How does decoupling capacitor provide necessary current with minimizing voltage drop of power line?

Thread Starter

Dong-gyu Jang

Joined Jun 26, 2015
115
Hello.

I guess decoupling capacitor has two aspect of its working principle so

1st is that it becomes shunt path from the power line to ground for high frequency noise. 2nd is that it is local energy storage to the active circuit like IC chip near to it so it provides necessary current when required but power supply is slow so can not response briefly.

I guess the capacitors plays both roles at the same time.

1st is easily understood. But I feel hard for 2nd. In order to draw current from the capacitor, according to I(t) = CdV(t)/dt, I(t) should be negative so is dV(t)/dt, means voltage must firstly be dropped. Is it conversational to 2nd aspects? In order to provide current, voltage drop on power line is inevitable?

Maybe if C is large enough compared to required current, then voltage drop rate will be minimized and it is what 2nd aspect actually want to say.

Could you please clarify my confusion?
 

wayneh

Joined Sep 9, 2010
16,398
In your mode 2, the cap is providing local power to the IC when that IC has a need for spike in current, for instance when switching an output. It reduces the noise that would otherwise be placed onto the power rail.

You might say it has a single function to reduce noise from two sources, both the power rail to the IC and vice versa.
 

Lestraveled

Joined May 19, 2014
1,946
Another way to look at what a capacitor does, is its impedance or Xc. The higher the frequency of the noise, the lower impedance the cap will present to counter/reject the noise.
 

crutschow

Joined Mar 14, 2008
25,116
..............
1st is easily understood. But I feel hard for 2nd. In order to draw current from the capacitor, according to I(t) = CdV(t)/dt, I(t) should be negative so is dV(t)/dt, means voltage must firstly be dropped. Is it conversational to 2nd aspects? In order to provide current, voltage drop on power line is inevitable?

Maybe if C is large enough compared to required current, then voltage drop rate will be minimized and it is what 2nd aspect actually want to say.
................
Yes, anytime the capacitor is delivering current its voltage has to drop.
And yes, you make C large enough so that this voltage drop is acceptable for the circuit voltage and current requirements.
 
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