Question

Test Your Understanding TYU 8.4 For the common-emitter output stage shown in Figure 8.16(a), let Vcc = 12 V and RL = 1 k12. Assume the transistor Q-point is in the center of the load line. (a) Determine the quiescent power dissipated in the transistor. (b) Assume the sinusoidal output voltage is limited to a 9V peak-to-peak value. Determine (i) the average signal power delivered to the load, (ii) the power conversion efficiency, and (iii) the average power dissipated in the transistor. (Ans. (a) Po = 36 mW; (b) (i) PL 10.1 mW, (ii) n = 14.1%, (iii) Pe = 25.9 mW)

8.4 a) b) Please solve

Answer 1

Given, VCC - 12v RL = 1kn a point is at the conter of load line. so VCERT Vec GV 12 2. 2. ovce vce - VCEO Ica RL Ico 7 VCE ovo 12-6 GMA Ikr a point = (UCED I Ica) vi =(6V, 6MA) (9) Quiscent power dissipated in transistor VCEQ X Ica 6UX GMA = 36mw 16 Given, Peak to peak output voltage = gu = Vap-p) delivered to load average signal power 2 81 mW PL VOLPP) 8 x 1kn 8 8 RL PL = 10.1 mW Pa PQ را XIDO (li) Power conversion efficiency n = Picoc) Input De power Picoes - Vcc IcQ = 12 x 6 mW PicaC) = 72 mW PL 10.1 mW x 100 M 72mw Piloc) 'n' = 14.1% x 100 =

..) Average Power dissipated in Transistor Po = Quiscent power dissipated in transistor -average sig power delivered to load Po = PQ-PL Pa = 36mw-101mw PQ = 25.9mw = 25.9 mw