Question 2.SP.25: In the small-signal circuit of Fig. 2-40, the capacitor mode......
In the small-signal circuit of Fig. 2-40, the capacitor models the diode diffusion capacitance, so that C = C_d = 0.02 μF, and v_{th} is known to be of frequency ω = 10^7 \text{rad/s}. Also, r_d = 2.5 Ω and Z_{Th} = R_{Th} = 10 Ω. Find the phase angle (a) between i_d and v_d and (b) between v_d and v_{th}.
(a) The diffusion capacitance produces a reactance
x_d = \frac{1}{\omega C_d} = \frac{1}{(10^7)(0.02 \times 10^{-6})} = 5 \Omega
so that Z_d = r_d\parallel (-jx_d) = \frac{(2.5)(5\underline{|-90^\circ} )}{2.5 – j5} = 2.236 \underline{|-26.57^\circ} = 2 – j1 \Omega
Thus, i_d leads v_d by a phase angle of 26.57°.
(b) Let Z_{eq} be the impedance looking to the right from v_{th}; then
Z+{eq} = Z_{Th} + Z_{d} = 10 + (2 – j1) = 12 – j1 = 12.04 \underline{|-4.76^\circ} \Omega
Hence, v_{th} leads v_d by an angle of 26.57^\circ – 4.76^\circ = 21.81^\circ.