# Determine: The r.m.s. ripple voltage

#### Ifan Davies

Joined Feb 29, 2016
5
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!

#### WBahn

Joined Mar 31, 2012
26,398
Your equations are utterly meaningless without a schematic of the circuit they apply to. We are NOT mind readers!

#### Ifan Davies

Joined Feb 29, 2016
5
I do not have a schematic. Just the question asked and the equations I thought were relevant........

#### WBahn

Joined Mar 31, 2012
26,398
I do not have a schematic. Just the question asked and the equations I thought were relevant........
The problem statement is fine. But your equations are specific to a particular solution to the problem (i.e., a particular schematic). Is it a full-wave rectifier? A half-wave rectifier? Is there a voltage regulator? Don't expect people to guess what type of circuit you have in mind.

Where are these equations coming from? Out of thin air? From some website?

#### Ifan Davies

Joined Feb 29, 2016
5
The previous question referred to a full wave bridge rectifier so I assume it is the same although it was a completely different question, no mention of a voltage regulator. The equations were from a lot of googling.

Im thinking this might be a better set of equations

Vrms= Vripple/(2sqrt3)

#### MrAl

Joined Jun 17, 2014
8,863
Hi,

Are you assuming a triangle or sawtooth wave? If so then the Rms voltage is just Vpk/sqrt(3).
Only if you measure the peak to peak voltage Vpp then it will be Vpp/(2*sqrt(3)).
That's because for a triangle wave the RMS value is Vpk/sqrt(3).
And:
Vpk is measured from zero to the positive peak.
Vpp is measured from the negative peak to the positive peak assuming the waveform is centered at zero.
If there is no negative peak then we use Vpk.
If there is a DC offset and we want the total RMS value then we have to do it a little differently.

#### Ifan Davies

Joined Feb 29, 2016
5
Thanks for you help
so

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

#### WBahn

Joined Mar 31, 2012
26,398
The previous question referred to a full wave bridge rectifier so I assume it is the same although it was a completely different question, no mention of a voltage regulator. The equations were from a lot of googling.

Im thinking this might be a better set of equations

Vrms= Vripple/(2sqrt3)

Disembodied equations are worthless -- they need to be the correct equations for the circuit being used.

The question is merely stating a problem -- they want a DC voltage output at that value that has a ripple voltage of no more than that amount while delivering that current.

There are many approaches that can achieve that goal and the equations that you need to work with are the ones that apply to that approach.

So pick an approach and then work out the equations that apply to THAT approach.

#### Ifan Davies

Joined Feb 29, 2016
5
Thanks
I reckon I've cracked it

Kr = Vrms/ Vdc

Kr x Vdc = Vrms

0.005 x 28 = 0.14 v