Beginner questions about bypass capacitors and power supplies - Page 1
Forum section: electronics / beginners

author: brendan-petrol
author_rank: Contributor
timestamp: August 07, 2012, 04:26:37 am
content: Hello!I have two questions regarding bypass capacitors and power supplies.Firstly: bypass capacitors!- In a DC system, a fully discharged capacitor will act as a short circuit. Is this correct? If so, how does that work with bypass caps, which span the power rails from the VCC to GND. Wouldn't this put too much load on the power supply as it starts?- How do I pick values for bypass capacitors? I've seen all the way from 0.1uF to 10uF used as bypass capacitors!Power supplies:- I would like to embed some regulation into a project I am about to begin. However, I would prefer to use something more efficient than a linear vreg. I began looking at a DC-DC switching regulator, however the recommended circuit (at least, for one) is a lot more complicated than I had imagined. In the Datasheet (http://datasheets.maxim-ic.com/en/ds/MAX730A-MAX744A.pdf) for one that I found on DigiKey, it has components like an inductor, etc. Do all DC-DC switching regulators require this? Are there any different types of inductors that I need to choose from? And how do I work out what values to use for the capacitors?Thanks,Brendan

author: Things
author_rank: Regular Contributor
timestamp: August 07, 2012, 04:47:42 am
content: Quote from: brendan-petrol on August 07, 2012, 04:26:37 amYes and no. For smaller power supplies, the capacitors are small enough values that they charge up very quickly, and don't really have much of an effect apart from maybe making the lights dim for a split second. On larger power supplies, they use what's known as "inrush limiting", which is usually a high power resistor and a relay. When you first turn on the power, the caps charge up slowly through the resistor, and then when they have charged enough to allow the PSU to run, the relay activates and bypasses the resistor.Capacitors are not dead shorts in a sense, in fact if you try measure them they'll appear basically open, this saying comes about because capacitors will try suck as much power as they can get as they charge up, basically acting like a short on the power supply until they are charged.Quote from: brendan-petrol on August 07, 2012, 04:26:37 amI'm probably not the best person to answer this, but usually their value is not critical at all.Quote from: brendan-petrol on August 07, 2012, 04:26:37 amYou will find that any decent power DC-DC converters will need external inductors. The inductors are the actual power handling components, and they can not be made small enough to embed right into the chip, at least for decent power handling.There are a few things you need to look for in an inductor. Of course, the inductance, but also the power handling, and the form factor. Apart from that, not really. It will change depending on the frequency you are working with, for example, ferrite core inductors will not perform well at all at low frequencies, such as mains.To work out the capacitor values, you need to read the datasheet. They will give you all the formulas needed to work out the component values. Dave/EEVBlog also has a youtube video on using a neat little DC-DC chip, you could use that if it's ratings are suitable to your project.Dan

author: JuKu
author_rank: Frequent Contributor
timestamp: August 07, 2012, 05:18:53 am
content: Quote from: brendan-petrol on August 07, 2012, 04:26:37 amNo. A capacitor is a short for AC, but open circuit for DC. And that is why bypass caps work, they short out the (AC) peaks on the supply rails, but leave the clean DC component for the circuit.QuoteYou'll will know when the values are critical (high speed circuits, picky ICs, need to pass EMI tests). But this is beginners forum: use 100n for every IC power pin, mounted as close to the pin as you can. The ground end is important, too. Use 10uF every few inches and one 47uF or 100uF where the power wire comes in to the circuit board. With this, you should be fine with any beginner level circuit.

author: brendan-petrol
author_rank: Contributor
timestamp: August 07, 2012, 05:35:57 am
content: Quote from: JuKu on August 07, 2012, 05:18:53 amAh, okay. I thought a fully discharged capacitor will act as a short, and the resistance would increase as the cap is charged - That's the impression I got from the other guy!

author: Things
author_rank: Regular Contributor
timestamp: August 07, 2012, 05:45:58 am
content: Quote from: Things on August 07, 2012, 04:47:42 am

author: T4P
author_rank: Super Contributor
timestamp: August 07, 2012, 02:56:16 pm
content: The key thing is Capacitors Block DC by charging up, but they allow AC to pass through ( they appear as shorts to AC ) that's why testing capacitors for ESR involves a AC source in a LCR Meter

author: billclay
author_rank: Regular Contributor
timestamp: August 07, 2012, 07:17:40 pm
content: Quote from: brendan-petrol on August 07, 2012, 04:26:37 amHere is an application note from Intersil about the topic:http://www.intersil.com/data/an/an1325.pdf

author: free_electron
author_rank: Super Contributor
timestamp: August 07, 2012, 09:03:37 pm
content: whoa... the truth is being trodden on in this topic like no other.first: Capacitors do NOT conduct electricity ( apart from leakage )An ideal capacitor is two pates separated by a nonconductive mediam called the dielectricum. Can be air, can be mica , plyethylene foil , polyester foil , ceramic or an oxide ( electrolitics and tantalums are oxide-based caps )When you stuff one of the plates with more electrons than the other you are 'charging' the capacitor. You create a potential difference between the plates. ( more electrons one one side and a lack of electrons on the other )While you can pull electrons from one plate and stuff them on the other plate it is important to note that the electrons do NOT run through the capacitor (the dielectricum)!Capacitance is determined by and proportional to the surface of the electrodes exposed to each other and inversely proportinal to the distance between them.- Make the plates larger and capacitance increases. make em smaller and capacitance decreases.- place plates closer together and capacitance increases , pull further away and capacitance decreases.There is another factor at play ( actually two) which is the dielectric constants, one from air and one from the dielectricum used. not important for the principles governing the caps.For DC capacitors cannot conduct any current ( in the sense of electrons flowing through the dielectricum ). For AC this stays in effect. Electrons can still not flow through the dielectricum. But if you measure the current flowing in and out you will see it. it is the current of electrons charging and discharging the plate. the electrons do not physically flow 'through' the capacitor.if i make a setup with an AC source , two capacitors and a load resistor you will see a current flow on both sides of the capacitorsCode:source-->---||->--LOAD---||->--sourceIs Ilyou will see an Is (I source )equal to Il (I load) but there is not a single electron that makes it from the source to the load, as on both sides there is an isolator in the way in the form of the dielectricum in the capacitor. al you are doing is applying an electrical field on the outside of the capacitors (the plate connected to the source ) . this field displaces electrons on the insides ( the plates connected to the load ) of the capacitors and pushes them back and forth. but they do not run THROUGH the capacitor.Capacitors can thus be used to make galvanic isolation between systems. And that is their function as COUPLING capacitors. like in coupling AC signals from one system to another.now, when we talk about 'decoupling' we are actually talking out of our 'behind'.. what we are doing is coupling the positive rail to the negative rail of a system.If a device decides to draw a peak current the capacitor will supply electrons (i'm using conventional electron flow here in this explanantion, not real electron flow , the principle is the same only the sign changes) from its charged plate to the device. this creates a point of attraction on the other plate ( the one connected to the ground ) and this will attrack electrons there. So in a sense the capacitor is 'helping' current flow through the device : by feeding electrons to the top , and pulling them out of the bottom. When the surge is gone the capacitor will re-balance itself by replenishing electrons from the source and spitting them out into the ground from its , now, charged bttom plate.So the capacitor is actually coupling the power to the ground rail. it is de-coupling the load. but it is coupling the power rail.How to dimension capacitors.the larger the capacitor the longer it takes to reach full charge. so , for fast signals , large capacitors do zilch.. they don't react fast enough.So this is why you see all these different values used for decoupling...Up to a 10MHz system a 0.1uf ( 100nF ) performs adequately. Hit 100Mhz and it becomes drab .... you need to go to 10nF.1/(2 PI F C) gives you the equivalent 'impedance'. for a CPU running at 1GHz .... 10nF does zilch...So why don't we put 1pF everywhere ? than we can run at terahertz range ?well, because such a small capacitor can only hold so many electrons... if your peak current is high the capacitor will be out of 'oompf' before the peak is over ...and that is why every device gets its own capacitor. each 'small' cap supplies peak to the local device. the bigger caps scattered aroudn the board then have time to 'kick in' and supply a bulk current. so that is why you see a 10uF in paralle with a 100nF. the 100nF react very fast and covers the transient. the 10uf is a bit slower to start feeding current but that's fine, there's just enough in the 100nF to get the ball rolling.Now, if you look at actual capacitors ( we are stepping out of the 'ideal' world now) there is other factors at play. ESr ESL , soakage , leakage , charge dispersal and many other 'problems'. ESr is bad as it dissipates energy in the capacitor and makes it hot plus causes a voltage drop across the ESR. the same for the ESL. Because of all these 'additional' parameters , capacitors actually drift tremendously. Even the DC voltage across them causes them to drop value.a 10uF , designed for 10 Volt capacitor in 0805 package made from X7R material will be 10uF at 0 volts. Put 5 volts on it and you have about 9uF left. put 10 volts on it and you have about 8uf. heat the thing upand it goes to 7 uF , cool it down and it goes back up.Apply an AC signal to it of 100MHz and it becomes a friggin inductor !Take another 10uF , 10 volt specced cap in 0805 made from Y5V or Z5U material ,put it at 70 degrees C ambient and apply 10 volts and you are lucky if you have 1uF left ....Temperature, dc bias ,frequency all have an impact.That is why serious capacitor manufacturers supply the actual 'waterfall' curves. TDK , epcos, murata and others have sotware ( free) where you can select the exact model of capacitor , and they will show you how it behaves. that software can help you pick the 'right mix' of capacitors. you tell it the peak current , rise times and peak duration, ambient temperature etc and they will advise which 2 or 3 or 4 caps to put in parallel to get to a system that covers your requirements.a schematic that is just full of 100nF scattered everywhere may have worked 30 years ago when we had slow rise times and slwo clocks... today ... not so. It is mindboggling to see people make a design with a fast cpu ( like a cortex at 120Mhz ) and still slap on the three 100nF and one 10uF bulk cap. that's just an indication they don't understand what they are doing. they know the equasions but fail to apply them. Because we've always used 100nF ....and i haven't even touched on where to place the capacitors in respect to the load and on how to connect them ... even the placement of the via (or better via's) is important. Keep in mind you are coupling the rails ! so it is better to plunk the capacitors straight into the planes than going first to the component pin and then into the plane ...