# How the output of Full-wave rectifier frequency will double ?

Discussion in 'Power Electronics' started by zakmen, Jan 18, 2018.

1. ### zakmen Thread Starter New Member

Jan 18, 2018
3
0
Hello,

Please, could you clarify my doubts which are discussed below.

1) How the frequency will be double at the output of the Full-wave rectifier ? Explain in detail please
2) The frequency will only depending on the positive half-cycles ?
3) Generally, we have sine-wave before the full-wave rectifier and after the rectification the signal will be DC right!. So, How the DC signal has the frequency ?

2. ### smooth_jamie Member

Jan 4, 2017
99
18
1. The voltage doesn't double. After the rectifier the DC voltage is 1.414 times (sqrt[2]) the peak AC input
2. DC doesn't have a frequency because the voltage doesn't alternate (like AC), but it does have a ripple frequency
3. Before the smoothing capacitors the frequency of the output is twice the input AC frequency. This is because the negative half cycles are rectified to positive so there are now double the amount of positive cycles (I think this is what you mean by double, hence 50Hz becomes 100Hz).

Pure DC doesn't have a frequency, but rectified AC to DC does actually have a ripple frequency even after smoothing. Lot of information on Google if you need elaboration.

EDIT:
Sorry jumped in a bit early and misread part 1 of your question.

3. ### ericgibbs Moderator

Jan 29, 2010
6,777
1,335
hi zakmen,
Is this a Homework or College assignment, if Yes, I can move it to the Homework Forum.?
E

4. ### shortbus AAC Fanatic!

Sep 30, 2009
6,069
3,480
I was under the idea that the 1.414 only applies AFTER a filter cap? That rectified AC without a cap is the same as the AC voltage LESS the diode drop.

5. ### wayneh Expert

Sep 9, 2010
15,554
5,763
Of course you are correct, but once we're talking about pulsing DC, using the peak voltage is more conventional than using the RMS voltage. Maybe it's just because the filter cap is assumed. It's sloppy but it's convention as far as I can tell.

shortbus likes this.
6. ### smooth_jamie Member

Jan 4, 2017
99
18
Yes that's what I meant. Poor clarity on my part.

shortbus likes this.
7. ### sghioto Active Member

Dec 31, 2017
656
92
The frequency is doubled because you are rectifying both the positive and negative sides of the waveform. After rectification and no filtering the signal is a DC pulsed waveform at twice the frequency as the input line voltage.

Steve G

8. ### zakmen Thread Starter New Member

Jan 18, 2018
3
0
Still, I am not clear my doubt fully.
Usually, when we talking about the frequency is depending on the time period right!
But, from the above statement your saying that the negative half cycle are rectified to positive so there are now double the amount of positive cycles. If the positive cycles are double the frequency will double? How does it happen?

The below waveform are described by sghioto, there are same time period of input and output waveforms. So, how the frequency will double?

Please could you explain detailed!

9. ### ericgibbs Moderator

Jan 29, 2010
6,777
1,335
Hi Z,
A definition of Frequency is how often the same EVENT occurs in a given period.
The EVENT does not have to be sinusoidal.

The positive peak voltage EVENT of the rectified pulse is at twice the frequency of the original frequency.
E

Added: positive peak voltage

Last edited: Jan 19, 2018
atferrari likes this.
10. ### wayneh Expert

Sep 9, 2010
15,554
5,763
Two peaks per input cycle = doubled frequency.

11. ### smooth_jamie Member

Jan 4, 2017
99
18
If you look at the helpful diagram zakmen has posted you will see a full AC cycle between 0 and 6 on the y-axis. This includes one negative peak and one positive peak.

Now the graph immmediately underneath now has two positive peaks in the same space on the y-axis.

The first graph has one positive peak per cycle, the second graph has two peaks per cycle. One cycle in this case being from 0 units to 6 units on the y-axis.

12. ### AnalogKid AAC Fanatic!

Aug 1, 2013
7,294
2,053
Not counting the diode drop(s), with or without filtering, the peak value of the waveform is the same before and after rectification. The filter capacitor current is severely distorted, which is how the capacitor can "fill in" the output voltage between the peaks.

ak

shortbus likes this.