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electronics electronics • About some fundamental semiconductor devices • Diode 半導體二極管 , transistor晶體管 & Op Amp 運算放大器 electronics • Semiconductor materials: – n-type (e-) & p-type (hole) • Using semiconductor, we can control the amount current flowing through the circuit. • Logical and mathematical operation can be carried out. Semiconductor diode ( 半導體二極管 ) I A B p n • To allow current flow in one direction only • A B: current can pass (VA > VB) [Forward biased] • B A: no current can pass [Reversed biased] Ideal diode R 6V = 0V 6V R 0V R 0V = 6V 0V R 6V I – V characteristics 特性曲線 I Ideal Real reverse forward Junction potential barrier Why V Vf = 0.7V Forward bias or forward voltage Vf ( 正向偏壓 ) Vf = 0.6V to 0.8V for silicon diodes Rectification 整流 • a.c. d.c. half-wave rectification 半波整流 Vin Vout Reverse bias: Vout = 0 Vout Vf Forward bias: Vout = Vin – Vf rectified t full-wave rectification 全波整流 • Centre-tapped full-wave rectifier ( 中插頭全波整流器 ) • Two diodes and one transformer are used D1 A main C Direction of current I B load at output D2 Va > Vb > Vc: A D1 E B Vc > Vb > Va: C D2 E B E Bridge full-wave rectifier 橋式整流器 • Four diodes are used. A D4 D1 Y X D3 D2 B I load at output Va > Vb: A D1 X R Y D3 B Vb > Va: B D2 X R Y D4 A T/2 = 0.01S • Bridge full-wave rectifier 橋式整流器 • Unidirectional but pulsating voltage • If the original frequency is 50Hz, the peak occurs at every 0.01s. Smoothing ( 濾波 ) Storage capacitor (reservoir) ( 存儲電容器 ) Simple smoothing circuit ( 平流電路 ) rectified + unsmoothed d.c 已整流但未平流 直流電 + C charge up discharge T/2 Without C R load at output with C ripple 紋波 • Time constant = RC • Larger time constant smoother output • in order to have a good smoothing effect, RC should be very large compared to T/2 (time between peaks occurs). Filter capacitor ( 濾波電容器 ) filter circuit ( 濾波電路 ) a.c. rectified + unsmoothed d.c 已整流但未平 流直流電 L + C + R C2 load at output d.c. L and C2 in series, acts as a potential divider L: read a.c. component C: read d.c. component a.c. rectified + unsmoothed d.c 已整流但未平 流直流電 L + C + R C2 load at output d.c. XL = L high reactance for a.c. (high f) low reactance for d.c. (f 0) XC = 1/ C low reactance for a.c. (high f) high reactance for d.c. (f 0) NPN transistor ( NPN 晶體管 ) Emitter Collector N-type p-type N-type p-type IC IB C B Base VCE E VBE IE NPN transistor ( NPN 晶體管 ) • B: base基極 C: collector集電極 E: emitter 發射極 • IB: base current • IC: collector current • IE: emitter current IC IB C B • Used as switches or amplifiers VBE • In an integrated circuit (IC): contains thousands of transistors VCE E IE http://micro.magnet.fsu.edu/electromag /java/transistor/index.html Making of a transistor Investigation RL ICRL IBRB Vin RB VCC VBE VCE Investigation RL ICRL IBRB Vin RB VCC VBE VCE Investigation RL ICRL IBRB Vin RB VCC VBE VCE Investigation RL IC IB Vin RB VCC VBE VCE Common emitter configuration IC • E is earthed (0V) IB C B VCE E VBE IE Two important relationships • IE = IB + I C IC • Current gain 電流增益 = IC / I B I B = IC / IB • Typically, > 100 • is transistor dependent. C B E IE Two important relationships = IC / IB IC = IB IC IB IC / mA C B E Slope = IB / mA IE Max IC is limited by VCC and RL RL ICRL IBRB VCC Max IC = VccRB/ RL Vin VBE Vcc = IC RL+VCE VCE Max IC is limited by VCC and RL IC / mA saturated Max IC = Vcc / RL Slope = RL IB / mA ICRL IBRB Vin RB VCC VBE VCE • Input ( base ) characteristic ( IBVBE curve ) ( 輸入特徵曲線 ) •Same as a diode IB IB ideal VBE VBE 0.7 V • Input ( base ) characteristic ( IBVBE curve ) ( 輸入特徵曲線 ) • Same as a diode •ideally, VBE = 0.7V I Vin 0.7V, VBE = VIN, IB =0 B ideal Vin > 0.7V, VIN = IB RB + VBE VBE IBRB VIN VBE 0.7 V Vin > 0.7V, VIN = IB RB + VBE IB IBRB VIN ideal VBE VIN VBE IN 0.7 V Vin > 0.7V, VIN = IB RB + VBE IBRB VIN IBIB IB ideal ideal VBE ideal VBE IN VIN VIN 0.7 V 0.7 V 0.7VV in Vin > 0.7V, VIN = IB RB + VBE IBRB VIN VBE VIN - VBE = IB RB IB = VIN / RB - VBE / RB IB ideal 0.7 V • IB = VIN /RB - VBE /RB or IB = VIN /RB - 0.7 /RB Vin Function of RB: limiting the IB & IC ( IBVIN curves ) VIN • IB = VIN /RB - 0.7 /RB Output ( collector ) characteristic ( ICVCE curves ) ( 輸出特徵曲線 ) VCC • VCE = VCC – IC RL Ideally IC 5.2 mA IB = 40 mA 3.9 mA IB = 30 mA 2.6 mA IB = 20 mA IC IB = 10 mA 1.3 mA 0.2 V 1V VCE IB = 0 IC = 0 VCE>0.2 V, IC becomes constant IC = IB Typically, > 100 4mA 4mA 4mA IB = 0.04mA IB = 0.04mA IB = 0.04mA Current transfer characteristic ( IC IB curve ) ( 電流傳輸特徵曲線 ) = IC / IB= IC / IB IC / mA Max IC = Vcc / RL Slope = IB / mA Input/output voltage characteristic ( 輸入/輸出電壓特徵曲線 ) Vout = VCE / V cutoff 6.0 Vout = VCE 0.2 * ideal linear region 0.5 0.7 saturation Vin /V http://www.ngsir.netfirms.com/englishhtm/Amplifier.htm Vout = VCE / V Cut-off region cutoff 6.0 • • • • • Vcc = 6 V Vin<0.7 V 0.2 Ib = 0 Ic = 0 By VCC = ICRL + Vout Vout = Vcc = 6 V ideal linear region Slope = A (voltage gain) 0.5 0.7 saturation Vin /V Linear region 線性區域,放大區 Vout Slope A voltage gain ( 電壓增益) Vin • Vout varies linearly with Vin • Input: Vin>0.7 V IB>0 and VBE = constant • also Vin = VBE + IBRB d [ Vin = VBE + IBRB ] Vin = RB IB ….. (1) • Output: VCC = ICRL + Vout d[ VCC = ICRL + Vout ] 0 = RLIC + Vout … (2) ( 2) (1) Vout RL I C A Vin RB I B RL A RB A = slope at the linear region Saturation ( 飽和 ) • Vout remains almost zero 0.1-0.2 V • Since IC has a maximum value Vcc / RL when saturated, given by Vout 0 saturation occurs when Ib > (Ib)saturation = (Ic)max / Vin> (Vin)saturated = IbRB VBE Transistor as a linear voltage amplifier 線性電壓放大器 IC Vi C Signal generator vi RL IB IB Vin Vo C VCC RB Vout vo IC Vout = VCE / V cutoff 6.0 ideal linear region Slope = A (voltage gain) 0.2 0.5 0.7 saturation Vin /V Vin = total input voltage Vi = bias (d.c.) voltage vi = input signal (to be amplified) Vin = Vi + vi Vout = total output voltage Vo = output voltage due to bias input voltage vo = output signal (amplified) Vout = Vo + vo voltage amplification use linear region only (suppose Vcc = 6V) Vout = VCE / V cutoff 6.0 Middle of linear region Vo 3V 0.2 linear region 0.5 0.7 Vi saturation Vin /V Proper biasing condition ( 最佳偏壓狀態 ) • adjust the potentiometer ( Vi & IB) steady output voltage Vo = 3 V (the middle of the linear region) • Without signal generator, (vi = 0) Vin = Vi & Vout = Vo (vo = 0) bias voltage Vi ( 偏壓 ) & bias current IB偏壓電流 voltage amplification use linear region only (suppose Vcc = 6V) Vout = VCE / V cutoff 6.0 Middle of linear region Vo 3V 0.2 linear region 0.5 0.7 Vi saturation Vin /V Vout Vout = VCE / V 6.0 Vo t Vin /V Vin t Vi The a.c. input vi can then be amplified with maximum possible amplitude before distortion. Distortion: range of input/output voltage is out of the linear region Vout Vo Vout = VCE / 6. V 0 t Vin Vin /V t Vi • 180o out of phase between vi and vo , A is -ve Vo & Vi: (constant)d.c. & vi & vo: a.c. Vout Vo v o v o peak value of v o A Vin Vi vi vi peak value of vi vo /vi = A = - (RL/RB) For calculating the biasing voltage & biasing current, use the following equation: Vin = VBE + IBRB VCC = ICRL + Vout IC = IB remarks l capacitor C: d.c. blocking capacitor ( 隔直流電容 ) large capacitance block d.c. a.c. only l capacitor C : blocking d.c. (output now) but it can be omitted if the CRO is set at AC mode l variation of vi are too large vo will move into the non-linear regions distortions l RL decreases or RB increases smaller A wider linear region P.6 example 1 Best biasing condition: set Vo = 3V A = - (RL/RB) = -180 (2.2k/15k) = -26.4 VCC = ICRL + VOUT 6 = IC(2.2k) + 3 IC = 1.3636mA IB‘ = IC / = 1.3636 / 180 = 0.0076mA Vin = (Vi + 0) = VBE + IB’RB Vi = 0.7 + (0.0076m)(15k) = 0.814V • In this example, we see that typically, RB (15k) > RL (2.2k) • Since IB << IC (IC = IB) Example 2 Let Vcc = 6V IB RB C Signal generator vi RB IC RL C VCC IB vo IB IB IC + IB • set Vo = 3V • Now Vcc = VO + ICRL Vcc = IB’ RB’ + VBE IC = IB IC = 1.3636 mA IB’ = 0.00909mA Vcc = IB’ RB’ + VBE 6 = (0.00909m x RB’) + 0.6 RB’ = 594k AL 01-41 Operational amplifier ( Op amp 運算放大器 ) V1: Inverting input ( 反相輸入 ) V2: Non-inverting input ( 非反相輸入 ) VS: +VS V1 V2 Vo + VS The voltage of the power supply For example: + VS = +15 V, VS = 15 V VO = Voltage output Properties of op amp 1. A very high input impedance ( 106-1012 ) It draws tiny current (about 0A) from the device or circuit supplying its input. 2. A very low output impedance ( 100 ) 3. The open-loop voltage gain AO 開環電壓增益 VO = Ao (V2 V1) AO is very high, of the order of 105 . Ideal op amp: • • • infinite large input impedance 無限大的輸入阻抗 zero output impedance 零輸出阻抗 infinite large open-loop voltage gain AO 無限大開環電壓增益 I/O voltage transfer characteristic Vo / V saturation + VS V (V2 V1) / mV V saturation VS +VS V1 V2 + Vo VS V = VS / AO • If (V2 – V1) > V VO = VS • If (V2 – V1) < -V VO = -VS • If [ -V < (V2 – V1) < V ] (linear region) VO = Ao (V2 V1) • Since Ao is very large, V is very small V << 1mV • Larger AO smaller V narrow linear region Comparator ( 比較器 ) as a switch • if V2 V1>V saturation • compare two nearly equal voltage • V2 > V1 Vo = +ve • V2 < V1 Vo = -ve 98-40 Use of open loop gain Ao is very limited since it is too large and the op amp is often saturated. Negative feedback amplifier ( 負反饋放大器 ) • Closed-loop gain A 閉環電壓增益 feedback resistor Rf V1 Vo V2 + Advantages of negative feedback (I) Predictable & constant voltage gain A (independent of the intrinsic property (Ao & Vs) of the op amp) (II) More linear amplification (linear region of greater width) (III) The voltage gain is stable for a wider range of frequencies of input a.c. Negative feedback • We always use the assumption (I) V2 V1 (II) the current i flow between the inverting &non-inverting input 0 Inverting amplifier 反相放大器 Ii V1 VP VQ VQ = 0 VP 0 Now Vi VP = Ii Ri Vi = Ii Ri & …….…(1) Rf If Ri VP i VQ + virtual ground ( 虛接地 ) & VP VO = If Rf VO = If Rf VO Vi = Ii Ri & VO = If Rf …….…(1) since i 0 Ii = If + i If ….……(2) (1) & (2), Vi/Ri = VO/ Rf closed loop voltage gain A 閉環電壓增益 : Since Ii If VO I f Rf Rf A Vi I i Ri Ri • A depends only on Rf and Ri & is independent of Ao & Vs • If Rf = Ri A = 1 • it acts as an inverter反相器 Summing amplifier 加法放大器 V1 V2 V3 R1 I1 R2 I2 R3 I3 Rf If VP i VQ Vout VP VQ = 0 Vi VP = Ii Ri (i = 1, 3, 3) & VP VO = If Rf Vi /Ri = Ii & VO/Rf = If Vi /Ri = Ii & VO/Rf = If i=0 SIi = If + i If (1) & (2), …..(1) …..(2) V1 V2 V3 VO Rf R1 R2 R3 If R1 = R2 = R3 = Rf, Vo = - (V1 + V2 + V3) 97- 42 +15 +15 -15 0 +5 0 -5 -15 +5 V -5 V vo [5 2 5] 5 V Non-inverting amplifier 同相放大器 VP VQ = Vi VP VQ + i Vi If Rf i I R VO VP VQ = Vi VO VP = If Rf & (VO Vi)/Rf = If & i=0 VP 0 = IR Vi/R = I I = If i = If Vi VO Vi R Rf VO Rf 1 Vi R closed loop voltage gain: Rf A 1 R If Rf = 0 and R = (very large) then A = 1 voltage follower ( 電壓跟隨器 ) Though the output voltage is the same as the input voltage, its output current is very large. Owing to its high input but low output impedance, a voltage follower can be constructed into an electrometer. Input-output current characteristic of a voltage follower VP Ii VQ Vi VO = VP = VQ = Vi r IO + RL VO V I R O O L For open voltage gain, V VQ VP A0 A0 For input resistance, V VQ VP I i r I O RL Iir A0 A0 r Ii IO RL For A0 105 , r 106 , RL = 10 k , IO/Ii 107 Though voltage gain A is 1, the current is amplified. 00 - 42