Question

+ lov w 250k $2.0 K HE lok 150k LOK w HA Problem 2: For the circuit shown above, assume that 8 200, Vbelon) = 0.7V and Vce(min) = 0.4V and Ic(min) = 0.1mA. 1. Find Vce and Ic. 2. Draw the small signal AC equivalent circuit and find ver in terms of is (note these are AC values), 3. Plot the DC and AC load lines and mark the Q-point (you must provide numerical values for salient points on the load lines). 4. Find Vac(mar) and Lac(max). 5. Find the peak peak value of the largest symmetrical output volt- age swing

Answer 1

# flou 32K to Reh - (250 11150) kan lok B= 2000 993.75kn ERA 110 = 3.75 V 150+250 Vih Ik I gevenVee 0.7V ; VEE=0,4V Ismin s OlmA For De analysis: lov KuL in loop ! ak 0 Мm - К% - Vae – It Re: Rish E Uth ilka 3.75-93.75K 73 -0.7 - (1+200) 32 (IK) = } IB = 0.0103 mA Ie = PTRE 2,0695 MA Answer UCE 10-1212)(2.0695m)-(k) (1+200) (0.0103 m) (0) VCE = 3.79 v Answer, AC lead line? 26 4.238 MA 3lol Rac = (21120) s 1.67 km + 240 V Ac land line Uce VCGU des I, g + 3.73 1,67 = 4, 318MA Ans 2.669 + Rac vies Uchide + Peace Race 3 3.79 +(2,0698m) (1.67%) 7.246 V Aus

3. Ie (2.0695MA, 3.790) Vcc Rc *5mA DCIags : Uc69) De load lines Vee Vec slov 5. Ucc For maximum swing, ucesy 2 3 5 V 2 Bibi ২২k ২tok Rish fig! small signel Ac equiestent Uces-pib(21120)k Aces - ic (1.676) Vce -1.67*103c Answee