Question 17.PE.MCQ.21: Given a transistor with the following specifications: gm = 3......
Given a transistor with the following specifications: g_m = 38 mmhos, r_{b′e} = 5.9 kΩ, h_{ie} = 6 kΩ, r_{bb′} = 100 Ω, C_{b′c}= 12 pF, C_{b′e} = 63 pF, f_T= 80 MHz and h_{fe} = 224 at 1 kHz. What is the value of α-cutoff frequency and the value of common-emitter short circuit gain at that frequency?
(a) 95.91 MHz, 0.838 ∠0.21°
(b) 85.91 MHz, 0.838 ∠0.21°
(c) 95.91 MHz, 0.838 ∠−0.21°
(d) 85.91 MHz, 0.838 ∠−0.21°
(a)
f_{\alpha}=\frac{h_{\mathrm{fe}}}{2\pi r_{\mathrm{b′e}}C_{\mathrm{b′e}}}=\frac{224}{2\times\pi\times5.9\times10^{3}\times63\times10^{-12}}
=95.91 MHz
f_{\beta}={\frac{g_{\mathrm{b^{\prime}e}}}{2\pi(C_{\mathrm{b^{\prime}e}}+C_{\mathrm{b^{\prime}c}})}}
where
g_{\mathrm{b^{\prime}e}}={\frac{1}{r_{\mathrm{b^{\prime}e}}}}={\frac{1}{5.9\times10^{3}}}=1.69\times10^{-4}
Therefore,
f_{\beta}={\frac{1.69\times10^{-4}}{2\times\pi\times(63\times10^{-12}+~12\times10^{-12})}}
358.63{\mathrm{~kHz}}
The common-emitter short-circuit current gain is given by
A_{\mathrm{i}}={\frac{-h_{\mathrm{fe}}}{1+j(f/f_{\beta})}}
For f=f_{\alpha},
A_{1}=-\frac{224}{1+j\left\{(95.91\times10^{6})/(358.63\times10^{3})\right\}}
=-{\frac{224}{1+267.43j}}
\left|A_{i}\right|={\frac{224}{\sqrt{1^{2}+(267.43)^{2}}}}=0.838
\begin{array}{c}{{\angle A_{\mathrm{i}}=90^{\circ}-\tan^{-1}(267.43)}}\\ {{=90^{\circ}-89.786^{\circ}=0.21^{\circ}}}\end{array}