University of Pennsylvania Department of Electrical and Systems Engineering ESE 319 Microelectronic Circuits Final Exam 10Dec08 SOLUTIONS This exam is a closed book exam. Students are allowed to use a calculator and a single page reference sheet (two sided). Please show all work, justify all approximations and give the units for all calculated parameters. 1. In the circuit given in Fig. P1 let V CC 1 V, I E1 10 ma, V R3 1.7 V, I R1 1 ma and R 1 kω. Transistors Q 1 and Q are matched and assumed to be forward active such that V BE1 V BE 0.7 V and β 1 β very large. Also, V CEsat 0 V for both Q 1 and Q. (40 pts) Fig. P1 a. Determine the values of resistors R E, R 1 and R 3. (10 pts) b. Determine the voltage gain of the Q amplifier stage v o /v i. (10 pts) c. Let v O1 V O1 + V o1 sin ωt, determine the value of V O1 that will permit the largest signal swing, and determine maximum value of V o1. d. Determine the value of R C that permits the largest peak-to-peak signal swing. e. Determine the voltage gain of the overall amplifier v o1 /v i. (10 pts) SOLUTION: a. Determine the values of resistors R E, R 1 and R 3. R 3 V R3 1.7V I R1 1mA 1.7k! V RE 1.7V! 0.7V 1.0V " R E V RE 1V I E 10mA 100# V R I R1 * R 1mA *1k! 1V " V B1 1.7V + 1V.7V
R 1 1V!.7V 1mA 9.3k" b. Determine the voltage gain of the Q amplifier stage v o /v i. G Q v o v i! " R in1 R E v o v e1!r " i b1 where i b1 i e1! + 1 i c! + 1 " v # r $ e1! + 1 i r i e1 e e1 R in1 v e1!i e1 r e " G Q! r e R E! V T / I E 100#!.5 100!0.05 V / V c. Let v O1 V O1 + V o1 sin ωt, determine the value of V O1 that will permit the largest signal swing, and determine maximum value of V o1. For Q to remain forward-active V CE1 V O1! V O > V CEsat 0 where V O V B1! V BE.7V! 0.7V V " V # v O1 # V CC 1V 1V! V V o1max 5V " v O1 7 + 5sin#t 1V! V To allow for maximum signal swing V O1 + V 7V d. Determine the value of R C that permits the largest peak-to-peak signal swing. 1V! V 1V! 7V V O1 + V 7V V CC! I C R C " R C 10mA 500# e. Determine the voltage gain of the overall amplifier v o1 /v i. i c i e1!i b!v i r " +! + 1 ( ) R E! v i R E
3 v o1!i c1 R C!i c R C " G Q1 v o1 v i! R C R E! 500 100!5 V / V. The following set of questions require brief, to the point answers. (0 pts) a. Draw the schematic for the Widlar current source, and relate its advantages to a current sources implemented using a conventional current mirror. b. A closed-loop amplifier with loop-gain A(s)! F is unstable for large values of Γ F where the closed-loop gain approaches unity. Describe a design solution to stabilizing the closed-loop amplifier. c. The all-pass (AP) filter plays an important role in data communications. Write the gain function G(s) for a nd order AP filter in terms of ω 0, Q and dc gain K. Sketch the gain G(jω) and phase φ(jω) vs. frequency d. The Sallen and Key LP stage where R 1 C 1 R C RC has a symbolic gain function ( ) K 1 / RC G(s) s + 3! K RC s + " 1 % # $ RC & ' design?. What are the practical advantages and disadvantages of this LP stage Qualitative Descriptions do not involve specific numerical calculations and are largely accomplished with text. Descriptive adjectives like small or large are appropriate to use. Also appropriate are comparative phrases that involve approximately, less than, greater than, etc. If helpful to your description, you are welcome to use an equation and/or a sketch. SOLUTION: a. Draw the schematic for the Widlar current source, and relate its advantages to a current sources implemented using a conventional current mirror. Advantages: 1. Much increased output impedance
4. Low current can be realized with smaller resistors. b. A closed-loop amplifier with loop-gain A(s)! F is unstable for large values of Γ F where the closed-loop gain approaches unity. Describe a design solution to stabilizing the closed-loop amplifier. Compensation of A(s) in order to pull the lowest frequency pole to a lower frequency. Miller compensation pulls the lowest frequency pole lower and the highest frequency pole higher, i.e. ploe splitting. High frequency gain is traded for stability. c. The all-pass (AP) filter plays an important role in data communications. Write the gain function G(s) for a nd order AP filter in terms of ω 0, Q and dc gain K. Sketch the gain G(jω) and phase φ(jω) vs. frequency G(s) # K s! " 0 Q s + " & 0 $ % ' ( s + " 0 Q s + " 0 d. The Sallen and Key LP stage where R 1 C 1 R C RC has a symbolic gain function ( ) K 1 / RC G(s) s + 3! K RC s + " 1 % # $ RC & ' design?. What are the practical advantages and disadvantages of this LP stage 1. Advantage: ω 0 1/RC, Q 1/(3-K) > component spreads independent of Q.. Disadvantage: sensitivities high for high-q, i.e. high Q > K -> 3 realized by the small difference of two large numbers.
3. Consider the power amplifier stage in Fig. P. (40 pts) 5 Fig. P a. Let V BB be a variable. Draw the v O vs v I voltage transfer characteristic (VTC), labeling all key points on the VTC. State the condition for zero cross-over distortion. (10 pts) b. The operation of the class AB stage is based on a basic conservation relationship between i N and i P. Derive this relationship from the constant base-to-base voltage condition. (10 pts) Design the quiescent current of the class AB power amplifier in Fig. P so that the small-signal voltage gain v O /v I near the origin of the VTC is greater than or equal to 0.99 for loads R L 100 Ω. Let I S 6.914 E-14 A for both Q N and Q P and V CC 1V. c. Determine the R out and quiescent current I Q. (10 pts) d. Determine the value of V BB. (10 pts) SOLUTION: a. Let V BB be a variable. Draw the v O vs v I voltage transfer characteristic (VTC), labeling all key points on the VTC. For v I > 0: v O v I! v BEN + V BB Q N saturates when v O > V CC! V CENsat For v I < 0: v O v I + v EBP! V BB Q P saturates when v O <!V CC + V ECPsat
6 The crossover distortion is nulled when V BB V BEN V EBP! V BB V BEN + V EBP b. The operation of the class AB stage is based on a basic conservation relationship between i N and i P. Derive this relationship from the constant base-to-base voltage condition.
7 c. Determine the R out and quiescent current I Q. R L G v o! 0.99 for R L! 100" # v i R out + R L 0.99R out 100! " 99! # R out 1! R L R out + R L 0.99 with R L 100 Ω. At v O! 0V i N i P I Q " R out r en r ep V T i N + i P V T I Q I Q V T R out 1.5mA d. Determine the value of V BB.
8 V BB V I Q I S e T! V BB V T ln I Q I S 0.05ln 1.5E-3 6.914 E-14 1.96V