# ECE 3050A, Spring 2004 Page 1. FINAL EXAMINATION - SOLUTIONS (Average score = 78/100) R 2 = R 1 =

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1 ECE 3050A, Spring 2004 Page Problem (20 points This problem must be attempted) The simplified schematic of a feedback amplifier is shown. Assume that all transistors are matched and g m ma/v and r ds. (a.) Where should the switch be connected for negative feedback? (b.) Use the method of feedback analysis to find v 2 /, in /i, and out v 2 /. (a.) A quick check of the ac voltage changes around the loop show that the switch should be connected to A. FINAL EXAMINATION SOLUTIONS (Average score 78/00) i M M2 V DD A B 3 M3 v 2 S04FEP (b.) This feedback circuit is seriesseries. The units of A are A/V and the units of β are V/A. F z 2f f f i f 0 3 The circuit for calculating the smallsignal openloop gain is, v gs3 ' v s ' g m3 v gs3 ' 2 v g3 ' 2v i 3 o ' gs3 ' 4 g m v gs ' S04FES A i o ' v s ' i o ' v gs3 ' v gs3 ' v g3 ' v g3 ' v gs ' v gs ' v s ' (g m3 ) g m3 (g 4 m 2 ) 2 A i o ' v s ' (ms)(0.5)(5) 2.5mS A F i o A vs AF 2.5mS ms v 2 v v 2 v v 2 s io i o v s 4 (0.74mS) 0.74 V/V v 2 vs 0.74 V/V is not influenced by feedback so i o 4 (/g m3 ) 2kΩ of 2kΩ(2.5) 7kΩ out v 2 ( of 4 ) 7kΩ 875Ω v 2 875Ω

2 ECE 3050A, Spring 2004 Page 2 Problem 2 (20 points This problem must be attempted) This problem deals with finding the openloop gain and it application. The following questions are independent of each other. (a.) Find the loop gain of the feedback circuit shown, T(s), if the amplifier is an ideal voltage amplifier with a gain of K. K K C skc (b.) If T(s) s 2 2 C 2 2Cs, find f osc and C the value of K necessary for oscillation. S04Q3P (c.) If T(s) has the following properties: T(0) 0 and two poles at s 00, what is the phase margin of this feedback circuit? (a.) Opening the loop gives, T(s) V K in (/sc) K(/sC) (/sc) K 2 Cs (sc)(sc) K 2 Cs s 2 2 C 2 s2c T(s) K 2 Cs s 2 2 C 2 s2c (b.) T(s) skc s 2 2 C 2 2Cs jωkc T(jω) ω 2 2 C 2 jω2c or T(jω) jωkc ω 2 2 C 2 jω2c j0 f osc 2πC and K 2 (c.) We could plot a Bode plot and estimate the phase margin or do the following: 0 T(jω) ω Find the unitygain frequency from, ω 2 c 00 2 (0 ) ω c 00 2 or ω c 300 rads/sec. The phase margin can be expressed as, PM 80 2tan (7.56 ) 36.87

3 ECE 3050A, Spring 2004 Page 3 Problem 3 (20 points This problem is optional) A BJT amplifier is shown. Assume that the BJT has the small signal parameters of β F 00, r π, and V A. a.) Find the midband voltage gain of this amplifier, /. b.) Find the value of the lower 3dB frequency, f L, in Hz, using any method that is appropriate. The smallsignal model for this problem is: S04FEP2 3 V EE V CC C 2 0.µF C 00µF I b rπ βi b C 2 (a.) If the capacitors are large, or the frequency large, 3 C 2 4 the midband gain is, V in I b I b V F04FES3 in V [β( in 2 4 )] r (500)(0.5) 250 V/V π (b.) Since the capacitors are independent, let us choose the shortcircuit time constant approach. C 3 r π β 2 0 kω 9.42Ω p C 55 rads/sec. C C2 2 20kΩ p 2 C rads/sec. C kΩ There is also a zero due to 3 and C given as /( 3 C ) 0 rads/sec. ω L (0 2 ) rads/sec. or f 3dB 4.2 Hz

4 ECE 3050A, Spring 2004 Page 4 Problem 4 (20 points This problem is optional) a.) If the g m of the MOSFET is 0.mA/V, find the midband gain and the location of all zeros and poles of the circuit shown. b.) If the amplifier above has two zeros at the origin and a pole at rads/sec and 4 rads/sec., what is the lower 3dB frequency in Hz?.) Smallsignal model: C µf G MΩ S04FEP4 V DD S C2 0µF L V SS C G C G g m (/g m ) s (/g m ) s L V g V s s gm V g g m V s V g gm V g x L sc 2 g m G G s s 3 s 5k C 2 s M C V s s C 2 C 2 L L F02Q09S 5k M sc 2 5k sc 2 M sc 3 s s 6.67 s s MGB 0.333, two zeros at 0 rads/sec. and poles at rad/sec and 6.67 rads/sec. 2.) ω L p 2 p2 2 2(z 2 z2 2 ) (0) rads/sec. f L Hz

5 ECE 3050A, Spring 2004 Page 5 Problem 5 (20 points This problem is optional) The FET in the amplifier shown has g m ma/v, r d, C gd 0.5pF, and C gs 0pF. (a.) Find the midband gain, /. (b.) Find the upper 3dB frequency, f H, in Hz. (Note: You cannot use the Miller's theorem on this problem because there is no bridging capacitor.) The small signal model for the high frequency range is shown where g m V gs 3 20kΩ C 20µF S04FEP5 V DD C 2 µf 20kΩ V SS Vin 2 V gs C gs C gd 34 Because the capacitors are independent, probably the best way to work this problem is by superposition (opencircuit time constants). Therefore, C gs : C gd : Cgs 2 (/g m ) K K K 333Ω ω Cgs Cgd 34 ω Cgd ω H C 200 Mrads/sec. gd 66 Mrads/sec Mrads/sec 2 200Mrads/sec f L 26.48MHz The midband gain is given as C 300 Mrads/sec. gs 333Ω MBG 2 g m 2 g m g m (0) 3.33V/V

6 ECE 3050A, Spring 2004 Page 6 Problem 6 (20 points This problem is optional). A feedback amplifier is shown. Use the methods of feedback analysis to find the numerical values of v 2 /, /i, and v 2 /. Assume that all transistors are matched and that V t 25mV, β (of the BJT) 00, I C I C2 i 00µA, and r o. An openloop version of the feedback amplifier is shown below. The openloop ac schematic is given as, i ' Q Q2i f ' S04FEP6 F g 2F i F ' v 2 ' ' ' 0 3 Small signal BJT parameters are: I C g m 25mV 4mS and r π β g 25kΩ m i ' i b2 ' ' r π2 i b ' βi b2 ' v 2 ' 3 r π βi b ' F04FES6A v 2 ' i ' v 2 ' i i b2 ' i b ' b2 ' i b ' i ' [(β)( 3 4 )] β 2 2 r π2 (β)( 3 4 ) 3 3 r π (0 0K 0K) 00 0K 540K 0K 0K25K (505K)(.852)(0.286) 267.2kΩ T v 2 ' i ' 267.2kΩ v 2 T i F 267.2KΩ T kΩ in in 4 (r π ) 0K 25K 7.4kΩ, inf F 7.4kΩ T Ω i inf Ω v 2 v v 2 i i v 9.64K V/V out 3 4 r π2 2 β ( )kΩ 324Ω v 2 out i2 F 324Ω T Ω ' v 2 ' 4 F04FES6 V CC Q 3 V CC Q2 v 2

7 ECE 3050A, Spring 2004 Page 7 Problem 7 (20 points, this problem is optional) An C oscillator is shown. Express the frequency of oscillation of this circuit in terms of the components and evaluate. What is the value of the voltage amplifier gain, K, necessary for oscillation? In words, how is the amplitude of oscillation determined? Open the loop as follows, C 0.0µF K 00kΩ C 0.0µF 00kΩ K 00kΩ C 0.0µF K S04FEP7 Now, V x K 00kΩ C 0.0µF T(s) V x V x ' K(/sC) (/sc) 3 T(jω) K 00kΩ C 0.0µF K 3 s 3 3 C 3 3s 2 2 C 2 3sC K 00kΩ C 0.0µF S04FES7 K 3 K 3 sc (sc) 3 K 3 ( 3ω 2 2 C 2 ) jω[3c ω 2 3 C 3 ] j0 V x ' 3C ω osc 2 3 C 3 ω osc 3 C and K 3 3ω 2 osc 2 C 2 K 3 9 K3 8 K 8/3 2 K 2 Substituting the values gives f osc 3 2π Hz (,732 rads/sec) The amplitude of the oscillation is determined by the amplifiers K having a nonlinear transfer function as illustrated. The slope for small values is greater than 2 while the slope for large values is less than 2. The amplitude of the oscillator will stabilize where the effective gain is exactly 2. K2 F04FES7A

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