Analog Electronics In a p⁺n junction diode under reverse bias , the magnitude of electric field is maximum at The edge of the depletion region on the n-side The center of the depletion region on the n-side The edge of the depletion region on the p-side The p⁺n junction The edge of the depletion region on the n-side The center of the depletion region on the n-side The edge of the depletion region on the p-side The p⁺n junction ANSWER DOWNLOAD EXAMIANS APP
Analog Electronics The base - emitter voltage is usually none of these. equal to the base supply voltage. more than the base supply voltage. less than the base supply voltage. none of these. equal to the base supply voltage. more than the base supply voltage. less than the base supply voltage. ANSWER DOWNLOAD EXAMIANS APP
Analog Electronics Which of these is the best description of a zener diode? It is a constant voltage device. It works in the forward region. It is a rectifier diode. It is a constant electric current device. It is a constant voltage device. It works in the forward region. It is a rectifier diode. It is a constant electric current device. ANSWER DOWNLOAD EXAMIANS APP
Analog Electronics A Schottky diode is a majority carrier device. both a majority and a minority carrier diode. fast recovery diode. minority carrier device. majority carrier device. both a majority and a minority carrier diode. fast recovery diode. minority carrier device. ANSWER DOWNLOAD EXAMIANS APP
Analog Electronics CMOS stands for active - load switching. complementary MOS. common MOS. p - channel and n - channel devices. active - load switching. complementary MOS. common MOS. p - channel and n - channel devices. ANSWER DOWNLOAD EXAMIANS APP
Analog Electronics An npn BJT has gm=38mA/v, cµ =10¯14 F, cπ =10¯13F and DC current gain β0=90.For this transistor fT & fβ are fT =1.33 x 1012 Hz & fβ = 1.47 x 1010 Hz fT =1.47 x 1010 Hz & fβ = 1.33 x 1012 Hz fT =1.47 x 1010 Hz & fβ = 1.64 x 108 Hz fT =1.64 x 108 Hz & fβ = 1.47 x 1010 Hz fT =1.33 x 1012 Hz & fβ = 1.47 x 1010 Hz fT =1.47 x 1010 Hz & fβ = 1.33 x 1012 Hz fT =1.47 x 1010 Hz & fβ = 1.64 x 108 Hz fT =1.64 x 108 Hz & fβ = 1.47 x 1010 Hz ANSWER DOWNLOAD EXAMIANS APP
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