1 University of California at Berkeley College of Engineering Dept. of Electrical Engineering and Computer Sciences EE 105 Midterm II Fall 2005 Prof. Borivoje Nikolić November 17, 2005 Your Name: ______________________________ Student ID Number: ________________________ Guidelines Closed book and notes; there are some useful formulas in the end of the exam. You may use a calculator. You can unstaple the pages with formulas, but do not unstaple the exam. Show all your work and reasoning on the exam in order to receive full or partial credit. Time: 80 minutes = 1 hour, 20 minutes. Score Problem Points Possible Score 1 20 2 14 3 18 Total 5221. MOS current source [20 points] For the current source shown in Figure 1, dimensions of MOS transistors M1 and M2 are (W/L)1 = 2, (W/L)2 = 10. VDD = 2.5V. μnCox = 100 μA/V2, VTn = 0.5 V, λn = 0.05 V-1, γ = 0. a) [4 points] Find the value of RREF such that IREF = 100μA. You can ignore channel length modulation in this part. VDDRREFvOUTM1M2R1IREFIOUTFigure 1. RREF =3(b) [4 points] Find the value of R1 that gives IOUT = 40μA? Assume that M2 is in saturation. You can ignore channel length modulation in this calculation. (c) [4 points] Find the lowest output voltage vOUT for which the circuit in Figure 1 still acts as a current source. You can ignore channel length modulation. R1 = vOUT, min =4(d) [6 points] Find the small-signal output resistance of the current source in Figure 1. Rout =5(e) [2 points] If body effect parameter, γ > 0, would it increase or decrease the value of output resistance from part (d)? Explain your answer. Rout increases / decreases (circle one)62. MOS amplifiers [14 pts] For the MOS amplifier in Figure 2, (W/L)1 = 10, (W/L)2,= 20, (W/L)3 = 10, IBIAS = 50μA. VDD = 2.5V. μnCox = 100 μA/V2, μpCox = 30 μA/V2,VTn = VTp = 0.5 V, λn = λnp = 0.05V-1, γ = 0. CGS2 = 2CGS1 = 2CGS3 = 100fF. CGD2 = 2CGD1 = 2CGD3 = 10fF. Input voltage vS has negligible input resistance and contains a DC and an AC component. Figure 2. (a) [2 points] Find the bias current of transistor M1. IM1 =7 (b) [4 points] Find the small-signal voltage gain Av = vout/vs. Av =8(c) [4 points] Find the maximum and the minimum voltage at the output of this amplifier. (d) [4 points] Find the frequency of the dominant pole of this amplifier. Vout, max = V;Vout, min = V ω = rad/s93. Amplifier frequency response [18 points] An amplifier has all of its poles and zeros in the left-hand frequency plane (it is a stable, minimum-phase system) and an amplitude frequency response, as shown in Figure 3. |A(j1011021031041051061070204060-20-40Figure 3 a) [6 points] Write the transfer function that produces this response. b) [6 points] Draw the phase response that corresponds to this amplitude response.10(c) [6 points] Write the transfer function that produces the response from Figure 4. Figure 4.11Some equations Mass-action law )(2Tnpni=× Resistivity: Resistance: Total current (e-): Gauss’s law: Depletion layer: pn depletion layer capacitance: pn diffusion current ⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛−⎟⎟⎠⎞⎜⎜⎝⎛+= 12kTDqvpanndpidiffeWNDWNDqnJ⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛−= 1kTDqvSDeIi Diffusion capacitance: τ=kTqICDd21 effdnnnNq,11μ=σ=ρ⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛⎟⎠⎞⎜⎝⎛ρ=ρ=WLRWLtWtLRsqdxdnqDnEqJJJnndiffdrift+μ=+=∫ε=⋅QdSECVQ=dxdEφ−=⎟⎟⎠⎞⎜⎜⎝⎛+φε=+=dabisnpdNNqxxX112000biDdDdVXVXφ−= 1)(0biDjbiDbipajVCVxqNCφ−=φ−φ=1120012Threshold voltage (NMOS) ()pasoxpFBTnNqCVV φ−ε+φ−= 2212 ipnNaqkTln−=φ ()ppSBTnTnVVV φ−−φ−γ+= 220 NMOS equations: TnGSDVI <= V ,0 () VV, VVVVvvVvCLWiTnGSDSTnGSDSDSDSTnGSoxD−<>λ+⎟⎠⎞⎜⎝⎛−−μ= ,12 ()() VV, VVVVVvCLWiTnGSDSTnGSDSTnGSoxD−>>λ+−μ= ,122 MOS capacitances in saturation MOS signal parameters: ()() ()TnGSoxDSTnGSoxDSVGSVGSDmVVLWCVVVLWCvig −μ≈λ+−μ=∂∂= 1, DSDSVGSVDSDoIvirλ≈⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛∂∂=−11, pBSmQBSDmbVgvigφ−−γ=∂∂=22 Power: Canonic transfer function: Bode plots Amplitude Phase: Second order TF: LC: ovoxgsCWLCC+=)3/2(oxDovWCLC=(){}*Recos2VIVIVIP
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