EasyPhysicsHigh School (11–12)ElectronicsNumerical

A half-wave rectifier has an ideal diode and a load of 470 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 an ideal diode and a load of $470\ \Omega$ . The transformer gives $12$ V rms at $50$ Hz. Find the peak voltage of the transformer, the direct voltage across the load, the direct current in mA and the rms value of the output voltage.

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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