Question 7.6: A BFY50 transistor is used in a common-emitter amplifier sta...
A BFY50 transistor is used in a common-emitter amplifier stage with R_L = 10 kΩ and I_C = 1 mA.
Determine the output voltage produced by an input signal of 10 mV. (You may ignore the effect of h_{re} and any bias components that may be present externally.)
The equivalent circuit (with h_{re} replaced by a short circuit) is shown in Fig. 7.30. The load effectively appears between the collector and emitter while the input signal appears between the base and emitter. First we need to find the value of input current, I_b , from:
I_b=\frac{V_{in}}{h_{ie}}=\frac{10 \ mV}{250 \ \Omega }=40 \ \mu ANext we find the value of current generated, I_f , from:
I_f = h_{fe} × I_b = 80 × 40 μA = 320 μAThis value of current is shared between the internal resistance between collector and emitter (i.e. 1/ h_{oe} ) and the external load, R_L . To determine the value of collector current, we can apply the current divider theorem (Chapter 3):
I_c=I_f\times \frac{\frac{1}{h_{oe}}} {\frac{1}{h_{oe}}+R_L}=320 \ \mu A\times \frac{\frac{1}{80\times 10^{-6}}} {\frac{1}{80\times 10^{-6}}+10 \ k\Omega }Thus:
I_c=320 \ \mu A \times \frac{12.5 \ k\Omega } {12.5 \ k\Omega +10 \ k\Omega }from which:
I_c=320 \ \mu A \times 0.555=177.6 \ \mu AFinally, we can determine the output voltage from:
V_{out} = I_c × R_L= 177.6 \ μA × 10 \ kΩ = 1.776 \ V