Quote:
Originally Posted by Mizerable /img/forum/go_quote.gif
here are the pics
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A few of those joints show signs of being "cold": the solder surface should always be shiny and smooth. Use desoldering braid to remove some of this solder (don't completely desolder it, just get rid of some of it) and put fresh solder back on the joint. Alternately, if you have some liquid flux around, dab a bit on the joint and reflow it with your iron.
Then,
clean the flux off the board.
Finally, there are several places where I can see stray wire strands or overly long component leads on the bottom side of the board. Clip these away.
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how does the 1/(2 pi R C ) formula work? |
A capacitor (C) in series with an AC signal (like music) followed by a resistor (R) to ground is called a high-pass filter. This means that it lets high frequencies pass through it [mostly] untouched. As frequency goes down, the filter attenuates the signal more and more.
There's a special frequency where attenuation changes from a very slow roll-off to a constant amount of significant roll-off, which is the -3 dB point: the point at which the signal coming out is attenuated by 3 dB relative to the input. (Negative dB is attenuation, or "down" from the input.) Below the -3 dB point, the signal is down an additional 6 dB with every octave, which is a frequency doubling or halving.
For example, if the -3 dB point is 12 Hz, the signal is 9 dB down at 6 Hz, 15 dB down at 3 Hz, etc. Because this is an exponential curve, at 0 Hz it's infinite attenuation, which is another way of saying that a capacitor blocks DC.
Quote:
how does the 1/(2 pi R C ) formula work? |
R is the resistor value in ohms, C is the capacitor value in farads, and pi you should know. So to rework the calculation by Pars:
C = 0.00000022 farads (0.22 microfarad caps, in series with the signal)
R = 120 ohms (your headphones, a resistance between signal and ground after the C)
f = 1/(2 × pi × R × C)
f = 1/(2 × 3.14159 × 120 × 0.00000022)
f = 1/0.00016587609211
f = 6029 Hz