Determine: The r.m.s. ripple voltage

In summary, an aircraft electrical system operates at 115 volts at 400 hertz and supplies an average d.c. voltage of 28 volts and an average current of 20 A at a ripple factor of 0.005.
  • #1
ifan davies
14
0
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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  • #2
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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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..
 
  • #6
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?
 
  • #7
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
 
  • #8
Again, you're all over the map. Write out the defining equation for the ripple factor. It's one equation.
 
  • #9
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
 

What is r.m.s. ripple voltage?

R.m.s. ripple voltage refers to the root mean square value of the AC component of a voltage signal that is superimposed on a DC voltage signal. It is a measure of the amount of fluctuation or ripple present in a signal.

Why is it important to determine the r.m.s. ripple voltage?

Determining the r.m.s. ripple voltage is important because it can affect the performance and lifespan of electronic devices. Excessive ripple voltage can cause damage to components and result in unstable or inaccurate readings.

How is r.m.s. ripple voltage calculated?

R.m.s. ripple voltage can be calculated using the formula: V_rms = V_p/√2, where V_p is the peak-to-peak voltage of the AC component of the signal. This formula takes into account the fluctuations of the signal and provides a more accurate measurement compared to simply taking the average value.

What factors can affect r.m.s. ripple voltage?

R.m.s. ripple voltage can be affected by factors such as the quality of the power supply, the circuit design, and the load on the circuit. It is also influenced by the type of capacitor used to filter the signal and the frequency of the AC component.

How can r.m.s. ripple voltage be reduced?

R.m.s. ripple voltage can be reduced by using a high-quality power supply, properly designing the circuit to minimize noise, and choosing the right type of capacitor for filtering. Adding additional filtering components, such as inductors, can also help reduce ripple voltage.

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