Smoothing capacitor calculation problems.

[cartman007]

Hey Guys.

Im having some trouble with a PSU question that I go the other day.
I cant seem to figure it out. I think the answer is suppose to be 94.3uF but the closest I get is 88uF

Question is:

Design and build a full-wave rectifier with a smoothing capacitor that can deliver 21V to a 10Kohm load with ripple voltage of 0.31V(p-p).
The secondary voltage of the transformer can be taken as V2 = 22V .

Components to be used:
4 x 1N4007
1 x 10Kohm Resistor
1 x Smoothing CAP

CALCULATE Vrl(p) ; Vdc ; Irl and a value for the capacitor.


Ok soo first thing that came to my mind was that getting 31V from 22V will probably not happen without some kind of regulater.
UNLESS the 22V is in Vrms...Made sense since it came from a Tranformers coil? Am I correct in assuming that its in V(rms).

Ok so now I take some formulas that they give me.

V(rms) = Vp/sqrt(2) = 22*sqrt(2) = 31.11V
Vp = 31.11V - 0.7V = 30.41V "Drop over diode"
Ip = Vp/Rl = 30.41/10000 = 3.041x10^-3 A
Idc = 2(Ip)/pi = 2(3.04x10^-3)/pi = 1.94x10^-3
Vdc = Idc*Rl = 19.4V

Now that I have Vdc I can use this forumla.

Vdc = Vp - Idc / ( 4*f*( Vdc - Vp ) ) = 881pF

Not that cant possible be right?
What am I doing wrong?

 

[cgchas]

Hey Guys.
..
Am I correct in assuming that its in V(rms).
..

My understanding, which doesn't necessarily come from a great deal of knowledge, is that power transformers generally specify AC RMS voltage as well as a current rating, so I would assume 22V(rms) in the absence of clarification.

To solve for the filter capacitor, assuming full-wave rectification, given the load resistance, supply voltage, and ripple voltage, I have been using this formula:

C = Vs / (2 * R * f * Vr)

where,
C is the smoothing capacitance
Vs is the supply voltage (peak voltage)
R is the load resistance
f is the AC frequency (I use 60 Hz)
Vr is the ripple voltage (peak voltage)

C = 30.41 / (2 * 10000 * 60 * .31) = 81.7uF

My answer is a bit different than yours.
Perhaps your electrical service is 50Hz.

C = 30.41 / (2 * 10000 * 50 * .31) = 98.1uF

Close enough?

 

[cartman007]

C = 30.41 / (2 * 10000 * 50 * .31) = 98.1uF

Close enough?

That is perfect. I think that is actually what its suppose to be !

So my books will have every formula in existence accept the one that works :/ .

Thanks allot!

 

[cgchas]

That is perfect. I think that is actually what its suppose to be !

So my books will have every formula in existence accept the one that works :/ .

Thanks allot!

Just out of curiosity, what formula does your book have?
Most I have seen have some form of C = I x t / dV
where,
C is capacitance
I is current
t is the time period
dV or deltaV is the change of voltage in relation to the time, or Vr which is ripple voltage

Assuming a 10% ripple voltage, the following transformation is useful.

If ripple voltage is 1/10th supply voltage, then in terms of supply voltage the following is true:
C = 10it/Vs

We know that frequency is the reciprocal of the time period, so in terms of frequency, the following is true:
C = (10*I)/(Vs*f)

Assuming full-wave rectification, that means the smoothing capacitance we are seeking is for the half-cycle (double the AC frequency), so the following is true:
C = (10*I)/(Vs*2*f)

a reduced form would be:
C = (5*I)/(Vs*f)

The above form is very commonly seen on various websites, but let's move to the form that I used for your answer.

To solve for a given ripple voltage and resistance, let's move back to this formula:
C = I x t / Vr

For full-wave rectification in terms of frequency:
C = I / (2 * f * Vr)

We substitute I for Vs using I = E/R:
C = Vs / (2 * f * Vr * R)

which is what I used for your answer.

I am guessing your book has one form of one of the above equations.

Cheers,
Charles