There's mention that at DC T=0, a capacitor should be considered as short circuit.
So how does it work when we use it as a filter capacitor?
Won't it be shorted to ground as the sketch in a flash?
Best Answer
The capacitor is considered a short-circuit for sufficiently high frequency components relative to its capacitance. That's how it acts as a filter. The lower frequencies see it as an open circuit and ignore capacitor, but the high frequencies (i.e. noise frequencies) see it as a short-circuit and take the detour through the capacitor and are short-circuited, preventing them from travelling through the load.
It kind of acts as an almost infinite ohm resistor for "really low" frequencies, and an almost zero ohm resistor for "really high frequencies". Then there is an intermediate range where the capacitor kind acts in between. It's not a hard rule that a capacitor always acts as a short-circuit (or an open-circuit) for all frequencies.
You are missing some conditions when you say that that the capacitor is appear as a short-circuit at t = 0. This statement assumes that you apply a DC voltage to the capacitor at t = 0. You are basically applying an infinitely fast vertical voltage edge to the capacitor (turning on the power supply). That infinitely fast vertical needs infintely high frequency components. It's those super high frequency components that see the capacitor as a short-circuit. The lower and intermediate frequencies in this edge charge up the capacitor to varying degrees until the capacitor is full charged.
That's how AC "flows" through a capacitor (by AC I mean non-zero frequency components like sinusoids, anything that is not just a steady flat 0Hz voltage. I do not mean only the AC wall voltage). It's no different than DC, except that a AC/sinusoid voltage is always changing so the capacitor's voltage is never allowed to catch up and match the applied voltage so charge is always being added or removed from the capacitor in AC which is equivalent to current flow through the capacitor. If a frequency is higher, a given capacitance will provide less impedance to it and it appears more as a short-circuit and will flow through the capacitor more easily.
Electrolytic capacitors are well known to pass a small DC current. How much they pass is a complex function of temperature, capacitance, the age of the capacitor, and the DC voltage across it.
This is because the insulating film is formed by electrolysis; as the current flows, it electrolytically deposits an insulator on the anode (which is aluminium : this is the same process as anodised aluminium) and the insulator eventually stops the current flow.
So for example an electrolytic that has been unused for several years will draw quite a large current (milliamps) for a while, then the current will reduce as the insulator re-forms itself. The capacitor is healing itself.
The result of this is:
do not use electrolytics in series with high impedances (say, as coupling capacitors between amplifier stages)
do not use them with no (or reverse!) DC across them (they will not heal, and reverse DC will break them down)
Ideally you want 2/3 their rated DC voltage across them to keep them formed (at least 50% anyway)
do not use them in circuits like R-C timer circuits where leakage will change the time constants
remember that at high temperature (usually 85C) their lifetime is rated at 8000 hours - which is only a year!
Tantalums can be a better choice but have their own problems
Don't use them across a high current supply. Across a low current supply they can self-heal too, but across a high current supply they can turn into fireworks...
Camera flash capacitors are constructed to have low resistance, and more importantly, low inductance, so that they can deliver their energy to the flash tube as quickly as possible — which means achieving a fast risetime on the pulse of current. The internal connections are also made more robust in order to avoid localized heating as a result of the high current.
Best Answer
The capacitor is considered a short-circuit for sufficiently high frequency components relative to its capacitance. That's how it acts as a filter. The lower frequencies see it as an open circuit and ignore capacitor, but the high frequencies (i.e. noise frequencies) see it as a short-circuit and take the detour through the capacitor and are short-circuited, preventing them from travelling through the load.
It kind of acts as an almost infinite ohm resistor for "really low" frequencies, and an almost zero ohm resistor for "really high frequencies". Then there is an intermediate range where the capacitor kind acts in between. It's not a hard rule that a capacitor always acts as a short-circuit (or an open-circuit) for all frequencies.
You are missing some conditions when you say that that the capacitor is appear as a short-circuit at t = 0. This statement assumes that you apply a DC voltage to the capacitor at t = 0. You are basically applying an infinitely fast vertical voltage edge to the capacitor (turning on the power supply). That infinitely fast vertical needs infintely high frequency components. It's those super high frequency components that see the capacitor as a short-circuit. The lower and intermediate frequencies in this edge charge up the capacitor to varying degrees until the capacitor is full charged.
That's how AC "flows" through a capacitor (by AC I mean non-zero frequency components like sinusoids, anything that is not just a steady flat 0Hz voltage. I do not mean only the AC wall voltage). It's no different than DC, except that a AC/sinusoid voltage is always changing so the capacitor's voltage is never allowed to catch up and match the applied voltage so charge is always being added or removed from the capacitor in AC which is equivalent to current flow through the capacitor. If a frequency is higher, a given capacitance will provide less impedance to it and it appears more as a short-circuit and will flow through the capacitor more easily.