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FREQUENCY RESPONSE OF AMPLIFIERS * Effects of capacitances within transistors and in amplifiers * Build on previous analysis of amplifiers from EENG 341 z Transistor DC biasing z Small signal amplification, i.e. voltage and current gain z Transistor small signal equivalent circuit * Use in Bipolar and FET Transistors Amplifiers and their analysis * Build on previous analysis of single time constant circuits z Review simple RC, LC and RLC circuits z Recall frequency dependent impedances for C and L z Review frequency dependence in transfer functions * Magnitude and phase * Examine origins of frequency dependence in amplifier gain z Identify capacitors and their origins; find the dominant C z Determine equivalent R and determine RC time constant z Use to describe approximately the amplifier’s frequency behavior z Examine effects of other capacitors * GOAL: Use results of analysis to modify circuit design to improve performance. 1 Analysis of Amplifier Performance * Previously analyzed z DC bias point z AC analysis (midband gain) * Neglected all capacitances in the transistor and circuit * Gain at middle frequencies, i.e. not too high or too low in frequency i B iC DC bias or quiescent point vBE vCE 2 Frequency Response of Amplifiers * In reality, all amplifiers have a limited range of frequencies of operation z Called the bandwidth of the amplifier z Falloff at low frequencies * At ~ 100 Hz to a few kHz * Due to coupling capacitors at the input or output, e.g. CC1 or CC2 z Falloff at high frequencies * At ~ 100’s MHz or few GHz 20 logT(ω) * Due to capacitances within the transistors themselves. Midband Gain Equivalent circuit for bipolar transistor ω 3 Frequency Response of Amplifiers * First approximation – describe the amplifier’s high Low-Pass Network and low frequency responses in terms of that of single time constant (STC) circuits z High frequency falloff – Æ Like that of a low pass filter Vi * Simple RC equivalent circuit * Shunting capacitor shorts signal at the output at high frequencies High-Pass Network z Low frequency falloff Æ Like that of a high pass filter * Simple RC equivalent circuit * Series capacitor blocks output signal at Vi low frequencies (acts like open circuit) * Amplifier frequency analysis z Determine equivalent R for each C z Compare and find the most important 1 1 ⎧0 (short) as ω → ∞ (dominant RC) combination ZC = = → ⎨ sC jωC ⎩∞ (open) as ω → 0 ÆFind the dominant one (RC) at high frequencies ÆFind the dominant one (RC) at low frequencies 4 Review of Complex Numbers * Complex numbers z General form a + bj where a = real part, b = imaginary part and j = −1 2 2 z Magnitude of complex number M = a + bj = a + b ⎛ b ⎞ z Phase of complex number θ = tan −1⎜ ⎟ ⎝ a ⎠ z Phasor form a + bj = Me jθ * Complex number math z Multiplication of two complex numbers a + bj = Me jθ and c + dj =Ne jφ ()()a + bj c + dj = (ac − bd) + j(bc + ad) OR ()()a + bj c + dj = ()()Me jθ Ne jϕ = MNe j()θ +ϕ z Reciprocal of a complex number 1 1 a − bj a − bj a b = = = − j a + bj a + bj a − bj a2 + b2 a2 + b2 a2 + b2 OR 1 1 1 = = e− jθ a + bj Me jθ M 5 Amplifier Transfer Function (Gain) - General Form * A (s) = Gain Function (general form of amplifier transfer function) A(s) = AM FH (s)FL (s) z AM = midband gain (independent of frequency) 1 F (s) = z FH(s) = high frequency function (acts like low pass filter) H 1+ s /ωH 1 z FL(s) = low frequency function (acts like high pass filter) FL (s) = 1+ωL / s Magnitude AM FL(s) FH(s) ωL ωH 6 Amplifier Transfer Function (Gain) - General Form A(s) = AM FH (s)FL (s) AM FL(s) FH(s) Coupling Transistor’s Capacitors Capacitors ωL ωH 7 Frequency Response of MOSFET vs BJT Amplifiers Equivalent circuit for MOSFET Equivalent circuit for bipolar transistor Similar equivalent circuits Common Source Common Emitter Amplifier Amplifier Corresponding amplifier circuits Gain Gain Similar frequency performance 8 Amplifier Transfer Function (Gain) - General Form A(s) = AM FH (s)FL (s) Now we consider the low frequency behavior. Coupling Transistor’s Capacitors Capacitors ωL ωH 9 Summary * Examined origin of falloff in amplifier gain at low and high frequencies. z Degradation in magnitude of the gain. z Shift in phase of output relative to input. * Due to presence of capacitors within the amplifier (Create poles and zeros). z Coupling capacitors limit gain at low frequencies. z Transistor’s capacitances limit gain at high frequencies. * Examined and quantified the falloff due to single and multiple poles and zeroes. z Bode plots of gain and phase shift with frequency. * Next, we will apply this method of analysis to transistor amplifiers. z Multiple capacitors so multiple RC combinations. z Investigate how to determine which capacitors are most important and limit the bandwidth. z Examine how to change the amplifier to get better frequency performance. Higher frequency operation before falloff (improved bandwidth). Better low frequency behavior. 10 Analysis of Bipolar Transistor Amplifiers * Single stage amplifiers z Common Emitter (CE) z Common Base (CB) z Emitter Follower (EF) (Common Collector) * DC biasing z Calculate IC, IB, VCE z Determine related small signal equivalent circuit parameters Transconductance gm Input resistance rπ * Low frequency analysis z Gray-Searle (Short Circuit) Technique Determine the pole frequencies ωPL1, ωPL2, ... ωPLn z Determine the zero frequencies ωZL1, ωZL2, ... ωZLn * High frequency analysis z Gray-Searle (Open Circuit) Technique Determine the pole frequencies ωPH1, ωPH2, ... ωPHn z Determine the zero frequencies ωZH1, ωZH2, ... ωZHn 11 CE Amplifier Frequency Analysis - Long and Difficult Way * DC analysis: IC , IB , VCE ; gm , rπ * Draw ac equivalent circuit * Substitute hybrid-pi model for transistor * Obtain KVL equations (at least one for each capacitor in circuit (5)) * Solve set of 5 simultaneous equations to obtain voltage gain AV = Vo /Vs * Put expression in standard form for gain AV(ω) = AVo FH(ω) FL(ω) * Identify midband gain AVo * Determine FH(ω) part and factor to z Determine high frequency poles ωPH1 and ωPH2 z Determine high frequency zeros ωZH1 and ωZH2 * Determine FL(ω) part and factor to z Determine low frequency poles ωPL1, ωPL2 and ωPL3 z Determine high frequency zeros (s + ω )(s + ω )(s + ω ) F (s) = Z1 Z1 Z 3 ω , ω and ω ⎛ s ⎞⎛ s ⎞ L ()()()s + ω s + ω s + ω ZL1 ZL2 ZL3 ⎜1+ ⎟⎜1+ ⎟ P1 P2 P3 ⎜ ω ⎟⎜ ω ⎟ ⎝ Z1 ⎠⎝ Z 2 ⎠ ⎛ ωZ1 ⎞⎛ ωZ 2 ⎞⎛ ωZ 3 ⎞ FH (s) = ⎜1+ ⎟⎜1+ ⎟⎜1+ ⎟ * Is there an easier way ? YES ⎛ s ⎞⎛ s ⎞ s s s ⎜1+ ⎟⎜1+ ⎟ = ⎝ ⎠⎝ ⎠⎝ ⎠ ⎜ ω ⎟⎜ ω ⎟ ⎛ ω ⎞⎛ ω ⎞⎛ ω ⎞ ⎝ P1 ⎠⎝ P2 ⎠ ⎜1+ P1 ⎟⎜1+ P2 ⎟⎜1+ P3 ⎟ ⎝ s ⎠⎝ s ⎠⎝ s ⎠ 12 CE Amplifier - Starting Point is DC Analysis * Q is quiescent point (DC bias point) * Q needs to be in the active region z IC = β IB * If Q is in saturation (VCE < 0.3 V) , then IC < β IB and there is little or no gain from the transistor amplifier * If the transistor is in the cutoff mode, there is virtually no IC so there is no gain, i.e. gm ¡ 0. * Q depends on the choice of R1 and R2 since they determine the size of IB. Active * Q point determines the size of the region small signal parameters Saturation Q point z Transconductance g = I / V region m C T VT = kBT/q = 26 mV z Input resistance rπ = β / gm Cutoff region 13 Example of CE Amplifier - DC Analysis * GIVEN: Transistor parameters: z Current gain β = 200 z Base resistance rx = 65 Ω z Base-emitter voltage VBE,active = 0.7 V z Resistors: R1=10K, R2=2.5K, RC=1.2K, RE=0.33K * Form Thevenin equivalent for base; given VCC = 12.5V z RTh = RB = R1||R2 = 10K||2.5K = 2K z VTh = VBB = VCC R2 / [R1+R2] = 2.5V z DC Base Current (use KVL base loop) IB = [VTh-VBE,active] / [RTh+(β +1)RE] IB = 26 µA * DC collector current IC = β IB IC = 200(26 µ A) = 5.27 mA * Transconductance gm = IC / VT ;VT = kBT/q = 26 mV gm = 5.27 mA/26 mV = 206 mA/V * Input resistance rπ = β / gm = 200/[206 mA/V]= 0.97 K * Check on transistor region of operation (Find V ) R1 = 10K CE z KVL collector loop R2 = 2.5K R = 1.2K z VCE = VCC -IC RC -(β +1) IB RE = 4.4 V C (okay since not close to zero volts, i.e. > 0.2V). RE = 0.33K 14 CE Amplifier - Midband Gain Analysis * Construct amplifier’s small signal ac equivalent circuit (set DC supply to ground) * Substitute small signal equivalent circuit (hybrid-pi model) for transistor * Neglect all capacitances z Coupling and emitter bypass capacitors become shorts at midband frequencies (~ 105 rad/s) Why? Their impedances are negligibly small, e.g. few ohms because CC1, CC2, CE are large, e.g.~ few µF (10-6F) 1 1 Z = ~ =10 Ω C ωC (105 rad / s)(1µF) Hybrid-Pi Model for BJT z Transistor capacitances become open circuits at midband frequencies Why? Their impedances are very large, e.g. ~ 10’s M Ω because Cπ , Cµ are very small, e.g. ~ pF (10-12 F) 1 1 Z = ~ =107 Ω C ωC (105 rad / s)(1pF) * Calculate small signal voltage gain AVo = Vo /Vs 15 CE Amplifier - Midband Gain Analysis r Io x RL = 9K Vo + RC =1.2K V V r V g V RE = 0.33K s Vii π π m π R1 =10K - - R2 = 2.5K R 5K RL||RC S = Vo Vo Vπ Vi Break voltage gain into a series of voltage ratios AVo = = Vs Vπ Vi Vs Vo − gmVπ ()RL RC = = −gm ()RL RC = −()206mA /V ()1.2K 9K = −218 Vπ Vπ V r 0.97K π = π = = 0.94 Vi rx + rπ 0.97K + 0.065K V []()r + r R []()0.97K + 0.065K 2.0K 0.68K i = x π B = = = 0.12 Vs Rs + []()rx + rπ RB 5K + []()0.97K + 0.065K 2.0K 5.68K AVo = (− 218 )()()0.94 0.12 = −24.6 V /V Negative sign means output signal is 180o out of phase with the input signal.
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  • High-Frequency Amplifier Response

    High-Frequency Amplifier Response

    Section H5: High-Frequency Amplifier Response Now that we’ve got a high frequency models for the BJT, we can analyze the high frequency response of our basic amplifier configurations. Note: in the circuits that follow, the actual signal source (vS) and its associated source resistance (RS) have been included. As discussed in the low frequency response section of our studies, we always knew that this source and resistance was there but we just started our investigations with the input to the transistor (vin). Remember - the analysis process is the same and the relationship between vS and vin is a voltage divider! Again, in some instances in the following discussion, I will be using slightly different notation and taking a different approach than your author. As usual, if this results in confusion, or you are more comfortable with his technique, let me know and we’ll work it out. High Frequency Response of the CE and ER Amplifier The generic common-emitter amplifier circuit of Section D2 is reproduced to the left below and the small signal circuit using the high frequency BJT model is given below right (based on Figures 10.17a and 10.17b of your text). Note that all external capacitors are assumed to be short circuits at high frequencies and are not present in the high frequency equivalent circuit (since the external capacitors are large when compared to the internal capacitances – recall that Zc=1/jωC gets small as the frequency or capacitance gets large). We can simplify the small signal circuit by making the following observations and approximations: ¾ Cce is very small and may be neglected.