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Chapter 10 Diodes 1. Understand diode operation and select diodes for various applications. 2. Analyze nonlinear circuits using the graphical load-line technique. 3. Analyze and design simple voltage-regulator circuits. 4. Solve circuits using the ideal-diode model and piecewise-linear models. 5. Understand various rectifier and wave-shaping circuits. 6. Understand small-signal equivalent circuits. 10. Diodes – Basic Diode Concepts 10.1 Basic Diode Concepts 10.1.1 Intrinsic Semiconductors * Energy Diagrams – Insulator, Semiconductor, and Conductor the energy diagram for the three types of solids 2 10. Diodes – Basic Diode Concepts 10.1.1 Intrinsic Semiconductors * Intrinsic (pure) Si Semiconductor: Thermal Excitation, Electron-Hole Pair, Recombination, and Equilibrium When equilibriu m between excitation and recombination is reached : electron density hole density ni pi 1.5 10 10 cm -3 for intrinsic Si crystal at 300 K ( Note : Si crystal atom density is ~ 5 10 22 cm -3 ) 3 10. Diodes – Basic Diode Concepts 10.1.1 Intrinsic Semiconductors *Apply a voltage across a piece of Si: electron current and hole current 4 10. Diodes – Basic Diode Concepts 10.1.2 N- and P- Type Semiconductors * Doping: adding of impurities (i.e., dopants) to the intrinsic semiconductor material. * N-type: adding Group V dopant (or donor) such as As, P, Sb,… n p constant for a semiconductor For Si at 300K n p ni2 pi2 1.5 10 10 2 In n - type material n N d the donor conceration n N d ni , p pi We call electron the major charge carrier hole the minor cahage carrier 5 10. Diodes – Basic Diode Concepts 10.1.2 N- and P- Type Semiconductors * Doping: adding of impurities (i.e., dopants) to the intrinsic semiconductor material. * P-type: adding Group III dopant (or acceptor) such as Al, B, Ga,… n p constant for a semiconductor For Si at 300K n p n p 1.5 10 2 i 2 i 10 2 In p - type material p N a the acceptor conceration p N a pi , n ni We call hole the major charge carrier electron the minor cahage carrier 6 10. Diodes – Basic Diode Concepts 10.1.3 The PN-Junction * The interface in-between p-type and n-type material is called a pn-junction. The barrier potential VB 0.6 0.7V for Si and 0.3V for Ge at 300K : as T ,VB . 7 10. Diodes – Basic Diode Concepts 10.1.4 Biasing the PN-Junction * There is no movement of charge through a pn-junction at equilibrium. * The pn-junction form a diode which allows current in only one direction and prevent the current in the other direction as determined by the bias. 8 10. Diodes – Basic Diode Concepts 10.1.4 Biasing the PN-Junction *Forward Bias: dc voltage positive terminal connected to the p region and negative to the n region. It is the condition that permits current through the pn-junction of a diode. 9 10.1.4 Biasing the PN-Junction *Forward Bias: dc voltage positive terminal connected to the p region and negative to the n region. It is the condition that permits current through the pn-junction of a diode. 10 10. Diodes – Basic Diode Concepts 10.1.4 Biasing the PN-Junction *Forward Bias: Current Voltage Forward direction Diode as an insulator Diode as a conductor 11 10. Diodes – Basic Diode Concepts *Reverse Bias: dc voltage negative terminal connected to the p region and positive to the n region. Depletion region widens until its potential difference equals the bias voltage, majoritycarrier current ceases. 12 10. Diodes – Basic Diode Concepts *Reverse Bias: majority-carrier current ceases. * However, there is still a very small current produced by minority carriers. 13 10. Diodes – Basic Diode Concepts 10.1.4 Biasing the PN-Junction * Reverse Breakdown: As reverse voltage reach certain value, avalanche occurs and generates large current. Diode is acting as a Conductor after the Breakdown voltage 14 10. Diodes – Basic Diode Concepts 10.1.5 The Diode Characteristic I-V Curve Switch ON region Switch OFF region 15 10. Diodes – Basic Diode Concepts 10.1.6 Shockley Equation * The Shockley equation is a theoretical result under certain simplification: vD 1 i D I s exp n VT where I s 10 -14 A at 300K is the (reverse) saturation current, n 1 to 2 is the emission coefficient, VT kT 0.026V at 300K is the thermal voltage q k is the Boltzman' s constant, q 1.60 10 -19 C v when v D 0.1V, i D I s exp D n VT This equation is not applicable when v D 0 16 10. Diodes – Load-Line Analysis of Diode Circuits 10.2 Load-Line Analysis of Diode Circuit dv di We can use v iR, i C , v L ,... dt dt vD 1 but when there is a diode : i D I s exp n VT It is difficult to write KCL or KVL equations. For the circuit shown, KVL gives : VSS R i D v D If the I - V curve of the diode is given, iD 0 we can perform the " Load - Line Analysis" vD 0 17 10. Diodes – Load-Line Analysis of Diode Circuits Example 10.1- Load-Line Analysis For the circuit shown, Given : VSS 2V, R 1kΩ , the I - V curve of the diode Find : the diode current and voltage at the operating point (Q - point) VSS R i D v D , i.e., 2 1000 i D v D perform load - line analysis at the operating point VDQ 0.70 V, i DQ 1.3 mA 18 10. Diodes – Load-Line Analysis of Diode Circuits Example 10.2 - Load-Line Analysis For the circuit shown, Given : Vss 10 V, R 10 k , the I - V curve of the diode Find : the diode current and voltage at the operating point VSS R i D v D , i.e., 10 10k i D v D perform load - line analysis at the operating point VDQ 0.68 V, i DQ 0.93 mA 19 10. Diodes – Zener Diode Voltage-Regulator Circuits 10.3 Zener-Diode Voltage-Regulator Circuits 10.3.1 The Zener Diode * Zener diode is designed for operation in the reverse-breakdown region. * The breakdown voltage is controlled by the doping level (-1.8 V to -200 V). * The major application of Zener diode is to provide an output reference that is stable despite changes in input voltage – power supplies, voltmeter,… 20 10. Diodes – Zener-Diode Voltage-Regulator Circuits 10.3.2 Zener-Diode Voltage-Regulator Circuits * Sometimes, a circuit that produces constant output voltage while operating from a variable supply voltage is needed. Such circuits are called voltage regulator. * The Zener diode has a breakdown voltage equal to the desired output voltage. * The resistor limits the diode current to a safe value so that Zener diode does not overheat. 21 10. Diodes – Zener-Diode Voltage-Regulator Circuits Example 10.3 – Zener-Diode VoltageRegulator Circuits Given : the Zener diode I - V curve, R 1k Find : the output voltage for VSS 15 V and VSS 20 V KVL gives the load line : VSS R i D v D 0 From the Q - point we have : vo 10.0 V for VSS 15 V vo 10.5 V for VSS 20 V 5V change in input 0.5V change in vo Actual Zener diode performs much better! 22 10. Diodes – Zener-Diode Voltage-Regulator Circuits 10.3.3 Load-Line Analysis of Complex Circuits * Use the Thevenin Equivalent 23 10. Diodes – Zener-Diode Voltage-Regulator Circuits Example 10.4 – Zener-Diode Voltage-Regulator with a Load Given : Zener diode I - V curve, VSS 24V, R 1.2k , RL 6k Find : the load voltage v L and source currents I S RL R RL Applying Thevenin Equivalent VT VSS 20V , RT 1k R RL R RL VT RT i D v D 0 v L -v D 10.0 V I S (V SS - v L )/R 11.67 mA 24 10. Diodes – Zener-Diode Voltage-Regulator Circuits Quiz – Exercise 10.5 Given : the circuit and the Zener diode I - V curve as shown. Find : the output vol tage v o for i L 0, i L 20mA, and i L 100mA 25 10. Diodes – Ideal-Diode Model 10.4 Ideal-Diode Model * Graphical load-line analysis is too cumbersome for complex circuits, * We may apply “Ideal-Diode Model” to simplify the analysis: (1) in forward direction: short-circuit assumption, zero voltage drop; (2) in reverse direction: open-circuit assumption. * The ideal-diode model can be used when the forward voltage drop and reverse currents are negligible. 26 10. Diodes – Ideal-Diode Model 10.4 Ideal-Diode Model * In analysis of a circuit containing diodes, we may not know in advance which diodes are on and which are off. * What we do is first to make a guess on the state of the diodes in the circuit: (1)For " assumed on diodes" : check if i D is positive; (2) For " assumed off diodes" : check if v D is negative ALL YES BINGO! not ALL YES make another guess.... iterates until " ALL YES" 27 10. Diodes – Ideal-Diode Model Example 10.5 – Analysis by Assumed Diode States Analysis the circuit by assuming D1is off and D2 on +10V +7V (1) assume D1 off, D2 on +3V Vd1 != +7V, because it should be in forward bias! (2) assume i D2 D1 on, D2 off +6V R1 0.5mA OK! v D1 7V not OK! +3V i D1 1 mA OK! v D2 -3 V OK! 28 10. Diodes – Ideal-Diode Model Quiz – Exercise 10.8c * Find the diode states by using ideal-diode model. Starting by assuming both diodes are on. (1) assume D3 on D4 on (2) assume D3 off and D4 on iD 3 -1.7 mA, not OK i D 4 6.7 mA, OK i D 4 5 mA, OK v D 3 -5 V, OK 29 10. Diodes – Piecewise-Linear Diode Models 10.5 Piecewise-Linear Diode Models 10.5.1 Modified Ideal-Diode Model * This modified ideal-diode model is usually accurate enough in most of the circuit analysis. 30 10. Diodes – Piecewise-Linear Diode Models 10.5.2 Piecewise-Linear Diode Models v R a i Va 31 10. Diodes – Rectifier Circuits 10.6 Rectifier Circuits * Rectifiers convert ac power to dc power. * Rectifiers form the basis for electronic power suppliers and battery charging circuits. 10.6.1 Half-Wave Rectifier Rewerse Forward 32 10. Diodes – Rectifier Circuits * Battery-Charging Circuit No current flow In this half period. Current Flow On the battery * The current flows only in the direction that charges the battery. 33 10. Diodes – Rectifier Circuits * Half-Wave Rectifier with Smoothing Capacitor * To place a large capacitance across the output terminals: Capacitor is discharging 34 10. Diodes – Rectifier Circuits 10.6.2 Full-Wave Rectifier Circuits Diode B OFF Diode A OFF * Center-Tapped Full-Wave Rectifier – two half-wave rectifier with out-ofphase source voltages and a common ground. * When upper source supplies “+” voltage to diode A, the lower source supplies “-” voltage to diode B; and vice versa. * We can also smooth the output by using a large capacitance. 35 10. Diodes – Rectifier Circuits 10.6.2 Full-Wave Rectifier Circuits * The Diode-Bridge Full-Wave Rectifier: A,B C,D Without Capacitor With Capacitor 36 10. Diodes – Wave-Shaping Circuits 10.7 Wave-Shaping Circuits 10.7.1 Clipper Circuits * A portion of an input signal waveform is “clipped” off. 37 10. Diodes – Wave-Shaping Circuits 10.7 Wave-Shaping Circuits 10.7.2 Clamper Circuits * Clamp circuits are used to add a dc component to an ac input waveform so that the positive (or negative) peaks are “clamped” to a specified voltage value. 38 10. Diodes – Linear Small-Signal Equivalent Circuits 10.8 Linear Small-Signal Equivalent Circuits * In most of the electronic circuits, dc supply voltages are used to bias a nonlinear device at an operating point and a small signal is injected into the circuits. * We often split the analysis of such circuit into two parts: (1) Analyze the dc circuit to find operating point, (2) Analyze the small signal ( by using the “linear smallsignal equivalent circuit”.) 39 10. Diodes – Linear Small-Signal Equivalent Circuits 10.8 Linear Small-Signal Equivalent Circuits * A diode in linear small-signal equivalent circuit is simplified to a resistor. * We first determine the operating point (or the “quiescent point” or Q point) by dc bias. * When small ac signal injects, it swings the Q point slightly up and down. * If the signal is small enough, the characteristic is straight. d iD i D d vD d iD d vD v D Q i D is the small change in diode current v D is the small change in diode voltage 40 Q 10. Diodes – Linear Small-Signal Equivalent Circuits 10.8 Linear Small-Signal Equivalent Circuits Define the dynamic resistance of the diode as : d i rd D d v D di i D D d vD Q d iD d vD 1 We will have : v D v D i D rd Q Replace i D and v D by id and v d denoting small changes, we have for ac signals : vd id rd Furthermore, by applying the Shockley equation, we have : rd n VT I DQ 41 Q 10. Diodes – Linear Small-Signal Equivalent Circuits 10.8 Linear Small-Signal Equivalent Circuits vd n VT id , rd rd I DQ * By using these two equations, we can treat diode simply as a linear resistor in small ac signal analysis. * Note: An ac voltage of fixed amplitude produces different ac current change at different Q point. 42 10. Diodes – Linear Small-Signal Equivalent Circuits 10.8 Linear Small-Signal Equivalent Circuits i D I DQ i d v D V DQ v d vd nV id , rd T rd I DQ (1) V DQ and I DQ represent the dc signals at the Q point. (2) v d and i d represent the small sc signals. (3) v D and i D represent the total instantaneous diode voltage and current. 43 10. Diodes – Linear Small-Signal Equivalent Circuits Voltage-Controlled Attenuator * The function of this circuit is to produce an output signal that is a variable fraction of the ac input signal. * Two large coupling capacitors: behave like short circuit for ac signal and open circuit for dc, thus the Q point of the diode is unaffected by the ac input and the load. ZC 1 j C 44 10. Diodes – Linear Small-Signal Equivalent Circuits Voltage-Controlled Attenuator First apply dc analysis to find the diode Q point, determine I DQ , then the rd of the diode : rd n VT I DQ Next,we perform small ac signal analysis : (note : the dc voltage source has an ac component of current but no ac voltage, the dc voltage source is equivalent to a short circuit for ac signal.) Rp vo 1 Rp , based on voltage divider : Av 1 1 RC 1 RL 1 rd vin R R p 45 10. Diodes – Linear Small-Signal Equivalent Circuits Exercise 10.20 - Voltage-Controlled Attenuator Given : R 100 Ω , RC R L 2kΩ , diode n 1 at 300K Find : the Q - point values assuming V f 0.6V and Av for VC 1.6 and 10.6V First apply dc analysis to find the diode Q point, VC - 0.6 nVT I DQ , rd withVT 0.026V RC I DQ Next, we perform small ac signal analysis : Rp vo 1 Rp , Av 1 RC 1 R L 1 rd v in R R p 46 Chapter 11: Amplifiers: Specifications and External Characteristics 11. Amplifiers – Basic Amplifier Concepts 11.1 Basic Amplifier Concepts 11.1.1 For Starting * Ideally, an amplifier produces an output signal with identical waveshape as the input signal but with a larger amplitude. v o t Av v i t Av is the Voltage Gain 48 11. Amplifiers – Basic Amplifier Concepts 11.1.1 For Starting * inverting and non-inverting amplifiers 49 11. Amplifiers – Basic Amplifier Concepts 11.1.1 For Starting * Often, one of the amplifier input terminals and one of the output terminals are connected to a common ground. * The common ground serve as the return path for signal and the dc power supply currents. 50 11. Amplifiers – Basic Amplifier Concepts 11.1.1 For Starting * Another example for common ground 51 11. Amplifiers – Basic Amplifier Concepts 11.1.2 Voltage-Amplifier Model * Amplification can be modeled by a controlled source. Amplifier Av > 0, non inverting amplifier Av < 0, inverting amplifier Av = 1, 1:1 buffer (but may Ai > 1) Ri : the input resistance (or impedance), is the equivalent resistance seen when looking into the input terminals. Ro : in series with the output terminals, is the output resistance (or impedance) Avo v oc /v i : the open - circuit voltage gain (note : the real gain is smaller tham Avo ) 52 11. Amplifiers – Basic Amplifier Concepts 11.1.3 Current and Voltage Gains i i is the current delivered into the input terminals of the amplifier; i o is the current flowing through the load. io The current gain Ai is the ratio between output and input currents : Ai ii Furthermor e, Ai i o v o RL R v Av i , where Av o is the voltage gain ii v i Ri RL vi The voltage gain Av is usually smaller than the open - circuit voltage gain Avo 53 11. Amplifiers – Basic Amplifier Concepts 11.1.3 Power Gain The power gain is the ratio of the output power to the input power : G Po Pi Since the average power is the product of the rms values Vrms and I rms , we have : G Po Vrms o I rms o 2 R Av Ai Av i Pi Vrms i I rms i RL 54 11. Amplifiers – Basic Amplifier Concepts Example 11.1 Calculating Amplifier Performance Find the voltage gains Avs Vo /V s and Av Vo /V i Also, find the current gain and the power gain Vo AvoVi R L /(R o R L ) RL 8 4 Av Av o 10 8000 Vi Vi Ro R L 28 Av s Vo Vo Ri Av 5333 V s Vi (Ri Rs )/R i Ri R s Note : due to the loading effect, A vs Ai Av Ri 2 10 9 , RL Av Av,o G Av Ai 16 10 12 55 11. Amplifiers – Cascade Amplifiers 11.2 Cascade Amplifiers v o 2 v o1 v o 2 v o1 v o 2 Av Av Av 1 Av 2 v i 1 v i 1 v 01 v i 1 v i 2 Av o vo c 2 vi 1 Av o 2 v i 2 vi 1 Av o 2 In addition, Ai Ai 1 Ai 2 , vo 1 vi 2 Av o Av 1 Av o 2 G G 1G 2 56 11. Amplifiers – Cascade Amplifiers Example 11.2 – Calculating Performance of Cascade Amplifier Ri 2 R L 1 Find the current gain, voltage gain, and power gain Av 1 Avo 1 Ri 2 RL 150 , Av 2 Avo 2 50 Av Av 1 Av 2 7500 Ri 2 Ro 1 R L Ro 2 Ri 1 Ri 2 5 Ai 1 Av 1 10 , Ai 2 Av 2 750 Ai Ai 1 Ai 2 75 10 6 Ri 2 RL G 1 Av 1 Ai 1 1.5 10 7 , G 2 Av 2 Ai 2 3.75 10 4 G G 1 G 2 5.625 10 11 57 11. Amplifiers – Cascade Amplifiers Example 11.3– Simplified Model for Amplifier Cascade Av 1 Av o 1 Ri 2 150 R i 2 Ro 1 Av o 2 100 Av o Av 1 Av o 2 15 10 3 58 11. Amplifiers – Power Supplies and Efficiency 11. 3 Power Supplies and Efficiency * The power gain of an amplifier is usually large, the additional power is taken from the power supply. The total average power supplied to the amplifier is Ps V AA I A V BB I B Conservation of energy : Pi Ps Po Pd Pi is the power of input signal Po is the power of output signal Pd is the power dissipated in internal carcuits The efficiency of an amplifier : Po η 100% Ps 59 11. Amplifiers – Power Supplies and Efficiency Example 11.4 Amplifier Efficiency 2 V Pi i 10 11 W 10 pW Ri RL Vo Avo V i 8 V rms R L Ro 2 Vo Po 8W RL Ps V AA I A V BB I B 22.5 W Pd Ps Pi Po 14.5 W η Po 35.6% Ps 60 11. Amplifiers – Additional Amplifier Models 11.4 Additional Amplifier Models Current-Amplifier Model Thevenin vs. Norton approach vi ii Ri and i osc Avo v i Ro the short circuit current gain : Aisc i osc R Avo i ii Ro Ri and Ro are the same for the two models. 61 11. Amplifiers – Additional Amplifier Models Example 11.5 – Transform Voltage-Amplifier to Current-Amplifier Model vi ii Ri Aisc and i osc Avo v i Ro i osc Ri Avo 10 3 ii Ro 62 11. Amplifiers – Additional Amplifier Models 11.4 Additional Amplifier Models Transconductance-Amplifier Model Gmsc iosc vi Transresistance-Amplifier Model Rmsc vosc ii 63 11. Amplifiers – Ideal Amplifiers 11.6 Ideal Amplifiers Voltage Amplifier : Ri , v i v s max. Ro 0, v o Avo v i max Maximum Voltage Gain! Current Amplifier : Ri 0, i i i i sc max. Ro , i o Ai sc i i max. Maximum Current Gain! 64 11. Amplifiers – Frequency Response 11.7 Frequency Response * The gain of an amplifier is a function of frequency. Av Vo Vi * Both amplitude and phase are affected. Example 11.8 The input for a certain amplifier is : v i t 0.1 cos2000 π t 30 the outpot is : v 0 t 10 cos2000 π t 15 Find the complex voltage gain and express the amplitude in dB Vo 1015 100 45 Vi 0.1 30 , Vo 1015 Av 0.1 30 Vi Note : there is a 45 phase shift. Av dB 20 log Av 20 log( 100 ) 40 dB 65 11. Amplifiers – Frequency Response Gain as a Function of Frequency 66 11. Amplifiers – Frequency Response AC Coupled Amplifiers The gain drops to zero at dc (low frequency). 67 11. Amplifiers – Frequency Response DC Coupled Amplifiers Amplifiers that are realized as integrated circuits are often dc coupled, because capacitors or transformers can not be fabricated in integrated form. 68 11. Amplifiers – Frequency Response High-Frequency Drop Off * The small amount of capacitance in parallel or inductances in series with the signal path in the amplifier circuit will cause the gain of the amplifier to drop at high frequencies. j2 πfL, open as f 1/j2 πfC, short as f 69 11. Amplifiers – Frequency Response Half-Power Frequencies and Bandwidth * Bandwidth is the distance between the half-power frequencies. * Half-power frequencies: A Amid / 2 3dB from Amid 20log(1/ 2 ) -3.01 dB * Wideband (Baseband) Amp. * Narrowband (Bandpass) Amp. 70 11. Amplifiers – Linear Waveform Distortion 11.8 Linear Waveform Distortion * Distortion may occur even though the amplifier is linear (i.e., obeys superposition principle). Amplitude Distortion If a signal contains components of various frequencies, the output waveform may be distorted due to the frequency response of the amplifier gain. 71 11. Amplifiers – Linear Waveform Distortion Phase Distortion * If the phase shift of an amplifier is not proportional to the frequency, phase distortion may occur. Assume we have an input like : v i t 3 cos2000 π t cos6000 π t and three amplifiiers having the gains : The output would look like : v oA t 30 cos2000 π t 10 cos6000 π t v oB t 30 cos2000 π t 45 10 cos6000 π t 135 v oC t 30 cos2000 π t 45 10 cos6000 π t 45 72 11. Amplifiers – Linear Waveform Distortion 11.8 Linear Waveform Distortion * In order to avoid distortion: (1) the magnitude of the gain must be constant against frequency. (2) The phase response must be proportional to the frequency. θ 1 : θ 2 ω1 : ω2 T2 : T1 θ1 θ T1 : 2 T2 2π 2π θ θ ( ω / ω2 ) 1 T1 : 1 1 T1 ( ω2 / ω1 ) 2π 2π 1:1 The same time delay! Δt 1 : Δt 2 73 11. Amplifiers – Pulse Response 11.9 Pulse Response 74 11. Amplifiers – Nonlinear Distortion 11.10 Nonlinear Distortion 75 11. Amplifiers – Differential Amplifiers 11.11 Differential Amplifiers * A Differential amplifier has two input sources, the output voltage is proportional to the difference between the two input voltages. v o t Ad v i 1 t v i 2 t Ad v i 1 t Ad v i 2 t Non-inverting input Inverting input Define the differential signal : v id v i 1 v i 2 and Ad the differentail gain We can write the output of a differentail amplifier as : v o Ad v id 76 11. Amplifiers – Differential Amplifiers Electrocardiogram * The desired waveform is given by the difference between the potentials measured by the two electrodes, i.e., the output of an ideal differential amplifier.v o Ad v id Ad (v i1 - v i2 ) * While both electrodes (also act like antennas) pick a commonmode signal (noise) from the 60 Hz power line: v i1 Vn cos(377t φn ), v i2 Vn cos(377t φn ) 77 11. Amplifiers – Differential Amplifiers Electrocardiogram v o Ad v id Ad (v i1 - v i2 ) v i1 Vn cos(377t φn ), v i2 Vn cos(377t φn ) * Ideally, the common-mode signal was nullified by the differential amplifier. * However, real differential amplifier responds to both differential-mode and common-mode signals: v o Ad v id Acm v icm where Acm is the common - mode gain, v id is the differential - mode signal and v icm the common - mode signal Common-Mode Rejection Ratio (CMRR): CMRR 20 log Ad Acm * At 60 Hz, CMRR of 120 dB is considered good. 78 11. Amplifiers – Differential Amplifiers Example 11.12 Determination of Minimum CMRR Specification * Find the minimum CMRR for an electrocardiogram amplifier if the differential gain is 1000, the desired differential input signal has a peak amplitude of 1 mV, the common-mode signal is 100 V-peak 60 Hz sine wave, and it is desired that output contain a peak common-mode signal that is 1% or less of the peak output caused by the differential signal. peak v i d 1 mV, Ad 1000 v o d 1 V peak v o cm 1 V 1% 0.01 V Acm CMRR 20 log Ad Acm 20 log 0.01V 10 4 100 V 1000 140 dB 4 10 Ans : CMRR 140 dB 79