Question

2. (20%). In the circult shown below, the DC current gain for the npn translator is 1 - 120 and for the pop transistor is 82 = 100. Find the Q-point for both transistors, i.e. Inq, low, and Voeg. MMM པ :༠ ནྟི་པ W MV) S 40 k2 $23 S2002 MacBook Air

Answer 1

solution: au 100- 322 & solst ovo & 40kr <2с 2002 Then, Applying Thevenin's theorem to this post. = 40k sokst 40kr 40k2 drop across Voltage Ven= .: Web = 3 V. aok x 80k 26.66ke MOct 0 ) 80kr // 4010-22 Rth = af. the Cirruit redrawn can be tay lood IB Rth 20 t Be=100 лм (26.6kn) ovo. VBE 2005 52 Van OV -- 0. Applying love to this loop. -3 +26.66k) 70+ 125+ Te (QK.) -0.

I . From the above De load line graph, VCEO = 3.47v Ico = 1.75mA 7.30 - RICO non transistor 125mm 0.0144mA IBO - 14MA tort & woor pool AICI • 1 of Rth mm clct , & 2002 wth CS circuit VCI - VB2 from the above 9- VOI 1.02m = 21 VCC-101 = 9 Тct 2 - - 9- ve 2.04 - Vo9-204 = 6,964. 1: V1 = 6.961. = VB2

- and part of The Il chowit can be rearown as 72-100 - 1 tav & 2002 6.964 IC. VEB - OZU outer loop. to the Applying love ag + (600) TEAVED + 6,96=0. 9-690-0.7 100 0.0134 A. 13.4 mA 150 TE= 13.4mA ) TB = 13.4m = TE = (1+B) IB = 13.4m - 0.132MA 1 IB2 0-132 mA , Ica Is-IB- 13-4m - 0.132m - 13-26ms. - 13.26ml

- From the Ô Loup of the lone eauation the clef, Can be written as - - 9 & 1oo Is & VCE + (Tol 2007 20.- (18.30A) (vero, Io) = ('ht, I ) = (5 83 919mA) Deload line of 02 (POR). VCE → Point ③ from oq 6 is. VCE - 9- VoolTE 9-(100) 013-4m) > 7.66 v. is Ic= 9- (200) IF - ② from en Point 200 7.66 00383 A 200 Tc. 38.3 mA NCELL= 3.834 Pnp transistor Ico - 19.15mA Irl = 0.19mA = IBCO B

non Pap 3.47 v 3.834 VCEO 1.75mA Ico 19.15mA 0.014 MA. Olama IBO.