Determine: The r.m.s. ripple voltage

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    Ripple Voltage
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Discussion Overview

The discussion revolves around calculating the root mean square (r.m.s.) ripple voltage in an aircraft electrical system that operates at 115 volts and 400 hertz, aiming to supply an average d.c. voltage of 28 volts with a ripple factor of 0.005. The conversation includes various approaches to the calculation, definitions of terms, and considerations for capacitor sizing.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant attempts to calculate the r.m.s. ripple voltage using a formula involving the average voltage and load resistance, but expresses uncertainty about the result.
  • Another participant notes that different definitions of ripple factor exist, suggesting that the definition used may affect the calculations.
  • Some participants clarify that the ripple factor is defined as the ratio of the r.m.s. ripple voltage to the average d.c. voltage.
  • A participant proposes a new calculation for r.m.s. voltage based on the peak voltage derived from the average voltage, leading to a significantly different value.
  • There are multiple calculations presented for the ripple voltage and capacitor sizing, with participants questioning and refining each other's approaches.
  • One participant suggests that the capacitor calculation may be incorrect and provides an alternative formula for peak-to-peak ripple voltage.
  • Discrepancies arise regarding the values used in calculations, particularly concerning the load current and resulting capacitance values.

Areas of Agreement / Disagreement

Participants demonstrate a lack of consensus on the correct approach to calculating the r.m.s. ripple voltage and the appropriate values for capacitance. Multiple competing views and calculations are presented without a clear resolution.

Contextual Notes

Some calculations depend on specific definitions of terms like ripple factor and average load voltage, which may not be uniformly understood among participants. There are also unresolved mathematical steps and assumptions regarding the waveform shape and rectification method.

Who May Find This Useful

This discussion may be of interest to students and professionals in electrical engineering, particularly those working with power supply design and ripple voltage calculations.

ifan davies
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moved into h/w help, so template is missing
An aircraft electrical system operates at 115 volts at 400 hertz. It is to supply an average d.c. voltage of 28 volts and an average current of 20 A at a ripple factor of 0.005.

Determine: The r.m.s. ripple voltage

Here is my attempt.

Vc= Vs-2Vd
=28-2(0.7)
= 26.6 Volts

Vr= (1/2 x fs x C x Rl )(Vc)

R=V/I 28/20= 1.4

(1/2x400x1.4)(26.6)
= 0.023

Vrms= Vr/2sqrt3 Vrms = 0.0063

Now there's a good chance I'm way of here...Thanks for any help!
 
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Hi ifan davies, Welcome to Physics Forums.

Different authors define the ripple factor slightly differently. In most cases it's considered to be equal to the ratio of the RMS ripple voltage to the DC component of the power supply output. Others use ratio of the peak to peak value of the ripple voltage to the DC component. Still others convert the result to a percentage ("percent ripple") by multiplying the ratio by 100.

What definition is being used in your textbook or course?
 
Hey hey, its the first one you mentioned where it's considered to be equal to the ratio of the RMS ripple voltage to the DC component of the power supply output.
 
ifan davies said:
Hey hey, its the first one you mentioned where it's considered to be equal to the ratio of the RMS ripple voltage to the DC component of the power supply output.
Okay, so write it out as an equation and plug in your givens. Solve for what you want.
 
right let's start again...

Vrms= Vripple/2sqrt3

Vripple= I(load)/2fc

Vripple= 2/(2x400)

= 0.0025

Vrms= 0.0025/(2sqrt3)

vrms= 0.00072

this seems low..
 
You're making it much more complicated than it needs to be. Work with the definition of the ripple factor. What terms in that defining equation are you given in the problem statement?
 
Vpk= Vaverage x (pi/2)

Vpk = 28 x (pi/2)

Vpk= 43.98v

Vrms= Vpk/sqrt(3)

Vrms = 43.98/ sqrt(3)

Vrms= 25.39
 
Again, you're all over the map. Write out the defining equation for the ripple factor. It's one equation.
 
Kv= rms value of the a.c. voltage component/ average value of the load
 
  • #10
What specifically do you mean by "average value of the load"?

Hint: See your own post #3.
 
  • #11
its the average load voltage according to my notes? Thanks for sticking with me!
 
  • #12
Okay, it's the average DC component of the load voltage. So the equation is:

##K_r = \frac{V_{rms}}{V_{DC}}##

where ##K_r## is the ripple factor, ##V_{rms}## is the RMS ripple voltage, and ##V_{DC}## is the average DC voltage.

Which values do you know from the given information?
 
  • #13
Vrms = 115/sqrt2

Vdc= 28

Kr= 81.3/28

Kr = 2.9

Please be right
 
  • #14
You're given the ripple factor (##K_r = 0.005##) and the DC voltage (##V_{DC} = 28~V##). You want to find the ##V_{rms}##.
 
  • #15
this has to be it ...

Kr x Vdc = Vrms

0.005 x 28 = 0.14
 
  • #16
Yup.

Be sure to include units on your answer.
 
  • #17
When you put it like that its easy! Your awesome!

Now to calculate a suitable value of smoothing capacitor

C= i load / frequency x V

20/ (400x0.14)

0.357 F
 
  • #18
I'm not sure about your capacitor calculation. For a full-wave rectifier you should have something like

##V_{pp} = \frac{I}{2 f C}##

where ##V_{pp} = 2 \sqrt{3} V_{rms}## is the peak-to-peak ripple voltage assuming a sawtooth waveform.

That would yield a somewhat smaller value for the capacitor.
 
  • #19
0.48= 200/2x400x C

0.00024= 2x 400 x C

0.0012= 400 x C

C= 3 uF
 
  • #20
What does the "200" value represent? The load current was given as 20 Amps.
 
  • #21
30 uF it is my bad! Thanks for the help, think il go and attempt some other questions now.
 
  • #22
I think you should recheck your calculation. The capacitance should be much larger than 30 μF .
 
  • #23
0.48 = 20/ 2 x 400 x C

0.48/20 = 2 x 400 x C

0.024 = 2x 400 x C

0.024/2= 400 x C

0.012= 400 x C

C = 0.00003 F or 30 uF

Not correct?
 
  • #24
ifan davies said:
0.48 = 20/ 2 x 400 x C

0.48/20 = 1/(2 x 400 x C)
Or,

20/0.48 = 2 x 400 x C
... etc.
 
  • #25
Or that yea lol

0.05 F
 
  • #26
That looks better.

If the marker is a stickler for significant figures be sure to carry sufficient digits through all calculations and only round values accordingly at the end for presentation of final values.
 
  • #27
I will keep that in mind. Thank you