Understanding Half-Wave Rectification in AC Circuits: V(dc)=0.318Vm Explained

[M. next]

Which occurs after supplying the circuit with AC voltage.
What does it mean when they write V(dc)=0.318Vm [IDEAL CASE]

What do we have to do with DC, if what we are applying is AC?

 

[truesearch]

Vdc = 0.318Vm Is another way of writing Vdc = Vm/∏
This is the average value for 1/2 wave sine wave

 

[tiny-tim]

the clue's in the title!

"half-wave rectification" means that you simply rub out the bottom half of the graph …

it still has the same frequency, but it only flows in one direction …

so it's direct current, but with frequency ​

 

[M. next]

Wow thank you, I thought about that for half an hour just staring at the thing :p!
Another thing about the same issue,
I ran across a sentence that says: The effect of removing two diodes from the bridge configuration is therefore to reduce the available dc level ..

How can that be explained [If can be done with an example]

 

[vk6kro]

That isn't always true, but we can get to that later.

Here is a diagram showing the different waveforms from half and full wave rectifiers.

http://dl.dropbox.com/u/4222062/rectifiers.PNG

If you omitted the diodes shown in grey, you would only get half wave rectification. You can follow the path of current along the red and blue lines.

Notice that full wave rectified output has a higher average voltage because it is supplying voltage for a greater proportion of the time. (You can picture the tops of the waveforms above the average lines being clipped off and put in the space between the waveforms below the average line. When this produces a constant level, you can call this the average voltage.)

Half wave and full wave rectified outputs both produce the same PEAK output, though, and they are usually used with a large capacitor across the output. This will charge up to the peak value and give the same output for both types of rectifier.
Even then, though, the full wave rectifier will perform better on load because it gives pulses of charge to the capacitor more often.

 

[M. next]

Thank you for your effort in placing the data, and in explaining it :)