Question: Design a simple logic inverter with an NPN transitor and a 1k collecter pull-up resistor. Choose the base resistor (E12) as large as possible (but as to ensure that the transistor saturates when the input is at + 5V)
My answer: I collecter = V / R collecter = 5 / 1 = 5 mA
i base = (Vin - 0 .7) / R base
B = i collecter / i base
i collecter = B * i base
i collecter = B * (Vin - 0.7) / R base
R base = i collecter / B * (Vin - 0.7) = 5/ (50 * (5 - 0.7)) = 0.02k
Right answer: With 1k pull-up saturation current will be (5-0) / 1k = 5mA so minimum base current required is 5mA / 50 = 100 micro A. so (5-0.7)R base >= 100 micro A, so R <= 4.3 / 100 micro A = 43k. We much therefore choose the next lower E12 value i.e. 39k.
I dont really understand the answer, could someone please help? My circuit diagram is attached.
My answer: I collecter = V / R collecter = 5 / 1 = 5 mA
i base = (Vin - 0 .7) / R base
B = i collecter / i base
i collecter = B * i base
i collecter = B * (Vin - 0.7) / R base
R base = i collecter / B * (Vin - 0.7) = 5/ (50 * (5 - 0.7)) = 0.02k
Right answer: With 1k pull-up saturation current will be (5-0) / 1k = 5mA so minimum base current required is 5mA / 50 = 100 micro A. so (5-0.7)R base >= 100 micro A, so R <= 4.3 / 100 micro A = 43k. We much therefore choose the next lower E12 value i.e. 39k.
I dont really understand the answer, could someone please help? My circuit diagram is attached.
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