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GEEZE!!!   All I added was a tad more resistance on the loop and it started working!!!

So at the fork -  I added a fair bit more resistance, then it started discharging,  but maybe its always discharging,  but its as if - if the discharge is too high, i get no reading.     

So u might be right about the current going back on itself, is deluminating the led -  and it doesnt matter if i have back to back diodes, it still cuts it off.

My theory said both leds should be full power,   but it may be the case that the current is 0 there,  so both leds are OFF.  instead...
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One last thing,  If I bypass the capacitor (touch its legs together)  the led goes on.   when the capacitor is in the circuit,  then no led.
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Yes, thats true but I actually meant the other way, from a to b. (sorry for confusing)  the current for the discharge goes the same way as the t-junction bypass current through the loop. So im saying the amount of water bypassing, actually is a complex number involving the neutralization of the capacitor,  plus the diverting current.   But the capacitor seems to just block off and stay blocked off.  doesnt make sense!
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Quote
And in my setup its as if there is a resistored leak from side b to side a - So it should always be releasing back to the central position a certain degree.

No.

The pipe is already full of water that is being pushed from A to B by the pump.  So even though the pipe is there, nothing can "leak" back from B to A -- it's blocked by all the water flowing the other way!  Once again, you can't have current flowing opposite directions in the same wire or component at the same time.  In your two-balloon version, any water that comes out of Balloon B is going to flow out the spout on the right.

Here's another way you know that no current is going to "leak" from B to A: the A side of the resistor is at a higher voltage than the B side (because it's closer to the positive terminal of your battery).  Current doesn't flow "uphill."

Finally, if you did somehow manage to get current to flow through the resistor in that direction, all it would do is charge the capacitor faster!
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Im reading and listening WriterofMinds, what u say makes sense...  I definitely agree on the t-fork your talking about.
But I still think im right about whats happening to me not making sense!

This is getting too complicated to explain...



In my thinking a single balloon isnt a perfect analogy to an electrical capacitor that has 2 connections, an anode and a cathode,  an electrical capacitor is better said (in my thinking) two balloons pressed into each other and the membrane in the middle pushing side b out when side a has pressure applied to it.   I think its a correct analogy as long as you both suck and blow at the same time, and its a perfect vacuum,  then I think its the same as electricity.

And in my setup its as if there is a resistored leak from side b to side a - So it should always be releasing back to the central position a certain degree.
[EDIT] sorry i meant from a, to b, not b to a. [/EDIT]

So im not convinced, but my analogy might not be correct, as well....  ive got it sitting on my desk till I work it out.

Im going to go to the heavenly courts and complain about this to the high magistrates.  :knuppel2:

Either that or my components are dodgy. :)
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Just because there's a path connecting the two terminals of the capacitor, does not mean the capacitor is going to discharge through that path.  You can only have a net flow of charge in one direction at a time!  If the capacitor is charging (current flowing into the positive terminal and out of the negative terminal), it cannot simultaneously be discharging (which would require current flowing into the negative terminal and out of the positive terminal).

I'll try to clarify with the water analogy.  Imagine you have water flowing down (being pushed by a pump) into a tee that connects to 1) a stretchy balloon and 2) a pipe.  At the tee the water stream splits; some of it goes into the balloon, and some of it goes down the pipe.

When the water falls into the balloon, is any of it going to come back out and go down the pipe instead?  No; it can't get back out, because more, new water is pushing into the balloon every second.  The flow only goes one direction at a time.

So will the balloon eventually fill up?  Yes.  If you stretch it enough, the pressure imposed by the walls of the balloon will equal the pressure exerted by the pump, and no more water will enter the balloon.

Will any water come back out of the balloon now?  No.  There's still water coming down from above and entering the pipe.  The pressure it exerts will keep all the water in the balloon.

But what if you turn the pump off?  Now there's no external pressure holding the water in the balloon, and the balloon wants to shrink, so it will push the water out of itself and down the pipe.

The balloon is your capacitor, the pipe is your resistor/diode/load/whatever, and the pump is your battery.
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Thanks for all the help everyone.

Writer of minds,  u sound like u know electronics well!
Yes, ur right about the capacitor filling up,  but it has a loop which is connecting its annode and cathode,  so it should be constantly discharging,  so it should never even get to full.

It doesnt make sense to me.
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Regarding your second example:

1. The two branches of your fork are not going to carry the same current.  As your capacitor charges, current through the branch containing the capacitor will decrease.  When the capacitor is fully charged, that current will drop to zero.  100% of the current will be going through the branch containing the single LED.  Imagine the battery pumping out charge like a water hydrant; no more charge can fit in the capacitor, so it all "overflows" down the other branch.  The cap will not be continuously discharging, either.

2. It is important to remember that an LED cannot turn on unless the voltage on its input side is X Volts higher than the voltage on its output side.  No voltage drop means nothing to push the electrons through the diode.  (X = 0.7 is a good rule of thumb, iirc.)  The voltage drop across your single LED branch must be at least X, so the drop across the capacitor branch is also at least X.  But!  In the capacitor branch, not all of that drop is across the pair of diodes!  Vdiodes = X - Vcapacitor.  As the capacitor charges, Vcapacitor gets bigger, until Vdiodes is too small to turn one of the diodes on.  The shut-down diode blocks further current flow through that branch.  (The other diode doesn't have a prayer of lighting up, because the voltage difference across it is *negative.*)

In short, what you're actually seeing happen in the video makes sense to me.  Capacitors block DC current.  If you want them to experience a charge/discharge cycle then your input voltage needs to oscillate.  A capacitor doesn't charge and discharge simultaneously; at any given time, you should think of it as doing one or the other (or else just sitting there at a constant level of charge).
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General Hardware Talk / Re: Something crazy is happening to me with my work
« Last post by WriterOfMinds on August 20, 2018, 11:58:11 pm »
Is your power supply DC?  If I may presume that your plus is "VCC" and your minus is "Ground," there's a constant voltage drop of VCC across your resistor/capacitor pair.

Your capacitor will charge until it is holding enough to create a potential of VCC* between the plates. Once it's charged, it will just sit there -- I do not expect it to discharge.  Your power supply is still, as it were, exerting "pressure" on it.  No more charge will go in (because it's "full"), but nothing will come out either. 

*Minus a bit for the requisite voltage drop across the bottom diode

Current (driven by your power supply) will continue to flow through the resistor, but you will not see anything through the diodes once the capacitor is charged (which may be happening within a fraction of a second).

Imagine it like a water tank filling up if that helps you.

Now, if you switch the power supply off, then you should see a brief current surge through the top diode as charge flows from the positive side of the capacitor around the loop to the negative side.

EDIT: All these comments are based on your first circuit as drawn in your initial post.
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General Hardware Talk / Re: Something crazy is happening to me with my work
« Last post by spydaz on August 20, 2018, 10:15:07 pm »
the little diode setup (rectification circuit) is switching current on/off ;

therefore the capacitor is getting no chance to charge perhaps change the Farads of the capacitor ..... allowing for the capacitor to charge quicker therefore dissipate its energy..

A lot of unnecessary diodes here...

MAYBE!
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