Question

You are going to design a dimmed light off your wall outlet (i.e., Vin- 120 Vrms, f 60 Hz). You would like to dim a 75 W lightbulb to 1/3 its rated power. Shown below is a controlled full-wave rectifier for this design. The turn-on of the SCRs is delayed by a time td. Assume ideal devices. Rbulb ssin ( tdi Determine the turn-on time td and the corresponding phase delay. Sketch the current drawn from the grid when the rectifier is operating as intended. a. b. C. Your roommate is a little concerned about the power factor of the circuit, and demands to know it. Help him/her calculate the power factor for this 1/3 power. You wish to cut the circuit cost by using only two SCRs. You have plenty of power diodes, however. How can you accomplish the same goal with just two SCRs and diodes? Draw this new circuit. d.

Answer 1

v_in = v_s/squareroot 2 = 120 v f = 810 Hz Therefore w = 2 pi f = 2 pi 60 rad/sec Rayed power of bulb = 75 w Rated power of balls p_rated = v62_ratyed/R_bulb where v_rated rightarrow rated voltage across bulb v_rated = 120 v_r m s Therefor 75 = 120^2/R_bulb rightarrow R_bulb = 120^2/75 = 192 ohm Therefore R_bulb = 192 ohm (a) During +ve half cycle of source voltage SCR s_1 & s_2 are on with a time delay of t_1 and in -ve half cycle of source voltage SCR s_3 & s_4 are on with time delay of t_d \r\n Therefore output voltage v_bulb we have: - v_s = 120 times squareroot 2 = 169.7 v Therefore v_bulb = 1/pi integral^pi_2 v_s sin (w t) dw t = v_s/pi Vertical-bar - cos (w t) Vertical-bar ^pi_alpha v_bulb = v_s/pi (1 + cos alpha), where alpha rightarrow phase delay angle w = 2 pi f = 2 pi 60 t_alpha rightarrow turn ON time t_d

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