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How to choose the right capacitor?

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LAN_403

twist2b

Tue May 05 2009, 10:20PM

Registered Member #2086 Joined: Tue Apr 21 2009, 02:33AM
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Posts: 117

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

I saw this, And I wanted to know how he choose the right capacitor o_O
Maybe I just don't understand the logic chip enough. It's obviously frequency based, but I think a potentiometer on R3 would be better, cause then you can change the frequency and personally set the pulsing speed right?

Not limiting to this schematic though, Can anyone point me in the right direction for choosing capacitors?

waffle

Wed May 06 2009, 03:06AM

Registered Member #642 Joined: Fri Apr 13 2007, 10:19PM
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Posts: 15

Choosing a capacitor generally depends on what exactly your making. Sometimes the value does not matter critically. For example if you were building a simple power supply and you wanted to put a filter capacitor in you would want to try and get the largest capacitor you could. Sometimes the value of the capacitor is critical though. In this case, there is generally a formula for the specific application you are applying the capacitor to. In the case of the 555 timer, the formula for the frequency in terms of resistor one, resistor 2, and a capacitor (which I just pulled off of some 555 info website) is:
F = 1.44/(R1 + 2*R2)*C
Using the formula, if you wanted a 1kHz frequency, and had 1000 ohms for R1 and 1000 ohms for R2,
1000 = 1.44/(1000 + 2*1000)*C or 1000 = 1.44 / 3000*C you would get a value for C of .00000048 Farads, or .48 microfarads.

twist2b

Wed May 06 2009, 04:01AM

Registered Member #2086 Joined: Tue Apr 21 2009, 02:33AM
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Posts: 117

Oh, OK, that makes perfect sense.

I understand that, its just a mathematical means to an end. And you can use a 555 timer for LED pulsing as well.

Some testing of my own, when I made a 30V regulator, It has mathmatics Vout based on 2 resistive values. Thats pretty simple, but there where capacitors to help regulate... I put two 10uF and it worked fine. I was not 100% if it was right, but I guess if you just have a large enough capacitor (and a voltage rating double what its dealing with) everything is A ok.

Nice, Thanks.

Still, whats the Voltage to Farad ratio? Like with my regulator, whats "enough"? I chose 10uF but could I go lower?

twist2b

Wed May 06 2009, 06:55PM

Registered Member #2086 Joined: Tue Apr 21 2009, 02:33AM
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Posts: 117

I figured i out... I had just forgotten for those that care I wrote myself a note so that I don't forget:
Based off a schematic I edited 50% of:

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 mathmatics:

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 10,000ÂµF 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

Now that we have a somewhat smoothed high output, the voltage should be regulated to give a continuous smooth output.

The LM338T voltage regulator has a formula to choose the output based off Resistance values.

Vout = 1.25V*( 1 + R2/R1) + Iadj*R2

The 1.25V is the nominal refrency voltage (Vref) between the output and adjustment terminal.

Through some tests, the Iadj has been verfied as 1/15Amps or 0.0667 Amps.

We need a 30V out, so through basic mathmatics we find the wanted Resistance values:

R2 = 400 Ohms
R1 = 300 Ohms

Due to the capacitors, diodes are added to protect from the voltage peaks effecting output results.

The other side with the LM7812 has a voltage ripple of a little over 5V which is ok because the peak is close to 6V over 12V
Then the regulator takes care of the rest...
However, how did he get the 680uF?

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