EE 321 Analog Electronics, Fall 2009Exam 4 December 9, 2009Rules: This is a op en book test. You may use the textbook as well as your notes. The examwill last 5 0 minutes. Each problem counts equally toward your grade. None of the problemsrequire long calculations.R1 RcR2Vcc1. For VCC= 10 V, R1= R2= 500 kΩ, and RC= 100 kΩ , what is the oper-ating mode of this circuit, VB, VC, IC, and IB? Since you will find it is insaturation, what is βforced? Ignore the E arly effect.Replace the input network with RB= R1||R2= 250 kΩ, VBB=R2R1+R2VCC= 5 V. Thenwe see the BE junction is on, so VB= VBE= 0.7 V. Then the base current isIB=VBB− VBRB=5 − 0.7250= 0.0172 mANext assume saturation, so we get VC= VCE= 0.2 V. ThenIC=VCC− VCRC=10 − 0.2100= 0.098 mAThen, βforced=ICIB=0.0980.0172= 5.7. Thus βforced< β = 100, so saturation mode isconfirmed.2. Now make RC= 1 kΩ. The circuit is now in active mode. What is thetransconductance and the raw voltage gain,vcvbe?If it is in active mode then IC= βIB= 0.017 × 100 = 1.17 mA. In that case, thetranconductance isgm=ICVT=1.7225= 0.0688 Ω−1The raw voltage gain isvcvbe= −gmRC= 0.0468 × 1 03= −68.83. If the active mode c ir cuit is attached to a signal generator with outputresistance Rsig= 10 kΩ, and a load with resistance RL= 1 kΩ, what is theoverall gain,vcvsig?This is a common emitter circuit. In the book we look up the formula for the overallvoltage gain,vovsig= Gv= −RB||rπRB||rπ+ Rsiggm(ro||RC||RL)with ro= ∞ and rπ=βgm= 1453 Ω,Gv= −250 × 103||1453250 × 103||1453 + 10 × 103× 0.06 88 ×103||103= −4.344. For the following circuit perform a simple DC analysis. For each transistordetermine the operating mode, the base voltage, and t he collector voltage.Assume β = 100, VCC= 15 V, RB= 1 MΩ, RC1= 5 kΩ, and RC2= 1 kΩ.Ignore t he Early effect.VccRBRC1RC2Q1Q2We see immediately that both transistors are on, so VB2= 0.7 V, and VB1= 1.4 V.Next find IB1,IB1=VCC− VB1RB=15 − 1.4106= 13.6 µANext assume active mode for transistor 1, we get IC1= βIB1= 1.36 mA. And thenVC1= VCC−IC1RC1= 15−1.36×5 = 8.2 V. Since this is greater than the base voltageactive mode is confirmed.For the second t ransitor I will assume saturation mode (because the base current isvery large). We then get VC2= 0.2 V. In that case, IC2=VCC−VC2RC2=15−0.21= 14.8 mA.Since this is only slightly more than 10 times IB2= IE1, saturation mode is
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