Electrical Engineering Problem Set: Transistor Amplifier, Logic Circuits, Diode Circuits, Exams of Microelectronic Circuits

Download Electrical Engineering Problem Set: Transistor Amplifier, Logic Circuits, Diode Circuits and more Exams Microelectronic Circuits in PDF only on Docsity! EECS 40, Fall 2008 Prof. Chang-Hasnain Final Exam Solutions Version A 8:00 am – 11:00 am, Saturday December 13, 2008 Total Time Allotted: 180 minutes Total Points: 300 1. This is a closed book exam. However, you are allowed to bring 4 pages (8.5” x 11”), double-sided notes. 2. No electronic devices, i.e. calculators, cell phones, computers, etc. 3. SHOW all the steps on the exam. Answers without steps will be given only a small percentage of credits. Partial credits will be given if you have proper steps but no final answers. 4. Remember to put down units. Points will be taken off for missed unit. Last (Family) Name:_____________________________________________________ First Name: ____________________________________________________________ Student ID: ___________________________ Discussion Session: ________________ Signature: _____________________________________________________________ Score: Problem 1 (116 pts) Problem 2 (36 pts) Problem 3 (58 pts) Problem 4 (90 pts) Total (300 pts) Page 2 of 17 Problem 1: Transistor Amplifier (116 pts) Note: The sub-parts are somewhat independent. Read through the entire problem even if you are not able to solve some parts. Good luck! In this problem you are about to analyze the amplifier circuit shown in the figure below. The threshold voltage Vt0 = 1 V and the capacitor impedance at signal frequency is negligible. D D 2 kOhm 1 kOhm C C 150 Ohm vs*sin(wt) vin vout 5 V D D Figure 1: Amplifier circuit for Problem 1 The diodes D in figure 1 have the following I-V characteristic: Idiode [mA] Vdiode [V]0.7 1.0 1.3 1 2 Figure 2: I-V characteristics of diodes in circuit shown in figure 1. Page 5 of 17 e) For the values found in d) and given in the problem, in Figure 1 and the I-V characteristic of the diode: Is the transistor in saturation? Why or why not? (12 pts) If IDS is 2mA the voltage drop across 1kOhm resistor is 2 V. From the diode characteristics the voltage drop across the diode is 1.3 V. That leaves 5 – 3.3 V for VDS which is 1.7 V. We have to check whether VDS > VGS – VT and since 1.7 > 2 – 1 V the transistor is in saturation. If one argues that VG is 3 V and since VGS is equal to 2 V the voltage drop across the diode has to be 1 V it is ok as well. In that case VDS would be 2V and the transistor still in saturation. f) Based on the values you got in part d), calculate gm (K of transistor is 2mA/V2) (assume the transistor is in saturation regardless of your answer in part e) ) (4 pts) gm = 2 sqrt (K * IDS) = 4 mS Page 6 of 17 g) Draw the small signal equivalent circuit for the circuit given in Figure 1. You can neglect rd of the transistor. If you don’t know how to deal with the diodes simply replace them with 100 Ω resistors in the small signal circuit (for partial credit). If you include the full diode model, explain how you do it. (27 pts if model includes diodes, 24 pts if you choose to replace diodes with 100 Ω resistors) 2 kOhm 900 Ohm(300 Ohm) 150 Ohm 300 Ohm (100 Ohm) 1 kOhm vS G D S + vIN - gm*vgs + vout - Diodes can be simply replaced by their equivalent resistance in the on region. The DC voltage sources of the diode model are replaced by a short. Values in parenthesis are values if 100 Ohm resistor used in stead of diodes Page 7 of 17 h) For this part of the problem assume a gm of 10 mS (again this might be different from the gm you found in part f)) Using small signal analysis and the circuit you drew in g), find the small signal voltage gain Av=vout / vin and the input impedance of the circuit without  Vout = - gm * 1 kOhm * vin / (1+300*gm) the AC source and the 150 Ω resistor connected to Vin (you can leave the numerical results in fractions). (24 pts) vout = - gm * vGS * 1 kOhm vGS = vin – gm*vGS * 300  vgs = vin / (1 + 300*gm)  AV = 10/(1+3) = -2.5 (-5 if 100 Ohm resistors are used instead of diodes) Rin = vin/iin = 2kOhm//900kOhm is approx. 600 Ohm (=1800000/2900) (6000000/2300 or approx 300 Ohm if 100 Ohm resistor used instead of diodes) i) Now connect the AC source and the 150 Ω resistance to the circuit again (as shown in figure 1) (8 pts) Calculate the AC output voltage vout for a source voltage amplitude vs of 0.1 V. Vout = vs * Rin / (150 + Rin) * AV = 0.1 * 4/5 * 2.5 = 0.2 V (0.33 V if 100 Ohm resistor is used) Page 10 of 17 Problem 3: Diode Circuits (60 pts) In this problem you are to analyze the circuit below. Assume a forward bias turn-on threshold voltage for each diode (Zener as well as regular diode) of 0.6 V. Further assume that the resistors R are big enough such that the capacitor C does not charge nor discharge noticeably through them within a period of Vin. If you don’t know how to solve the problem with both Zener diodes, you can replace the 100 V Zener diode with a regular diode (for partial credit vin C R 10 V R 4.4 V 100 V + - Vx + Vout - + VC - ). Clearly state that you do so though! Using the 100 V Zener diode or a regular diode does not make a difference here since we never reach 100 V here a) What is Vx and Vout if Vin = 0 V? (6 pts; (4 pts if regular instead of 100 V Zener diode is used) VX = 4.4 + 0.6 V = 5 V (4.4 V Zener diode in reverse breakdown, 100 V Zener diode forward biased) Vout = Vx – 0.6 V = 4.4 V Page 11 of 17 b) Now assume that Vin has been 0 for a while before it gets turned on at t=0 and has an amplitude of 2 V. Vin is given in the graph below. Complete the graph with VC, Vx, and Vout. Make sure you label your graph! (Derive VC at t=0 from part a)) (24 pts) VC = -5 V at the beginning but charges to -3 V within the first quarter of T (since VX is clamped to 5 V and Vin raises to + 2V  Capacitor charges). From there on, since Vx is never larger than 5 V again, the zener diode branch is basically always off and the voltage across the capacitor cannot change anymore (no charge/discharge path) and Vx = Vin - VC for t > T/4. Vout is always Vx – 0.6 V due to the voltage drop across the diode. V t -2 V -4 V -5 V 4 V 2 V 5 V 3 V 1 V -3 V 0.6 V __ Vx __ Vout __ VC Page 12 of 17 c) Repeat part b), but now with an Vin = 3 V (28 pts) VC = -5 V at the beginning but charges to -2 V within the first quarter of T (since VX is clamped to 5 V and Vin raises to + 3V  Capacitor charges). From there on, since Vx is never larger than 5 V again, the zener diode branch is basically always off and the voltage across the capacitor cannot change anymore (no charge/discharge path) and Vx = Vin - VC for t > T/4. Vout is Vx – 0.6 V due to the voltage drop across the diode if Vx > 0.6 V and zero otherwise (after all it is a simple half-wave rectifier) V t -3 V -6 V -2 V 6 V 3 V 5 V 1 V -5 V 0.6 V __ Vx __ Vout __ VC 0.6 V Page 15 of 17 d. Write the differential equation to find VA (30 pts) V2 = Itotal * 3 + VL + VA = 3 * Itotal + L dItotal/dt +VA =5 = 3 * Itotal + 0.5 dItotal/dt +VA Itotal = IA + IB = C * dVA / dt + VA / 1 = 0.5 * dVA / dt + VA Plugging in Itotal in first equation  5 = 3 * VA + 1.5 dVA/dt + 0.25 *d2VA/d2t + 0.5 dVA/dt + VA 20 = d2VA/d 2t + 8 dVA/dt + 16 VA Page 16 of 17 e. Solving for VA i. Give the forced response for VA (4 pts) VA = 1.25 V (DC solution) ii. Find the natural response for VA ( 9 pts) damping ratio can be found from differential equation found in part f  ζ=1  Critical damping  Natural response: VA (t) = K1 e-4t + K2 * t * e-4t iii. Using the above and the initial conditions given, find the complete solution for VA (9 pts) VA (t) = K1 e -4t + K2 * t * e -4t + 1.25 VA(t=0) = 1 V = K1 + 1.25  K1 = - 0.25 V IA (t=0) = 0.25 = C * dVA / dt = 0.5 * ( -4*K1 +K2) = 0.5 + 0.5K2  K2 = -0.5 V VA (t) = -0.25 e -4t + -0.5 V * t * e-4t + 1.25 Page 17 of 17 f. After a long time we flip the switch back i. What happens to the capacitor and the voltage source V1? (5 pts) Right after the switch gets flipped back VC stays at 1.25 V and the capacitor starts getting discharged through the 10 kOhm resistor to 1 V. During that time current flows from the capacitor to the voltage source. It basically gets “charged” ii. What happens to the inductor and the voltage source V2? (5 pts) Nothing! Removing the charged capacitor does not change anything to the circuit on the right, since the capacitor was not drawing any current anymore anyway