Question 19.1: Determine the quiescent collector current and the quiescent ......
Determine the quiescent collector current and the quiescent output voltage of the following circuit, given that the h_{FE} of the transistor is 100.
The base-to-emitter junction of the transistor resembles a forward-biased p_{n} junction, therefore we will assume that the base-to-emitter voltage V_{BE} is 0.7 V.
From a knowledge of V_{BE} , we also know the voltage across R_{B} as this is simply V_{CC} – V_{BE} , which in turn enables us to calculate the base current I_{B} . In this case
I_{B} = \frac{V_{CC} – V_{BE}}{R_{B}} = \frac{10 – 0.7 V}{910 kΩ} = 10.2 μAThe collector current I_{C} is now given by
I_{C} = h_{FE}I_{B} = 100 × 10.2 μA = 1.02 mAThe quiescent output voltage is simply the supply voltage minus the voltage drop across R_{C} and is therefore
V_{o} = V_{CC} – I_{C}R_{C} = 10 – 1.02 × 10^{-3} × 4.7 × 10^{3} ≈ 5.2 V
Thus, the circuit has a quiescent collector current of about 1 mA and a quiescent output voltage of approximately 5.2 V.
Note: In this circuit, the quiescent collector current and the quiescent output voltage are both determined by the value of h_{FE} , which varies considerably between devices (and with temperature). For this reason, this simple circuit arrangement is rarely used. However, it is useful to look at the disadvantages of this arrangement before progressing to more suitable techniques.