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From what I understand, when there is a drop in the voltage the current is drawn out of the capacitor and so the balance is maintained. But when there is an over voltage from the power supply or when the load draws more current how does the decoupling capacitor balance it. The capacitor is already fully charged. So when the load draws more current, doesn't that current get drawn from the capacitor/power supply?

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You can think of a decoupling capacitor as a toilet tank. If you just had the water pipe feeding the toilet you would never have enough volume of water to flush the toilet. The is a local storage area for extra water. When you flush it, it gives a large supply of water at once and allows a flush to happen. Then when the toilet is done flushing, the tank gets filled back up.

It's not quite the same with a capacitor, but similar. When your load needs more current, the capacitor will source some extra current. This causes a slight dip in voltage but not a huge one. If there wasn't a capacitor there, if the load needs a lot of current for a moment, the voltage will droop a lot because the load resistance would have dropped and power supply resistance would consume too much of the voltage.

Now for the other case for voltage spikes. A capacitor is never really "full". There's a maximum voltage it can handle, but usually that should be at least 25% higher than the normal operating voltage. Let's say the supply voltage is 5V, the capacitor should be able to handle at least 6.25 volts. That means that when there's a voltage spike coming down the line, the capacitor will absorb some of the extra current caused by the voltage and quench the incoming voltage spike to be much less than it would be otherwise. In this case, you should think of a capacitor like a flexible membrane attached to the side of a water hose. If there's ever a higher pressure transiently in the water hose, the flexible membrane (kind of like a shock absorber) would absorb the excess pressure and allow anything further down the line to not see the excess pressure (voltage) spike because it would have been absorbed by the membrane. Bypass capacitors work in a similar way.

Also note that when the load needs more current transiently, it will absorb current from both the capacitor and the power supply. It's just that the transient seen from the power supply will be distributed over time more.

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    \$\begingroup\$ -1 You literally don't mention anywhere that a capacitor stores ENERGY, not current. \$\endgroup\$ – Funkyguy Aug 8 '14 at 2:39
  • \$\begingroup\$ @Funkyguy It stores a charge differential. Movement of charge is current. Sorry to disappoint you. \$\endgroup\$ – horta Aug 8 '14 at 2:40
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    \$\begingroup\$ @Funkyguy I don't see a need. It stores the charge differential. The circuit cares about the current it provides by releasing the energy. Getting into specifics confuse people just starting to understand the basics of how to use a capacitor. \$\endgroup\$ – horta Aug 8 '14 at 2:43
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    \$\begingroup\$ @Funkyguy Please feel free to post your own answer. That's what this forum is for. I provided my answer, others can provide theirs. Better answers are voted up. You've obviously stated that you don't like my answer. Furthermore, the OP asked how a capacitor is used, not how the EM fields work in it. \$\endgroup\$ – horta Aug 8 '14 at 2:46
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    \$\begingroup\$ @DanD. Voltage will droop, resistance dropped. Two different things. Let's keep it productive here... en.wikipedia.org/wiki/Voltage_droop \$\endgroup\$ – horta Aug 8 '14 at 3:28

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