paragraf keberapa kang membahas input impedance, rumus nilai Zb ngga ketemu

So'ir, Zb itu bahasa saya, itu Vin/Input Current

* Input Stages
The input stages of all transistor amps must be able to supply the base current of the Class-A driver. This time, a margin of between 2 and 5 times the expected maximum base current is needed. If the Class-A driver operates with a quiescent current of 8mA, the maximum current will be 12mA (quiescent + driver base current. Assuming a gain of 50 (again), this means that the input stage has to be able to supply 12 / 50 = 240uA, so it must operate at a minimum current of 240uA * 2 = 480uA to preserve linearity.

* Input Current
The input current of the first stage determines the input impedance of the amplifier. Using the above figures, with a collector current of 480uA, the base current will be 4.8uA for input devices with a gain of 100. If maximum power is developed with an input voltage of 1V, then the impedance is 208k (R = V/I).

Since the stage must be biased, we apply the same rules as before - a margin of between 2 and 5, so the maximum value of the bias resistors should be 208 / 2 = 104k. A lower value is preferred, and I suggest that a factor of 5 is more appropriate, giving 208 / 5 = 42k (47k can be used without a problem).

Quote :

Using the above figures, with a collector current of 480uA, the base current will be 4.8uA for input devices with a gain of 100

gain 10 atau gain 100 kang ? puyeng juga bahasa linggis pas-pasan...

"figure"-nya bukan berarti gambar tapi bisa berarti "pernyataan" atas perhitungan-perhitungan sebelumnya.
"gain" disana merujuk ke gain/beta/hFE Tr Q1, bukan gain voltage amplinya mas..


480uA adalah arus di kaki Collector Q1, dan karena dia(Rod Elliott) mengasumsikan gain/beta/hFE dari Q1 adalah 100x, maka Ib (Q1) = 480/100 = 4,8uA.

iya yah 400 dibagi seratus itu 4 perlu pakai kacamata nih...
thx kang sue..