ECSE 334 Winter 2009 Midterm

2 assuming r o 6 by what factor is the small signal

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Unformatted text preview: f the CMOS amplifier in Fig. 1, the simple MOS current mirror (Q5 - Q6) can be replaced by a cascode MOS current mirror (Q5 - Q8), as shown in Fig. 2. Assuming R o 6 = ∞ , by what factor is the small-signal differential voltage gain A d improved in Fig. 2, compared to Fig. 1? State why. Assuming R o 6 = ∞ , the output resistance of the CMOS amplifier in Fig. 2 is: R o = r o 2 || R o 6 ≅ r o 2 = 40 k Ω Therefore, compared to Fig. 1, the small-signal differential voltage gain A d in Fig, 2 is improved by a factor of: R o in Fig. 2 ro2 A d in Fig. 2 40 k Ω --------------------------- = ----------------------------- ≅ ------------------- = ------------------- = 2.6 R o in Fig. 1 r o 2 || r o 6 A d in Fig. 1 15.4 k Ω ECSE334 - Prof. Anas Hamoui Midterm - SOLUTION page 3 of 7 ECSE334 Introduction to Microelectronics + VDD VSG - Q7 Q5 Winter 2009 + VSG - Q8 VD8 = V DD - VSG = VDD - (VOV + |Vt|) VSG + - + Q6 VOV - Ro6 = ∞ Q2 Q1 + vo v id IREF = 50μA - Q3 Q4 Fig. 2 e) Write expressions for the maximum output voltage V o, max of the differential amplifiers in Figures 1 and 2. Here, V o, max denotes the largest value of output voltage V o for proper operation of the current mirror (Q5-Q6 in Fig. 1 and Q5-Q8 in Fig. 2), such that all its transistors are in saturation. Find their values, assuming transistors Q5 - Q8 operate at V OV = 0.25V and V DD = 2.5V . Since all PMOS transistors have the same overdrive voltage, let V GS ≡ V GS 5 = V GS 6 = V GS 7 = V GS 8 = V OV + V tp = 0.25V + 0.65V = 0.9V In Figure 1, for Q6 to remain in saturation: V SD 8 = V DD – V o ≥ V OV ⇒ V o, max = V DD – V OV = 2.5V – 0.25V = 2.25V In Figure 2 (see circuit schematic): V D 8 = V DD – V GS = V DD – ( V OV + V tp ) Therefore, for Q6 to remain in saturation: V SD 8 = V D 8 – V o ≥ V OV ⇒ V o, max = V DD – ( 2 V OV + V tp ) = 1.35V ECSE334 - Prof. Anas Hamoui Midterm - SOLUTION page 4 of 7 ECSE334 Introduction to Microelectronics Winter 2009 IBIAS1 = 1.125mA X R1 CL1 RS vS g m v IN IN Q1 Vbias Q2 R2 OUT IBIAS2 = 0.75mA vo CL2...
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