# HW5_Soln - March 24, 2008 ECE315 Homework # 5 Solutions...

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March 24, 2008 ECE315 Homework # 5 Solutions (Rev. 0) Spring 2008 1. Textbook Chapter 7, Problem 4. MOSFET Bias Circuit. 200 mV drop on R s means that I Q D = 2 mA. (a) Assume M 1 is saturated, then I D = 1 2 μ n C ox ( W L )( V GS - V T ) 2 -→ W L = 2 I D μ n C ox ( V GS - V T ) 2 . In the drain circuit V Q DS = V DD - I Q D ( R D + R S ) = 1 . 8 - 0 . 2(0 . 6) = 0 . 6 volts . This corresponds to a maximum value of V GS - V T to remain at edge of active region (saturation). Hence V Q GS = V T + 0 . 6 = 0 . 4 + 0 . 6 = 1 volts . W L = 2 I D μ n C ox ( V GS - V T ) 2 = (2)2(10 - 3 ) 200(10 - 6 )(0 . 6) 2 = 55 . 5 . (b) R in = R 1 k R 2 = 30 k Ω , voltage divider V DD R 1 / ( R 1 + R 2 ) = V Q GS = 1 volts . Then R 2 R 1 + R 2 = R 1 k R 2 R 1 = V Q GS V DD = 1 1 . 8 = 0 . 555 -→ R 1 = 30 . 5555 = 54 k Ω . 1 . 8 R 2 = R 1 + R 2 -→ R 2 = R 1 0 . 8 = 54 0 . 8 = 67 . 5 k Ω . 2. Textbook Chapter 7, Problem 14. Assume V DD = 1 . 8 volts. Scaled Current Sources. V B = V G = 1 volt , so V Q SG = V DD - V G = 1 . 8 - 1 = 0 . 8 volts . Then: I x = 1 2 μ p C ox W L ( V SG + V T ) 2 = 1 2 (100)(10 - 6 ) 20 0 . 25 (0 . 8 - 0 . 4) 2 = 0 . 64 mA , I y = 2 I x = 1 . 28 mA . r o 1 I D -→ r o 1 = 2 r o 2 . 3. Textbook Chapter 7, Problem 16. Another CMOS Inverter.

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ECE315 March 24, 2008 Page 2 (a) V x = 0 . 9 volts -→ V DS 1 = 0 . 9 volts , V GS 1 = V B . Then: I D = 1 2 μ n C ox ( W L ) 1 ( V B - V T ) 2 (1 + λ 1 V x ) , -→ V B = V T + s 2 I D μ n C ox ( W L ) 1 (1 + λ 1 V x ) . V B = 0 . 4 + s (2)0 . 5(10 - 3 ) 200(10 - 6 )( 5 0 . 18 )[1 + (0 . 1)0 . 9] = 0 . 8064 volts . (b) 1. For V x < V B - V T then M 1 is ohmic so I D 1 < I D 2 so I x = I D 1 - I D 2 < 0. 2. For V DD - V x < V DD - V B + V T then M 2 is ohmic so I D 2 < I D 1 so I x = I D 1 - I D 2 > 0 3. For values of V x in between these limits both device are saturated. I D 1 I D 2 near V x = 0 . 9 volts . The sketch. .. 4. Textbook Chapter 7, Problem 23. nmos vs. pmos. For higher voltage gain the circuit in Figure (a) is preferred as the nmos device usually has a higher g m (higher electron mobility) for the same dimensions. The eﬀective load is r o 1 k r o 2 in either case. 5. Textbook Chapter 7, Problem 32. Voltage Gain Drill.
ECE315 March 24, 2008 Page 3 (a) Common Source with R S replaced by a diode connected M 3 and R D replaced by a diode connected M 2 . We derived in class for r o → ∞ : A CS V = - g m R D k R L 1 + g m R S = - R D k R L 1 g m + R S = - r o 2 k 1 g m 2 1 g m 1 + r o 3 k 1 g m 3 ≈ - 1 g m 2 1 g m 1 + 1 g m 3 . (b) Common Source with R S replaced by a diode connected M 3 and R D replaced by a diode connected M 2 and the R D combo. The equivalent resistance of this combination is just 1 /g m 2 - draw the circuit if you do not see this. Therefore, A V ≈ - 1 g m 2 1 g m 1 + 1 g m 3 k r o 3 . (c)

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## This note was uploaded on 09/28/2008 for the course ECE 3150 taught by Professor Spencer during the Spring '07 term at Cornell University (Engineering School).

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HW5_Soln - March 24, 2008 ECE315 Homework # 5 Solutions...

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