Question

In the Common-Emmitter, Bipolar Junction Transistor Cicuit,

Vbe(on) = 0.7 V, Vce(sat) = 0.2 V, Beta=200.

Vcc = 9V, Rc =1K and Rb = 10K.

a.) Calculate the maximum collector current, ic.

Answer 1

The maximum collector current (Ic) in a common-emitter bipolar junction transistor circuit, with Vcc = 9V, Vce(sat) = 0.2V, and Rc = 1KΩ, is calculated to be 8.8 mA.

Step-by-step explanation:

To calculate the maximum collector current Ic in a common-emitter bipolar junction transistor circuit, we can use Ohm's Law and the transistor characteristics. The maximum collector current occurs when the transistor is in saturation. The collector-emitter saturation voltage Vce(sat) is given as 0.2 V, and the supply voltage Vcc is 9V. The collector resistor Rc is 1K ohms.

First, we find the collector-emitter voltage (Vce) when the transistor is saturated, which is approximately the saturation voltage Vce(sat), and subtract it from Vcc to find the voltage across the collector resistor (Vrc):

Vrc = Vcc - Vce(sat)

Vrc = 9V - 0.2V = 8.8V

Then we use Ohm's Law to find the maximum collector current (Ic):

Ic = Vrc / Rc

Ic = 8.8V / 1KΩ

Ic = 0.0088 A or 8.8 mA

Thus, the maximum collector current (Ic) is 8.8 mA.