I have a power source which supplies 1A at 5V. I am using a switching regulator which supplies 1.3A at 3.3V. The load requires a burst of 2A without reducing the voltage across it than 3.2 V for 477 usec after every 5 minutes. The normal operating point is 500 mA at 3.3 V.
I want to use a capacitor to provide that extra current for the required duration without reducing the voltage.
How to choose proper capacitor value with low ESR?
Electronic – Deciding capacitor value to use in the circuit
capacitordecoupling-capacitorpower supplyswitch-mode-power-supply
Related Solutions
The most likely reason why that is done is because, in real life, capacitors do not have infinite bandwidth. Generally, the higher the capacitance of the capacitor, the less it will be able to react to high frequencies, while small-valued capacitors react better to higher frequencies, as seen in the graph below. Using two different-valued capacitors together is just done to improve the response of the filtering.
Though it depends on your specific caps, all the very low leakage caps I have bought for precision low drift integrators (time constants of 1 second or more) were metal polyester film. They were also very stable over temperature compared to ceramic disk, which often stated +80 -20% over the commercial temperature range.
They are good for high frequency as well and the ones you show are quite high manufacturing quality. Just don't use the physically big ones to bypass. The leads will be too long or the distance from the part will defeat the purpose.
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Best Answer
You start with $$ Q = CV $$ and differentiate both sides, $$ \dfrac{dQ}{dt}= C\dfrac{dV}{dt} $$, which can be represented by $$ I = C \dfrac{dV}{dt}$$, rearranging variables you get $$ \dfrac{I}{\frac{\Delta{V}}{\Delta{t}}} = C $$
Using your values from above, $$ \dfrac{2 [A]}{\frac{(3.3 - 3.2)[V]}{477 [us]}} = 9.54 [mF] $$