Lo, Im currently building my self a basic 2 way cross over for my car, comprimising of 1 inductor and what was going to be 1 capacitor, however I cant get the capacitor required. I was thinking about wiring several caps in parrallel to add up to the correct capactiance, but I dont know if this will filter frequencies properly, my logic says no, but my logic's also carp, so if anyone knows help would be appreciated. 🙂
Of course you can put several capacitors in parallel to get the value you want. For example two 10uF 50V caps in parallel will make a 20uF 50V cap. And two 10uF 50V in series will make a 5uF 100V cap.
/Freddie
/Freddie
I dont think i made myself clear in my first post, I know about capactiance i.e. in parrallel C+C=CT and series 1/C+1/C=1/CT, ill explain how my logic's working right now. When a capacitor has AC thru it i.e. a music signal passing through it, it has a reactance, reactance is the same as resistance only it varies with freqency of the AC/signal, with a capacitor the higher the frequency the lower the reactance/resistance, so *in an AC circuit when a capacitor is acting as a filter it is in effect a resistor*. When these are in series and parrallel they behave in the complete opposite way to 'capacitors'
*This is the bit im not sure of*
Sorry about not being clear
*This is the bit im not sure of*
Sorry about not being clear
Beggar
Sorry about the problem you are having with your logic. The formula is Z=1/2*3.14159*F*C. Z = impeadance F = Frequency and C = capacitance. By using this formula we can find that increasing the value of C will lower the value of Z at a frequency of F, OR increasing the value of C will lower the frequency F for a give value of Z.
All this means is, Freddie is right. Using caps in parrallel to get a larger value or some special value works just fine.
Good luck
Sorry about the problem you are having with your logic. The formula is Z=1/2*3.14159*F*C. Z = impeadance F = Frequency and C = capacitance. By using this formula we can find that increasing the value of C will lower the value of Z at a frequency of F, OR increasing the value of C will lower the frequency F for a give value of Z.
All this means is, Freddie is right. Using caps in parrallel to get a larger value or some special value works just fine.
Good luck
capacitors in parallel
Dont think twice, its all right :
You can obtain any value of capacitance by wiring several capacitors in parallel, simply adding individual values.
Regards, P.Lacombe
Dont think twice, its all right :
You can obtain any value of capacitance by wiring several capacitors in parallel, simply adding individual values.
Regards, P.Lacombe
Thanks everyone, nice to see im wrong as usual (it is actually nice as it saves me time and money!). I just found a book that had why caps filter and stuff and now i can see i was being a prune, and have adjusted my logic accordingly 🙂
Why sometimes there are 2 capacitors in parallel, one with bigger value (usualy electrolitic) and one smaller (usualy film)? I found this configuration alot as a coupling caps. Thanks
The smaller cap is "faster", less parasitic resistance, less parasitic inductance, so it can discharge and recharge faster.
So the small cap start acting faster, then after a few nanoseconds, the bigger cap wake up and start working before the smaller cap is drained up.
That's more complicated in reality, but you can see it that way to make things more simple.
So the small cap start acting faster, then after a few nanoseconds, the bigger cap wake up and start working before the smaller cap is drained up.
That's more complicated in reality, but you can see it that way to make things more simple.
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