440_hmk5_soln

440_hmk5_soln - ECE 440 Electronics II Solutions to...

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Unformatted text preview: ECE 440 Electronics II Solutions to Homework #5 1. DC analysis: For Q1, taking the Thevenin equivalent of the base circuit and writing KVL: 20 K 20 K + 80 K (10) = (20 K ||80 K ) I c 1 + 0.7 + 1 KI c 1 I c 1 = 1.12 mA r 1 = 2320 . Similarly for Q2: 15 K 15 K + 85 K (10) = (15 K ||85 K ) I c 2 + 0.7 + 500 I c 2 I c 2 = 1.27 mA r 2 = 2039 . We form the A circuit by redrawing the amplifier at AC. The loading effects are found by severing the feedback at the emitter of Q2. This results in a series 10K + 500 in parallel with the 80K||20K at the input of Q1. Also the input side of the feedback is shorted to ground resulting in 10K||500 at the emitter of Q2. We define the emitter current of Q2, the feedback current as i o . Using this circuit we find the current gain: A = i o ' i i = + 1 ( ) 2 K ||85 K ||15 K 2 K ||85 K ||15 K + r 2 + + 1 ( ) 500 ||10 K ( ) 10 M ||80 K ||20 K ||10.5 K 10 M ||80 K ||20 K ||10.5 K + r 1 = 264 . The circuit consists of the 10K and 500 with the feedback current flowing through the 10K. Thus, i f i o ' = 500 10500 = 0.0476 a. With feedback: A if = A 1 + A = 19.47 = i o ' i i ; i o = i o ' 2 A i = i o i i = A if 2 = 9.74 b. The input impedance of the A circuit is: 10M||80K||20K||10.5K||r 1 = 1698. Thus R if = 1698 1 + A = 125 . R if = R in ||10 M R in = 125 c. R o = 4K 2. Since this is shunt-series feedback, we need to begin with the current gain. Thus, we transform the input voltage and 1.2K series resistance to a current source with value i s in parallel with the 1.2K resistor. The output current i o is defined as the into the emitter. We form the A circuit by redrawing the amplifier at AC. The loading effects are found by severing the feedback at the emitter of Q2. This results in a series 1.2K + This results in a series 1....
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440_hmk5_soln - ECE 440 Electronics II Solutions to...

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