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Unformatted text preview: amplifier circuit. Using the dependent source equivalent, we get the circuit shown in Figure (b). To find I B , we can write KCL at the base node (labeled B). Note that V B = 0.7V (assume that the emitter is used as the reference node)-I 1 + I 2 + I B = 0-(2-0.7)/100k + 0.7/200k + I B = 0 I B = 9.5 A To find we can write KVL for the output loop:-16 + 1k * 150I B + Vo = 0 Vo = 14.575V Extra- credit problems: Problems 3.89, 3.90, 3.91, 3.92...
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