Question

BJT amplifier.pur Page < 5 > of9 | 0 ANALYSIS Pracuce Problem yx Determine the low cutoff frequencies fub, fic, fle, the dominant cutoff frequency fi, and Ay/mid) in dB for the amplifier circuit of Figure 9-17. Assume B= 200. Vcc o 15 V Ri luf 30k12 Rc 33.6 kN - 2N2222 1 HF IL CC wo 1.5 k 2 Di CB RL > 4.7 k22 R2 7.5 ks 2 СЕ RE 31.5 k 2 7100 uF = = Figure 9-17: A common-emitter amplifier Answers: fl = 43 Hz, fic= 20 Hz, fue = 69 Hz, fu=85 Hz, Ayſmid) = 41.41 dB

Answer 1

hiver circuit is to Vec : 150 B: 200 'Re: 3.6kr. R.-3034 lur both Rs:1.5 2 23 : 4.3 4.0 (2H JuF R: 7.580 9 TCe lour Re:1.51² _ Determine de for dc conditions To 15v 3.6km 30 x 7.5 Ş 3.75 30K B200 6kn 3156 15X7.5K 37.5 * _Vc:15U Vc:15U : 30 § 3.6kr Apply KV2 3V - BIC 6K -0.7 - Le Reto 21:51 ļ

IE =) . 2.7 - 1.5k + b 200 => Te = 1.5mA 1, 52 46% W 26mv re: 2 . 26mv. -> Zeme -) 17.330 The input Impedence h; = R1 1122 11 Bre pre= 3.46k2 =) 668 11 200x17.33 =) 2. 1969 kr hip - 35 (R$ +R;) CS 201 1.51+ 2.1969%) IUF fue : 63.05 42

271 ( hin + 2). Co 27 (2196.9 +4 700) INÓ 23.07 H 2 :: File : 23.07 HZ fire 2де Се Re: Rell ( Ril11 +70) Roll Re111g : GUN]) 4.5kn Re: Rell ( 4200 + 17.332 ) 900 Re: Re 1123.339 -) 12.971

1 FLE : 27 (19.97) (100418) 1 69.28112 . Pre = 69. 29542 The dominant Cut at frequency : 8512 Au mid hand gain = -) -207.73 = 20 log (+207.73) re+ RE : PLE: 69.28 HZ PLC = 23.07 HZ fie : 43,05112 Av: -'41. AldB