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Question Number 66249 by Umar last updated on 11/Aug/19
$$\mathrm{The}\:\mathrm{npn}\:\mathrm{transistor}\:\mathrm{in}\:\mathrm{the}\:\mathrm{voltage}\: \\ $$$$\mathrm{amplifier}\:\mathrm{circuit}\:\mathrm{operates}\:\mathrm{satisfctorily} \\ $$$$\mathrm{on}\:\mathrm{a}\:\mathrm{quiescent}\:\mathrm{collector}\:\mathrm{current}\:\mathrm{of}\:\mathrm{3mA}. \\ $$$$\mathrm{if}\:\mathrm{the}\:\mathrm{battery}\:\mathrm{supply}\:\left(\mathrm{V}_{\mathrm{cc}} \right)\:\mathrm{is}\:\mathrm{6v},\:\mathrm{calculate} \\ $$$$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:; \\ $$$$\left(\mathrm{a}\right)\:\mathrm{the}\:\mathrm{load}\:\mathrm{resistor}\:\mathrm{R}_{\mathrm{L}} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{the}\:\mathrm{base}\:\mathrm{current}\:\mathrm{for}\:\mathrm{the}\:\mathrm{quiescent} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{collector}−\mathrm{emitter}\:\mathrm{voltage}\:\mathrm{V}_{\mathrm{ce}} \:\mathrm{to} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{be}\:\mathrm{half}\:\mathrm{the}\:\mathrm{battery}\:\mathrm{voltage}. \\ $$$$\mathrm{Th}\:\mathrm{transistor}\:\mathrm{dc}\:\mathrm{current}\:\mathrm{is}\:\mathrm{100}. \\ $$$$ \\ $$$$\mathrm{please}\:\mathrm{help}. \\ $$
Answered by peter frank last updated on 11/Aug/19
$${V}_{{cc}} ={I}_{{c}\:\:} {R}_{{L}\:} +{V}_{{ce}} \\ $$$${V}_{{cc}\:} ={V}_{{B}} +{V}_{{BE}} \\ $$$${V}_{{ce}\:} =\frac{{V}_{{cc}} }{\mathrm{2}} \\ $$
Commented by Umar last updated on 29/Sep/19
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$