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In a single state transistor amplifier,When the signal changes by 0.02 V,the base current change by 10 muA and collector current by 1mA.If collector load R_(C)=2 kOmega and R_(L)=10kOmega ,Calculate ,(i)Current Gain (ii)Input impedance ,(iii)Effective a.c load (iv)Voltage gain and (v)Power gain. |
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Answer» Solution :(i)CURRENT GAIN `BETA=(deltai_(c))/(deltai_(h))=(1MA)/(10muA)=100` (ii)Input impedance `R_(i)=(deltaV_(BE))/(deltai_(b))=(0.02)/(10muA)=2000Omega=2kOmega` (iii)Effctive (a.c)load `R_(AC)=R_(C)||R_(L)=(2xx10)/(2+10)=1.66 kOmega` (iv)Voltage gain `A_(v)=betaxx(R_(AC))/(R_(in))=(100xx1.66)/(2)=83` (v)Power gain ,Ap=Current gain `xx` Voltage gain `=100x83=8300` |
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