34 x pn0 np0 Quasi-neutrality demands that at every point in the quasi-neutral regions: excess minority carrier concentration = excess majority carriers concentration source: Howe & Sodini
talarico@gonzaga.edu 35 Minority and Majority carriers concentrations in a forward biased diode Quasi-neutrality demands that at every point in the quasi-neutral regions: excess minority carrier concentration = excess majority carriers concentration NOTE: at this point we did not put much thinking in the profile of the minority carriers diffusing in the QNRs, however if we assume the diffusion occurs over a short distance we can approximate the curves with straight lines ! 0 −xp −Wp p(−xp) p p-QNR Na qPp qNp n(−xp) ni 2 Na x source: Sodini n
• Under the low injection condition we can approximate the Law of the Junction as follows: talarico@gonzaga.edu 36 pn (xn ) = pp (−xp ) eΦB /VT ⋅eVD /VT ≈ pp0 eΦB /VT ⋅eVD /VT = pn0 ⋅eVD /VT np (−xp ) = nn (xn ) eΦB /VT ⋅eVD /VT ≈ nn0 eΦB /VT ⋅eVD /VT = np0 ⋅eVD /VT pn (xn ) ≈ pn0 ⋅eVD /VT ≈ ni 2 ND ⋅eVD /VT np (−xp ) ≈ np0 ⋅eVD /VT ≈ ni 2 NA ⋅eVD /VT The concentration of minority carriers at the edges of the space charge region are an exponential function of the applied bias Diode’s Current/Voltage Characteristic source: G.W. Neudeck
Minority carriers in a forward biased PN junction • The excess concentration of minority carriers at the edges is: • As the injected carriers diffuse through the quasi-neutral regions they recombine with the majority carriers and eventually disappear. The excess minority carriers decay exponentially with distance, and the decay constant is called diffusion length L. talarico@gonzaga.edu 37 Δnp ! np0 ⋅eVD /VT − np0 = np0 1− eVD /VT ( ) Δpn ! pn0 ⋅eVD /VT − pn0 = pn0 1− eVD /VT ( ) np (x) = Δnpe (x+xp ) Ln pn (x) = Δpne −(x−xn ) Lp source: Sedra & Smith p+ n It can be shown through the current continuity equation that under the low injection condition the concentration’s profile of the minority carriers diffusing in the quasi neutral regions is exponential
Minority carrier concentrations Equilibrium vs. Forward Bias talarico@gonzaga.edu 38 source: G.W. Neudeck
Aside (a simple trick to simplify the math) talarico@gonzaga.edu 39 • Change the axis selection for the bulk regions as follows: • This way we can write the minority carriers concentrations as follows: np (x'') = Δnpe − x'' Ln pn (x') = Δpne − x' Lp x'' ! 0'' ≡ −xp x' ! 0' ≡ xn Ln = Dnτn Lp = Dpτ p source: G.W. Neudeck source: G.W. Neudeck Ln = Diffusion length for electrons [m] Lp = Diffusion length for holes [m] Dn = Diffusion coefficient for electron [m2/s] Dp = Diffusion coefficient for holes [m2/s] τn = mean lifetime for electrons [s] τp = mean lifetime for holes [s] VA = voltage applied ≡ VD
Physical interpretation of L, D, and τ • Physically Lp and Ln represents the average distance minority carriers can diffuse into a “sea” of majority carriers before being annihilated • Physically τp and τn represents the average time minority carriers can diffuse into a “sea” of majority carriers before being annihilated • Physically the Dp and Dn represents the ease with which holes and electrons diffuse talarico@gonzaga.edu 40 Lp ! xΔpn (x) 0 ∞ ∫ dx Δpn (x) 0 ∞ ∫ dx Ln ! xΔnp (x) 0 ∞ ∫ dx Δnp (x) 0 ∞ ∫ dx Dp ! µp KT q Dn ! µn KT q
PN Junction current • The current that flows through any “transverse” section x must be the same and is formed by both holes and electrons • If we look at the current through a section x in the quasi-neutral regions, since in the quasi-neutral regions there is no electric field there will be no drift talarico@gonzaga.edu 41 J(x) = Jn (x)+ Jp (x) Jn (x) = Jn,drift (x)+ Jn,diff (x) Jp (x) = Jp,drift (x)+ Jp,diff (x) Jn (x) = Jn,diff (x) = qDn dnp (x) dx = q Dn Ln np0 eVD /VT −1 ( )e (x+xp )/Lp for −Wp ≤ x ≤ −xp Jp (x) = Jp,diff (x) = −qDp dpn (x) dx = q Dp Lp pn0 eVD /VT −1 ( )e −(x−xn )/Lp for xn ≤ x ≤ Wn injected electrons diffusing on P-side (minority carriers) injected holes diffusing on N-side (minority carriers) +(x+xp)/Ln
PN Junction currents • The easier sections for evaluating the diffusion currents are the edges of the space charge region (x=−xp and x=xn) • The total current through a section of the diode can be found as follows: • If we make the simplifying assumption that there is no recombination-generation in the space-charge region, this means that the electrons/holes current at x=−xp and at x=xn do not change in the space-charge region and therefore we can write: talarico@gonzaga.edu 42 Jn (−xp ) = Jn,diff (−xp ) = q Dn Ln np0 eVD /VT −1 ( ) Jp (xn ) = Jp,diff (xn ) = q Dp Lp pn0 eVD /VT −1 ( ) J = Jn (−xp )+ Jp (−xp ) = Jn (xn )+ Jp (xn ) Jn (−xp ) = Jn (xn ) Jp (xn ) = Jp (−xp ) We do not know have this value (yet) ! We do not know have this value (yet) !
PN Junction current • Therefore we can add together the diffusion currents at x=−xp and x=xn 43 J = Jn (−xp )+ Jp (xn ) = q Dn Ln np0 eVD /VT −1 ( )+ q Dp Lp pn0 eVD /VT −1 ( ) J = q Dn Ln np0 + q Dp Lp pn0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ eVD /VT −1 ( ) ID = J × A = In,diff + Ip,diff = Aq Dn Ln np0 + Aq Dp Lp pn0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ eVD /VT −1 ( ) ID = Aq Dn Ln ni 2 NA + Dp Lp ni 2 ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ eVD /VT −1 ( )= IS eVD /VT −1 ( ) scale current (a.k.a. saturation current)
Minority concentrations for a forward biased short base diode • If the diode is short the exponential decays can be approximated with straight lines talarico@gonzaga.edu 44 −xp 0 xn np0 pn0 −Wp Wn VD A GND K WidthN WidthP x ID = Aq Dn WidthP np0 + Dp WidthN pn0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ eVD /VT −1 ( )= IS eVD /VT −1 ( ) dnp dx ≈ np (−xp )− n(−Wp ) WidthP = np0eVD /VT − np0 WidthP dpn dx ≈ pn (xn )− pn (Wn ) −WidthN = − pn0eVD /VT − pn0 WidthN
Current densities in the forward bias case talarico@gonzaga.edu 45 source: G.W. Neudeck No matter which section x we consider Jp and Jn must add to the same total value J
Assumptions used to derive the Ideal diode equation • Depletion approximation (in depletion region: NA >> pp0(x), ND >> nn0(x)) • Negligible fields in the quasi neutral regions (QNR). This means that the applied voltage VD is dropped entirely across the space charge region (SCR) • Quasi-neutral regions that are so long that the injected carriers all recombine before reaching the end contacts • Low levels of injection • No recombination-generation in the space-charge region. This means that the electron current and the hole current are constant in the space- charge region ( that is Jn(−xp)=Jn(xn) and Jp(xn)=Jp(−xp) ) • Real diodes may deviate considerably from the ideal equation. 46 ID = I* S (e VD nVT −1) n = emission coefficient (a.k.a. ideality factor) Typically the ideality factor n varies from 1 to 2
Ideal diode talarico@gonzaga.edu 47 ID IS VD VD log10|ID| IS y = log10 IS ⋅e VD VT ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = log10 IS ( )+ log10 (e)⋅ VD VT ≈ log10 IS ( )+ 0.43 VD VT x =VD dy dx ≈ 0.43 VT at room temperature (300K): VT 0.43 ≈ 26mV 0.43 ≈ 60mV 10 source: Sodini every 60 mV the current changes by a decade
Deviations from the ideal diode talarico@gonzaga.edu 48 source: Hu Ideal with IS VDIODE (V)
Ideal diode equation • Although we derived the ideal diode equation by subjecting the diode to a forward bias, the steps followed can also be applied for the case of reverse bias (the only difference is that now VD < 0). • The equation works for both forward and reverse region. talarico@gonzaga.edu 49 ID VD ID = IS eVD /VT −1 ( ) source: Streetman ≈ ISeVD /VT ≈ −IS
Minority concentrations for reverse biased diode • In forward bias, minority carriers are injected into quasi-neutral regions • In reverse bias, minority carriers are extracted (swept away) from the quasi- neutral regions • In reverse bias there is a very small amount of carriers available for extraction – ID saturates to small value talarico@gonzaga.edu 50 Minority-carrier density distribution as a function of the distance from the junction. (a) Forward biased junction (b) Reverse biased junction (The depletion region is assumed so small relative to the diffusion length that is not indicated in the figure) source: Millman & Halkias
Minority carrier concentrations Equilibrium vs. Reverse Bias talarico@gonzaga.edu 51 source: G.W. Neudeck
Small-signal equivalent model for diode • Examine the effect of a small signal overlapping the bias: talarico@gonzaga.edu 52 Figure 4.14 Development of the diode small-signal model. source: Sedra & Smith vD (t) ≡VD + vd (t) iD = IS e vD VT −1 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = IS e VD+vd VT −1 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ iD ≅ iD VD + diD dvD VD (vD −VD ) = ID + diD dvD VD vd iD is a function of vD: Q ≡ (ID,VD ) diD dvD VD = d dvD IS e vD VT −1 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ VD = ISe VD VT VT = ISe VD VT + IS − IS VT = = ID + IS VT iD − ID ≅ gdvd ⇔ id ≅ gdvd ≡ gd * Applied Taylor *
Small-signal equivalent model for diode • From a small signal point of view the diode behaves as a (non linear) resistance of value: • rdiode depends on bias: • In forward bias: In reverse bias: • Under the assumption that the signal overlapping the bias is small the i-v relationship of the diode can be linearized: talarico@gonzaga.edu 53 Figure 4.14 Development of the diode small-signal model. source: Sedra & Smith rdiode = 1 gd = VT ID + IS iD ≅ ID + gdvd = ID +id vD =VD + vd gd = ID + IS VT gd ≈ ID VT gd ≈ 0 rdiode = 1/gd
Small-signal equivalent model for diode • Besides having current flowing through it (resistive behavior) the diode also exhibit charge displacement mechanisms that can be modeled as a capacitive behavior • The charge response to the voltage applied across the diode is non-linear • As long as the voltage signal superimposed to the bias voltage is small, the relationship between charge and voltage can be linearized. • Complete small signal model for diode talarico@gonzaga.edu 54 vD =VD + vd ≡ Cdiode qD ≈ QD VD + dqD dvD VD (vD −VD ) = QD VD +Cdiode ⋅vd rdiode Cdiode
talarico@gonzaga.edu 55 Capacitive effects in the PN junction • Diodes have two capacitive effects • But … please careful: there are way too many letters “D”: – depletion capacitance = junction capacitance = CJ – diffusion capacitance = storage capacitance = CS – Cdiode = CJ + CS • We already know quite a bit about the junction cap. that develops when the diode is in reverse bias (… but what about forward bias?) • So far we did not even think about the existence of storage (diffusion) capacitance • We need to dig more into the capacitive effects … Cdiode A K CJ A K CS
Junction Capacitance talarico@gonzaga.edu 56 Behavior of junction capacitance as a function of bias voltage VD (Chawla-Gummel) ≡ ΦB ≡ ΦB 2 SPICE ≈ 2×CJ 0 For Hand Calculation source: Gray, Hurst, Lewis and Meyer
Junction Capacitance • Earlier, we derived the capacitance associated with depletion region (junction capacitance) for a reverse biased diode as: • Under significant forward bias VD > ΦB/2 the approximation that there are no mobile carriers in the depletion region becomes invalid (this was the basic assumption behind the derivation !!) • Therefore, under forward bias a common approximation is to take the value of CJ at VD = ΦB/2 (or being slightly more pessimistic) talarico@gonzaga.edu 57 Cj VD = CJO 1− VD ΦB CJO = zero-biasjunctioncapacitance = A εsiq 2 ⋅ NA ND NA + ND ⋅ 1 ΦB Cj VD= ΦB 2 = CJO 1−1/ 2 = 2 ×CJO ≈ 2×CJO
Diffusion capacitance • The diffusion capacitance is due to charge carrier storage occurring in QNRs (to maintain quasi-neutrality) talarico@gonzaga.edu 58 0 −xp −Wp p(−xp) p p-QNR Na qPp qNp n(−xp) ni 2 Na x Majority (+) Minority (−) Majority (−) Minority (+) qPn = qA 2 Wn − xn ( ) ni 2 Nd e VD VT − ni 2 Nd ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = −qNn n(−xp ) = ni 2 Na e VD VT What happen to the majority carriers in the QNRs ? minority carriers charge in QNRs? p(xn ) = ni 2 Nd e VD VT Let’s compute charge in minority carriers “triangles”: qNp = − qA 2 Wp − xp ( ) ni 2 Na e VD VT − ni 2 Na ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = −qPp Quasi-neutrality demands that at every point in the quasi-neutral regions: excess minority carrier concentration = excess majority carriers concentration n
Diffusion capacitance • Let’s examine what happen when we apply a small increase in voltage (vD=VD+vd=VD+ΔvD) talarico@gonzaga.edu 59 Small increase in VD Small increase in qPn Small increase in qNn Behaves as a capacitor of capacitance: n(xn) [for VD+vd] n(xn) [for VD] n p p(xn) [for VD+vd] p(xn) [for VD] Cdn = dqPn dvD vD=VD = qA 2 Wn − xn ( ) ni 2 Nd 1 VT e VD VT
• If we recall the equation of the short-base diode: • using some math manipulation we can rewrite the diffusion capacitance of the n-QSR region as a function of the portion of diode current due to the holes in n-QNR (ID,p) : • We define transit time of the holes through the n-QNR the average time for a hole to diffuse through N-QNR: talarico@gonzaga.edu 60 Cdn = qA 2 Wn − xn ( ) ni 2 Nd 1 VT e VD VT Wn − xn ( ) Wn − xn ( ) Dp Dp = 1 VT Wn − xn ( ) 2 2Dp qA Wn − xn ( ) Dp ni 2 Nd e VD VT ≅ 1 VT Wn − xn ( ) 2 2Dp ID,p ID = Aq Dn WP − xp ni 2 Na + Dp WN − xn ni 2 Nd ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ eVD /VT −1 ( )= ID,n (−xp )+ ID,p (xn ) Diffusion capacitance τTp ≡ Wn − xn ( ) 2 2Dp
• Then in summary: • Aside: – we already had a name for the average time it takes a hole to diffuse through an n-region. It is the hole’s lifetime τp – In fact, if we compute the amount of excess charge given by the holes in n-QSR accurately (see slide 39): talarico@gonzaga.edu 61 Diffusion capacitance τp = Lp 2 Dp Cdn ≅ τTp ID,p VT qPn = qA pn (x') 0 ∞ ∫ ⋅dx' = qA Δpne − x' Lp 0 ∞ ∫ ⋅dx' = qA pn0 eVD /VT −1 ( )e − x' Lp 0 ∞ ∫ dx' = = qApn0 eVD /VT −1 ( ) e − x' Lp 0 ∞ ∫ dx' = qApn0 eVD /VT −1 ( )Lp
– Therefore: – Similarly for the excess charge and capacitance in the p-QNR – The total excess charge stored and the associated total capacitance is: talarico@gonzaga.edu 62 Diffusion capacitance τp ≡ Lp 2 Dp Cdn = dqPn dvD vD=VD = qA ni 2 Nd Lp VT eVD /VT = Lp 2 Dp 1 VT qA ni 2 Nd Dp Lp eVD /VT ≈ τp ID,p VT qNp = −qA np (x'') 0 ∞ ∫ ⋅dx'' = −qA Δnpe − x'' Ln 0 ∞ ∫ ⋅dx'' = −qAnp0 eVD /VT −1 ( )Ln Cdp = −dqNp dvD vD=VD = dqNp dvD vD=VD = Ln 2 Dn 1 VT qA ni 2 Na Dn Ln eVD /VT ≈ τn ID,n VT qS = qPn + qNp = τpID,p +τnID,n Note: The increment in stored electron charge is negative, but results from electrons that are injected from the cathode (N-side) to the anode (P-SIDE) CS = dqS dvD vD=VD = Cdn +Cdp ≈ τn ID.n VT +τp ID. p VT Note: The increment in the stored hole charge comes from injected holes from the P-SIDE (anode) to the N-side (cathode)
Diffusion capacitance talarico@gonzaga.edu 63 0 −xp −Wp p(−xp) for VD p p-QNR Na −dqNp n(−xp) for VD ni 2 Na x n p(−xp) for VD+vd +dqPp n(−xp) for VD+vd n(xn) [for VD+vd] n(xn) [for VD] n p p(xn) [for VD+vd] p(xn) [for VD] −dqNn +dqPn
Diffusion capacitance: physical intuition • If we had looked more carefully at the expressions of the charge stored in the QNRs rather than immediately starting to crunch equations we would have noticed that: • Therefore, the stored charge qS, is proportional to the diode’s current ID, which is also proportional to exp(VD/VT)−1 • The constant of proportionality is called charge-storage time or transit time talarico@gonzaga.edu 64 −qNp = qPp ∝ e VD VT −1 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ qPn = −qNn ∝ e VD VT −1 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ qS = qPn + qNp ∝ e VD VT −1 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟∝ ID ⇔ qS = τT ID CS ≅ τT ID VT ≅ τT gd = τT rdiode
• There is a very simple explanation to this proportionality, ID is the rate of minority charge injection into the diode. • In steady state. This rate must be equal to the rate of charge recombination, which is qS/τT • In a one-sided junction the transit time τT is the recombination lifetime on the lighter-doping side, where most of the charge injection and recombination take place • In general, τT is an average of the recombination lifetimes on the N-side and P- side • In any event, ID and qS are simply linked through a charge storage time τT (a.k.a. τS) talarico@gonzaga.edu 65 Diffusion capacitance: physical intuition
Example • NA >> ND – N-side is more lightly doped than the P-side – Diffusion of minority carriers on N-side dominates – The excess charge stored is: – And finally: talarico@gonzaga.edu 66 Diffusion capacitance: physical intuition ID ≈ ID,p (xn ) ≈ qA Dp Lp pn0e VD VT = qA Dp Lp ni 2 Nd e VD VT τp ≡ holelifetime ≡ Lp 2 Dp τS ≈ τp CS ≈ τp ID VT ≈ τp × gD qS ≈ qPn ≈ τpID,p (xn ) ≈ τpID
Bias Dependence of CJ and CS • CJ (associated with charge modulation in depletion region) dominates in reverse bias and small forward bias • CS (associated with charge storage in QNRs to maintain quasi-neutrality) dominates in strong forward bias talarico@gonzaga.edu 67 source: Sodini Cdiode Cdiode CS CJ Vdiode CS ∝e VD VT CJ ∝ 1 ΦB −VD
source: Sedra & Smith Summary: B = 5×1015 cm−3K−3/2 ni = 1×1010 cm−3 Eg = 1.12 eV
Summary: source: Sedra & Smith
The Five Semiconductor Equations • Poisson’s Equation: • Electron current density: • Hole current density: • Continuity of electrons: • Continuity of holes: talarico@gonzaga.edu 70 d2 Φ dx2 = − ρ ε = − q ε p − n + ND − NA ( ) E(x) = − dΦ dx Gauss' Law Jn = qµnE + qDn dn dx Jp = qµpE − qDp dp dx ∂n ∂t = 1 q ∂Jn ∂x + Gn − Rn ( ) ∂p ∂t = 1 q ∂Jp ∂x + Gp − Rp ( ) d2 Φ dx2 = − dE dx = − ρ ε
SPICE Diode model talarico@gonzaga.edu 71 source: Antognetti & Massobrio CJ(0) τT ΦB
SPICE characteristic of diode talarico@gonzaga.edu 72 SPICE source: Antognetti & Massobrio Example: D1 1 2 diode_1N4004 .model diode_1N4004 D (IS=18.8n RS=0 BV=400 + IBV=5.00u CJO=30 M=0.333 N=2) D2 3 4 idealmod_d .model idealmod_d D + IS=10f + n = 0.01 + IBV=0.1n * By default BV = inf General Sxntax: D[name] <anode_node> <cathode_node> <modelname> .model <modelname> D (<paramX>=<valueX> <paramY>=<valueY> ...)

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ch3_pnjunction.pptx.pdf

  • 1.
    Electronics I -PN Junction talarico@gonzaga.edu 1 p++ n NA >> ND A K Chapter 3 Fall 2017
  • 2.
    PN Junction (Diode) talarico@gonzaga.edu2 • When N-type and P-type dopants are introduced side-by- side in a semiconductor, a PN junction or a diode is formed. source: Razavi JUNCTION
  • 3.
    Diode I/V characteristics •Our goal is to find out the current-voltage characteristics of the device • Diode operation regions: – Equilibrium = open circuit terminals = no bias – Forward bias – Reverse bias – (Breakdown) talarico@gonzaga.edu 3 D D Equilibrium D D “smoke” ΦB
  • 4.
    PN Junction inEquilibrium (no Bias applied) • Because each side of the junction contains an excess of holes or electrons compared to the other side, there exists a large concentration gradient. Therefore, a diffusion current flows across the junction from each side. talarico@gonzaga.edu 4 source: Razavi nn0 pn0 np0 pp0 + ND < NA nn0 ≈ ND pn0 ≈ ni 2/ND pp0 ≈ NA np0 ≈ ni 2/NA
  • 5.
    PN junction inEquilibrium (no bias applied) • Diffusion across the junction causes a lot of electrons and holes to recombine leaving a depletion region in proximity of the junction • The fixed ions left behind create an electric field • The electric field opposes the diffusion of holes in the N region and free electrons in the P region • Once the electric field is strong enough to stop the diffusion currents the junction reaches equilibrium – A “barrier” opposing diffusion forms 5 − − − − − − − − − − − − − − − − − − + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + − − − − + + + + Free Electrons Positive Donor Ions Negative Acceptor Ions Free Holes Depletion Region P N ND < NA
  • 6.
    Depletion approximation • Whathappens inside the PN junction in equilibrium is not quite as black and white as depicted talarico@gonzaga.edu 6 ρ(x) = q p0 (x)− n0 (x)+ ND − NA ( ) • On the N-side of the transition region (0 < x < xn0), since there are no acceptors (NA = 0) and the hole concentration is negligible (p0 ≈ 0) • Since n0(x) < ND on the N-side of the transition (and even more so on the P-side of the transition) there is a roll-off in electron concentration ρ(x) = q p0 (x)− n0 (x)+ ND − NA ( )≈ q −n0 (x)+ ND ( ) source: Howe & Sodini
  • 7.
    Depletion approximation talarico@gonzaga.edu 7 •The depletion approximation assumes that the mobile carriers (n and p) are small in number compared to the donor and acceptor ion concentrations (ND and NA) in the depletion region (SCR), and the device is charge-neutral elsewhere (QNR) • Depletion approximation – NA >> pp hence ρ ≈ −qNA for −xp ≤ x ≤ 0 – ND >> nn hence ρ ≈ qND for 0 ≤ x ≤ xn – The charge density is negligible in the bulk regions; that is ρ ≈ 0 for x > xn and x < − xp • On the P-side of the transition region (−xp < x < 0), since there are no donors (ND = 0) and the electrons concentration is negligible (n0 ≈ 0) • Since p0(x) < NA on the P-side of the transition region (and even more so on the N-side of the transition) there is a roll-off in hole concentration ρ(x) = q p0 (x)− n0 (x)+ ND − NA ( )≈ q p0 (x)− NA ( ) source: Howe & Sodini That is, there are no free mobile charges in depletion region
  • 8.
    One more time… • aa talarico@gonzaga.edu 8 source: Sodini (P-side) (N-side) = 0 = 0 ≈ 0 (minority carriers) ≈ 0 (minority carriers) Na >> p0 (depletion approximation) Nd >> n0 (depletion approximation) P-type N-type NOTE: for the example in this picture NA < ND
  • 9.
    Depletion approximation talarico@gonzaga.edu 9 Depletionapproximation means we assume the charge density in each transition region to be uniform Charge density x qND −qNA − + source: G.W. Neudeck
  • 10.
    PN junction inEquilibrium (no bias applied) • Since the circuit is open no currents exists • Therefore in equilibrium the drift currents resulting from the electric field generated by the fixed ions in the depletion region must exactly cancel the diffusion currents • This must hold for both types of charge carriers: Jn = Jp = 0 • Let’s take one of the two equations and try to solve it: 10 − − − − − − − − − − − − − − − − − − + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + − − − − + + + + Free Electrons Positive Donor Ions Negative Acceptor Ions Free Holes Depletion Region P N ND < NA Jp = 0 = pqµpEx − qDp dp dx Jn = 0 = nqµnEx + qDn dn dx 0 nn0 pn0 pp0 np0 JUNCTION x xp0 −xn0 −pµp dV dx = Dp dp dx pqµpEx = qDp dp dx Ex = − dV dx
  • 11.
    Built-in voltage (equilibrium) talarico@gonzaga.edu11 −pµp dV dx = Dp dp dx − dV V (−xn0) V (xp0) ∫ = Dp µp dp p p(−xn0) p(xp0) ∫ V(−xn0 )−V(xp0 ) =VT ln pp0 pn0 ΦB =VT ln NA ni 2 / ND ! ΦB =VT ln NA ND ni 2 nn0 pn0 pp0 np0 x xp0 −xn0 P N + + − − Ex 0 ΦB + − + + − − Example Find built in voltage of a silicon PN junction at room temperature and with NA ≈ 1018 and ND ≈ 1017 ΦB ≅ 26mV × ln 1018 1017 1020 ≅ 26mV × ln(10)× Log10 1018 1017 1020 ≅ ≅ 60mV ×15 ≅ 900mV depletion region
  • 12.
    PN Junction inequilibrium (no bias applied) talarico@gonzaga.edu 12 − − − − − − − − − − − − − − − − − − + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + − − − − + + + + Depletion Region P N + − • Some of the thermally generated holes in the N region (minority carrier) moves toward the junction and reach the edge of the depletion region. There, they experience the strong electric force present in depletion region, and get swept all the way into the P region (where they become majority carriers). • The same happen for the thermally generated minority carriers (free electrons) in the P region. • At equilibrium the flow of carriers swept by the electric field (drift) and the flow of carriers due the gradient in carriers concentration (diffusion) cancel out ID = Idiff – Idrift =0 Ex Idrift Idiff + − ΦB P N ID − − − − − − + + + + + + + + + VD
  • 13.
    Width of theDepletion region in equilibrium 13 − − − − − − − − − − − − + + + + + + + + + + + + + + + + + + + + + + + + − − − − + + + + P N −xn0 +xp0 0 E metal contact A K metal contact −xn0 +xp0 x concentration [cm−3] pp0=NA np0≈ni 2/NA nn0=ND pn0≈ni 2/ND W=Xd0 Idrift Idiff −xn0 x +xp0 ρ = charge density [Cb/cm−3] −qNA +qND |Q −|= qNAAxp0 |Q +|= qNDAxn0
  • 14.
    Width of theDepletion region in equilibrium talarico@gonzaga.edu 14 −xn0 x +xp0 ρ = charge density [Cb/cm−3] −qNA +qND |Q −|= qNAAxp0 |Q +|= qNDAxn0 Charge conservation: |Q+ |=|Q− | ! QJ 0 " AqND xn0 = AqNAxp0 " ND xn0 = NAxp0 E(x) x −xn0 +xp0 E(0) = −qNAxp0 εsi = −qND xn0 εsi ! Emax PoissonEquation: dE dx = ρ εsi ! dE = ρ εsi dx As long as we know how to compute the area of a rectangle we are fine with the math !!! Permittivity of crystalline silicon: εsi ≈ 11.7×8.85×10−12 F/m εr ε0
  • 15.
    • Let’s gofrom the electric field to the voltage talarico@gonzaga.edu 15 Width of the Depletion region in equilibrium − E(x) x −xp0 +xn0 qNAxp0 εsi = qND xn0 εsi E(x) = − dΦ0 dx ⇔ − E(x)⋅dx = dΦ0 Φ0(x) x Area “red” triangle 1 2 qND xn0 2 εsi Area “blue” triangle = Area “blue” triangle 1 2 qNAxp0 2 εsi Area “red” triangle = 0 As we move in the x direction (from –xp0 to xn) the area under –E(x) keeps growing NOTE: for convenience I flipped the orientation of the p-side and the n-side with respect to the x direction so the name of the depletion region’s edge at the p-side becomes –xp0 and the name of the depletion region’s edge at the n-side becomes +xn0 ΦB=Φ0 (xn0)−Φ0 (−xp0) ΦB = 1 2 qNAxp0 2 εsi + 1 2 qND xn0 2 εsi ΦB + − E − + + parabola N parabola P source: Howe & Sodini −
  • 16.
    Metal-Semiconductor Contact Potential •At first glance would seem strange that the potential drops at the contacts would have to add up to the built in potential of the junction. Were this is not the case, KVL would imply that current would flow in thermal equilibrium when a resistor is connected across the metal contacts to the p and n region. This conclusion violate our basic understanding of the meaning of equilibrium: namely the net current must be zero in equilibrium talarico@gonzaga.edu 16 source: Howe & Sodini
  • 17.
    talarico@gonzaga.edu 17 Width ofthe Depletion region in equilibrium W = xp0 + xn0 W = xn0 1+ ND NA ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = xn0 NA + ND NA ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ W = xp0 1+ NA ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = xp0 NA + ND ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ NAxp0 = ND xn0 Charge neutrality: xn0 = W NA NA + ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ xp0 = W ND NA + ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ΦB = qNA 2εsi xp0 2 + qND 2εsi xn0 2 ΦB = qNA 2εsi W 2 ND NA + ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 + qND 2εsi W 2 NA NA + ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 ΦB = q 2εsi W 2 NAND 2 + ND NA 2 NA + ND ( )2 W 2 = 2εsi q ΦB NA + ND ( )2 NAND 2 + ND NA 2 = 2εsi q ΦB NA + ND ( )2 NAND (ND + NA ) W = 2εsi q ΦB NA + ND NAND ! Xdo xn0 = 2εsi q ΦB NA + ND ( ) NAND NA 2 (NA + ND )2 = 2εsi q ΦB NA ND (NA + ND ) xp0 = 2εsi q ΦB ND NA (NA + ND )
  • 18.
    Charge on eitherside of the depletion region (in equilibrium) talarico@gonzaga.edu 18 QJ 0 !|Q+ |=|Q− |= AqND xn0 xn0 x −xp0 ρ = charge density [Cb/cm−3] −qNA +qND |Q −|= qNAAxp0 |Q +|= qNDAxn0 QJ 0 = AqNDW NA NA + ND = Aq NAND NA + ND W QJ 0 = Aq NAND NA + ND 2εsi q ΦB NA + ND NAND = A 2εsiq NAND NA + ND ΦB Junction Charge xn0 = W NA NA + ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ xp0 = W ND NA + ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = xn0 = W=Xd0
  • 19.
    Effect of biasvoltage on the PN Junction talarico@gonzaga.edu 19 source: Streetman ΦB Φp Φn ΦJ=ΦB−VF ΦJ=ΦB+VR VF=VD VR=−VD + VD − + VD − + VD= 0 − Idiff Idrift Idiff Idrift Idrift Idiff ID = 0 ID > 0 ID < 0 ID = Idiff − Idrift The drift current has the same direction of the electric field (and it is proportional to it) (a) Open circuit (b) Forward Bias (c) Reverse Bias (Equilibrium) Jp,drift = pqµpE Jn,drift = nqµnE ΦB ΦJ ΦJ
  • 20.
    PN Junction underReverse Bias • The external voltage VR is directed from the N side to the P side (same as ΦB), therefore the electric field (ER) due to the bias VR enhances the built in field E. The depletion region W is widened • Since the “barrier” potential rises higher than in equilibrium it gets harder for the majority carriers to cross the junction and easier for the minority carrier to be swept (drifted) across the junction. • The current carried under reverse bias is negligible the minority carriers are too few !! talarico@gonzaga.edu 20 VR = −VD VR = −VD ID − ΦB + − ΦB + source: Millman & Halkias source: Razavi Equilibrium (open circuit) Reverse Bias + ΦB − E E ER + ΦJ − ΦJ = ΦB + VR
  • 21.
    PN junction underreverse bias talarico@gonzaga.edu 21 KVL: VD + Φmn + ΦJ + Φmn = 0 ΦJ = −VD − Φmn − Φmp = ΦB−VD = ΦB potential barrier in equilibrium potential barrier in non-equilibrium source: Howe & Sodini
  • 22.
    PN junction underreverse bias 22 Example: PN junction under reverse biases VD=−2.4 V and VD=−6.4V Assuming the potential barrier in equilibrium is ΦB = 0.8V As the PN junction gets more reverse biased • the depletion region Xd = xp + xn becomes wider • the potential barrier becomes stronger (a) charge density (b) electric field (c) potential source: Howe & Sodini
  • 23.
    Application of ReverseBiased PN Junction: Voltage Dependent capacitor • A reverse biased PN junction can be used as a capacitor. By varying VR, the depletion width changes, and so does the capacitance associated. Therefore a reverse biased PN junction can be seen as a voltage- dependent capacitor. talarico@gonzaga.edu 23 Varactor p n Qualitatively we see that if VR goes up then CJ goes down source: Razavi
  • 24.
    Capacitance of ReverseBiased Diode talarico@gonzaga.edu 24 Xd @VD = 2εsi q ΦB −VD ( ) NA + ND NAND = 2εsi q ΦB 1− VD ΦB ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ NA + ND NAND = Xd0 1− VD ΦB CJ @VD = εsi ⋅ A Xd @VD = qεsi 2 NAND NA + ND ( ) 1 ΦB −VD ( ) = qεsi 2 NAND NA + ND ( ) 1 ΦB 1 1−VD / ΦB ( ) = = CJ0 1−VD / ΦB width of depletion region in equilibrium CJ0 = junction capacitance at zero bias (in equilibrium) = CJ0 = εsi A Xdo careful: cap per unit area
  • 25.
    Junction capacitance ofreverse biased diode talarico@gonzaga.edu 25 CJ0 VR=−VD VD= VF Reverse Voltage Forward Voltage CJ source: Streetman CJ (VD ) = CJ0 1− VD ΦB ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ M = CJ0 1+ VR ΦB ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ M • M=0.5 for the abrupt PN junction we modeled • Typical values for M range from 0.3 to 0.5 • SPICE uses VJ (or PB) to denote the built in voltage in equilibrium (ΦB) The junction capacitance is non linear
  • 26.
    talarico@gonzaga.edu 26 QJ @VD =A 2εsiq NAND NA + ND ΦB −VD ( ) = A 2εsiq NAND NA + ND ΦB 1− VD ΦB ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = QJ 0 1− VD ΦB = = QJ 0 1+ VR ΦB charge stored in either side of the depletion region in equilibrium Junction capacitance of reverse biased diode • The relation between charge QJ and voltage (VR=−VD) across the diode is not linear • It is difficult to define a capacitance that changes whenever VR changes source: Sedra & Smith P VP QJ = A 2εsiq NAND NA + ND ΦB +VR ( ) = ς ΦB +VR = ς 2 ΦB 1+ VR ΦB ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = CJO 1+ VR ΦB CJ @VP = dQJ dVR VR =VP = d dVR ς ΦB +VR ( ) @VP = ς 2 ΦB +VR @VP = = ς @VP @VP square root law 𝐶"|@%& = 1 𝐴 𝑑𝑄" 𝑑𝑉- . /01/& /A = 2 3 /A /A
  • 27.
    talarico@gonzaga.edu 27 Junction capacitanceof reverse biased diode source: Sedra & Smith P VP CJ @VP = dQJ dVR VR =VP ≠ QJ (VR ) VR VR =VP Important result to remember when using non linear devices /A
  • 28.
    Application of areverse biased diode talarico@gonzaga.edu 28 LC fres 1 2 1 p = source: Razavi
  • 29.
    PN Junction inForward Bias talarico@gonzaga.edu 29 ID VD=VF VD=VF − ΦB + − ΦB + source: Millman & Halkias • The electric field (EF) due to the bias (VF) is opposite to the built in field E. The depletion region W is shortened • Since the “barrier” potential becomes lower than in equilibrium it gets easier for the majority carriers to diffuse through the junction and harder for the minority carrier to be swept (drifted) across the junction. • The majority carriers are a lot, so we expect a considerable current flow: ID = Idiff − Idrift Equilibrium (open circuit) + ΦB − + ΦJ − Forward Bias EF E ΦJ = ΦB − VF source: Razavi p-n ID
  • 30.
    Diode’s Current/Voltage Characteristic talarico@gonzaga.edu30 ID = IS (e VD VT −1) ≈ −IS NOTE: Consider |VD/VT| ≈ 2.3 exp(2.3) ≈ 9.97 ≈ 10 exp(−2.3) ≈ 0.1 At room temperature VD = 2.3 × VT ≈ 59.8 mV≈ 60 mV source: Razavi source: Streetman −I(drift) ID VD VD ID ID = I(drift) e qVD KT −1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ very steep !
  • 31.
    Diode’s Current/Voltage Characteristic 31 •How can we quantify the I/V behavior of the diode ? • We need to look at how electrons and holes move under the effect of the bias voltage • In equilibrium: • With an external bias voltage applied the situation is similar but now the potential barrier is ΦB – VD instead of only ΦB ΦB = VT ln pp0 pn0 ⇔ pn0 = pp0 ⋅e−ΦB /VT ⇔ pn0 = pp0 eΦB /VT holes at xn0 holes at −xp0 −xp0 and xn0 are the edges of the depletion region in equilibrium pn = pp ⋅ eVD /VT eΦB /VT holes at xn holes at −xp −xp and xn are the edges of the depletion region with non zero bias x −Wp −xp 0 xn Wn WP WN P-side N-side + + − − VD GND A K quasi-neutral region quasi-neutral region Law of the Junction np = nn ⋅ eVD /VT eΦB /VT electrons at −xp electrons at xn The Law of the Junction relates the minority carrier concentration on one side of the junction to the majority carrier concentration on the other side of the junction np0 = nn0 eΦB /VT electrons at −xp0 electrons at xn0
  • 32.
    Diode’s Current/Voltage Characteristic talarico@gonzaga.edu32 • When we apply a forward bias VD the potential barrier is reduced from ΦB to ΦB−VD. This reduces the drift field and upsets the balance between diffusion and drift that exists at zero bias. • Electrons can now diffuse from the N side (where they are majority carriers) to the P side (where they become minority carriers). Similarly, holes diffuse from the P side into the N side. This migration of carriers from one side to the other is called injection. • As a result of injection, more electrons are now present at –xp and more holes appear at xn than when the barrier was higher – There is an excess of minority carriers at the edges of the depletion region • Let’s make some simplifying assumptions to help quantifying pn(xn), pp(−xp), nn(xn) and np(xp) – There is electric field only in the space charge region (that is between –xp and xn). The regions from –Wp to –xp (a.k.a. P bulk region) and from xn to Wn (a.k.a. N bulk region) are quasi-neutral (it is like they were perfect conductors and in perfect conductors there is no electric field inside)
  • 33.
    Diode’s Current/Voltage Characteristic talarico@gonzaga.edu33 • The requirement that the bulk regions on either side remain charge-neutral even under applied bias (remember: E(x) = ρ(x)/ε, so no field means no charge density) implies that any increases in the minority carriers due to injection must be counter balanced by in an increase of majority carriers – On the P side ρ(x=−xp) = 0 implies that: – On the N side ρ(x=xn) = 0 implies that: • Assuming the level of injection is low we can neglect the slight increase in majority carrier concentration due to carrier injection and write: pp (−xp ) = NA + np (−xp ) ≈ pp0 + np (−xp ) nn (xn ) = ND + pn (xn ) ≈ nn0 + pn (xn ) pp (−xp ) ≈ pp0 ≈ NA nn (xn ) ≈ nn0 ≈ ND np (−xp ) << pp0 pn (xn ) << nn0
  • 34.
    Minority and Majoritycarriers concentrations in a forward biased diode • Concentrations Howe & Sodini talarico@gonzaga.edu 34 x pn0 np0 Quasi-neutrality demands that at every point in the quasi-neutral regions: excess minority carrier concentration = excess majority carriers concentration source: Howe & Sodini
  • 35.
    talarico@gonzaga.edu 35 Minority andMajority carriers concentrations in a forward biased diode Quasi-neutrality demands that at every point in the quasi-neutral regions: excess minority carrier concentration = excess majority carriers concentration NOTE: at this point we did not put much thinking in the profile of the minority carriers diffusing in the QNRs, however if we assume the diffusion occurs over a short distance we can approximate the curves with straight lines ! 0 −xp −Wp p(−xp) p p-QNR Na qPp qNp n(−xp) ni 2 Na x source: Sodini n
  • 36.
    • Under thelow injection condition we can approximate the Law of the Junction as follows: talarico@gonzaga.edu 36 pn (xn ) = pp (−xp ) eΦB /VT ⋅eVD /VT ≈ pp0 eΦB /VT ⋅eVD /VT = pn0 ⋅eVD /VT np (−xp ) = nn (xn ) eΦB /VT ⋅eVD /VT ≈ nn0 eΦB /VT ⋅eVD /VT = np0 ⋅eVD /VT pn (xn ) ≈ pn0 ⋅eVD /VT ≈ ni 2 ND ⋅eVD /VT np (−xp ) ≈ np0 ⋅eVD /VT ≈ ni 2 NA ⋅eVD /VT The concentration of minority carriers at the edges of the space charge region are an exponential function of the applied bias Diode’s Current/Voltage Characteristic source: G.W. Neudeck
  • 37.
    Minority carriers ina forward biased PN junction • The excess concentration of minority carriers at the edges is: • As the injected carriers diffuse through the quasi-neutral regions they recombine with the majority carriers and eventually disappear. The excess minority carriers decay exponentially with distance, and the decay constant is called diffusion length L. talarico@gonzaga.edu 37 Δnp ! np0 ⋅eVD /VT − np0 = np0 1− eVD /VT ( ) Δpn ! pn0 ⋅eVD /VT − pn0 = pn0 1− eVD /VT ( ) np (x) = Δnpe (x+xp ) Ln pn (x) = Δpne −(x−xn ) Lp source: Sedra & Smith p+ n It can be shown through the current continuity equation that under the low injection condition the concentration’s profile of the minority carriers diffusing in the quasi neutral regions is exponential
  • 38.
    Minority carrier concentrations Equilibriumvs. Forward Bias talarico@gonzaga.edu 38 source: G.W. Neudeck
  • 39.
    Aside (a simpletrick to simplify the math) talarico@gonzaga.edu 39 • Change the axis selection for the bulk regions as follows: • This way we can write the minority carriers concentrations as follows: np (x'') = Δnpe − x'' Ln pn (x') = Δpne − x' Lp x'' ! 0'' ≡ −xp x' ! 0' ≡ xn Ln = Dnτn Lp = Dpτ p source: G.W. Neudeck source: G.W. Neudeck Ln = Diffusion length for electrons [m] Lp = Diffusion length for holes [m] Dn = Diffusion coefficient for electron [m2/s] Dp = Diffusion coefficient for holes [m2/s] τn = mean lifetime for electrons [s] τp = mean lifetime for holes [s] VA = voltage applied ≡ VD
  • 40.
    Physical interpretation ofL, D, and τ • Physically Lp and Ln represents the average distance minority carriers can diffuse into a “sea” of majority carriers before being annihilated • Physically τp and τn represents the average time minority carriers can diffuse into a “sea” of majority carriers before being annihilated • Physically the Dp and Dn represents the ease with which holes and electrons diffuse talarico@gonzaga.edu 40 Lp ! xΔpn (x) 0 ∞ ∫ dx Δpn (x) 0 ∞ ∫ dx Ln ! xΔnp (x) 0 ∞ ∫ dx Δnp (x) 0 ∞ ∫ dx Dp ! µp KT q Dn ! µn KT q
  • 41.
    PN Junction current •The current that flows through any “transverse” section x must be the same and is formed by both holes and electrons • If we look at the current through a section x in the quasi-neutral regions, since in the quasi-neutral regions there is no electric field there will be no drift talarico@gonzaga.edu 41 J(x) = Jn (x)+ Jp (x) Jn (x) = Jn,drift (x)+ Jn,diff (x) Jp (x) = Jp,drift (x)+ Jp,diff (x) Jn (x) = Jn,diff (x) = qDn dnp (x) dx = q Dn Ln np0 eVD /VT −1 ( )e (x+xp )/Lp for −Wp ≤ x ≤ −xp Jp (x) = Jp,diff (x) = −qDp dpn (x) dx = q Dp Lp pn0 eVD /VT −1 ( )e −(x−xn )/Lp for xn ≤ x ≤ Wn injected electrons diffusing on P-side (minority carriers) injected holes diffusing on N-side (minority carriers) +(x+xp)/Ln
  • 42.
    PN Junction currents •The easier sections for evaluating the diffusion currents are the edges of the space charge region (x=−xp and x=xn) • The total current through a section of the diode can be found as follows: • If we make the simplifying assumption that there is no recombination-generation in the space-charge region, this means that the electrons/holes current at x=−xp and at x=xn do not change in the space-charge region and therefore we can write: talarico@gonzaga.edu 42 Jn (−xp ) = Jn,diff (−xp ) = q Dn Ln np0 eVD /VT −1 ( ) Jp (xn ) = Jp,diff (xn ) = q Dp Lp pn0 eVD /VT −1 ( ) J = Jn (−xp )+ Jp (−xp ) = Jn (xn )+ Jp (xn ) Jn (−xp ) = Jn (xn ) Jp (xn ) = Jp (−xp ) We do not know have this value (yet) ! We do not know have this value (yet) !
  • 43.
    PN Junction current •Therefore we can add together the diffusion currents at x=−xp and x=xn 43 J = Jn (−xp )+ Jp (xn ) = q Dn Ln np0 eVD /VT −1 ( )+ q Dp Lp pn0 eVD /VT −1 ( ) J = q Dn Ln np0 + q Dp Lp pn0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ eVD /VT −1 ( ) ID = J × A = In,diff + Ip,diff = Aq Dn Ln np0 + Aq Dp Lp pn0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ eVD /VT −1 ( ) ID = Aq Dn Ln ni 2 NA + Dp Lp ni 2 ND ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ eVD /VT −1 ( )= IS eVD /VT −1 ( ) scale current (a.k.a. saturation current)
  • 44.
    Minority concentrations fora forward biased short base diode • If the diode is short the exponential decays can be approximated with straight lines talarico@gonzaga.edu 44 −xp 0 xn np0 pn0 −Wp Wn VD A GND K WidthN WidthP x ID = Aq Dn WidthP np0 + Dp WidthN pn0 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ eVD /VT −1 ( )= IS eVD /VT −1 ( ) dnp dx ≈ np (−xp )− n(−Wp ) WidthP = np0eVD /VT − np0 WidthP dpn dx ≈ pn (xn )− pn (Wn ) −WidthN = − pn0eVD /VT − pn0 WidthN
  • 45.
    Current densities inthe forward bias case talarico@gonzaga.edu 45 source: G.W. Neudeck No matter which section x we consider Jp and Jn must add to the same total value J
  • 46.
    Assumptions used toderive the Ideal diode equation • Depletion approximation (in depletion region: NA >> pp0(x), ND >> nn0(x)) • Negligible fields in the quasi neutral regions (QNR). This means that the applied voltage VD is dropped entirely across the space charge region (SCR) • Quasi-neutral regions that are so long that the injected carriers all recombine before reaching the end contacts • Low levels of injection • No recombination-generation in the space-charge region. This means that the electron current and the hole current are constant in the space- charge region ( that is Jn(−xp)=Jn(xn) and Jp(xn)=Jp(−xp) ) • Real diodes may deviate considerably from the ideal equation. 46 ID = I* S (e VD nVT −1) n = emission coefficient (a.k.a. ideality factor) Typically the ideality factor n varies from 1 to 2
  • 47.
    Ideal diode talarico@gonzaga.edu 47 ID IS VDVD log10|ID| IS y = log10 IS ⋅e VD VT ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = log10 IS ( )+ log10 (e)⋅ VD VT ≈ log10 IS ( )+ 0.43 VD VT x =VD dy dx ≈ 0.43 VT at room temperature (300K): VT 0.43 ≈ 26mV 0.43 ≈ 60mV 10 source: Sodini every 60 mV the current changes by a decade
  • 48.
    Deviations from theideal diode talarico@gonzaga.edu 48 source: Hu Ideal with IS VDIODE (V)
  • 49.
    Ideal diode equation •Although we derived the ideal diode equation by subjecting the diode to a forward bias, the steps followed can also be applied for the case of reverse bias (the only difference is that now VD < 0). • The equation works for both forward and reverse region. talarico@gonzaga.edu 49 ID VD ID = IS eVD /VT −1 ( ) source: Streetman ≈ ISeVD /VT ≈ −IS
  • 50.
    Minority concentrations for reversebiased diode • In forward bias, minority carriers are injected into quasi-neutral regions • In reverse bias, minority carriers are extracted (swept away) from the quasi- neutral regions • In reverse bias there is a very small amount of carriers available for extraction – ID saturates to small value talarico@gonzaga.edu 50 Minority-carrier density distribution as a function of the distance from the junction. (a) Forward biased junction (b) Reverse biased junction (The depletion region is assumed so small relative to the diffusion length that is not indicated in the figure) source: Millman & Halkias
  • 51.
    Minority carrier concentrations Equilibriumvs. Reverse Bias talarico@gonzaga.edu 51 source: G.W. Neudeck
  • 52.
    Small-signal equivalent modelfor diode • Examine the effect of a small signal overlapping the bias: talarico@gonzaga.edu 52 Figure 4.14 Development of the diode small-signal model. source: Sedra & Smith vD (t) ≡VD + vd (t) iD = IS e vD VT −1 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = IS e VD+vd VT −1 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ iD ≅ iD VD + diD dvD VD (vD −VD ) = ID + diD dvD VD vd iD is a function of vD: Q ≡ (ID,VD ) diD dvD VD = d dvD IS e vD VT −1 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ VD = ISe VD VT VT = ISe VD VT + IS − IS VT = = ID + IS VT iD − ID ≅ gdvd ⇔ id ≅ gdvd ≡ gd * Applied Taylor *
  • 53.
    Small-signal equivalent modelfor diode • From a small signal point of view the diode behaves as a (non linear) resistance of value: • rdiode depends on bias: • In forward bias: In reverse bias: • Under the assumption that the signal overlapping the bias is small the i-v relationship of the diode can be linearized: talarico@gonzaga.edu 53 Figure 4.14 Development of the diode small-signal model. source: Sedra & Smith rdiode = 1 gd = VT ID + IS iD ≅ ID + gdvd = ID +id vD =VD + vd gd = ID + IS VT gd ≈ ID VT gd ≈ 0 rdiode = 1/gd
  • 54.
    Small-signal equivalent modelfor diode • Besides having current flowing through it (resistive behavior) the diode also exhibit charge displacement mechanisms that can be modeled as a capacitive behavior • The charge response to the voltage applied across the diode is non-linear • As long as the voltage signal superimposed to the bias voltage is small, the relationship between charge and voltage can be linearized. • Complete small signal model for diode talarico@gonzaga.edu 54 vD =VD + vd ≡ Cdiode qD ≈ QD VD + dqD dvD VD (vD −VD ) = QD VD +Cdiode ⋅vd rdiode Cdiode
  • 55.
    talarico@gonzaga.edu 55 Capacitive effectsin the PN junction • Diodes have two capacitive effects • But … please careful: there are way too many letters “D”: – depletion capacitance = junction capacitance = CJ – diffusion capacitance = storage capacitance = CS – Cdiode = CJ + CS • We already know quite a bit about the junction cap. that develops when the diode is in reverse bias (… but what about forward bias?) • So far we did not even think about the existence of storage (diffusion) capacitance • We need to dig more into the capacitive effects … Cdiode A K CJ A K CS
  • 56.
    Junction Capacitance talarico@gonzaga.edu 56 Behaviorof junction capacitance as a function of bias voltage VD (Chawla-Gummel) ≡ ΦB ≡ ΦB 2 SPICE ≈ 2×CJ 0 For Hand Calculation source: Gray, Hurst, Lewis and Meyer
  • 57.
    Junction Capacitance • Earlier,we derived the capacitance associated with depletion region (junction capacitance) for a reverse biased diode as: • Under significant forward bias VD > ΦB/2 the approximation that there are no mobile carriers in the depletion region becomes invalid (this was the basic assumption behind the derivation !!) • Therefore, under forward bias a common approximation is to take the value of CJ at VD = ΦB/2 (or being slightly more pessimistic) talarico@gonzaga.edu 57 Cj VD = CJO 1− VD ΦB CJO = zero-biasjunctioncapacitance = A εsiq 2 ⋅ NA ND NA + ND ⋅ 1 ΦB Cj VD= ΦB 2 = CJO 1−1/ 2 = 2 ×CJO ≈ 2×CJO
  • 58.
    Diffusion capacitance • Thediffusion capacitance is due to charge carrier storage occurring in QNRs (to maintain quasi-neutrality) talarico@gonzaga.edu 58 0 −xp −Wp p(−xp) p p-QNR Na qPp qNp n(−xp) ni 2 Na x Majority (+) Minority (−) Majority (−) Minority (+) qPn = qA 2 Wn − xn ( ) ni 2 Nd e VD VT − ni 2 Nd ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = −qNn n(−xp ) = ni 2 Na e VD VT What happen to the majority carriers in the QNRs ? minority carriers charge in QNRs? p(xn ) = ni 2 Nd e VD VT Let’s compute charge in minority carriers “triangles”: qNp = − qA 2 Wp − xp ( ) ni 2 Na e VD VT − ni 2 Na ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ = −qPp Quasi-neutrality demands that at every point in the quasi-neutral regions: excess minority carrier concentration = excess majority carriers concentration n
  • 59.
    Diffusion capacitance • Let’sexamine what happen when we apply a small increase in voltage (vD=VD+vd=VD+ΔvD) talarico@gonzaga.edu 59 Small increase in VD Small increase in qPn Small increase in qNn Behaves as a capacitor of capacitance: n(xn) [for VD+vd] n(xn) [for VD] n p p(xn) [for VD+vd] p(xn) [for VD] Cdn = dqPn dvD vD=VD = qA 2 Wn − xn ( ) ni 2 Nd 1 VT e VD VT
  • 60.
    • If werecall the equation of the short-base diode: • using some math manipulation we can rewrite the diffusion capacitance of the n-QSR region as a function of the portion of diode current due to the holes in n-QNR (ID,p) : • We define transit time of the holes through the n-QNR the average time for a hole to diffuse through N-QNR: talarico@gonzaga.edu 60 Cdn = qA 2 Wn − xn ( ) ni 2 Nd 1 VT e VD VT Wn − xn ( ) Wn − xn ( ) Dp Dp = 1 VT Wn − xn ( ) 2 2Dp qA Wn − xn ( ) Dp ni 2 Nd e VD VT ≅ 1 VT Wn − xn ( ) 2 2Dp ID,p ID = Aq Dn WP − xp ni 2 Na + Dp WN − xn ni 2 Nd ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ eVD /VT −1 ( )= ID,n (−xp )+ ID,p (xn ) Diffusion capacitance τTp ≡ Wn − xn ( ) 2 2Dp
  • 61.
    • Then insummary: • Aside: – we already had a name for the average time it takes a hole to diffuse through an n-region. It is the hole’s lifetime τp – In fact, if we compute the amount of excess charge given by the holes in n-QSR accurately (see slide 39): talarico@gonzaga.edu 61 Diffusion capacitance τp = Lp 2 Dp Cdn ≅ τTp ID,p VT qPn = qA pn (x') 0 ∞ ∫ ⋅dx' = qA Δpne − x' Lp 0 ∞ ∫ ⋅dx' = qA pn0 eVD /VT −1 ( )e − x' Lp 0 ∞ ∫ dx' = = qApn0 eVD /VT −1 ( ) e − x' Lp 0 ∞ ∫ dx' = qApn0 eVD /VT −1 ( )Lp
  • 62.
    – Therefore: – Similarlyfor the excess charge and capacitance in the p-QNR – The total excess charge stored and the associated total capacitance is: talarico@gonzaga.edu 62 Diffusion capacitance τp ≡ Lp 2 Dp Cdn = dqPn dvD vD=VD = qA ni 2 Nd Lp VT eVD /VT = Lp 2 Dp 1 VT qA ni 2 Nd Dp Lp eVD /VT ≈ τp ID,p VT qNp = −qA np (x'') 0 ∞ ∫ ⋅dx'' = −qA Δnpe − x'' Ln 0 ∞ ∫ ⋅dx'' = −qAnp0 eVD /VT −1 ( )Ln Cdp = −dqNp dvD vD=VD = dqNp dvD vD=VD = Ln 2 Dn 1 VT qA ni 2 Na Dn Ln eVD /VT ≈ τn ID,n VT qS = qPn + qNp = τpID,p +τnID,n Note: The increment in stored electron charge is negative, but results from electrons that are injected from the cathode (N-side) to the anode (P-SIDE) CS = dqS dvD vD=VD = Cdn +Cdp ≈ τn ID.n VT +τp ID. p VT Note: The increment in the stored hole charge comes from injected holes from the P-SIDE (anode) to the N-side (cathode)
  • 63.
    Diffusion capacitance talarico@gonzaga.edu 63 0 −xp −Wp p(−xp)for VD p p-QNR Na −dqNp n(−xp) for VD ni 2 Na x n p(−xp) for VD+vd +dqPp n(−xp) for VD+vd n(xn) [for VD+vd] n(xn) [for VD] n p p(xn) [for VD+vd] p(xn) [for VD] −dqNn +dqPn
  • 64.
    Diffusion capacitance: physicalintuition • If we had looked more carefully at the expressions of the charge stored in the QNRs rather than immediately starting to crunch equations we would have noticed that: • Therefore, the stored charge qS, is proportional to the diode’s current ID, which is also proportional to exp(VD/VT)−1 • The constant of proportionality is called charge-storage time or transit time talarico@gonzaga.edu 64 −qNp = qPp ∝ e VD VT −1 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ qPn = −qNn ∝ e VD VT −1 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ qS = qPn + qNp ∝ e VD VT −1 ⎛ ⎝ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟∝ ID ⇔ qS = τT ID CS ≅ τT ID VT ≅ τT gd = τT rdiode
  • 65.
    • There isa very simple explanation to this proportionality, ID is the rate of minority charge injection into the diode. • In steady state. This rate must be equal to the rate of charge recombination, which is qS/τT • In a one-sided junction the transit time τT is the recombination lifetime on the lighter-doping side, where most of the charge injection and recombination take place • In general, τT is an average of the recombination lifetimes on the N-side and P- side • In any event, ID and qS are simply linked through a charge storage time τT (a.k.a. τS) talarico@gonzaga.edu 65 Diffusion capacitance: physical intuition
  • 66.
    Example • NA >>ND – N-side is more lightly doped than the P-side – Diffusion of minority carriers on N-side dominates – The excess charge stored is: – And finally: talarico@gonzaga.edu 66 Diffusion capacitance: physical intuition ID ≈ ID,p (xn ) ≈ qA Dp Lp pn0e VD VT = qA Dp Lp ni 2 Nd e VD VT τp ≡ holelifetime ≡ Lp 2 Dp τS ≈ τp CS ≈ τp ID VT ≈ τp × gD qS ≈ qPn ≈ τpID,p (xn ) ≈ τpID
  • 67.
    Bias Dependence ofCJ and CS • CJ (associated with charge modulation in depletion region) dominates in reverse bias and small forward bias • CS (associated with charge storage in QNRs to maintain quasi-neutrality) dominates in strong forward bias talarico@gonzaga.edu 67 source: Sodini Cdiode Cdiode CS CJ Vdiode CS ∝e VD VT CJ ∝ 1 ΦB −VD
  • 68.
    source: Sedra & Smith Summary:B = 5×1015 cm−3K−3/2 ni = 1×1010 cm−3 Eg = 1.12 eV
  • 69.
  • 70.
    The Five SemiconductorEquations • Poisson’s Equation: • Electron current density: • Hole current density: • Continuity of electrons: • Continuity of holes: talarico@gonzaga.edu 70 d2 Φ dx2 = − ρ ε = − q ε p − n + ND − NA ( ) E(x) = − dΦ dx Gauss' Law Jn = qµnE + qDn dn dx Jp = qµpE − qDp dp dx ∂n ∂t = 1 q ∂Jn ∂x + Gn − Rn ( ) ∂p ∂t = 1 q ∂Jp ∂x + Gp − Rp ( ) d2 Φ dx2 = − dE dx = − ρ ε
  • 71.
    SPICE Diode model talarico@gonzaga.edu71 source: Antognetti & Massobrio CJ(0) τT ΦB
  • 72.
    SPICE characteristic ofdiode talarico@gonzaga.edu 72 SPICE source: Antognetti & Massobrio Example: D1 1 2 diode_1N4004 .model diode_1N4004 D (IS=18.8n RS=0 BV=400 + IBV=5.00u CJO=30 M=0.333 N=2) D2 3 4 idealmod_d .model idealmod_d D + IS=10f + n = 0.01 + IBV=0.1n * By default BV = inf General Sxntax: D[name] <anode_node> <cathode_node> <modelname> .model <modelname> D (<paramX>=<valueX> <paramY>=<valueY> ...)