Question

Q1) LED. An LED circuit has an optical 3 dB bandwidth of 725 kHz. Determine the modulation frequency at which the LED output is reduced to 20% of the low-frequency value of he ow reen v

Answer 1

From given data, An LED circuit has an optical 3dB bandwidth of 725k HA P_0 = P_dc/squareroot 1 + (w tau_i)^2 equation (1) For optical 3dB P_0/P_dc = 1/2 = 1/squareroot 1 + (w_5dB tau_i)^2 = 1/2 1 + (w_5dB tau_i)^2 = 4 let us consider, f_3dB = squareroot 3/2 Pi tau_i tau_i = squareroot 3/2 Pi f_3dB tau_i = squareroot 3/2 times Pi times.725 times 10^3 tau_i = 380.22 ns \r\n The modulation frequency at which LED output is reduced to 20% of the low frequency value is, P_0 (w) = 20% P_dc From equation (1) 20/100 P_dc = P_dc/squareroot 1 + (w tau_i)^2 1/5 = 1/squareroot 1 + (w tau_i)^2 squareroot 1 + (w tau_i)^2 = 5 Squaring on both sides then we get 1 + (w tau_i)^2 = 25 implies (w tau_i)^2 = 25 - 1 (2 Pi f tau_i)^2 = squareroot 24 f = squareroot 24/2 Pi tau_i f = squareroot 24/2 times Pi times 380.22

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