Re: Has my PIC blown?

Silverfox wrote: 
Ok I think I am understanding the capacitor thing to smooth out the
voltage so stop it from being eratic. Am I right in saying that voltage
will run through the capacitor to ground until the capacitor is fully
charged then it will block the voltage allowing the voltage to the 'In'
pin of the voltage regulator. Now the part I am getting lost on is, if
the voltage now coming to to the 'In' pin is a spike, how does the
capacitor smooth this out and I'm also lost if the voltage dips to the
'In' pin.

You have the sort of general idea, but the actual mathematical description may be some help. The relation between voltage and current for a capacitor is I=C*(dv/dt). In English, that is, Current through a capacitor (in amperes) is equal to the capacitance (in farads) times the rate of change of voltage (in volts per second).

So a capacitor passes current any time the voltage changes, and in proportion to the speed of the change. A "spike" implys a fast rate of change, so a large current. The capacitor doesn't hold the voltage still, but it does pass current in the direction that will help reduce the speed and magnitude of voltage change. It is, in effect, voltage inertia. If the spike in in the direction of increasing voltage, the capacitor will reduce the rate of change and lower the peak voltage by absorbing some of the energy in the spike. If the spike is is the direction of less voltage, the capacitor will reduce the rate of change, and raise the minimum voltage by dumping some of its stored energy out into the circuit.

If the voltage is in the dip, does the capacitor feed voltage back up
to the 'In' pin?


Yes, something like a rechargeable battery.

I hope you understand what I am trying to say. 

I hope you can tell whether or not I did.

Thank you very much for your help 

You're welcome.
.

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