Beginner's capacitor question

Tony Duell ard at
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 
open-collector output. 
> 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                      |
 Speaker                         /
                                 \ 1k
  o------------------------------+------ Output ground
PC 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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