ECE PN Junctions and Diodes


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1 ECE PN Junctions and iodes Jose E. SchuttAine Electrical & Computer Engineering University of Illinois ECE 342 Jose Schutt Aine 1
2 B: material dependent parameter = for Si E G : Bandgap energy = 1.12 ev k: Boltzmann constant= ev/k n i : intrinsic carrier concentration At T = 300 K, n i = carriers/cm 3 J p : current density A/m 2 q: electron charge p : iffusion constant (diffusivity) of holes p : mobility for holes = 480 cm 2 /V sec n : mobility for electrons = 1350 cm 2 /V sec N : concentration of donor atoms n no : concentration of free electrons at thermal equilibrium N A : concentration of acceptor atoms p po : concentration of holes at thermal equilibrium n p kt Einstein Relation : VT : thermal voltage q n efinitions p ECE 342 Jose Schutt Aine 2
3 PN Junction When a p material is connected to an ntype material, a junction is formed Holes from ptype diffuse to ntype region Electrons from ntype diffuse to ptype region Through these diffusion processes, recombination takes place Some holes disappear from ptype Some electrons disappear from ntype A depletion region consisting of bound charges is thus formed Charges on both sides cause electric field potential = V o ECE 342 Jose Schutt Aine 3
4 PN Junction Potential acts as barrier that must be overcome for holes to diffuse into the nregion and electrons to diffuse into the pregion Open circuit: No external current Junction builtin voltage From principle of detailed balance and equilibrium we get: NAN Vo VT ln 2 ni For Si, V o is typically 0.6V to 0.8V Charge equality in depletion region gives: qx AN qx AN p A n A: crosssection of junction x p : width in p side x n : width in n side s :silicon permittivity F/m s o x x 2 s 1 1 W x x V q NA N n p N N dep n p o A ECE 342 Jose Schutt Aine 4
5 Example Find the barrier voltage across the depletion region of a silicon diode at T = 300 K with N =10 15 /cm 3 and N A =10 18 /cm 3. Use V o NAN VT ln 2 300K, ni /cm V V T 10 3 V o ln 0.026ln o V o volts o V o o volts ECE 342 Jose Schutt Aine 5
6 PN Junction under Reverse Bias When a reverse bias is applied Transient occurs during which depletion capacitance is charged to new bias voltage Increase of space charge region iffusion current decreases rift current remains constant Barrier potential is increased A steady state is reached After transient: steadystate reverse current = I S I (I is very small) reverse current ~ I S ~1015 A Under reverse bias the current in the diode is negligible ECE 342 Jose Schutt Aine 6
7 epletion Layer Stored Charge q q qn x A j N n A: cross section area q j : stored charge Let W dep = depletionlayer width N N A qj q AWdep NA N The total voltage across the depletion layer is V o + V R 2 s 1 1 W V V q NA N dep o R ECE 342 Jose Schutt Aine 7
8 epletion Capacitance C j dq dv j R V R V Q Q is bias point V R is reverse voltage C j s A W dep C jo V 1 V R o C jo C sq NAN 1 C j A V 2 NA N Vo 1 V jo R o m m is the grading coefficient and depends on how the concentration varies from the p side to the n side 1/3 <m <1/2 For an abrupt junction, m=0.5 ECE 342 Jose Schutt Aine 8
9 ForwardBiased Junction Carrier istribution p n,n p pn(xn ) P Region n p (x p ) epletion Region N Region Excess concentration p n (x) n p (x) p no Thermal Equilibrium Value n po x p 0 x n x N A >> N Barrier voltage is now lower than V o In steady state, concentration profile of excess minority carriers remains constant ECE 342 Jose Schutt Aine 9
10 iode equation: ForwardBiased PN Junction S I I e V/ nv T 1 I S Aqn 2 p n i LN p LN n A since n 2 i is a strong function of temperature; thus I s is a strong function of temperature n has a value between 1 and 2. iodes made using standard IC process have n=1; discrete diodes have n=1 In general, assume n=1 / T If V V, we can use VV T I ISe ECE 342 Jose Schutt Aine 10
11 iode Characteristics Three distinct regions The forwardbias region, determined by v > 0 The reversebias region, determined by v < 0 The breakdown region, determined by v < V ZK ECE 342 Jose Schutt Aine 11
12 iode IV Relationship Breakdown Electric field strong enough in depletion layer to break covalent bonds and generate electronhole pairs. Electrons are then swept by Efield into the n side. Large number of carriers for a small increase in junction voltage ECE 342 Jose Schutt Aine 12
13 The iode + I V  iode Properties Twoterminal device that conducts current freely in one direction but blocks current flow in the opposite direction. The two electrodes are the anode which must be connected to a positive voltage with respect to the other terminal, the cathode in order for current to flow. ECE 342 Jose Schutt Aine 13
14 Ideal iode Characteristics + I + I V V   V < 0 I > 0 OFF ON ECE 342 Jose Schutt Aine 14
15 Ideal iode Characteristics ECE 342 Jose Schutt Aine 15
16 iode Models Exponential Piecewise Linear ConstantVoltagerop 16 ECE 342 Jose Schutt Aine 16
17 Idealdiode iode Models Smallsignal 17 ECE 342 Jose Schutt Aine 17
18 PiecewiseLinear Model for v : 0 V0 i 1 for v V : i v V 0 0 r ECE 342 Jose Schutt Aine 18
19 PiecewiseLinear Model ECE 342 Jose Schutt Aine 19
20 ConstantVoltagerop Model for i 0: v 0.7V ECE 342 Jose Schutt Aine 20
21 ConstantVoltagerop Model ECE 342 Jose Schutt Aine 21
22 iodes Logic Gates OR Function AN Function Y ABC Y ABC ECE 342 Jose Schutt Aine 22
23 iode Circuit Example 1 Assume both diodes are on; then V 0 and V 0 I B ma 10 At node B IEAL iodes 0 ( 10) I 1 I 1 ma, V 0V 5 1 is conducting as originally assumed ECE 342 Jose Schutt Aine 23
24 IEAL iodes iode Circuit Example 2 Assume both diodes are on; then 10 0 I2 2 ma 5 At node B 0 ( 10) I 2 I 1mAwrong 10 original assumption is not correct assume 1 is off and 2 is on V B I V 0 and V 0 2 B 10 ( 10) ma V 1 is reverse biased as assumed ECE 342 Jose Schutt Aine 24
25 Example The diode has a value of I S = ma at room temperature (300 o K) (a) Approximate the current I assuming the voltage drop across the diode is 0.7V (b) Calculate the accurate value of I (c) If I S doubles for every 6 o C increase in temperature, repeat part (b) if the temperature increases by 40 o C (a) The resistor will have an approximate voltage of = 5.3 V. Ohm s law then gives a current of I ma (b) The current through the resistor must equal the diode current; so we have I 6 V 2 ( resistor current) VV / I I T Se diodecurrent ( ) ECE 342 Jose Schutt Aine 25
26 6 V 2 10 e 12 V /0.026 Example (cont d) Nonlinear equation must be solved iteratively Solution: V = V Using this value of the voltage, we can calculate the current I 6 V ma When the temperature changes, both I s and V T will change. Since V T =kt/q varies directly with T, the new value is: V T 340 (340) VT(300) ECE 342 Jose Schutt Aine 26
27 Example (cont d) The value of I s doubles for each 6 o C increase, thus the new value of I s is 40/6 10 IS(340) IS(300) ma The equation for I is then Solving iteratively, we get I 6 V e 2 10 V / V 0.640V and I 2.68 ma ECE 342 Jose Schutt Aine 27
28 Example Two diodes are connected in series as shown in the figure with I s1 =1016 A and I s2 =1014 A. If the applied voltage is 1 V, calculate the currents I 1 and I 2 and the voltage across each diode V 1 and V 2. The diode equations can be written as: I V1 / VT V2 I e / VT 1 S1 I I e 2 S2 S1 1 2 T ln 0.12 IS 2 V 1 V2 1 from whichv V V I I I V1V2 S1 V I 1 e T I S2 2 Using KVL, we get from which V V and V V 1 I 10 e 0.22 A= I / ECE 342 Jose Schutt Aine 28
29 Approximation  valid for small fluctuations about bias point Small Signal Model r d i I e i v v d 1 / nv i I T nv I T i I i d Total C applied (small) v V v d ECE 342 Jose Schutt Aine 29
30 iodes as Voltage Regulators Objective Provide constant dc voltage between output terminals Load current changes c power supply changes Take advantage of diode IV exponential behavior Big change in current correlates to small change in voltage ECE 342 Jose Schutt Aine 30
31 Voltage Regulator  Example Assume n=2 and calculate % change caused by a ±10% change in powersupply voltage (a) with no load (b) with 1k load I r Nominal value of current is: d ma 1 Incremental resistance for each diode: nvt I 7.9 Resistance for all 3 diodes: r 3r d 18.9 Voltage change r v o mv 18.5 mv 0.9% rr ECE 342 Jose Schutt Aine 31
32 Voltage Regulator Example (con t) When 1k load is connected, it draws a current of 2.1 ma resulting in a decrease in voltage across the 3 diodes given by v 2.1r o v o mv ECE 342 Jose Schutt Aine 32
33 iode as Rectifier While applied source alternates in polarity and has zero average value, output voltage is unidirectional and has a finite average value or a dc component ECE 342 Jose Schutt Aine 33
34 iode as Rectifier v s is a sinusoid with 24V peak amplitude. The diode conducts when v s exceeds 12 V. The conduction angle is 2 where is given by 24cos The conduction angle is 120 o, or onethird of a cycle. The peak value of the diode current is given by Id A The maximum reverse voltage across the diode occurs when v s is at its negative peak: 24+12=36 V ECE 342 Jose Schutt Aine 34
35 HalfWave Rectifier ECE 342 Jose Schutt Aine 35
36 FullWave Rectifier ECE 342 Jose Schutt Aine 36
37 Bridge Rectifier ECE 342 Jose Schutt Aine 37
38 Bridge Rectifier Properties Uses four diodes. v o is lower than v s by two diode drops. Current flows through R in the same direction during both half cycles. The peak inverse voltage (PIV) of each diode: PIV v 2v v v v s s ECE 342 Jose Schutt Aine 38
39 Peak Rectifier Filter capacitor is used to reduce the variations in the rectifier output ECE 342 Jose Schutt Aine 39
40 Rectifier with Filter Capacitor ECE 342 Jose Schutt Aine 40
41 Rectifier with Filter Capacitor Operation iode conducts for brief interval t Conduction stops shortly after peak Capacitor discharges through R CR>>T V r is peaktopeak ripple i v / R L o i i i C dv i dt I C L L v o V e p T tcr / V I V / R p L Vr Vp CR fcr fc L I p i I 1 2 V / V av L p r i max I 12 2 V / V L p r ECE 342 Jose Schutt Aine 41
42 iode Circuits  Rectification V Asint in Rectification with ripple reduction. C must be large enough so that RC time constant is much larger than period ECE 342 Jose Schutt Aine 42
43 iode Circuits V out V I I e S V / V T 1 V RI V RI ( V ) V S Nonlinear transcendental system Use graphical method I iode characteristics V s /R Load line (external characteristics) V out V S V Solution is found at itersection of load line characteristics and diode characteristics ECE 342 Jose Schutt Aine 43
44 iode Circuits Iterative Methods Wish to solve f(x)=0 for x NewtonRaphson Method V out V Use: 1 x k 1 x k f '( xk) f( xk) ( k1) ( k) ( k) 1 ( k) x x f '( x ) f( x ) V VS V / VT f( V) ISe 10 R 1 IS V / VT f '( V ) e R V V V ( k1) ( k) V ( k ) V S IS e R 1 IS V e R V T Procedure is repeated until convergence to final (true) value of V which is the solution. Rate of convergence is quadratic. V ( k ) ( k ) / V / V T T T 1 ( Where k ) is the value of V at the kth iteration V ECE 342 Jose Schutt Aine 44
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