15 Apr
2008
15 Apr
'08
9:53 p.m.
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 :
1uF
+
|| | 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
damage.
-tony