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How can I calculate decoupling capacitor value? Which amount of noise we can filter use it. Why use here 10 µF, 0.1 µF capacitor?

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  • \$\begingroup\$ most likely "because the IC datasheet says so" \$\endgroup\$ Commented May 16, 2018 at 20:43
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    \$\begingroup\$ Is your symbol correct? Usually Vdd and Vss would be positive supply and ground respectively. \$\endgroup\$ Commented May 16, 2018 at 21:17
  • \$\begingroup\$ @Colin__s checked the datasheet, turns out it says Vs in there, not Vss, meaning Supply Voltage. OP, you should totally fix that symbol, it's confusing. :) \$\endgroup\$ Commented May 16, 2018 at 21:46
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    \$\begingroup\$ Also, a quick Eagle tip: If you want to write signals as negated, you could write them prefixed with a !, such as !CS. If you say want to negate only a part of it, you can quit negation with another ! such as !IO!/MEM. \$\endgroup\$ Commented May 16, 2018 at 21:48
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    \$\begingroup\$ Another quick Eagle tip: If you want to have multiple pins named the same in a part, add @somerandomnumber at the end, such as NC@0 and NC@1. The part of the name after the @ would be invisible in the schematic. \$\endgroup\$ Commented May 16, 2018 at 21:49

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It's not for amount of noise. It's to supply current (the big one) and to react quickly (the small ones). So to size the big- consider current. Analog circuits, like sensors and stuff, would want 1uF and more. Digital- depends, but normally you put 0.1uF on each VCC input. For really fast stuff it would also be 1nF for low ESL.

In most cases you will not feel a difference. When you will, it will take long hours to understand why the hell your circuit behaves weird way.

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Suppose you have 1 amp surges, every 1 microsecond. What size Cbypass to use?

What if the 1 amp surge exists for 1 nanosecond. What amount is charge is consumed during that 1 nanosecond? \$Q = C V = I T\$

\$Q = I T\$

Q = 1 amp * 1 ns = 1 nanocoulomb

So what? Can your circuit tolerate 1 millivolt of VDD sag?

\$C V = Q\$

\$C = Q / V\$ = 1 nanocoulomb / 1 millivolts = 1 microfarad.

Rinse the math, bring in your own assumptions, and repeat the math.

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    \$\begingroup\$ You've got V and Q backwards in a derived formula. It should be C = Q / V instead. \$\endgroup\$ Commented Aug 6, 2019 at 17:56

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