HW2: Slew Rate, Compensation and Time Constant Analysis
advertisement
HW2: Slew Rate, Compensation and Time Constant Analysis ECEN 607 Advanced Analog Circuit Design By Luis A. Tellez-Estrada Prob1. A resistive inverting amplifier with an Op Amp with a GB = 4.1 MHz and SR = 3.0 V/μs is connected as shown in Fig. 1. Consider two cases for the input: I. II. A step of magnitude Vm A sinusoidal v(t) = Vpsin wt 2R1 GBW = 4.1MHz SR = 3.0 V/µs R1 Vi Vo Fig. 1 Determine: a) The maximum value of Vom yielding a bandwidth limited amplifier. The maximum value of Vom can be calculated from equation (1). πππ(ππππ‘) = ππ 2ππΊπ΅π½ (1) Therefore, πππ(ππππ‘) = 3π₯106 1 2π (4.1π₯106 3) = 349 ππ (2) And the maximum input voltage is, ππππππ₯ = 174 ππ (3) b) The maximum frequency for a sinusoidal input with a peak magnitude of Vp = 0.2 V so output is bandwidth limited. The maximum frequency can be calculated from equation (4). ππππ₯ = Therefore, ππ 2πππ π΄π£ππ (4) ππππ₯ = 3π₯106 = 1.19 ππ»π§ 2π(0.2 β 4) (5) c) The 3dB cutoff frequency of the inverting amplifier. The cutoff frequency can be calculated from equation (6). π»(π ) = π΄π£ππ 1 1 = π΄π£ππ π½⁄ π π½⁄ 1+ π΄ 1+ πΊπ΅ (6) π3ππ = πΊπ΅/π½ (7) 4.1π₯106 = 1.36 ππ»π§ 3 (8) Therefore, π3ππ = Prob. 2 Two Stage Miller with Source Follower Compensation Due to the direct path between the first and second stage in common two stage Miller amplifier a RHP zero is created. However, there are different methods to eliminate or minimized the effect of this zero. A source follower (voltage buffer) is implemented in this work as is shown in Fig. 2. Voltage Buffer M8 M4 M3 M9 Cc V- M5 M1 M2 M6 Vo V+ M10 Fig. 2 M7 With the implementation of the source follower the RHP zero is eliminated by eliminating the direct path between the first and second stage as shown in the small signal model presented in Fig. 3. Moreover pole splitting is still generated due to the compensation capacitor and Miller effect. The final transfer function is presented in equation (9). Vo V1 Cc gm1Vi R1 C1 Vo gm8V1 C2 R2 Fig. 3 π΄π£ππ = ππ1 ππ2 π 1 π 2 π π (1 + π€ ) (1 + π€ ) π1 π2 (9) π1 ≈ −1 ππ2 π 1 π 2 πΆπ (10) π2 = −ππ7 πΆπ πΆ2 (πΆπ + πΆ1 ) (11) We can observe that the zero is not presented in the transfer function and the same poles remains almost at the same place. Buffer Design Ideally a voltage buffer has a gain of 1 v/v. However, in practice this value is impossible. Since the voltage buffer is implemented in a source follower configuration, the gain of this stage depends on the transconductance of M9 and the output resistance as shown in equation (12). π΄π£ππ = ππ9 ( 1 1 || ||π ||π ) ππ 9 ππ9 ππ 9 ππ 10 (12) Assuming that ππ9 β« ππ 9 then for an ideal buffer the drain to source resistances must be large enough to maintain a gain approximately off 1v/v. According to the description the voltage buffer was designed to have a current equal to the tail current. As a result the power consume is not increased too much and the slew rate remains the same. However little mismatch between these two branches might produce a slightly different performance. The characterization of the voltage buffer is presented in Fig 4. Buffer Gain Fig. 4 We can observe that the gain is approximately 0.82 V/V. The final implementation and the total performance of the amplifier is presented in the next section (plot section) and summarized in Table 1. Plot Section AC Response CMRR Phase Margin VS CLoad GBW VS CLoad Positive Slew Rate Negative Slew Rate PSRR Final Results The final results presented in the following table summarized the three design techniques from homework 1 and this work. Two Stage OTA Specs QUADRATIC ACM ACM + Ahuja ACM + Buffer VDD 1.8 V 1.8 V 1.8 V 1.8 V VSS 0V 0V 0V 0V Gain 59.8 dB 64 dB 53.62 dB 64.02 dB CMRR 61.68 dB 111.2 dB 68.25 dB 111.27 dB PSRR @100 Hz -61 dB -63 dB -62 dB -63.66 dB PSRR @100 kHz -34 dB -33 dB -61.92 dB -38.03 dB GBW 4.96 MHz 4.06 MHz 8.94 MHz, 4.8MHz** 7.06 MHz PM 60.42° 60.37° 89.8°, 63.2° ** 62.03° CL 20pF 20pF 20pF 20pF SR+,SR3.78μV/s, 4.3μV/s, 6.3μV/s, 7.22μV/s, -2.94μV/s -3.34μV/s -3.22μV/s -3.96μV/s SettlingTime +,- 0.21μs, 0.4μs 0.3μs, 0.5μs 0.15μs, 0.3μs 0.156μs, 0.34μs Cc 4 pF 3.5 pF 1.5pf 2pf IQ 232 μA 101.7 μA 113.0 μA 108.5 μA PD 417.6 μw 183.1 μw 203.4 μW 195.3 μW Min if 6 6 6 Max if 10 13 10 FOM* 1.29 2.41 7.1 4.05 *FOM (MHz)(°)/( μA) ** Values for CL = 200 pF Table 1 We can observe that ACM + Ahuja and ACM + Buffer compensation are much better amplifiers over the normal OTA amplifier with Miller Compensation. Moreover we can say that the Ahuja OTA handle higher values of load capacitance while providing better stability. However the gain is compromised and the power consumption is increased due to the biasing branch for the NMOS and PMOS active loads in the current buffer. One point to take into account is the performance of PSRR at high frequencies. Once again the Ahuja compensation has the better performance over the other designs. A FOM (figure of merit) was selected according to the GBW, the phase margin and the quiescent current. The higher the FOM the better the amplifier performance. According to FOM The OTA + Ahuja compensation exhibits the best performance. Dynamic Range Performance The following plots presents the dynamic response of the ACM+ Ahuja and ACM + Buffer amplifiers previously designed and summarized in Table 2. Voltage Buffer Compensation 1 dB Compression plot IP3 - Third Order Interception plot Harmonic Distortion Freq. 10kHz THD for 1% Distortion plot Ahuja Compensation 1 dB Compression plot IP3 - Third Order Interception plot Harmonic Distortion Freq. 10kHz THD for 1% Distortion plot Dynamic Performance Summary Dynamic Range Performance 1 dB Compression (dB) IP3 (dBm) THD 1% Distortion (10khz) (dBm) OTA +Ahuja -38.09 -18.63 -46.53 Table 2 OTA + Buffer -40.93 -39.99 -56.33 According to the results, the OTA + Ahuja compensation shows better linearity over the OTA + Buffer. Prob. 3 The following work implements the time constant analysis to determine the expression for a1 and a2 of the transfer function in (13) for the Ahuja amplifier shown in Fig. 5. Current Buffer M4 M3 M8 V1 M9 IB V- M1 M2 V+ Vo V3 V2 Cc M5 M7 M6 Fig. 5 π π΄π (1 + π€ ) πΎπ(1 + π€π§ )π€1 π€2 π€3 π»(π ) = = 2 3 π π π (1 + π€ ) (1 + π€ ) (1 + π€ ) 1 + π1 π + π2 π + π3 π π1 π2 π3 π§ (13) The equivalent small signal model is presented in Fig. 6 to simplify the calculation of the coefficients. V1 Gm1Vi R1 Cc V3 C1 Gm3V3 R3 C3 V2 Gm2V1 C2 R2 Fig. 6 πΊπ1 = ππ1 πΆ1 ≈ πΆππ 8 + πΆππ9 π 1 = πππ 1 ||πππ 3 πΊπ2 = ππ8 πΆ2 ≈ πΆπΏ + πΆππ7 π 2 = πππ 8 ||πππ 7 πΊπ3 = ππ9 πΆ3 ≈ πΆππ 9 π 1 ≈ 1 ππ9 Calculation of a1 The coefficient of a1 is given by, π1 = πΆ1 π 011 + πΆ2 π 0 22 + πΆ3 π 0 33 + πΆπ π 0 23 Where π 0 π₯π₯ is the resistance seen by terminal x to ground and π 0 π₯π¦ is the resistance seen by terminal x to y where all capacitors in the network are open-circuit but the one between the two terminals is replaced by a current source. Following these rules π 011, π 0 22 and π 0 33 can be extracted from the small signal model in an easy way. V1 1 R1 V2 1 V3 R2 π 011 = π 1 π 0 22 = π 2 π 0 33 = π 3 For π 0 23 a nodal analysis was implemented. 1 R3 1 V2 V1 R1 Gm3V3 R3 R2 Gm2V1 π1 = πΊπ3 π 3 π 1 π3 = π 3 β 1π΄ π 0 23 = π3 − π2 = πΊπ2 πΊπ3 π 2 π 1 + π 2 + π 3 The final value of a1 can be expressed in terms of its coefficients by π1 = πΆ1 π 1 + πΆ2 π 2 + πΆ3 π 3 + πΆπ (πΊπ2 πΊπ3 π 2 π 1 + π 2 + π 3 ) (14) Calculation of a2 The coefficient of a2 is given by, π2 = πΆ1 πΆ2 π 011 π 1 22 + πΆ1 πΆ3 π 011 π 1 33 + πΆ1 πΆπ π 011 π 1 23 +πΆ2 πΆ3 π 0 22 π 2 33 + πΆ2 πΆπ π 0 22 π 2 23 +πΆ3 πΆπ π 0 33 π 3 23 Where π π§ π₯π₯ is the resistance seen by terminal x to ground and π π§ π₯π¦ is the resistance seen by terminal x to y where z is short-circuited to ground and all capacitors in the network are opencircuit but the one between the x to y terminal is substituted by a current source of 1A. The same procedure of previous work is implemented to determine the rest of the resistances. An example to calculate π 1 22 is presented. V1 = 0 R1 C1 V2 V3 = 0 Gm3V3 R3 Gm2V1 R2 According to the small signal π 1 22 = π 2 . π 1 33, π 2 33, π 2 23 and π 3 23 are straight forward to calculate following the same procedure, however π 1 23 needs to be calculated with the same procedure shown in π 0 23 analysis. The final values are, π 1 22 = π 2 π 2 33 = π 3 π 1 33 = π 3 π 2 23 = π 3 π 1 23 = π 2 + π 3 π 3 23 = π 2 As a result a2 can be expressed as follow. π2 = πΆ1 πΆ2 π 1 π 2 + πΆ1 πΆ3 π 1 π 3 + πΆ1 πΆπ π 1 (π 2 + π 3 ) +πΆ2 πΆ3 π 2 π 3 + πΆ2 πΆπ π 2 π 3 +πΆ3 πΆπ π 2 π 3 (15) Finally H(s) can be expressed in terms of equation (14) and (15). Applying approximations 1 πΊπ π β« 1, π 3 ≈ πΊ , and π 1,2 β« π 3 then the coefficients are determined by, π3 π1 ≈ πΊπ2 π 1 π 2 πΆπ π2 ≈ π 1 π 2 πΆ1 (πΆ2 + πΆπ ) The next plot shows the approximated AC response of the Ahuja OTA by implementing the coefficients extracted in the time constant analysis. π1 ≈ 14.1π₯10 − 5 π2 ≈ 2.5π₯10 − 13 π΄π£π ≈ 1.2π₯103 We can observe that the values approximate to the ones obtained in cadence simulation. Avo GBW PM Simulation Matlab 53.62 dB 61 dB 8.94 MHz 8.32 MHz 89.8° 89.2° Some discrepancy occurred due to the parasitic capacitances that are not taking into account. References [1] Ahuja, B.K., "An improved frequency compensation technique for CMOS operational amplifiers," in Solid-State Circuits, IEEE Journal of , vol.18, no.6, pp.629-633, Dec. 1983 doi: 10.1109/JSSC.1983.1052012 [2] P.E. Allen – 2001, “LECTURE 430 – COMPENSATION OF OP AMPS-II”, Online: http://www2.ece.gatech.edu/academic/courses/ece4430/Filmed_lectures/OAC2/L430OpAmpCompII.pdf [3] Dr. Sanchez Sinencio, ECEN 607, “Conventional OpAmps”, online: http://ece.tamu.edu/~sanchez/607-Lect%201A%20Conventional%20Op%20Amps%20-2015.pdf [4] Coitinho, R.M.; Spiller, L.H.; Schneider, M.C.; Galup-Montoro, C., "A simplified methodology for the extraction of the ACM MOST model parameters," in Integrated Circuits and Systems Design, 2001, 14th Symposium on. , vol., no., pp.136-141, 2001
