Quiz on Common Emitter Amplifier - Microelectronic Circuits | ECE 3040, Quizzes of Electrical and Electronics Engineering

Download Quiz on Common Emitter Amplifier - Microelectronic Circuits | ECE 3040 and more Quizzes Electrical and Electronics Engineering in PDF only on Docsity! ECE 3040 Microelectronic Circuits Quiz 8 July 7, 2004 Professor Leach Name Instructions. Print your name in the space above. The quiz is closed-book and closed-notes. The quiz consists of one problem. Honor Code Statement: I have neither given nor received help on this quiz. Initials 1. The figure shows the ac signal circuit of a common-emitter amplifier. It is given that IE = 2mA, β = 99, r0 =∞, and VT = 0.025V, gm = IC/VT , rπ = VT/IB, re = VT/IE , and r0e = Rtb/ (1 + β)+re. (a) Let Rte = 0 and Rtb = 3.3 kΩ. Replace the transistor with the π model and use i0c = gmvbe. Write the appropriate equations and solve for the value of Rtc such that vo/vtb = −200. (b) Let Rte = 0 and Rtc = 3.3 kΩ. Replace the transistor with the T model and use i0c = aι0e. Write the appropriate equations and solve for the value of Rtb such that vo/vtb = −50. (c) Let Rtb = 1kΩ, Rte = 50Ω, and Rtc = 10kΩ. Use the simplified T model to solve for vo/vtb. Solutions: (see the class notes for the small-signal circuits) gm = IC/VT = αIE/VT = 0.079S, rπ = VT/IB = (1 + β)VT/IE = 1.25 kΩ, α = β/ (1 + β) = 0.99, re = VT/IE = 12.5Ω, r0e = Rtb/ (1 + β) + re = 22.5Ω (a) vo = −i0cRtc = − [gm (vbe)]Rtc = − · gm µ vtb rπ Rtb + rπ ¶¸ Rtc =⇒ Rtc = Rtb + rπ gmrπ µ − vo vtb ¶ = 3300 + 1250 0.079 ∗ 1250 (200) = 9.192 kΩ (b) vo = −i0cRtc = − [α (i0e)]Rtc vtb = ibRtb + i0ere = i0e 1 + β Rtb + i 0 ere = i 0 e µ Rtb 1 + β + re ¶ =⇒ i0e = vtb Rtb 1 + β + re =⇒ vo = − α  vtbRtb 1 + β + re  Rtc =⇒ Rtb = (1 + β) " αRtc µ vo vtb ¶−1 − re # = (1 + 99) " 0.99× 3300 µ 1 50 ¶−1 − 12.5 # = 5.284 kΩ (c) vo = −i0cRtc = − [α (i0e)]Rtc = − α  vtbRtb 1 + β +Rte + re  Rtc =⇒ vo vtb − αRtc Rtb 1 + β +Rte + re = − 0.99× 10000 1000 1 + 99 + 50 + 12.5 = −136.5 1