Determine: The r.m.s. ripple voltage

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SUMMARY

The discussion focuses on calculating the RMS ripple voltage in an aircraft electrical system operating at 115 volts and 400 hertz, supplying an average DC voltage of 28 volts and a current of 20 A with a ripple factor of 0.005. The correct formula for the ripple voltage is derived, leading to the conclusion that the RMS ripple voltage (Vrms) is 0.14 volts when applying the ripple factor to the average DC voltage. Additionally, the participants discuss the importance of using the correct definitions and equations, such as the relationship between peak voltage and average voltage, to arrive at accurate results. The final calculations suggest a required smoothing capacitor value of approximately 0.05 F.

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  • Understanding of electrical engineering concepts, specifically ripple voltage and ripple factor.
  • Familiarity with AC and DC voltage relationships in power supply systems.
  • Knowledge of capacitor sizing for smoothing applications in rectifier circuits.
  • Proficiency in using formulas related to voltage, current, and frequency in electrical systems.
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  • Study the derivation of ripple voltage formulas in power supply design.
  • Learn about the impact of ripple factor on power supply performance.
  • Research capacitor selection criteria for smoothing in rectifier circuits.
  • Explore the differences between full-wave and half-wave rectification and their effects on ripple voltage.
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Electrical engineers, aerospace engineers, and students studying power electronics who are involved in designing or analyzing aircraft electrical systems and power supply circuits.

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