A half-wave rectifier has an ideal diode and a load of 470 Ω…
Contributed by StemPy LabTry 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 . The transformer gives V rms at 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, , and the ripple frequency is . For an ideal full-wave rectifier, , and the ripple frequency is . A capacitor filter gives and . A silicon diode drops V when it conducts.
The worked solution
Have you tried it?
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 peak. V.
Step 2. The direct voltage. V.
Step 3. The direct current. mA.
Step 4. The rms output. V.
Answer: V, V, mA and V.
Check. The direct voltage is only of the peak. A full-wave rectifier would double it.
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