Most ICs have large capacitors to ground tied to power supply, why?

Question

i know its to filter noise on the power input, but what other reasons are there to have this cap to ground ?

Answer 1

To locally filter higher frequency noise or power disturbances to ground. At the positive terminal of the cap you should have a clean DC level. The reason you want it to go to ground is because if you didn't then you'd be transferring the ripple to your power rail, and then there'd be no point.

Answer 2

Your question is stated incorrectly.

Most IC's have very small capacitors located as close as is practical between the IC's power and ground terminals. These are called decoupling capacitors and they make the IC much more resistant to local voltage transients, by insuring that the high frequency noise generated by other nearby digital components is not coupled into the IC's power inputs.

All linear power supplies utilize large capacitors to store electrons and thus to filter 120 Hz noise from their DC outputs. A type of power supply called "switching" operates at much higher than line frequency and uses medium sized capacitors to filter its output to pure DC.

Another good reason to have a large filter capacitor in a stereo amplifier is that it provides "headroom." Stored energy is available for loud transient tones that would've been clipped (distorted) had the extra capacitance not been available.

Some very large computer power supplies have enough energy stored in extra capacitance that, in the event of a power failure, there is enough energy to allow the computer to perform its critical "shutdown operations" before the capacitor is fully discharged.

Answer 3

ICs generally run at milliamps of current (at 5V, or 3.3V. etc), but when an output switches from low-to-high, the output driver must supply extra current, sometimes 10's of milliamps. If the IC is, say, an octal buffer, and all outputs switch simultaneously, then thats possibly 100's of milliamps of current.

If the switching rise-time is 1 nanosecond, that's a change in current (di) over a change in time (dt) of 10^8 Amps/second! What power supply can handle that kind of dI/dt?? answer: none that I know of. Even the smallest inductance in series with the power supply pin of the IC will produce huge voltages (L* di/dt) at 10^8A/s

So how does the system handle this? By placing capacitors near the power pin of the IC, inductance is cut way down, so voltage drop from the "main" power supply is minimized, you don't have power supply droop.

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Answer 4

In DC power supplies, capacitors are used smooth out the rectified direct current so that it's a more constant voltage.