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

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 $R_L = 900\ \Omega$ . 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 $r_f = 100\ \Omega$ when it conducts and is otherwise ideal. The current is $i = I_m \sin \omega t$ in the positive half cycle. Find the efficiency now. The direct power is $I_{dc}^2 R_L$ with $I_{dc} = I_m/\pi$ . The power from the transformer is $I_{rms}^2 (R_L + r_f)$ with $I_{rms} = I_m/2$ .

Need a reminder? The key ideaShow

For an ideal half-wave rectifier, $V_{dc} = V_m/\pi$ , $V_{rms} = V_m/2$ and the ripple frequency is $f$ . For an ideal full-wave rectifier, $V_{dc} = 2V_m/\pi$ , $V_{rms} = V_m/\sqrt{2}$ and the ripple frequency is $2f$ . A capacitor filter gives $\Delta V = \dfrac{I}{f_r C}$ and $V_{dc} \approx V_p - \Delta V/2$ . A silicon diode drops $0.7$ V when it conducts.

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.

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