EE 140 ANALOG INTEGRATED CIRCUITS SPRING 2009 C. Nguyen PROBLEM SET #9 Issued: Tuesday, April 7, 2009 Due: Tuesday, April 14, 2009, 5:00 p.m. in the EE 140 homework box in 240 Cory 1. An amplifier has low-frequency forward gain of 40,000, and its transfer function has three negative real poles with magnitudes 2 kHz, 200 kHz, and 4 MHz. (a) If this amplifier is connected in a feedback loop with a constant feedback factor f and with low-frequency gain A0=400, estimate the phase margin; (b) Repeat (a) if A0=200 and then 100. 2. The amplifier a(s) is has DC gain of 10,000 and three real negative poles. The pole frequencies of the first and the third pole are 1 kHz and 200 MHz, respectively. (a) If R1=R2 find the location of the second pole such that the feedback amplifier shown in Figure PS9-2 is stable with a phase margin of 60o. Neglect the input impedance of the amplifier. (b) Write the transfer function in as a function of the complex variable s and draw Bode plots for the open-loop amplifier gain a(s) and the closed-loop gain A(s). (c) What is the new phase margin if: i. In addition to the three poles the amplifier a(s) has one real right half-plane zero at the frequency of the second pole. ii. In addition to the three poles the amplifier a(s) has one real left half-plane zero at the frequency of the second pole. iii. The closed-loop amplifier A(s) is configured as a unity gain buffer. iv. R1=9R2. A(s)R1R2vINvOUT Figure PS9-2 a(s)EE 140 ANALOG INTEGRATED CIRCUITS SPRING 2009 C. Nguyen 3. Razavi, Chapter 10: Problem 10.4. 4. In the amplifier shown in Figure PS9-4 transistors M3-M8 are biased with Vov=200mV. The gates of M3 and M4 are biased to allow the maximum undistorted sinusoidal signal at the output. Calculate all currents, channel widths and the value of capacitor Cc so that the amplifier has a DC gain of 20, a unity gain frequency of 50MHz, and a phase margin of 60o when placed in unity gain closed-loop feedback. All transistors have the same channel length. Neglect all parasitic capacitances in this problem. 0,0,0,5.0,250,5.05,10,32,0=======Ω==doxnnthLLDDLmLVACVVpFCkRVVγλμμμ Figure
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