Question

In the following problem. Vcc =9.3 v. 1g=0.6 mA, R1=2.9 ko, R2= 6.6 k0. Assume B=58 and VBE=0.7V Find the operating point for each transistor (VCE, IC) In the answer box, write Ic3 only 30kn OB ER Η 1kΩΣ 80ks Riš 3kn -Vcc

Answer 1

16.940 IE2 1 0.58971 – ② colve er ander DE 2= 0.3 (MA) = 0.584 (MA) 8022 & DE2 - 3022 ( 11 ) TE I (2= (so) x 0.33 = 0.324 (MA) (2= vcc- 6.6 4103 IC2 v 9.3 – 6.64 0.324 = 1.189 can VEz = -903 + 348 x 8E2 = -9.3 + 340.33 VE2z - 8.31 (vy VCE2 = 6-VE2= 7,1597831 = 15:469 con (15.469, 0.329 (mA) LE2, I(2) = kch at ca 12 apply kch at wcz IF Icat TR03 12 0.3zut TE B+1) 259 er _ 1.6.98 103 283 + I2 IB3 and Is in (MA) «Рply kuю оор = 0.324 ③ 9.3 = 6.6 I2 + 0.7 +1x I E3-93 lfedes 17.92 1E3 +66 I2 -

folue ea and ea Та TE3 = 14. (6 (mn) J5 2 оос 64 (m) Tess = " - ч - о-эла (mn) s 4 » Тез кіѕ, ѕ8 40-> 42 (.4ч me) VC 3 - С. 3 (v) — 4*3+ 14-18 2 ч. 38 VE3 - |EXTE3 — 92 Vc E3 = Nc3- VE32 • 3- 4:33 - 4-4- (E3 Tes) - Си-чXvi, 14.1% (m) ) vc2: vсс — 6 6 *1, а 9.3- bx os Su vсэг с. 48 () ? ТЕ, -9-36 3х0. 33 VE2 . – С. 3 + 3 VE) - - 8-31 vст: vto - ve2: _28+%-31 NCE» (2.84 w) СЕ» (1с2 ) = ( 13. 84, б•3?ч бme )) Ve: 2.6 x + кві 2 – 3 = 24x102x0 3416 2 3 V4 - 16ttv) appy Ev to LaoP 4:3 Зоя + 60 Тя – 4.3. go Tx + by - (8.6 -2@ ket at NBI - Затта му бъs )

--In try= 0.0107 © are en and ea 293 0.162(mn) By= 0.173 (mm) V ve- UBIQ 3oxta - 4044 913- 30x 0.162 NB12 NERIE 0.7 0.7t u.uy VEZ 0.7+VB, VEL= 5.14 (ul VLEE VEGE 5.141 ara VECI= 5:14 +7061 VECI= 12.75W) (VEG, IC )= (12075, 0.6 (MAD)