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  1. Problems
  2. /A half-wave rectifier has a load RL = 900 Ω…
MediumPhysicsHigh School (11–12)Electronics › Semiconductors and DiodesNumericalConceptualBoard ExamAEEE (Amrita)AMUEEEAP EAPCETBCECECOMEDK UGET
+15 more exams
CUET UGGUJCETJCECEJKCETKCETKEAMKIITEELPUNESTMET (Manipal)MHT-CETOJEESRMJEEETS EAPCETUPESEATVITEEE

A half-wave rectifier has a load RL = 900 Ω…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A half-wave rectifier has a load RL=900 ΩR_L = 900\ \OmegaRL​=900 Ω. First take an ideal diode. Find the efficiency, which is the direct power in the load divided by the alternating power that the transformer gives. Then take a diode that has a constant forward resistance rf=100 Ωr_f = 100\ \Omegarf​=100 Ω when it conducts and is otherwise ideal. The current is i=Imsin⁡ωti = I_m \sin \omega ti=Im​sinωt in the positive half cycle. Find the efficiency now. The direct power is Idc2RLI_{dc}^2 R_LIdc2​RL​ with Idc=Im/πI_{dc} = I_m/\piIdc​=Im​/π. The power from the transformer is Irms2(RL+rf)I_{rms}^2 (R_L + r_f)Irms2​(RL​+rf​) with Irms=Im/2I_{rms} = I_m/2Irms​=Im​/2.

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The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. The ideal diode. η=(Im/π)2RL(Im/2)2RL=4π2=0.405=40.5 %\eta = \dfrac{(I_m/\pi)^2 R_L}{(I_m/2)^2 R_L} = \dfrac{4}{\pi^2} = 0.405 = 40.5\ \%η=(Im​/2)2RL​(Im​/π)2RL​​=π24​=0.405=40.5 %.

Step 2. The real diode. η=(Im/π)2RL(Im/2)2(RL+rf)=4π2×RLRL+rf\eta = \dfrac{(I_m/\pi)^2 R_L}{(I_m/2)^2 (R_L + r_f)} = \dfrac{4}{\pi^2} \times \dfrac{R_L}{R_L + r_f}η=(Im​/2)2(RL​+rf​)(Im​/π)2RL​​=π24​×RL​+rf​RL​​.

Step 3. The numbers. RLRL+rf=900 Ω1000 Ω=0.90\dfrac{R_L}{R_L + r_f} = \dfrac{900\ \Omega}{1000\ \Omega} = 0.90RL​+rf​RL​​=1000 Ω900 Ω​=0.90. So η=0.405×0.90=0.365=36.5 %\eta = 0.405 \times 0.90 = 0.365 = 36.5\ \%η=0.405×0.90=0.365=36.5 %.

Answer: 40.5 %40.5\ \%40.5 % and 36.5 %36.5\ \%36.5 %.

Check. A diode with a larger resistance wastes more power as heat, so the efficiency falls. A load much larger than rfr_frf​ makes the loss small.

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For an ideal half-wave rectifier, Vdc=Vm/πV_{dc} = V_m/\piVdc​=Vm​/π, Vrms=Vm/2V_{rms} = V_m/2Vrms​=Vm​/2 and the ripple frequency is fff. For an ideal full-wave rectifier, Vdc=2Vm/πV_{dc} = 2V_m/\piVdc​=2Vm​/π, Vrms=Vm/2V_{rms} = V_m/\sqrt{2}Vrms​=Vm​/2​ and the ripple frequency is 2f2f2f. A capacitor filter gives ΔV=IfrC\Delta V = \dfrac{I}{f_r C}ΔV=fr​CI​ and Vdc≈Vp−ΔV/2V_{dc} \approx V_p - \Delta V/2Vdc​≈Vp​−ΔV/2. A silicon diode drops 0.70.70.7 V when it conducts.

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