Dependent Sources - Introduction to Microelectronic Circuits - Solved Exam, Exams of Microelectronic Circuits

Download Dependent Sources - Introduction to Microelectronic Circuits - Solved Exam and more Exams Microelectronic Circuits in PDF only on Docsity! EE40 Midterm 2 Solutions Spring 2000 Problem #1: Circuits with Dependent Sources [20 points] a) Find V0. [4 pts] Current divider formula: ix = (10 kohm/(10 kohm + 40 kohm))*5 mA = 1 mA V0 = (-10ix)(10 kohm) = (-10*1 mA)(10 kohm) = -100 V b) In the circuit below, the independent source values and resistance are known. Use the nodal analysis technique to write 3 equations sufficient to solve for Va, Vb, and Vc. To receive credit, you must write your answer in the box below. [6 pts] DO NOT SOLVE THE EQUATIONS! Write the nodal equations here: Note that the only unknowns in these equations are Va, Vb, Vc node a: IAA + Va/R1 + (Va-Vb)/R2 = 0 supernode: (Vb-Va)/R2 + (Vb+VAA)/R3 + Vc/R4 = 0 relationship due to dependent source: Vc-Vb = 10Va/R1 EE40 Midterm 2 Solutions EE40 Midterm 2 Solutions Spring 2000 1 c) Consider the following circuit: i) Find the voltage Vab [5 pts] i1 = Vx/7 kohm Applying KCL to node x: 70-Vx/3 kohm = i1 + 20i1 = 21i1 = 21Vx/7 kohm 70-Vx = 9Vx => Vx = 7 i1 = 7 V/7 kohm = 1 mA; ia = 0 => Vab = Vy = 20i1(3 kohm) = 60V ii) What is the current ia when the terminals a and b are shorted together? [3 pts] Current divider formula: ia = (3 kohm/(3 kohm + 1 kohm))(-20 mA) = -15 mA iii) Draw the Thevenin Equivalent Circuit. [2 pts] Vth = Voc = Vab from part (i) Rth = -Voc/Isc = Vab from part (i)/ia from part (ii) = 60 V/-15 mA = 4 kohm Problem #2: Transient Response [30 points] EE40 Midterm 2 Solutions Problem #1: Circuits with Dependent Sources [20 points] 2 Assume the op-amps in this problem are ideal. a) Consider the following circuit: i) Find an expression for V0 as a function of Va. [6 pts] This is a difference amplifier circuit (which you've studied in the lab) with Ra/Rb = Rc/Rd = 1/2 V0 = (Rb/Ra(6-Va) = 2(6-Va) = 12 - 2Va ii) Find V0 for Va = 2 V. [3 pts] V0 = 12 - 2(2) = 8 V iii) For what values of Va will the op-amp be saturated? [6 pts] V0 = 12 - 2Va => Va = (12 - V0)/2 V0 saturates at 15 V: Va <= (12-15)/2 = -3/2 V V0 saturates at -15 V: Va >= (12 - (-15))/2 = 27/2 V Values of Va for which the op-amp will be saturated: Va <= -1.5 V; Va >= 13.5 V b) In the following circuit, the op=amps are operating linearly. EE40 Midterm 2 Solutions Problem #1: Circuits with Dependent Sources [20 points] 5 Find Vout in terms of V1, V2, R1, R2, R3, R4 [10 pts] (Hint: The superposition method might be helpful here.) Find the individual contributions of each voltage source: i) Set V2 to 0 V: Vx = 0, so the circuit simplifies to a simple inverting amplifier ii) Set V1 to 0 V: circuit simplifies to simple non-inverting amplifier Add the contributions of each source together: Vout = Vout' + Vout'' = -(R2/R1)(V1) - (R4/R3)(1+(R2/R1))(V2) Problem #4: Semiconductor properties; p-n diodes [25 points] a) Consider a silicon sample maintained at 300K under equilibrium conditions, uniformly doped with 1*1016 cm-3 phosphorus atoms. The surface region of the sample is additionally doped uniformly with 5*1016 cm-3 boron atoms, to a depth of 1 microm, as shown in the figure below. EE40 Midterm 2 Solutions Problem #1: Circuits with Dependent Sources [20 points] 6 i) In the figure above, indicate the type of regions (I and II) by labelling them as "n" or "p" type. [2 pts] ii) What are the electron and hole concentrations in Region I? [5 pts] NA > ND, and NA >> ni so p = NA - ND = 5*1016 - 1*1016 = 4*1016 pn = ni2 => n = ni2/p = (1.45*1016)2/4*1016 = 5256 n = 5256 cm-3; p = 4*1016 cm-3 iii) What is the sheet resistance of Region I? [5 pts] rho = 1/(qmunn+qmupp) ~ 1/qupp >From plot on Page 2, mup ~ 350 cm2/Vs for NA + ND = 6*1016 cm-3 Rs = rho/t = 1/qmuppt = [(1.602*10-19)(350)(4*1016)(10-4)]-1 = 4458 ohm/square iv) Suppose any voltage between 0 V and 5 V can be applied to Region I. What fixed voltage ("bias") would you apply to Region II, to guarantee that no current would ever flow between Region I and Region II? Briefly explain your answer. [3 pts] To prevent current from flowing, we need to ensure that the p-n junction will never be forward biased. Thus, the n-type region must be biased at 5 V or higher. b) If a diode is operated only within a small range of forward-bias voltages, its behavior can be accurately modeled by a resistor, whose value is dependent on the bias voltage. Derive an expression for the diode "small-signal" resistance: Rdiode = (dI/dV)-1 in terms of the saturation current Is, the bias voltage V, and the absolute temperature T. [5 pts] I = Is(eqV/kT - 1) dI/dV = Is(q/kT)eqv/kT Rdiode = (kT/qIs)e-qV/kT c) Plot VL vs. VIN for -10 V < VIN < 10 V on the axes provided, for the circuit below. Note that the diode is a perfect rectifier. Label the axes. [5 pts] EE40 Midterm 2 Solutions Problem #1: Circuits with Dependent Sources [20 points] 7