I have not done much work on this topic and I am not sure if this is how I am supposed to answer this past paper question.The mark scheme is not very useful as it does not give a diagram of the answer.

 

What you have looks good so far, what are you confused about? Also, with the capacitor added in, this would become a half wave peak rectifier.

 

What you have looks good so far, what are you confused about? Also, with the capacitor added in, this would become a half wave peak rectifier.

The mark scheme is just this

Is the ripple voltage also the same as the voltage drop?
Also can you put the resistor and capacitor in series and not in parallel?
Surely it all depends on what resistor value you have for the load, as this effects the Ripple voltage.

 

The mark scheme is just this

Is the ripple voltage also the same as the voltage drop?
Also can you put the resistor and capacitor in series and not in parallel?
Surely it all depends on what resistor value you have for the load, as this effects the Ripple voltage.

For a peak rectifier you would put the capictor in parallel with the loading resistor.

To calculate the peak to peak ripple voltage you can apply the following formulae,

\(V_{r} = \frac{V_{p}}{fRC}\)

Where Vr is the peak to peak ripple voltage, and Vp is the peak voltage across the load.

Attached to this post I have posted figures containing both half wave and peak rectifiers.

Vr is indicated on the output graph.

 

I believe that the 0.7V drop is the forward voltage drop of the diode. This results in the rectifier peak output voltage being less than the input voltage. This is separate from the ripple voltage.

You would not expect to put the resistor and capacitor in series.

 

I believe that the 0.7V drop is the forward voltage drop of the diode. This results in the rectifier peak output voltage being less than the input voltage. This is separate from the ripple voltage.

You would not expect to put the resistor and capacitor in series.

Does that mean it should look something like this then because the peak rectified output would be 0.7 less because of the diode voltage drop.
Forgive the scale, by the way this is E2 WJEC so I have not seen anything similar to these questions, understanding the new terminology is daunting at first.

 

Does that mean it should look something like this then because the peak rectified output would be 0.7 less because of the diode voltage drop.

I believe this is correct. Sorry I did not catch your mistake at first. Hopefully Adjuster can confirm my thoughts.

 

I believe this is correct. Sorry I did not catch your mistake at first. Hopefully Adjuster can confirm my thoughts.

Good, regarding your earlier post I am not expected to calculate the ripple voltage, so I will just leave it till I cover it in class.

 

That looks better. The marking scheme says nothing about finding the actual size of the ripple, beyond it being "substantial", so this should be sufficient.