How can i calculate decoupling capacitor value? Which amount of noise we can filter use it.why use here 10uf , 0.1uf capacitor?enter image description here

  • \$\begingroup\$ most likely "because the IC datasheet says so" \$\endgroup\$ – PlasmaHH May 16 '18 at 20:43
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    \$\begingroup\$ Is your symbol correct? Usually Vdd and Vss would be positive supply and ground respectively. \$\endgroup\$ – Colin May 16 '18 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\$ – Richard the Spacecat May 16 '18 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\$ – Richard the Spacecat May 16 '18 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\$ – Richard the Spacecat May 16 '18 at 21:49

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.


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

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

Q = I * T

Q = 1amp * 1nS = 1 nanoCoulomb

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

C * V = Q,

C = V / Q = 1e-3 volts / 1 nanoCoulomb = 1microFarad.

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

  • \$\begingroup\$ You've got V and Q backwards in a derived formula. It should be C = Q / V instead. \$\endgroup\$ – PF4Public Aug 6 at 17:56

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