ECE 342 2. PN Junctions and iodes Jose E. Schutt-Aine Electrical & Computer Engineering University of Illinois jschutt@emlab.uiuc.edu ECE 342 Jose Schutt Aine 1
B: material dependent parameter = 5.4 10 31 for Si E G : Bandgap energy = 1.12 ev k: Boltzmann constant=8.6210-5 ev/k n i : intrinsic carrier concentration At T = 300 K, n i = 1.5 10 10 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
PN Junction When a p material is connected to an n-type material, a junction is formed Holes from p-type diffuse to n-type region Electrons from n-type diffuse to p-type region Through these diffusion processes, recombination takes place Some holes disappear from p-type Some electrons disappear from n-type 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
PN Junction Potential acts as barrier that must be overcome for holes to diffuse into the n-region and electrons to diffuse into the p-region Open circuit: No external current Junction built-in 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: cross-section of junction x p : width in p side x n : width in n side s :silicon permittivity 8 11.7 1.0410 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
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 ni @ 300K, ni 1.510 /cm V 0.026 V T 10 3 V o 18 15 13 10 10 10 0.026ln 0.026ln 2 20 2.25 o 1.5 10 V o 0.02629.12 0.7571 volts o V o o 0.7571 volts ECE 342 Jose Schutt Aine 5
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: steady-state reverse current = I S -I (I is very small) reverse current ~ I S ~10-15 A Under reverse bias the current in the diode is negligible ECE 342 Jose Schutt Aine 6
epletion Layer Stored Charge q q qn x A j N n A: cross section area q j : stored charge Let W dep = depletion-layer 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
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
Forward-Biased 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
iode equation: Forward-Biased 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
iode Characteristics Three distinct regions The forward-bias region, determined by v > 0 The reverse-bias region, determined by v < 0 The breakdown region, determined by v < -V ZK ECE 342 Jose Schutt Aine 11
iode I-V Relationship Breakdown Electric field strong enough in depletion layer to break covalent bonds and generate electron-hole pairs. Electrons are then swept by E-field into the n- side. Large number of carriers for a small increase in junction voltage ECE 342 Jose Schutt Aine 12
The iode + I V - iode Properties Two-terminal 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
Ideal iode Characteristics + I + I V V - - V < 0 I > 0 OFF ON ECE 342 Jose Schutt Aine 14
Ideal iode Characteristics ECE 342 Jose Schutt Aine 15
iode Models Exponential Piecewise Linear Constant-Voltage-rop 16 ECE 342 Jose Schutt Aine 16
Ideal-diode iode Models Small-signal 17 ECE 342 Jose Schutt Aine 17
Piecewise-Linear Model for v : 0 V0 i 1 for v V : i v V 0 0 r ECE 342 Jose Schutt Aine 18
Piecewise-Linear Model ECE 342 Jose Schutt Aine 19
Constant-Voltage-rop Model for i 0: v 0.7V ECE 342 Jose Schutt Aine 20
Constant-Voltage-rop Model ECE 342 Jose Schutt Aine 21
iodes Logic Gates OR Function AN Function Y ABC Y ABC ECE 342 Jose Schutt Aine 22
iode Circuit Example 1 Assume both diodes are on; then V 0 and V 0 I B 2 10 0 1 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
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) 15 1.33 ma 10 101.33 3.3V 1 is reverse biased as assumed ECE 342 Jose Schutt Aine 24
Example The diode has a value of I S = 10-12 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 6-0.7 = 5.3 V. Ohm s law then gives a current of I 5.3 2 2.65 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
6 V 2 10 e 12 V /0.026 Example (cont d) Nonlinear equation must be solved iteratively Solution: V = 0.744 V Using this value of the voltage, we can calculate the current I 6 V 6 0.744 2 2 2.63 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) 0.0295 300 ECE 342 Jose Schutt Aine 26
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) 2 1.01610 ma The equation for I is then Solving iteratively, we get I 6 V 1.01610 e 2 10 V /0.0295 V 0.640V and I 2.68 ma ECE 342 Jose Schutt Aine 27
Example Two diodes are connected in series as shown in the figure with I s1 =10-16 A and I s2 =10-14 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 2 0.44 V and V 1 0.56 V 1 I 10 e 0.22 A= I 16 0.56 / 0.026 1 2 ECE 342 Jose Schutt Aine 28
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
iodes as Voltage Regulators Objective Provide constant dc voltage between output terminals Load current changes c power supply changes Take advantage of diode I-V exponential behavior Big change in current correlates to small change in voltage ECE 342 Jose Schutt Aine 30
Voltage Regulator - Example Assume n=2 and calculate % change caused by a ±10% change in power-supply voltage (a) with no load (b) with 1-k load I r Nominal value of current is: d 10 2.1 7.9 ma 1 Incremental resistance for each diode: nvt 2 25 6.3 I 7.9 Resistance for all 3 diodes: r 3r d 18.9 Voltage change r 0.0189 v o 2 2 37.1 mv 18.5 mv 0.9% rr 0.0189 1 ECE 342 Jose Schutt Aine 31
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 2.118.9 39.7 o mv ECE 342 Jose Schutt Aine 32
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
iode as Rectifier v s is a sinusoid with 24-V peak amplitude. The diode conducts when v s exceeds 12 V. The conduction angle is 2 where is given by 24cos 12 60 The conduction angle is 120 o, or one-third of a cycle. The peak value of the diode current is given by Id 24 12 100 0.12 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
Half-Wave Rectifier ECE 342 Jose Schutt Aine 35
Full-Wave Rectifier ECE 342 Jose Schutt Aine 36
Bridge Rectifier ECE 342 Jose Schutt Aine 37
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
Peak Rectifier Filter capacitor is used to reduce the variations in the rectifier output ECE 342 Jose Schutt Aine 39
Rectifier with Filter Capacitor ECE 342 Jose Schutt Aine 40
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 peak-to-peak 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
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
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
iode Circuits Iterative Methods Wish to solve f(x)=0 for x Newton-Raphson 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