Beginner's capacitor question
ard at p850ug1.demon.co.uk
Tue Apr 15 16:53:34 CDT 2008
> I have a need to record the output of a (5150) speaker. Although I
I would think the characteristics of the (small, low quality) speaker in
a 5150 would affect the sound, and that if you want the result to sound
like a 5150, you should proably record the sound produced by said speaker
using a microphone. But that leads to all sorts of other problems
> thought that I could just alligator-clip a positive lead to one speaker
> terminal and the negative to the case/ground, the output was decidedly
> "buzzy" (I assumed it was too "hot" and overmodulating). I routed it
> into a mixer and turned it down (speaker is 5v, not sure what line input
> is) but it still didn't sound right.
Line input is about 0.75V to 1V RMS IIRC.
The 5150 speaker, IIRC, is connected between the +5V line and an
> While I have read the wikipedia entry on capacitors, I'm missing
> something obvious. My question: Why the 4.7uf capacitor? Does it
> serve to limit the signal? Reduce it's voltage? (or increase it?)
> Filter the signal in some way?
One property of a capacitor is that it'll pass AC (the changes in a
signal) while 'locking' DC (the steady, average, voltage of the signal).
And I think that's the main use here, to remove that 5V standing voltage.
I think I would try the following circuit if I wanted to record my PC
speaker output :
|| | 10k
o------|| |----------/\/\/----+----- Output 'hot' (centre of RCA plug)
Hot || | |
Side of PC |
o------------------------------+------ Output ground
Where the hot side of the PC speaker is the terminal _not_ connected to
+5V. If youy get that wrong, you'll get no sound, but it shouldn't do any
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