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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Chapter 13 Frequency Response Microelectronic Circuit Design Richard C. Jaeger Travis N. Blalock

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Chapter Goals Review transfer function analysis and dominant-pole approximations of amplifier transfer functions. Learn partition of ac circuits into low and high-frequency equivalents. Learn short-circuit and open-circuit time constant methods to estimate upper and lower cutoff frequencies. Develop bipolar and MOS small-signal models with device capacitances. Study unity-gain bandwidth product limitations of BJTs and MOSFETs. Develop expressions for upper cutoff frequency of inverting, non-inverting and follower configurations. Explore gain-bandwidth product limitations of single and multiple transistor circuits.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Chapter Goals (contd.) Understand Miller effect and design of op amp frequency compensation. Develop relationship between op amp unity-gain frequency and slew rate. Understand use of tuned circuits to design high-Q band- pass amplifiers. Understand concept of mixing and explore basic mixer circuits. Study application of Gilbert multiplier as balanced modulator and mixer.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Transfer Function Analysis A mid is midband gain between upper and lower cutoff frequencies. for,i =1…l for,j =1…k

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Low-Frequency Response Pole P2 is called the dominant low-frequency pole (> all other poles) and zeros are at frequencies low enough to not affect L. If there is no dominant pole at low frequencies, poles and zeros interact to determine L. For s=j, at L, Pole L > all other pole and zero frequencies In general, for n poles and n zeros,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Transfer Function Analysis and Dominant Pole Approximation Example Problem: Find midband gain, F L (s) and f L for Analysis: Rearranging the given transfer function to get it in standard form, Now, and Zeros are at s=0 and s =-100. Poles are at s= -10, s=-1000 All pole and zero frequencies are low and separated by at least a decade. Dominant pole is at =1000 and f L =1000/2 = 159 Hz. For frequencies > a few rad/s:

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl High-Frequency Response Pole P3 is called the dominant high- frequency pole (< all other poles). If there is no dominant pole at low frequencies, poles and zeros interact to determine H. For s=j, at H, Pole H < all other pole and zero frequencies In general,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Direct Determination of Low-Frequency Poles and Zeros: C-S Amplifier

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Direct Determination of Low-Frequency Poles and Zeros: C-S Amplifier (contd.) The three zero locations are: s = 0, 0, -1/(R S C 3 ). The three pole locations are: Each independent capacitor in the circuit contributes one pole and one zero. Series capacitors C 1 and C 2 contribute the two zeros at s=0 (dc), blocking propagation of dc signals through the amplifier. Third zero due to parallel combination of C 3 and R S occurs at frequency where signal current propagation through MOSFET is blocked (output voltage is zero).

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Short-Circuit Time Constant Method to Determine L Lower cutoff frequency for a network with n coupling and bypass capacitors is given by: where R iS is resistance at terminals of ith capacitor C i with all other capacitors replaced by short circuits. Product R iS C i is short- circuit time constant associated with C i. Midband gain and upper and lower cutoff frequencies that define bandwidth of amplifier are of more interest than complete transfer function.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Estimate of L for C-E Amplifier Using SCTC method, for C 1, For C 2, For C 3,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Estimate of L for C-S Amplifier Using SCTC method, For C 1, For C 2, For C 3,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Estimate of L for C-B Amplifier Using SCTC method, For C 1, For C 2,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Estimate of L for C-G Amplifier Using SCTC method, For C 1, For C 2,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Estimate of L for C-C Amplifier Using SCTC method, For C 1, For C 2,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Estimate of L for C-D Amplifier Using SCTC method, For C 1, For C 2,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Frequency-dependent Hybrid-Pi Model for BJT Capacitance between base and collector terminals is: C o is total collector-base junction capacitance at zero bias, jc is its built-in potential. Capacitance between base and emitter terminals is: F is forward transit-time of the BJT. C appears in parallel with r. As frequency increases, for a given input signal current, impedance of C reduces v be and thus the current in the controlled source at transistor output.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Unity-gain Frequency of BJT The right-half plane transmission zero Z = + g m /C occurring at high frequency can be neglected. = 1/ r (C + C ) is the beta-cutoff frequency where and f T = T /2 is the unity gain bandwidth product. Above BJT has no current gain.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Unity-gain Frequency of BJT (contd.) Current gain is o = g m r at low frequencies and has single pole roll- off at frequencies > f, crossing through unity gain at T. Magnitude of current gain is 3 dB below its low-frequency value at f.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl High-frequency Model of MOSFET

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Limitations of High-frequency Models Above 0.3 f T, behavior of simple pi-models begins to deviate significantly from the actual device. Also, T depends on operating current as shown and is not constant as assumed. For given BJT, a collector current I CM exists that yield maximum f Tmax. For FET in saturation, C GS and C GD are independent of Q-point current, so

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Effect of Base Resistance on Midband Amplifiers Base current enters the BJT through external base contact and traverses a high resistance region before entering active area. r x models voltage drop between base contact and active area of the BJT. To account for base resistance r x is absorbed into equivalent pi model and can be used to transform expressions for C-E, C-C and C-B amplifiers.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Direct High-Frequency Analysis: C-E Amplifier The small-signal model can be simplified by using Norton source transformation.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Direct High-Frequency Analysis: C-E Amplifier (Pole Determination) From nodal equations for the circuit in frequency domain, High-frequency response is given by 2 poles, one finite zero and one zero at infinity. Finite right-half plane zero, Z = + g m /C > T can easily be neglected. For a polynomial s 2 +sA 1 +A 0 with roots a and b, a =A 1 and b=A 0 /A 1. Smallest root that gives first pole limits frequency response and determines H. Second pole is important in frequency compensation as it can degrade phase margin of feedback amplifiers.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Direct High-Frequency Analysis: C-E Amplifier (Overall Transfer Function) Dominant pole model at high frequencies for C-E amplifier is as shown.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Direct High-Frequency Analysis: C-E Amplifier (Example) Problem: Find midband gain, poles, zeros and f L. Given data: Q-point= ( 1.60 mA, 3.00V), f T =500 MHz, o =100, C =0.5 pF, r x =250 C L Analysis: g m =40I C =40(0.0016) =64 mS, r = o /g m =1.56 k. Overall gain is reduced to -135 as v th =0.882v s.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Gain-Bandwidth Product Limitations of C-E Amplifier If R th is reduced to zero in order to increase bandwidth, then r o would not be zero but would be limited to approximately r x. If R th = 0, r x <

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl High-Frequency Analysis: C-S Amplifier

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Miller Multiplication Total input capacitance = C(1+A) because total voltage across C is v c = v i (1+A) due to inverting voltage gain of amplifier. For the C-E amplifier,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Miller Integrator Assuming zero current in input terminal of amplifier, where For frequencies >> o, assuming A>>1, which is the transfer function of an integrator.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Open-Circuit Time Constant Method to Determine H At high frequencies, impedances of coupling and bypass capacitors are small enough to be considered short circuits. Open-circuit time constants associated with impedances of device capacitances are considered instead. where R io is resistance at terminals of ith capacitor C i with all other capacitors open-circuited. For a C-E amplifier, assuming C L =0

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Gain-Bandwidth Trade-off Using Emitter Resistor for and gain decreases as emitter resistance increases and bandwidth of stage will correspondingly increase. To find bandwidth using OCTC method:

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Gain-Bandwidth Trade-off Using Emitter Resistor (contd.) Test source i x is first split into two equivalent sources and then superposition is used to find v x =(v b - v c ). Assuming that o >>1 and

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Dominant Pole for C-B Amplifier Using split-source transformation Assuming that o >>1 and r x << r Neglecting first term of order of 1/ T and since last term is dominant.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Dominant Pole for C-G Amplifier

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Dominant Pole for C-C Amplifier A better estimate is obtained if we set R L =0 in expression for R o.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Dominant Pole for C-D Amplifier Substituting r as infinite and r x as zero in expression for emitter follower,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: Differential Amplifier C EE is total capacitance at emitter node of the differential pair. Differential mode half-circuit is similar to a C-E stage. Bandwidth is determined by the product. As emitter is a virtual ground, C EE has no effect on differential-mode signals. For common-mode signals, at very low frequencies, Transmission zero due to C EE is

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: Differential Amplifier (contd.) Common-mode half-circuit is similar to a C-E stage with emitter resistor 2R EE. OCTC for C and C is similar to the C-E stage. OCTC for C EE /2 is: As R EE is usually designed to be large,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: Common- Collector/ Common-Base Cascade R EE is assumed to be large and neglected. Sum of the OCTC of Q 1 is:

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: Common-Collector/ Common-Base Cascade (contd.) Sum of the OCTC of Q 2 is: Combining the OCTC for Q 1 and Q 2, and assuming that transistors are matched,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: Cascode Amplifier OCTC of Q 1 with load resistor 1/ g m2 : As I C2 = I C1, g m2 = g m1, gain of first stage is unity. Assuming g m2 r o1 >>1, OCTC of Q 1, a C-B stage for r o1 >> R L and f >>1: Assuming matched devices,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: MOS Current Mirror For matched transistors,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: Multistage Amplifier Problem:Use open-circuit and short-circuit time constant methods to estimate upper and lower cutoff frequencies and bandwidth. Approach: Coupling and bypass capacitors determine low-frequency response, device capacitances affect high-frequency response. At high frequencies, ac model for multi-stage amplifier is as shown.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: Multistage Amplifier (Estimate of L ) SCTC for each of the six independent coupling and bypass capacitors has to be determined.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Frequency Response: Multistage Amplifier (Estimate of H ) OCTC for each of the two capacitors associated with each transistor has to be determined. For M 1, For Q 2, For Q 3,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Single-pole Op Amp Compensation Frequency compensation forces overall amplifier to have a single-pole frequency response by connecting compensation capacitor around second gain stage of the basic op amp.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Three-stage MOS Op Amp Analysis Input stage is modeled by its Norton equivalent- current source G m v dm and output resistance R o. Second stage has gain of g m5 r o5 = f5 and follower output stage is a unity-gain buffer. V o (s) = V b (s) = - A v2 V a (s) For large A v2

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Transmission Zeros in FET Op Amps Incorporating the zero determined by g m5 in the analysis, This zero cant be neglected due to low ratio of transconductances of M 2 and M 5. Zero can be canceled by addition of R Z =1/ g m5.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Bipolar Amplifier Compensation Bipolar op amp can be compensated in the same manner as a MOS amplifier Transmission zero occurs at too high a frequency to affect the response due to higher transconductance of BJT that FET for given operating current. Unity gain frequency is given by:

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Slew rate of Op Amp Slew-rate limiting is caused by limited current available to charge/discharge internal capacitors. For very large A v2, amplifier behaves like an integrator: For CMOS amplifier, For bipolar amplifier,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Tuned Amplifiers Amplifiers with narrow bandwidth are often required in RF applications to be able to select one signal from a large number of signals. Frequencies of interest > unity gain frequency of op amps, so active RC filters cant be used. These amplifiers have high Q (f H and f L close together relative to center frequency) These applications use resonant RLC circuits to form frequency selective tuned amplifiers.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Single-Tuned Amplifiers RLC network selects the frequency, parallel combination of R D, R 3 and r o set the Q and bandwidth. Neglecting right-half plane zero,

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Single-Tuned Amplifiers (contd.) At center frequency, s = j o, A v = A mid.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Use of tapped Inductor- Auto Transformer C GD and r o can often be small enough to degrade characteristics of the tuned amplifier. Inductor can be made to work as an auto transformer to solve this problem. These results can be used to transform the resonant circuit and higher Q can be obtained and center frequency doesnt shift significantly due to changes in C GD. Similar solution can be used if tuned circuit is placed at amplifier input instead of output

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Multiple Tuned Circuits Tuned circuits can be placed at both input and output to tailor frequency response. Radio-frequency choke(an open circuit at operating frequency) is used for biasing. Synchronous tuning uses two circuits tuned to same center frequency for high Q. Stagger tuning uses two circuits tuned to slightly different center frequencies to realize broader band amplifiers. Cascode stage is used to provide isolation between the two tuned circuits and eliminate feedback path between them due to Miller multiplication.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Mixers: Conversion Gain Amplifiers discussed so far have always been assumed to be linear and gain expressions involve input and output signals at same frequency. Mixers are nonlinear devices, output signal frequency is different from input signal frequency. A mixers conversion gain is the ratio of phasor representation of output signal to that of input signals, the fact that the two signals are at different frequencies is ignored.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Single-Balanced Mixer Eliminates one of the two input signals from the output. No signal energy appears at 1, but 2 appears in output spectrum, so circuit is single-balanced. Up-conversion uses component ( ) and down-conversion uses ( ) component.

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2 Microelettronica – Circuiti integrati analogici 2/ed Richard C. Jaeger, Travis N. Blalock Copyright © 2005 – The McGraw-Hill Companies srl Double-Balanced Mixer/ Modulator: Gilbert Multiplier Double-balanced mixers dont contain spectral components at either of the two input frequencies. Modulator applications give double sideband suppressed carrier output signal. Amplitude-modulated signal can also be obtained if

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