Hi, I know that it's good practice to include a pulldown resistor R2 between the base  emitter junction to make sure that the transistor remains off when small leakage current is present, and understand that it has to be much larger than the base resistor R1, but I don't fully understand the effect of adding R2 to R1.
When calculating the base current, are R1 & R2 in parallel? E.g. if Vcc = Vin = 5V, R1 = 1K, R2 = 10K.
Is it correct that the input impedance = (1/R1+1/R2)^(1) = 909, so Ib = 4.3V/909 = 4.73mA? So if we have a smaller R2, it's effectly reducing the base current?
Thanks!
When calculating the base current, are R1 & R2 in parallel? E.g. if Vcc = Vin = 5V, R1 = 1K, R2 = 10K.
Is it correct that the input impedance = (1/R1+1/R2)^(1) = 909, so Ib = 4.3V/909 = 4.73mA? So if we have a smaller R2, it's effectly reducing the base current?
Thanks!
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