1999-2009 [Subscribe to Daily Digest] |
An EE may be able to explain this better but the cap is a Low ESR cap. If I remember my EE classes (I am a ME, but dabble with EE stuff) ESR stands for Equivalent Series Resistance. The ESR is given at a particular frequency. If the resistance is high at the frequency that you are trying to attenuate the cap won't work effectively since you are trying to pass the AC signal to ground and the resistance is high.
Again recalling what I know from EE classes please feel free to correct me if I error... An ideal capacitor is an open circuit for DC and allows AC to pass through at different attenuation levels depending on the frequency and the capacitance value.
Your alternater does not naturally make DC currnet. It makes AC that is rectified? via a diode bridge. The diode bridge basically takes an input of sin(x) and makes an output of |sin (x)|. What you get is not true DC but DC with AC noise since |sin (x)| is not a straight line. Fortunately the average value of |sin(x)| is a constant. This is a bit of an idealization that does not take into acount any filtering or modifying of the signal from the alternater due to inductance or capacitance of the system.
An appropriately sized cap at the alternater will allow the DC component from the alternater to pass while grounding out (and thuss reducing) the AC component. The cap is only a first order filter and has a roll off rate of 6dB per octave. That is a fairly weak roll off rate, but if the cap is sized appropriately the frequency that we are worried about will be many octaves below the cut off.
Again I am no EE, but have taken a few EE courses in Undergrad and have always played with the stuff. The above is a very basic explination.
-Joe
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