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CE Frequency Response the Miller Approximation CE Frequency Response The Miller Approximation n The “exact” analysis is worked out on pp. 639-641 of H&S. n Therefore, we consider the effect of Cm on the input node only ærp ö –gmç------------------÷ (ro rocRL)(1 – jw¤ wz) V èrp + RSø ---------out- = -------------------------------------------------------------------------------------------------- Vs (1 + jw¤ wp1)(1 + jw¤ wp2) ærp ö The low-frequency voltage gain (loaded) is –gmç------------------÷ (ro rocRL) èrp + RSø The zero is higher than the transition frequency wT = gm ¤ (Cp + Cm) The first (lowest frequency) pole is (approximately) neglect the feedforward current Im in comparison with gm Vp ... a good –1 approximation wp1 = {(RS rp)[Cp + (1 + gmRout')Cm ]+ Rout'Cm } It = (Vt - Vo) / Zm ’ Vo = - gmVt RL / (RL + Rout ) = AvCm Vt where Rout' = ro rocRL where AvCm is the low frequency voltage gain across Cm I = V (1- A ) / Z The second pole is (approximately) t t v m Zeff = Vt / It = Zm / (1 - Av) (RS rp)(Rout') w = ---------------------------------------------------------------------------------------------------------- p2 1 1 1 1 (RS rp)[Cp + (1 + gmRout')Cm]+ Rout'Cm Z =-------------æ-------------------- ö=--------------------------------------------= -------------- effjwC è1 – A ø jw(C (1 – A ) ) jwC m vCm m vCm M n Brute force analysis is not particularly helpful for gaining insight into the frequency response ... C = (1 – A )C is the Miller capacitor M vCm m EE 105 Fall 2000 Page 1 Week 12 EE 105 Fall 2000 Page 2 Week 12 Generalized Miller Approximation Voltage Gain vs. Frequency for CE Amplifier Using the Miller Approximation n An impedance Z connected across an amplifier with voltage gain AvZ can be replaced by an impedance to ground ... multiplied by (1 - A ) vZ n The Miller capacitance is lumped together with Cp, which results in a single- pole low-pass RC filter at the input n Common-emitter and common-source: n Transfer function has one pole and no zero after Miller approximation: A = large and negative for C or C --> capacitance at the input is magnified vZ m gd –1 w3dB = (rp RS)(Cp + CM) n Common-collector and common-drain: A » 1 --> capacitance at the input due to C or C is greatly reduced –1 vZ p gs w3dB = (rp RS)[Cp + (1 + gmro rocRL)Cm] –1 –1 w3dB » w1 from the exact analysis (final term Rout¢Cm is missing) n The break frequency is reduced by the Miller effect ... which results in a large Miller capacitor at the input. EE 105 Fall 2000 Page 3 Week 12 EE 105 Fall 2000 Page 4 Week 12 Common Collector Frequency Response Common-Collector Frequency Response Basic approach: two-port model from Chapter 8, “decorated” with device n Voltage buffer two-port model has input at base and output at emitter: capacitances Note carefully where the capacitances are connected! RL “Millerize” the base-emitter capacitor: gain across it is Avp = ------------------------ RL + Rout The DC output voltage VOUT is selected to be 0 V, so that the output voltage is just the small-signal voltage vout (phasor representation Vout) ææRL ö ö æRout ö CM =Cp(1 – Avp)=Cpç1 – ç----------------------- ÷ ÷ = Cpç------------------------÷ èèRL + Routø ø èRL + Routø The break frequency of the common-collector is: 1 w–3dB = --------------------------------------------------- (RS Rin )(Cm + CM) can approach the transition frequency for small source resistances EE 105 Fall 2000 Page 5 Week 12 EE 105 Fall 2000 Page 6 Week 12 Common-Base Frequency Response Common-Base Frequency Response Similar approach: start with two-port and add capacitors .... n Poles are separate (no coupling between input and output) 1 wp, in = ------------------------------- (RS Rin )Cp 1 wp, out = ---------------------------------- (RoutRL)Cm n The input pole is beyond the transition frequency, due to the low value of Rin = 1/gm n The output pole is a function of RL since the output resistance is so large. For small load resistances, the output pole can approach the transition frequency Two-port model: Summary of single-stage amplifiers: CE/CS: Miller effect decreases the bandwidth severely CC/CD: wideband CB/CG: wideband EE 105 Fall 2000 Page 7 Week 12 EE 105 Fall 2000 Page 8 Week 12 Multistage Amplifiers Example 1: Cascaded Voltage Amplifier n Single-stage transistor amplifiers are inadequate for meeting most design n Want Rin --> infinity, Rout --> 0, with high voltage gain. requirements for any of the four amplifier types (voltage, current, Try CS as first stage, followed by CS to get more gain ... use 2-port models transconductance, and transresistance.) Rin n Therefore, we use more than one amplifying stage. The challenge is to gain Rout insight into when to use which of the 9 single stages that are available in a modern BiCMOS process: CS Bipolar Junction Transistor: CE, CB, CC -- in npn and pnp* versions 1 CS2 MOSFET: CS, CG, CD -- in n-channel and p-channel versions * in most BiCMOS technologies, only the npn BJT is available n How to design multi-stage amplifiers that satisfy the required performance goals? n solve for overall voltage gain ... higher, but Rout = Rout2 which is too large * Two fundamental requirements: 1. Impedance matching: output resistance of stage n, Rout, n and input resistance of stage n + 1, Rin, (n+1), must be in the proper ratio Rin, (n+1) / Rout, n --> ¥ or Rin, (n+1) / Rout, n --> 0 to avoid degrading the overall gain parameter for the amplifier 2. DC coupling: we will directly connect stages: effect on DC signal levels must be considered, too EE 105 Fall 2000 Page 9 Week 12 EE 105 Fall 2000 Page 10 Week 12 Three-Stage Voltage Amplifier Cascaded Transconductance Amplifier n Fix output resistance problem by adding a common drain stage (voltage buffer) n input resistance should be high; output resistance should also be high initial idea: use CS stages (they are “natural” transconductance amps) Rin Rout CS1 CS2 n Output resistance is not that low ... few kW for a typical MOSFET and bias --> could pay an area penalty by making (W/L) very large to fix. n Overall Gm = - gm1 (ro1 || roc1) gm2 = Av1 gm2 ... can be very large BUT, output resistance is only moderately large ... need to increase it EE 105 Fall 2000 Page 11 Week 12 EE 105 Fall 2000 Page 12 Week 12 Improved Transconductance Amplifier Two-Stage Current Buffers n Output resistance: boost using CB or CG stage n since one CB stage boosted the output resistance substantially, why not add another one ... Rin R CS1 CS2 CB out n high-source resistance current sources are needed to avoid having roc3 limit the nd resistance n The base-emitter resistance of the 2 stage BJT is rp2 which is much less than the 2nd stage source resistance = 1st stage output resistance RS2 =Rout1 = bo1ro1 roc1 n Therefore, the output resistance expression reduces to Rout» gm2ro2rp2 roc2 = bo2ro2 roc2 ... no improvement over a single CB stage EE 105 Fall 2000 Page 13 Week 12 EE 105 Fall 2000 Page 14 Week 12 Improved Two-Stage Current Buffer: CB/CG n The addition of a common-gate stage results in further increases in the output resistance, making the current buffer closer to an ideal current source at the output port n The product of transconductance and output resistance gm2 ro2 can be on the order of 500 - 900 for a MOSFET --> Rout is increased by over two orders of magnitude ... practical limit ... on the order of 100 MW Of course, the current supply for the CG stage has to have at least the same order of magnitude of output resistance in order for it not to limit Rout. n General “resistance matching” ... try not to lose much in doing a current divider or a voltage divider. Which of these is appropriate depends on whether the signal is current or voltage at the port. EE 105 Fall 2000 Page 15 Week 12.
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