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Support Forum » Question about Caps that came with the kit

August 08, 2011
by Alexp
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So on page 27 it says:

"Inside your parts bag, you should see three yellow capacitors. The largest one is 0.1uF, or 0.1 microFarads. "

But in my parts kit it looks like the smallest one is 0.1uF. I have 3 caps, 2 are identical and say "22" on them (I assume 22uF?), and the smallest one has lettering too small to read. Am I correct in thinking that this smallest one is the 0.1uF?

Also, a totally offtopic question, but how do you determine how much bypass capacitance that you need? Is it a function of the clock speed?

August 08, 2011
by Rick_S
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Actually I think they are talking largest in value not physical size. The one's labeled 22 are actually 22 pico farad or 0.000022 microfarad. The .1 microfarad cap would be 100,000 pico farad.

I'm not too sure on the answer to your second question, but I've seen Noter (another NK member) talk about this. He may chime in and help with that one.


August 08, 2011
by Ralphxyz
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I believe [quote] how much bypass capacitance [/quote] is determine by how much "noise" your circuit/application can tolerate.

The bypass capacitors furnished in the Nerdkit do the job of keeping the breadboard reasonable quite.

You could experiment with other values and even try not using them, to see what happens.

I believe you would need a oscilloscope to see what was really happening.

There are various "recommendations" on the web, but it really is determined by what your circuit needs.

Which of course leads to the next logical question of:

What is to much bypass filtering?

If I put X capacitor in my circuit as a bypass capacitor will it hurt or help?

There has to be formulas to figure all of this out but probable not practical to spend to much time on it as you are just learning

the basics and using a breadboard.


August 09, 2011
by mongo
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Bypass capacitors are generally small in value because they are more apt to shunt the high frequency noise more readily. A lot of it depends on how much noise there is and what ranges it happens to be in. Most of them that I have seen in use are .01 to .1 uF.

I have used small electrolytics for similar purposes, but not to stop noise so much as to reduce power fluctuation. I did that on the servo circuits, as I noticed that when the servos would run, it would cause a momentary drop in the supply voltage and made things act funny.

August 09, 2011
by bretm
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How much bypass capacitance do you need? Let's try to take some of the mystery out of that question.

The MCU contains thousands of switches. As long as they're not switching, they leak almost no current. But when they switch from one state to another, they draw a lot of current (relatively speaking).

The power supply is far, far away, and those long wires are inductive. So when the MCU ticks over, which is does 14745600 times per second, the sudden current draw causes the voltage supply lines to sag. That's why the capacitor should be as close as possible to the MCU, with short, widely-spaced leads--to minimize inductance.

So it's the MCU itself that is causing the noise. The bypass capacitor decouples the MCU part of the circuit from other parts of the circuit so that the MCU doesn't mess with the power supply for the rest of the circuit. It also provides a local supply for the spikes of current that the MCU demands every clock cycle.

How big should it be? As small as possible, to minimize inductance. As big as possible, with as small an ESR as possible, to supply these current spikes.

Is 100 nF the right amount? Well, a farad is one coulomb per volt. Let's say we want to limit the voltage sag to 10 mV. Let's also estimate that switching time is 10 ns (total clock cycle is 68 ns, and switching happens faster than that). So we have 100 nF = current x 10 ns / 10 mV. Therefore 100 nF is enough to supply 100 nF x 10mV / 10 ns of transient current. The 10's cancel out, the nano's cancel out, so that's 100 mA. This is in the ballpark of what the MCU can source or sink, so 100 nF is in the right ballpark for a bypass capacitor.

August 09, 2011
by Ralphxyz
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That is just for the MCU, if you had other components in/on your circuit like 5 transistors or even a bunch of LEDs, everything contributes to the noise or to the voltage sag.

I think "experiment or investigate" is àpropos.

Thanks bretm that is once again an understandable, concise explanation.


August 09, 2011
by Alexp
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Thanks guys, that is a solid discussion of bypass caps!

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