2
\$\begingroup\$

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

| improve this question | |
\$\endgroup\$
  • \$\begingroup\$ most likely "because the IC datasheet says so" \$\endgroup\$ – PlasmaHH May 16 '18 at 20:43
  • 1
    \$\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
  • 1
    \$\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
  • 1
    \$\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
2
\$\begingroup\$

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.

| improve this answer | |
\$\endgroup\$
0
\$\begingroup\$

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.

| improve this answer | |
\$\endgroup\$
  • 1
    \$\begingroup\$ You've got V and Q backwards in a derived formula. It should be C = Q / V instead. \$\endgroup\$ – PF4Public Aug 6 '19 at 17:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.