Presentation on theme: "Chapter 13 Frequency Response"— Presentation transcript:
1Chapter 13 Frequency Response Microelectronic Circuit DesignRichard C. JaegerTravis N. Blalock
2Chapter GoalsReview 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.
3Chapter 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.
4Transfer Function Analysis for ,i =1…lAmid is midband gain between upper and lower cutoff frequencies.for ,j =1…k
5Low-Frequency Response Pole wP2 is called the dominant low-frequency pole (> all other poles) and zeros are at frequencies low enough to not affect wL.If there is no dominant pole at low frequencies, poles and zeros interact to determine wL.Pole wL > all other pole and zero frequenciesIn general, for n poles and n zeros,For s=jw, at wL,
6Transfer Function Analysis and Dominant Pole Approximation Example Problem: Find midband gain, FL(s) and fL forAnalysis: Rearranging the given transfer function to get it in standard form,Now,andZeros are at s=0 and s =-100. Poles are at s= -10, s=-1000All pole and zero frequencies are low and separated by at least a decade. Dominant pole is at w=1000 and fL =1000/2p= 159 Hz. For frequencies > a few rad/s:
7High-Frequency Response Pole wP3 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 wH.Pole wH < all other pole and zero frequenciesIn general,For s=jw, at wH,
8Direct Determination of Low-Frequency Poles and Zeros: C-S Amplifier
9Direct Determination of Low-Frequency Poles and Zeros: C-S Amplifier (contd.) The three zero locations are: s = 0, 0, -1/(RS C3).The three pole locations are:Each independent capacitor in the circuit contributes one pole and one zero. Series capacitors C1 and C2 contribute the two zeros at s=0 (dc), blocking propagation of dc signals through the amplifier. Third zero due to parallel combination of C3 and RS occurs at frequency where signal current propagation through MOSFET is blocked (output voltage is zero).
10Short-Circuit Time Constant Method to Determine wL Lower cutoff frequency for a network with n coupling and bypass capacitors is given by:where RiS is resistance at terminals of ith capacitor Ci with all other capacitors replaced by short circuits. Product RiS Ci is short-circuit time constant associated with Ci.Midband gain and upper and lower cutoff frequencies that define bandwidth of amplifier are of more interest than complete transfer function.
11Estimate of wL for C-E Amplifier Using SCTC method, for C1,For C2,For C3,
12Estimate of wL for C-S Amplifier Using SCTC method,For C1,For C2,For C3,
13Estimate of wL for C-B Amplifier Using SCTC method,For C1,For C2,
14Estimate of wL for C-G Amplifier Using SCTC method,For C1,For C2,
15Estimate of wL for C-C Amplifier Using SCTC method,For C1,For C2,
16Estimate of wL for C-D Amplifier Using SCTC method,For C1,For C2,
17Frequency-dependent Hybrid-Pi Model for BJT Capacitance between base and emitter terminals is:tF is forward transit-time of the BJT. Cp appears in parallel with rp. As frequency increases, for a given input signal current, impedance of Cp reduces vbe and thus the current in the controlled source at transistor output.Capacitance between base and collector terminals is:Cmo is total collector-base junction capacitance at zero bias, Fjc is its built-in potential.
18Unity-gain Frequency of BJT The right-half plane transmission zero wZ = + gm/Cm occurring at high frequency can be neglected.wb = 1/ rp(Cm + Cp ) is the beta-cutoff frequencywhereand fT = wT /2p is the unity gain bandwidth product. Above BJT has no current gain.
19Unity-gain Frequency of BJT (contd.) Current gain is bo = gmrp at low frequencies and has single pole roll-off at frequencies > fb, crossing through unity gain at wT. Magnitude of current gain is 3 dB below its low-frequency value at fb.
21Limitations of High-frequency Models Above 0.3 fT, behavior of simple pi-models begins to deviate significantly from the actual device.Also, wT depends on operating current as shown and is not constant as assumed.For given BJT, a collector current ICM exists that yield maximum fTmax.For FET in saturation, CGS and CGD are independent of Q-point current, so
22Effect of Base Resistance on Midband Amplifiers To account for base resistance rx is absorbed into equivalent pi model and can be used to transform expressions for C-E, C-C and C-B amplifiers.Base current enters the BJT through external base contact and traverses a high resistance region before entering active area. rx models voltage drop between base contact and active area of the BJT.
23Direct High-Frequency Analysis: C-E Amplifier The small-signal model can be simplified by using Norton source transformation.
24Direct 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, wZ = + gm/Cm > wT can easily be neglected.For a polynomial s2+sA1+A0 with roots a and b, a =A1 and b=A0/A1.Smallest root that gives first pole limits frequency response and determines wH. Second pole is important in frequency compensation as it can degrade phase margin of feedback amplifiers.
25Direct High-Frequency Analysis: C-E Amplifier (Overall Transfer Function) Dominant pole model at high frequencies for C-E amplifier is as shown.
26Direct High-Frequency Analysis: C-E Amplifier (Example) Problem: Find midband gain, poles, zeros and fL.Given data: Q-point= ( 1.60 mA, 3.00V), fT =500 MHz, bo =100, Cm =0.5 pF, rx =250W, CL =0Analysis: gm =40IC =40(0.0016) =64 mS, rp = bo/gm =1.56 kW.Overall gain is reduced to -135 as vth =0.882vs.
27Gain-Bandwidth Product Limitations of C-E Amplifier If Rth is reduced to zero in order to increase bandwidth, then rpo would not be zero but would be limited to approximately rx.If Rth = 0, rx <<rp so that rx = rpo and
29Miller Multiplication For the C-E amplifier,Total input capacitance = C(1+A) because total voltage across C is vc = vi(1+A) due to inverting voltage gain of amplifier.
30Miller Integrator where For frequencies >> wo, assuming A>>1,which is the transfer function of an integrator.Assuming zero current in input terminal of amplifier,
31Open-Circuit Time Constant Method to Determine wH 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 Rio is resistance at terminals of ith capacitor Ci with all other capacitors open-circuited.For a C-E amplifier, assuming CL =0
32Gain-Bandwidth Trade-off Using Emitter Resistor for andgain decreases as emitter resistance increases and bandwidth of stage will correspondingly increase.To find bandwidth using OCTC method:
33Gain-Bandwidth Trade-off Using Emitter Resistor (contd.) Test source ix is first split into two equivalent sources and then superposition is used to find vx =(vb - vc).Assuming that bo >>1 and
34Dominant Pole for C-B Amplifier Using split-source transformation Assuming that bo >>1 and rx << rpNeglecting first term of order of 1/ wT and since last term is dominant.
36Dominant Pole for C-C Amplifier A better estimate is obtained if we set RL =0 in expression for Rpo.
37Dominant Pole for C-D Amplifier Substituting rp as infinite and rx as zero in expression for emitter follower,
38Frequency Response: Differential Amplifier CEE 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, CEE has no effect on differential-mode signals.For common-mode signals, at very low frequencies,Transmission zero due to CEE is
39Frequency Response: Differential Amplifier (contd.) As REE is usually designed to be large,Common-mode half-circuit is similar to a C-E stage with emitter resistor 2REE. OCTC for Cp and Cm is similar to the C-E stage. OCTC for CEE/2 is:
40Frequency Response: Common-Collector/ Common-Base Cascade REE is assumed to be large and neglected.Sum of the OCTC of Q1 is:
41Frequency Response: Common-Collector/ Common-Base Cascade (contd.) Sum of the OCTC of Q2 is:Combining the OCTC for Q1 and Q2, and assuming that transistors are matched,
42Frequency Response: Cascode Amplifier OCTC of Q1 with load resistor 1/ gm2 :As IC2 = IC1, gm2 = gm1, gain of first stage is unity. Assuming gm2 rpo1 >>1,OCTC of Q1, a C-B stage for ro1 >> RL and mf>>1:Assuming matched devices,
43Frequency Response: MOS Current Mirror For matched transistors,
44Frequency 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-stageamplifier is as shown.
45Frequency Response: Multistage Amplifier (Estimate of wL) SCTC for each of the six independent coupling and bypass capacitors has to be determined.
46Frequency Response: Multistage Amplifier (Estimate of wH) OCTC for each of the two capacitors associated with each transistor has to be determined.For M1,For Q2,For Q3,
47Single-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.
48Three-stage MOS Op Amp Analysis Input stage is modeled by its Norton equivalent- current source Gmvdm and output resistance Ro. Second stage has gain of gm5ro5= mf5 and follower output stage is a unity-gain buffer.Vo(s) = Vb(s) = - Av2Va(s)For large Av2
49Transmission Zeros in FET Op Amps Incorporating the zero determined by gm5 in the analysis,This zero can’t be neglected due to low ratio of transconductances of M2 and M5. Zero can be canceled by addition of RZ =1/ gm5.
50Bipolar Amplifier Compensation Bipolar op amp can be compensated in the same manner as a MOS amplifierTransmission 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:
51Slew rate of Op AmpSlew-rate limiting is caused by limited current available to charge/discharge internal capacitors. For very large Av2, amplifier behaves like an integrator:For CMOS amplifier,For bipolar amplifier,
52Tuned AmplifiersAmplifiers 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 can’t be used.These amplifiers have high Q (fH and fL close together relative to center frequency)These applications use resonant RLC circuits to form frequency selective tuned amplifiers.
53Single-Tuned Amplifiers RLC network selects the frequency, parallel combination of RD, R3 and ro set the Q and bandwidth.Neglecting right-half plane zero,
54Single-Tuned Amplifiers (contd.) At center frequency, s = jwo, Av = Amid.
55Use of tapped Inductor- Auto Transformer CGD and ro 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 doesn’t shift significantly due to changes in CGD.Similar solution can be used if tuned circuit is placed at amplifier input instead of output
56Multiple 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.
57Mixers: 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 mixer’s 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.
58Single-Balanced Mixer Eliminates one of the two input signals from the output.No signal energy appears atw1 , but w2 appears in output spectrum, so circuit is single-balanced.Up-conversion uses component (w2-w1) and down-conversion uses (w2+w1) component.
59Double-Balanced Mixer/ Modulator: Gilbert Multiplier Double-balanced mixers don’t 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