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An NPN BJT having reverse saturation currents I_S=10^(-15) A is biased in the forwared active region withV_BE=700mV and the current gain (beta)may vary from 50 to 150 due to manufacturing variation. What is the maximum emitter currents(inmuA). |
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Answer» SOLUTION :`I_S=10^(-15)A` `V_BE=0.7` `V_T=25mV` `beta` RANGE from 50 to 150 `I_C=I_o e^(V_BE//V_T)` `I_E=(beta+1)/(beta)I_C` `I_E=(beta+1)/(beta)I_Se^(V_BE//V_T)` `I_E`be MAXIMUM when `beta` is 50 `=1.02xx10^(-15)xxe^(700xx10^(-3)//25xx10^(-3))` `I_E=1475muA` |
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