Determine: The r.m.s. ripple voltage

1. Feb 29, 2016

ifan davies

• 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 theres a good chance I'm way of here.....Thanks for any help!

2. Feb 29, 2016

Staff: Mentor

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. Feb 29, 2016

ifan davies

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. Feb 29, 2016

Staff: Mentor

Okay, so write it out as an equation and plug in your givens. Solve for what you want.

5. Mar 1, 2016

ifan davies

right lets start again...

Vrms= Vripple/2sqrt3

Vripple= 2/(2x400)

= 0.0025

Vrms= 0.0025/(2sqrt3)

vrms= 0.00072

this seems low..

6. Mar 1, 2016

Staff: Mentor

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. Mar 1, 2016

ifan davies

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. Mar 1, 2016

Staff: Mentor

Again, you're all over the map. Write out the defining equation for the ripple factor. It's one equation.

9. Mar 1, 2016

ifan davies

Kv= rms value of the a.c. voltage component/ average value of the load

10. Mar 1, 2016

Staff: Mentor

What specifically do you mean by "average value of the load"?

Hint: See your own post #3.

11. Mar 1, 2016

ifan davies

its the average load voltage according to my notes? Thanks for sticking with me!

12. Mar 1, 2016

Staff: Mentor

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. Mar 1, 2016

ifan davies

Vrms = 115/sqrt2

Vdc= 28

Kr= 81.3/28

Kr = 2.9

14. Mar 1, 2016

Staff: Mentor

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. Mar 1, 2016

ifan davies

this has to be it ....

Kr x Vdc = Vrms

0.005 x 28 = 0.14

16. Mar 1, 2016

Staff: Mentor

Yup.

17. Mar 1, 2016

ifan davies

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. Mar 1, 2016

Staff: Mentor

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. Mar 1, 2016

ifan davies

0.48= 200/2x400x C

0.00024= 2x 400 x C

0.0012= 400 x C

C= 3 uF

20. Mar 1, 2016

Staff: Mentor

What does the "200" value represent? The load current was given as 20 Amps.