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VCC=12 V HW #5 R1=47kΩ Common Emitter Amp AV = -RC/RE voltage gain ZIN ~ β RE input impedance ZOUT ~RC RC=4.7kΩ VC C VB B ΔVCE = 0.7V E VOUT VIN R2=4.7kΩ RE=470Ω GND #1 A common-emitter amplifier is displayed above. The input VIN and output VOUT share a common ground at the emitter. The transistor base is biased to a voltage VB to allow the input signal VIN to always lie above 0.7V . Use the voltage divider theorem to determine VB in the circuit above (47kΩ and 4.7kΩ chain). What is the maximum amplitude signal |V0 | that can be amplified such that VIN ! 0.7V ? What is the gain of the amplifier AV ? What is the input impedance ZIN and output impedance ZOUT? (β~100) | VIN VB = (4.9/4.9+49)12V=1.1V |V0 | <= 0.25V |V0 | (amplitude) VB AV =-RC/RE = 10 V=0.7V V V=0.0V ground ZIN = 100 x 4700 = 470000Ω ZOUT = RC =4.7kΩ 10 (2Vo+0.7V) < = 12V or 2Vo=1.2V-0.7= 0.5V Vo<= 0.25V #2. Assume we use a C=1uF capacitor on the input of our amplifier, and there is 4 nF of stray capacitance in the circuit. Determine the fLO and fHI cutoffs for this transistor amplifier. flo = (2πZinCin)-1= ( 6.28x47000Ωx0.000001F)-1=3.9Hz fhi = (2πZoutCout)-1= ( 6.28x4700Ωx4.0e-9F)-1=8470Hz #3. On slide 5 we see the display of the collector current IC of a transistor versus the collector-emitter voltage VCE. At about VCE =5V the transistor is operating in its linear range. Graph IC vs IB and determine the β of this transistor. ( IC = β IB ). #4. Derive the feedback equation AV ! = where AV ! 1 " ! AV AV is the open loop gain and β is the feedback fraction. Av = ( ) Vout Vout Vout Av# Vin !! != = !!!!!!! Vin Vin " #Vin 1 " # Vout 1 " # Av# Vin ( )