9.3 For the differential amplifier specified in Problem 9.1 let v G2 =0 and vG1 = vid. Find the value of vid that corresponds to each of the following situations: (a) iD1 =iD2 =0.08 mA; (b) iD1 =0.12 mA and iD2 =0.04 mA; (c) iD1 =0.16 mA and iD2 =0 (Q2 just cuts off); (d) i D1 =0.04 mA and iD2 =0.12 mA; (e) iD1 =0 mA (Q1 just cuts off) and iD2 =0.16 mA. For each case, find vS, vD1, vD2, and (vD2 – vD1).

9.3 - 9.3 For the differential amplifier specified in Problem 9.1 let v G2 =0 and vG1 = vid. Find the value of vid that corresponds to each of the following situations: (a) iD1 =iD2 =0.08 mA; (b) iD1 =0.12 mA and iD2 =0.04 mA; (c) iD1 =0.16 mA and iD2 =0 (Q2 just cuts off); (d) i D1 =0.04 mA and iD2 =0.12 mA; (e) iD1 =0 mA (Q1 just cuts off) and iD2 =0.16 mA. For each case, find vS, vD1, vD2, and (vD2 – vD1).

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images - 9.3 For the differential amplifier specified in Problem 9.1 let v G2 =0 and vG1 = vid. Find the value of vid that corresponds to each of the following situations: (a) iD1 =iD2 =0.08 mA; (b) iD1 =0.12 mA and iD2 =0.04 mA; (c) iD1 =0.16 mA and iD2 =0 (Q2 just cuts off); (d) i D1 =0.04 mA and iD2 =0.12 mA; (e) iD1 =0 mA (Q1 just cuts off) and iD2 =0.16 mA. For each case, find vS, vD1, vD2, and (vD2 – vD1).

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