ECSE 334 Winter 2009 Midterm

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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online