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Chapter 14 Single-Transistors Amplifiers. Microelectronic Circuit Design Richard C. Jaeger Travis N. Blalock. Chap 14 - 1. Chapter Goals. Detailed study of three broad classes of amplifiers
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Chapter 14Single-Transistors Amplifiers Microelectronic Circuit Design Richard C. Jaeger Travis N. Blalock Microelectronic Circuit Design McGraw-Hill Chap 14 - 1
Chapter Goals • Detailed study of three broad classes of amplifiers • Inverting amplifiers- that provide high voltage gain with a 1800 phase shift, common-emitter and common-source configurations, • Followers- that provide nearly unity gain similar to op amp voltage follower, common-collector and common-drain configurations, • Noninverting amplifiers- that provide high voltage gain with no phase shift, common-base and common-gate configurations. • Detailed design of voltage gain, input voltage range, current gain, input and output resistances, coupling and bypass capacitor design and lower cutoff frequency for each type of amplifier. • Understand differences between SPICE ac (small-signal), transient (large-signal) and transfer function analysis modes. Microelectronic Circuit Design McGraw-Hill Chap 14 - 2
Signal Injection and Extraction: BJT • In forward-active region, • To cause change in current, vBE = vB - vE must be changed. Base or emitter terminals are used to inject signal because even if Early voltage is considered, collector voltage has negligible effect on terminal currents. • Substantial changes in collector or emitter currents can create large voltage drops across collector and emitter resistors and collector or emitter can be used to extract output. Since iB is a factor of bF smaller than iC or iE currents, base terminal is not used to extract output. Microelectronic Circuit Design McGraw-Hill Chap 14 - 3
Signal Injection and Extraction: FET • In pinch-off region, • To cause change in current, vGS = vG - vS must be changed. Gate or source terminals are used to inject signal because even with channel-length modulation, drain voltage has negligible effect on terminal currents. • Substantial changes in drain or source currents can create large voltage drops across drain and source resistors and drain or source can be used to extract output. Since iG is always zero, gate terminal is not used to extract output. Microelectronic Circuit Design McGraw-Hill Chap 14 - 4
Amplifier Families • Constraints for signal injection and extraction yield three families of amplifiers • Common-Emitter (C-E)/Common- Source (C-S) • Common-Base (C-B)/Common- Gate (C-G) • Common-Collector (C-C)/Common- Drain (C-D) • All circuit examples here use the four-resistor bias circuits to establish Q-point of the various amplifiers • Coupling and bypass capacitors are used to change the ac equivalent circuits. Microelectronic Circuit Design McGraw-Hill Chap 14 - 5
Inverting Amplifiers: Common-Emitter and Common-Source Circuits AC equivalent for C-E Amplifier AC equivalent for C-S Amplifier Microelectronic Circuit Design McGraw-Hill Chap 14 - 6
Inverting Amplifiers: Terminal Voltage Gain For C-S Amplifier, take limit of voltage gain of C-E amplifier as and If RE and RS are set to zero, the voltage gain has an upper bound of If RE and RS are large, the voltage gain has a lower bound of Using test source vb to drive the base terminal of the transistor, neglecting ro, Assuming R is the unbypassed resistor in the emitter or source. Microelectronic Circuit Design McGraw-Hill Chap 14 - 7
Inverting Amplifiers: Input Signal Range For small-signal operation, magnitude of vbe developed across rp in small-signal model must be less than 5 mV. If , vb can be increased beyond 5 mV limit. In case of FET, magnitude of vgs must be less than 0.2(VGS - VTN). Presence of RS increases permissible value of vg Microelectronic Circuit Design McGraw-Hill Chap 14 - 8
Inverting Amplifiers: Condition for gmR >>1 • This condition simplifies the gain expression and is used to stabilize voltage gain, achieve high input and output resistances and increase input signal range. • For BJT: • For MOSFET: Microelectronic Circuit Design McGraw-Hill Chap 14 - 9
Inverting Amplifiers: Input Resistance and Overall Voltage Gain Overall voltage gain is For C-S Amplifier, Input resistance looking into the base terminal is given by For C-S Amplifier, Microelectronic Circuit Design McGraw-Hill Chap 14 - 10
Inverting Amplifiers: Voltage Gain Calculations (Example) • Problem: Find overall voltage gain. • Given data: Q-point values and values for RI, R1, R2, R3, R7 ,for both BJT and FET as well as values for RE and RS . • Assumptions: Small-signal operating conditions. • Analysis: For C-E Amplifier, Microelectronic Circuit Design McGraw-Hill Chap 14 - 11
Inverting Amplifiers: Voltage Gain Calculations (Example contd.) • Analysis: For C-S Amplifier, Microelectronic Circuit Design McGraw-Hill Chap 14 - 12
Inverting Amplifiers: Output Resistance But Rout = ro when RE = 0, not infinite. Now, we also include ro in our analysis. Microelectronic Circuit Design McGraw-Hill Chap 14 - 13
Inverting Amplifiers: Output Resistance (contd.) Assuming and , with …….for Finite current gain of BJT places an upper limit on size of output resistance. rp appears in parallel with RE if Rth is neglected. If we let RE be infinite, maximum value of output resistance is Output resistance of C-S amplifier is given by, Microelectronic Circuit Design McGraw-Hill Chap 14 - 14
Inverting Amplifiers: Output Resistance (Example) • Problem: Find output resistance. • Given data: Q-point values and values for RI, R1, R2, R3, R7 ,for both BJT and FET as well as values for RE and RS . • Assumptions: Small-signal operating conditions. Small -signal values are known. • Analysis: For C-E Amplifier, For C-S Amplifier, Microelectronic Circuit Design McGraw-Hill Chap 14 - 15
Inverting Amplifiers: Current Gain • Terminal current gain is the ratio of the current delivered to the load resistor to the current being supplied to the base terminal. Microelectronic Circuit Design McGraw-Hill Chap 14 - 16
Inverting Amplifiers: Summary • C-E and C-S amplifiers have similar voltage gains. • C-S amplifier provides extremely high input resistance but that of C-E is also substantial due to the mfRE term. • Output resistance of C-E amplifier is much higher than that of C-S amplifier as mf is much larger for BJT than for FET. • Input signal range of C-E amplifier is also higher than that of C-S amplifier. • Current gains of both are identical to those of individual transistors. • Following transformation is used to simplify circuit analysis by absorbing RE (or RS ) into the transistor (For FET, current gain and input resistance are infinite). Microelectronic Circuit Design McGraw-Hill Chap 14 - 17
Follower Circuits: Common-Collector and Common-Drain Amplifiers AC equivalent for C-C Amplifier AC equivalent for C-D Amplifier Microelectronic Circuit Design McGraw-Hill Chap 14 - 18
Follower Circuits: Terminal Voltage Gain For C-S Amplifier, take limit of voltage gain of C-E amplifier as and In most C-C and C-D amplifiers, Output voltage follows input voltage, hence theses circuits are called followers. BJT gain is closer to unity than FET. Mostly, ro can be neglected as gain<< mf Neglecting ro, Assuming Microelectronic Circuit Design McGraw-Hill Chap 14 - 19
Follower Circuits: Input Signal Range For small-signal operation, magnitude of vbe developed across rp in small-signal model must be less than 5 mV. If , vb can be increased beyond 5 mV limit.Since only small portion of input signal appears across base-emitter or gate-source terminals, followers can be used with relatively large input signals without violating small-signal limits. In case of FET, magnitude of vgs must be less than 0.2(VGS - VTN). Microelectronic Circuit Design McGraw-Hill Chap 14 - 20
Follower Circuits: Input Resistance and Overall Voltage Gain Overall voltage gain is For C-S Amplifier, Input resistance looking into the base terminal is given by For C-S Amplifier, Microelectronic Circuit Design McGraw-Hill Chap 14 - 21
Follower Circuits: Voltage Gain Calculations (Example) • Problem: Find overall voltage gain. • Given data: Q-point values and values for RI, R1, R2, R4, R7 ,for both BJT and FET. • Assumptions: Small-signal operating conditions. • Analysis: For C-C Amplifier, Microelectronic Circuit Design McGraw-Hill Chap 14 - 22
Follower Circuits: Voltage Gain Calculations (Example contd.) • Analysis: For C-D Amplifier, Microelectronic Circuit Design McGraw-Hill Chap 14 - 23
Follower Circuits: Output Resistance Current is injected into emitter of BJT. Current aoi coming out of collector must be supported by veb = aoi/gm, given by first term. ib =-i/bo+1creates voltage drop in Rth given by second term In case of FET, Thus equivalent resistance looking into emitter or source of a transistor is approximately 1/ gm. Microelectronic Circuit Design McGraw-Hill Chap 14 - 24
Follower Circuits: Output Resistance (Example) • Problem: Find output resistance. • Given data: Q-point values and values for RI, R1, R2, R4, R7 ,for both BJT and FET. • Assumptions: Small-signal operating conditions. Small -signal values are known. • Analysis: For C-C Amplifier, For C-D Amplifier, Microelectronic Circuit Design McGraw-Hill Chap 14 - 25
Follower Circuits: Current Gain • Terminal current gain is the ratio of the current delivered to the load resistor to the current being supplied from the Thevenin source. Microelectronic Circuit Design McGraw-Hill Chap 14 - 26
Follower Circuits: Summary • Both C-C and C-D amplifiers have voltage gains approaching unity. • C-D amplifier provides extremely high input resistance because of infinite resistance looking into gate terminal of FET as compared to C-C amplifier. • Output resistance of C-C amplifier is much lower than the C-D amplifier due to higher transconductance of BJT than an FET for given operating current. • Both C-C and C-D amplifiers can handle relatively large input signal levels.. • Current gains of FET is inherently infinite, whereas that of BJT is limited by its finite value of bo. Microelectronic Circuit Design McGraw-Hill Chap 14 - 27
Noninverting Amplifiers: Common-Base and Common-Gate Circuits AC equivalent for C-E Amplifier AC equivalent for C-S Amplifier Microelectronic Circuit Design McGraw-Hill Chap 14 - 28
Noninverting Amplifiers: Terminal Voltage Gain and Input Resistance For C-S Amplifier, take limit of voltage gain of C-E amplifier as Polarities of vbe and dependent current source gmvbe are both reversed, signal source is transformed to its Norton equivalent ro is neglected. Microelectronic Circuit Design McGraw-Hill Chap 14 - 29
Noninverting Amplifiers: Input Signal Range For small-signal operation, In case of FET, Relative size of gm and RI determine signal-handling limits. …for RI >> R4. Microelectronic Circuit Design McGraw-Hill Chap 14 - 30
Noninverting Amplifiers: Overall Voltage Gain Overall voltage gain is For C-S Amplifier, For RI >> R4, For , This is the upper bound on gain. For , ro can be neglected as gain<< mf Microelectronic Circuit Design McGraw-Hill Chap 14 - 31
Noninverting Amplifiers: Voltage Gain Calculations (Example) • Problem: Find overall voltage gain. • Given data: Q-point values and values for R1, R2, R3, R7 ,for both BJT and FET, RI =2 kW, R4 =12 kW. • Assumptions: Small-signal operating conditions. • Analysis: For C-E Amplifier, For C-S Amplifier, Microelectronic Circuit Design McGraw-Hill Chap 14 - 32
Noninverting Amplifiers: Output Resistance Desired resistance is that looking into collector with base grounded and resistor Rth in emitter. The redrawn equivalent circuit is same as that for C-E amplifier except resistance in base is zero and resistance in emitter is relabeled as Rth. Using Microelectronic Circuit Design McGraw-Hill Chap 14 - 33
Noninverting Amplifiers: Current Gain • Terminal current gain is the ratio of the current delivered to the load resistor to the current being supplied to the base terminal. Microelectronic Circuit Design McGraw-Hill Chap 14 - 34
Noninverting Amplifiers: Summary • C-B and C-G amplifiers have similar voltage and current gains. Numerical differences occur due to difference in parameter values of BJT and FET at similar operating points. • C-B amplifier can achieve high output resistance due to higher amplification factor of BJT. • C-B amplifier can more easily reach low levels of output resistance due to higher transconductance of BJT for a given operating current. • Input signal range of C-G amplifier is inherently larger than that of C-B amplifier. Microelectronic Circuit Design McGraw-Hill Chap 14 - 35
Selecting Amplifier Configuration • A single-transistor amplifier with a gain of 80 dB and input resistance of 100 kW. • Av = 1080/20 = 10,000. For even best BJTs, gain< mf = 40VA = 40(150) = 6000 and FET typically has much lower intrinsic gain. Hence such large gain can’t be achieved by single-transistor amplifier. • A single-transistor amplifier with gain of 52 dB, input resistance of 250 kW. • Av = 1052/20 = 400. Since we need large gain and relatively large input resistance, we can use C-E amplifier. Av = 20VCC , so, VCC =20 V. which is small but acceptable. For FET, even with small gate overdrive, VDD =100 V which is too large Microelectronic Circuit Design McGraw-Hill Chap 14 - 36
Selecting Amplifier Configuration (contd.) • A single-transistor amplifier with gain of 30dB and input resistance of 5MW. • Av = 1030/20 = 31.6. Since we need large input resistance and moderate gain, we can use C-S amplifier. Input resistance of can be set by our choice of gate-bias resistors. For C-E amplifier, required high input resistance could be attained but values of base bias resistor could be a limiting factor. • A single-transistor amplifier with gain of 0dB and input resistance of 20MW with load resistor of 10kW. • Gain of 0 dB implies a source follower, Rin=boRL=100(10kW)=1MW, so BJT can’t meet input resistance requirement, A source follower can be used easily. Microelectronic Circuit Design McGraw-Hill Chap 14 - 37
Coupling and Bypass Capacitor Design • Since impedance of a capacitor increases with decreasing frequency, coupling and bypass capacitors reduce amplifier gain at low frequencies. • To choose capacitor values, short-circuit time constant method is used: each capacitor is considered separately with all other capacitors replaced by short circuits. • To neglect a capacitor, the magnitude of capacitive impedance must be much smaller than the equivalent resistance appearing at its terminals. Microelectronic Circuit Design McGraw-Hill Chap 14 - 38
Coupling and Bypass Capacitor Design: C-E and C-S Amplifiers For C-E amplifier, For C-S amplifier, For coupling capacitor C1, For coupling capacitor C3, w is chosen to be lowest frequency for which midband operation is needed in given application. Microelectronic Circuit Design McGraw-Hill Chap 14 - 39
Coupling and Bypass Capacitor Design: C-E and C-S Amplifiers (contd.) In this case, we can neglect impedances of capacitors C1 and C3 , the find the equivalent resistance looking up into emitter or source of amplifier. Microelectronic Circuit Design McGraw-Hill Chap 14 - 40
Coupling and Bypass Capacitor Design: C-E and C-S Amplifiers (Example) • Problem: Choose values of coupling and bypass capacitors. • Given data: f = 1000Hz, values of all resistors and input and output resistances for both C-E and C-S amplifiers. • Analysis: For C-S amplifier: For C-E amplifier: Microelectronic Circuit Design McGraw-Hill Chap 14 - 41
Coupling and Bypass Capacitor Design: C-C and C-D Amplifiers For C-E amplifier, For C-S amplifier, For coupling capacitor C1, For coupling capacitor C3, Microelectronic Circuit Design McGraw-Hill Chap 14 - 42
Coupling and Bypass Capacitor Design: C-C and C-D Amplifiers (Example) • Problem: Choose values of coupling and bypass capacitors. • Given data: f = 1000Hz, values of all resistors and input and output resistances for both C-E and C-S amplifiers. • Analysis: For C-D amplifier: For C-C amplifier: Microelectronic Circuit Design McGraw-Hill Chap 14 - 43
Coupling and Bypass Capacitor Design: C-B and C-G Amplifiers For C-E amplifier, For C-S amplifier, For coupling capacitor C1, For coupling capacitor C3, Microelectronic Circuit Design McGraw-Hill Chap 14 - 44
Coupling and Bypass Capacitor Design: C-B and C-G Amplifiers (contd.) In this case, we can neglect impedances of capacitors C1 and C3 , the find the equivalent resistance looking up into emitter or source of amplifier. Microelectronic Circuit Design McGraw-Hill Chap 14 - 45
Coupling and Bypass Capacitor Design: C-B and C-G Amplifiers (Example) • Problem: Choose values of coupling and bypass capacitors. • Given data: f = 1000Hz, values of all resistors and input and output resistances for both C-E and C-S amplifiers. • Analysis: For C-B amplifier: Microelectronic Circuit Design McGraw-Hill Chap 14 - 46
Coupling and Bypass Capacitor Design: C-B and C-G Amplifiers (Example contd.) For C-G amplifier: Microelectronic Circuit Design McGraw-Hill Chap 14 - 47
Lower Cutoff Frequency of an Amplifier • We can choose capacitor values to set the lower cutoff frequency of the amplifier at desired value. • Pole associated with a capacitor occurs at the frequency at which capacitive reactance is equal to resistance at the capacitor terminals. • In discussed amplifiers, there are several poles and a bandwidth shrinkage occurs at low frequencies. • A transfer function with n identical poles at wo is given by • Lower cutoff frequency is higher than frequency corresponding to individual poles. Microelectronic Circuit Design McGraw-Hill Chap 14 - 48
Dominant Pole Design • Instead of having the lower cutoff frequency set by the interaction of several poles, it can be set by the pole associated with just one of the capacitors. The other capacitors can be chosen to have their pole frequencies much below fL. • Capacitor associated with emitter or source part of the circuit tends to be the largest due to low resistance presented by emitter or source terminal of transistor and is commonly used to set fL. • Values of other capacitors are increased by a factor of 10 to push their corresponding poles to much lower frequencies. Microelectronic Circuit Design McGraw-Hill Chap 14 - 49
Follower Design Example • Problem: Design an amplifier with given specifications. • Given data: Av>0.95, Rin>20MW, Rout<3 kW. • Analysis: Gain is approximately unity, high input resistance is required, relatively low load resistance implies that low output resistance is required. Se we can choose between emitter or source-follower configurations. Emitter follower’s input resistance is limited by bo RL and current gain bo required to meet the input resistance specification with given load resistance is > 6600 (beyond range of normal BJTs). So we choose source-follower configuration. Microelectronic Circuit Design McGraw-Hill Chap 14 - 50