removing ripple -- determining capacity of a smooth capacitor --


Japanese version here.


Electrical appliances which manage DC will often convert AC into DC.
That is because there are AC in the outlets of our houses.

It is difficult to completely convert AC into DC.
There are irregularities through this transform in any way.

I expect that lessening them will lead us to excellent sounds.

listmarkdeciding capacity of smooth capacitor

The first step of converting AC into DC is commutation.
There are some kinds of rectifier such as Half-wave rectification, Full-wave rectification and other rectifications.

(This dotted line is after passing through Bridge rectification. Almost every switching AC adopts this method.)

For further information, please refer to wikipedia.

Only after passing through rectification, the wave can't be called DC because of its shape like mountains.

The next step is smoothing.
We can remove such mountains by smoothing.
There are also some kinds of smoothing circuit such as capacitor input method, choke coil input method, PI method and so on.

This is capacitor input method.
C: smoothing capacitor
RL: speaker impedance

By charging C, the output voltage Ei gets like solid line in upper figure.

The mechanism of this is following.
Input voltage ED is the wave of dotted line.
The term of charging C by rising ED is t1-t2.
The speed of charging capacitor is generally very rapid, so Ei follows ED when C is being charged. (the term of t1-t2)

In contrast, the term which the wave is decreasing linearly is that smoothing capacitor is discharging.
The speed of discharging is determined by the time constant, C*RL.
If the capacity of C is high, the time constant will get to be high.
So the speed of discharging will be slow.
As the result of this, ED will be the shape of solid line.

So if we choose smoothing capacitor which has high enough capacity, output voltage will get to be clean.

Then I'd like to introduce the rate of including ripple.
"Ripple" means small waves.
In electrical words, it is small swing of voltage.
I definite it as the letter "gamma".

V: the average value of volutage
Va: the size of voltage swings

If it is small, we can say that the voltage is almost DC.

And if it is under 3%, the power supply will be qute a good one.

Then we solve the fomula which determines the capacity of smoothing capacitor in order to lessen it under alpha*100[%].
In Japan, the frequencies of power supply are following.
In East: 50Hz
In West: 60Hz

So the time cycles are following.

In East: T=0.020sec
In West: T=0.0167sec

The response of discharging is index function, but we approximate it as linear because we consider the term of that T/2 is much smaller than time constant.
And we get following formula by approximating the value of point that discharging curve meets sin curve as 3T/8.

By transforming this, we get following.

Then I'd like to try caluculating the value of smoothing capacitor for my environment.

I live in west side of Japan: T = 0.016sec
the impedance of speaker is 8ohm: RL = 8
I want to lessen ripple under 5%: alpha = 0.05

So substituting these values for the formula.

I find that I should choose a smoothing capacitor which has capacity of 10,000uF.

If we increase the capacity of smoothing capacitor, ripple will be lessen.
But we must not increase it in vain because we have to consider the charging current.
If so huge charging current runs, it will damage precision instruments.

It will be dangerous to use capacitor which has over 10,000uF capacity.

listmarkprogram of determining the capacity of smoothing capacitor

Select the frequency of your power supply: 50Hz or 60Hz
speaker impedance: R = [ohm]
the rate of including ripple: alpha = (If you want to suppress it under 5%, please input "0.05".)

C = [uF]

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