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 β + 0.7 + 1 KI c 1 ⇒ I c 1 = 1.12 mA ⇒ r π 1 = 2320 Ω . Similarly for Q2: 15 K 15 K + 85 K (10) = (15 K ||85 K ) I c 2 β + 0.7 + 500 I c 2 ⇒ I c 2 = 1.27 mA ⇒ r π 2 = 2039 Ω . 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 = − β + 1 ( ) 2 K ||85 K ||15 K 2 K ||85 K ||15 K + r π 2 + β + 1 ( ) 500 ||10 K ( ) × β 10 M ||80 K ||20 K ||10.5 K 10 M ||80 K ||20 K ||10.5 K + r π 1 = − 264 . The β circuit consists of the 10K and 500 Ω with the feedback current flowing through the 10K. Thus, i f i o ' = − 500 10500 = − 0.0476 a. With feedback: A if = A 1 + β 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 π 1 = 1698. Thus R if = 1698 1 + β A = 125 Ω . R if = R in ||10 M ⇒ R in = 125 Ω c. R o = 4K Ω 2. Since this is shunt-series feedback, we need to begin with the current gain. Thus, we transform the input voltage and 1.2K Ω series resistance to a current source with value i s in parallel with the 1.2K Ω 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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