Question

Electrical Engineering Quiz Review 03 Topic: Analog Electronic Design 9.(a). Explain the significance and utility of N/8, /4 and 1/2 lines. (2pts) (b). For the two-stage RC coupled amplifier circuit shown, calculate the individual stage voltage gains and the overall voltage gain. Input impedance of individual stage is given as 2.4 k... And ß of individual transistor as 80. (5pts) _ V18 V R6 10 kohm 3.6 kohtra R 10 kohm 10 kohm 922 T . I kohm 3900 ohm Figure (c). In a tuned amplifier circuit C = 500 pF, L = 204H, RL = 1.5 k., and the transistor has hfe = 50 and input resistance of 2001. The coil used has Q factor = 30. Calculate, (i) Resonant frequency of the tuned circuit (ii) Impedance of the tuned circuit (iii) Voltage gain of the stage (5pts)

Answer 1

Ans @ The significance and utility of an is given below. if a transmission line of length , and Characteristic impedance ?z? is terminated by a load impedance 'Z& then the impedance at the Sending end is given by. ſze + j2o tan (9 x ] T 2o + Zin Zo zotZR - Zol | Zrt j Zo i ZR (1) When ZR=0, then Zin =jzo (ii) When 2p=00, then Zin = -32, The Significance and utility of 1/4 line is given below. For a quarter - wavelength line, with characteristic impeclance zo and terminated with 'Zť The input impedance 'z'in is given as,

24 +12. -1.0 (6271933440) 7.-2, =z, Zelten (12) +372 2,1 ton ( 12) - Z Z Ž content log The quarter - bisselength can be used for impedence The inversion ie, the normalized impedance of a quarter. save length linu, is equal to the normalized adenittance at the receiving and (ii) The quarter- wavelength line can be used for impedance, matching Zin - 25 Zin ZR

The above equation shows that the impedance at the input of a quarter wave line depends on two quran- tities. These are the load impedance (which is fixed for any load at a constant frequency) and the characteristic impedance of the interconnecting transmission line. If the Zo can be Varied, the impedance seen at the input to the (1/4) line will be varied accordingly and the load may thus be matched to the characteristic impedance of the main line. The Significance and utility of Cal2) line is n below The input impedance of a half-wavelength line C212) with characteristic impedance 'z's terminated with impedance 'Zľ is given by. C 2o+jzo tan | 27 x ² Zinazol . ZoXZR IS COM> Zo Zin = Z R' Thus, the input impedance of a 1/2 line is equal to the load impedance independent of 'zó

Measurement of Load Impedance It is very often not practical to measure the impedance of a load directly . This being the case ; the impedance may be measured along a transmission line Connected to the load, at à distance which is half. wavelength from the load. Sometimes it is necessary to short - Circuit a transmission line at a point that is not physically accessible. 1. The same result will be obtained if the Short - Circuit is placed a half - wavelength away from the load.

Ans:b.Given that, For a two stage RC coupled amplifier shown in figure (1).

34 Kobo & R6 Plzo, Plokohin all & R3 Jo 1 2 3 6 Kohm E10 kohm that si Lau t Vin . 22 Kohm R2 I 900 ohm T r akohm į R8 R4 Lich T ve I kohal . Figure (1) Input impedance of first stage, Rn = 2.4612 Input impedance of second stage, Re = 2:46.12 Current gain of transistor, B = hfe = 80 Voltage gain of first stage, Avp = ? Voltage gain of second stage, Av2 = ? Overall voltage gain, Avs: ? The h - parameter equivalent Circuit of above two stage RC Coupled amplifier is shown in figur (2).

Bima ---pow to Y t hele but ERRURI @ She hallu Figure (2): h- parameter Eavuivalent Circuit Since Rs is not present in the Circuit, considering it to be Ik slie., Rs = 1ks Vollage gain for and stage is given by. Arz=1 1. Aiz XRL 1. Riz Where Current gain (4.a). A12=-hf = - 80

Input resistance (Riz). Riz = hie = 2.4k TR = Raheksa :: RL = h k On Substituting equations (2), (3) and (H) in equation (1), We get, Av2 = 80XHX 103 2.4 % 103 .: Av2 = -133-33 gain for 1st stage is given by, (ii) Voltage Aviso de Where Current gain, Aid = chose [Ail = -80) ----..169

Ri = Roll Rell Rell Riz = 3.6k ll lok 11 2.2 k 112.44k RC = 800.64627 Input resistance , Rip = hope ---... ) Ril = 2.462 .....® on Substituting corresponding values in equation 6). we get Avi -80 X 800.6H L 2.4 % 103 -64051.2 3400 Ay = - 26-688 Overall voltage gain (hu) (11) Ave - Arx Ril 1 Riltrs Where -Av=Awix Ayz = 26.688 X - 133.33 Ay = 3558-3) Rin = R3 Il Rull Ril ----.... (0) = 10K 11 2.2k || 2.4k

" Ril = 1.03 k2 ----....10 corresponding values in equation (9), On Substituting we get, Ais = 3558.31*1.03 x103 1.03 x 103 +1x103 - 3665060.371 2030 . Aus = 1805. Hy

Ang : C. Given that, For o tuned Omplifier Cunut, C = 500 pF L = 20 21 Pe=15 ka he = 50 P, = 20062 Q = 30 0 Resonant frequency of the turned Circuit, 1,- ? (11) Impedance of the tuned Circuit, pe = ?

(ii) Voltage gain of the stage, Av = ? 1 1 The resonant frequency of a tuned amplifier circuit is given by € - zobac 207 / 20x10-6x 500 x 10-12 = 1.59 X 106 . fp = 1.59 MHZ (11 The impedance of the tuned circuit is obtained as, Rp = QgW, L = Qo: 2Tful = 30 x 20 x 1.59 x 106 x 20 x 10-6 = 5.99 x 1035 1: Rp = 5.99k

(11 The voltage gain of a Slage is obtained as, A = A.RL Ri =-hfi (Rp 11 (2) - Ri = -50(5.99111.5) 200 1. A - -50 x 1199.6 200 = - 299.9 1... Aya 300