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24.16.182.213
In Reply to: RE: Cap "breaker-inner" posted by 1973shovel on October 17, 2020 at 09:32:16
The (polarized) 22uF capacitor is in the circuit backward. I don't think I would trust much about that circuit, or the person who designed it. :)
Dave.
Follow Ups:
...a good 9VDC source will have low impedance effectively shorting out the line level signal if both are connected to the circuit simultaneously. So the caps being burned-in see no signal, only 9VDC...assuming the polarized cap hasn't been damaged from reverse polarity.
why do you guys think that cap is backwards?
with regards,
As drawn, the negative terminal of the 22uF cap is tied to the positive terminal of the DC voltage source. When an electrolytic cap is reversed biased as this one would be, the insulating oxide layer breaks down and the cap becomes a resistor allowing DC current from the voltage source to flow thru it. This situation typically damages a polarized cap and may cause the cap to fail with a spectacular and messy explosion.
'the negative terminal of the 22uF cap is tied to the positive terminal of the DC voltage source'
yes, that's the direction of the current flow ... the whole DC circuit 'flows' to the positive of the line level source, which actually carries the 'negative' audio signal right? the line level's negative flows to the negative terminal of the battery ... are you saying the battery should be reversed as well?
when you say it converts that bi-polar cap to a resistor I agree, but that circuit is also treating the battery as if it was a 'pre-charged' capacitor, and the whole thing is dumping DC into the line level source in reverse ... I was thinking that bi-polar cap should just be a high value resistor instead ... on the other hand it doesn't make much sense to reverse charge the line level output ... that seems like 'bad ju-ju' to me
regards,
Do you see DC current flowing anywhere in that circuit????
Dave.
is that not a 9VDC battery in circuit? is there any switching inverter circuit there?this thread took off because 1973Shovel was concerned that 9 volts weren't enough correct?
what am I missing?
Edits: 10/19/20
Assuming you had the correct capacitor installed....and the correct direction....this is essentially just a phantom power circuit. So the signal would couple via the capacitor and ride on top of the 9VDC on the "burn in" capacitors. So, you wouldn't damage your signal source.
But the whole concept of it is ridiculous to begin with and a waste of time.
Dave.
...behaves something like a true voltage source (approaching near zero AC impedance), it will shunt any AC signal to ground leaving only 9VDC across the caps being broken. The circuits I've seen attempting to passively superimpose an AC signal on a DC source do so by placing a large value resistor in series with the DC supply.
I think the assumption is the 9 volt battery does not have zero internal resistance. There really should be a resistor between it and the rest of the circuit.
The whole thing is a joke and "designed" by somebody who had no clue what they were doing.
Dave.
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