Lecture%2020_%20Mar%209%202011

Lecture%2020_%20Mar%209%202011 - 23 . 1 ) 1 ( 2 = + = DS V...

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Chapter 13 Small-Signal Modeling and Linear Amplification Lecture 20 Covered Section: 13.10 – 13.12, 14.2.3 – 14.2.6 20-1
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Small-Signal Analysis of Complete C-S Amplifier: ac Equivalent ac equivalent circuit is constructed by assuming that all capacitances have zero impedance at signal frequency and dc voltage sources represent ac grounds. Assume that Q-point is already known. R G = R 1 R 2 20-2
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Small-Signal Analysis of Complete C-E Amplifier: Small-Signal Equivalent Terminal voltage gain between gate and drain is: Overall voltage gain: 20-3
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C-S Amplifier Voltage Gain: Example Problem: Calculate voltage gain Given data: Κ n = 0.5 mA/V 2 , V TN = 1V, λ = 0.0133 V -1 , Q-point is (1.45 mA, 3.86 V), R 1 = 430 k , R 2 = 560 k Ω, R 3 = 100 k , R D = 4.3 k Ω, R I = 1 k Ω, R S = 0 k . Assumptions: Transistor is in active region. Signals are low enough to be considered small signals. Analysis: dB 4 . 13 69 . 4 = - = + - = G R I R G R L R m g v A mS
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Unformatted text preview: 23 . 1 ) 1 ( 2 = + = DS V DS I n K m g k 5 . 54 1 = + = D I DS V o r k 243 2 1 = = R R G R k 83 . 3 3 = = R D R o r L R V 48 . 2 2 . 2 . = 2245- n K D I TN V GS V i v 20-4 Small-Signal Model Simplification If we assume R I << R G If g m R s >> 1, This requires a large voltage drop over Rs. If R s = 0, and generally R 3 >> R D and load resistor << r o . Hence, total load resistance on drain is R D . For this case, common design allocates half the power supply for voltage drop across R D and ( V GS- V TN ) = 1V Also, if load resistor approaches r o , ( R D and R 3 infinite), voltage gain is limited by amplification factor, f of the MOSFET itself. s R m g L R m g A A vt v +-= 2245 1 A v 2245-g m R D =-I D R D V GS-V TN 2 2245-V DD 20-5 s R R A A L vt v-= 2245...
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Lecture%2020_%20Mar%209%202011 - 23 . 1 ) 1 ( 2 = + = DS V...

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