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For a CE transistor amplifier , the audio signal voltage across the collector resistance of 2.0 k Omega is 2.0 V . Suppose the current amplification factor of the transistor is 100 . What should be the value of R_(R) in series with V_(BB) supply of 2.0V if the dc base current has to be 10 times the signal current . Also calculate then de drop across the collector resistance . Take VBE=0.6V |
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Answer» Solution :The output ac voltage is 2.0 V , So the ac COLLECTOR current `i_(c) = (2.0)/(2000) = 1.0 mA` . The signal current through the BASE is , therefore given `i_(B) = (i_(C))/(beta) = (1.0 mA)/(100) = 0.010 mA` . The DC base current has to bc `10 XX 0.10 = 0.10` mA We have , `R_(B) = ((V_(BB) - V_(BE)))/(I_(B))` . Assuming `V_(BC) = 0.6 V , R_B = ((2.0 - 0.6))/(0.10)= 14 k Omega` The de collector current `I_(C) = 100 xx 0.10= 10 mA` |
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