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I am on a chapter of my electronics beginner’s book where the author talks about the PICAXE microcontroller. One of the first things the author mentions in this chapter is to make sure that the PICAXE microcontroller power source capacitor configuration looks something like this:

enter image description here

Right of the bat this configuration looks strange to me because of the need for using two different types of capacitors in parallel. My first thought was why not just combine the two capacitors and use one bigger one instead?

Also, interesting to note is that the bigger capacitors sit further away from the LM7805 than the smaller capacitors. Why not place them the other way around? There must be a good reason for this.

Could someone offer an explanation for this awkward capacitor configuration need?

Thanks.

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  • \$\begingroup\$ "Decoupling capacitor" is the search term. \$\endgroup\$
    – Eugene Sh.
    Jul 14 '15 at 14:51
  • \$\begingroup\$ I would say the placement of the caps is plain wrong, but having multiple ones is answered multipled times already, it might even be on the right unter related questions. \$\endgroup\$
    – PlasmaHH
    Jul 14 '15 at 14:51
  • \$\begingroup\$ Please see this answer to a somewhat-related question. It should help with some of the reasoning. \$\endgroup\$
    – bitsmack
    Jul 14 '15 at 15:24
  • \$\begingroup\$ The small capacitors are decoupling capacitors and the big ones are bulk capacitors \$\endgroup\$
    – Matze Strawberrymaker
    Jul 14 '15 at 15:51
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Indeed it's a pretty horrible layout, though almost anything is acceptable with a 7805 and a digital circuit.

The ceramic disk capacitors (last-century single layer version of today's ubiquitous MLCC chip capacitors) have lower impedance at high frequencies than the aluminum electrolytics (which have a certain amount of inductance and ESR), however they are limited in value so they can't cover the whole frequency range. So you get a lower overall impedance (and some damping) at a lower overall cost by combining two different types and values of capacitors (a 100uF ceramic disk capacitor would probably be the size of a dinner plate).

The ground trace layout is the most egregious- there should be a star connection at the input filter capacitors. 100uF on the output is unnecessarily large, but it won't hurt anything (a 100uF/10V cap is pretty small physically). 100uF on the input may or may not be sufficient.

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  • \$\begingroup\$ Given the formula Xc=1/2πfC, the reactance of a 100 µF capacitor at any given frequency is much lower than the reactance of a 0.1 µF. So in theory, this should mean that a 100 µFcapacitor should be able to better filter higher frequencies than a 0.1 µFbecause its reactance is lower which means that it would shunt those higher frequencies to ground better right? But based on what you are saying, it looks like this formula really only applies when comparing the same type of capacitor because different types of capacitors have more or less inductance and ESR skewing the formula. Is that correct? \$\endgroup\$
    – T555
    Jul 14 '15 at 16:03
  • \$\begingroup\$ Yes, that's correct. If you look at the data for an inexpensive 100uF/10V capacitor it might have an equivalent of 4 or 5 ohms ESR, so the ceramic takes over at ~400kHz. Since the output impedance of the 7805 is < 5 ohms at lower frequencies you can pretty much say the 100uF on the output is virtually worthless, but if it makes you feel good.. \$\endgroup\$
    – Spehro Pefhany
    Jul 14 '15 at 16:51
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The capacitors are carrying out different functions. An unregulated supply may be quite ripply, so the high-value electrolytic capacitors are there primarily for charge storage - to smooth out the DC. However, electrolytic capacitors have quite a high impedance to high-frequency signals - this is not just down to the capacitance, but also the inductance and resistance of the materials they are made of.

The low-value capacitors hold little charge, but have a lower impedance to high-frequency signals. This can help keep the supply stable while digital circuits are switching (which may only take nanoseconds). They also effectively short out any noise on the power lines noise, giving a cleaner supply and reducing electromagnetic interference.

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There's a small value cap and a large value cap.

The large one filters some frequencies but not all, it is not so good at filtering high frequency noise.

That's where the small value cap comes in, it takes care of filtering the high frequency noise.

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  • \$\begingroup\$ I thought about this but then I plugged capacitance values on the formula Xc=1/2πfC and saw that a higher capacitor capacitance yields a lower reactance value. Would this not mean that the 100 microfarad filter higher frequencies than the 0.1 microfarad capacitor making the 0.1 microfarad capacitor pointless? What am I missing? \$\endgroup\$
    – T555
    Jul 14 '15 at 15:06
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    \$\begingroup\$ What you are missing is that IDEAL capacitors do not exist ! There is always a bit of (parasitic) series resistance and inductance in a capacitor making it non-ideal. So instead of Xc=1/2πfC you get Xc=1/2πfC + 2πfL + R. This L and R have a value depending on the value and type of capacitor C. Have a look at the practical impedance of some capacitors here: murata.com/en-global/support/faqs/products/capacitor/mlcc/char/… What happens above 1 MHz ? Xc goes UP with frequency !!! A 100uF electrolytic cap as in your example will be even worse ! \$\endgroup\$
    – Bimpelrekkie
    Jul 14 '15 at 19:39

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