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?
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?