What is the difference between these two formula of ripple voltage

In summary: For a darlington or SCR the capacitor charges every cycle of the waveform, so the ripple frequency is f.
  • #1
null void
102
1
I found 2 equation for calculating the peak to peak ripple voltage,

this one is the common one i see,

Vripple p-p = I / 2fCAnd i also see this one in my book, but can't really find any other source use this formula,

Vripple p-p = Vrect / fRLC

Vrect is the peak voltage of unfiltered voltage
f is frequency,
C is capacitance
RL is the load voltageWhat is the difference between these two formula? Are they really for counting the same parameter?
 
Engineering news on Phys.org
  • #2
null void said:
I found 2 equation for calculating the peak to peak ripple voltage,

this one is the common one i see,

Vripple p-p = I / 2fC


And i also see this one in my book, but can't really find any other source use this formula,

Vripple p-p = Vrect / fRLC

Vrect is the peak voltage of unfiltered voltage
f is frequency,
C is capacitance
RL is the load voltage


What is the difference between these two formula? Are they really for counting the same parameter?

That's Resistance, I think.
The ripple is very much dependent upon the load resistance. If the load has infinite resistance then there will be no ripple at all because the capacitor will hold its charge from one positive supply peak to the next. The smaller the R, the faster the C will discharge and the time constant will be RC. The "I" in the first formula is a result of I = V/R, so both formulae are equivalent. They are approximations.
See this link, as an example.

The two formulae assume that the source resistance is low enough to ignore; the volts on the C at each peak need to 'follow the nominal sinusoidal input wave' for the simple approximation. If you use a cheapo transformer and over-load it, the output volts will never reach the open circuit maximum value. The volts can 'sag' because of the potential divider effect of the source and load resistances and you'll get less ripple and less DC.
 
  • #3
from the second equation,

Vripple p-p = Vrect / fRLC

= Vrect / fC x I/Vrect

= I / fC

but the first equation has 1/2
 
  • #4
I think the 1/2 factor must come from the peak to peak of the Voltage waveform and the average value of current (due to the average voltage). They are both approximations, in any case.
 
  • Like
Likes 1 person
  • #5
Get it, thanks
 
  • #6
the 1/2 factor comes from the type of rectifer used in the circuit.
For a full wave or bridge rectifier the capacitor charges every half cycle of the waveform, so the ripple frequency is 2f.
 

What is the difference between these two formula of ripple voltage?

The first formula is used to calculate the ripple voltage in a half-wave rectifier circuit, while the second formula is used for a full-wave rectifier circuit.

What is ripple voltage?

Ripple voltage is the small fluctuation in the output voltage of a rectifier circuit due to the imperfect conversion of alternating current (AC) to direct current (DC).

How do the two formulas differ in their calculations?

The first formula takes into account the peak value of the input voltage, while the second formula takes into account the RMS (root mean square) value of the input voltage. Additionally, the second formula accounts for the fact that a full-wave rectifier circuit produces a smoother output compared to a half-wave rectifier circuit.

Which formula should be used in different scenarios?

The first formula should be used for a half-wave rectifier circuit, while the second formula should be used for a full-wave rectifier circuit. Additionally, the second formula should be used when calculating the ripple voltage for a circuit that has a capacitor filter.

Why is it important to calculate ripple voltage?

Ripple voltage can have a significant impact on the performance of a circuit, especially in electronic devices that require a constant and stable power supply. By calculating the ripple voltage, scientists and engineers can determine the effectiveness of a rectifier circuit and make necessary adjustments to ensure a smoother and more stable output.

Similar threads

  • Electrical Engineering
Replies
13
Views
2K
  • Electrical Engineering
Replies
6
Views
844
Replies
2
Views
1K
Replies
38
Views
3K
  • Electrical Engineering
Replies
22
Views
2K
Replies
2
Views
629
  • Electrical Engineering
Replies
5
Views
2K
  • Electrical Engineering
Replies
10
Views
2K
  • Electrical Engineering
Replies
9
Views
1K
  • Electrical Engineering
Replies
7
Views
1K
Back
Top