Question

Change the frequency to 50kHz. Only do the first part to find the -45° shift. What is the value of I that results in a -45° phase shift?

SIM *4.53 In the circuit shown in Fig. P4.53, is a de current and v, is a sinusoidal signal with small amplitude (less than 10 mV) and a frequency of 100 kHz. Representing the diode by its small-signal resistance r which is a function of I, sketch the small-signal equivalent circuit and use it to determine the sinusoidal output voltage V, and thus find the phase shift between V, and V.. Find the value of I that will provide a phase shift of-45°, and find the range of phase shift achieved as Iis varied over the range of 0.1 times to 10 times this value. Do 10 nF

Choose correct answer from the following:

a. 69.68E-6

b. 771.88E-6

c. 74.03E-6

d. 76.25E-6

e. 78.54E-6

f. 80.90E-6

g. 83.32E-6

h. 85.82E-6

Answer 1

v_o/v_i = 1/swc/rs + 1/jwc = 1/1 + jwcrd Phase shift = -tan^-1 ([wcrd]) = -45 degree implies [wcrd] = tan (45 degree) = 1 implies 2pi (50k) (10n) [V_T/I_D] = 1 implies I_D = 2pi (50k) (10n) (26m) implies I_D = 81.681 times 10^-6 A (V_T = 26mV) (not in options) (or) I_D = 2pi (50k) (10n) (25mV) = 78.54 times 10^-6 A (V)T = 25mV) Therefore (e) is correct