Sedra/Smith Microelectronic Circuits 6/E
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Sedra/Smith Microelectronic Circuits 6/E Chapter 1: Introduction to Electronics S. C. Lin, EE National Chin-Yi University of Technology 1 【Outline】 1-1 Signals 1.2 Frequency Spectrum of Signal 1.3 Analog and Digital Signals 1.4 Amplifiers 1.5 Circuit Models for Amplifier 1.6 Frequency Response of Amplifiers 1.7 Intrinsic Semiconductors 1.8 Doped Semiconductors 1.9 Current Flow in Semiconductors 1.10 The P-N Junction with open-circuit Terminals 1.11 The PN Junction with Applied voltage 1.12 Capacitive Effects in the PN Junction S. C. Lin, EE National Chin-Yi University of Technology 2 1-1 Signals RTh vTh (t ) + − (a) the Thévenin form vTh (t ) = iN (t ) ⋅ RN RTh = RN RN iN (t ) (b) the Norton form. vTh (t ) iN (t ) = RTh RN = RTh S. C. Lin, EE National Chin-Yi University of Technology 3 Example: Find the source transformation for the circuits shown in below Figures 14Ω 28V + RN iN (t ) − iN (t ) = vTh (t ) 28 = = 2A, RTh 14 RN = RTh = 14Ω RTh 3A vTh (t ) + 5Ω − vTh (t ) = iN (t ) RN = 3A × 5Ω = 15V RTh = RN = 5Ω S. C. Lin, EE National Chin-Yi University of Technology 4 1.2 Frequency Spectrum of Signal Sinusoid: v(t ) v(t ) = Va sin ( ωt + φ1 ) v(t ) = Va sin ωt v(t ) = Va sin ( ωt − φ2 ) Va ϕ1 ϕ 2 1 f = , ω = 2πf rad / sec T φ1 = 90o ⇒ v(t ) = va sin(ωt + 90o ) = va cos ωt S. C. Lin, EE National Chin-Yi University of Technology 5 v T +V t −V Figure 1.5 A symmetrical square-wave signal of amplitude V. The symmetrical square-wave signal in Fig.1.5 can be expressed as: 4V v(t ) = π 1 1 ⎛ ⎞ ⎜ sin ωot + sin 3ωot + sin 5ωot + " ⎟ 3 5 ⎝ ⎠ S. C. Lin, EE National Chin-Yi University of Technology 6 4V v(t ) = π 1 1 ⎛ ⎞ ⎜ sin ωot + sin 3ωot + sin 5ωot + " ⎟ 3 5 ⎝ ⎠ 4V π 4V 3π ωo 3ωo 4V 5π 5ωo 4V 7π 7ωo ... ω (rad/s) Figure 1.6 The frequency spectrum (also known as the line spectrum) of the periodic square wave of Fig. 1.5. S. C. Lin, EE National Chin-Yi University of Technology 7 Frequency Spectrum Va (ω ) in volts ω (rad/s) Figure 1.7 The frequency spectrum of an arbitrary waveform such as that in Fig. 1.3. S. C. Lin, EE National Chin-Yi University of Technology 8 1.3 Analog and Digital Signals v(t ) t t0 t 2 t 4 t6 v(t ) t0 t 2 t 4 t6 t Figure 1.8 Sampling the continuous-time analog signal in (a) results in the discrete-time signal in (b). S. C. Lin, EE National Chin-Yi University of Technology 9 v(t ) +5V t Figure 1.9 Variation of a particular binary digital signal with time. S. C. Lin, EE National Chin-Yi University of Technology 10 + b0 b1 vA − } bn −1 Figure 1.10 Block-diagram representation of the analog-to-digital converter (ADC). S. C. Lin, EE National Chin-Yi University of Technology 11 Example: Dual-Slop Analog-to-Digital Converter Analog input signal (Vin ) sw C C Completer CLK R OPA −VREF OPA R n Counter Q0 Q1 Q2 Q3 Q4 Q5 Q6 Q7 D0 D1 D2 D3 D4 D5 D6 D7 Latch Control Logic Q0 Q1 Q2 Q3 Q4 Q5 Q6 Q7 Binary output S. C. Lin, EE National Chin-Yi University of Technology 12 1.4 amplifiers Voltage gain Input Output Input Output vo Av = vi Figure 1.11 (a) Circuit symbol for amplifier. (b) An amplifier with a common terminal (ground) between the input and output ports. S. C. Lin, EE National Chin-Yi University of Technology 13 1.4.3 Voltage Gain vo Av io ii RL vI (t ) + vo (t ) − (a) 1 vi (b) Figure 1.12 (a) A voltage amplifier fed with a signal vI (t) and connected to a load resistance RL. (b) Transfer characteristic of a linear voltage amplifier with voltage gain Av. S. C. Lin, EE National Chin-Yi University of Technology 14 . Current gain Power gain io Ai ≡ ii Load power PL vo ⋅ io Ap ≡ .........(1.7) = = Input power PI vi ⋅ ii 1.4.5 Expressing Gain in Decibels log 2 = 0.3 log 3 = 0.477 Ap ≡ log10 P2 P1 (Bel) P2 (dB) = 10 log10 P1 log 4 = 0.6 log 5 = 0.7 S. C. Lin, EE National Chin-Yi University of Technology 15 . AdB m P2 = 10 log10 1mw (dBm ) R = 600 Ω ⎛ V2 ⎞ ⎛ R1 ⎞ P2 V22 / R2 AdB = 10 log10 = 10 log10 2 = 20log ⎜ ⎟ + 10log ⎜ ⎟ P1 V1 / R1 ⎝ V1 ⎠ ⎝ R2 ⎠ ⎛ I2 ⎞ ⎛ R2 ⎞ P2 I 22 ⋅ R2 AdB = 10 log10 = 10 log10 2 = 20log ⎜ ⎟ + 10log ⎜ ⎟ P1 I1 ⋅ R1 ⎝ I1 ⎠ ⎝ R1 ⎠ For the series system the total gain as AVT = AV1 × AV2 × AV3 × " × AVn AVdB = 20 log AVT = 20 log AV1 + 20 log AV2 + " + 20 log AVn ⇒ AdBT = AdB1 + AdB2 + " + AdBn (dB) S. C. Lin, EE National Chin-Yi University of Technology 16 1.4.6 The Amplifier Power Supplies VCC I CC ii vI (t ) I CC V+ + v (t ) RL − o V − V+ ii VCC vI (t ) + I EE V −v o I EE VEE −VEE (t ) − Figure 1.13 An amplifier that requires two dc supplies (shown as batteries) for operation. S. C. Lin, EE National Chin-Yi University of Technology 17 RL Example1.2: Consider an amplifier operating from power supplies ±10V.It is fed with a sinusoidal voltage having 1V peak and delivers a sinusoidal voltage output of 9V peak to a 1kΩ load. The amplifier draws a current of 9.5mA from each of its two power supplies. The input current of the amplifier is found to be sinusoidal with 0.1mA peak.(a) Find the voltage gain, (b) the current gain, (c) the power gain, (d) the power drawn from the dc supplies,(e) the power dissipated in the amplifier, and (f) the amplifier efficiency 9 (a ) A = = 9 V/V V1 1 I1 AvdB = 20log 9 ≈ 19.1 dB ▲ + V ii 9V = 9mA (b) lI o = 1kΩ + vI (t ) I V − v (t ) RL 9mA I o o Ai = = = 90 A/A ▲ I i 0.1mA −V2 − AidB = 20log 90 = 39.1 dB ▲ 2 S. C. Lin, EE National Chin-Yi University of Technology 18 (c ) ⎫ = 40.5mW ⎪ PL 2 2 ⎪ = 810 W/W ⎬ Ap = 1 0.1 PI ⎪ PI = vi ( rms ) ⋅ ii ( rms ) = ⋅ = 0.05mW ⎪⎭ 2 2 Ap dB = 10log810 = 29.1 dB ▲ PL = vo ( rms ) ⋅ io ( rms ) = 9 ⋅ 9 (d ) Pdc = 10 × 9.5 + 10 × 9.5 = 190mW ▲ (e) Pdissipated = Pdc + PI − PL = 190 + 0.05 − 40.5 = 149.6mW PL × 100% = 21.3% (f) η= Pdc ▲ ▲ S. C. Lin, EE National Chin-Yi University of Technology 19 1.4.7 Amplifier saturation o Output peaks clipped due to saturation L+ n L+ AV L− AV L− n o L− L+ ≤ vI ≤ Av Av Figure 1.14 An amplifier transfer characteristic that is linear except for output saturation. S. C. Lin, EE National Chin-Yi University of Technology 20 1.4.8 Symbol Convention iC iC (t ) = I C + ic (t ), ic (t ) = I c sin ωt ic Ic IC 0 t S. C. Lin, EE National Chin-Yi University of Technology 21 1.5 Circuit Models for Amplifier . Vi Ii AV Gm Rm AI Vo Io Av :Voltage Amplifier Rm :Transresistance Amplifier AI :Current Amplifier Gm :Transconductance Amplifier S. C. Lin, EE National Chin-Yi University of Technology 22 1.5.1 Voltage Amplifier . ii Rs vs + vi io Ro Ri Avo vi − + vo + vL RL − − Ri vi Ri vi = vs ⇒ = Rs + Ri vs Rs + Ri RL vL RL vL = Avo vi ⇒ = Avo ⋅ RL + Ro vi RL + Ro Ri vL vL vi RL ⋅ Av = = ⋅ = Avo ⋅ vs vi vs RL + Ro Rs + Ri S. C. Lin, EE National Chin-Yi University of Technology Ideal characteristic Ri = ∞, Ro = 0 23 . (b) Current Amplifier ii is Rs + vi io Ri Aisii iL Ro − + vo + vL RL − − Rs ii Rs ii = is ⇒ = Rs + Ri is Rs + Ri Ro iL Ro iL = Aisii ⇒ = Ais RL + Ro ii RL + Ro iL iL ii Ro Rs = ⋅ = Ais ⋅ is ii is RL + Ro Rs + Ri S. C. Lin, EE National Chin-Yi University of Technology Ideal characteristic Ri = 0, Ro = ∞ 24 .(c) Transconductance Amplifier ii Rs io iL + vs vi + + Ri Gmvi Ro vo − − RL vL − Ri vi Ri ⇒ = vi = vs Rs + Ri vs Rs + Ri Ro Ro iL iL = Gm vi ⇒ = Gm ⋅ RL + Ro vi RL + Ro Ro Ri iL vi iL = ⋅ = Gm ⋅ ⋅ vs vs vi RL + Ro Rs + Ri S. C. Lin, EE National Chin-Yi University of Technology Ideal characteristic Ri = ∞, Ro = ∞ 25 (d) Transresistance Amplifier . ii is Rs + vi Ri Rmii − ii = is io Ro + vo − RL + vL − Rs i Rs ⇒ i = Rs + Ri is Rs + Ri vL = Rmii RL v RL ⇒ L = Rm ⋅ RL + Ro ii RL + Ro Rs vL vL ii RL = ⋅ = Rm ⋅ ⋅ is ii is RL + Ro Rs + Ri S. C. Lin, EE National Chin-Yi University of Technology Ideal characteristic Ri = 0, Ro = 0 26 1.5.5. Determining Ri and Ro vi n Input resistance Ri ≡ ii where vi is the input voltage and ii is the input current. o The output resistance found by eliminating the input signal source (then ii and vi will be zero ) and applying a voltage signal vx to the output of the amplifier, we have a current ix , vx then Ro ≡ ix ix + − vx vx Ro ≡ ix S. C. Lin, EE National Chin-Yi University of Technology 27 ib + + vbe rπ g mvbe − ro vs (c) An alternative small-signal circuit model for the BJT. + vbe − ro βib − (a) Small-signal circuit model for a bipolar junction transistor (BJT). Rs rπ vbe + rπ g mvbe ro RL vo − (b) The BJT connected as an amplifier with the emitter as a common terminal between input and output (called a common-emitter amplifier). S. C. Lin, EE National Chin-Yi University of Technology 28 1.6 Frequency response of amplifiers ii vi = Vi sin ωt Linear Amplifier + vo (t ) = Vo sin(ωt + φ) − Figure 1.20 Measuring the frequency response of a linear amplifier. At the test frequency f, the amplifier gain is characterized by its magnitude (Vo/Vi) and phase φ . vo T (ω) = , vi ∠T (ω) = φ S. C. Lin, EE National Chin-Yi University of Technology ▲ 29 20log T (ω ) Bandwidth ω1 ω2 ω Figure 1.21 Measuring the frequency response of a linear amplifier. At the test frequency f, the amplifier gain is characterized by its magnitude (Vo/Vi). S. C. Lin, EE National Chin-Yi University of Technology 30 1.6.4 Single-time constant networks (a) a low-pass network R2 vi (t ) + − C2 1/ jωC2 vo = vi R2 + (1/ jωC2 ) + vo 1 vo = vi 1 + jωR2C2 − Let R2C2 be given time constant,hence ω2 = 1/ R2C2 1 1 T (ω) = = ∠ − tan −1 (ω / ω2 ) 1 + jω / ω2 1 + (ω / ω2 ) 2 ∠T ( ω ) T (ω) S. C. Lin, EE National Chin-Yi University of Technology 31 T (ω) = 20log 1 1 + (ω / ω2 ) 2 ⎧ ω << ω2 ⇒ T (ω) ≈ 1 ⎪ ⎨ ω = ω2 ⇒ T (ω) ≈ 1/ 2 = 0.707 ⎪ω >> ω ⇒ T (ω) ≈ 0 2 ⎩ T ( jω ) (dB) K 3dB −6dB/Octave or −20dB/decade ω (log scale) ω2 S. C. Lin, EE National Chin-Yi University of Technology 32 ⎛ω⎞ ∠T (ω) = ∠ − tan ⎜ ⎟ ⎝ ω2 ⎠ −1 ⎧ ω << ω2 ⇒ ∠T (ω) ≈ 0o ⎪ o ω = ω ⇒ ∠ T ( ω ) ≈ − 45 ⎨ 2 ⎪ω >> ω ⇒ ∠T (ω) ≈ −90o 2 ⎩ φ (ω ) ω (log scale) ω2 5.7 O −45o −45o /decade −90 o S. C. Lin, EE National Chin-Yi University of Technology 5.7 O 33 (b) A High-Pass Network. C1 vi (t ) + − R1 + vo − R1 vo = vi R1 + (1/ jωC1 ) 1 vo = vi 1 + (1/ jωR1C1 ) Let R1C1 be given time constant,hence ω1 = 1/ R1C1 1 1 −1 T (ω) = = ∠ tan (ω1 / ω) 1 − jω1 / ω 1 + (ω1 / ω) 2 ∠T ( ω ) T (ω) S. C. Lin, EE National Chin-Yi University of Technology 34 . T (ω) = 20log 1 1 + (ω1 / ω) 2 ⎧ ω << ω1 ⇒ T (ω) ≈ 0 ⎪ ⎨ω = ω1 ⇒ T (ω) ≈ 1/ 2 = 0.707 ⎪ω >> ω ⇒ T (ω) ≈ 1 1 ⎩ T ( jω ) (dB) K +6dB/Octave or +20dB/decade 3dB ω (log scale) ω1 S. C. Lin, EE National Chin-Yi University of Technology 35 ⎧ω << ω1 ⇒ ∠T (ω) ≈ 90o ⎪ −1 ∠T (ω) = ∠ tan (ω1 / ω) ⎨ ω = ω1 ⇒ ∠T (ω) = 45o ⎪ ω >> ω ⇒ ∠T (ω) ≈ 0o 1 ⎩ φ (ω ) 90o 5.7 O 45o −45o /decade 5.7 O 0o ω (log scale) ω1 S. C. Lin, EE National Chin-Yi University of Technology 36 Transfer Function T (s) K 1 + ( s / ω2 ) Ks = 1 + (ω1 / s ) s + ω1 Transfer Function (for physical frequencies) T ( jω ) K K Megnitude Response T ( jω ) Phase Response∠T ( jω ) K 1 + j ( ω / ω2 ) K 1 + ( ω / ω2 ) 2 − tan −1 (ω / ω2 ) 1 − j (ω1 / ω ) K 1 + (ω1 / ω ) tan −1 (ω1 / ω ) Transmission at ω = 0 (dc) K 0 Transmission at ω = ∞ 0 K 3-dB frequency 2 ω0 = 1/ τ ≡ time constant τ = RC or τ = L / R S. C. Lin, EE National Chin-Yi University of Technology 37 Example 1.5 Below figure shows a voltage amplifier. Let Rs = 20kΩ, Ri = 100kΩ, Ro = 200Ω, RL = 1kΩ, Ci = 60pF, µ = 144. Find vo (t ) for each of the following inputs: (a)Vs = 0.1sin102 t (V), (b)Vs = 0.1sin106 t (V), (c) Vs = 0.1sin108 t (V). ii Rs vs Ri + vi Ci + vo µvi + Ri vs Rs + Ri vi − RL ii io Ro Ci µvi + vL − − − RS // Ri io Ro + vo − S. C. Lin, EE National Chin-Yi University of Technology RL + vL − 38 Sol: vi = vs 1/ jωCi Ri Ri 1 ⋅ = vs ⋅ Rs + Ri ( Rs // Ri ) + (1/ jωCi ) Rs + Ri 1 + jωCi ( Rs // Ri ) Ri 1 = vs ⋅ ∠ − tan −1 (ω /106 ) Rs + Ri 1 + (ω /106 ) 2 Ri RL RL 1 vo = µvi = µvs ⋅ ∠ − tan −1 (ω /106 ) Ro + RL Ro + RL Rs + Ri 1 + (ω /106 ) 2 vs 1kΩ 100kΩ ⋅ ∠ − tan −1 (ω /106 ) = 144 ⋅ ⋅ 100kΩ + 20k Ω + 0.2k Ω 1 + (ω /106 ) 2 Ω 1k 100 vo = 100 ⋅ vs 1 + (ω /106 ) 2 ⇒ AV = 100 ⋅ ∠ − tan −1 (ω /106 ) 1 1 + (ω /106 ) 2 , ∠AV = ∠ − tan −1 (ω /106▲ S. C. Lin, EE National Chin-Yi University of Technology 39 (c) Vs = 0.1sin108 t (V ) (a)Vs = 0.1sin102 t (V ) AV = 100 ⋅ 1 AV = 1 + (10 /10 ) 2 6 2 ≈ 100 ⋅1 = 100 ∠AV = − tan −1 (102 / 106 ) ≈ 0o Vo (t ) = 10sin102 t (V) ▲ 100 1 + (10 /10 ) 8 6 2 ≈1 ∠AV = − tan −1 (108 /106 ) ≈ −89.4o Vo (t ) = 0.1sin108 t (V ) ≅ 0.1sin(108 t − 89.4o )(V) ▲ (b)Vs = 0.1sin106 t (V ) AV = 100 ⋅ 1 1 + (10 /10 ) 6 6 2 ≈ 70.7 ∠AV = − tan −1 (106 /106 ) ≈ −45o Vo (t ) = 7.07 sin(106 t − 45o ) (V) ▲ S. C. Lin, EE National Chin-Yi University of Technology 40 dB 20dB/decade 40 20 0 1 10 10 2 10 3 4 10 104 105 10 5 10 6 10 7 8 10 10 9 φ (ω ) 101 102 103 106 107 108 109 ω −45o / decade S. C. Lin, EE National Chin-Yi University of Technology 41 ω Example : Sketch Bode plots for the magnitude and phase of the transfer function 106 ⋅ s 100 s (1) T ( s ) = = 3 ( s + 10)( S + 10 ) (1 + s /10)(1 + s /103 ) ( −1 20log T ( s ) = 20log100 + 20log s + 20log(1 + s /10) + 20log 1 + s /10 ) 3 −1 100 1 10 2 10 3 10 (1 + ( s /10) ) 4 10 −1 5 10 10 (1 + ( S /10 ) ) 3 6 7 10 8 10 109 −1 S. C. Lin, EE National Chin-Yi University of Technology 42 ω s 100 ⎡⎣1 + ( s /10 ) ⎤⎦ −1 ⎡1 + ( s /103 ) ⎤ ⎣ ⎦ −1 S. C. Lin, EE National Chin-Yi University of Technology 43 ω 104 ⋅ (1 + s /105 ) (2) T ( s ) = (1 + s /103 )(1 + s /104 ) ( 4 5 ) ( ⇒ 20log T ( s ) = 20log10 + 20log 1 + s /10 + 20log 1 + s /10 ( + 20log 1 + s /10 1 2 3 4 ) 3 −1 ) 4 −1 5 6 S. C. Lin, EE National Chin-Yi University of Technology 7 8 9 44 ω 1 2 3 4 5 6 S. C. Lin, EE National Chin-Yi University of Technology 7 9 8 ω 45 The Structure of Nucleus 7s 7p 7d 7f S. C. Lin, EE National Chin-Yi University of Technology 46 高 能 階 低 電殻層 S. C. Lin, EE National Chin-Yi University of Technology 47 S. C. Lin, EE National Chin-Yi University of Technology 48 鍺 S. C. Lin, EE National Chin-Yi University of Technology 49 已知銦(In)的原子序為49，試繪出其電子在電殼層上的分佈情形？ S. C. Lin, EE National Chin-Yi University of Technology 50 1.7 Intrinsic Semiconductors Eg (Si) = 1.21 − 3.6 × 10−4 T (ev) Eg (Ge) = 0.785 − 2.23 ×10−4 T (ev) S. C. Lin, EE National Chin-Yi University of Technology 51 ⎧ p : concentration of hole ⎪ p = n = ni ⎨ n : concentration of free electrons ⎪ n : Intrinsic concentration ⎩ i S. C. Lin, EE National Chin-Yi University of Technology 52 ni2 = BT 3e − Eg / kT (1.26) where B is a meterial-depend parameter = 5.4 × 1031 (Si) k is Boltzmann's constant = 8.62 × 10-5 eV/K T is an absolute temperature = 273+oC At T = 300o K ⇒ ni = 1.5 × 1010 carriers/cm3 .(P.73, Table 1.3) pn nn = ni2 ni2 ni2 pn = nn ND (1.29) ni2 ni2 np = pp NA (1.31) S. C. Lin, EE National Chin-Yi University of Technology 53 1.8 Doped Semiconductors --- N-type: P, As, sb 銻的第五個價電子 (磷 ,砷 ,銻 ) 銻雜質 Figure 1.30 A silicon crystal doped by a pentavalent element. Each dopant atom donates a free electron and is thus called a donor. The doped semiconductor becomes n type. S. C. Lin, EE National Chin-Yi University of Technology 54 P-type: B,Al,In ( 硼 ,鋁 ,銦 ) 電洞 銦雜質 Figure 1.31 A silicon crystal doped with a trivalent impurity. Each dopant atom gives rise to a hole, and the semiconductor becomes p type. S. C. Lin, EE National Chin-Yi University of Technology 55 1.9 Current Flow in Semiconductors 1.9.1 Drift current E + − + Potential gradient Holes Electrons V − x Drift current S. C. Lin, EE National Chin-Yi University of Technology 56 The holes acquire a velocity v p − drift given by v p − drift = µ p E (1.32) The holes component of the drift current flowing through the bar. I p = Aqpv p − drift = Aqpµ p E The current density: J P = Ip A = qpµ p E (1.35) The free electrons acquire a velocity vn − drift given by vn − drift = − µn E (1.33) The free electrons component of the drift current flowing through the bar. I n = − Aqnvn − drift = Aqpµn E In Similar: The current density : J n = = qnµn E A S. C. Lin, EE National Chin-Yi University of Technology (1.36) 57 The total drift current density: J = J P + J n = q ( p µ p + nµ n ) E (1.37) Hence the current density J is given by J =σE σ = q(nµn + pµ p ) conductivity (1.40) E ρ = (resistivity) = σ J (1.42) 1 S. C. Lin, EE National Chin-Yi University of Technology 58 1.9.2 Diffusion and drift + + +++ + + + ++ ++ + ++ + + + + ++ + + + + + + ++ x x 0 Concentration gradient Diffusion current Electron-current density J p ( J n ) dp ( x) ⎫ q = 1.6 × 10−19 C dx ⎪⎪ 2 : hole's diffusion constant are 12cm /s D ⎬ p dn( x) ⎪ J n = − qDn Dn : electrons diffusion constant are 34cm 2 /s dx ⎪⎭ J p = − qD p S. C. Lin, EE National Chin-Yi University of Technology 59 The total current in semiconductor (1) The net hole current density dp J p = qµ p pE − qD p (A/m 2 ) dx drift diffusion (2) The net electrons current density dn (A/m 2 ) J n = qµn nE − qDn dx drift diffusion (3)The total current density in semiconductor J total dp dn = J p + J n = qE ( µ p p + µn n) − q ( D p + Dn ) (A/m 2 ) dx dx drift diffusion S. C. Lin, EE National Chin-Yi University of Technology 60 1.9.3 Relationship Between D and µ Einstein relationship equation: Dp Dn = = VT µp µn KT ≅ 25mV(300o K ) VT = q S. C. Lin, EE National Chin-Yi University of Technology 61 Example 1.9 Consider a bar of silicon in which a hole concentration profile described by p ( x ) = p0 e − x / Lp is estabilished. Find the hole-current density at x = 0. Let p0 = 1016 / cm3 and LP = 1 µm. If the cross-sectional area of the bar is 100µm 2 , find the current I p . Solution: ( ) dp( x) d −x/ L = − qD p p ( x) = p0 e p dx dx 2 Dp 12cm /s −19 16 3 2 × = J p (0) = q p0 = 1.6 × 10 × 10 / cm 192A/cm ▲ Lp 1×10−4 cm J p = − qD p I p = J p × A = 192A/cm 2 × 100 × 10−8 cm 2 = 192µA S. C. Lin, EE National Chin-Yi University of Technology ▲ 62 Example A: An intrinsic silicon bar is 3mm long and has a rectangular cross section 50µm ×100µm at 300o K determine the electric field intensity in the bar, and the voltage across the bar When a steady current of 1µA is measured. (ρ =2.3 ×105Ω-cm) Sol: J 1 I I (1) ξ = = × = ρ σ σ A A 10−6 A 5 = × ⋅ 2.3 10 Ω-cm −4 −4 50 ⋅10 cm × 100 ⋅10 cm = 4.60 ⋅103 V/cm ▲ 3mm 50µm 100µm (2) Vbar = ξL = 4.60 ⋅103 V/cm × 0.3cm = 1380V ▲ The result obtained that an extremely large voltage is needed to produce a small current 1µA S. C. Lin, EE National Chin-Yi University of Technology 63 Example B: An N-type silicon bar is 3mm long and has a rectangular cross section 50µm ×100µm. The donor concentration at 300o K is 5 × 1014 cm -3 and corresponding to impurity atom for 108 silicon atoms, a stead current of 1µA exists in the bar, determine the electric and hole concentrations the condctivity, and the voltage across the bar. (in 300o K the ni = 1.45 × 1010 , µn = 1500) Sol: (1)n ≅ N D = 5 ×1014 cm −3 ni 2 (1.45 ×1010 ) 2 5 −3 = = 4.2 × 10 cm p= ▲ 14 5 ×10 n (2)σ = q (nµn + pµ p ) ≈ qnµn (n >> p) = 1.6 × 10−19 × 5 ×1014 × 1.5 × 103 = 0.12(Ω ⋅ cm) −1 ▲ J I ⋅L (3)Vbar = L = σ A⋅σ 10−6 (A) ⋅ 0.3(cm) = = 0.05V ▲ −3 −2 −1 5 ⋅10 (cm) ⋅10 (cm) ⋅ 0.12(Ω×cm) S. C. Lin, EE National Chin-Yi University of Technology 64 1-10 The P-N junction with open-circuit Terminals 1.10.1 Physical structure p − type silicon n − type silicon Figure 1.35 Simplified physical structure of the junction diode. S. C. Lin, EE National Chin-Yi University of Technology 65 1.10.2 Operation with Open-Circuit Terminals I = 0A I = 0A Barrier voltage Vo Figure 1.36 (a) The pn junction with no applied voltage (open-circuited terminals). (b) The potential distribution along an axis perpendicular to the junction. S. C. Lin, EE National Chin-Yi University of Technology 66 The Junction Built-In Voltage (Barrier Voltage)Vo With no external voltage applied, the total current density J = 0. ( J n , J p = 0) dp =0 For J p ( net ) = 0 ⇒ pqµ p ξ − q D p dx dp p qµ p ξ − qµ p VT =0 dx dVo VT dp VT dp ∴ξ = (V / m ) ⇒ − = p dx dx p dx dp x Vo = ∫− x −VT = −VT ln p −x x p pp p( x = − x p ) Vo = VT ln = VT ln p ( x = xn ) pn n n p p 0 0 S. C. Lin, EE National Chin-Yi University of Technology 67 where p p : Thermal-equilibrium holes concentration in p side 0 pn : Thermal-equilibrium holes concentration in n side 0 n p : Thermal-equilibrium electrons concentration in p side 0 nn : Thermal-equilibrium electrons concentration in n side 0 p p : N A (3 − 46), 0 pn : ni2 / N D ⇒ N D = ni2 / p (3 − 45) 0 ⎡ N AND ⎤ Vo = VT ln ⎢ 2 ⎥ ⎣ ni ⎦ S. C. Lin, EE National Chin-Yi University of Technology (1.46) 68 Width of the depletion region −xp xn qN D+ Q+ = qAxn N D (+) −xp xn (─) Q− = − qAx p N A x x − qN A− −xp dξ 1 = (− qN A ) dx ε W xn x ξ(0) dξ 1 = (qN D ) dx ε S. C. Lin, EE National Chin-Yi University of Technology 69 Since the net charge must be zero then N A xn q A N A xP = q A N D xn ⇒ = N D xP Q+ Q− ⎧ ρV : charge density. G ρ ⎪ By Gauss's law ⇒ ∇ξ = V ⎨ εs : electrical permittivity of silicon. εs ⎪ −12 ε ε ε ε 1 1.7 1.04 10 F/cm. = = = × s 0 r 0 ⎩ G d ξ ρV q p − n) ∴ = = (ND − N A + N ε dx ε s s neglect For 0 < x < xn JJJJG JJJJG q 0 xn d ξ ( x) q = N D ⇒ ∫ d ξ ( x) = N D ∫ dx 0 ξ( 0 ) dx εs εs ⇒ ξ ( 0) q = − N D xn εs S. C. Lin, EE National Chin-Yi University of Technology 70 For − x p 0 < x < 0 JJJJG ξ( 0 ) JJJJG 0 d ξ ( x) q q = − N A ⇒ ∫ d ξ ( x) = − N A ∫ dx 0 − xp dx εs εs ⇒ ξ (0) ξ (0) q = − N Axp εs NA NA xn = xp = (W − xn ) ND ND q q = − N D xn = − N A x p εs εs 1 −Vo = ∫ ξ ( x ) dx ⇒ Vo = − ξ (0)W − xp 2 1q 1 q ND N A Vo = N D xnW = W2 2 εs 2 εs N D + N A xn 2ε s ⎛ 1 1 ⎞ W= + ⎜ ⎟Vo q ⎝ N A ND ⎠ (1.50) ⎛ NA ⎞ NA xn ⎜1 + W ⎟= ⎝ ND ⎠ ND N AW xn = (1.51) ND + N A Typically, W is in the rang of 0.1～1 µm S. C. Lin, EE National Chin-Yi University of Technology 71 Example 1.10 Consider a pn junction in equilibrium at room temperature for which the doping concentrations are N A = 1018 cm −3 and N D = 1016 cm −3 and the cross-sectional area A = 10−4 cm 2 .Calculate p p , n p 0 , nn , pn 0 , V0 , W , xn , x p , and QJ . Use ni = 1.5 × 1010 cm −3 , ε s = 11.7ε 0 = 1.04 × 10−12 F/cm. Solution p p ≈ N A = 1018 cm −3 ▲ np = 0 ni2 pp 1.5 × 10 ( ≈ 10 cm ) −3 2 1018 cm −3 = 2.25 × 102 cm −3 ▲ nn = N D = 1016 cm −3 ▲ pn = 0 ni2 nn 1.5 × 10 ( = 10 cm 1016 cm −3 ) −3 2 = 2.25 × 104 cm −3 ▲ S. C. Lin, EE National Chin-Yi University of Technology 72 ⎡ 18 −3 16 −3 ⎡ N AND ⎤ × 10 cm 10 cm Vo = VT ln ⎢ 2 ⎥ = 26mV × ln ⎢ ⎢ 10 −3 2 n ⎣ i ⎦ ⎢⎣ 1.5 × 10 cm ( W= ) ⎤ ⎥ = 0.814V ▲ ⎥ ⎥⎦ 2εs ⎛ 1 1 ⎞ 2 × 1.04 × 10−12 ⎛ 1 1 ⎞ V + = + 0.814 ⎜ ⎟ o ⎜ 18 −3 −19 16 −3 ⎟ q ⎝ N A ND ⎠ 1.6 × 10 10 cm ⎠ ⎝ 10 cm = 3.27 × 10−5 = 0.327µm ▲ N AW 1018 xn = = 0.324µm ▲ = 0.327µm 18 16 ND + N A 10 + 10 1016 N DW = 0.327µm 18 = 0.00324µm ▲ xp = 16 ND + N A 10 + 10 18 16 ⎛ N AND ⎞ −4 −19 ⎛ 10 × 10 ⎞ −4 10 1.6 10 0.327 10 = × × × × × QJ = Aq ⎜ W ⎜ 18 ⎟ 16 ⎟ + N N 10 10 + A ⎠ ⎝ D ⎝ ⎠ = 5.18 × 10−12 C = 5.18pC ▲ S. C. Lin, EE National Chin-Yi University of Technology 73 1.11 The PN junction in the breakdown region Reverse breakdown can be occur by two mechanisms. (1) Zener effect : operating at low voltage (Vz < 5V) , (2) Avalanche effect: operating at higher voltage (Vz >7V) , For junction that breakdown between 5V and 7V the breakdown mechanism can be either the zener or the avalanche effect or a combination of the two. Zener breakdown When a heavily doped junction is reverse biased the energy bands become crossed at relatively low voltages. The ξ ↑ →it can break covalent bands. →generate electron-hole pairs. → the electrons will be swept into the N side and the holes swept into the P side, thus the electron-hole pairs constitute a reverse current. (ξ ≅ 106 V/cm) S. C. Lin, EE National Chin-Yi University of Technology 74 Avalanche breakdown S. C. Lin, EE National Chin-Yi University of Technology 75 I 0 V The pn junction I-V characteristic with the breakdown region S. C. Lin, EE National Chin-Yi University of Technology 76 1.11 The PN junction under forward-bias conditions ID Is p n − VR + I Figure 1.38 The pn junction excited by a constant-current source I in the reverse direction. To avoid breakdown, I is kept smaller than IS. Note that the depletion layer widens and the barrier voltage increases by VR volts, which appears between the terminals as a reverse voltage. S. C. Lin, EE National Chin-Yi University of Technology 77 Minority-carrier distribution in a forward-biased pn junction. It is assumed that the p region is more heavily doped than the n region; NA >> ND. pn ⋅ n p p − region n − region Depletion region pn ( xn ) pn ( x) n p (− x p ) n p ( x) pn 0 np0 −xp 0 x xn S. C. Lin, EE National Chin-Yi University of Technology 78 1.11.2 The Current-Voltage Relationship of the Junction The concentration of minority carriers at the edge of the depletion region. denoted by Pn ( xn ) in Fig.1.39 for forward voltage V Pn ( xn ) = Pn 0 eV / VT (1.57) (The low of the junction) The distribution of excess hole concentration in the N region is an exponentially decaying function of distance we can expressed as Pn ( x) = Pn 0 + [ Pn ( xn ) − Pn 0 ] e − ( x − xn ) / L p − ( x − xn ) / L p = Pn 0 + ⎡⎣ Pn 0 (eV / VT − 1) ⎤⎦ e S. C. Lin, EE National Chin-Yi University of Technology (1.59) 79 where L p is a constant, It is called the diffusion lenth of holes in the N-type silicon. ⎧ D p : diffusion constant for holes in ⎪ the N-type silicon ⎪ Lp = Dp τ p ⎨ ⎪ τ p : excess-minority-carrier lifetime 1 µm ↔100 µm ⎪ (1ns ↔ 10000ns ) ⎩ The hole current density(the holes diffusion in the N-region) dp J p = − qD p dx d ⎡⎣ Pn 0 ( eV / VT − 1) ⎤⎦ e − ( x − xn ) / L p = − qD p dx Dp − ( x − xn ) / L p =q Pn 0 ( eV / VT − 1) e Lp S. C. Lin, EE National Chin-Yi University of Technology 80 That J p is largest at the edge of the depletion region (x = xn ) and decays exponentially with distance at x = xn , the current density due to hole injected is given by Dp Jp = q Pn 0 ( eV / VT − 1) (1.61) Lp Similar,the current density due to electrons injected is given b y Dn (1. 62) Jn = q n p 0 ( eV / VT − 1) Ln S. C. Lin, EE National Chin-Yi University of Technology 81 The total current I = ( J p + J n ) A (since J p and J n are in the same direction ) ⎡ qD p Pn 0 qDn n p 0 ⎤ V / V I = A⎢ + ⎥ ( e T − 1) Ln ⎥⎦ ⎢⎣ L p ni2 Substituting for Pn 0 = (1.29), ND ⎡ Dp Dn ⎤ V / VT 2 I = Aqni ⎢ + − 1) ⎥ (e ⎢⎣ L p N D Ln N A ⎥⎦ N p0 ni2 = (1.31) NA (1.63) Is ⎡ Dp Dn ⎤ I s = Aqn ⎢ + ⎥ ⎢⎣ L p N D Ln N A ⎥⎦ 2 i S. C. Lin, EE National Chin-Yi University of Technology (1.65) 82 I Is 0 V Figure 1.40 The pn junction I-V characteristic S. C. Lin, EE National Chin-Yi University of Technology 83 S. C. Lin, EE National Chin-Yi University of Technology 84
