sol1 - ECSE 330 Introduction to Electronics Solutions to...

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Unformatted text preview: ECSE 330 Introduction to Electronics Solutions to suggested problems Problem 1.43 i R v i R o vo A v i R L v o s R v s i i Voltage gain = v o v s = A vo R i R i + R s R L R L + R o (a) v o v s = A vo 10 R s 10 R s + R s 10 R o 10 R o + R o = 10V/V × 10Ω 11Ω × 10Ω 11Ω = 8 . 26 V/V v o v s dB = 20log 10 v o v s = 20log 10 (8 . 26) = 18 . 34 dB (b) v o v s = A vo R s R s + R s R o R o + R o = 10V/V × 1Ω 2Ω × 1Ω 2Ω = 2 . 5 V/V = 7 . 96 dB (c) v o v s = A vo ( R s / 10) ( R s / 10) + R s ( R o / 10) ( R o / 10) + R o = 10V/V × 1Ω 11Ω × 1Ω 11Ω = 0 . 083 V/V =- 21 . 66 dB Problem 1.44 i R v i R o vo A v i R L v o R = 1M i Ω A = 40 dB o R = 10 Ω Ω L R = 100 vo Using the voltage-divider rule we get: v o = ( A vo v i ) R L R L + R o = ⇒ A v = v o v i = A vo R L R L + R o 1 Note that the above equation is the same as equation (1.12) and is a useful rule to remember. A v = 100 V/V × 100Ω 100Ω + 10Ω = 90 . 91 V/V = 39 . 17 dB where A vo is calculated as follows: 20log 10 | A vo | = 40 dB = ⇒ A vo = 100 V/V The power gain is calculated as follows: A p = P o P i = v o 2 /R L v i 2 /R i = A 2 v R i R L = A 2 vo × 10 4 = 82 . 6 × 10 6 W/W A p | dB = 10log 10 (82 . 6 × 10 6 W/W) = 79 . 17 dB If the amplifier has a peak output-current limitation of 100 mA, then the peak undistorted output-voltage is...
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sol1 - ECSE 330 Introduction to Electronics Solutions to...

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