Ok, so I understand HOW capacitors work, but when to choose the right values?

I realize it depends on the application too, like how a 555 Timer has a formula where the capacitor value changes the frequency.

But for instance:

I have a 25.2V,2A center tapped transformer.

After rectifying it, I got about 27V.
THen I wanted to regulate it at 30V with a LM338T regulator...

so it has its own mathematics to get the out that I wanted.... so I first used a 22000uF capacitor in parallel with the source. This helped get me 36-37V out. (more about this later)

Then I used the regulator to flatten the voltage even more and get the exact 30V I wanted....
Turned out perfect.. I get the exact voltage I want, and the LM338T does not even get hot (YUS, one less heatsink!)

Even though everything turned out AWESOME,
I don't konw if I REALLY needed 22000uF capacitor,
so I was wondering, in THIS application, what value could I scale down to?

Like, so we have rectified voltage, which is just a constant positive sine wave: y=|sin(x)|
So lets say the amplitude is 27V, which capacitor would be best to up the voltage to 35V? What Farad value should I use. I know the voltage value should be double the value going in, so like 60V would be a safe side to the capacitor.

Times like these I really wish I had an oscilloscope. -_-

Thanks for any help you guys can give

 

Hello,

Take a look at the attached PDF.
It is about calculating the capacitor value versus the ripple voltage.

Greetings,
Bertus

 

The above response... plus the definition of Farad.

A Farad is the value that one Amp for one Second will change the voltage across it by one Volt. Now you have to account for the load current (probably in amps), the capacitance (probably microF) and the time between voltage peaks (50Hz? 60Hz? or twice that for full-wave). The capacitor will discharge between peaks at the rate determined by this math. It is a relatively minor task to calculate how far it will decline in this much time. You must do the calculation at the highest current you will expect to use.

Another note... the fact that "the LM338T does not even get hot (YUS, one less heatsink!)" must also be determined at the max load. No heat at no load is easy. As the current load goes up, so does the heat. Also, as the input voltage goes up, so does the heat. The ideal is to have the lowest point of the input voltage slightly above the "drop-out" voltage of the regulator. This is about 2.1V above the output voltage for a standard linear regulator, somewhat less for a "low drop-out" regulator.

Then the heat is (input voltage - output voltate) times output current. Heat-sinks are cheap insurance and greatly increases the available output current of a regulator.

 

Ahhh Thanks guys... actually some intense googling got me the answer before I refreshed this page, but that pdf is PERFECT for not forgetting anymore....

http://www.majhost.com/gallery/BZPchronicler/GIMPCreations/first_schematic.png
I made the lm338t part, the other part i understand now though...
I wrote my self a note if your interested:

How to measure the voltage out:
When the AC is rectified and filtered with a large capacitor, the voltage is the peak voltage minus two diode voltage drops and minus some ripple if the filter capacitor has a value that is too low.

TO start, we have a 25.2VAC and 2A Transformer

So if we where to rectify and regulate it, to find the final outcome, we will use mathematics:

The peak voltage of a sine-wave is the RMS voltage times the root of 2. The root of 2 is 1.414.
So the 25.2VAC has a peak of 35.64V and the full load current at this point is double the original RMS value devided by the peak voltage:
50.4/35.64 = 1.41 Amps
Diodes have voltage drops of .7-1.7V so the rectifier would make the voltage out of the rectifier:
35.64V - 1V - 1V = 33.64V

Keep in mind that the voltage 33.64V is the output value only if a capacitor, in parrallel with the rectifiers positive and negative, has a large enough value.

A general rule of thumb is for 1V of ripple the capacitor needs to be 10,000µF for every amp of loading, if you want to halve the ripple then double the capacitor size or halve the capacitor size if you don't mind double the ripple.

Here's the simple formula:

C = I/(2F*V).

C is the capacitance in Farads.
I is the current in Amps.
F is the mains frequency, normally 50Hz or 60Hz.
V is the maximum ripple.

A reasonable value for a rectifier's filter capacitor has about 1V per period of ripple. Then the voltage regulator circuit drops it to only 1mV per period.

SO, in this case:
C = 1.414A / (2 * 60Hz * 1V) ≈ (.012F = 12,000uF)

I only had a 22,000uF capacitor, so i had a ripple of ≈.5V before it was regulated, though a 12KuF would been fine.

As for the LM7812, the capacitor still leaves a ≈5.36V ripple but with a 17.819 peak voltage, the ripple never goes under the 12V demand. Then the regulator does its work and its down to 5mV.

Reading the LM7812's pdf and I am still puzzled how the author of THAT part derived the 680uF capacitor.
Help?



DonQ - OOOOOOOOOOOOOOHHHHHHHh that makes sense... well like I said I had a larger capacitor then nessisary. Thanks for giving the def. though, I never knew!