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In a transistorconnected in the common base configuartion, alpha=0.95,I_E=1mA.Calculate the value of I_C and I_B. |
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Answer» SOLUTION :`ALPHA=(I_C)/(I_E)` `I_C=alphaI_E=0.95xxI=0.95mA` `I_E=I_B+I_C` `:. I_B=I_C-I_E=1-0.95=0.05mA` |
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