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Why have a small neon lamp in power supply input circuitry? A couple of times when taking apart older bits of electronics, I've seen a small neon lamp about the size of a fuse (but its definitely not a fuse) positioned near the power supply circuitry. What is its purpose? Is it used as some kind of input protection? Does it illuminate under fault conditions? Why not use a MOV or some other purpose designed component? <Q> It is used as discharger for overvoltage conditions - in case of overvoltage a discharge starts through the lamp and that protects the main circuit from overcurrent. <S> A neon lamp is used because it is relatively cheap, very reliable <S> and there's zero current through the lamp until the discharge actually starts. <A> Another perspective would be for safety during troubleshooting. <S> From the venerable Bob Pease: <S> So, whenever I start work on a high-voltage circuit, I tack in a neon lamp in series with a 100k resistor across the high-voltage busses. <S> Then when I see the neon's glow, I'm graphically reminded that this really is a high-voltage circuit, and that the power is still ON <S> (I don't care what the power switch says) and I should revert to the mode of High-Voltage Cautions. <S> If I grab onto a really hot wire, the shock might not injure me, but I might convulse and jerk backwards. <S> That's not a good idea if I'm standing on top of a ladder, for instance. <S> So, looking for the glow of a neon lamp is a way to remind me to be serious, and I recommend it for you, too. <S> From http://electronicdesign.com/electromechanical/whats-all-reflex-response-stuff-anyhow <A> On some old electronics (especially during vacuum tube era) the neon bulb was put in parallel with a fuse and used as a blown fuse indicator.
| Other times it was merely a power applied indicator (pilot lamp).
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Capacitor on USB cable? I was messing around with a USB camera. I wanted to change the cable because is trimmed but as I was cutting it I saw a small thing (image below). As I was trying to cut the casing and see what it has inside I saw a spark. I tried to connect it with another cable but I don't know if I'm gonna damage the camera or the USB port. What is it and can I connect the camera without it? <Q> I guess you're talking about the ferrite bead : <S> Source: <S> Wikipedia - Ferrite bead <S> Its purpose is to prevent interference at high frequencies signals. <S> It may also work without it and with a cable as short as possible. <A> It is a ferrite bead and acting as a choke, or very small valued inductor. <S> It is acting on all the conductors of the cable together, so it is blocking common-mode noise from entering or exiting the connected device. <S> The USB connection will absolutely work fine without the ferrite bead present. <S> USB cables are normally sold without it. <S> It was likely packaged with your device because the device was found to be emitting EMI, and the cable was determined to be the source of the emission. <S> It was cheaper to add a choke to the cable than to do any additional work on the design of the camera itself. <S> A single ferrite bead acts as a low-pass filter, but it rolls off at frequencies well above the highest frequency used by USB. <S> Furthermore, the effect is applied only to common-mode signals. <S> The USB data is on a differential pair, and both conductors are passing through the same core, along with the power conductors. <S> So the USB data will see very little effect from the bead, while EMI will be blocked by the bead. <S> Lots more than you want to know can be found by reading data sheets. <S> Here is some documentation from one maker. <S> Discussion of common mode starts around page 8 in that PDF. <A> <A> It's a ferrite bead choke , which is used to suppress unbalanced high-frequency signals, typically caused by EM interference. <S> It is essentially a form of passive low-pass filter . <S> CCD sensors are often quite susceptible to noise. <S> Bottom line is that, while the USB connection will probably work fine even without the filter, <S> any pictures that you take while connected to the PC may suffer a degree of degradation in quality.
| Imaging devices such as scanners and cameras often have such filters on their USB cables because otherwise the EM interference from the cable may introduce noise in the sensor and, thus, the stored image. Its a ferrite bead , designed to suppress high frequency noise on the cable. You can use it without it, but you might get issues with USB data transfer.
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Turn on a 7812 Voltage regulator using an Arduino? I'm using an Arduino in a project, but this is not an Arduino question! I'm able to switch 5+ volt high or low using a digital write pin. I need to use a 7812 voltage regulator to regulate voltage coming in to some other power transistors, but it does not need to be on at all times consuming power and creating heat. I'm wondering if I can use an NPN transistor to turn on/off the GND to the 7812. Is it just as simple as hooking up the NPN to the 7812? Also, will I need a high power NPN since the 7812 will be drawing upwards of 1.4 amps? Is there a much better way to switch the 7812 on and off? <Q> I dont think there is really a clean safe way to do that, you could instead chose other regulators in the market which have a specific input pin to enable/disable the regulator like this one for example, http://www.linear.com/solutions/1703 . <A> You could switch the voltage into or out of the 7812 using a MOSFET or BJT high-side switch. <S> The latter choice means the regulator will continue to draw a few mA even when the output is off. <S> Don't muck with the GND pin. <S> You'd need a couple transistors (one power) a zener and few resistors. <S> If you want a single-chip solution there are very few choices in through-hole linear regulators with shutdown. <S> The LT1764AET#06PBF is only good for 20V <S> (abs max) input. <S> If that's okay, it should work for you- <S> the /SHUTDOWN input is compatible with your control signal. <S> Be sure to pay attention to ALL the information on capacitor ESR, they are not as easy to work with as the 78xx regulators. <S> You'll also need a couple of resistors to set the 12V output voltage. <S> If you don't mind a SMT TO-263 package, the TL1963AKTTR is about half the price and has similar limitations and requirements to the LT1764AET#06PBF. <S> It's stable with low-ESR ceramic capacitors, without any series resistor so it's a bit more user-friendly than the LT product. <S> Again, pay attention to the capacitors, particularly the output capacitor, application information for these (and all LDOs and negative regulators). <S> They're not like 78xx regulators. <A> The current through the ground pin of a 78xx regulator is under 10 mA, so you wouldn't need a high-current transistor there. <S> However, disconnecting the ground pin probably won't turn off the 7812 output - I would expect the output voltage to rise to within a volt or two of the input (but I could be wrong). <S> What exactly do you mean by "turn off the 7812"? <A> By the way, 1.2 amp with a linear regulator would require a beefy heatsink.
| I would use a power pnp transistor, or better yet a p channel mosfet to switch off the regulator.
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Does the position of the Ferrite bead matter? Often I will see a ferrite bead a couple of inches from one end of a cable. Does the position of a ferrite bead really matter? <Q> My answer hopefully applies to beads but also ferrite clamps on cables. <S> If it's preventing susceptibility problems due to interference coming down a cable then its position is not that critical. <S> However, if it's stopping emissions getting out from a device, a better position is as close to the interfering source as possible. <S> There is every chance that interference getting onto a cable (such as from a switched mode power converter) can radiate from the cable so, best to keep the ferrite up as close to the source of noise as much as possible. <A> The position is important. <S> There's two cases I can imagine. <S> In the first case, a device is radiating through the cable, and this is unwanted. <S> This reduces the length of the radiating element. <S> In the second case, a device is receiving unwanted signals picked up on a cable. <S> In this case, again, the bead should be as close to the device as possible. <S> The reason we don't use two beads on a cable is that usually it's not required. <S> Either the communication link is unidirectional, only one device is susceptible, or one bead is enough to change the electrical length enough to attenuate the unwanted signal. <A> Beads provide suppression by impeding (developing a voltage across the bead) and absorbing (converting to heat) <S> magnetic fields (resulting from to currents on the wire). <S> Beads provide the most suppression at VHF and UHF current maxima (as-in standing waves). <S> The most likely position for a current maxima for all frequencies is near a chassis, bulkhead or other electromagnetic boundary. <S> Undesired RF energy can form standing waves bilaterally when leaving a chassis or entering it. <S> Placing a bead at a current maxima will shift that maxima to another yet weaker frequency dependent location where a second bead may provide some additional improvement for a particular frequency (test equipment required).
| In this case, the bead should be as near to the transmitting device as possible.
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How to tell if an electronic component is counterfeit? I've been looking into sites (like Alibaba, etc...) for components, but they are usually at the price of the rain. How can someone know if a company or factory is not a scam and that those components really belong to said factory? <Q> The best way to avoid this problem is to use reputable suppliers, like Mouser, DigiKey, Element-14, etc. <S> There is no way I'd ever buy parts from Alibaba. <S> There is a reason some sites have such low prices. <S> Think about it. <S> How can someone sell you small quantities of parts for less than places like Mouser that buy in bulk directly from the manufacturers? <S> These are parts they got as "surplus" from companies that supposedly bought more than they needed (sometimes that's legit, but sometimes these parts were abused or failed tests, which is why the company that bought them is dumping them), knockoff parts made to lower quality standards to cut costs, outright frauds, or even parts that "fell off a truck" in Asia someplace. <A> Except for the obvious it will be very hard to tell. <S> But one of the best ways is to keep an open mind and don't fall into the trap of painting everyone with the same brush. <S> That is the start of stereotyping which leads to worse things. <S> The fact is, if you purchase through Western distribution your are suffering from huge mark ups. <S> The reason why costs in some other locations are much much lower is that there is less collusion and more competition and in some cases counterfeit parts. <S> As a case in point, in ramping production for large volumes we got quotes from around the world (through local vendors/partners) and the price difference was staggering ... <S> and these where the same parts, made in the same factory (In Europe) and drop shipped from the same factory. <S> When faced with the proof that they were trying to fleece us with the North American Quotes they matched prices that the factory gave us else where. <S> So things are not as simple as they might first appear. <S> Certainly there will be bad people, selling bad products, wherever you are. <S> The best strategy is to mix your supply chain, order from multiple locations and let every vendor know what you are doing. <S> For the North America supply chain they will know they have non colluding competition <S> so will sell you parts that aren't as marked up, with your Asian suppliers they will realize that you have direct comparatives for quality and so will not attempt to mix in re-worked parts etc. <S> If you can't do this, then order from the most secure source you know of but know that North American vendors often deliberately mark up prices and can keep them high because of FUD <S> (Fear Uncertainty and Doubt). <S> And in all things, even this is not uniformly applied so do your research. <S> There is no single answer. <A> In Europe, or in the EU to be more precise, there are much larger labor cost variations between countries than there are between US states. <S> And labor costs do affect quite a bit the distribution business mainly through packaging (even if automated, someone has to at least watch the machines) and then shipping (driver's cost etc.) <S> This moreso for small quantities than by the truckload. <S> So a (say) Poland-based distributor can well undercut the more established (typically UK) ones... sometimes the prices are 2/3 or nearly half and no fakes. <S> Some of the Central/Eastern Europe distributors have now been in business for nearly two decades and have built up enough reputation, have websites in all or most EU-member languages (this may sound like a silly metric, <S> but it's fairly expensive to maintain a coherent website in that many languages), have warehouses in several countries (mostly for fast shipping times to all member states), etc. <S> I can give concrete examples, but they may sound like advertisement. <S> The metrics I mentioned are fairly good at discriminating between the wheat and the chaff when looking at EU-based distributors without giving any names.
| Except for blantantly obvious counterfeits where the manufacturer name is misspelled (yes, that has actually happened), it is difficult for you to tell.
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Capacitance between Earth and Moon Is there a capacitance between the Earth and the Moon, and if there was enough potential difference, could a discharge strike occur? <Q> I remember that - in one of his columnes in "Electronic Design" - <S> the late Bob Pease has shown how to calculate this capacitance. <S> Just now I have found an addendum to the original contribution: Here it comes Quotation <S> R.A.Pease : <S> I received a lot of answers after asking the question, <S> "What is the actual capacitance from the earth to the moon? <S> " There were a few odd ones at 0.8µF or 12µF. <S> But about 10 guys said it was 143 or 144µF. <S> They used the formula: $$C = <S> 4x(\frac{l}{r_1} + \frac{1}{r_2} <S> − \frac{2}{D})−l$$ <S> valid for \$r_l <S> , r_2 << D\$ . <S> NOW, my original estimate of 120µF was based on this approximation: <S> The capacitance from the earth to an (imaginary) metal sphere surrounding it, 190,000 miles away, would be 731µF. <S> (If that surrounding sphere were pushed out to 1,900,000 miles away, the capacitance would only change to 717µF — just a couple percent less. <S> If the "sphere" moved to infinity, the C would only decrease to 716µF.) <S> Similarly, the C from the moon to a surrounding sphere 48,000 miles away would be 182.8µF. <S> If the two spheres shorted together, the capacitance would be 146.2µF. <S> I guessed that if the spheres went away, the capacitance would drop by perhaps 20% to about 120µF, so I gave that as my estimate. <S> But removing those conceptual "surrounding spheres" would probably only cause a 2% decrease of capacitance. <S> That would put it in close agreement with those 10 guys that sent in the 143µF figure. <S> But THEN 6 readers wrote in LATER — from Europe — all with answers of 3µF. <S> I checked their formulae, from similar books, in several different languages. <S> They were all of the form: $$C = \frac{4\pi <S> \times \epsilon <S> \times <S> ( r1 \times r2 )}{D}$$ multiplied by a correction factor very close to 1.0. <S> If you believe this formula, you'll believe that the capacitance would be cut by a factor of 10 if the distance D between the earth and moon increased by a factor of 10. <S> Not so! <S> Anybody who used a formula like that, to arrive at 3µF, should MARK that formula with a big X. Finally, one guy sent in an answer of 159µF. <S> Why? <S> Because he entered the correct radius for the moon, 1080 miles rather than 1000. <S> That's the best, correct answer! / <S> RAP Originally published in Electronic Design, September 3, 1996. <A> I believe the answers are 1) Edit: <S> see other answer about Bob Pease 2) <S> There's no theoretical reason why not, but there are a number of practical reasons: <S> It requires a colossal amount of charge. <S> Wikipedia claims the breakdown voltage of vaccum is 20 MV/meter. <S> The moon is 384,400,000 meters from the earth. <S> That puts the minimum voltage at 7,688,000,000,000,000 volts. <S> Where would this charge come from? <S> The "solar wind" contains a constant stream of charged particles moving at speed. <S> On entering Earth's atmosphere this results in the Northern Lights. <S> On encountering a planet with a very large non-neutral charge it will tend to attract opposite charges and repel like charges, gradually reducing the net charge to zero. <A> It is straightforward to calculate the capacitance of any two conductors. <S> Place equal and opposite amounts of charge on each conductor then calculate the voltage between them. <S> By definition, C=Q/V. <S> In the case of the Earth and Moon the calculation is difficult because the charges are not distributed over perfect spheres but oblate spheroids. <S> To a reasonable approximation though we can assume that they are spheres. <S> With this approximation, the electric potential difference is roughly (to about 0.3%) equal to the difference of potential of each body at its own surface. <S> This is a bit strange, but because the Moon is so far away the electric potential of say the Earth at the Moon is very small when compared to the electric potential of the Moon itself. <S> The self capacitance of the Earth is about 709 microFarads and that of the Moon is about 193 microfarads. <S> The effective capacitance of the pair is 1/709+1/193=1 <S> /Ceq, so Ceq=152 microfarads. <S> Again, it is odd that the capacitance between the Earth and Moon is not dependent upon the Moon's orbital radius, but that is the answer. <S> To do this problem exactly requires you to integrate the electric field between the Earth and Moon over any path between them then divide this voltage into the charge that you used to create the field. <S> This will show a small dependence upon separation. <S> As a last comment, this is a nice problem in that it shows that the conductors themselves hold charge and store energy in their respective electric fields. <S> Capacitance must account for all of this energy. <S> Normally, the mutual capacitance dominates as in a parallel plate capacitor with a small gap between the plates. <S> But the capacitance of a parallel plate capacitor, where the plates-size-to-gap ratio is very small, is just the sum of the capacitance of each plate in isolation!
| The mutual capacitance is quite small compared with the self capacitance of the Earth and Moon separately.
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How can I construct my own capacitor? I want to construct my own capacitor, primarily as an experiment in different plate configurations and orientations, so I can better understand field theory. (In other words, I'm playing, and not looking for an off-the-shelf solution. The building of it is half the point.) I've built small PCB units on the order of a few nanofarads. I'd like to build something larger, on the order of 5-10 uF, which seems to make PCBs impractical. I also want the cap to be unpolarized, so I can run AC across it. I'm thinking I need a couple rolls of foil, a roll of some insulator, and some way to attach conductors to the foil. But before I just go to Wal-Mart and buy all their aluminum foil, I wanted to ask: what might I be missing? Are there standard techniques for hand-building capacitors? Are there particular materials I should use (or avoid)? <Q> You need to do some basic calculations first. <S> The formula for capacitance is $$C = \epsilon_R \epsilon_0 \frac{A}{d}$$ <S> Let's say you want to use 1-mil (25.4 µm) waxed paper as a dielectric Note that this will determine the voltage rating of your capacitor. <S> The dielectric strength of waxed paper is about 35-40 MV/m, so this will give you a capacitor that can theoretically handle almost a kilovolt, but be conservative in how you use it! <S> The relative permittivity of waxed paper is about 3.7, the permittivity of free space is 8.854e-12 F/m. <S> Solve for the area required: $$A = <S> \frac{C <S> \cdot <S> d}{\epsilon_R \epsilon_0}$$ <S> $$A = <S> \frac{5 \mu <S> F \cdot 25.4 \cdot <S> \mu <S> m}{3.7 \cdot 8.854 <S> \cdot 10^{-12} F/m <S> } = 3.8767 <S> m^2$$ <S> If you get aluminum foil and waxed paper that's about 12" (30 cm) wide <S> , you can probably get an overlap of, say, 25 cm, which means that you'll need a length of about 15.5 m to get the area you need. <S> If you then roll up your capacitor (using a second layer of waxed paper), the capacitance will be doubled, or about 10 µF. <S> Obviously, this will be physically rather large, over a foot long and several inches in diameter. <A> Welcome to the world of capacitor building. <S> I have built quite a lot of capacitor from stacked to rolled type. <S> The only thing is that I build them for high voltage AC (4kVAC to 15kVAC). <S> I'm not going into the details of physics or ESR etc.... <S> If you are still unsure, just head down to any kitchen department and look for the resealable plastic bag, LDPE are quite stretchy. <S> the ones used for trash and dustbin have loads of contaminants from recycled stuffs so try avoiding them. <S> Usually carpet (yes, carpet) rolls comes bagged in long plastic wrap/tube. <S> They are LDPE as well. <S> Try to avoid high density polyethylene (HDPE) as they tend to rip easier. <S> HDPE is also the same stuff normal used in supermarket plastic carrier bag. <S> If you want to splash some cash then you can try purchasing Poplypropylene (PP) rolls from florist <S> but they are quite expensive. <S> I have no experience on this as I build giant high voltage capacitor at tiny microfarads (0.01 to 0.2uF) and PP would be uneconomical. <S> The thicker the dielectric, the higher the voltage it will resist and the smaller the uF <S> it will be, but for 240VAC, I think you can get away with thin stuff, I dont have the list of dielectric breakdown voltage on hand right now, try googling for it. <S> Some info: <S> http://www.richieburnett.co.uk/parts.html http://deepfriedneon.com/tesla_cap1.html Tesla coil are dangerous but funn at the same time. <A> If the whole point of the exercise is to experiment with configurations that are mathematically tractable for you , probably the simplest is to have two large and very flat circular metal plates, in air, securely anchored at a fixed distance (use standoffs around the perimeter) that is adjustable and won't short out. <S> Are you planning to stick probes between the plates to measure fields, voltages, or anything? <S> From there, you can try various separations, different shapes (square, rectangular, etc.), lateral offsets, putting different dielectrics between the plates, and whatnot. <S> I would suggest sticking to reasonably low voltages. <S> Then you can try rolled-up designs. <S> I'm not sure how the resulting interleaved plates will behave differently than one large pair of plates. <S> Remember the old variable capacitors from radio tuners? <S> Have fun (and stay safe)! <A> For the order of 5-10uF you'll need electrolytic type capacitors. <S> Check for instance <S> http://www.youtube.com/watch?v=LXf3SUQ8F5g <A> You could go Old School with a Simple Leyden Jar
| The best bang for the bucks of all the tried and tested material is low density polyethylene (LDPE), which sounds very technical until you realize that it is basically the same stuff used in resealable freezer bags. You can work out what capacitances you should expect from a reasonably-sized apparatus, and adjust your circuit from that. Foil-covered discs might also work (you might need to glue down the foil).
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Why does my DIY "solar panel circuit" show battery voltage and not solar panel voltage? I've got two solar panels set up in series as shown on the below schema. I connect positive on panel A to positive on battery, and negative on panel B to negative on battery. The two panels produce 9V each, and 18V in series on a good sunny day. That's what my multimeter says. If I disconnect cabling between panels and battery, my multimeter shows that the panels produce about 15V (a bit cloudy today). However, when I connect the cables between panels and battery, my multimeter shows about 12.2V. I believe this is the battery voltage. Why is it that when I connect my multimeter at point A and B it shows 12.2V (when solar panels are connected to the battery). As you can see on the schema, there's a diode in front of point B. Shouldn't that diode make sure that current from the battery not flows back to the panels? If this is how it is supposed to work, how do I check the voltage my panels produce when the panels are connected to the battery. <Q> If you connect the voltmeter leads to the terminals of the battery, you will read the voltage across the battery. <S> There really isn't much more than can be said about that. <S> The battery is essentially modelled as an ideal voltage source in series with a relatively small resistance. <S> The solar panels, on the other hand, are more like a current source in parallel with a relatively large resistance. <S> When the solar panels are connected across the battery, a simple model of the circuit is simulate this circuit – <S> Schematic created using CircuitLab <S> The current \$I_{sc}\$ is the short circuit current produced (for a given illumination) by the panels when the panels are short circuited. <S> The voltage \$V_{oc}\$ is the open circuit voltage produced by the battery when the battery is open circuited. <S> Elementary circuit analysis tells us that the voltage across the battery terminals with the solar panel connected is $$V_{bat} = <S> V_{oc}\frac{R_p}{R_p + R_b} <S> + I_{sc}R_p||R_b$$ <S> Since, typically, the resistance \$R_b\$ is much less than the resistance <S> \$R_p\$, the voltage across the battery is approximately $$V_{bat} = V_{oc} <S> + I_{sc}R_b \approx <S> V_{oc}$$ <A> When you connect the voltmeter between points A and B, you're connecting it directly across the battery, so if the solar panels put out less than the battery voltage the diode will be reverse biased and will, in effect, disconnect the battery from discharging into the solar panels and you'll be measuring only the battery voltage. <S> On the other hand, if the voltage from the solar panels is higher than the battery voltage and the drop across the diode, the solar panels will force current into the battery, charging it. <S> However, because the impedance of the battery is so low, it'll drag the voltage of the solar panels down close to the battery voltage, even though the solar panels will be continually pumping current into the battery. <S> As time goes by and the battery becomes more fully charged, you'll notice that its voltage will rise, but never to your solar panels' open-circuit full-sun value, because the battery chemistry won't allow it. <S> The proper way to monitor your battery's charging is to measure the voltage across it and the current into it, and never let either rise above the manufacturer's recommendations. <A> Solar panels have quite high output resistance and therefore any appreciable load will lower the output voltage of the panels. <S> A lead-acid battery, on the other hand can supply a load of several amps without hardly changing its output voltage. <S> This is largely the same when charging current goes into the battery so, in effect the battery will dominate and the voltage you'll read across it will be largely the same whether solar cells are connected or not. <A> If you put a momentary, normally-closed switch in the loop to the battery side of where you connect the voltmeter, you can read the open-circuit panel voltage while you hold the button. <S> Edit: <A> I've read that I should use a voltage regulator to make sure that I don't provide more than about 13.8V <S> (trickle charge) from the solar panels to the battery Depends on the panel... <S> you don't show them in your photos, but I assume you are using the smaller ~12V panels, and not the more common 60-cell ~35V ones?
| In other words, unless the panels are providing a relatively large current and / or the internal resistance of the battery is relatively large, you will measure approximately the open circuit battery voltage with the solar panel connected .
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Make a custom plastic case I have worked on many electronic DIY projects like MIDIbox ( http://www.ucapps.de/ ), it always was fun and working but at the end, the problem was that I ended with a cool gear, but in a poor casing , for example : wood very roughly cut (by myself!). What are the best solutions in 2014 in order to make a custom plastic case, with labelling, some round and square holes ? Example : Is 3D-printing adapted for this purpose ? Are there other solutions ? As the desired quantity = 1 unit, are there some companies that propose such a service ? <Q> If you're only making one <S> it's entirely reasonable to just order a plastic project enclosure <S> (all sizes are available) and then drill and cut holes yourself. <S> The issue it seems you're trying to overcome is a lack of craftsmanship, you can help alleviate that problem by purchasing the correct tools for the job. <S> The other keys are practice and patience. <S> Literally measure twice, cut once. <S> These ones , for about $30, will fit in to a small drill press (just for the press portion), which you can get for about $130. <S> For drilling precision I use a cross-slide vice mounted to my drill press. <S> This improves the linearity of the holes I drill significantly. <S> So, get yourself some of the right tools and the quality of the project cases you build will be much better. <S> You'll also only pay ~$10 for a case rather than a couple hundred for a one-off. <S> Labeling can be easily achieved by printing a large stick on label (or paper, like this guy <S> image below) <S> that you can cut holes in using the case you've already made. <S> You can also order some custom stickers with common symbols/markings. <S> 3D printing is a good option too, but is also a lot of design work for a single case. <S> Take your time, you'll end up making some very quality enclosures that you can be proud of for years to come. <A> I have used Front Panel Express for custom aluminum panels and boxes. <S> I think they also do some work with plastics. <S> See http://www.frontpanelexpress.com/ <S> There is also an affiliated German company: http://www.schaeffer-ag.de/en/ <A> It's no problem getting a 3D printed part made. <S> There are plenty of service bureaus from crazy expensive to quite cheap. <S> Sometimes even the public library or your local hackerspace . <S> The physical characteristics of the printed material and the quality of the printing vary greatly, and often there's a lot of work done post-printing to make it look more like an injection molded part (smoothing, filling, painting). <S> You pay for some combination of cubic cm of material and cubic cm of work envelope or time on the machine (often mostly the volume of material used). <S> One company with lots of information available online is Quickparts . <S> Designing the part can be done easily if you're familiar with any 3D parametric modelling software. <S> It's way easier than design for injection molding because you can ignore many of the guidelines that are necessary to get a high end part (it is not going to look that great anyway, and many of the rules are related to filling and heat transfer so they don't apply to 3D printing). <S> You can modify existing 3D models (add holes, etc.). <S> I use a popular professional program (Solidworks) which would be considered not high end by many compared to Pro Engineer or Catia, but it's still out of the range of most hobbyists and some small businesses. <S> Perhaps something like Sketchup could be used (some version of that is or was free). <S> There are probably others. <S> You'll want to produce a .stl file for printing. <S> You can also consider laser cutting something like acrylic, which again is pretty easy to arrange, but places significant design limitations on your housing, since it's basically 2D, so a .dxf file can be used to describe the cuts. <S> Printing markings can be done by screen printing (which tends to be a bit messy) or (onto metal) toner transfer methods used for PCB printing. <S> Electrical specifications and fire retardancy may not be guaranteed, so care should be taken if those are requirements. <S> For metal cases, where cost is not too important, Protocase does some nice work, highly professional work including color printing of panels <S> (.AI format files are suitable for the artwork where fonts and colors and precision positioning is important). <S> You could certainly put one of their beautifully cut and printed metal panels into an off-the-shelf plastic housing to get a very professional result at a reasonable price. <A> Protolabs can make machine plastics and metals for reasonable cost. <S> I was able to get good looking aluminum parts for less than 100$. <S> They have an interactive quoting tool online, if you have the CAD files for your design.
| For instance you can buy some corner punches for cutting out nice square holes. Or simply order the case with labels/markings printed directly on to it .
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Position tracking system for frisbee I'm trying to build a smart frisbee for use in Ultimate . I'd like to be able to track the position of the frisbee in a playing field with pretty good accuracy (within ~2ft). GPS is a possibility, but I'm not sure if it is accurate enough. Are there any other technologies that would help, such as some system where I set up transmitters on the four corners of the field and place a receiver on a frisbee? <Q> You may be able to do this with a TrackR like low energy bluetooth tag attached to the center of the frisbee, and then four readers or more readers, along the side lines. <S> The readers would then use the power level and triangulation to determine the position of the frisbee. <A> GPS has a relatively low accuracy for every sample (in the order of 5-10 meters 2sigma [95% of values]) at 10Hz. <S> A frisbee is not slow, so averaging (which reduces the position fix rate) to reduce the confidence interval is not going to make it suitable. <S> Inertial measurement units are much faster and much more accurate but drift over time (also called "random walk", due to the double integration of noise on accelerations). <S> It may be acceptable for you, especially if you reset them regularly using an absolute position sensor or by putting the frisbee at a location which position is precisely known. <S> Data fusion is doing this automatically on-the-fly, that's what GPS navigators do on top of shifting the position to the nearest known road. <S> For accurate absolute measurement, you can use differential GPS (also known as d-GPS) which will require a fixed beacon somewhere on the field and a wireless connection. <S> This can give sub-meter accuracy down to 15cm. <S> Though d-GPS is perfect for you, it's usually complex and/or expensive, so I'd recommend turning yourself to IMUs with at least 6 degrees of freedom. <S> You may want to decouple the system from the frisbee spin (i.e. bearing), in which case you'll need an IMU which integrates a magnetometer to take care of the angular motion of the accelerometers frame of reference while decoupled. <A> You can use ultra wide-band ( UWB ), but I think if you're asking the question it's not something you actually want to get into. <S> Plus a robust frisbee system there could be marketed to disc golf players. <S> It will suffer the precision issues you mentioned. <S> Likely the best solution is computer vision . <S> It won't be easy. <S> If you're willing to add electronics to the disc you may <S> as well place UV LEDs on it and track it with a Wii type system in addition to standard video tracking.
| GPS is a good bet, both in ability to implement and low cost to experiment with.
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Single phase from 3 phase generator I got a good deal on a 3-phase generator, but everything I have is single phase. There's a sticker on it that says 230v 60hz. The output is a 4-wire plug. I did some testing with my multimeter and it seemed I had the following: Green wire: Ground Red, Black, and White wires: Somewhere around 110v I believe (hard to read on the analog multimeter I had handy). I'm used to having 110 house wiring have a hot wire, a neutral, and a ground, so I'm a little confused on what I'd have to do here. Would I wire one of the 110 wires to hot, the green to ground, and a common wire between all of my outlets as the neutral? Second question is what to do if I wanted 220v? Two of the hot wires, green to ground, and common wire between all outlets? And finally, I've read that it's best to distribute load across all the phases. So would that mean if I wanted 3 110v outlets, I'd want to wire one up to each of the hot wires? <Q> Your 1st question, yes and common up earth and neutral, BUT are you qualified to do such work? <S> 2nd question. <S> If you measured the voltage between two hot wires you won't get 220 volts. <S> You should see about 190 volts because line voltage is phase voltage (110 volts) <S> x square root of 3 (1.732). <S> Question 3. <S> I'm thinking about this one because potentially (possible apt word) <S> appliances on one phase may cone into contact with appliances on a different phase and this would cause sparks if not properly insulated. <S> Also, when you are old and infirm and you get an electrician round to fix or modify your AC distribution or ring mains he may be in for a big shock! <A> Are you sure it is three-phase. <S> You should measure about 120 volts between red and white, and between black and white, and about 240 volts between red and black, if it is 120/240 single phase. <S> The white wire would be neutral, and would normally be connected to the green safety ground. <S> If it was three phase, delta, I'd expect 120 volts between red and black, and between red and white, and also between black and white, with no neutral. <A> I think Peter is right, if it was three phase you would have black red and blue, and it would be 208v. <S> It is possible to have a high leg delta configuration (like on older buildings) where the high leg gives you 208v 1ph. <S> You probably have 120/240v, and yes you want to balance your loads between the two phases (split the wattage between your 120v loads on red to white and black to white). <S> This will balance the amperage on your breakers. <S> But like they have said, an electrician should be wiring up these voltages.
| From the wire colours (and from your voltage readings, if I interpret them properly), it should be 120/240 volt, single phase.
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Why harvesting energy why not harvesting power? I am wondering why is it called Harvesting energy and why not harvesting power? Energy is measured in Joules/s and power is Watt but the result is watt. For example if we harvest energy from sun or from wind or from heat or from vibration all of these energy sources gives a output in the form of Volt and Current. So basically what we get is power. So why don't we call harvesting power from sun or from other sources of energy? <Q> Most such systems run intermittently, so a key parameter is how much energy the the application consumes each time it runs (output power times run time). <S> This then determines how often the application can run, because it takes a certain amount of time to collect at least that much energy (input power times collection time) in the energy storage device (capacitor or battery). <A> Power is the rate at which energy is being consumed. <S> An energy harvester stores energy rather than consuming power. <A> Power is the amount of energy in a specified time, usually in the form Joules (energy) per second (time). <S> Let's say you have a 60 W light bulb. <S> That is its power. <S> Let's say you turn it on for a day (24 hours). <S> It needed 60 x 24 = 1440 Wh = 1.44 kWh energy. <S> Energy is the capacity to do work while power is the rate at which work is done ( diffen.com ). <S> Let's say you can lift a 10 kg weight easily <S> but you don't do that. <S> That is your power. <S> Let's say you lifted a 10 kg weight 30 times. <S> In this situation you spent energy.
| To answer the question, it's called energy harvesting because in most systems the rate of collection (input power) is very different from the rate of use (output power).
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Help Finding Transfer Function Of Circuit What would be the transfer function of the following circuit, assuming an ideal op-amp? <Q> This looks like a homework problem, so I'm going to demonstrate the setup, and leave the algebra to you. <S> As a note, you've got two resistors called \$R_1\$, so I'm going to denote them by \$R_{1L}\$ <S> and \$R_{1R}\$ for left and right, respectively. <S> You have negative feedback, so (assuming an ideal op amp as stated) <S> the input voltages to the op amp are equal. <S> This is a natural consequence of negative feedback on an op amp, and part of what makes it so useful. <S> If the positive input is higher than the negative input, the output goes up, but that pulls the negative input up too. <S> The only stable point is when the inputs are equal. <S> $$V_a = <S> V_b$$ <S> \$R_2\$ and \$C\$ make a voltage divider. <S> We use the complex impedance of the capacitor in the standard voltage divider equation. <S> $$V_b <S> = V_{in}\frac{\frac{1}{sC}}{R_2 + \frac{1}{sC}}$$ <S> We know the voltage on both sides of the left <S> \$R_{1L}\$ <S> , so we know the current through that \$R_{1L}\$. $$ <S> I_1 = \frac{V_a- <S> V_{in}}{R_{1L}}$$ <S> The op amp is assumed to have infinite input impedance, so all the current flowing through \$R_{1L}\$ must also flow through \$R_{1R}\$. <S> It has nowhere else go to! <S> $$I_2 = <S> I_1$$ <S> We know the voltage on the left side of \$R_{1R}\$, and the current through \$R_{1R}\$, so we know the voltage on the other side of \$R_{1R}\$. $$V_{out} = <S> V_a + I_1R_2$$ <S> (Note: the transfer function of the voltage divider at point b can be problematic for DC signals. <S> You could take the limit of the expression I gave as \$s \to 0\$. <S> However, you should know what a capacitor looks like in a DC circuit, and be able to write the equation directly from that.) <S> You can run the algebra yourself. <S> But don't just take the answer and turn it in! <S> Learn from the steps so you can do it yourself, next time or ten years from now. <S> You want to be a good engineer, right? <A> This is a Phase Shifter, with transfer function given by: $$\frac{V_o(s)}{V_i(s)} = H(s) = <S> \frac{-R_2Cs+1}{R_2Cs+1}$$ <S> Gain : <S> \$1 <S> V/V\$ (passes all signals without altering their amplitude). <S> Phase lag : \$0^0\$ to \$-180^0\$, with \$-90^0\$ <S> at \$\omega = <S> \omega_0=\frac{1}{R_2C}\$. <A> What you have drawn is an allpass filter (see http://www.analog.com/media/en/training-seminars/tutorials/MT-202.pdf ) which actually mimics a 1st-order Padé approximate of a pure delay. <S> This is <S> a 1st-order circuit (only 1 energy-storing element) and the generalized transfer function linking \$V_{out}\$ to \$V_{in}\$ is given by: \$H(s)=\frac{H_0+sH^1\tau_1}{1+s\tau_1}\$ \$H_0\$ is the gain linking \$V_{out}\$ to \$V_{in}\$ determined for \$s=0\$ in which the capacitor is open. <S> The time constant \$\tau_1\$ is determined by reducing the excitation voltage \$V_{in}\$ to 0 V and "looking" at the resistance driving capacitor \$C_1\$. Then, place this capacitor in its hi-frequency state (a short circuit) and calculate the gain \$H^1\$. <S> Once you have these elements, assemble them according to the generalized transfer function expression. <S> All the steps are drawn in the below figure: <S> Once you have captured the component values and expressions into a Mathcad sheet you can see how combining a left-half-plane pole (LHPP) \$\omega_p\$ with a right-half-plane zero <S> (RHPP) \$\omega_z\$ produces a constant 0-dB gain but with a phase lagging down to 180°. <S> This is a good way to model a delay in a transfer function. <S> A delay for instance incurred to an A/D conversion time or a comparator transition time. <S> The dynamic response is here: <S> The FACTs helped me derive the transfer function in a very short time without manipulating complex expressions. <S> You can find an introduction to the technique via the below links: http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf and http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf
| You can derive the transfer function either by using superposition (you split \$V_{in}\$ in \$V_{in1}\$ and \$V_{in2}\$) or using the fast analytical circuits techniques or FACTs.
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Unused Op-Amp in Photodetector setup, but only occasionally I am designing a modular system that will take 9 or 16 photodiode inputs, depending on its setup, and put them through an op amp. The diodes and the op amps are on different PCBs, connected by pin headers and sockets. The schematic looks like this: However, some of the time some diodes will not be present (i.e. D1 will be missing), because on the modular diode board there may be only 9 diodes (if configured that way). The remainder of the components will still exist, as they are all on the same PCB. The datasheet says to do this with the unused op amps: I can see no way to make either of these circuits by only modifying the schematic at the point where D1 is inserted, because the design of the PCB means all the + inputs are always tied to ground. I effectively have one place to make changes, so the question is what to do with the - input. Leave it floating, connect it to a voltage divider, connect it to ground or supply via a resistor (or no resistor). I realise the situation is not ideal. Any suggestions? <Q> R2 provides DC feedback so that the output will be driven to the same as the positive input, which is ground. <S> Think of it this way. <S> Your circuit is basically a transimpedance amplifier. <S> By disconnecting the diode, you are feeding 0 current into the transimpedance amplifier. <S> 0 current is a valid signal in this case, and no different from the diode being in the dark. <S> Also, the datasheet is being too specific about what to do with unused opamps. <S> You want to drive them so that they don't pick up noise or oscillate, but otherwise use little current. <S> There are various ways to achieve this. <S> They only show some examples. <A> You can leave the opamp with it's feedback components and with no diode connected. <S> This won't be a problem at all. <S> The ideas suggested by the data sheet are just typical suggestions. <A> On your modular board that holds the PD's , if the PD is not present, then substitute it for a resistor. <S> Connecting it to something that prevents activity is all that is necessary.
| What you are looking to do is to prevent the open input from being coupled to stray fields and causing the op-amp to consume excessive power and to cause cross talk on other channels. There is nothing more you need to do. Your circuit without the diode is stable all by itself.
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Is quality of VGA Horizontal and Vertical sync Pulse relevant to be able to display on sensitive projectors I am currently trying to diagnose a VGA problem on hardware A. Hardware A has some trouble sending display to certain projectors (while it is able to send display on most LCD monitors). Hardware B connects successfully on both projectors and monitors (Nvidia quadro 2000). The question here is : could there be a link with the spikes I see on both horizontal and vertical sync pulses of Hardware A and the fact that Hardware A has more problems with certain projectors ? Hardware B seems to have a much clearer signal. Also, does Horizontal and vertical sync pulse need to align (at both beginning and end of a vertical sync pulse). I see here that Hardware B aligns perfectly, but Hardware A shows a 800 nsec delay. Vertical sync pulse and horizontal sync pulse - Hardware A Vertical sync pulse and horizontal sync pulse - Hardware B <Q> Overshoot or delay are not the cause of the problem. <S> Overshoot should not be more than 30% of the logical pulse. <S> Here, we don't see the exact measure but it is no more than 24 %. <S> As for the delay, it should not be that large, but internal loops in the design make it that way. <S> However, it doesn't prevent the display to work. <S> What was causing the problem was color levels being 1V instead of 0.7 V (R, G, B) pins. <A> Different projector vendors use different approaches for determining which scan line should be considered the "first". <S> I would not expect the spikes to be an issue; a bigger issue would likely be the relative timing of the horizontal and vertical sync pulses. <S> If it samples vertical sync precisely on the rising edge of horizontal sync, the first would work but the second would be problematical (since the vertical sync could appear to happen before or after the horizontal). <S> If it tries to avoid problems with the second by slightly delaying HSync internally, then the state of VSync should be well-established in either case <S> well before it's sampled. <S> If, however, if delays VSync, then it would be possible that in the first case it might sample VSync signal just as the change is propagating through the delay circuit. <S> This could potentially cause one scan line of vertical positioning jitter. <S> In practice, I would expect most modern digital projectors to determine at what point within a scan line vertical sync seems to change (it should be consistent) and then select some other point in the scan line to sample it. <S> On the other hand, I have certainly seen projectors which have more trouble than it would seem they should with sync signals that aren't exactly "perfect", so it's hard to guarantee that any deviation from the norm wouldn't cause problems with some equipment somewhere. <A> Upon redesigning VGA output for HSYNC and VSYNC to be in different process (enabling them to be simultaneous if they have to be) we found that the 850 ns is the cause. <S> There was a color problem too (impedance mismatch) <S> but the delay is what was causing the problem. <A> Suggestion: Regarding your hypothesis that the vertical sync pulse goes low "too early" <S> , you could make a very simple circuit to check this. <S> With two gates, and an R-C delay in between, you can create a delayed version of VSync. <S> Use a variable R to adjust the delay. <S> Add a diode to delay just one edge. <S> A 2/4 of a 4093 or 2/6 of a 74AC14 would be good candidates as they have Schmitt trigger inputs, so work more cleanly with analog-ish input from the R-C "analog" delay. <S> With minimal additional logic, you could compose your own VSync pulse that looks more like the one that works reliably.
| If a projector uses the state of the vertical sync line on the falling edge of a horizontal sync pulse as its cue for whether a particular scan line is "the first", either waveform should work.
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Why data buses should support 1024 bits, when processor are working with 64bits? If the processors are working with 64bits, then why do they need Bus protocols to be as high as 1024bits. Is it for re usability? <Q> The internal ALU width of a CPU might be 32 or 64 bits, and this represents the amount of data that can be processed per CPU clock cycle. <S> On a modern processor, the CPU(s) are running at clock speeds of several GHz. <S> If each operation involves 1-2 input operands and a result, this represents a raw data bandwdith of tens of GB per second. <S> Processor chips have internal memories, called caches — often in several levels, that handle much of that raw processor bandwidth, taking advantage of the fact that data tends to be reused. <S> Caches are used as buffers for the contents of external memory, and so data frequently needs to be transported between the caches and the external memory. <S> The external memory interface runs at a much slower rate than the processor core, usually by a factor of about 10 (i.e., hundreds of MHz). <S> In order to have as much bandwidth as possible over this interface, burst transfers are used to transfer entire cache "lines" at a time. <S> Since the ability to adderess individual words within an external transfer is not important, it makes sense to make the data bus as wide as practical in order to minimize the number of clock cycles required to complete a burst transfer. <S> An external bus of 1024 bits (128 bytes) will be able to transfer 1kB of data in 4 clock cycles (DDR). <S> If the bus is running at, say 500 MHz, this represents a raw bandwidth of 128 GB/s, which makes it a good match for a multi-core processor. <A> To get a high bandwidth (number of bits transferred per second). <S> Roughly the situation is: the consumer needs 64-bit chunks of data very fast. <S> the producer can produce data, but much slower. <S> solution <S> : use 16 produces, that each produce 64 bits (at a slow pace), 1024 bits in total. <S> Buffer these bits. <S> the consumer consumes from the buffer. <S> If all goes well it consumes the entire buffer in the time that the producers can produce one new buffer worth of data, and everybody is happy. <A> A typical processor can execute hundreds of instructions in the time required for a typical DRAM ("bulk" memory chip) to fetch a single piece of data. <S> If every piece of data required by the processor had to be fetched separately from RAM, even today's fastest processors would be little better than those of ten years ago. <S> In many cases where a processor wants a piece of data, however, it will also want other data that's nearby. <S> In a modern system, when the processor requests a piece of data that needs to be fetched from slow memory, the memory subsystem won't just fetch enough data to satisfy the request, but will instead copy into faster RAM a large chunk of data which contains the requested piece. <S> If the request is followed reasonably soon by requests for any of the other data in that large chunk, the requests can be satisfied using the high-speed RAM rather than the slower main memory. <S> Note that faster RAM is much more expensive than bulk DRAM. <S> While it might be possible to build a system with many gigs of extremely fast memory, the added performance would in most cases be insufficient to justify the cost compared with a system that combines a large amount of slower main RAM with smaller sections of faster memory. <S> Because not all memory operations are sequential, doubling the transaction size won't cut the number of transactions in half, but system designers have apparently determined that expanding the chunk size from 512 bits to 1024 bits offered a sufficient performance increase to justify the cost.
| For operations which require processing large amounts of sequential data, increasing the size of chunk that can be read in each transaction will reduce the number of separate transactions required.
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Resistance, and learning for beginners I am building an electrified "fence" of sorts to keep slugs out of my raised garden beds. The "fence" consists of two 18ga galvanized steel wires, stapled to wood and connected to a 9V battery. The circuit is open until the slug touches both wires, at which point he completes the circuit and gets a little shock. This works great in small scale tests, but my garden bed as a circumference of 30' (9.1 meters) and I'm concerned that a little 9V battery will have a hard time working at full power over that distance of wire. Because I don't really know how resistance works (I just know how to get my volt meter to measure resistance so that I can make sure my solders are done well) I'm not sure how to do the math to see what kind of a drop in voltage I'm going to get. If someone can answer this question directly, great. But I'd also love a website that I can use to help learn about this stuff. How can I find out the resistance of this wire? (Doesn't resistance change with voltage/amperage?) How does that resistance compound over distance? Are there simple parts (transistors, resistors, whatever) that I can go get at Radio Shack or the like that will help get over this resistance or should I just wire a second 9V battery into the system? Bonus question: While 9V is a very light charge, if you imagine I'm building a much more powerful electric fence, what would I have to do to protect the battery if the circuit were completed for an extended period of time? (A slug dies on the line or something.) I wouldn't want some grass completing the circuit and having my battery explode. <Q> The wire suggested wire is adequate- <S> the resistance is about 130 ohms/km. <S> I got that from kinda a long route. <S> I used the table for copper (AWG) wire resistance per km, adjusted that for the ratio between copper and steel resistivity , and then for the ratio (squared) between the diameters for AWG and the gauge system used for steel wires (whew!) <S> which is "Washburn & Moen; Roebling; or American Steel and Wire". <S> So only maybe 2.5 ohms for your circuit (based on two wires, 9.1m each, connected at one end. <S> You might want to use something other than wood for the supports, insulate the wire from the posts, or at least seal the wood with epoxy. <S> If, say, you were in <S> Portland OR , it will (on average) rain 144 days per year and the leakage through the wet wood would tend to drain the battery. <S> Here are some insulators used on commercial electric fences. <S> In terms of $ per kWh , AA batteries are a much better deal than 9V batteries, and six AA cells in series would give you 9V. Energy capacity being a lot higher, it also means that you won't need to replace the batteries as often. <S> There is a good reason why 9V batteries are seldom found in modern electronics. <S> AA cells have a short-circuit current in the amperes, so a current limiting resistor as Andy suggests would be a very good idea, or you could use a blinking LED in series such as this Lumex one, available from distributors such as Digikey for $1 each. <S> If you see the LED blinking, the wire is shorted. <A> 18 gauge wire (assuming AWG) is about 21 ohms per kilometre - see this table for other gauges. <S> With 9.1 metres your total resistance is going to be 0.189 ohms and won't have a problem with the circuit you described. <S> Resistance changes as wire heats up but on your circuit there'll be no heating up <S> and I'd employ a current limit resistor in series with the battery to make sure - probably 1 kohm is large enough to prevent rapid discharge of the battery <S> should a short occur but no so large as to prevent the little garden friends receiving a tingle or two. <A> The resistance of the wire can be measured using the ohm or R setting of the multimeter. <S> If you touch one probe to the one end and the other to the other end, you'll get the resistance in-between. <S> Note that if it's touching itself (curled up) then the resistance won't be accurate, it would need to be not in contact with itself if it's uninsulated. <S> Since the resistance of the wire is so low, your multimeter may not measure a resistance at all. <S> It doesn't change with the voltage or current, however it does determine how the current will change relative to the voltage. <S> Resistance increases proportionately with length. <S> A wire of double the length will have double the resistance. <S> Half the length will have half the resistance. <S> Since the resistance of the wire is so low, you won't really have an issue with the 9V supply.
| Resistance is a physical property of the wire. However it will be a good idea to put a resistor in line with the 9v battery just so that if something causes a short (like if it rains) then there will be a current limiting mechanism in place.
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Replacing usb speaker circuit with LiPo battery & Bluetooth. Not sure what watt audio amp to use I have a set of Logitech usb speakers which I wish to convert to bluetooh speakers. A usb has the the power, ground, data+, and data- connections. Which i believe cannot be simply replaced with a battery & bluetooth data connections(no drivers etc). So I intend to build a chip myself but the the tutorial is for 3W audio amp chips BUT the newegg specs for my speakers show total power of 1.3W. Is 1.3watt the correct value I am looking for? or would be that the power prior to an amp chip in the original circuitI currently intend to build this circuit for each amp chip I need (2) Is there some way i could test for the correct wattadge without breaking the speaker? Edit: Thanks for the advise I will try my best to apply it tomorrow.And for the images here it is.. ( will update with part numbers at a later time ) <Q> Easy enough, you need to look at the circuit board. <S> Typical USB Powered speakers can be broken down into three parts. <S> Part 1 is the USB Sound Card. <S> This is powered by 5V and accepts a USB connection and decodes the audio. <S> Part 2 is the amp stage. <S> As most USB soundcards have internal pre-amps, there will only be one or two identical 5V amplifiers. <S> Most amps can power two speakers in single ended mode, or one speaker in dual/double ended mode. <S> Finally, Part 3 is the speakers themselves. <S> To know what wattage the speakers are designed for, get the part number of amplifier IC(s) and look for the datasheet. <S> That will tell you the maximum wattage. <S> And then sketch out the circuit. <S> 9 <S> out of 10 times, the circuit on your speakers will match the Typical Application/Recommended Design/Layout Schematic in the Amplifier datasheet. <S> The actual resistor/capacitor values and layout will tell you the actual wattage it is designed for. <S> Again, 9 out of 10 times, the circuit will match the typical application for both the usb sound card IC and the amplifier ic, be a simple single or double layer board, and be fairly easy to figure out. <S> Shouldn't take more than an hour, even for a beginner. <S> To add in your bluetooth control, replace the USB sound card IC with your bluetooth audio driver. <S> Leave the amplifier in place unless you need to change it for some pointless reason :D. <S> Update: The posted circuit board only has one IC , the TP6902 (or clone) . <S> It is an old (2005) <S> all-in-one USB Audio Controller, USB HID Buttons (Maybe), and two D-Class amplifiers built in. <S> Each channel can support up to 5v * 346mA or 1.75W. <S> The typical being 200mA or 1W. <S> The only thing that you would be using is the case and the speakers themselves. <A> The logitech speakers you have linked to both derive power and digital-audio data from the USB. <S> If you are looking to feed them with an analogue signal you'll need to strip them down and do some rework (if you can get in without breaking the plastic box). <S> Assuming you do that then it's easy to limit the peak power going into the speakers by adding a series resistance - probably about 10 ohms <S> but it depends on the amplifier spec and the internal speaker impedances. <A> The speakers you have listed appear to be made by Jiaxing Zhongyong Electronics Co and they have a max rating of 5W RMS at 4ohms which means you should be able to use them in the instructable without any issues. <S> It looks like the 1.3W rating is the limit of what the Logitec internal power amp can provide to the speakers but is not the max limit of the speakers themselves. <S> It's quite typical for engineers to design products with this sort of overhead to prevent damage at max power use cases.
| So those speakers are probably not more than 2.5W each, and the only way to mod the speakers for bluetooth is... completely replace the IC with a custom Bluetooth Audio receiver + D-Class Amplifier board + LiPo Battery + LiPo power regulator.
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Is there a type of Switch that is open until Armed? I'm designing a circuit that will need to be off until the user decides to 'arm' the circuit. It should then always be closed (and hard to hack into the on position). Does a certain component exist for something like this purpose? If not can you suggest a strategy? <Q> Here's a method to create a permanent "latching" action from a switch and a relay: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Once the button (SW1) is pressed, it charges the relay coil and closes the relay contacts. <S> As usual, this bring power to the load. <S> But, in this case, it also brings power back to the relay coil. <S> So, even when the button is released, the coil is still energized and the load stays powered. <S> The load will stay powered until you remove the voltage source. <S> This could easily be modified to use BJTs or MOSFETS, if you would like a solid-state solution. <S> Is this what you're looking for? <A> Since you used words like 'arming' the circuit and 'hacking' into, the feeling I got is that you want to be able to arm the device (turn it on) and have it impossible to be turned on/off by outsiders. <S> Again I'm not sure if this is actually your intention, but if it is you could use a simple key switch. <S> - ie something like this. <S> Its just a normal on/off switch, but you need a key to operate it. <A> I'm a little confused by your use of "off", "on", "armed", and "closed". <S> Perhaps an "emergency off" button as used in machine control systems might meet your needs. <S> These buttons are normally on, but when pressed, open the circuit, and must be twisted to return to the normal position. <S> Some require a key to reset the switch to normal. <S> Some may include an SPDT switch, so could be used as either normally open or as normally closed. <S> edit: such buttons are also called Emergency Stop (or E-stop) buttons. <S> They provide a mechanical solution for turning something off quickly and intuitively, then they keep the off state and making accidental turn on unlikely. <S> ( datasheet ) <S> There's nothing special about this particular model. <S> There are plenty of buttons like this.
| If you need a method to turn off the circuit, you could add a normally-closed pushbutton in the path from the load back to the relay coil.
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Powering 16 super bright blue LEDs with Arduino What is the best way to power 16 super bright LEDs using an Arduino Uno to achieve max brightness? Each LED has the following characteristsics: IF Typical (mA) = 20VF Typical (V) = 3.2IV Typical (mcd) = 3700IF Max Continuous (mA) = 30Preferred Series Resistor (ohms)5VDC = 919VDC = 30012VDC = 430 ( http://www.jaycar.com.au/productView.asp?ID=ZD0132 ) Output from Arduino pins = 5V. Power suppply is 12V, while the final will be an Automotive ~12V rail. I've used two LED calculators: http://led.linear1.org/led.wiz this site says I should use 16 LEDs in parallel with 100Ohm resistor on each one. http://www.hebeiltd.com.cn/?p=zz.led.resistor.calculator this site says I should use 16 LEDs in parallel with just one 5.625Ohm resistor. The Arduino Uno has 14 digital pins outputting 5V.I would like to use the least amount of 5V digital outputs, and the least amount of resistors necessary, while maintaining maximum brightness. Hope you can help :) <Q> 8 parallel strands of 2 LEDs each and a 150ohm resistor. <S> All loose anodes should be connected to V+. <S> The loose cathodes should be connected to the collector of a TIP31 NPN transistor. <S> The emitter should be connected to ground. <S> The base should be connected to a 100ohm resistor, and the other side of the resistor should be connected to a digital output. <S> 16 LEDs, 9 resistors, 1 transistor, 1 digital output. <S> (12V−(3.2V⋅2+1.2V))/30mA <S> = <S> 146.66667 <S> Ω 30mA⋅8 = <S> 240 mA (5V−1.8V)/(30mA⋅8/10) = <S> 133.33333 <S> Ω simulate this circuit – <S> Schematic created using CircuitLab <A> You need to read the data sheet. <S> For an ATMega 48, http://www.atmel.com/Images/Atmel-8271-8-bit-AVR-Microcontroller-ATmega48A-48PA-88A-88PA-168A-168PA-328-328P_datasheet_Complete.pdf <S> page 320 gives typical output under load of an output as 4.2 volts at 20 mA. <S> At best, you can directly drive 14 LEDs, each with a 50 ohm series resistor. <S> A "5V" output is only nominal. <A> Two LEDs in series, dropping about 3.2V each with 30 mA through them, added to about 0.3 V for the transistor's Vce(sat) comes out to about 6.7 volts. <S> Subtracting that from the 9V supply leaves about 2.3 volts which needs to be dropped through a current-limiting resistor with 30 mA through it, <S> so from Ohm's law, the value of the resistor needs to be: R= <S> E/ <S> I = <S> 2.3V/30mA ~ <S> 77 ohms. <S> For a forced beta of ten and a collector current of 30 mA * 8 = 240 mA, the base current needs to be 24 mA. With a 5V output from the Arduino and a Vbe(sat) of about 1V max for the transistor, the value of the resistor needs to be Rb = <S> Vin <S> - Vbe(sat)/Ib <S> = 5V - 1V <S> / 24 mA <S> ~ <S> 167 ohms. <S> 9V | [LED] <S> | [LED] <S> | <S> [75R] | C I/ <S> O>--[160R]---B <S> NPN E | GND
| So the short answer is that, without external power boosters such as transistors, you cannot do what you want.
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Why can't one battery power an infinite number of leds (or anything else)? If voltage is equal across parallel resistors, then why can't one battery supply power to a infinite number of leds, if they are all in parallel? <Q> Because a battery, like any real power source, has an output impedance. <S> The more current you draw, the lower the voltage output. <S> Granted, for a battery the output is not simply a fixed resistor, but the principle remains the same. <S> As an example, let's pretend that a 12-volt battery has a 0.1 ohm output impedance. <S> If you were to short the outputs of this notional battery, the current would be 12 v / 0.1 ohms, or 120 amps. <S> By the same token, if a load were to draw 60 amps, the output of the battery would be 12 - (60 * 0.1), or 6 volts. <S> Another way to put it is that the output impedance sets an upper limit on the amount of power a battery (or any power source) can provide. <S> For DC, this upper limit is (V * V / R) /4.In <S> the case of our notional battery, this upper limit is 360 watts. <A> Because it doesn't store an infinite amount of energy, nor can it move the energy it does store at an infinite rate. <S> A chemical process is happening when the battery provides current <S> and it can't happen faster or for longer than the chemicals can proceed. <S> Even a huge voltage supply powered directly from the electrical grid couldn't power an "infinite" number of LEDs. <S> Infinity is a number beyond the capacity of electrical engineers to design around. <S> In a simulator where you can define a voltage source to have infinite capacity to deliver current it might work fine, but never in the real world. <A> A single LED requires a finite amount current. <S> To have infinite LEDs means you need infinite current. <S> This is not possible. <S> Incidentally, batteries do not contain infinite energy. <S> You can not hook up an infinite number of LEDs because you can not do impossible things. <S> Sorry about that. <A> In this case more LEDs in parallel, with the same battery, will imply less current available to each branch of the circuit. <S> Bellow a certain point the LEDs will stop producing light and the current will tend to zero: <S> 3 LED - <S> > <S> I/3 <S> A <S> 5 LED -> <S> I/5 <S> A <S> INF LED -> <S> I/INF -> ~0 <S> A
| Aside from the fact that batteries can not provide infinite current (they have internal impedance), to supply infinite current requires infinite power, to supply infinite power for any duration at all requires infinite energy.
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Planning to make a pen-like device which has a roller rolling on a surface and measured its linear distance travelled I am planning to use a microcontroller to do the brain works of receiving pulses from the encoder and compute it to output linear distance and at the same time indicating the measured distance as short, medium or long (of course that means I'll write a simple program for classifications/ranges of distances). The encoder I chose is of reflective type. My idea is similar to the case of a digital vernier caliper where the distance measured are constantly displayed on the lcd screen. I suppose the way to do it will be something like to count the number of pulses over the entire length of signal? How to make the microcontroller to do that? will the microcontroller reads a binary number of 10101010 ... as the signal? if so, all I have to do is to translate these binary number to decimal then subsequently extract the angular displacement? please tell me whether am i right lol <Q> You can take apart a mouse you have laying around or buy the sensor separately. <S> Here's a link to one such sensor though there are many more out there: https://www.sparkfun.com/products/12907 <S> Key specs are running at 5V and 400 cm per inch resolution. <A> You can use an opto-coupler like QVE11233 and a disk like the one in following picture. <S> When you roll the disk, it would simply ON/OFF the signal between QVE11233. <S> If you simply implement a counter, and multiply it with the distance between two counts, you can get the distance travelled. <S> This might not be the best way as the accuracy is the distance between two solts. <S> You can have a bigger disk with a transmitter and receiver diode implemented separately, to achieve better accuracy. <S> This can be done like <S> This is the most simplest way to calculate the distance with an accuracy of 2mm. <A> The encoder will give you a pulse (the digital pin will read a 1) after the ball has rotated a specified amount in degrees, so you will have to mathematically calculate exactly how much linear distance the ball has travelled per pulse. <S> and you will have the total distance travelled. <S> You will need to setup one of the pins on the microcontroller to use an interrupt and program an interrupt service routine which is the code that will execute each time a pulse is detected, you wont really be reading in 10101010... <S> There is a lot of info online on how to use interrupts, counters, etc.
| If you're still working on this, this sounds like the perfect case to use the optical sensor in a mouse. It would probably be best to use an interrupt to increment a counter each time a pulse is detected, then just multiply the number of pulses seen by the linear distance travelled per pulse
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Do pick & place machines require reels of components, or can cut tape be used? If I wanted to have a PCB assembled, would I have to give the manufacturer reels of components, or will pick & place machines also work with cut tape? I ask because I'm thinking of having a low quantity of boards assembled (100-200), but most reels on Digi-Key have several thousand components or ICs on them. I would rather buy cut tape with a few hundred ICs (even if it means a higher cost per component), since buying a reels with thousands would be a much higher upfront cost. Tried Googling this but couldn't find an answer anywhere.. <Q> Talk to your assembler to find out what their capabilities are. <S> You could always get the distributor to re-reel the tape though. <S> They may add a surcharge for this. <S> Look for a "Digi-Reel", "MouseReel", etc. packaging option. <A> As Ignacio points out, cut tape and mini-reel configurations are often acceptable. <S> That said, an assembler I used warned me to be a bit careful with all of these mini-reel options, as they've had some experiences with spliced-together fragments of cut-tape causing feed problems. <S> For things like resistors, where not much money buys you a full reel, just go with the full reel. <S> My preference is to make turnkey arrangements, and let my assembler order through whatever mechanism has been successful for them. <S> If I don't like the quote, I call my assembler and see what can be done. <S> If you choose to fully or partially kit, your assembler might well demand a part count a certain percentage above the boards to be delivered. <S> Again, talk to your assembler to get this right BEFORE the job is started. <A> At least on the machines i've seen the reel itself is just there to keep the tape neat. <S> However the feeders are almost certain to need a lead-in on the cover part of the tape and may also need a lead-in on the base of the tape. <S> They may also need a trailer if they are to successfully use the last few components on the tape. <S> When component manufacturers put their components and reels they include a lead-in and lead-out as a continous part of the tape. <S> If you order a "re-reel" service then the vendor will splice on leaders and trailers for you. <S> None of the prototype assembly houses i've worked with have expressed any concern about components supplied on cut tape. <S> I have sometimes had the components come back with extra cover tape still attatched. <S> Splices can break or jam <S> so full manufacturer reels are preffered. <S> My rule of thumb is if I think I will use more than 100 of a low cost component in the forseeable future <S> I get a manufacturers reel. <S> Obviously for high cost components that is not so practical. <S> I'm not sure I see much point in re-reels. <S> If there has to be a splice <S> i'd <S> rather it be done by the same people who are going to run the components through their machine to avoid the possibility of blame games. <S> With small components it is very likely that some components will be lost when setting up the real. <S> So allow generous overages. <A> You want to have a look at this series from Dave: http://www.eevblog.com/2010/11/15/eevblog-127-pcb-design-for-manufacture-tutorial/ <S> He goes into manufacturing, and, more importantly, the in's and out's of tape, reel, etc. <S> Starts at about 5:30 in the video above.
| Some P+P machines can use cut tape, others must use reels. Components bought as "cut tape" will not have a leader or trailer so the assembly house may have to attatch one themselves. For more expensive parts, do what you have to do. It's not really needed if the ammount of tape in the machine is small.
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Why are rectifiers-inverters used to drive AC motors instead of using the rectified current to drive DC motors? I understand the advantages of using a rectifier-inverter system to drive an AC motor instead of simply plugging it into mains power, as it allows much better control of its speed and performance; but what I don't understand is: since the original AC power has to be converted to DC in order to feed the inverter circuit, why is this DC not directly sent to a DC motor, instead of converting it back to AC and then sending it to an AC motor? <Q> Because AC motors are generally much more efficient than DC motors, and since they don't require electrical contacts to the rotor, are more reliable as well. <S> Remember, a BLDC motor is really an AC motor with the drive circuitry built-in. <S> At higher power levels, it makes sense to separate the control and drive circuitry from the motor itself. <S> Also, motors with permanent magnet (PM) rotors have limited power-handling capability. <S> At higher power levels, AC induction motors are used, even in electric vehicles. <A> DC motors have effectively one variable: how much power are you feeding the motor? <S> AC motors have two variables: power, and frequency. <S> I'm not an expert in motors, but I'd expect that AC motors would thus allow independent control of speed and torque, while DC motors do not. <S> Directional control is also a concern. <S> An AC motor's direction can be controlled by the rotational direction of the power being fed to it. <S> A DC motor's direction isn't so easily controlled. <S> More broadly, all motors operate because there's a rotating magnetic field somewhere. <S> That rotation is either generated inside the motor (self-commutating) or because the power feed to the motor is itself rotating (externally commutated). <S> DC motors must be self-commutating; DC is definitionally not rotating. <S> How do you achieve commutation inside the motor? <S> Typically, either there are brushes , or there's an inverter built into the motor . <S> Brushes wear out, and I suspect have other disadvantages. <S> And if you're going to build an inverter into the motor, why not put it outside the motor and get better control of it? <A> With many kinds of AC motors, the rate of rotation will be strongly correlated with the frequency of the driving current. <S> In many cases, the rotational speed in revolutions per second will either be an exact fraction of the drive frequency in cycles per second (e.g. 1/3), or else an exact fraction minus a certain amount of "slip" which depends upon the drive voltage. <S> While it may be possible to control the speed of some AC motors by varying the drive voltage and thus allowing varying amounts of slippage, it's more efficient to vary the drive frequency and try to minimize slippage. <S> Note also that nearly all motors that are capable of doing a non-trivial amount of work require that the polarity of current in some of the coils be periodically switched. <S> This is just as true of DC motors as AC ones. <S> Most DC motors use a mechanical commutator and brushes to perform such switching; these tend to have a limited useful lifetime before requiring service or replacement. <S> Some use electronics to switch the actual motor current, but that essentially turns them into an "inverter-plus-AC-motor" combination. <A> There could be lots of reasons. <S> The most obvious is that the brushes in PMDC motors wear out and need to be replaced after 2000-5000 hours, depending on the environment. <S> Whereas AC motors (both induction and PMSM aka brushless motors aka BLDC motors) can last 20,000 hours. <S> So if maintenance-free operation is important, you might want an AC motor. <S> Second, if you are doing any sort of speed or torque control you aren't going to <S> just have DC for a DC motor. <S> You are going to have PWM DC. <S> And once you have the electronics to do that, it isn't that much different to go to PWM AC. <S> Third, a lot of modern induction motor and PMSM controls operate using a technique called field oriented control. <S> This type of control allows you to control both the operate your motor smoothly at low speed and high speed <S> and it gives you independent control over your torque and your magnetizing field. <S> You can't do this with a PMDC control because your brushes/commutator mechanically align the field. <S> So if that is important to you, you might choose an AC over a DC motor. <A> Another advantage of AC motors is that they do not use brushes and commutators, like DC motors do. <S> These generate a lot of sparking and broadband EM noise. <S> There are environments where such actions are a really, really undesirable :) <A> AC motors are more reliable than DC motors. <S> DC motors produce output power from the current flowing in the armature. <S> The DC motor transfers the current to the armature with commutator and brushes. <S> The electrical inductance of the armature causes arcing as each brush breaks connection from each successive contact bar of the armature. <S> This pits the armature and brushes, making them rough. <S> The roughness wears both armature and brushes. <S> When AC motors use electromagnet rotors, current connects to the rotor with slip rings and brushes. <S> There is no switching at the brushes on slip rings. <S> This avoids the pitting from arcing suffered by DC motors. <S> Slip rings and brushes last many times longer than DC motor brushes and commutators. <S> Most AC motors operate without brushes and slip rings by using inductive coupling, hysteresis, or permanent magnets in the rotors. <S> Service life of brushless motors might be limited only by bearing life. <S> DC motor controllers can change the magnetic field from the stator or the voltage or current applied to the armature. <S> AC motor controllers can change stator voltage, current, frequency, or phase, or rotor current. <S> Some AC motors can change the number of magnetic poles on the stator. <S> This makes AC motors able to efficiently convert electricity to motive power over a wider range of operating speeds than DC motors of equivalent power levels. <A> Another aspect that I don't see mentioned is that a 3-phase AC motor fed with pure sine wave input produces uniform torque through all 360 degrees of rotation. <S> A simple DC motor will experience torque variation as each rotor pole rotates past its counterpart stator pole. <S> This can be an important consideration in, for example, precision machining.
| AC motors can be more controllable than DC motors.
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Step response to feedback loop with disturbance When using step response to a feedback loop with disturbance. Why do we analyze the step response of the disturbance D and the step response of the input R separately. Instead of considering them together like with $$Y = \frac{GR}{1+HG} + \frac{GD}{1+HG}$$ given that \$Y\$ is the output. I'm assuming we can't take the step response with both together in the function as we want the transfer function to be in terms of \$\cfrac{Y}{R}\$ or \$\cfrac{Y}{D}\$, but I just wanted to confirm that this was the case. <Q> Mainly because it's easier. <S> As long as your system is both linear and time-invariant (LTI), the principle of " superposition " allows the probelm to be decomposed this way and still get valid results. <A> The input step response and load step response are often different when the drivers are not linear or constant impedance throughout the entire load curve. <S> Often power supplies are tested 10 to 90% or 50 to 100% or some other different step size due to this characteristic. <S> The step size in fact will rarely give an identical time response because designs are rarely ideal linear current vs voltage and time invariant over the whole range. <S> Disturbances can come from either inputs or outputs so <S> it best to understand how they can be different and change with step size and direction, for both inputs and outputs. <S> This is the essence to understanding the non-linearities of circuits. <A> I believe that using state-space methods (MISO - Multiple Input, Single-Output), you can create an input vector having the reference and disturbance as components. <S> Alternatively, "Robust Control" and QFT (Quantitative Feedback Theory) are methods that emphasizes command / disturbance treatment.
| Due to the property of superposition that holds for linear systems, the response to the reference and disturbance applied simultaneously is equal to the sum of those ones.
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Using a transistor as a switch I am attaching a motor to an ATmega32 chip. I want to build an H-bridge but before doing that, I want to control the motor using the chip. I bought few NPN Transistors from RadioShack. I have 5V running though the breadboard. I attached Vcc to the Collector, Base to the Pin, and Emitter to the motor. And other wire of motor is hooked up to Ground. Unfortunately, the power coming out of the transistor is too low, the motor is barely moving (As opposed to when I hook up my motor straight to 5V). What is wrong? Do I need a different transistor or is my setup wrong? If it is the transistor, can you please recommend the right one, I am not too good at those. <Q> As already noted, the direct connection of a NPN BJT (or N-channel Mosfet) in high-side configuration is not suitable (emitter follower no saturates with equal voltages Vb and Vc). <S> A proper driver is required. <S> For example these ones on figure below (B, instead uses a PNP): (A) Input <S> Low -> load <S> On <S> ; Input High -> Load Off <S> (B) <S> Input <S> Low -> load Off <S> ; Input High -> Load <S> On <S> The problem with (A) is the initial state under power-on or reset by watchdog activation. <S> Maybe, in some application, can be a bonus. <A> With the connections you described, the motor will see only about 4v because of the vbe diode drop; maybe that's the problem? <S> The usual way to do this (base THROUGH RESISTOR to pin, collector to motor and emitter to ground, so motor between vcc and collector) is better since the vce saturation voltage is much lower; the transistor will also dissipate less heat. <S> Choose the resistor high enough to not fry your mcu pin (40mA max for atmega328 iirc), but low enough to turn the transistor fully on - that is, the current to base must be more than collector current / hfe. <S> One tenth of the motor current would be probably plenty; check the transistor datasheet. <A> Your configuration is wrong. <S> In a nutshell with an NPN transistor you need to connect one lead of the motor to Vcc and the other on the collector. <S> Then the emitter goes to the ground and the gate to an arduino pin. <S> Then there are other issues to consider depending on the transistor you're using and for protecting the micro. <S> I suggest to read this nice article for more specific instructions <S> http://bildr.org/2012/03/rfp30n06le-arduino/ <A> You have two problems. <S> First, you connected the transistor in the emitter follower configuration instead of common emitter. <S> That means the available drive voltage is reduced by the 750 mV or so base-emitter drop. <S> Second, your transistor may not have enough gain. <S> You didn't say how much current the digital output can deliver, and how much the motor requires to spin well. <S> Try this: <S> This assumes the digital output can deliver 10 mA. <S> If your motor needs more than that, then this simple circuit won't work. <S> However, without real specs it's pointless to go into more detail. <S> You say it spins well when conntected to the 5 V supply directly. <S> Measure its voltage to make sure you still have 5 V when the motor is connected. <S> If not, there is no guarantee what the processor might do.
| When the signal on the Arduino pin is high, the transistor will let a significant current pass from its collector to its emitter, thus also through the motor, turning it on. Also make sure your 5 V supply can really support the motor. This will allow most small transistors to support a motor current up to a few 100 mA.
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Power 4.5V Lego motor by 5V PSU, how? It's a long story. Please bear with me as I try to describe what I intend to accomplish. I've got a Raspberry-PI controlled Lego Technics (pre-Mindstorms era stuff) construction. It uses a couple of 1980's Lego motors, that are normally be powered by 3 C-size 1.5V batteries in series: Effectively some 4V to 4.5V. The motors appear to be just that: No fancy electronics, overload protection, rectifier circuity, at all. (There may be an internal capacitor in parallel, like in the newer 9V Lego motors, but that would be just about it.) The battery-box only contains a mechanical switch with neutral, left and right positions. Single 2 wire lead from battery-box to motor. (Polarity gets reversed to change direction of motor.) I'm not directly switching the power to the motors from the PiFace. I found a relay with 2 inputs (cutoff and polarity) which mimics the behavior from the mechanical switch on the battery-box and which completely isolates the "controlling" circuitry from the "controlled" circuitry. (I know I'm not using the proper terminology for relays, but I hope I'm clear.) It made sense for me to use that, to make sure the rather delicate Pi/PiFace doesn't get sprayed with electric noise from the motors, when the polarity reverses. The entire setup is now working, using the original battery-boxes, to power the motors. As they eat batteries at an alarming rate I'm interested in powering these motors with a 5V feed from the power-supply that currently feeds the Raspberry-Pi and PiFace that control the Lego motors. The PSU is actually a normal 450W computer PSU, where one of the 5V leads is brought outside the case to feed the hobby-project. The computer itself is always on (drawing abotu 250W), so there is no issue with using an unloaded or severely under-loaded switching PSU. My days of analog electronics are 25 years behind me. I was never any good at it either. (I'm a digital guy.) Even so I'm quite certain I better not wire this up "as is". (I do know enough to know when I need a bit of help.) Will 5V straight from the PSU fry the motor? I guess some sort of voltage (and maybe current) limiting circuit is desired. (I don't have any spare motors, so this really worries me.) Will the induction of the motor-coils cause stability issues with the PSU when suddenly reversing polarity to the motor. I can, in software, introduce a "not-powered" period, between reversing the polarity if needed, but how long should that period be? If a limiting circuit (1st point) is used how will that affect point 2? Has someone tried something like this before? Any guidance will be appreciated. <Q> In my opinion you are good to go. <S> .5V more for a dc brushed motor that normally accepts 4.5V is not an issue at all. <S> That's a bit over 10% increase in voltage, that's not much. <S> You can assume that the lego guy that designed the system choose a motor with a maximum voltage much higher than its normal operating voltage. <S> If there's a cap inside <S> it's probably (certainly) rated for much more than 4.5V About the EM emissions/disturbance on the power lines <S> remember that: the Pi has its own filtering stage that is probably designed to handle much worse disturbances <S> the psu is so oversized that it would barely notice a change in load <S> there's plenty other disturbances going around coming from your pc <S> My advice is: try it out. <S> It's not like you are going to blow up anything. <S> Keep the motors on and see if they get hot (that is very unlikely), then sketch down a program that switches polarity (no off time) <S> let's say once a second or so, and see if the pi is concerned about that. <S> If something go wrong come back here <S> and we can help you to design a filter circuit that can reduce the disturbances on the power lines. <S> I'd like to see the circuit you are using to power the relays from the Pi <S> , that's because disturbances are much more like coming from there. <S> disclaimer <S> I think my advice applies here because it seems we are talking of a guy who wants to hack together a robot or something similar. <S> If you want to design a reliable thing that will be produced you better choose carefully the motors and their power sources. <A> Will 5V straight from the PSU fry the motor? <S> I guess some sort of voltage (and maybe current) <S> limiting circuit is desired. <S> (I don't have any spare motors, so this really worries me.) <S> 5V will probably be OK. <S> The amount of work the motor does depends on the mechanical resistance. <S> If you hold the motor to stop it turning then it will draw more current. <S> The current is what causes the heat in the coils. <S> Will the induction of the motor-coils cause stability issues with the PSU when suddenly reversing polarity to the motor. <S> I can, in software, introduce a "not-powered" period, between reversing the polarity if needed, but how long should that period be? <S> DC motors draw a lot of current when they start. <S> The motion of the motor generates a field which opposes the field in the coil which drives it, so the moving of the motor is what limis the current. <S> No movement means no limit. <S> The good side of this is that you get enormous amounts of power when the motor is just starting. <S> That's really useful if you are a tram or a railway locomotive <S> and you need to start moving a heavy load. <S> It is bad news if you are a power supply. <S> A delay will not completely solve this. <S> A big capacitor can help supply that initial current. <S> If a limiting circuit (1st point) is used how will that affect point 2? <S> A current limit will kick in as the motor starts for a breif period. <S> It would also protect the motor if it were to become mechanically jammed <A> Brand new Alkaline batteries have a charge of 1.65V. <S> So 3 in series is 4.95V. That quickly drops to 1.5V each in use or over time. <S> 1.5V is the nominal voltage. <S> Even with 4.5V, 5V is only .5V more. <S> That is exactly 11.11% more. <S> Most motors have a typical voltage, as a function of the required current and coil resistance (and rpm and load), but have a wide range of what voltage ranges will allow it to start spinning (trigger voltage), and what voltage range will keep it spinning once started (Hold voltage). <S> 11% is within even the most sensitive motor's range. <S> Keep in mind that increasing the voltage increases the current pulled, and the speed and torque of the motor. <S> Or an H-bridge to drive the motor, as you see some drops from the transistors controlling the motor. <S> As for emi or motor surge, a flyback diode across the motor helps. <S> And computer psus are fairly robust. <S> They are used to powering cd and harddrive motors, as well as many hobbyist projects without any emi concerns.
| If you really wanted to drop the voltage, a single silicon diode in series would work (but makes it complicated to reverse the direction).
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Avalanche breakdown voltage, how does it start? I would like to learn about the avalanche effect. I understand what goes on in the avalance effect, but I can't understand how it starts. For the avalanche effect to start, an electron (the first) must have enough energy to go to the depletion zone and knock the electron from the atom. For me it is a little hard to understand. How did that electron get enough energy to do that? <Q> Electrons are not all the same energy distribution - some are higher than others. <S> Same with speed of molecules in air at a given temperature. <S> What we see as voltage and current flow (or temperature in air) is an average. <A> Every diode has a (possibly tiny) leakage current when reverse biased. <S> For typical silicon diodes, the reverse leakage is on the order of a few femtoamps (10 -15 's A). <S> I don't know the typical leakage current for an avalanche diode off the top of my head, but it is likely within a couple of orders of magnitude of this value. <S> And even a femtoamp is still something on the order of 6000 carriers passing through the depletion region each second. <S> As the reverse biased is increased, the electric field in the depletion region increases, and this field accelerates the leakage carriers as they pass through the depletion region. <S> Avalanche begins when the acceleration due to the field gives the carriers enough energy to break new carriers loose ( impact ionization as is mentioned in another answer) when they interact with bound electrons in the depletion region. <A> OK. <S> I found the answer on this link . <S> Here's the citation:
| Avalanche breakdown is caused by impact ionization of electron-hole pairs.
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Resonant Frequency from Bode plot If we have a transfer function that shows no peaking in the magnitude bode plot (Starting from a flatline and then rolling off). Does this mean that there is no resonant frequency? Or do we consider the point at which the curve begins to roll off the resonant frequency? I understand that resonant frequency is the location at which we have the maximum value so I'm assuming that there isn't a resonant frequency in this case but I wanted to be sure. <Q> My answer applies to higher-than-1st-order systems. <S> There will always be a resonant point even if you can't see it. <S> You need to understand how "poles" work. <S> Take a look at this: - Even if there doesn't appear to be a resonance in the bode plot there will be a "pole" that is present and this pole represents the resonant frequency even though the "dampening" is causing it not to appear in the bode plot. <S> Here is what a 2nd order low pass filter looks like with varying degrees of Q (where Q = \$\dfrac{1}{2\zeta}\$) <S> : - If you could determine the phase angle where the output shifted by 90 degrees from the input you would find the resonant frequency even if it doesn't appear to have a "peak" in the bode plot. <A> Peaks in the frequency response can only exist in systems with conjugate complex poles. <S> For an underdamped (\$\zeta<1\$ or \$Q > 0.5\$) second-order system, the peak appears specifically for \$\zeta<1/\sqrt{2}=0.707\$. <S> $$H(s)=\frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}$$ where \$\omega_n\$ is the natural frequency (also called corner frequency when considering assymptotes), the peak <S> $$ <S> M_p=\frac{1}{2\zeta\sqrt{1-\zeta^2}}$$ occurs at resonant frequency <S> $$\omega_p=\omega_n\sqrt{1-2\zeta^2}$$ <S> Note on figure below: When varying the damping ratio \$\zeta\$, the peak follows a specific curve. <S> In filter theory, that special value for \$\zeta=0.707\$ corresponds to a Butterworth response. <S> The magnitude curve is sais to be maximally flat (no peak). <S> The meaning of \$w_n\$ for the Butterworth response is the same as for the first-order case, that is, \$w_n\$ represents the -3 dB frequency, also called cuttoff frequency. <S> Only in this case . <S> Also, \$w_n=w_p\$, causes an infinite response (undamped system - oscillator). <A> Is there a mathematical threshold for damping constant below which the system will just have no resonance? <S> Yes. <S> D = 2^(-0.5) = 0.707 approx. <S> ( D = damping constant) <S> For a general second order system with transfer function as : <S> Wn^2/{s^2 + 2*s D Wn + Wn^2} <S> Wn = undamped natural frequency. <S> You can compute the resonance frequency Wr by differentiating w.r.t Wn and equating the result to 0. <S> The result will be :Wr = <S> Wn*sqrt{1-2D^2} which can only be real if D > <S> 1/sqrt{2} <S> Hope I have answered your query to satisfaction.
| It is absolutely correct to say that there is no resonant frequency defined if the system is sufficiently underdamped.
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Can I replace a general-purpose electrolytic capacitor with a low-ESR capacitor? I have a dead PSU . I opened it up, and I found three bulging 1000 µF general purpose-type crappy electrolytic capacitors. I've replaced capacitors like this on several motherboards in the past, so this will be fairly trivial to do, but the only capacitors I have handy are low- ESR types. Could a swap like this work out, or should I grab a few general-purpose capacitors from a supplier? <Q> For example, the LM1117 is a very common semi-LDO regulator (mostly because it's cheap and you can get them very easily). <S> It has requirements as follows (from the datasheet ): <S> The ESR of the output capacitor should range between 0.3Ω - 22Ω <S> A lot of low-ESR 1000uF parts are better than that (by more than 10:1 in some cases). <S> You could always add a series resistor to degrade the low ESR parts. <A> I'm not saying it is the case in your power supply but some switching regulators rely on the non-zero <S> (or low) ESR of the smoothing capacitor(s) to produce a distinct ripple voltage at the output. <S> This ripple voltage is fundamental to how some switching regulators work. <S> I'm thinking of the Texas Instruments LM5009 - <S> it requires some series resistance with the smoothing cap for it to work properly because it is of the "constant-on" topology: - <A> High power SMPS usually benefit from the lowest ESR capacitors, as currents are high enough to provide 50mV sense current feedback (or feed-forward) from a shunt. <S> Small SMPS can be quite unstable with extremely low or high ESR as there is not enough voltage ripple from wide control range current feedback if no minimum load is specified. <S> Datasheets will specify. <S> So I would hazard a guess that high ESR and thermal fatigue caused these failures from use of GP caps. <S> You can observe regulation with no load or nominal load for verification. <S> 10% minimum pre-load is common on PC PSU 5V only.
| Generally, low ESR can be used to replace general purpose capacitors, but there are situations where the low ESR capacitor could cause oscillation due to the use of a finicky regulator.
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Serial chaining of LED strips I need to confirm / refute my assumption about current in LED strips. I know they are usually wired in paralel blocks of several serial connected diodes. This means that current consumed by the LED strip flows thru the entire strip power wiring, right? So if I connect for example 2 strips in series, I will double the current flowing thru both strips. Is this right? If it is, how much current can typical LED strip handle? I could not find this information for every single LED strip I tried, I only assume this is given by the width and thickness of the copper wire of the flexible PCB used. <Q> Be careful about what you mean by "in series". <S> If you mean that you're connecting the two inputs of one strip to the two outputs of another, you're really connecting them in parallel electrically, and yes, your concern about the maximum current that the first strip in the sequence (the one closest to the power supply) can handle is valid. <S> On the other hand, if you connect the positive input of one strip to the negative input of the other, you have them in series electrically. <S> In this case, you would need twice the voltage, but at the same current as a single strip. <A> So if I connect for example 2 strips in series, I will double the current flowing thru both strips. <S> Is this right? <S> No that is not correct. <S> For any given voltage that is powering a load of (say) 10 ohms, if you put another 10 ohms in series, the current will halve. <S> Basic ohms law. <S> For each 10 ohms you put in parallel the current will increase so if the first 10 ohms took 1 amp, adding a 2nd 10 ohms in parallel means there is another 1 amp taken from the power supply. <S> LEDs can be viewed this way but they have a non-linear resistance that could mean if 1 strip of LEDs took 1 amp, putting two in series might take nearly 0.5 amps or it could be much lower at just a handful of milli amps. <S> Here's another picture that shows how a series resistor can affect things: - No two LEDs can be regarded as the same. <S> Some LEDs have a limit of a few milliamps while others can take an amp. <S> Read the data sheet for the device - this is very important and if you can't locate a data sheet for the device then the product is probably cheap and crappy. <A> Let's say that this is a LED strip: simulate this circuit – Schematic created using CircuitLab <S> The voltage at the ends of D1... <S> Dnv line is the same as the voltage across D1a.. <S> Dna and D1x <S> ... <S> Dnx and is equal to the supply voltage because they are connected in parallel . <S> The current drawn by the whole strip is the sum of currents drawn by each series strip like D1... <S> Dn. <S> Only the voltage is lower because they are connected in series . <S> So, to supply similar devices with the normal voltage/current: <S> parallel: voltage remains the same across every element of the circuit and it is the voltage of the supply. <S> The supply must be able to provide a current which is at least the sum of all currents drawn by each parallel device series: current remains the same for every element of the circuit. <S> If the devices draw different currents then they shouldn't be connected in series. <S> The voltage of the power supply must be equal to the sum of voltages required by each device. <S> If you connect LED stripes in series, then you'll need the same current as for a single LED strip but a voltage n times greater, where n is the number of stripes you are using. <S> Only connect in series identical LED strips with the same ratings! <A> Your premise is built on a false assumption. <S> While Voltage across parallel circuits is the same, and Current across Series circuits is the same, you are forgetting one thing. <S> Each of these 12V led strip segments are composed of 3 leds AND ONE RESISTOR set for ~20mA @ 12. <S> In fact, 6 3.2V diodes in series would require more than at least 16V to even light up. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> In essence, you have Doubled the Voltage & Doubled the Resistance, thus keeping the current the same. <S> I = <S> V/R is the same as I = <S> 2V/2R. <S> But total power is the same. <S> 12V <S> * 0.04A <S> = 48mW. <S> 24V <S> * 0.02A <S> = 48mW. <S> As for the current rating of the FPC, keep in mind that RGB led strips have 60mA per segment (3 Colors x 20mA), and a typical 5 meter roll has 300 leds, so 100 segments. <S> 6 Amps are full power. <S> 12V <S> * 6A = 72 Watts. <S> But keep in mind, that FPC has a high resistance per meter. <S> You quickly start seeing color fading at the far end of the strip. <S> Ideally, any long runs are have power injected at 2.5 meter increments.
| If you were to place two segments in series, you would not get the same current if you do not double the voltage. The current drawn by a single LED is equal to the current drawn by the other LEDs in series with it.
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Grounding a Device And/Or Room Hoping this is the correct exchange. At the moment I'm currently renting my house in Canada (120V), and it seems the landlord has attempted to try pull a fast one. From what I understand, or possibly before the previous tenant, there were only two prong outlets installed in the house. The landlord decided to install three prong outlets with of course not having a ground connected. This is obviously terrible for pretty much everything in my house, but unfortunately the rental board won't do anything about it and I can't afford to do anything myself for the entire house, so instead I'd like to attempt to ground a single outlet or atleast myself (then possibly move on to the rest of the house slowly). I'm putting together a recording studio in the basement, only to find I get zapped by my instruments and can hear popping and the "sound" of electrical current on my tracks. Would anyone happen to have a suggestion to help clear this up? I used to build computer for a living, so wiring I can get away with, but this I'm a bit stuck on. Any help is appreciated! EDIT: Just to make sure I'm not crazy, here's a few images of one of my outlets. Got hot and neutral, but no ground. https://i.imgur.com/TX7Lyxc.jpg https://i.imgur.com/IKBMANe.jpg https://i.imgur.com/zXc9fyc.jpg <Q> The boxes should be grounded. <A> The recording studio's hum and click <S> problems may be ground loops and surges on the line rather than lack of a safety ground. <S> Get everything plugged into a single circuit and sharing a ground, even if that ground doesn't go to earth. <S> Getting zapped by your instruments shouldn't happen. <S> Something in your setup is damaged and unsafe. <S> Isolate it and fix it. <A> This usually involves capacitors to earth and the chassis. <S> So, on your unearthed chassis you have approximately half the AC supply voltage and this is easily felt. <S> What I'd do is use an AC power isolation transformer and connect one of the outgoing feeds to an earthed metal rod - <S> this creates a neutral wire and the other wire can be regarded as "hot" or "live". <S> Take care though <S> - you can get a nasty "belt" from 120V.
| Getting small zaps from unearthed equipment is normal because modern EMC rules "suggest" the use of AC input filtering. In the old days people would jumper the ground to the box screw (potentially dangerous if the box is not properly grounded).
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Heat sink, how much does fin and base width matter? I have lot of aluminum sheet 0.5mm thickness at home, normally heat sinks that I have seen had 1mm thickness. Now I am wondering should I get that sheet welded as heatsink or do I absolutely have to invest in thicker sheet for efficient (at taking away and disposing heat) heat sinks ? <Q> Rather than ask about how such a heat sink would work, why don't you try actually computing it? <S> http://www.heatsinkcalculator.com/calculator.html will let you design your own. <S> Note that I entered "heat sink design software" into Google, and that was the third or fourth item on the list. <S> Let's take your example. <S> You did not specify the number of fins, so I picked 10. <S> The calculator says your 40 x 40 mm, 50 watt Peltiers will operate at about 160 C. <S> Is that a good heat sink or not? <S> Since do not provide a quantified performance goal, there is no way tell for sure, but most likely Not. <A> three ways for heat transfer: <S> conduction- <S> usually a good way of transferring heat. <S> convection- <S> fluid movement...caution: <S> stagnant air is a good heat insulator. <S> Radiation- often overlooked. <S> Heat transfer to the air through use of a heat sink starts with conduction from the heat source through the source-sink interface. <S> For best results, the interface should be bonded or as intimately made close to the source as possible. <S> Some bonding materials, if used correctly, aid in heat transfer. <S> The selection of the sink should be based on its conductivity. <S> Metals such as copper and brass are better than aluminum. <S> Some porcelains are good under discrete devices to spread heat to the heat sink on top where the fins are... <S> If possible position the device near a metal part of the container at a lower temperature so that the heat sink can be designed to contact it. <S> Others noted elsewhere have good ideas on the importance of fin design and relationship to total area exposed to air and fans to increase airflow. <S> To use radiation, the distance and field of view to a cooled surface plus the surface <S> field of view are important. <S> Radiation is a function of the surface absorptivity and emissivity. <S> For most cases, a black surface with high a/e will create radiant flow depending on the temperature of the cooler surface. <S> I hope this will give a general perspective in planning to cool a hot spot. <A> Thin metal will distribute the heat poorly; you will get a hotspot near the heat source and the outer regions of the metal will be relatively cool. <S> (And 1mm already sounds thin to me, all the heatsinks I have seen approaching 28cm long have been considerably thicker) <A> Cooling electronics it is not an easy matter. <S> First please look at here Designing a copper plate heat sink to have an idea about the relevant calculations. <S> I don’t recommend to invest in metal sheets as well, <S> but -roughly- if you have a sheet of 0.5mm thick alum. <S> and 20x20cm side, do not expect to dissipate more than 5W and to keep your heatsink at the temperature of 70 degrC. About the same final temperature you can achieve with a piece of 12x12 cm alum. <S> 2mm thick, or you can mount the semiconductor to one side of your (iron) metal enclousure, <S> 2mm thick, if it is sized 18x18 cm Finally the lower the thermal resistance needs your system, the most critical the calculations are, and the most difficult the selection process is. <A> Since you seem pretty attached to using your thin sheet metal, there is a possibility you can try. <S> Make a series of U channels with your .5 mm sheet, each channel the same length and about the same height, but with gradually decreasing width. <S> Stack them together, and weld the areas in contact. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This figure shows the idea in cross-section. <S> If you use, let's say, 10 pieces of aluminmum, the maximum base thickness will be 5 mm. <S> If you can produce small differences in width you can produce a lot of fin area for a given overall width. <S> There are a couple of problems you need to think about. <S> 1 - You must keep the bottom of each channel absolutely flat during welding, or the finished heatsink will not have a flat base, and you will not be able to make good thermal contact with your Peltiers. <S> This probably means you will need a spot welder, rather than an arc or gas welder, and will require real attention to what you are doing. <S> 2 - Because each channel is connected to the others at separate points, the thermal conductivity of your final base will be less than if you made it out of solid aluminum. <S> It may not be too bad, though, since each sheet is pressed against its neighbors. <A> Aluminium heat sinks are always made with the fins too thick. <S> Because optimum-thickness aluminium heat sink fins are too soft and fragile. <S> Notice that any (old) steel heat sink you see will have thinner fins than any (new) aluminium heat sink: even though the thermal conduction is worse, the toughness of steel allows thinner fins. <S> Fortunately, the exact thickness of fins is not a critical parameter for a convective-flow heat sink. <S> Making fins twice as thick only slightly reduces the fin spacing, which only slightly reduces the airflow, which only slightly changes the thermal resistance of the system.
| Nobody wants to spend much money on heatsinks : if the commercial ones could be made thinner and cheaper, they would be. It is best to have the sink enclose the device as much as possible.
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Why peltier elements have smaller heatsink on cold side? I have seen in nearly all configurations that Peltier elements are mounted with huge heat sinks on hot side as compared to tiny ones on small side. Is this just a way of explaining the need of heat sink on hot side? Or are there any real benefits of smaller heat sink on cold side? Would an element work better(produce lower temperature on cold side) with equally sized heat sinks? <Q> The Peltier cell generates far more heat on the hot side than it consumes on the cold side; the difference being simply the power you are pumping it with. <S> At a minimum, about 2.5x more, and at large temperature differences, more than that. <S> The different sizes of heatsink simply compensate for that. <A> This is a simplified explanation, however it should help explain why the cold sink is smaller than the heat sink: A peltier cooler pumps heat from the cold side to the hot side, and in addition a peltier cooler generates heat in addition to the heat its pumping . <S> A peltier unit's Qmax (the amount of Watts it can pump) is rated at zero delta T (the difference in temperature between the cold and hot sides). <S> Additionally, a peltier cooler's ability to pump heat diminishes as the delta T rises. <S> At some point (depending on the peltier unit design, etc.) <S> the peltier will pump zero watts (btu) of heat when delta T rises too high. <S> So, to get the coldest temperatures possible on the cold side, you want to be able to remove as much heat as possible on the hot side. <S> This needs to include any heat created by the peltier. <S> This coupled with the diminishing heat pumping ability when delta T rises, means that the cold sink is smaller than the heat sink, so the system has a high (or higher) coefficient of performance (COP) than a system with same size cold and hot heat sinks. <S> Hope this explains it. <A> A Peltier element creates a heat flux from the cold side to the hot side of it. <S> This means that: it is "absorbing" heat energy from the cold side it is transferring heat energy to the hot side <S> it is taking in electrical energy to do <S> that Heat and current are the input energy sources. <S> In order to work it must dissipate (output) at least what gets in. <S> Otherwise, the remaining extra energy will overheat and destroy the element. <S> That's why the heatsinks on the cold side are smaller than the ones on the hot side. <A> If you make the cold side heat sink larger, it will be able absorb more heat. <S> To emit that heat, the hot side heatsink needs to be larger again. <S> There will be a limit somewhere where the cold side won't absorb more heat, but assuming someone designs a minimum cost, you'd want the smallest heatsink that still works. <S> This means that the hot-side heatsink will be larger than the cold side one. <A> There's really no justification for it. <S> The two heatsinks could be any size at all. <S> It's just a question of how much thermal resistance you want to have between each surface of the Peltier device and its respective "ambient" environment. <S> Similarly, the hot face is the hottest point in the system. <S> The size of each heatsink determines its thermal resistance. <S> The thermal resistance multiplied by the heat flow determines the delta-T between the Peltier device and the environment. <S> Generally, you want that delta-T on each side to be as small as possible, so you use the biggest heatsink that fits. <S> It often comes down to the fact that the space being cooled is fairly compact, and so only a small heatsink fits on that side. <S> Sometimes a small fan is used on that side to improve its effectiveness. <S> A secondary issue is that if you have large heatsinks on both sides of the device, it can be difficult to insulate the space between them effectively — if heat from the hot-side heatsink flows directly to the cold-side heatsink, the overall efficiency of the system is reduced.
| The coldest point in a Peltier-based system is the cold face of the device itself; the thing that it is cooling must necessarily be warmer; otherwise there would be no flow of heat into the face.
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What are the pins for in this Relay module? I am trying to figure out how to use this relay module. There is no datasheet that I could find for the whole module, there however are for individual box. Here is the picture below, there are three jumper pins which read VCC, VCC, GND. And there are four activator pins (I guess) they read GND, IN1, IN2, VCC At the output end, I got current coming in connected to Middle terminal and receiving wire connected to the one showing Isolated line. To activate it, I am applying 3V DC -ve on the GND (on the activator pins set) and +ve on the IN1. Am I doing it wrong ? Here is the URL for further info: http://www.miniinthebox.com/2-channel-5v-high-level-trigger-relay-module-for-arduino-works-with-official-arduino-boards_p727426.html <Q> Typical relay module clone, they are all similar. <S> Below schematic is per channel. <S> Keep in mind that RY-VCC (JD-VCC in the schematic) is for the Relay section, while VCC on the 4 pin connector is for the opto-isolated side. <S> https://arduinoinfo.mywikis.net/wiki/ArduinoPower for a full wiki on using these. <A> See passerby his answer for the whole schematic of your module per channel. <S> For the data sheet of just the relay check sbell's answer. <S> However, to activate it you should provide the VCC with +5V. <S> The GND connected to the GND and give a +5V signal to the IN1 or IN2 to activate it. <S> IN1 and IN2 should be connected to one of your output pins, they will be able to switch your relay. <S> But when I look at the schematic passerby send you it might also be possible that you'll have to put your output pin to LOW in order to activate it. <A> I just wired up one of these modules a few minutes ago to confirm the info I'm posting below. <S> From your controller (eg Arduino or Raspberry Pi) connect Ground (0V) to the relay module's GND and Power to the module's Vcc. <S> In most Arduinos the power will be 5V, in a Raspberry Pi it will be 3.3V - but these relay modules will work OK with 3.3V even though they are officially rated at 5V. Connect one of your I/ <S> O pins (in output mode) to IN1 and the value on that I/ <S> O pin will control Relay #1. <S> Connect <S> a second I/ <S> These modules have the circuitry described above built-in <S> so you don't need to add it - <S> that's only needed if you are driving a stand-alone relay directly. <S> All that's usually needed is a few Dupont jumper wires. <S> The other three pins (GND/Vcc/JD-Vcc on a separate connector) are normally not used and the jumper is left in place. <S> They are used to power the relay switching and pick up the Vcc and Gnd off the data connector. <S> By removing the jumper you can power that other Vcc pin separately. <S> If you need to do this or need the optical isolation feature to keep high voltages away from your controller, you need more expert advice than we are giving here; in fact if you don't already know what to do then you shouldn't be doing it as there is a good chance of damaging both people and devices. <A> I just came across this as I am using the same modules, and they don't work for me. <S> I need a relay that comes 'ON' with a 'high' input (I am using one of the contacts to power off the pi running the show)I have modified the module to work this way. <S> To do this :- Get rid of the jumper (between JD-VCC and the left VCC) <S> Cut the track on the back of the PCB connecting the two VCC's together Solder a link across 4 contacts; GND, GND, IN1 & IN2 (they are adjacent) <S> Connect power (3.3 to 5v) at JD-VCC and 0v to anywhere on the joined contacts (from 3.) <S> Use the left VCC as IN1 (+ve input trigger) <S> Use the right VCC as IN2 (+ve input trigger) Just tested this using the Pi 5v rail for JD-VCC and GPIO outputs (3.3v) to drive the triggers and it works fine. <S> The inputs sink about 0.47-0.48mACheersMuttley <A> Based on the identifying marks in your picture, you should be powering the relays with 5Vdc to activate them (per the data sheet ) <A> You can activate the relay by using npn transistor(bc547) connect the output of the microcontroller to the base of the transistor via 10k resistor, collector to ground ( - negative of your supply ) and emitter to the relay input. <S> This relay are different from the normal relays they operate by ground connection.
| O pin to IN2 to control Relay #2.
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Convert 3.3v to 14v? I want to use a 4D Systems OLED display with an ATmega32u4 based PCB. I'm still working on designing the PCB and have hit a problem. In the data sheet it says that one of the power supply pins for the display requires 13.5-14.5 volts (which is VDDh). The problem is I will be powering the ATmega32u4 from a lipo battery with a 3.3V regulator. How do I convert 3.3V to 14V? Would I need to use a boost converter? <Q> You will need to use a boost convertor. <S> For best efficiency, you'll want to use a dedicated boost IC, rather than spin your own. <S> It's not the best in performance, but I like the LM3578 for use by inexperienced people. <S> It's adjustable, and the output voltage is set through a pair of resistors. <S> As long as you put a reasonable LC network (see the datasheet) on the output and use a Schottky diode, it will regulate. <S> It's also available in a DIP package for easy prototyping. <A> There are ways of doing this with discrete components (which I'm not really qualified to comment on), but another alternative would be an all-in-one package, such as MEE1S0315DC , which can convert 2.9-3.6V to 15V (but only ~1W, ~60mA). <S> Looking at the datasheet for the OLED, that should be enough current, and the voltage can be dropped to ~14V with an inline diode (ie a 1N4001). <A> As others have said, you need a boost converter. <S> For the best efficiency you want to go directly off of your battery voltage, not off of the 3.3V. <S> Also, the 3.3V regulator might not be able to supply enough current. <S> It would probably be worth it if you are making your own board already and going into production, but for prototypes there are numerous boards available with adjustable output voltages. <S> Mostly they're on sites like Ebay or Alibaba. <S> None of them are "name brands" or anything, you're buying direct from the factory, but I've had pretty good experiences with them. <S> Here's a few examples . <S> You obviously want to make sure they cover your whole battery voltage range, can output the voltage you need, and have enough current capacity.
| What you will need is a Boost converter, to convert the voltage from 3.3V to 14V. You could spin your own design with an existing chip like Matt suggested, but I've heard that even the circuit board layout is difficult for switched mode power supplies.
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What will I see in oscilloscope when attempting to analyze Bluetooth or Wi-Fi signals? I would like to know what will show up if I put bluetooth or wifi carrier signal (2.4GHz family) in a say 50MHz oscilloscope? Am I going to see something like a rectangular block? And the only information I will get is the amplitude? <Q> Nothing. <S> That is, you'll just see a flat line, as if there's no signal. <S> It's too high frequency to see. <S> The input stage on a 50 MHz scope is not configured to pass a 2.4 GHz signal. <S> It'll be filtered out, and you'll be left looking at the (probably nonexistent) low-frequency components of the signal. <A> If it's a good scope, you'll see nothing at all, just a flat line representing DC.. <S> Ironically if it's an inferior scope with incompetently designed input amplifiers and the signal strength is high enough, you may see a step in the flat line when the poorly filtered RF carrier is demodulated by nonlinearities in the amplifier, but that will only tell you the presence of carrier and nothing else. <S> And if the carrier is continuous, you won't see the steps, <S> so you'll never know it's there. <S> This is the same phenomenon as AM radios and landline telephones picking up a distinctive "dud - dududud - dududud" noise from a GSM phone on the same table. <A> (You are then viewing relative power across a frequency band and not the signal itself -- it is important that you realize the difference, I think.) <S> None of the signals in that band are continuous -- they are not dedicated simplex channels. <S> What you will see on a spectrum analyzer is transient power spikes when devices transmit. <S> As those bands are for low powered public use, they are not designed for continuous output because the band would become useless quite quickly and the devices would consume enormous amounts of power in an attempt to communicate in all of the cacophony. <S> There are inexpensive spectrum analyzers available to use for this type of signal analysis, but oscilloscopes are useless for nearly all RF work. <S> Once you have demodulated the RF signal and extracted an Intermediate Frequency, you can use a storage scope to view the intelligence, or IF, depending on the signal. <S> A good signal to work up your bench skills is something like USB -- because you can easily create a setup to capture and analyze that stream. <S> RS-232 is even better because the framing is very apparent and the data rate is relatively low.
| The signal you have described is what you would see in on a spectrum analyzer, which is used for RF signals like the ones you are asking about.
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Object detection using sensors and then reporting the location I am trying to prototype a sensor + location reporting chip that will do the following: Sense if a object is present over the top of it (a vehicle to be specific) Reporting the location of the chip (Not of vehicle) The second option may be hardcoded into the chip and should be reprogrammable. These chips would needed to be installed in large numbers and there will be several such chips installed in close proximity. Each will report its own location to a centralized server (via wifi?) That centralized server will then send the locations and busy/empty slots to a remote server which will then update some database. I know the above description is very broad in terms of technical aspects but I need a starting point. Another thing to consider here is that I need to keep the cost of such chips lower considering that there will be 100's of such chips needed. <Q> Detecting vehicles above a roadway is usually done by what is essentially a large metal detector. <S> A coil of wire is embedded in the pavement. <S> Together with other circuitry, a oscillator is created such that the oscillation frequency is sensitive to the coil inductance. <S> Since frequency can be measured very accurately (50 ppm is cheap), even small changes of inductance can be detected. <S> Detecting a steel car chassis 1 foot above the loop is routine. <S> This is how most sensors for automatic traffic lights work, for example. <S> As for how to get the data back to a central location, you haven't given us enough details. <S> You mentioned parking spots, which implies a fixed installation that needs to run for a long time, so running wires for power seems appropriate. <S> While you're providing the wired power, you can run some communication lines at the same time. <S> There are various options, but CAN sounds quite appropriate as a first reaction. <S> Your bandwidth requirements are quite low. <S> CAN running at 125 kbits/s can easily cover something the size of a parking lot. <S> The limit with CAN will be the number of nodes the bus can handle electrically, which is usually 80-95 nodes per bus, depending on exactly what transceivers are used. <S> Even with a separate CAN bus for each 80 parking spots, the extra master nodes will be a tiny fraction of the overall cost. <A> It appears as though you would like a device that is cheaper and has a smaller footprint than what Olin suggested. <S> I am answering based on your provided specs. <S> I would not expect any RF sensors to be as weather-resistant or accurate as what is currently done all over the world to detect stationary vehicles. <S> Since you are discussing building a 'prototype', I would encourage you to experiment with several sensors from different manufacturers -- the solution you desire may require more than a single sensor per device depending on your desired precision and accuracy. <S> Possibilities may include ultrasonic, infrared proximity, or pressure/vibration sensors that can all be purchased on Sparkfun. <S> With regard to to the wireless transmission of sensor information, there are several things to keep in mind. <S> First, range is an issue if you're talking about large parking lots which may prohibit 802.11 (wifi). <S> You may wish to consider Zigbee or 802.15.4 radios (as are used in electrical power smart meters). <S> Additionally, you may have problems trying to transmit with a large vehicle on top of the radio -- I would suggest either isolating the transmitter and/or antenna away from the vehicle (say, on a light pole). <S> I agree with Olin that running power may be necessary. <S> Depending on your access, you may be able to pull this from electrical units local to the parking lot (which may also be associated with light pole). <S> However, if you are looking into low power units (as it seems you are), batteries with mounted solar chargers may suffice. <S> As you said, your question is very broad and opens up a lot of possibilities and potential limitations. <S> Please post or message if you have more specific questions. <A> If you want to detect only the presence of any vehicle there, i would suggest Giant Magnetoresistive Sensor or short range Radar sensor (used for crane collision avoidance) instead of IR or ultrasonic transceivers. <S> As far as reporting of the location is concern, use the digital addressing to interface all the sensors with a main station. <S> As Lathrop sir suggested 802.15.4 would also be very efficient to make it ease mobile setup. <S> Thanks <A> For the problem of monitoring a lot of car parking slots, I'd guess you don't want to have to dig up every one to install an induction loop or other sensor. <S> IP CCTV cameras are cheap now, and the infrastructure is easy & cheap. <S> One or two HD cameras mounted on a single pole could monitor hundreds of spaces, and that makes the wiring even easier. <S> For object recognition they don't need to be great cameras <S> and you can in fact get better results by using them in night-vision mode (IR sensitive, mono picture) as the contrast is much better. <S> ANPR cameras typically run in this mode with fairly bright IR illuminators as number plates are highly reflective <S> and you don't care about colour data - it's possible the floor of your car park is quite bright (concrete) or dark ( <S> tarmac) in the IR spectrum, which makes the recognition easier. <S> As you are detecting stationary objects <S> your image capture & processing requirements are pretty low, you can get away with very slow updates from the cameras and a central PC or whatever doing the processing & spitting out updates. <S> If it updated every few seconds it'd be plenty. <S> To make the software task even easier you could paint marker dots on the floor, if they are done in reflective road-marking paint and you use an IR-sensitive camera & illuminator it pretty much becomes a job of counting the bright dots against a dark background.
| I'd look seriously at using a few cameras and running OpenCV or similar to recognise when there's something in a space or not.
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What happens when you fall off the end of main() in a PIC? Another question included this program: void main(void){ TRISD = 0x00; PORTD = 0xFF;} Let us assume that it does exactly what was intended up through the last line of main() . My question is, what happens in a PIC18F4550 when the flow of control falls off the end of main() ? I assume any decent PIC programmer will fill the instruction space after main() with either all-1s or all-0s, but I don't know which. I can see from the PIC instruction set docs that an all-zeroes instruction is NOP , which is harmless in itself, but what happens when the program counter hits the end of the instruction space? Does it wrap around, halt, catch fire, emit monkeys...? If PIC programmers fill program space with all-1s instead, as in an empty EEPROM, it appears that that will be an ADDLW instruction that adds 0xFF to W . If that is correct and we don't care about what happens to W , is this the same as the NOP case? Do the answers to the above questions argue for doing something else, such as running an infinite loop to prevent execution from leaving main() ? I don't think we can sleep the microcontroller here, since that won't keep the port D lines up indefinitely, as the OP wanted. <Q> You can easily answer this for yourself by looking the disassembly of the compiled code in MPLAB, for whatever compiler you happen to be using (and by stepping through the program using MPSIM). <S> This is considered a "freestanding environment" because there is no operating system. <S> Officially, from the ISO/IEC 9899:1999 (E) C standard, "The effect of program termination in a freestanding environment is implementation-defined " (emphasis added). <S> It is probably not a good idea to invoke implementation-defined behavior if it is not necessary. <A> main() is a function just like any other. <S> When you "run off the end", it returns to whoever called it. <S> For a discussion of what happens then, see this question . <A> If it were my intent to run the processor in an infinite loop at that point, I'd write it explicity and not count on whatever the C/C++ initialization code does if it ever gets control back (though one likely scenario is that it re-runs the application, initialization and all).
| Typically it will start back again at the beginning rather than executing random stuff in the program memory, because the program will be entered with a 'call' in assembly language and will end with a 'return 0'.
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Grounding between enclosures in a closed system Currently we are developing enclosures in an isolated "system of systems". Our overall system is fed AC power and converted to 28V with an off-the-shelf converter. Currently the enclosures we are designing each have a PCB internally and currently these PCB's power supplies are tied to the same ground (dashed red lines shown in the below diagram). My project manager is worried that this connection will cause problems when we go to test these items for EMI and he wants to remove these connections and have isolated lower voltages. I'm contending that the grounds should be connected because there won't be any common reference for the single ended connections between each enclosure. Also, some of these circuits use the 28V input as a reference as some of the signals are 28V. Do these lower voltages need to be isolated from the 28V? <Q> Since you are showing analog and signal grounds being connected between the two boxes, connecting the power grounds together is not a great idea. <S> You are inviting ground loops. <S> That said, if your boxes are at all separated physically, and your interconnecting signals are low voltage/ high impedance (10 volts, a few k), you will be very well advised to use differential transmission for your interbox signals. <S> Particularly when dealing with 28 volts systems, the IR drop on the power lines can produce significant voltage offsets. <S> Since your 28 volt return is labeled ground, if your boxes are likewise grounded you need to be extremely careful of your internal ground routing. <S> Make sure that your signal and digital grounds are isolated from case ground by several k. <A> Set your system up so that you can easily swap between several different connection schemes. <S> (like with disconnectable grounding straps, resistors you can easily clip or solder on, etc) Try them all out at home and see how well each one works. <S> When you go to test EMC, have your wits about you and test the most preferred configuration first. <S> If you fail, go to #2. <S> If you fail, go to #3. <S> If you keep failing, start throwing ferrites in the system. <S> This is the way it's done in 'real life'. <S> A lot of engineers get a sense for things, but without being able to actually LOOK at your setup and consider all the random factors you've probably left out, any specific advice is kinda worthless. <S> Besides, even engineers who know what they're doing screw it up and have to reconfigure slightly at the lab. <S> The previously mentioned advice about ground loops is legit, so you should take that into consideration if you're having noisy signal issues. <A> I second Daniel's scheme, which is to prepare for contigencies. <S> Then, each enclosure will form a Star ground to the external supply. <S> The supply should then be earth grounded. <S> You should carry ground between the enclosures along with the IO lines so all references are common. <S> This also reduces the loop area for the ground return, thus reducing emissions.
| Ideally, all blocks of each enclosure will form a Star ground to the chassis.
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Set a constant high signal to low I have a keypad circuit, when I press and hold a key, the signal "key_pressed" is always high, as long as I keep the key pressed, which is normal, when I leave it, it gets low again. But I don't want this signal to be constantly high, no matter how long I press the key, I would like to be detected as a single short press. In other means I should find a way to set this key_pressed to low after lets say 10ms. I'm doing everything in VHDL, no resistors capacitors... I also can't wait for a falling edge then it makes no sense, no key will be displayed on the LCD until the key is released which does not make sense. <Q> The way I do this in firmware is to detect a change in the input (after rejecting noise and bounce) and then generate an event based on key press or an even based on key release for each key. <S> You can then use the key press and release events, in conjunction with timers to do all the usual functions you associate with keys on modern electronics. <S> Hold the key for 3 seconds to get into a special mode, Hold the key, then release it to lock in an encoder setting or whatever. <S> Edit: <S> Based on your changes, in the case of using VHDL (hardware) <S> this can be accomplished with a clock and flip-flops. <S> Simply compare the input X to a delayed version of the input X'. <S> That will give you a one-clock pulse for the input rising edge (and, if you want it, a one-clock pulse for the input falling edge). <S> It works just the same in an HDL as in sequential code. <A> You need a rising edge detector. <S> This is done by generating a delayed input that is 1 clock cycle later than the actual input. <S> For example: ____ ____ ____ ____ ____ ____ ____ ____ CLock ___/ <S> \___/ \___/ <S> \___/ <S> \___/ <S> \___/ <S> \___/ <S> \___/ <S> \ <S> _______________________________Input <S> ___/ <S> \ <S> ____________________________________ _______________________________Input_z ____________ <S> / <S> \ <S> ___________________________ <S> ^ <S> You want to detect this point, where the signals are not the same. <S> Input has gone high, but because Input_z is delayed, it isn't high yet <S> The delayed input is generated as follows: gen_input_z : process(clk,rst) begin if (rst = '1') <S> then Input_z <= '0'; elsif (rising_edge(clk)) then Input_z <= <S> Input; <S> end if;end process <S> Now you want to detect your rising edge: gen_edge_det : process(clk,rst) <S> begin if (rst = '1') <S> then Edge_detect <= '0'; elsif (rising_edge(clk)) <S> then if (Input = '1' and Input_z = '0') <S> then Edge_detect <= '1'; else <S> Edge_detect <= '0'; end if; end if;end process <S> But now our edge detect is only one clock cycle: ____ ____ ____ ____ ____ ____ ____ ____ CLock ___/ <S> \___/ \___/ <S> \___/ <S> \___/ <S> \___/ <S> \___/ <S> \___/ <S> \ <S> _______________________________Input <S> ___/ <S> \ <S> ____________________________________ _______________________________Input_z ____________ <S> / <S> \___________________________ ________Edge_det <S> ____________/ <S> \ <S> __________________________________________________ <S> To modify that, add in a counter which only makes edge detect fall after a certain number of clock cycles: -- <S> Port Declarationssignal clk : in std_logic;signal rst : in std_logic;signal input : <S> in std_logic;-- Signal declarationssignal input_z : <S> std_logic;signal edge_detect : std_logic;signal counter : unsigned(31 downto 0); -- include numeric_std for thisgen_edge_det : process(clk,rst) begin if (rst = '1') <S> then Edge_detect <= '0'; counter <= '0'; input_z <= '0'; <S> elsif (rising_edge(clk)) then input_z <= <S> input; <S> if (Input = '1' and Input_z = '0') <S> then Edge_detect <= '1'; counter <= <S> (others => '0'); elsif (counter < 2) then <S> -- we want an edge detect of 2 clock cycles Edge_detect <= '1'; counter <S> <= counter + "1"; -- declare counter as unsigned. <S> else <S> Edge_detect <= '0'; end if; end if;end process <S> Now it's doing what we want: ____ ____ ____ ____ ____ ____ ____ ____ CLock ___/ <S> \___/ \___/ <S> \___/ <S> \___/ <S> \___/ <S> \___/ <S> \___/ <S> \ <S> _______________________________Input <S> ___/ <S> \ <S> ____________________________________ _______________________________Input_z ____________ <S> / <S> \___________________________ _____________Edge_det <S> ____________/ <S> \_____________________________________________ <A> I'd say you're looking for a monostable multivibrator. <S> You can make them from a 555 timer, or use a dedicated chip. <S> For TTL, it's available as the 74xx123 component. <S> They work by detecting an edge and outputting a programmable duration pulse. <S> Pull up a datasheet. <S> They should contain example circuits for you to cut your teeth on.
| You can set the pulse parameters with external resistors and caps, if I remember correctly.
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DC or (AC + DC) coming out of a full wave rectifier DC or (AC + DC) coming out of a full wave rectifier Is there a truly correct answer to this? Considering the representation as a waveform I would say its an AC wave form offset by some DC amount. Using the standard electronics notion I would say it is purely DC. A quick Wikipedia such on Rectifier returns, "A rectifier is an electrical device that converts alternating current (AC) , which periodically reverses direction, to direct current (DC), which flows in only one direction." I know that they both can be true but, I am curious if any consensus exists for the green waveform below. The question originated from can you feed a, "dc voltage with a bad ripple" into a dc to dc buck converter.... At first I assumed no using the notion that DC meant constant, but then thinking about the output of the buck is frequency driven I started to have second thoughts.Thanks,Josh <Q> What is DC, anyway? <S> If we have a square pulse of 1V for 1ns, is that DC? <S> What if it's for 1ms? <S> 1s? <S> 1 day? <S> 1 year? <S> "It's only ever positive -- it doesn't alternate", I hear you thinking. <S> OK, what if we have a square wave that alternates between +1V and -1V <S> every 1 ms. <S> Clearly this is AC, right? <S> What if the edge transitions occur only every 100 years? <S> Is it DC now? <S> What if the device has only been on for 99 years and you don't know that it's going to change yet? <S> Point being, the notion of "DC" is arbitrary, and usually chosen to be something convenient for a particular circuit or situation. <S> When voltage or current changes so slowly or so insignificantly that we can consider it as not changing it at all, it's DC. <S> For example, if we are discussing microwave circuits, then the 60 or 50Hz AC on your wall might as well be DC. <S> The rate of change in that waveform is so slow compared to the reactive components in a microwave circuit that it might as well not be changing at all. <S> In other contexts we might define DC in terms of Fourier analysis . <S> In this case, "DC" is the average value. <S> However, the average value depends on the span of time we choose to consider in the analysis. <S> For example, in the case of our +/- <S> 1V alternating every 100 years, the average is 0V, but if we look at just one particular day, the average is likely to be 1V or -1V. <S> In the case of a power supply such as you have drawn, an engineer is likely to call that "DC". <S> An engineer understands that a real power supply will necessarily have some ripple, so you don't have to say "this is a DC with some ripple" unless the ripple is unacceptable for your application. <S> If the engineer wants to quantify the ripple (usually to keep it under some specified maximum), then it will be treated as an AC component, and the DC component will be calculated as an average taken oven a span of time sufficiently long to average out all the variations. <A> A quick Wikipedia such on Rectifier returns, "A rectifier is an electrical device that converts alternating current (AC), which periodically reverses direction, to direct current (DC), which flows in only one direction." <S> See my answer here . <S> ' <S> The answer' is that it depends on what one means by DC. <S> I believe that it is safe to say that, for most, "DC" no longer stands for Direct Current which is defined as a current that does not alternate in direction as opposed to Alternating Current which does. <S> In most contexts, "DC" is a synonym for constant. <S> For example, a (good) 5 VDC power supply produces a (more or less) constant 5V and not, for example, a varying but positive voltage. <S> So, you are correct. <S> However, the output is not DC in the sense that the output voltage is not constant but, rather, varies with time <S> (though, it is not alternating either). <S> Some would call this output " pulsating DC " while others would consider that a contradiction in terms. <S> What can be said, without equivocation, is that the output of the rectifier is non-alternating , i.e., the voltage and current do no alternate in polarity or direction. <S> One can also say, without equivocation, that the output can be decomposed into a constant (not time varying) component and one or more alternating components. <A> The difference between AC and DC is determined by its average. <S> An AC voltage supply has an average voltage equal to zero; a DC voltage supply has an average unequal to zero. <S> In real life, most DC sources have some AC disturbances, called 'ripple'. <S> The larger the load (R1), the larger the ripple. <S> Take R1 away, and you would see an almost perfect DC voltage.
| The output of a rectifier is DC in the sense that the current is unidirectional.
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How do I invert a signal without deforming it, and without using a logic gate? I have an IC which outputs pulses over an open collector pin. I need both these pulses and their inverses in my circuit. I want to do the inversion by using as few as possible transistors and without using any ICs. First, I tried simply inverting the signal by a single BJT as seen below. But this won't work. Because, when Qin is on, SIG will be low, and when Qin is off, again SIG will be low since this time base of Q1 will ground the signal. The signal will be lost in any case. Connecting a resistor to the base of Q1 may seem to be a solution, but this way, the signal will be attenuated by this resistor network (it will be less than 12V). Next, I tried first generating a copy of the signal by and emitter follower circuit (by Q1 in the image below), then do the inversion by this copy of this signal. But again the same thing happened; now the signal is lost through the bases of Q1 and Q2. In my final attempt, I connected R3 to prevent the loss of signal. By doing so, the signal is not lost, but it is attenuated by the voltage divider network built up with R1 and R3. I added Q3 to pull it up to 12V level as seen below. I think this final circuit will do what I want, but this, time the circuit looks too complex to me. Is there a simpler way to do this with at most two external transistors? (Note: Only 12V is available in the circuit. The frequency of the signal (clock pulse) is 20kHz. The signal and its inverse will drive separate totem-pole transistor pairs. They should be pulled up by 10k\$\Omega\$ resistor at most when high, and should be directly shorted to the ground when low.) EDIT: I rebuilt up my circuit according to both Brian's and Vladimir's suggestions. I implemented two methods in one circuit. Then I simulated them. Here is the schematic: Results of Brian's circuit: (Results are almost similar without the totem pole stage.) Results of Viladimir's circuit: (Results are much more terrible without the totem pole stage.) I'm loosing my hopes on obtaining crystal clear signals. It is only 20kHz and I'm having this much of troubles. Maybe I should give up my stubbornness and use a logic gate IC. EDIT 2: After Viladimir's comment about IRF530 being a wrong choice of MOSFET, I started to try out different transistor types. I got a very good result when I changed Q1 with 2N2222 and M1 with BSS138. The new and better signal timings are in the image below. And these timings are without a totem pole drive; I removed it; it doesn't make a difference now. However, this was just a simulation. Will I get the same (or similar at least) results in the real life? If yes, what was the problem with Q1 and M1? What features of 2N2222 and BSS138 made this circuit run better? I will be glad if someone makes a short comment on this. <Q> Your first circuit is perfect... <S> If you use a mosfet instead of a bjt. <S> A mosfet does not pull anything low when it is on, it indeed is a capacitive load but you are going at 20kHz <S> so that should not be an issue at all. <S> Just be careful: your mosfet should support a quite high \$V_{GS}\$, i.e. 12V, <S> that's not something all MOS are happy to do. <S> Your circuit should look like this: simulate this circuit – <S> Schematic created using CircuitLab <A> What you're looking for is a single-input, differential output buffer. <S> There are lots of examples for this online, for example the last on on this page. <S> It uses 2 transistors. <S> http://www.interfacebus.com/Transistor-Differential-Amplifier-Circuit-Description.html <A> Your first circuit was almost there. <S> The problem was your analysis. <S> But this won't work. <S> Because, when Qin is on, SIG will be low, and when Qin is off, again SIG will be low since this time base of Q1 will ground the signal. <S> The signal will be lost in any case. <S> When Qin is on, yes, SIG is low. <S> This means INV is high. <S> When Qin is off, the base emitter junction of Q1 is held low, but INV is low. <S> What you need is a resistor between SIG and the base of Q1. <S> As you point out, doing this will attenuate the value of SIG, but this is not a problem. <S> Replace R1 with 1 k, and put a 100k resistor from SIG to the base of Q1 (and, just to be safe, another 100 k from the base of Q1 to ground. <S> SIG will be reduced by 1%. <S> Is that really a problem? <S> Otherwise, the circuit will work just fine. <S> By comparison, your last circuit will drive SIG appropriately, except that the output voltage will be no more than about 11 volts. <S> Your currents are low enough that the emitter-base voltages will actually be less than the standard assumption of 0.7 volts, or the result would be even worse. <S> As it stands, assuming 2N3904 transistors, the voltage drop across R1 is about 0.1 volts, and the drop across Q1 and Q1 about 0.45 volts each. <S> Final SIG output is about 11 volts. <S> If a 1% drop (.12 volts) is unacceptable, why is a drop of 1 volt OK?
| Connecting a resistor to the base of Q1 may seem to be a solution, but this way, the signal will be attenuated by this resistor network (it will be less than 12V).
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When to use pull-down vs. pull-up resistors After learning and experimenting with microcontrollers, I've understood the concept of pull-up- and pull-down resistors. I now understand when and how to use them, and how they work. I've mainly used pull-ups because I was taught to, but it has always seemed a little backwards to me, as closing the switch sets the MCU input to LOW. I think it would make more sense to use a pull-down resistor, so that the input is LOW when the switch is open, but that is just my way of thinking. Should I pull my single-throw inputs up or down? When is pulling down preferred over pulling up and vice versa? <Q> The answer depends on what you want the "default" configuration to be. <S> For example, say you have a down-stream N-channel MOSFET, and you want it default off. <S> Then you would use a pull-down resistor to ensure this behavior if the input becomes high impedance. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> On the other hand, suppose you have an upstream P-channel MOSFET, and want it default off. <S> This time a pull up resistor is required to create this behavior. <S> simulate this circuit <S> There's also the alternative case where you want a device to be default-on, in which case the above two cases would be reversed (pull-up for the N-channel MOSFET, pull-down for the P-channel MOSFET). <S> A few other considerations: <S> I2C lines specify pull-up resistors because devices are "expected" to have an open-drain to ground, and thus need some way to raise the line potential. <S> Analog comparators are usually configured as open-drain devices, and thus also need pull up resistors to get a high potential output. <S> Either configuration could works equally well in your application (i.e. there's no significant advantage one way or the other). <S> ... <S> And any number of very application-specific reasons why one configuration may be preferred. <A> If the signal doesn't already have a specification, use whichever one makes the most sense to you. <S> It is your choice to make an input active-high or active-low . <S> If it's buttons, make sure to use a debounce circuit (or do it in software). <S> simulate this circuit – <S> Schematic created using CircuitLab <A> If your circuit design is such that you can choose - in other words you aren't required by the rest of the circuit to use a pull up or pull down - then you should consider safety and security in the case of failure. <S> If your microcontroller fails, or just that output fails, the pull up or pull down will be in effect. <S> How will this change the operation of your device? <S> Will it put the user in danger - for instance by forcing a heating element on? <S> Will it affect security, such as disabling a door lock? <S> Pull up/down resistors determine the default state of the wire. <S> Deciding what the default state should be depends on safety, security, and finally the desired functionality of the circuit. <A> If you are working with an Arduino/ATmega328 you can use the built in pull-up resistor . <S> There are 20K pullup resistors built into the Atmega chip that can be accessed from software. <S> These built-in pullup resistors are accessed by setting the pinMode() as INPUT_PULLUP. <S> This effectively inverts the behavior of the INPUT mode, where HIGH means the sensor is off, and LOW means the sensor is on. <S> The value of this pullup depends on the microcontroller used. <S> On most AVR-based boards, the value is guaranteed to be between 20kΩ and 50kΩ. On the Arduino Due, it is between 50kΩ and 150kΩ. <S> For the exact value, consult the datasheet of the microcontroller on your board. <S> When connecting a sensor to a pin configured with INPUT_PULLUP, the other end should be connected to ground. <S> In the case of a simple switch, this causes the pin to read HIGH when the switch is open, and LOW when the switch is pressed. <S> The Raspberry Pi has them as well . <A> You often want pull-ups or downs - often downs - on outputs of programmable devices like microcontrollers to define their state during power up sequences. <S> Such outputs are often high impedance on power on, and connected devices can get unintended signals if this is not done. <S> If for example multiple supplies are involved it's best to design each section as safe with zero volts on the inputs and to use pull-downs. <S> Bit more obscure than the other answers, but I've seen examples involving blue smoke and threats of legal action.
| You may draw more current using pullup/pulldown resistors, depending on what's hooked to the input/output.
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Why is PFC done 'equipment-side'? I have only a basic understanding of power factor correction, but I am curious why it is necessary for manufacturer's and their customers to be concerned with ensuring a high power factor, other than to avoid larger bills* from the electricity supplier. Why does the supplier not - that is why can they not, I assume - perform this correction with some large capacitance on their end? *To clarify, I mean manufacturers of large equipment, and their industrial customers buying that equipment. <Q> Look at the two scenarios: - simulate this circuit – Schematic created using CircuitLab <S> In the top scenario a reactive load is power factor corrected back at the source of power and <S> the current flowing down the wires (dotted box) will be, for UK AC and a 0.1 henry load this: - Current <S> = \$\dfrac{230V}{2\pi\cdot <S> 50Hz\cdot <S> 0.1H}\$ = 7.32 amps <S> In the lower scenario the power factor correction is done right up at the load and, because the capacitor and inductor are pure components (and perfectly in antiphase) <S> the current flowing down the wires is ZERO. <S> This means cable power dissipation is zero in scenario 2. <A> The problem with non-power factor corrected loads is that they present an impedance other than purely resistive to the power line, with the reactive part of the impedance storing charge from the source during one part of the AC cycle and returning it during another. <S> This charge isn't used to do work in the load but - since it moves back and forth through the resistance of the mains wiring - is converted to heat (as an I²R loss) and is wasted. <S> Small commercial and residential customers don't ordinarily pay for the reactive power they waste, so the concern isn't about larger bills; rather it's about the waste affecting the utilities' capacities and its cost to the environment. <S> The problem with trying to correct power factor at the source is that even if all of the loads' reactances could, somehow, be summed and presented to the source as a single impedance, compensating for that lumped reactive part of the impedance at that point would lead to wild variations and fluctuations of voltage at the various load sites. <S> Compensating at each of the load sites, however, would render all the loads resistive, with the effect that as they were connected and disconnected from the mains, in parallel, all that would happen is that the source would see only a changing resistance and would supply the current required to satisfy the changing demand. <A> Power factor correction goes well beyond just making the voltage and current in phase. <S> The use of switching devices has increased the harmonics and distortion on the power grid. <S> Most electric utilities in U.S. now have maximum amounts of harmonic distortion permitted. <S> So, it is not just a matter of paying for the extra usage of imaginary power. <S> Electrical power service can and will be denied to large industrial users who exceed the maximums permitted for harmonic distortion. <S> Europe (EU) has general required standards for power factor correction (including harmonics) for all mains connected devices.
| Large energy-consuming customers are charged for reactive power by the energy supplier and, as a consequence, tune their loads to be resistive and thereby save that expense if it's substantial.
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What microcontroller has 20 or more ADC pins? I'm looking for a microcontroller with at least 20 ADC pins. Most things I'm used to using like an Atmel or Arduino have only a handful of ADC. I need to decode 20 analog wave forms being used to drive a segmented LCD screen. Any suggestions of where to find such an MCU or an alternative approach to solve this problem? Edit: The LCD screen I'm looking at doesn't have a part number and it is custom to the product it is in but I believe it is using a scheme like the one described as "multiplexed" in this blog http://www.circuitvalley.com/2012/02/pic16f917pic16f1907-7segmentlcddriverex.html with 16 SEG pins and 4 COM pins <Q> Many PIC microcontrollers have a lot of ADC inputs. <S> PIC18F45K50 with 25 analog inputs <S> Here you can find other PIC microcontrollers: <S> All 8-bit PIC microcontrollers parametric table <S> You can sort this by "Total of A/D channels". <S> However you should know, that all these inputs are multiplexed. <S> These microcontrollers have 2 or 3 converters connected with multiplexers, so your sampling rate will be at least 8x slower than ADC sampling rate. <S> Alternatively you can use multiple ATmega8 microcontrollers and sync them somehow. <S> If you need high sample rate - you can use some 16-bit dsPIC, some of them have more advanced ADC (500ksps in 12 bit mode, 1.1 Msps in 10-bit mode). <A> An alternative approach: an FPGA, reading 20 high-speed single-channel ADC's in parallel. <S> I suspect you're not going to have much choice about using the FPGA because if you really want to simultaneously read 20 ADC's at 16Msps each then that's 320Msps, or 5.12Gbps of throughput with 16-bit ADC's. <S> That's a serious flow of data to deal with; capturing it is all very well, but then you have to <S> do something with it, presumably store it for analysis. <S> You'd need a very fast SATA disk array to have a hope of storing the stream in its entirety. <S> Fortunately with the right FPGA that's not impossible. <S> In short, this is do-able but a significant challenge. <S> If all you've deal with before is Arduino then you're out of your depth. <A> First off, have you looked into a monitor driver IC for your specific display? <S> Most displays like that have ICs for such purposes. <S> Look up the datasheet for the display. <S> Its not too difficult. <S> Here are a couple of tutorials that should get you going. <S> Tutorial for SPI on Arduino Tutorial for I2C on Arduino <S> First things first, you need to find out what bit-depth you need for your ADCs, 10-bit should probably do becasue its probably just a byte being transmitted at a time. <S> You don't want an 8-bit ADC though if you are decoding 8-bit values because you will get errors. <S> You need at least one bit extra because your LSb is going to have some noise in it. <S> Second you need to find out how many ADCs are on you microcontroller development board. <S> Most common ones have 2-channel 10-bit ADCs multiplexed out through 16-pins through an analox MUX like the 74HC4051. <S> If you need more than 8 extra channels, you can always chain together multiple ADCs. <S> If you want to find some 8-channel ADCs, you can find them on either Mouser and Digikey. <S> Just go to their website, and navagate through their index to find the mutli-channel ADCs. <S> Digikey website <S> > <S> Product Index > Integrated Circuits (ICs) <S> > <S> Data Acquisition - Analog to Digital Converters (ADC) <S> mouser website <S> > <S> All Products > Semiconductors > Integrated Circuits - ICs > Data Converter ICs > <S> ADC / DAC Multichannel <A> The Cypress PSoC5LP might do what you're after. <S> I just dropped a sequencing SAR ADC on to a test project and told it I wanted 50 single ended analog inputs... <S> no problem. <S> They're nice little devices: <S> ARM Cortex-M3 with programmable digital and analog blocks. <S> It's also got an LCD driver and capacitive switch hardware built in. <S> I am pretty happy with both the PSoC4 (Cortex-M0, more restrictive digital/analog) and the 5LP and find myself dropping them into more and more projects. <A> We often need many analog readings, not necessarily simultaneously, fortunately there are integrated analog / digital switching circuits, such as the CD4067 and CD4051, which in cascade can expand the number of analog inputs of most microcontrollers economically.
| One way is to add an external ADC onto your current microcontroller board via either an I2C or SPI bus.
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Emulating a switch with open drain outputs I need to replace mechanical switches with some kind of digital switch. Specifically I want to sent control inputs to various arcade games that use standard JAMMA inputs, which are basically switches to ground for each button/joystick. The inputs will come from a 3.3V microcontroller, but most games use pull-ups 5V so a buffer is required. The obvious solution is to use open drain outputs. I'm trying to decide what type is best. The two candidates I have are the SN74LVC07 hex buffer and a ULN2003 or similar Darlington array. The SN74LVC07 is MOSFET based with open drain outputs and rated up to 5.5V, where as the ULN2003 will accept higher voltages (arcade systems do use 12V, although I doubt any use it for control inputs) and is probably a bit more robust. Any comments on which of these would be best, or would some other option entirely be better? What about unused buttons? I can't see any issue with leaving those outputs disconnected and the datasheets don't say anything. Thanks. <Q> Since you'll be driving the switches with an MCU, the 2003's inversion won't matter, so it looks to me: <S> Vce Vcc IC <S> $/pkg <S> $/dev V V mA <S> n <S> /pkg <S> DKEY1 <S> DKEY1 <S> |------|-----|-----|-------|-------|-------| <S> ULN2003 50 0 <S> 500 <S> 7 0.59 0.084 <S> 74LVC07 <S> 6.5 <S> 6.5 <S> 50 <S> 6 <S> 0.45 <S> 0.075 <S> like the extra penny per device for the 2003 is well worth its extra robustness and versatility. <S> Here are the data sheets for the ULN2003 and for the 74LVC07 . <A> The simplest method is to simply replace (or bypass) <S> each button with a single NPN transistor. <S> Connect the ground of your controller to the ground of the target device, pull down the base of the NPN with a 10KΩ resistor, and drive the base from your controller's IO port via a 470Ω resistor. <S> When you drive HIGH the NPN will turn on connecting the device's signal to ground. <S> No need for fancy chips. <A> Firstly, thanks to EM Fields for talking this out. <S> That conversation lead me to the answer. <S> The SN74LVC07A looks like the best option. <S> It will handle the required voltages and have a low V OL at 3.3V supply. <S> Sadly the datasheet doesn't have any curves but the numbers given suggest it will be okay. <S> Alternatively the LM3046 also looks similar and suitable. <S> At low currents in the tends of milliamperes range they have a Vce(sat) of only 0.2-0.3V, which should be low enough for almost any type of logic down to 3.3V. <S> The down side is that they will require base resistors and only come in quad or quintuple packages. <A> Since you stated that integration is an important requirement (at least 30 lines should be available), I recommend you get rid of these older integrated circuits and move to a more modern and effective solution. <S> MCP23016 <S> (16 bit \$I^2C\$) <S> From Microchip - MCP23017/MCP23S17 (16 bit \$SPI\$) <S> From ST - STMPE2401 (24 bit \$I^2C\$) <S> From NXP - <S> PCA9505/06 (40 bit \$I^2C\$) <S> The main advantage is the great saving of microcontroller I/ <S> O pins, which remains also using more than one of them. <S> In addition, there are other unique features - the pin behavior is totally programmable, like open drain outputs , PWM outputs (in some part numbers), interrupt capability , capture register (see the datasheets for the details). <S> If you do not have \$SPI\$ or \$I^2C\$ pins available on your microcontroller, that can always be emulated by software <S> - the bonus is that can be done on any MCU, since the speed is not a limiting point (remember that you are dealing with buttons). <A> This is a pretty simple problem. <S> Either of those chips will work. <S> Why you'd want to use a buffer over a standard open drain is beyond me. <S> It's not like you need the buffer driving your signals high fast unless you have not added in the pull-up resistors. <S> You can use a MOSFET in open drain topology, BJT in open collector topology, or a Darlington pair in the same config. <S> If you wanted to remove the pull-up resistors from the circuit, you could use a buffer. <S> Otherwise, that just feels like overkill. <S> A mux chip would save you pins on your MCU. <S> Others have recommended expanded IO chips. <S> I also wonder why you haven't gone a different route entirely. <S> Obtain a similar version of your MCU that can handle 5V with lots of IO pins to handle your 30 lines. <S> Then you don't need an extra chip(s) at all. <S> Alternately, you could look into analog switch arrays. <S> Here's one for instance. <S> If you were to actually use that chip, you would have to add a single level shifter because it's a 5V chip.
| As long as the devices can sink as much current as your pull-ups provide, anything will work. For example, use one of the following " GPIO expanders ": From Microchip - Another option would be the classic 2N3904 NPN transistor, which is available in a quad as either the MMPQ2222A or MMPQ3904.
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Why is propagation delay of ethernet cables unaffected by cable length? I did some searching on how propagation delay in ethernet cables is affected by the cable length. The answer seems to be that the delay is negligible until the cable length reaches a kind of threshold value which imposes a limit on the maximum length of ethernet before it becomes too unreliable. However, propagation delay is distance/speed, so technically a longer cable should have a longer delay. I am assuming the cables have the same diameter. My guess is that since information is continuously sent in a stream, while it takes longer to initiate the transfer, the rate of data transfer is unaffected. However, the latency would still be slower, which would be a problem. Why is the cable length rarely discussed? <Q> As Andy aka already wrote, any length of cable produces a delay. <S> However by limiting the length you know what the maximum delay is and you can correct for that. <S> Ethernet does this by specifying a minimum frame size in combination with a maximum cable length. <S> Thus ensure that even two computers on the ends of the cable (and thus the furthest away from each other as allowed) will see that the other is transmitting before they finish their own frame. <S> Imagine the following setup with two computers and ancient coax: <S> Both computer #1 and #3 want to transmit something. <S> Both check the cable and see noone else transmitting. <S> Both start to transmit their own small frame which will travel down the cable at limited speed. <S> Neither computer 1 or computer 3 will see the others signal while they are transmitting. <S> Both will receive a readable frame from the other. <S> however computer 2 will see an unidentifiable mess rather than one or more valid frames. <S> The solution employed by Ethernet is to limit cable length to about 200 meters and make a framesize of at least 64 char mandatory. <S> (If the frame is shorter it will be padded). <S> That way any transmitting from one edge of the network will reach the other side of the network. <S> Any computer (well, any NIC) wanting to transmit needs to check if the line is clear and stays clear from the start of transmission until it has send at least the min. frame size. <S> Finally getting to your question on <S> Why is the cable length rarely discussed? <S> (in combination with the tag Ethernet): <S> From ancient times the min. frame size and max. <S> cable length have been defined for Ethernet. <S> This is a standard. <S> Even with UTP we still adhere to those standards. <S> This makes the values always the same and not worth discussing. <A> No, there is no threshold as you put it; any length of cable produces a delay and that delay is proportional to cable length. <S> Different cables propagate slower of course and this is largely down to the dielectric of the material between the two conductors (or centre and screen in coax). <S> The higher the capacitance, the slower the speed of propagation. <S> Regarding maximum length that can be used for a certain data rate <S> yes, there is a "kind of" threshold - basically data gets misshaped the further it has to travel down a cable due to cable losses (resistive and dielectric). <S> Time to get a better cable or different modulation scheme! <S> Cable length is quite commonly discussed where I work for the reasons highlighted above. <A> Any length of cable imposes a delay which is roughly proportional to it's length. <S> There will also be a degradation in signal integrity, again getting worse with length. <S> If the signal integrity degrades too much then the receiver will not be able to correctly interpret the signals. <S> For coax and twisted pair Ethernet this is what limits the length of a single "segment". <S> Large propagation delays can cause incorrect collision detection in a CSMA/CD system. <S> Specifically there are problems if the propagation delays grows larger than the minimum packet length. <S> With a single segment of coax or twisted pair Ethernet that meets the length requirements for signal integrity you won't run into this problem. <S> However you can run into it in two cases. <S> chains of repeaters/hubs (NOT switches): With 10 megabit ethernet the rule <S> was you should have no more than 4 repeaters between the most distant segments in a collision domain. <S> With 100 megabit Ethernet this reduced to 2[1]. <S> fibre optic links: The fibre physical layers can maintain signal integrity over a much longer distance. <S> Therefore long fibre links cannot be correctly operated in half duplex mode. <S> Full duplex links do not use CSMA/CD. <S> Switch ports running in half duplex mode do use CSMA/CD but the collision domain does not cross the switch. <S> So basically as long as you make sure the length limits for the physical layer are respected, make sure you use switches rather than hubs and make sure any long distance fibre links are running in full duplex mode then you are fine. <S> [1] With gigabit the rules stayed the same as with 100 megabit because they increased the minimum packet size for half duplex gigabit links but it was irrelevant in practice because gigabit hubs are like hens teeth.
| At some cable-length and at some data-rate the cable can be deemed to be at the "point of no return" in that statistically the number of data errors incurred are too many to warrant further error correction/detection.
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Why did my fuse not blow? I purchased a few fuses to get familiar with their workings, and I was surprised to see that the 100mA fast-blowing fuses I bought happily conducted up to 215mA (6V power supply, 10\$\Omega\$ resistor), where the filament just started glowing. I could reproduce this with a second fuse. Am I gravely misunderstanding something here, or is this an issue with the fuses? They are Bel Fuse Inc. 5SF 100-R parts. <Q> Even fast fuses don't respond immediately after the rated current is reached. <S> This is an obvious requirement since its all about heat, and minor heating (a raise ambient temperature) shouldn't affect the fuse too much (or at least not cause it to "blow"). <S> Judging from the data-sheet, your 100mA fuse should blow after about 80 seconds @215mA <S> @25 <S> °C. <S> How long did you wait? <A> Glass fuses, especially at very low ratings, aren't really meant to be that precise. <S> You cannot, for example, just inline a 2N3904 with a 200 mA fuse and expect it to protect the transistor in all circumstances. <S> You would, after all, expect it to NOT blow at 190 mA. <S> In this example, you would probably want to over-specify the transistor and use a fuse at double the maximum expected load. <S> If the user does something silly, like shorting the output, the fuse will do its job. <A> This may give some clue, only posting here as I happened to be reading the source: <S> There were two main types of rating used for fuses. <S> The nominal rating which is the current the fuse will carry for 1000 hours without failing at a temp of 20 degrees centigrade. <S> The service rating is the current the fuse will carry for 1000 hours without failing at a temperature of 100 degrees centigrade. <S> The service rating is 80% of the nominal rating. <S> E.g. a fuse with a nominal rating of 10 amps has a service rating of 8 amps. <S> Confusion arose when the equipment in use before 1962 would quote either of the ratings. <S> (EMER General O 001, 1962). <S> Of course, there is a relatively unknown online reference known as Wikipedia which has some potentially helpful hints too.
| Most fuses require a significant over current to fire almost instantly.
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4 wire resisitive touch screen not working I am having a problem with making 4 wire resisitive touch screen work. : This is the touchscreen I am using.I also have 100 Ohm resistors in series with all the pins. If I understood correctly: Y+ \$\rightarrow\$ +5V, X- \$\rightarrow\$ GND, X+ \$\rightarrow\$ Voltmeter. Which should show me diffrent values, depending on where I press on the screen but it shows me 5V no matter where I press it on. I tried out 3 same touchscreens. On one of them, I even tried out every possible combination of +5V, GND, Voltmeter, but always the same result. Any help is greately appreciated! <Q> What information that I did find for the LGM12864B touchscreen does not seem to match the one you have because the datasheets explanation for how the screen works is through a protocol and doesn't use a resistive response. <S> So I'll answer for how the resistive touchscreen works in your case. <S> This image may be useful for understanding: You can see on page 3 of the document that Gurn64 provided that you can read the touchscreen in 2 steps: <S> First, to determine the horizontal position, set Y+ to be 5V and Y- to be ground. <S> X- is essentially floating, and X+ is now your output. <S> Measure X+ and the ratio of X+ to 5V is the ratio of the distance from the bottom of the screen to the length of the height of the screen. <S> The vertical sensing is similar. <S> Y+ and Y- are no longer 5V and ground respectively. <S> The explanation is the above picture is pretty clear. <S> Also, another note is to remove the 100Ohm resistors that you currently have hooked up, they don't aid in the procedure of determining touch positions. <A> You can't measure X and Y positions at the same time with this kind of touch screen. <S> You have to measure one and then the other. <S> To measure X, apply a voltage between x+ and x-. <S> For example connect 5v to x+ and 0v to x-. <S> Then measure the voltage at y- or y- <S> (it should be the same) to get your X position. <S> Then to measure Y you must apply a voltage between Y+ and Y- and then measure from X+ or X-. <S> A 4-wire touch screen like this is made of 2 squares of resistive material. <S> So when you apply voltage between X+ and X- the voltage will gradually drop across one of the squares. <S> When you touch the screen you connect the two squares together at the point where your finger is. <S> Because you are only sensing a very small current from the Y+/Y- sqaure there is no voltage drop there <S> so all of the Y square will be at the same voltage as the point on X where the two are touching. <A> You can test touchscreen with with only ohmmeter. <S> First you should check resistance between X+ and X− and also Y+ and Y−. Mine (AMT 98245) have about 500 ohm at X± and 400 ohm at Y± <S> (i name it <S> Rx <S> anr Rx for later use). <S> When you touch the screen these values can lower a bit (less than 5%) but not much. <S> All other combinations (between any X and any Y) should give ∞ if screen is not touched. <S> Then connect ohmmeter between one X and one Y and touch the screen. <S> You should get values somewhere between Rz (interlayer resistance) and Rx + Ry + Rz. <S> I got values from 350 ohm to 1 kohm with mine. <S> Y'll get low value when touch near electrodes you mesure -- <S> e.g. if you measure between X+ and Y+ then lowest value you'll get at upper right corner, and highest value at lower left. <S> When you ensure youe touchscreen is ok <S> , you should follow other's recomendations here to measure touch position correctly. :)
| X+ and X- are 5V and ground respectively, and Y+ becomes your output like X+ was before. When you press on a resistive touchscreen, the pressure will introduce a dynamic voltage divider between the two X nodes, and the two Y nodes.
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Voltage Drop and Safe Current Load on CAT5 Cable I am going to use this passive PoE injector cable set to power a Raspberry Pi model B. The Raspberry Pi model B consumes between 700mA-1000mA (mine will also be powering the RPi Camera Module). Using this cable set, I plan to power the Raspberry Pi and provide Ethernet data at the same time. My questions are: For a 60 meter CAT5 cable, how much DC voltage should I supply to account for voltage drop and still meet the needs of the Raspberry Pi? For a 60 meter CAT5 cable, and let's say a 6V DC power supply - the above passive injector cable will split the current between pins 7 and 8 (meaning each of the two 24 AWG wires will carry between 350mA-500mA). Does this breach the zone of how much current the CAT5 cable can safely carry? I don't want to start an electrical fire. <Q> According to a reference from the Handbook of Electronic Tables and Formulas for American Wire Gauge a conservative estimate for 24AWG power transmission is 0.577 amps. <S> Current-wise, I'd expect you to be fine. <S> I'd still recommend testing it and monitoring the temperature for a while to make sure. <S> If you've got that cable in a tight bundle of some kind, it will get warmer than if it was in free space. <S> On the bright side, wire that small won't take long to reach its ultimate temperature! <S> As for voltage drop, according to this AWG table , 60 meters of 24AWG has a resistance of about 5 ohms. <S> 500 mA will drop 2.5V over that distance. <S> You also have to consider that the current has a return path, so that's 2.5V drop in <S> each of the positive and negative legs <S> , 5V drop total. <S> If, on the other hand, the draw is on the low end of your spec, you'll only get 3.5V drop total. <S> That means that the voltage on your load (the Raspberry Pi) will vary by 1.5V depending on how much current it's drawing. <S> That's quite some variation, and you'll need to make sure it can handle that. <S> If it was me, I'd come up with a different plan. <A> I can't comment on your answer but have to give you an advice, hope somebody turns it into a comment. <S> When powering things over long lines, it's best to put more volts into it and put a DC-DC at the end where the device is, setting it to 5 volts (or having a fixed-voltage DC-DC). <S> LM2976 or XL6009 converters are the best bet, since they're switching-mode and thus much more efficient. <S> In the end, you have less current running through your wires and constant voltage powering the device, with no need to worry about all the things that could happen be the current high or voltage unstable. <S> I have a couple of routers powered like that and currently am in the process of making a big system with 30 devices powered like this <S> =) <S> I'm sure that's your best bet in this situation. <A> I powered a raspberry pi with an old printer power adapter <S> I believe it was around 1.7amp approx 28 volts. <S> Then I used a LM2956 module on the other side to reduce the voltage to 5v. <S> You probably will want to use both blue and brown <S> so you have 4 wires transmitting power. <S> It worked fine providing power and internet. <S> Very reliable it ran for 4 years outdoors in Alaska 365 days a year. <S> The Cat-5 was at least 50 meters or more.
| You're running really close to the edge of what's reasonable.
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Does the input polarity matter for a 12V DC adapter? I bought a car cigarette DC adapter yesterday with the following specs: Input: DC 12VOutput: DC 17V I'm planning to connect it to a 12V battery I have at home, but I was wondering if the adapter input polarity matter or could damage the adapter if reversed? <Q> Yes, it matters. <S> Check it carefully or you probably will damage the device. <A> I bet you have this kind of plug: <S> The tip is the positive contact while the side plates should be connected to the negative pole of the battery. <S> For extra safety, and since you are working with a somewhat big battery, add a fuse in series with your adapter: if you mess something up the fuse blows and the wires don't catch fire. <S> The fuse should be rated a bit above the maximum input current that should be written somewhere on your adapter. <S> Please note that it's well possible that there already is a fuse and that it's inside the above plug. <A> The polarity of the adapter matters and it should be mentioned on the device itself. <S> If not, you may first need to check whether it works with your car cigarette socket. <S> Then you can find the polarity of your car socket using a multimeter to be certain. <S> Please have a look at this answer too. <S> And as Vladimir has mentioned, add a fuse of proper rating when you are powering from an external battery. <S> The DC socket in your car is also 'fuse'd to limit the current drawn by the device attached.
| The adapter might have some sort of protection inside, but it will not work (and possibly it will fry) if you reverse the polarity.
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Using timer to power on/off a high current circuit I'm working on a Arduino project that takes measurements from the environment and sends them to a remote server over WiFi. The circuit should be battery powered and I've measured that what I have at the moment takes about 160mA of current, which makes it impossible to run on battery longer than couple days. The good thing is that it is enough to power the circuit once an hour and run it for approximately one minute and then power it off. So I was thinking of creating a very low power timer circuit that would power my main circuit for one minute every hour. Unfortunately I don't have any idea how to do that :). I tried to research and I think it might be possible with IC 555, but if I understood the data sheet correctly it draws around 3-6mA of current which is little bit too much. Does anyone have any ideas to point me to right direction? <Q> An hour is too much for a 555 timer or any other oscillator. <S> A better solution could be a 1 Hz multivibrator (a 4011 chip consumes less than 0.5 mA at 1 MHz, and even less at 1 Hz) and a counter to count to 3600. <A> If you have an Arduino and are after such lengths of time, why not purchase a RTC addon for the Arduino. <S> They are not that much and that way you can get decent accuracy over long periods of time. <S> http://playground.arduino.cc/Main/DS1302 <A> The ATmega on the arduino board has some sleep functions , which will make it use practically no power. <S> The ATMega328 has a watchdogtimer build in. <S> You can set it to up-to 8 seconds. <S> Just put it in SLEEP_MODE_PWR_DOWN , and the watchdog will wake it up again in 8 second. <S> Do this 450 times, and you got an hour worth of sleep, with only a few ms of CPU time. <S> The other thing is getting the Wifi module to sleep too.
| And maybe desolder some of the leds on the arduino board to reduce power usage.
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Current doesn't equal what I expected I'm making a simple LED circuit. I have a 6V power source, and a LED with 2 forward voltage and that requires 18mA. I calculated the resistance by dividing the voltage with the required amps (6-2 / 0.018). This gave me 222, which I rounded to 220 Ohms. Now I have a 220 Ohms resistor. Now I have a 6 volt battery, 220 Ohms resistor, and a LED. Now I can calculate the current. So I divided the total volt (6v) with the total resistance (220 Ohms). But this gave me 6v / 220 Ohms = 27mA. 27mA does not equal 18mA, which the LED needs to work. With my understanding, these two numbers should equal. What am I doing wrong? <Q> Now I have a 6 volt battery, 220 Ohms resistor, and a LED. <S> Now I can calculate the current. <S> So I divided the total volt (6v) with the total resistance (220 Ohms). <S> But this gave me 6v / <S> 220 <S> Ohms = <S> 27mA. <S> 27mA does not equal 18mA, which the LED needs to work. <S> Your math is wrong. <S> Mainly, the resistance of the led is negligible, so it can be ignored, but the voltage drop can not. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The voltage across the resistor is what sets the current, not the total voltage. <S> This can be tested with a multimeter. <A> For low frequency operations, a practical diode can be modeled as an ideal diode in series with a voltage source and a resistance as shown below. <S> D1 is an ideal diode. <S> Vd is the forward cut-in voltage and Rd is the dynamic resistance. <S> But usually Rd is neglected in most of the cases as its value will be small (in \$\Omega s\$). <S> Since LED is a diode you can apply the same model. <S> And Vd is 2V in your case. <S> Hence when LED is on, your circuit has a 6V source, 220\$\Omega\$ resistance and a 2V voltage source (drop), in opposite polarity as 6V source as shown below. <S> $$ <S> \therefore I = \frac{6V-2V}{220\Omega} = 18.18mA$$ <S> You missed the Vd in your calculations. <A> So I divided the total volt (6v) with the total resistance (220 Ohms). <S> To find the current through the resistor (and thus, the current through the diode), according to Ohm's Law $$i_R = \frac{v_R}{R} $$ <S> where \$v_R\$ is the voltage across the resistor. <S> But, by KVL, we have $$ <S> v_R = 6V - <S> v_D$$ <S> thus, $$i_R = \frac{6V - v_D}{220 \Omega} = i_D$$ <S> Assuming \$v_D <S> = 2V\$ <S> then $$i_D = \frac{6V - 2V}{220 \Omega} = <S> 18.2mA$$ <A> The LED has a dynamic, and no-so-constant resistance of R = <S> V/I during steady state, <S> and you have said it drops 2V and supposedly at 18mA. <S> It therefore has an R value of 111 Ohms. <S> Therefore, your circuit is: 6V / (111 + 222) <S> Ohms = <S> 0.01801 Amps. <S> Edit: <S> Just so it's clear, LEDs and other diodes don't actually have a "resistance" value, and if you give them their forward voltage <S> they will effectively short circuit and blow up <S> so please don't assume an LED always has a resistance. <S> They do not limit current at all. <A> What you did wrong is basic arithmetic, $$\frac{6-2}{.018} \approx 222$$ <S> Thus, working backward to the .018 has to be: <S> $$.018 \approx \frac{6 <S> - 2}{222}$$ and not: <S> $$.018 \neq <S> \frac{6}{222}$$ <S> You calculated (correctly) <S> the resistor value for the desired current using a voltage drop of 4 for the resistor. <S> So the same value must be used to work backward from the resistor to the current. <A> If you look at a plot of LED Vf VS <S> If: you'll see that a very small change in Vf will effect a very large change in <S> If. <S> Then, When you look at an LED data sheet and you see the specifications for Vf and <S> If, you may think that <S> If is specified by putting Vf across the LED and measuring If. <S> That's not the case however, and Vf is derived by forcing If through the LED and then measuring the voltage it drops. <S> That way, with a given current forced through a large number of samples, a range of measured Vfs will emerge which will allow the current-limiting resistor to be easily calculated by subtracting the LED voltage (at the current specified) from the supply voltage and dividing by the specified <S> If. <S> For example, let's say you have a 5 volt DC supply and a red LED with a typical Vf of 1.9V when there's 20mA <S> If through it, and that you want to drive it at 20mA. <S> Then you could say: Vs - Vf 5V - 1.9V <S> Rs = <S> --------- = <S> ----------- <S> = <S> 155 ohms <S> If 20mA <S> 155 ohms isn't a standard 5% value, but 160 is, and if you wanted to, you could figure out the drop in current for the higher value resistor by rearranging the formula and solving for If. <S> Plus, there's a pretty wide latitude of Ifs allowable, from very dim with a small If to, generally, 30 mA for an LED spec'ed at 20mA nominal.
| In your calculation for the LED ballast resistor value, you seemed to have included the fact that the LED has internal "resistance" because it consumes a certain amount of voltage for the given forward current, but then you ignore that fact later when trying to determine the full loop current.
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How does a temperature sensor work? How does a temperature sensor gather temperature? What is the mechanism used? More specifically I want to understand how the TMP102 sensor works. <Q> (from p.1 in the datasheet ) <S> The principle is that of a diode temperature sensor, which is also called Silicon Bandgap Temperature Sensor . <S> Forward voltage of a silicon diode is temperature-dependent. <A> The data sheet shows that it is a "diode" temperature sensor. <S> If you operate a diode forward biased with a known current, the voltage across it has a predictable change with temperature, about -2mV/ <S> K. <S> More likely they use a diode-connected transistor (which behaves more like an ideal diode) and probably operate it at two currents (alternating and typically about a decade apart) and look at the difference in voltage, which is more predictable over a wide range (but has output about 10% of the direct method, so about -200uV/K). <S> They also probably calibrate the temperature using EEPROM or resistor trimming at a single test temperature. <S> Unless they've published details of the above somewhere, they're IP of the manufacturer. <A> This kind of sensor uses a diode as a temperature sensor. <S> The value of the forward voltage is read by an analog to digital converter which makes digital data available through a serial interface.
| A diode can be used as a temperature measuring device, since the forward voltage drop across the diode depends on temperature, as in a silicon bandgap temperature sensor ( Wikipedia ).
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Is a digital bit represented by any two discrete signals? I am trying to understand computers from the building blocks up. I know computers use transistors to amplify voltages, and this is used for arithmetic, such as a MOSFET. However, what really makes up a typical computer bit? Is it just two input currents, one higher and one lower? Two possible states? Is movement a bit (run or walk)? How can I make a bit? Do I just need an emitter, current, and multiple passages? Would I have a bit if it couldn't be amplified (i.e. a two-series circuit with no amplification of current)? <Q> A computer bit can be considered the transmission of a signal from one place to another over a wire. <S> This could be defined as specific voltage level ranges, current amount ranges or even direction of current flow. <S> The most common scheme used in digital circuits is to use voltage levels. <S> One level pulls the signal line toward the GND whilst the other state of the signal line is determined by pulling the line toward the supply voltage. <S> Bit signal levels transmitted through a circuit on signal wires always need to originate from some place. <S> These can originate from electro mechanical components that hole the lines at the particular defined voltage levels. <S> They can also originate from special circuits that hold or store the signal level. <S> Such circuit, commonly referred to as a latch or flip-flop, are designed so that the output can drive an output to a signal wire to establish the pair of defined digital states that represent the '1' and '0' values. <A> A bit is the basic unit of information in computing and digital communications. <S> A bit can have only one of two values, and may therefore be physically implemented with a two-state device. <S> These values are most commonly represented as 0 and 1. <S> It is not represented by any two discrete signals. <S> It is represented by two well defined signals: <S> Low signal or bit value 0 is in logic electronics a voltage not greater than 25-30% of the supply voltage. <S> Ideally it should be 0 volts (ground). <S> High signal or bit value 1 is in logic electronics a voltage not smaller than 75 % of the supply voltage. <S> Ideally it should be equal to supply voltage. <S> A light switch can be considered a "1 bit memory" because it has two states (open and closed). <S> A light dimmer is not a bit because has an infinite number of states between open and closed. <S> Usually, bit value is defined by voltage not current. <A> "Would I have a bit if it couldn't be amplified (i.e. a two-series circuit with no amplification of current)?" <S> Yes, you can create an AND gate or an OR gate with two (or more) diodes. <S> There is no amplification.
| The "digital" nature of a bit definition can be any two states that make sense relative to how the circuitry that drives the bit level onto the wire and the circuit that detects the level of the bit signal on the other end of the wire. Movement is not a bit.
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How would one attach an optical fibre to a PCB for a display? For a model train layout, I am looking to route a light to a difficult part of the layout - even the 1.6mm LEDs are too big for the area. It is for display, not data communications. One person has solved the problem on their layout using optical fibre, which seems quite sensible. Rather ingeniously, the light is transferred from an LED the fibre via thimble with a hole drilled in the top. The fibre does not need to be connected/disconnected from the PCB - a permanent connection is fine. I am aware of solutions for POF in ethernet , but this is far beyond what I need. I can't be the first one to have this problem since fibre optic trees are cheap and plentiful. So, some questions: If I have a surface mount LED (or an array thereof), how do I attach an optical fibre to transmit the light for this purpose? I assume there is some sort of plastic mounting I can purchase off-the-shelf. What is the attachment called (i.e. does it have a standard name)? How are they assembled? (e.g. manual, pick and place?) What considerations are there in selecting the type of fibre to use? (it will be no longer than 1m) (Note to moderators, "optical-fibre" would be a good new tag for this question) <Q> Fiber Optics, as a light source, and not a data transmission platform, is a "shoe-fits" type application. <S> Fiber optic trees use heat shrink or electrical tape or tightly molded plastic as the method of optical coupling. <S> Data Transmission Optical Fiber Cable on the other hand, has closely designed standardized connectors, such as TOSLINK for consumer audio connections or SFP (small form-factor pluggable) connector used in some Cisco router Fiber Optic connections. <S> Using a 5mm or 3mm led with matching heat shrink tubing and enough fiber to fit is very very common in hobbyist modeling applications ( which a well done scale model requires much of to look good (Model Enterprise Project) ). <S> If you need just a single thin strand, you could drill a hole in a led and hotglue the strand in (Model Train Project) . <S> (The ends are mushroomed out using a lighter or other heat source, then hot glued in place). <S> Commercial applications use Light Pipes made of molded or extruded plastic, to channel a pcb mounted led (almost always smd) to an external case opening. <S> I can't think of a single commercial/consumer/industrial product that would use fiber optic strands as internal light redirecting method, due to how unwieldily that would be in manufacturing, let alone <S> technical support issues as connections are fragile due to the thinness. <A> I have had fair success using individual plastic fibers from an inexpensive decorative lamp like this one to create headlights for N-scale model railroad vehicles: . <S> For coupling, making a clean flat cut on the LED side of each fiber using a surgical knife, and then heat-shrinking or hot-gluing the fibers tightly to the LED, works well. <S> At the other end of each fiber, as described by Passerby, create a small lens-like melted blob using a heat source. <A> If the LED was flat, I'd use a length of 1/16" diameter extruded acrylic rod for a light pipe, polish one end of it flat, and secure that end to the "sweet spot" on the face of the LED with superglue. <S> Once the glue set I'd clean off its uncured remains and, for more structural support, run a fillet of epoxy around the end of the rod where it contacts the LED. <A> When you're talking about fibres you're usually talking about high speed transmission of data. <S> What you are looking for is a light guide. <S> They come in all shapes and can be bent easily. <S> I would guess you're looking for something similar to this: http://www.farnell.com/datasheets/1739879.pdf
| If the LED was domed, I'd polish a flat a little larger than the diameter of the rod at its very peak and then superglue and epoxy the light pipe to it. Fiber Optics are also used for Star Maps or Star Ceilings
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How to create an electronically controllable resistance to a physical movement? What is the best way to create an electronically controllable resistance to a physical movement? For example, at the gym you control the resistance to movement by putting more load, more weight. Is there a way to control that resistance electronically? I was thinking of a motor, it's possible to use it? Or through electromagnetism? <Q> A permanent magnet DC generator (or a generator with constant field excitation) can be modeled as a voltage source proportional to velocity (angular velocity in the case of a rotary generator) in series with some coil resistance. <S> Since viscous damping requires a force proportional to velocity, simply loading the output of the generator with a variable resistance RL will create a variable damping factor proportional to RL+RG (where RG is the winding resistance). <S> We want the torque to be -c\$v\$ where \$v\$ is the motor rotation speed to simulate viscous damping, so in this case c = \$k(R_L+R_G)\$, where k is a constant that depends on the generator construction. <S> An electronically-variable resistance can be created by literally switching resistances in and out, or it can be done with a MOSFET or BJT electronic load that simulates a resistance. <S> The maximum damping with a passive load is limited by the internal resistance of the generator. <S> With an active load and external power supply it should be possible to simulate a negative resistance externally to further reduce the total equivalent resistance. <S> For linear motion, a linear PM motor could be used, or a rack and pinion used to convert linear to rotational motion. <S> A second method, most useful if motion is guaranteed, would be to use coulomb friction (for example as a caliper and brake pads dragging on a rotor) and control the force applied to the brake pads using a torque motor or other method (for example change the position of an actuator that is spring loaded, so use Hook's law to determine the force). <S> This works because dry friction is proportional to the normal force applied, and the proportionality factor \$\mu\$ is a function of the materials involved <S> However, this method will have stiction (nonlinear behavior) before motion starts. <A> look at this figure: <S> Is it necessary to explain about the mechanical mechanism of it? <S> I hope it will help you. <A> Pump a fluid through a closed loop that includes an electrically operated variable valve. <S> Depending on how much fluid is in the reservoir, how long it must run, how long it rests before you run it again, and how hard you'll be driving it, you might need a radiator in the system as well. <S> Edit <S> : Look for indoor bicycle training stands. <S> They use either fans, generators, or pumps (CycleOps makes one) for resistance, and the resistance units are pretty compact. <S> Some feature a remote (push-pull cable) resistance adjuster; some just depend on the bike's own gear to adjust it. <S> But something along that line should get you started. <A> What you are looking for is a damper mechanism. <S> The previously shown system (the stationary bike) is an adjustable magnetic damper. <S> This patent explains its operation. <S> If you want to control the damping level electronically you might want to look at eddy current brakes, which produce the magnetic field from a winding, so can be controlled rapidly without moving parts. <S> You can of course attach a conventional DC motor to the apparatus, and emulate a damper or spring-mass-damper system electronically, however this has limitations over the above methods (due to the limited update rate of the software implementation when dealing with high frequency input signals, and the difficulty of estimating velocity at low speeds). <A> Well on most servomotors and motors the torque depends on the power supply current. <S> However, It's probably not best practice <S> and I don't think there actually is specific actuators allowing precise force control. <A> You can use a control system to do this for you. <S> In essence you have a motor with feedback for the force you are trying to provide. <S> This would allow you to use the motor to spin the crankset at a speed faster or slower than the rider is trying to maintain in order to produce the strain - or replicate the force - of either wind resistance, a hill, or freewheeling down a hill. <S> Another option is to trade the strain gauge for a rotary encoder, then model the physical system in the digital world. <S> You can model springs, or mass, or anything else you can imagine in terms of where they move the cranks to and how the physical world would really respond. <S> So you might have a high mass you're modeling, as though they are really riding a bike and they weigh 180lb, so when they aren't moving the crank, the motor puts no force on the crank. <S> But as they attempt the move the crank, the rotary encoder senses a few pulses in a given direction, and the motor will push back with nearly equivalant force, as it models the rider gaining momentum. <S> As the digital model of the system gains momentum, the motor will reduce the push-back, and eventually it will take very little real energy to maintain a given speed. <S> You can then model hills up and down, wind resistance, etc in the digital model, and have the motor present those forces to the crankset.
| For instance, you might attach a motor to a bicycle crank set, and put a strain gauge on the crank arms. I think you can find the answer of your question via search about the mechanical and electronically mechanism of Stationary bicycle .
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Setting up a non-ESD area in a lab this question is regarding standards and/or best practice guidelines for ESD control in a professional lab environment. We are currently tightening up our (admittedly too lax) ESD control in the labs, in preparation for an audit. We've had a chance to see the audit results from another one of our labs and are working on all the things they failed on. One of the issues we have is how to manage non-ESD areas in a Lab. The lab itself is marked off as an ESD protected area, and has suitable flooring. The lab in question is our 'construction' lab, which has a few soldering benches and general tools/equipment. The majority of it will be ESD protected and grounded as appropriate, but there is going to be a mechanical rework bench in the corner which will not be ESD protected. The bench will be a metal frame, with a thick piece of ply on top so you can hammer/drill/bash/whatever. Does anybody have an idea as to what best practice/standards would say about non-ESD working areas in an ESD room? Would it be sufficient to just mark out the area? It has the same dissipative floor tiles underneath, is that an issue? Would you ground the metal frame of the mech. workbench? It terms of geography, the room is isolated from the rest of the office by walls, has ESD flooring throughout the room and ESD walkway to the other labs. In order to walk to the mech. working area, you would need to walk through the ESD protected area. It is likely that workers will swap between the mech. work bench and the ESD-safe soldering bench. Any thoughts? <Q> Just to complete the circle, we passed the audit and it was acceptable to have a non-esd area in an other ESD protected lab. <S> The only requirements were that the area is suitably marked out (we have tape with arrows on which indicated which side was ESD protected), and that anything highly charged is a good distance from EPA work areas. <S> A side comment regarding ESD audits and protection is everybody has a different idea what is ESD good, and what is not, what is required and what is overkill. <S> By sticking to a fixed, official standard (in our case IEC 61340), we avoided conflicts of so called common sense and had black and white standards to adhere to. <S> Also training is an on going thing which often organisations can improve, but the biggest issue we had is that ESD was not really considered except when directly handling sensitive products. <S> We had to replace or refurb a lot of racking shelves, furniture, tools, workbenches etc because ESD was not considered when they were bought. <A> Earthing everything just provides a path for discharges. <S> The whole idea is that you try and prevent discharge via/thru the sensitive equipment and this involves <S> , amongst other precautions, a mindset <S> i.e. if you go into the area there is no guarantee that your body hasn't accumulated a few hundred or thousands of volts. <S> then they are able to work on equipment or move it around. <S> Use of earth wristbands helps of course <S> but somebody bringing some equipment in and placing it on an ESD bench is more than likely not to have been able to put on a wrist strap and if they just "plonk" the circuit or module down on the bench then all the charge accumulated flows through the electrical item and it could suffer. <S> Training, training and training. <A> Beware of the flooring materials, which can create static ie no carpets! <S> Also the overall environment should be air conditioned to ensure that humidity is kept fairly high . <S> Static becomes a problem in very dry environments
| You have to train people not to touch anything sensitive - instead they should touch the grounded areas first
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Identify key in a multiple keys, single lock system I have a hobby / office project underway. We want to issue a key to multiple users each with a different pin pattern. Upon turning the key we want to identify the key and thus the user. My google fu has so far failed me, does anyone have any ideas for craftinig a solution or a product that could be repurposed? <Q> You will find that multi-keyed locks become ridiculously complex (and expensive) <S> the more keys it is made to accept. <S> And there are practical limits. <S> An easier, non "wireless" solution would be two factor authentication. <S> Well, an idea similar to it anyway. <S> An electronic keypad lock paired with a regular lock (that cannot be left unlocked if the key is removed). <S> Each person gets a unique keycode, but still requires the physical key to get in. <S> (Something you know, something you have). <S> Regular key could be copied and the keypad code can be spied on, so it's not foolproof. <S> Other options are a Mag-Stripe door lock. <S> Easy to make unique cards, but easy to copy as well. <S> Slightly harder to clone are Smart card locks (They have contacts like a SIM card does). <S> Frankly, RFID is the way to go. <A> Doing this with a conventional lock sounds like quite a tricky project. <S> Suppose that we can fashion the tumblers out of some reasonably resistive material (yet which is tough enough to endure in that role). <S> Clearly, multiple keys are accommodated in the lock by means of tumblers that have multiple sections. <S> When the lock is turned successfully, the barrel holds some of the sections, while others remain in the casing. <S> Suppose you could somehow measure the combined series resistance of the tumbler sections which move with the barrel. <S> Or else the sections that are pushed into the casing and remain in place. <S> That could be converted to a code distinguishing the keys. <S> Another idea is to simply have multiple contacts in the passageway inside the casing, which can detect the extent to which the tumbler sections protrude. <S> Perhaps only one tumbler guideway needs to have these sensors. <S> The keys just have to be different in the first tumbler. <S> You need to then drill some precise holes through the casing from somewhere, and fashion some kind of contacts that are otherwise electrically isolated from the casing. <S> No matter how you look at it, is a project that requires cunning, skill and perhaps some luck (the configuration of parts being suitable for such and such approach). <A> Hotels have the same problem. <S> The most common solution I've seen to automated access control at hotels is the keycard -- a flat plastic card more or less the same size as a credit card,with some sort of bar code, magnetic stripe, RFID chip, etc. <S> on the card,and a card reader at each door. <S> Every card has a unique ID number. <S> Each card reader is programmed toidentify the card inserted into it and thus the user desiring access,and optionally log the access attempt. <S> If that number matches any of the numbers on the list of authorized numbers stored inside that particular card reader,then the card reader unlocks the door. <S> As Dave Tweed already pointed out, the iButton would work just as well. <A> We did some projects like hotel cards, there are some choice, one is use magnetic stripe cards, each reader on the door, encoding different information on the magnetic stripe, there are three tracks, each cards information connected with the reader.
| the other choice is ID cards, each cards encode a UID number, when the card reader recognize the UID number the door can open it.
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Technique to drive seven segment displays with external components I'd like to ask for some advice on the circuit described below: I'm building a matrix of multiplexed seven segment displays, with common anode. The ports D1-D3 and segments A-DP will be driven by I/O pins from an LPC1114 ARM microcontroller. As I understand the datasheet specs for this part I cannot exceed 4mA of source or sink current in the standard purpose I/Os. See this question for more clearance on this. The display datasheet says that the forward voltage of each segment of the display is about 2.10V. They will be feed with 3.3V. I want use 10mA of current in each segment (I believe it will give a nice brightness with this current). With all this in mind, and for what I understand of electronics, I aks these questions: 1 - I cannot connect the segments A-DP directly to I/O ports since it will exceed the sink capabilities of the I/O. Is this correct? 2 - Almost all schematics I found on the web on multiplexed displays do that. I did not found anyone that deals with this problem. What suggestions do you have to make it possible to sink more current in each display segment? 3 - I thought that maybe I could sink segments A-DP to ground using a MOSFET controlled by I/Os. Is this a good idea? What do you think? Any schematics on this? Thanks <Q> 1) <S> Yes that's correct. <S> If your GPIOs can source or sink 4mA <S> you can't ask them to source or sink more, they probably are overcurrent protected but they won't work properly. <S> 2) That may be because other designs use "stiffer" GPIOs that can source/sink all the current that is needed. <S> 3) <S> That is a great idea. <S> You can use a mosfet or a transistor, just like you do on the high side. <S> Why are you using a NPN bjt for the high side anyway? <S> A p-mos or a PNP bjt would be better in my opinion. <S> The schematic is simple: simulate this circuit – <S> Schematic created using CircuitLab <S> Remember that when sizing the limiting resistor you should take into account Q1 saturation voltage and Mx on resistance (that's quite low actually). <S> Pay attention to maximum mosfet \$V_{GS}\$, 5V might be too much, and be sure that Q1 is fully saturated also when every segment is on, i.e. when its collector current 80mA. <S> Of course its collector current rating must be high enough. <S> As I said you should probably use a p-mos for the high side switching, currents are low so you can find something suitable for your application. <S> Please note that a NPN bjt might not be suitable since the micro controller output is lower than 3.3V, then you have a \$V_\gamma\$, that's another 0.7V, then the LED, then the N-MOS <S> ... <S> If the micro output is too low your leds might not light at all. <A> You can use a bunch of NPN transistors that act as pull-downs on the 8 common lines feeding the displays. <S> You'll still need a resistor in series with each of course. <S> Your anode voltage is going to be about 2.7 volts (emitter follower loses ~0.6 volts) and assuming your LED drops 2 volts when being powered at 10mA, this leaves a resistor value of 0.7/0.01 = 70 ohms. <S> If you used PNP transistors on the high side, you'd probably only lose about 0.1 volts up there leaving 3.2 volts on the anode and about 1.2 volts on each current limiting resistor. <S> This makes the resistance 1.2/0.1 = 120 ohms (as you have shown). <S> Obviously if you use 8 more transistors you don't exceed the IO spec but there is an inversion in logic to fix. <S> Ditto if you swap to a PNP transistor as common emitter on the anodes. <A> This could be a more difficult question than it first appears, especially if you're planning on driving a blue or white LED display. <S> The forward voltage of those can approach the 3.3V you have available, so you cannot afford the ~0.9V voltage drop of your NPN emitter-follower transistors (you could lose the 330R resistors in that configuration, by the way). <S> For example: N-channel FDY3000NZ (dual MOSFET) for segment drivers common anode P-channel <S> AO3415 (single MOSFET) <S> for digit driver common anode <S> That combination will result in a total voltage drop of less than 100mV at reasonable currents for a 3-digit display, meaning most of the voltage will end up across the LED (making light) and the resistor (equalizing the light between segments). <S> I suggest gate resistors on all MOSFETs (1K should be fine for sensible multiplex rates) to control the current flowing from the outputs to the gates. <S> You may have to blank briefly between digits to prevent ghosting.
| You could use low voltage P-channel MOSFETs for digit drivers) and N-channel low-voltage MOSFETs to drive the segments.
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What do I need to start with AVR UC3 Microcontrollers? I am just beginning to learn how to use microcontrollers, and I have decided I am going to learn on the UC3B 32-bit avr. I will also be learning C programming in the process. What do I need in terms of programmers and other necessities, while staying pretty cheap, and where are some tutorials to help me get started? <Q> Microcontrollers are available with a wide variety of capability and performance. <S> If you're just starting out, it might be wise to begin with something simpler (like the tinyAVR series ). <S> Atmel makes evaluation kits (such as the EVK1101 ) which I would recommend when working with a specific microcontroller platform. <S> Such kits allow you to fully explore the feature set of the product without having to hook up external support circuitry (as much). <S> As for programming, you have a number of options again. <S> If you look at the UC3 Tools page at Atmel , various compatible programmers are listed, such as the AVR Dragon. <S> You can also probably use third-party programmers such as the USB Tiny from Adafruit , but I haven't personally used it for UC3. <S> Finally, to learn embedded C for microcontrollers, there are many resources online. <S> There is a lot of good references to programming at AVRFreaks.net . <S> If you're really new to programming, you should probably look at taking an introductory programming class (many community colleges offer this) and also at some books about the subject: <S> Embedded C Programming and the Atmel AVR Make: <S> AVR Programming: <S> Learning to Write Software for Hardware <A> Nice choice actually, been working on those things for the past months, both in the low power area and for general purpose, including USB and stuff like that. <S> If you've already decided on the B series (it actually doesn't matter, only few components change between the series such as the ADC) get the EVK1101 and a JTAGICE3. <S> That's it, nothing else needed, <S> no adapters, nothing. <S> Plug both into USB <S> and you're good to go (ok, you should download Atmel Studio). <S> First thing to note <S> : Start with an already existing example (the ASF (framework) comes with many examples for nearly every component). <S> Understand, which functions to execute. <S> Also, make sure you read the chapters on clocks and the power manager thoroughly. <A> This might not be the answer you're looking for but please consider this: You might have a better learning experience starting out with the 8bit AVRs, also from Atmel. <S> All of Atmel's MCU are supported by Atmel's IDE AVR Studio <S> so you learn the IDE once and keep on using it when you move on to more powerful MCUs. <S> The IDE is fantastic, really got better with version 6 and comes with visual assist for free!!! <S> Regarding a debugger / programer. <S> I would invest a little money and buy the JTAGICE MKII and avoid all 3rd party programmers and debuggers! <S> When I started out I used a 3rd party programmer with AVRDude and all kind of 3rd party apps and wasted a lot of time setting up the environment just to get going. <S> Save yourself the trouble, that extra $100 - $200 is worth so much less than your time. <S> Of course the JTAGICE MKII also supports all the protocols Atmel MCUs use (JTAG, ISP, debugWire). <S> Your best source for anything <S> Atmel - www.avrfreaks.net <S> which is a developers and support community like no other.
| You can use Atmel Studio to write and debug your code, but you can also use any text editor (i.e. Notepad++ ) and AVRDude to compile (I recommend you get WinAVR ).
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What is CV & CC in power supplies? In many power supplies, there's CV & CC indicators. What does they mean? <Q> They are acronyms and stand for "Constant Voltage" or "Constant Current". <S> They usually are associated with a LED or an indicator of some kind, as you suggest. <S> When you use a power supply you usually set the desired voltage and the maximum current. <S> When you connect the load two things can happen: The load requires more current than the maximum you set The load requires at most the maximum current you set <S> In the first case the PSU become a current source <S> : the current is limited to what you set and voltage drops accordingly, that's CC for you. <S> In the second case what is costant is the voltage, so that's CV. <S> As an example, consider this case: you set the voltage at 10V and set the maximum current at 1A, then you hook up a load that is over \$10\Omega\$ . <S> As you know that requires at most 1A, so the voltage is constant while the current can vary between 0A and 1A. <S> If you then hook up a lower impedance load it would require a higher current, but now the current protection kicks in so the current is limited to 1A, and it is constant, while voltage varies between 10V and 0V. <A> Most types of load need constant voltage to operate, so if the "CV" LED is lit, it means the PSU works fine with your load. <S> The PSU has a physical limit on how much current it can supply. <S> If the load attempts to draw more, the PSU decreases the output voltage to keep the current consumed at its maximum, not beyond it. <S> This mode is called "constant current". <S> The PSU behaves as a current source, so that it changes its output voltage to keep its current constant. <S> Some types of rechargeable batteries require mixed mode charge, where the charge starts at constant current and finishes at constant voltage. <S> Li-Ion ones are a good example. <S> So the battery chargers designed for such batteries can have this kind of indication. <A> There is an excellent presentation on CC-CV by Maria Cortez of Texas Instruments with visuals, sound, and printed transcript that really helped me wrap my head around the concepts. <S> (If you want to jump to the discussion, try starting at minute 1:37.) <S> https://training.ti.com/introduction-battery-management-part-3-li-ion-battery-charging
| "CV" stands for "constant voltage", and "CC" stands for "constant current".
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Does a pin jumper adds noise? When designing a PCB, is it sensible to use a pin jumper in the pathway of a small current signal (around 0.8 uA) or could there be issues with noise? I would use the jumper to select between different transimpedance amplifiers. <Q> Jumpers are no more noisy than any other simple male/female connector. <S> The general structure of a jumper, being wide and flat, means it's not much of an antenna like a raised loop of wire <S> would be, so it won't pick up much in the way of induced noise. <S> The main issues are with badly fitting jumpers. <S> Bad electrical contacts can result in additional resistances, which themselves can generate noise, or in a worst case scenario you get a diode effect, which can not only itself corrupt your signal, but under certain conditions act like a radio receiver (aka cats whisker) <S> picking up EM interference from the environment. <S> But in general with a good fitting jumper you get no more noise than say a jack plug, or a Molex KK connector, which is to say a pretty much immeasurable amount. <A> It sounds like there will just be a steady DC value on it. <S> You may have a small amount of noise introduced temporarily while the person fiddles with the jumper, but otherwise, it'll be a very quiet area. <S> If you can handle a little extra noise for short periods of time, then go for it. <S> If you're needing no noise all the time, even when you're selecting the other amplifier, then you'll want move the pin jumper elsewhere. <A> If you're switching the input to a transimpedance amplifier you may need to be a bit careful as to the capacitance and coupling you're adding. <S> You don't mention what the impedance is at that node- <S> I've seen megohms, hundreds of megs, and G ohms, as well as lower values. <S> Also frequencies can be Hz or VHF. <S> If the jumper is horizontal to the board and has a ground plane under it you shouldn't have troubles on that account, but it might add a few pF to the input, which could be enough to destabilize the amplifier.
| As long as the jumper is a nice and tight fit and there is a good electrical contact made there will be no connector noise.
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Best way to drop 12V fan to 7-9V? So my Xbox one PSU died and microsoft doesn't sell official ones yet, so I bought a third party one and the fan that is in it is a 12V blower fan with 2 wires running right to the board (it says 12V right on the fan). So obviously one of the wires is ground and the other is 12V. So I need help determining what would be the best method to lower the voltage to around 7-9V. (The reason I'm trying to drop it is because this thing is extremely loud and pretty much unbearable, it runs cold all the time so I do not think heat will be an issue here, if it is I'll buy a new PSU) So here is a photo of the set up: Album with pictures in it How do you think I should go about doing this (resistors, diodes, I have no problem installing or trying anything, I was thinking a 25 ohms resistor with a rating of at least 5 watts). <Q> At 12V 0.2 amps, you have 60 ohms equivalent motor resistance at speed (Ohm's law). <S> Using voltage division: $$V_{out} <S> = <S> V_{in} \frac{R_1}{R_1+R_2} $$$$\frac{V_{out}}{V_{in} \cdot <S> R_1} = <S> \frac{1}{R_1+R_2} <S> $$$$R_1 + <S> R_2 = \frac{V_{in} \cdot R_1}{V_{out}} $$$$R_2 = <S> \frac{V_{in} <S> \cdot R_1}{V_{out}} <S> - R_1 <S> $$$$R_2 = <S> R_1 <S> \cdot <S> (\frac{V_{in}}{V_{out}}-1) $$$$R_2 = <S> 60(\frac{12}{8} - 1 <S> ) = 60 \cdot 0.5 <S> = 30 <S> \Omega$$ $$30 \Omega <S> \cdot <S> (0.2 A)^2 <S> = 1.2 <S> W$$ <S> A 1.5 or 2 watt resistor will suffice if it is 30 ohms. <S> You can go bigger if you like, but it won't do much more. <S> Make sure your fan still spins well though. <S> If it doesn't you could be drawing more current than you want. <S> In that case, up the voltage some. <A> You could just put a bunch of cheap 1N4001 diodes in series until you get what you want in terms of speed. <S> About 5 should do it, since each would drop about 800mV when warmed up and conducting 150mA-ish. <S> More like 1W total dissipation than 5W. Heating will make the fan go a bit faster, but not much. <S> Less tendency to stall than with a resistor. <S> Note: <S> You're obviously asking for trouble- high voltage stuff in there that can be hazardous- <S> if you compromise isolation the game could be dangerous to users, fire hazard, reducing the cooling could cause a very early failure, and a lot of the aftermarket stuff doesn't have proper overvoltage protection, so you could kill your game box. <S> So I don't recommend you do it. <A> Another option would be to place a rheostat. <S> It's like a potentiometer, but simpler in that it's a variable resistor. <S> I have done exactly what you are doing with some pc fans that were a bit too loud. <S> At least this way, you can always adjust the fan speed if your xbox gets too hot.
| I would find a high power Potentiometer to test this out with. Or perhaps get a couple different resistance values to try out what's the optimal amount of fan cooling vs sound.
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Avalanche Pulse Generator: Common-collector alike setup with no input and fairly high voltage on the Common I am wondering what would be purpose of a setup like this? simulate this circuit – Schematic created using CircuitLab If you notice there is no input in this setup and the voltage is fairly high on the Common leg which exceeds the tolerance level of 2N3904. The output with the CRO shows a spike of roughly 10v high in approx. 300pS rise. Would appreciate it if somebody point me into the right direction. <Q> The 2N3904 is rated to at most 40 Volts collector-emitter, and we have a 100 Volt source here. <S> There are avalanche BJTs designed to be used this way, but any ordinary BJT will work. <A> From the clue I have been given by @motoprogger, I did a bit of a reading and this is what I ended up with. <S> This circuit is possibly an Avalanche Pulse Generator and one of many uses can be to measure the bandwidth for oscilloscopes. <S> This article, Avalanche Pulse Generator Build Using <S> 2N3904 by Kerry D.Wong gives a pretty good discretion of how this simple circuit operates. <S> and here is the pulse he generates in his experiment <S> Apparently this is a typical Jim William design :) <A> Use this type of circuit as a test pulse to check for open circuit or short and to measure cable length. <S> Time domain reflectometer. <S> See Mr Carlson''s lab videos
| It looks like it is an avalanche-breakdown oscillator.
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Best choice for low-power, short-range wireless communication I'm working on a project which relies on wireless communication, and I'm looking for the best alternative. These are the design requirements: Is easy to interface with Arduino Works over a small range (5-6 metres at most) Consumes little power - the least possible Devices mustn't interfere with each other Doesn't need to be real-time, the delay can be in the order of hundreds of milliseconds In general, the devices aren't within sight of each other (meaning IR and similar solutions won't work) The scenario is: About 10 devices in the same room, communicating with each other, sending short (30 characters at most) messages to each other once in a while, for a few hours (without changing batteries). The best I could find is Bluetooth (BLE, specifically), mostly because there are a few ready-made Arduino+Bluetooth modules and the promising "Low Energy" label (at most half the energy consumption of normal Bluetooth), but I'd love to get an opinion from people with more experience in this field. <Q> Take a look the XBee line of tranceivers. <S> There are various types depending on the transmission range and desired power rating. <S> These devices also interface with Arduino projects nearly trivially. <S> The Serial library makes use of the AVR's UART hardware which is directly compatible with the Rx pin of the XBee. <S> Also note that no level shifters are needed because the XBee <S> Rx/Tx pin work with 5V! <S> DO note , however, that the XBee must be powered with 3.3V, so a simple 3.3V LDO voltage regulator can be used. <A> "Best" is hard to define, but I would look into an IEEE 802.15.4 transceiver. <S> I used an MRF24J40MA for a project ( https://github.com/briksoftware/gradusnik ). <S> You can try to see if there is some code you can use there. <S> The project is for PIC, but many things are actually platform independent (especially the other projects it depends from). <S> The module consumes about 20mA in rx/tx, which is not much. <S> However, to get long battery time you need to put the module to sleep most of the time. <S> You could use a beacon enabled network for this, with one device acting as the coordinator. <S> The module consumes some µA in sleep mode (check the datasheet for exact numbers) <S> The only problem interfacing that with an Arduino is that you need a 5->3.3 level shifter (if Arduino still uses 5v). <A> I don't see it as a wireless problem <S> but more of a protocol problem. <S> If battery conservation is the name of the game and if one device can, thru appropriate programming, adopt the role of temporary "master" then each other device can be allocated a timeslot. <S> Once the allocation is done then the temporary master can return to being a peer but the important thing is that a framework of time slots has been created and all peers will have bought into this framework. <S> What does this do? <S> Having a timeslot means you can shut down the radio for hundreds of milliseconds and wake up to see if there is a transmission in progress intended to be received. <S> When a peer wakes up it has to wait to see if any of the other 8 peers are sending a message to it. <S> Timing is critical but if you want long battery life then concentrate on the protocol. <S> This answer is just a whiff of an idea. <S> The wake-up timeslot will be sub-divided into 8 other slots that each of the other 8 peers are allocated to transmit on and, if one is transmitting the others can listen first to check whether they can transmit. <S> Something like this: - The "next" slot allows new peers to join the group. <S> Low power transceivers are common place <S> so I'm not going to look into this. <A> Try the cheap 433 mhz txrx modules. <S> they come in handy and can be easily interfaced with arduino. <S> I use them for my personal hobby robotics projects. <A> Just curious why the NRF24L01 2.4GHz <S> Wireless <S> Transceiver Module hasn't been mentioned? <S> It is a low power unit, can be switched into a standby mode for even lower draw. <S> It operates off of 3.3v, so can use either a low power arduino or a splitter, and best of all is pretty cheap? <S> Here is the nRF24L01+ product page along with a nRF24L01 How- <S> To that contains information about interfacing one to an Arduino along with sample code. <A> The RFM12B comes to mind.. <S> http://www.hoperf.com/rf/fsk_module/RFM12B.htm <S> Some features: <S> SPI compatible interface <S> High data rate (up to 115.2 kbps in digital mode) 2.2V-3.8V power supply <S> Automatic antenna tuning 16-bit RX data FIFO Programmable TX frequency deviation (from 15 to 240 kHz) <S> Programmable receiver bandwidth (from 67 to 400 kHz) <S> Analog and digital signal strength indicator Clock and reset signal output for external MCU use <S> The SPI comms interface should be fairly to set up with Arduino. <S> The RFM12B-S2 model sells for under $7 these days.
| Another popular transceiver for the IEEE 802.15.4 protocol is the XBee module, but it is much more expensive.
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What is the difference between a 30 V power supply and a ±15 V power supply? I need to get a ±15 V power supply and I am wondering where to find it. Should I get a 30V DC power supply in its place? <Q> A 30V DC power supply probably isn't what you're looking for. <S> You would need a way to provide a low-impedance mid-point at 15V to substitute for 'ground'-- assuming that it's isolated, or that you're very careful about how you interface it to external parts. <S> The best way to go is to get a ±15V supply, unless you really know what you're doing. <S> Asking the question probably means you should just buy (or build) one! <A> In a ±15V power supply, you can get 3 different voltages without the need of any external circuitry -15V , 0V and +15V. <S> In a 30V power supply, you can get either 0V (ground by reference) or 30V. <S> I'd suggest you get a ±15V power supply. <A> First of all you need to determine what current demand will be placed on the power supply. <S> A ±15V supply is 2 X 15V supplies connected in series. <S> Each supply will essentially be operating independently. <S> It is unusual to find non-specialized power supplies with different current ratings for each rail. <S> Depending on the load of whatever you’re connecting to it, the highest current requirement of either rail will determine the whole power supply rating. <S> Some supplies are quoted as so many volts at X watts or Y amps. <S> For power value multiply total voltage, 30V, by the required current to give power in watts. <S> A 30V power supply will not suffice if your application needs + and – supplies. <S> Once you have determined the rating of the supply you can seek one out. <S> I have used http://www.maplin.co.uk/
| A ±15V supply will have 3 connections (+15, -15, 0), whereas a 30V supply will only have two (+30, 0).
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DC-DC step up from 3V to over 100V I'm currently building a DC-DC step up circuit to step up 3V to 100V. I don't want to use transformer because i want to keep the circuit as small as possible. My thought is ... building a 30V-output circuit using MIC2250, then build a boost circuit to boost the voltage 3-4 times. The 30V output has been achieved, but the boost doesn't work. The diagram is as below. I was thinking of repeating this circuit 2 to 3 times in order to step up the overall voltage 2-3 times. but it doesn't work. My another design is to use IC solely. some ICs like LT3757 and LT3783 seem to fit my design, but I just don't find any suggested circuit to boost from 3V to 100V. I tried MCP1652 before by following the circuit in the datasheet. Although the datasheet clams that the circuit can step up the input to 100V, I just can't do it. The maximum I got was only 40V.(I emailed the technician from MCP...he asked me to find the solution on the internet myself...poor aftersale service...) the only suggested circuit from 5V to 350V in LT3757 requires transformer, but i just try to avoid transformer. Can any body share you idea with me about the circuit stepping up from 3V to 100V? Iout = 0.1A will be good enough. slightly smaller will be acceptable too. <Q> Avoiding a transformer to keep the circuit small ... doesn't work so well. <S> See Spehro's answer : to avoid saturating the inductor in a flyback convertor, you need to keep the current low, or use a big inductor. <S> The problem is that you are storing energy in the inductor core at low voltage, and releasing that energy (in a fast pulse) at high voltage. <S> And it will only hold so much energy before it saturates. <S> With a transformer, while the primary winding is storing energy, the secondary is already pulling energy out (at higher voltage but lower current) <S> so very little energy is actually stored in the core. <S> This means you can transfer much more energy for the same size of core. <S> Or use a smaller core for the same power delivery. <S> Something like this might be close to what you want, about 1.6*1.6*1.6cm to deliver about 6W (0.06A at 100V). <S> You could drive its 3.3V secondary from an H-bridge at 100kHz and extract power from its primary, with a bridge rectifier made from those fast diodes. <S> Or wind a custom transformer on a core like this or this smaller one to get exactly what you need. <S> I haven't done the math to design a 10W supply around either of these cores. <A> 1N4004 diodes are inappropriate for this application because they have a very long \$ t_{rr}\$ reverse recovery time. <S> Use an ultra fast UF4004 here. <S> Assume output voltage is <S> 100V and current is 100mA. <S> Then ton/toff is about 2.5, and peak inductor current will be 700mA <S> (the inductor must not saturate - so that's a big inductor). <S> On time will be about 30usec and off time 12usec, so about 24kHz. <S> It's feasible if the parts are properly rated- <S> but the most likely culprit if it does not work when fed with a high capacity 30V bench supply <S> is the inductor- <S> use one that will not saturate with 1A or more. <A> Im guessing that your 3V power source was unable to deliver enough power/current <S> and thats why you had 40V output instead of 100V on converter built on MCP1652. <S> You should measure input voltage when converter is running.
| It may be better to look at application notes for some of the ICs you mentioned for guidance on winding suitable transformers.
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Maximum current for a single-core cable? I want to connect the cables of a LiPo 7.4 V 2200 mAh battery, to a breadboard. However, the only way to make this happens is to solder the ends of the battery cables with a single-core cable with solid conductor. Is this a safe option or there is the risk of damaging the single-core cable? As far as I know, the current draw will be about 700 mA. <Q> Theoretically - for 700mA you should use AWG 23 wire. <S> You can use this table for choosing wires. <S> AWG 23 is <S> 0.57404mm core diameter (or 0.0226 inch core diameter). <S> Practically - if wire is short (few cm/inches) <S> - you can use AWG24 or AWG25 wire and there will be no damage at all. <S> Even thinner wire probably could survive 700mA without damage. <S> Note, that wire current ratings are usually determined by voltage drop/resistance. <S> This is not maximum current without damage. <S> For relatively high currents (>500mA) I'm using double wires sometimes - to make sure that connection is good enough. <A> The optimum size for solid-core wire to fit snugly into a solderless breadboard is AWG 22 (0.64mm diameter), and that will be fine. <S> 12 <S> " of AWG 22 wire (two 6" leads) will drop only about 11mV at 700mA, which is pretty much negligible. <S> 700mA is towards the upper end for a breadboard, and you may see some significant voltage drop and heating if the contact turns out to be flaky. <S> Do not use the plug wire style unless you verify the wire size is substantial- <S> I've measured them at about AWG 28 (0.3mm), which is a little light. <S> The actual current rating of wire will depend on the insulation used and the environment (other wires and temperature) but ignoring voltage drop, a couple hundred mA would be a reasonable limit for cheap PVC wire in a lab environment. <A> A solid core is more likely to fatigue over time at stress points like where it enters a solder joint. <S> That is why in general flying leads (like power cables, network cables, etc) are stranded cable. <S> Conversely, solid core cable performs better when being used with things like screw connectors, or with individual-wire IDC connectors (such as RJ-45 "punch down" wall sockets), where the cable is fixed in place and doesn't move. <S> The current carrying capacity of stranded vs solid core is (virtually) the same for the same cross-sectional area of copper.
| Many breadboards on the market are poor quality and 700mA current may cause some voltage drop on connection between wire and breadboard conductive part. You should worry more about wire<->breadboard connection resistance. Solid core cables lack the flexibility and mechanical robustness that stranded cable gives you.
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SIM908 GSM and GPS modules individual power control I've been searching around google and even in manufacturer site for this information and found nothing relevant.Is it possible to turn off GSM module and keep turned on the GPS? How can it be done? This point is critical to my application since I need to keep GSM module OFF almost all the time and the GPS always ON. edit: Command manual and product page for Simcom SIM908. <Q> [This started as a comment, but I ran out of room.] <S> Look into the hardware design guide for SIM908 (4.2MB .doc) . <S> Section 4.3 describes the low power modes for GSM and how to enter them. <S> Section 5.2 describes power control for the GPS. <S> Somebody who had first-hand experience with this module might be able to provide a more definitive answer. <S> E-mailing a carefully crafted question to Simcom 's application support might also be a good idea. <S> (And if they tell you the answer, make sure to post it here on EE.SE for posterity.) <A> There is no way you can turn off GSM while using GPS. <S> As hardware manual say in section Section 5-GPS Application Interface <S> The GPS engine is controlled by GSM engine, so when it is necessary to run GPS, the GSM engine must be powered ON and not in SLEEP mode. <A> there is a method. <S> But this is not formal or SIM recommended. <S> Inbetween the data signal of the sim card and the SIM908 simcard data input pin, insert a MOSFET which is controlled by your MCU. <S> MOSFET must have not greater than 22 Ohms resistance. <S> When the mosfet is off, SIM908s' GSM engine will not consume as much as power it would, if it had access to the sim. <S> Whenever you need GSM connectivity, then enable the SIM through the MOSFET. <S> hope this helps.
| Having the GPS on while GSM is off (or in a low power mode) may be possible.
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900 MHz vs 433 MHz for urban environments At ranges of a few hundred meters, will there be an appreciable difference in performance (ie range, bit rate) between 900 MHz and 433 MHz for digital applications in an urban environment? Forgot to mention, I've read that some frequency ranges are more crowded than others, is that relevant here? Also I live in the US. <Q> 433MHz will tend to be a bit better IMHO. <S> It's the ability of the two frequencies to sneak between all the obstacles in our modern environment that is probably about equal <S> but, at approximately half the frequency 433MHz will have more penetration due to the free-space link-loss formula. <S> Link loss (attenuation in dB between transmitting antenna and receiving antenna) is: - 32.45dB + 20log\$_{10}\$(MHz) + 20log\$_{10}\$(kilometres) <S> At 500m, the link loss for 915MHz will be 32.45dB + 59.2dB + 0 = <S> 91.7dB <S> At 500m, the link loss for 434MHz will be 32.45dB + 52.7dB + <S> 0 = 85.2dB <S> Not much in it <S> but 5.5 dB is better than a kick in the crotch! <A> The 400MHz band has much lower power limits in most countries than the ISM bands at 900 MHz/2.4GHz/5GHz. <S> So right off the bat 900MHz will be better. <S> If transmission powers are equal then it depends entirely on your structure. <S> Are you propagating through outside walls to the inside? <S> Between floors? <S> Only on one floor? <S> Outside walls are usually concrete with steel re-enforcement. <S> The steel rebar really attenuates RF signals and this makes propagation poor. <S> The same can be said for propagating between floors. <S> However on the same floor many office spaces have metal grids holding a false ceiling. <S> This structure can be used as a successful waveguide and can help carry the signal around the building and through dry wall barriers. <S> Polarization types can also make a big difference between poor propagation and good wave guide conduction. <S> And in many general cases <800MHz <S> the propagation isn't as good through these barriers. <S> Unless you are doing some sort of application specific install where you can optimize the antennas to the layout and structure of the building, all this info won't do you much good. <S> If you're just looking for a generic solution that might work, then all you have to do is buy some cheap WIFI routers and do a site evaluation. <S> Test them on 900MHz (new 802.11ah) or 2.4Ghz (802.11b and 802.11g) and even 5GHz (802.11a) in the areas you want to install. <S> EDIT: <S> I've done number of research projects in 100MHz/200MHz/400MHz/900MHz...5.8GHz for in building propagation. <S> I can't really share the details, but here's an example of the type of testing and simulation that many people have done to try to better understand in-building propagation. <S> http://stakeholders.ofcom.org.uk/binaries/research/spectrum-research/in_building.pdf <S> In general the power limits on 400MHz isn't going to work well for you. <S> We did several narrow band data comparisons and the performance at the legal power levels for all the ISM bands (and a few other bands like FM/VHF/etc.) <S> is not very good for a variety of reasons. <S> Go spread spectrum in one of the higher frequency ISM bands. <S> But if you could tell us more about what you're trying to do it would help. <A> In the US, the power limit for FHSS transmissions is 1W (30dBm) <S> Non FHSS narrow band <S> (3dB BW < 500KHz) is limited to -1dBm, wide band (3dB bandwidth >500KHz) is limited to 8dBm PSD, for 433MHz <S> you can transmit at ~-5dBm, depends on your duty cycle. <S> Also, the 900MHz band offers far more bandwidth, from 902 to 928MHz than the 433 band. <S> Another thing to consider <S> , it is generally easier to find equipment for the 900MHz band, ie modems, antennae and various amplifiers than for the 433MHz band.
| In general I've found that 900MHz tends to get better coverage than the lower frequencies (for the same power levels).
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Do I need pull-down resistor on Enable pin cd74hc4051 Can I directly connect enable pin to ground? I am not going to use it. datasheet Also can I connect all unused pins of LM324 op-amp to ground through single resistor or what configuration does it need? <Q> Can I directly connect enable pin to ground? <S> I am not going to use it. <S> Yes, generally it is good practise to tie all unused inputs of digital devices to eiter GND or positive supply voltage. <S> Also can I connect all unused pins of LM324 op-amp to ground through single resistor? <S> You can, but that would waste power; instead tie the negative input of each unused opamp to its output (which forms a voltage follower) and the positive inputs of all unused opamps to halfway between positive and negative supply voltages, using a resistor divider (or GND in the case of split supplies). <S> E.g. like this (shown for a single opamp): <S> For more details about opamp termination, see Properly terminating an unused op amp and <S> What shall we do with the unused opamp . <A> Yes, you can connect an input directly to ground. <S> Inputs are high impedance so they're like a big resistor anyway. <S> As for the op-amp, that depends on the pins you're not using. <S> As a general rule of thumb: Inputs that are allowed to be the same potential at the same time can be connected directly together. <S> Never ever directly connect outputs together. <S> So you can connect your op-amp inputs together and ground them through a resistor ( <S> though I'd be inclined to only group the same type of inputs; one resistor for all the inverting, and one for all the non-inverting), but each output must have its own resistor. <A> The op amp output is (ideally) a voltage generator. <S> If you connect both inputs to ground, or just tie them together, the output should be zero and there should be no issues. <S> The output won't be zero though because of noise, offset, production imperfections or whatever, so you have this output that wants to drive ground, a power rail, and that's bad. <S> It can break, but if it doesn't it dissipates power.
| You can tie the enable pin directly to ground, for the op amp just leave the unused outputs not connected and connect the unused inputs to ground.
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Should you try and minimize via quantity? I have a 6.5" x 4.5" double sided board that is mixed analog and digital. I have partitioned the analog and digital grounds from each other and have zero ohm resistors that bridge the two grounds together. But my layout involves a lot of vias. When I looked at the drill output file, its about ~400 vias. The bottom traces are kept as short as possible so as to just allow routing on the top layer, but because of that, I have a lot of vias. Is this common to have so many vias? Should I look to make longer traces to reduce via count? Update Added top and bottom layers <Q> For basic low frequency work vias are pretty much "meh". <S> The only thing you need to watch for is when running larger currents - the vias tend to have a higher resistance, so dissipate more heat. <S> However, the interesting things start happening at higher frequencies. <S> Vias start to become antennae. <S> They can radiate EMI like nobody's business. <S> And by "high frequency signals <S> " I don't just mean if you're intentionally working on RF systems. <S> Digital systems have some very high frequency components too. <S> For instance, a 10MHz SPI bus - the clock is running at 10MHz, with harmonics at 30MHz, 50MHz, 70MHz etc (depending on slew rate of course). <S> Also you have the question of impedance to look at. <S> A <S> via has a different impedance than a normal trace, so if your circuit is impedance sensitive (if you are doing impedance matched traces, PCB antennae, etc) then you have to take the vias into account in your calculations. <S> So for general power distribution, vias should be noted and thought about. <S> For high frequencies, vias are anything from frowned upon to down right no-nos. <S> Everything in between is pretty much irrelevant. <A> One thing to be aware of is that when you're doing very large scale production, or optimizing heavily for cost vias do increase board cost marginally. <A> For a two sided board the ground plane is best kept as intact as possible and if this means via count goes up then so be it. <S> This is a generalism so it can change depending on the type of circuit. <S> Having said that, 400 vias seems a lot. <S> EDIT due to OP showing pictures. <S> I see the artwork you've done <S> but there is a lot more you can do to minimize the impact on the ground layer. <S> Look at the following small section I copied: - If the three vertical red traces (left of diagram) were routed closer to each other, the blue tracks that break the ground plane would be miniscule in comparison. <S> You've got to try harder in to minimize the discontinuities in the ground. <S> The vertical blue track could possibly stay as a red track and sneak around the two red traces it crosses.
| So for high frequency signals, keep the vias to a minimum. The more vias, the more time your board has to spend in the drilling CNC. There are lots of examples on your PCB where improvements can be made and yes, don't worry too much about via count - EMI performance is more important.
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Do I need a heatsink for my voltage regulator? I use L7805CV voltage regulator (TO-220) to get 5 V from a Li-Po 7.4 V 2200 mAh battery. However, after a while, voltage regulator gets really hot. Current measurement showed 200 mA. P=I*V=0.2*(7.4-5)=0.428 W. From datasheet: Thermal resistance: junction-case = 5 °C/W, junction-ambient = 50 °C/W, total 55 °C/W. In this case, temperature rise will be 0.428*55=23.55 C. Seems low to me, but I can't touch voltare regulator for more than one sec. Do I need a heatsink or it isn't necessary. Datasheet refers there is thermal overload protection. In this case, can I calculate how long does it take for voltage regulator to shut off (if this will happen)? <Q> That should be okay, in a TO-220 case with no heatsink, assuming moderate environmental temperature. <S> You don't need to add junction to case to junction to ambient, all you need is junction to ambient, which should heat it to less than 48°C (assuming 25°C ambient). <S> Remember, it's the temperature <S> rise <S> you are calculating. <S> Too hot to touch is around 60°C, at least for my calibrated finger. <S> If you're getting it too hot to touch there are several possibilities: a) <S> Your 200mA measurement is not accurate (perhaps the meter drops the voltage) b) <S> Your 7.2V measurement is not accurate c) <S> You have really sensitive fingers d <S> ) <S> It's really hot where you are testing it (30-40°C) <S> To answer your question about the maximum temperature and thermal shutdown- <S> the data sheet specifies an abs. <S> max. <S> junction temperature of 150°C, but to be conservative I like to see no more than 100°C. <S> That would imply a maximum ambient temperature of 75°C, which is reasonable. <S> The actual thermal shutdown is kind of a last ditch thing, and does not come on until you've well exceeded the absolute maximum junction temperature, typically around Tj <S> = 170 <S> °C. <S> It's very unhealthy for the chip to experience this, but it does prevent immediate destruction. <S> That would correspond to an ambient temperature of 145°C which seems mighty hot, even for southern Europe. <A> Your best bet is to get a heat sink installed for that regulator. <S> Last thing you want is for it to get too hot and cutout, effectively shutting down your electronics. <S> When the power level in the regulator is in the area that you describe it could turn unto a thermal oscillator wherein it gets too hot, cuts out and cools down some and then comes back on to restart the cycle. <S> This can get very annoying when trying to use your electronics for anything productive. <S> It is possible to find modules that have pinning similar to the 7805 that you are using now. <A> A heat sink (also commonly spelled heatsink[1]) is a passive heat exchanger that transfers the heat generated by an electronic or a mechanical device to a fluid medium, often air or a liquid coolant, where it is dissipated away from the device, thereby allowing regulation of the device's temperature at optimal levels. <S> In computers, heat sinks are used to cool central processing units or graphics processors. <S> Heat sinks are used with high-power semiconductor devices such as power transistors and optoelectronics such as lasers and light emitting diodes (LEDs), where the heat dissipation ability of the component itself is insufficient to moderate its temperature.
| An alternative to the heatsink idea is to purchase a switching regulator module that would operate more efficiently and cooler.
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Generating various reference voltages in a circuit How does one go about accurately setting reference voltages at various points in a given circuit (e.g. various threshold voltages). Obviously we can have a potential divider from the main power supply but that generally tends to fluctuate a lot, not to mention noise from other parts of the circuit. I would assume a sort of single voltage reference from which we can derive the rest of the references. Does one use potential dividers after that? Or are there any smarter methods? <Q> One smarter solution over the use of resistor dividers is to apply a shunt regulator. <S> A part such as a TLV431 can be biased from a higher voltage supply to produce an accurate reference voltage using a couple of resistors. <S> An additional resistor is used to limit the overall current through the shunt regulator. <S> If you are familiar with the operation of zener diodes you can think of the shunt regulator as being an nice accurate zener. <S> And then two resistors allow you to set its voltage. <S> For multiple references in one project you could use multiple shunt regulators or apply voltage dividers across one shunt regulator. <S> Today's shunt references are in small packages and something like the TLV431 is rather low cost so using multiple units on one board is usually not a big deal. <A> One reference can be used, or several, it depends what you need. <S> If you don't care much about the ratio between references, several might be easier or cheaper. <S> On the other hand, on a recent design I did, the reference was more than $100 in small quantity, so it was advantageous to generate several other precision reference voltages from the master reference (high precision resistor networks are expensive, but not that expensive). <S> You can use a voltage divider from a reference and buffer the resulting voltage, if required (but pay attention to maximum capacitive loading if you're using an op-amp, and decouple the load if necessary). <S> There are also "rail splitter" chips that can be useful (though they're generally not high precision). <A> For more precision use, there are dedicated "Voltage Reference" chips. <S> These are like a small linear voltage regulator, but favouring precision over current. <S> They can typically provide only a few milliamps, but stability and accuracy can go down to the milli- <S> (or even micro- <S> if you pay enough) volt level. <S> With voltage dividers, even taking an accurate reference as the source, you are at the mercy of both the accuracy of the resistors used, the temperature stability of the resistors, and even the ratio of resistors to the input impedance of whatever you're feeding the reference voltages into.
| You can invert a reference voltage to generate a negative, or generate a reference current, which can level-level shift the reference voltage to another supply rail.
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Is a protection diode needed for a resistive load? In some applications a flyback diode is added parallel to the load. Is that because the switching causes reverse voltage for inductive loads? What if we have a resistive load instead? Do we still need to take care of switching? In the below figure there is a read switch operating on an inductive load. Would we still need it if the load were nothing but a resistor? <Q> If the load is a resistor you should be safe without a diode. <S> Resistors have a parasitic inductance of some sort but the inductance value is usualy very, very small unless you are dealing with high power wire wound resistors that usually have a (somewhat) high parasitic inductance. <S> Have a look here too. <S> So the answer is no <S> , you won't need it in almost all practical cases. <A> I know this has been answered already but thought I'd throw this in there as well-- <S> You only need a diode on inductive loads, such as a transformer or a motor. <S> This is because when the magnetic field collapses, it causes back-EMF that can damage your circuitry. <S> The diode provides a safe path so that the EMF can dissipate without causing damage, but will only conduct when back-EMF is present. <S> Purely resistive loads do not generate any back-EMF, though in reality every load has some inductance and capacitance as well as resistance. <A> In some applications a flyback diode is added parallel to the ... <S> inductive load. <S> ... <S> Would we still need it if the load were nothing but a resistor? <S> If you can control your environment so that the transients at switch turnoff cause acceptably low damage and other problems then the diode can be omitted. <S> If you wish to control your situation so that "how did that happen?" unexpected factors in your environment do not generate transients at switch turnoff that cause unacceptable damage and other problems then adding a diode is usually a very very good idea. <S> A suitably rated diode is usually very compact and low cost compared to other circuit elements involved. <S> In a design 'that mattered' I'd tend to use a diode "as of right" unless there were compelling reasons not to. <S> A compelling reason may eg be guaranteed pure resistive load adjacent to switch (e not flexible wiring connected), no other connections, no risk of relevant long term contamination or degradation occurring, or (getting a bit keen now) conducted or radiated effects adding energy to the circuit, and cost is critical and room <S> limited and diode failure probability higher than problem probability, ... . <S> Such "overkill" usually has minimal overall cost. <S> When "underkill" happens unexpectedly in real world volume designs the cost is often far in excess of anticipating it. <S> You cannot and should not protect against every hypothetical problem. <S> But if the chances of protection being needed is only "very low" and the cost of needing but not having it is 'very very very high" then there will be a point where it is a good idea. <A> Generally speaking, at least in an industrial environment, the only time shunt diodes are necessary is when a relay coil is being driven by a Solid State component. <S> Diodes are there to prevent spikes created by the collapsing field on the relay coil when it is switched off. <S> I cant for the life of me <S> see any reason to protect a resistive load. <S> Well maybe, if its hooked to a very high impedance solid state output, a.k.a. <S> A CMOS driver driving through a low resistance. <S> But generally new CMOS outputs have protection diodes internally. <S> In the old days of CMOS merely handling the chips without grounding the arm you pick it up with could damage the chips outputs (and inputs) from static electricity.
| However, for all intents and purposes it is generally safe to leave out the protection diode unless you have a significant inductance for the load.
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Ensuring that the a circuit breaker in a secondary elec. distribution board will go off before a circuit breaker in the main board I want to create a small portable single-phase secondary electrical distribution board to safely power up some lights in my back yard. The board will be powered from a wall socket, and will contain 3 circuit breakers (perhaps a main breaker too) and three output sockets (to connect chains of lights). Is there a way to ensure that if I get a "short circuit" (between Live and Neutral ) or "leakage" (between Live and ground ), the circuit breaker that will trip will always be the one in the secondary board? I basically would like to avoid tripping breakers in the main board (which would shut down PC, refrigerator etc.). The circuit breaker of the wall socket is 16Amps, and I was planning to install 10Amp circuit breakers in the secondary board (perhaps, if you think it wise, I'll add a "main" circuit breaker of 16 amps in the 2nd board, too). But will this suffice? If not, is there another way to go about doing this? The "path" would look like: main board (c.b. 16A) ->wall socket ->schuko plug ->secondary board ->main c.b. 16A (optional) ->3x c.b 10A ->3x output female schuko plugs Note that the secondary box will be located inside the house (and not outside which would render it a hazard on its own!) <Q> If everything downstream is rated for lower breaking capacity compared to upstream then you've done the best you can to achieve your aims. <S> Regards the tripping of a breaker due to leakage to earth use a "residual current breaker" like this: - <S> See wiki page for details <A> Perhaps you are looking for something like this? <S> The above is a portable socket-outlet assembly intended for use in the construction industry. <S> This particular model is made by Clipsal for the Australian market. <S> It includes its own 30mA-sensitivity residual current device (ELCB for you folks in the US.) <S> As a nice extra feature, if you have an electrician terminate matching screw-on plugs to your lights, the assembly will be weatherproof to IP66 when the plugs are in the sockets. <S> Regarding how you want the sub-distribution board to trip off before the main circuit breaker does - this is hard to do. <S> E.g. if your main circuit breaker is 32A, a 16A breaker should provide acceptable discrimination for overload faults. <S> For full-bore, short circuit faults - these happen so fast that the circuit breakers all race with each other. <S> Typical tripping times for a miniature circuit breaker, under short-circuit fault conditions, are 10ms or less (see below.) <S> You will not be able to obtain discrimination between miniature circuit breakers for a short circuit fault. <S> From Clipsal 4-series MCB catalogue. <S> Note the sub board should be on a dedicated circuit. <S> In this drawing I have shown the connection between the main board and the sub board via a socket-outlet, like you specified, but I would highly recommend hardwiring this instead. <A> If you're simply going to plug your panel into a receptacle in your house I would install a 15amp gfi in the receptacles placed. <S> A gfi device is going to trip first versus a breaker nearly everytime. <S> While its not required by the NEC, I always like to put residential outdoor lighting, especially landscape lighting on gfi protection.
| Regarding topology, if you were going to go with the 'custom built' approach, rather than buying a portable socket-outlet assembly off the shelf, it should look something like this. For overload faults (putting a little bit too much load on one circuit), you should be able to obtain acceptable discrimination by installing a dedicated circuit breaker for this circuit, with a rating half that of the main circuit breaker.
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Thermocouple reading different from Commercial Datalogger when using Arduino A K-type thermocouple is connected to a thermocouple breakout board that outputs 1.2V to 8.8V that covers the entire -260C - 1380 °C of K-type thermocouple. This output is than divided by 2 using a potential divider made of two 10k ohm 1/4W 5% resistors, which is fed into an analog pin A0 of an Arduino Mega. The k-type thermocouple is not in contact with anything, essentially measuring the ambient temperature. Problem: Using the Arduino sketch below, Arduino is measuring the temperature of about 10.6 deg C. However using a commercial temperature datalogger Omega HH309A, the measurement is 26.7 deg C. Measuring the output of the thermocouple breakout board using a multimeter gives 2.19 V . What could be the reason for this significant difference in measurement? Sketch void setup() { pinMode(tcPin, INPUT); Serial.begin(9600);}void loop() { // Take the average of multiple readings tempSum = 0; for(int i = 0; i < tempSamples; i++) { tc1 = analogRead(tcPin); voltage = tc1 / 1024.0 * 5000 * 2; // in mV after correcting for V_divider tempSum += (voltage - 2.05) * 0.005; delay(100); } temp = tempSum / tempSamples; Serial.print("tc1: "); Serial.println(tc1); Serial.print("voltage: "); Serial.println(voltage); Serial.print("temp: "); Serial.println(temp); delay(1000);} Serial Monitor Output tc1: 218voltage: 2128.91temp: 10.62-tc1: 217voltage: 2119.14temp: 10.60-tc1: 218voltage: 2128.91temp: 10.61-tc1: 218voltage: 2128.91temp: 10.63- Updated Arduino Sketch void loop() { // Take the average of multiple readings tc1Sum = 0; for(int i = 0; i < tcSamples; i++) { tc1 = analogRead(tcPin); tc1Sum += tc1; delay(100); } tc1Average = tc1Sum / tcSamples; // Arduino ADC has 1024 steps // 10V max output from TC breakout board halved to 5000 mV before Arduino reads it voltage = tc1Average * 5.0 * 2.0 / 1024.0; // in V after correcting for V_divider temp = (voltage - 2.05) * 0.005; Serial.print("tc1Average: "); Serial.println(tc1Average); Serial.print("voltage: "); Serial.println(voltage); Serial.print("temp: "); Serial.println(temp); Serial.println("-");} Output -tc1Average: 218.20voltage: 2.13temp: 0.00-tc1Average: 218.40voltage: 2.13temp: 0.00-tc1Average: 218.20voltage: 2.13temp: 0.00-tc1Average: 218.40voltage: 2.13temp: 0.00- <Q> A ten degree change would induce 50 millivolts of change on the output of the thermocouple board (or 25 mvolts after your divider). <S> This is a relatively small number. <S> If your 210kohm resistors don't match perfectly, what would your output voltage be. <S> Start by assuming one resistor is perfect, and the other is worst case off. <S> If those are 5% tolerance resistors, assume one is at 210Kohms, and the other is at 220kohms (for a roughly 5% error), and figure out what your divider really outputs. <S> Thumbnail calcs show that if you're measuring across the 220Kohm, a 5 volt input would give you 2.558 volts out, and there's your 50 mvolts! <S> Next, you may be having input impedance problems on your A/D input. <S> You have a relatively high impedance source (>100kOhms) going into a cheap A/D with modest input impedance. <S> Your A/D could very well be loading your voltage divider. <S> Precision measurement in this case calls for precision techniques. <S> If you build an op amp circuit (these have low output impedance) with an amplifier that has a very low input offset voltage, give it a gain of 0.5 (yes-- you'll have to invert to get a gain less than 1.0), and use pots to trim all the resistances and maybe offsets, you'll do better. <S> If you still want to do things this way, try using 1% tolerance (or better) resistors of much lower value, maybe 5kOhms. <S> As a last possibility, maybe there's an offset somewhere on your thermocouple board that you have little control over. <S> Thermocouples can be tricky to use correctly, and maybe their cold junction compensation isn't what it should be. <S> Also, you'll learn more about your error if you actually chart it across a temp range rather than a single point. <S> I recommend a slow transition between an ice/water mix at 0 degrees, to hot tap water at around 40. <A> voltage = tc1Average * 5.0 * 2.0 / 1024.0; <S> // <S> in V after correcting for V_divider <S> Your calculations are assuming that your board (and hence your Vref) is exactly 5V. <S> It is highly unlikely that this is the case, especially if you are running from USB. <S> Secondly, you should be dividing by 1023, not 1024. <S> This, coupled with the other problems stated in the other answers, could account for the amount of drift. <A> The formula you are using in your program does not agree with that given in the specifications for your thermocouple board. <S> There is a 2.05 volt offset and a scale factor of 5 millivolts/deg <S> C. <S> Therefore to convert the board output voltage to deg C, the following should be done: deg C = <S> (voltage - 2.05)/.005 <S> Thus if the board voltage is 2.19 volts, the temperature calculates out to 28 deg C, close to your other temperature measuring device. <S> In your program you are multiplying by .005 instead of dividing. <S> That is why in your sample printout, you get a temperature of 0.00 for 2.13 volts instead of the correct value of 16 deg C. <A> There is a possibility that the thermo couple interface also provides an output for cold junction compensation. <S> I have a suspicion that this should also feed into an analogue input of the arduino and that you may need to add the two readings together in software. <S> In case you didn't know cold junc comp is needed to compensate for the electric potential that occurs when the thermo couple wires are connected to copper tracks of the PCB. <S> This is called the cold junction and it generates an error voltage that needs to be accounted for.
| I suspect a number of things that can cause errors of this size First bet would be resistor mismatch. I suggest getting a DMM and measuring the exact voltage of the 5V rail of the Arduino and using that in your calculations.
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What's the point of a "flipped" opamp power supply (going from -9V to GND)? I was looking at the attached schematic and realized that all opamps and transistors are "flipped" in respect to their power pins. I realize that this is only another way of writing the same thing with +9V on the positive rail and GND on the negative one, so what's the point of this? Bonus question: Why are the headphones and the Line Out in rerence to GND (positive rail) and the built-in speaker in reference to -9V (negative rail)? <Q> A possible reason is that some of the IC's, like that IG021610 part, seem to be designed for a negative supply (the power pins are actually labeled GND and VEE). <S> So the other ones that can be used either way are just adapted to that supply. <S> Their signal inputs and outputs are capactively coupled, and so the absolute voltage at which those chips operate doesn't matter. <A> There is an interesting Q & A here about that. <S> I suspect that these devices (or some subset of the design) was designed using parts that were originally intended for telephony designs. <S> Either that or this was designed to be operated in the shower ... ;) <A> Assuming this op-amp is a differential amplifier, it likely contains an emitter-coupled pair (also known as a "long-tailed pair"). <S> If you only have a single supply to work from, these pairs work best with ground as the positive-most point in the circuit because it provides better noise immunity. <S> I could go into further detail about how this configuration improves noise figure, but you are likely to find a better explanation in an analog electronics book. <S> One book that goes into more detail is Microelectronic Circuit Design by Richard Jaeger , although I am sure there are cheaper books which cover the same topics.
| Telephony systems originally operated at negative voltages for cathodic protection reasons.
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Are acid traps real? (2014) Some people online say that acute angles in trace routing will cause these problems: "Acid Trap" - PCB etchant gets stuck in the corner, and eats away too much, causing open circuit "Peelable" - Narrow strip of photoresist falls off the board before or while the etchant is applied, causing too much etching and thus open circuit Some people online say these are real problems, others say they are old wives' tales. Are they real or not? Or are they only real if you use a low-cost fab house that might use older processes? Please avoid answers like "ask your fab house". I'm looking to learn in general, not just get one design out the door. <Q> Are they real or not? <S> These are general design guidelines that reduce the incidence of problems. <S> However, today's PCB fabrication processes have improved to the point where good PCB fab lines can manage any trace angle that doesn't violate any of their other design rules, such as trace width. <S> Today's masks also provide substantial mechanical support to traces, so you can add some confidence that this won't happen to your design if you specify a mask. <S> are they only real if you use a low-cost fab house that might use older processes? <S> I'm sure there are fab houses with processes that will have a slightly higher failure rate with sharp angled traces than otherwise. <S> It doesn't matter whether this is due to poor process, materials, or skills. <A> The US-based assembly house I work with will complain and ask for corrections in PCBs with acute angles. <S> Whether their complaints are over-cautious or not, I do not know. <S> They contract to several different PCB vendors both US and overseas, so it could be an issue of making sure that ANYONE can build the part. <S> I finally just set an Altium rule and it hasn't been a problem since. <S> I typically spend 10 minutes or so cleaning up acute angle errors to pad entries at the end of a moderate sized layout if I haven't been watching it along the way. <A> In order to combat acid traps, underetching of thick copper layers and other such problems with traditional thermally activated etchants, a lot of board houses have switched to photoactivated etchants. <S> These etchants are much more active under the influence of light than just by themselves, which means you can get cleaner edges on thick copper layers. <S> Also, etchant trapped in sharp corners in the design doesn't eat away at the sides nearly as much as it would with traditional methods. <A> One disparity <S> I notice in the explanations <S> is the Peelable. <S> If too much enchant is applied, how could it possibly result in a short circuit? <S> It would be etching away MORE than it should be, not not etching what it should.
| Generally problems of this nature can occur in really cheap fab houses in china, although they are rare otherwise.
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op amp as comparator/switch - strange behaviour I have the following schematic (extract): The purpose of this is to enable a pin (chip enable) and to disable the 5V supply ("5V consumer") if the threshold at the signal IN is above a certain threshold (VRef=0.5V). So far so good, the circuit works fine under "normal" circumestances. The problem is that when I use the PCB with different power supply and input signals (for testing purposes), th op amp does not switch any more. Here are the differences: power supply: normally: USB(5V) -> battery charger IC -> 4.2V testing: 5V from Raspberry Pi -> battery charger IC -> 4.2V. In this case I noticed that the 5V are actually about 4.8V, but that's still enough.In both cases, no battery is connected so the battery charger is acting as a voltage regulator. signal IN: normally: analog audio signal with a bias voltage of 1.3V and no more than +/- 0.4V peak-to-peak voltage. testing: 1.3V DC voltage So when testing, I measured the 1.3V at the signal IN pin and the 0.5V at the VRef pin of the op amp, but it did not switch, i.e. the chip enable pin remained low and the 5V consumer voltage was still enabled. The total power consumption of the PCB is under 50mA, so it should not be a problem for the Raspberry Pi to power it. The power supply unit is also rated at 1000mA. Any idea why this happens? Am I overseeing something?Thanks! UPDATE Maybe it's relevant, so I added the circuit which switches the ~1.4V at the signal IN pin: One thing I noticed: while sweeping the input signal ("signal IN") from 0V to about 2V, I noticed that the op amp does switch above the threshold of about 0.55V, but the output voltage jumps from about -15mV (input under 0.55V) to only 60mV (input over 0.55V). UPDATE 2 I disconnected the output of the op amp (no load) and now it does not switch any more at all (using the testing method). The output remains low. I also measured the output of the op amp with an oscilloscope (both with and without load) and it does not switch fast high and low (so it averages to 60mV), as Spehro Pefhany suggested. The voltages are pretty clean, i.e. no spikes or large AC variations. UPDATE 3 In the last update I said that the op amp does not switch without a load. That was a mistake from my part: it still does switch, but the ON-voltage is at about 30mV. UPDATE 4 If I power the board (first schematic) normally (5V from USB) and use the Raspberry only to switch the signal (second schematic), the op amp switches correctly! So I tend to think that there is a problem in my power supply. I power the board with the Raspberry very similar as I switch the signal IN (second schematic), except without the R18 and R21 (so I basically feed-through the 5V). The 5V (at Q8 ) are coming directly from the Raspberry pin header. <Q> What you describe SHOULD work, so it seems you may be just crossing over a limiting boundary when you change modes. <S> Try it with NO load on the op-amp. <S> If that still doesn't work then circuit is not as you believe OR opamp is damaged. <S> LM321 data sheet <S> Presumably the chip enable <S> pin does not provide a massive pullup? <S> Or does it?Try it with nothing connectd to LM321 output pin. <S> Things to look at include: Supply voltage. <S> 3V in datasheet so 4V2 or 5V <S> OK. <S> Max Vcommon_mode = <S> Vdd - 1.5V or 2V depending on conditions. <S> In either case you SHOULD be well below that. <S> Massive output pullup - see above. <S> The LM321 and LM314 (quad big brother) data sheets provide sink currents at high supply voltages but not at low voltage. <S> IF these decrease with supply voltage you may be below the 1 to 2 mA needed by the LED etc. <A> Sweep the input beyond your normal ranges (but within the allowable maximums from the datasheet) while watching the output of the op-amp <S> : If it moves with the input then it's acting like an amplifier instead of a comparator. <S> If it switches, but at a much higher voltage, you have hysteresis due to positive feedback. <A> The ~60mV differential implies that that "chip enable" input is actually an output and is effectively shorting the LM321 output to ground. <S> Or something like that. <S> Another possibility is that it is very briefly switching above ground and back again <S> and it averages to ~60mV. <S> If you don't have access to an oscilloscope, one way to check that is to connect another LED + resistor from the output to ground. <S> You wouldn't be able to see the slight reduction in brightness of D1 if the output was switching above ground with a low duty cycle, but you would be able to see the difference between dark and dimly lit for an LED connected from the output to ground (with a few hundred ohms in series). <S> Using the LED to level shift the output a bit is clever, and should work, but your system should have some kind of UVLO lockout protection so if the 4.2V battery drops too low, the battery will be protected for one thing, and secondly so the MOSFET won't damage or unsolder itself when the gate voltage becomes insufficient to fully turn it on. <S> Maybe you've got that elsewhere. <A> I found the problem: it was a mean, little shortcut to GND at the output of the op amp. <S> So I was looking in the wrong place at first. <S> The circuit as-is works fine. <S> Thanks for all the answers!
| The output impedance of the RPi might be high enough to give you some negative (or positive) feedback through the VCC line to the op-amp.
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LED persistence of vision (POV) project I recently saw an article with a LED cube 8x8x8. I saw that it relies on an optical phenomenon called persistence of vision (POV). So I wanted to create a simple circuit with 4 lines each one with one LED, that will light up with a frequency that will give the illusion that they light up all the time. (I know how to light up 4 LEDs but I want it simple now to catch the point) Can someone help me and tell me what parts should I need to create this?I have an Atmega 16L microcontroller. Can I complete this without any other chip?Can I achieve this also with 555 and flip flops? EDIT: Sorry if I cant be understandable. I want a simple circuit like the one in the below picture that will light up each strip for a certain time.This time must be sort enough to see a constant lighting. The black box is whatever parts it will need because I dont know them. <Q> Yes, you can do a POV application with your ATmega16L and a bunch of LEDs and resistors and some clever programming. <S> This particular board has a set of 20 inline LEDs that you can program to display the POV effect. <S> You can then achieve the POV effect by waving it around in the dark. <S> So, it's mechanical and human powered (yes, your arm will get tired after playing with it for a while). <S> It's based on Arduino Uno and related boards, but you can easily build your own standalone version and even modify it to use your ATmega16L. <S> Here's a few pictures of the POV effect in action: <S> The idea behind the circuit is quite simple, you just have to wire a LED and its corresponding series <S> current limiting resistor to every digital output pins you have available (as in the schematic below). <S> The rest is programming . <S> The board can do a few other tricks as well, such as The Night Rider effect and more. <A> As asked for, here's one way to do it in hardware: <S> R1C1 is an integrator used to reset U1 and U2 on power-up, U3B and U3C are leftover gates wired to make an astable multivibrator with C2 and R6 determining the frequency of oscillation, and they generate the clock for U1 and U2, a couple of dual "D" type flip-flops which generate a positive going pulse used to drive the LEDs. <S> The pulse is mutually exclusive, so it only lights one LED at a time, and it circulates through the dflops, one at a time, from U1A to U1B, to U2A, to U2B, and then back to U1A to repeat the cycle anew. <A> The black box can be a four bit shift register. <S> You just need: a configuration such that the bit shifted out of the most significant bit feeds into the least significant bit. <S> ensure that the shift register is loaded with 0001 on power up. <S> feed a clock signal to drive it. <S> For instance, take a look at the datasheet for the SN54 . <S> This IC has parallel inputs which are read when the S pin is held low, and the CP1 pin transitions from high to low. <S> So you can just tie these pins to high or low voltage to create the pattern 0001, and ensure that your circuit generates a power-on-reset pulse which holds S slow for a while, and have a clock pulse going on CP2. <S> The clock can be generated with some multivibrator circuit, for instance. <S> Power-on-reset pulses can be created with a RC circuit and comparator (charging of a capacitor until a threshold is reached).
| The simplest POV application I've seen and that I have incidentally built is an Arduino shield called Blinkenlight .
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Why do they place a high current device in a low current package? For instance, it is said that the practical current limit of the TO-220 device is 75A. But I frequently see MOSFET devices for which the specified maximum continuous drain current is above 75A. ( Examples ) Is there any practical difference between two TO-220 devices such that, one of them has a maximum continuous current of 75A and the other one has a limit of 150A, and all other specifications are same? If there is no difference, why do they bother producing high current devices in TO-220 package and advertise it with current ratings above the practical limit? <Q> The link you had goes to a supplier, who, it appears, doesn't read the data sheet very carefully. <S> I would argue that they should be listing these exact parts at the lower current in the packaging provided. <S> The internal silicon is capable of higher currents, but the data sheet clearly shows that, as packaged, these parts are not capable of such high currents. <S> Refer to figure 9 on page 5 of the data sheet: <S> They could manufacture a 75A silicon part just for these packages, but that would result in significant part proliferation and stocking issues. <S> As it is, they manufacture the one part, store it as dice for a short time, and package it according to customer demand. <S> This reduces the load on their plants and storage facilities, and they can focus on improving the reliability and lowering the defect rate of one part rather than two, or more, different parts. <A> There are a number of factors which limit the maximum current a device can handle. <S> These include DC resistance, pinch-off current, lead bonding, and thermal dissipation. <S> Some characteristics, such as having a low DC resistance, will allow a device to handle more current than it might otherwise be able to, but may also be useful in other ways. <S> For example, suppose one is trying to power a device which requires 50 amps at one volt (i.e. 50 watts). <S> If one were using a device with an internal resistance of 0.01 ohms, it would have to dissipate 25 watts when passing 50 amps. <S> A <S> TO-220 would have no trouble dissipating 25 watts, but the device which would require 50 watts would instead require 75 (of which a third would be wasted in the transistor). <S> Using a transistor with a lower internal resistance would reduce the amount of power wasted in the transistor--not necessarily a benefit to the transistor (which would have perfectly happily dissipated 25 watts all day) but to whatever is supplying the power. <S> Because there is often an inverse correlation between current-handling capacity and internal resistance, the silicon transistors which can handle higher currents are often useful in situations which don't need high currents but do need a low internal resistance. <S> In such cases, there would be no need to construct the transistor's case or bonding wires in such a way as to handle the the full current and head dissipation of which the transistor might be capable, since the part would not actually be exercised anywhere near those limits. <A> Usually, current rating is correlated inversely with the ON-Resistance of the device. <S> If the internal device has a lower ON-Resistance, then less power will be dissipated within the device at a given current. <S> The cooler a device runs the lower the ON-Resistance remains and the less power is wasted in a device. <S> Higher rated devices will likely last longer, run cooler and run more efficiently under the same current conditions even if the package doesn't allow the silicon to be used to the full potential. <S> Another benefit would be the ability to run at higher pulse currents in a higher rated device as the 75 Amps stated is a continuous current rating rather than a pulse current rating. <S> Another reason this may be done is because it is cheaper to make one silicon device and package it in many different cases.
| This will allow the device overall to run cooler.
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How to wire 12 cameras to trigger simultaneously In the documentation of a camera it says: "To activate a GPI trigger, momentarily short pin 2 to 1." But I have 12 cameras and I want to trigger 12 cameras at the same time, how should the circuit be for that? On the pins, there are 3.6 V. I am really a beginner. To give more information: this is from the handbook and its a red-one camera. <Q> If they're all at 3.6 V , you should be able to short all pins 2 together. <S> Then short all pins 1 together. <S> Then put one switch that shorts all pins 2 to pins 1. <S> This assumes there's no safety issue with this, and all of the cameras operate at the same ground voltage. <S> To ensure there isn't some compatibility problem and you don't damage them, I'd wire up a BJT or a MOSFET to each camera. <S> Then send one control signal to all of the BJTs or MOSFETs. <S> If you want to go low-tech or you want isolation between the cameras, you could use 12 relays instead of the BJTs or MOSFETs. <S> EDIT: <S> After reading the manual on it, it looks like they're all <S> battery powered cameras and the GPI is just an input pin to a microcontroller that has a pull-up resistor on it (educated guess). <S> You should be completely safe with my first suggestion, wire <S> all Pins 1 <S> (Gnd) <S> together. <S> Wire all pins 2 together. <S> That will connect all of the pull-ups together. <S> Then use one BJT or MOSFET to connect these two together if you want an electronic signal triggering them all. <A> Maybe, use something like SN74ABT5402 12bit buffer, and short all the input pins (as horta suggested) of the buffer, instead of the pins of all cameras? <S> Also, it will ensure synchronicity as opposed to separate relays. <A> Relays or MOSFETs or ... . <S> It is quite likely that a diode to ground will work. <S> If so you can use a single switch or relay or MOSFET and 12 diodes to GPI pins. <S> Using a diode is easily trialled and will do no harm <S> *. <S> If grounds can not be connected then optocouplers or relays will work. <S> It is highly likely that optocouplers with bipolar transistor outputs will be OK and this gives you isolation against unforeseen circumstances. <S> Ask if more detail wanted. <S> * <S> I'm sure it won't cause harm, BUT this advice is liability freee <S> - I can't afford one Red One, let alone 12.
| Just wire them all together as suggested and then add a momentary push switch between the set of Pins 2 and the set of Gnd's (Pins 1). If ground can be commoned you can use 12 switches of your choice. If you just want a push button switch, you don't need any electronics at all.
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Does altitude affect wifi signals? What is the maximum height of a wifi signal in atmosphere? I am not mentioning range and also I dont want answers in milliWatt. I want to know answer in kilometers or meters. <Q> It's difficult (almost impossible) to answer this question. <S> Why? <S> Because the range (vertial or not) depends a lot on factors that you did not specify. <S> First you should have a look at notions such as "Link budget" Wikipedia $$ P_{RX} = P_{TX} + G_{TX} <S> - L_{TX} <S> - L_{FS} <S> - L_M + G_{RX} <S> - L_{RX} \,$$ <S> where: \$P_{RX}\$ = received power (dBm) <S> \$P_{TX}\$ = transmitter output power (dBm) <S> \$G_{TX}\$ = transmitter antenna gain (dBi) <S> \$L_{TX}\$ = transmitter losses (coax, connectors...) <S> (dB) <S> \$L_{FS}\$ = free space loss or path loss <S> (dB) <S> \$L_M\$ = miscellaneous losses (fading margin, body loss, polarization mismatch, other losses...) <S> (dB) <S> \$G_{RX}\$ = receiver antenna gain <S> (dBi) <S> \$L_{RX}\$ = <S> receiver losses (coax, connectors...) <S> (dB) <S> You can see that the antenna gain is part of the equation. <S> Using the stock "omni" antenna sold with the WiFi access point will not achieve the same distance that using a very directional antenna such as a Yagi Uda or a parabolic one. <S> Without specifying this, it's impossible to give you a number for the range. <S> Then it comes to the transmitter power and receiver sensitivity. <S> Each access point and WiFi card is different. <S> But, there is also another factor not related to the strength of the signal: the time delay. <S> The WiFi protocol (which one? <S> this has to be defined) defines the maximum delay between a sent packet and the related acknowledge. <S> If the distance is too big, a timeout will occur and this will not work. <S> But without knowing the specific WiFi norm you plan to use, it's impossible to give you a number. <A> It varies by Wifi version, and speed/data rates: <S> The concept of range for wireless LANs is pretty basic. <S> Range is usually thought of as a measure of distance -- the distance between two wireless stations at which they can effectively communicate. <S> For typical Wi-Fi systems, range is the distance between a laptop client and the access point (AP) <S> it is using. <S> Any discussion of wireless range also has to include data rate (not to be confused with the actual, physical throughput). <S> In wireless systems, you can trade off data rate for range – at lower data rates, it is possible to communicate at greater distances. <S> The 802.11 standard supports multiple data rates at the physical layer. <S> 802.11 has an automatic rate fallback mechanism that reduces the data rate when communication degrades in order to maintain a better link between a wireless client and AP. <S> And by antenna type : <S> And as you can see in the picture above, a standard wifi antenna like you normally get, has a donut/torus shaped radiation pattern. <S> That means if the antenna is vertical/straight up and down/perpendicular to the ground/90° to the floor, it has basically 0 radiation at a 90° angle while weaker at other angles. <S> If you make the antenna parallel to the ground/lay it flat, it will have better vertical (compared to us) performance. <S> And the farther away the two nodes are (The Wifi router/access point, and the computer/laptop), the slower the wifi speed is. <S> 802.11n improves on 802.11b/g by using a second antenna, spaced a half wavelength away. <S> 802.11n is shown to be 250m effective range, outdoors, with no obstructions but the data rate and type of antenna used is unknown . <S> And that's before interference. <S> The more wifi traffic on the same channel, the slower and weaker the effective signal will be. <S> In the end it depends on A, the antenna used, B, the radio power, C, obstructions, D, the direction of the antenna, E, interference from other sources on the same radio bandwidth/channel, F, Speed required, <S> so there is no single answer . <A> I don't know what you mean by "I am not mentioning range." <S> AFAIK that is the only way to answer this question. <S> The range of Wi-Fi depends on the protocol. <S> The range of 802.11n is about twice that of 802.11b and g . <S> The typical outdoor range for 802.11n is approximately 820 feet . <S> Although this is assumed to be horizontal, it should apply to a vertical distance also. <S> Although not explicitly stated, I am going to assume this figure of 820 feet assumes the the use of an omnidirectional dipole antenna, since it is the most common type (i.e. the "rubber duck" antenna that comes with routers). <S> However by using a high-gain directional antenna, ranges of several kilometers can be achieved. <S> The examples given are all point to point horizontally, but it again should apply vertically also.
| According to Wikipedia page about long range WiFi using modified firmware in order to increase the timeout value and dedicated antennas, ranges up to 304 km is possible without using additional amplifiers.
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Are any USB-to-SPI *slave* bridge chips readily available? I would like to interface a battery-powered device to USB, without requiring users to install drivers. Data transfer requirements are moderate, so the ~60kbytes/sec of HID mode would be fine, but I don't want to be much slower than that. The main processor is probably going to be powered by batteries even when a USB cable is plugged in, so I would like to have that processor sleep any time the PC is not actively communicating. I've been considering using a USB-to-serial bridge, but the ones I've seen would with most microcontrollers require the data rate to be slow enough to accommodate the worst-case serial-port-interrupt response time. Further, while all of the ones I've seen allow a controller to indicate whether it's ready to receive data, none have any means of asking an "unready" controller to get ready. All the USB-to-SPI bridge chips I've seen are limited to acting as SPI masters, which would make things even worse for any microcontroller connected to them. A USB interface which acted as an SPI slave with an interrupt pin would seem ideal, since it could communicate very quickly when the main processor wanted to communicate, and could tell the main processor when it needed attention, but wouldn't lose data even if the PC were to send it just as the main processor was wanting to go to sleep (the interrupt would trigger a wakeup within a 100 microseconds or so, whereupon the controller could ask what happened and then receive data). Unlike USB-to-serial or USB-to-SPI-master bridges, which need non-volatile configuration storage, a USB-to-SPI-slave bridge could be configured by the SPI master to which it is connected, possibly reducing silicon requirements. It should, if nothing else, be possible to use a cheap USB-compatible microcontroller and program it to act as a bridge; if someone has done such a thing and released it as open-source, I'd just as soon use that as re-implement it. It might be possible to use a UART instead of SPI if the device supported handshaking and had a pin which would indicate when it had data to send (even when held off); such an approach might be necessary when using a microcontroller as the bridge, since most microcontrollers make lousy SPI slaves. On the other hand, if such a thing is available as an inexpensive off-the-shelf part, that would seem easier than having to buy microcontroller chips and then program them. Does anyone have any recommendations? Addendum I should perhaps have clarified that I'm trying to find something really cheap, and am evaluating the possibilities of either using a main controller with integrated USB versus using some sort of bridge. Using a bridge may allow the use of a cheaper main processor than would otherwise be required, and may simplify power supply design (since the bridge could be USB powered without having to ensure that the processor would receive power from the +5V USB rather than the +6V battery when both supplies were available); even if bridge+CPU ends up being a dime or two more than a USB-capable main CPU, separation of the USB from the other CPU functions could still end up being a "win". Of the designs I've built so far, my favorite used an FT245 in combination with a CPLD located in a keyboard/display/USB board which communicated to the main board via 3-wire SPI-style interface (two wires from the main board; one wire going back). When the clock and data were both in the idle state, the CPLD would use the data return to indicate when the main CPU needed to service it (because a button was pushed, data arrived on the USB, the USB was plugged/unplugged, etc.) The main CPU could then, at its relative leisure, wake up, ask the CPLD what exactly it wanted, and do whatever needed to be done. I really liked that style of communication, but the FT245 and CPLD together cost too much. Further, being able to use HID rather than CDC could improve the end customer experience. Conceptually, it would seem that a chip that was designed to serve as a bridge between USB and a CPU that was designed to talk to it could be electrically simpler and cheaper than the USB-to-other-things interfaces I've seen since it wouldn't need any sort of non-volatile configuration storage and wouldn't require very many configuration options--mainly enables for the various endpoints, possibly some timeouts, and a memory for data to feed the host on the control endpoint. Most other logic would be handled by the connected processor. Unfortunately, I don't know of any such devices that are actually for sale. <Q> FTDI USB to Serial converter followed my a MCU in SPI slave mode. <S> Firmware would simply transfer data between the SPI in port to the Tx line to be sent via the FTDI to the USB port. <S> Similarly firmware would cache Rx data from the USB/serial port and place in the SPI Tx buffer waiting to be read back. <S> Two IC simply solution to achieve USB to SPI slave. <A> You could customize it to whatever protocol you want your main CPU to see, it comes with a few hundred bytes of RAM to buffer while the wake-up interrupt is processed, and it's a SOP-8 <S> that costs literally a few cents. <A> This FTDI chip works as an I2C slave, and seems to be relatively cheap: http://www.ftdichip.com/Support/Documents/DataSheets/ICs/DS_FT200XD.pdf <S> And this one works as an SPI slave: http://www.ftdichip.com/Support/Documents/DataSheets/ICs/DS_FT220X.pdf
| I think, from a cost and board space point of view, a USB-capable uC (e.g. the ATTiny85) would be the best approach.
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Can several CS pins be multiplexed with a shift register? When sharing a SPI data line between several ICs and an Arduino, what exactly are the frequency constrains of the CS pin? Can I use a shift register to multiplex the CS pins (thus only requiring one I/O pin, the Shift Register CS, out of the Arduino)? Or does CS must be synced carefully with SCK (in order to avoid garbage if CS is pulled high too "late") <Q> Data sheets are usually pretty clear on the timing issues related to chip select lines <S> so I urge you to use the relevant data sheet for the device. <S> Some devices can be activated at the same point at which the data is sent to them; other devices need clear space between CS going low and the SPI data being applied. <S> Yes, you can use a shift register to select just one particular device but this shift register may need flushing regularly to ensure that over time it hasn't "accumulated" an extra low that could cause two chips o be selected simultaneously. <S> In the main I'd use either a 3 to 8 line decoder for chip selects or individual GPIO lines for each peripheral chip. <A> It isn't quite that simple. <S> Think about the sequence of what happens: <S> Assert the CS to the shift <S> register Clock <S> the device CS into the shift register Negate the CS to the shift <S> register Clock the data to the device <S> Assert the CS to the shift <S> register Clock <S> the new device CS data into the shift register ... <S> That last step is a potential problem. <S> Since the CS for the first device is still asserted while this is going on, it's going to try to interpret the "new device CS data" as data intended for it. <S> the selected line(s). <A> If you want to use a single GPIO to control all the CS lines you can use a straight ring counter . <S> It has a single clock input and <S> n outputs (depending on the size). <S> Of the n outputs, only one is high at any one time, depending on whether the CS are active low or high use <S> NOT gates accordingly. <S> All you have to do is keep count of how many times you have clocked the ring counter. <S> If you want to be able to program the chips in any order that's fine as well, because on most IC's if there are not the correct number of clocks before the CS line is deactivated then the input buffers are cleared. <S> So all you do: Clock ring counter x times to activate required CS Clock in data Clock Ring counter once to deactivate current CS Rinse. <S> Repeat. <S> I suggest getting a ring counter with at least one more state that you need so that you have a state where no CS is active, after each data transfer, just clock the ring counter to that position. <S> As regards your original question, just make sure you put a small delay between clocking the ring counter and starting the data transfer.
| So, while you can use a shift register to select among an arbitrary number of device CS lines, you'll still need a separate "device CS" signal from the Arduino to actually assert (and negate)
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Should mounting holes be plated? When putting mounting holes for screws on a PCB, should the holes be plated or unplated? I've read plating the holes provides better support for the screw and helps prevent the PCB from being damaged, but looking at some PCBs I have around (including the extremely popular Arduino UNO), most of them don't seem to have plated mounting holes. <Q> If the screw is used to ground the board (for example, PC motherboards) <S> then plated is the way to go- <S> this one has the top, bottom and internal ground plane tied together with many small vias radially around the hole. <S> There is no thermal relief on the vias. <S> The vias ensure that even if the hole plating is damaged by the screw thread, the top and bottom pads are solidly connected to the ground plane. <S> Otherwise it doesn't matter much, though if you don't have any other unplated holes in the board, as Scott Seidman says, it can add cost to the board. <S> Whether there is a pad or not (and the finish on the pad after assembly) may affect what kind of lockwasher or screw you choose to use on the PCB, since solder has a tendency to cold flow. <S> If the board is multilayer, there should be a large clearance between non-connected internal planes and (particularly) an unplated hole, because the thread can damage the internal surface (and sometimes people do things like drilling out holes that don't quite match the mating surface), and you don't want the screw shorting to (say) an internal power plane. <A> I don't see why a plated hole should provide better support or protection. <S> A mounting hole should be a bit bigger than the screw that it will fit, the coating material of the hole's internals does not matter at all. <S> As Scott says non plated holes could add an extra step: think of it, you etch your pcb, drill all the holes, plate them <S> and you are good to go. <S> If you want to have non plated holes you need to make another drill run after the plating, and that can be quite time and money expensive. <S> So here's my guess: some PCB manufacturers use different drilling machines for different hole sizes: the tiny, through-hole like holes are drilled with one machine while the bigger, some five to ten mm holes are drilled on another. <S> If you drill the bigger holes after the plating you don't add a step at all <S> and you save coating material, which is good. <S> Now that's why your arduino has non plated mounting holes. <S> All that said, that applies only if with plating you mean plating. <S> If we are speaking of the pads around a mounting hole now that's a completely different story. <A> A few of things to consider: The plating thickness is not something that the board house can control to super tight tolerances, but <S> 0.001" is a good rule of thumb. <S> You're not going to get any extra support there. <S> You should be calling out finished hole sizes in your design. <S> Since the plating thickness varies, you can end up with tighter mounting holes than you spec'd. Plan accordingly. <S> If you're relying on the plating to make a connection to an internal ground plane, and an assembler over torques a screw, you run the risk of damaging the plating and breaking that plane connection. <S> This is only an issue if tolerances are not carefully considered. <S> As an aside, some board houses do not like to plate holes that do not have traces going to them. <S> We had some boards with Tag Connect headers, and the guide holes were plated. <S> Until the footprint was fixed, we got a phone call on that one every time. <A> For the small chance that the PCB builder may mistakingly not-plate all the holes (disaster city) <S> I'd make all the holes plated. <S> Plus what @scottseidman said as a comment - it's an extra process. <S> I'd also add this - "should all mounting holes be connected to local PCB 0V? <S> " - some will benefit and some won't of course. <S> I attach them and put little 0603 pads to link them to the ground plane then this covers all eventualities. <S> This mainly applies to circuits inside metal boxes but also, when bench testing it gives a nice copper area to join 0V wires from inputs etc. <A> I've read many advantages/reasons on different forums about these mounting holes with vias <S> but I did not see the reason that I am using them for. <S> Using non-plated mounting hole with vias eliminates the placing/removing kapton tape step assuming that these non-plated mounting holes have their bottom pad covered with soldermask. <S> This was suggested to me many years ago from a board assembler contractor. <S> Since then, that's how I make my mounting holes.
| With plated mounting holes, if the board is going through a wavesolder machine during the assembly process, all these holes need to be masked with a piece of kapton tape to prevent the solder to get into these holes.
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What is a ripple clock? I am reading Chapter 12. Recommended Design Practices from the Quartus II Handbook Version 13.1Volume 1: Design and Synthesis which states (p. 8): Ripple counters use cascaded registers, in which the output pin of one register feeds the clock pin of the register in the next stage. This cascading can cause problems because the counter creates a ripple clock at each stage. These ripple clocks must be handled properly during timing analysis, which can be difficult and may require you to make complicated timing assignments in your synthesis and placement and routing tools. What is a ripple clock? Why is timing analysis difficult on a ripple clock? <Q> This is a ripple counter: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> It is an asynchronous counter that will divide the input clock by 2 each stage. <S> It is an asynchronous counter because each stage will change at different times and each flip-flop has a different clock input. <S> The time difference between each stage is determined by the CLOCK->Q delay of the flip-flop used. <S> The simulated result is shown below, showing that each stage delays the output transition by the clock-to-output delay. <S> Now to put the significance of this in perspective of an FPGA, the timing analysis tool wants to make sure everything is getting clocked at the right time. <S> Part of that is make every signal that enters the CLK pin of a flip-flop a system clock that should be synchronized with all the other clocks. <S> As such, if I entered the above schematic into an FPGA synthesis tool, it would consider the nets CLK_IN , DIV_2 , DIV_4 , and DIV_8 to be "clock" nets regardless of whether they are used to drive any other clocks. <S> This will probably work fine as a counter (there is a possibility for a hold-time violation on each flip-flop), but it's not made in the synchronous logic method. <S> If you're using this to take a fast input clock and derive a single, slower clock (e.g. make DIV_8 a master clock for the system) then you're probably fine. <S> The trouble comes in when you want to have fast circuits clocked with CLK_IN interacting with slow circuits clocked with DIV_8 . <S> In this case, you want the rising edges of the clock to be synchronized, but you will have a large clock skew between these clock nets. <S> The amount of clock skew generated by one stage could be enough to cause synchronization errors, and more stages will almost guarantee it. <S> If you want to create two synchronized clocks inside an FPGA, your best bet is to use a synchronous clock generator, or a clock module that is internal to the FPGA, such as a PLL/DCM block. <A> Couple of issues with ripple clocks: <S> The last clock will have a delay than the input clock, because it goes through a number of flops. <S> So what is the problem with this delay? <S> You'll have problem when your design have cross-domain paths between these two clocks. <S> If any path have a launching clock from the input clock domain and capturing clock coming from that derived clock domain, that path will have a large skew. <S> So you'll have a hard time to meet the timing. <S> Another problem is with writing SDC constraints. <S> You have to write the clock definitions on each stage even if they aren't used. <S> See an example here on page 18. <A> This issue affects not only simple binary counters, but also more complicated ones like decade counters (such as the 74HCT4017 ) where each counter internally counts from 0-9, and are wired to reset back to 0 on the tenth pulse. <S> Suppose you have a number of decade counters, one for <S> the units position, tens position, hundreds position etc. <S> Each of the decade counters has a clock input. <S> The clock of the units counter is fed from the main clock source, which can be turned on and off presumably. <S> The clock of the tens counter is connected to carry output of the units counter. <S> When the units counter counts from 9 to 10, two things happen: the counter is reset back to 0 <S> (so there really never is a valid 10 output), and a clock pulse is propagated to the clock input of the next counter, in this case the tens place. <S> The reason it is called a ripple clock, is that the clock going into the tens counter will be delayed at least one propagation delay from the original clock going into the units place. <S> This will then have what is called a "ripple effect", e.g. <S> if you have a 6-place counter, the clock going into the sixth place on a 099999 to 100000 transition will be delayed five times. <S> This can create timing problems, for example if one were to try to compare the output of the counters to a particular value, the counters do not change all at once so the comparison circuitry may fire at the wrong time -- there is no signal that says: all outputs are stable. <A> From what I have seen, Ripple Conters, as they are sometimes called, are digital timers (counters) that are used when precision is not required and simplicity is the goal. <S> They can additionally be used as clock-dividers to prescale an input clock signal by an order of 2 per stage. <S> Basically, you have a series of connected FlipFlops where the output of the previous stage becomes the clock of the next stage. <S> See: Ripple Counter
| In Quartus II, Ripple clock is any clock driven by the output of another register.
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How to determine annular ring width for thru-hole pads? When placing thru-hole components on a PCB, how do you determine what the annular ring width should be? For example, I'm making a PCB with a thru-hole switch whose contacts are 1.2mm in diameter. To me, an annular ring width of .4mm (so an outer diameter of 2mm and drill diameter of 1.2mm) looks "about right" for this hole size, but is there a general rule of thumb or formula that should be used to determine the width? <Q> Of course the annular ring should be at least as large as the minimum annular ring specified by the PCB house, but generally you'll want to make it much larger, especially on heavy parts or parts (like your switch) <S> that are exposed to force. <S> At least 10 mils (0.25mm) <S> annular ring is better for manufacturability regardless of any strength considerations, and a bit bigger than that is even better. <S> If the board has plated-through holes , exceeding a 1mm annular ring is probably not of much benefit for pins up to 1.5mm or so. <S> If the board is single-sided , the adhesive under the pad has to provide a lot of strength, so you probably want to go as high as 2mm annular ring or more (bigger is better for strength) on a part requiring strength such as a switch, terminal block, coil, etc. <S> If there is not enough room to get appropriate clearances you can use a non-round (oblong or rectangular) pad and still get some of the benefit. <S> I don't know of a specific formula or rule of thumb. <A> According to the standard IPC-7251 "Generic Requirements for Through-Hole Design and Land Pattern Standard" (was a working draft as of Feb'17), for radial leaded parts with perpendicular mounting, recommended annular ring width is from 250 um for Maximum Material Condition (MMC, means the most robust solder joint) to 150 um for Least Material Condition (LMC, means the least robust solder joint). <S> In terms of diameters to be added to the hole diameter, these numbers should be multiplied by two (500 um and 300 um respectively). <A> I normally go for 12 thou (0.3mm) but have been known to sneak em down to 10 thou (0.25mm). <S> Anything larger is fine. <S> This is for low to medium current (up to 1A) - anything greater requires a progressive increase in the hole diameter and a bit more meat on the annular ring. <S> It also depends on the accuracy of your PCB supplier.
| For parts under great mechanical stress or with mechanical electrical contact (not soldered) (e.g. power supply banana jacks mounted with screws), annular rings should be widened to provide necessary mechanical strength of the joint or required electrical contact respectively.
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Counter is missing an output signal I used a 4040 counter to count the pulses from an encoder. I then connected two counter output signals to an AND gate, and after that one way is go to reset 4040 counter. The issue is that when I connect the gates in this way, there is no logic-high output signal from the 2 pins of the counter. If I disconnect them like the image below, I can read the output signal again from a multimeter, and other pins always have a signal. I measured the AND gate and NAND gate; there are no short circuits. Why there is no output signal when connect these gate? <Q> I think you do have an output signal <S> but it only lasts for maybe 10 nano seconds before the reset in the chip has activated and cleared everything down again. <S> Try using a scope with a trigger function and a fast time base. <A> Your circuit, from a logic point of view, ought to work. <S> Barely. <S> The most obvious suggestion is this: are all the unused inputs to the 74HCT08 tied to ground? <S> And do you have a 0.1 uF ceramic capacitor connecting the 74HCT08 power to ground, connected right at the pins (or at least within less than an inch)? <S> If the answer to either is no, I suspect that your 08 output has spikes on it that are keeping the 4040 reset. <S> A similar capacitor on your 4040 power is also are REALLY good idea. <S> So is a solid ground set of ground connections. <S> In principle, you do not need your resistors. <S> If you do want a little insurance on the pulse width of your reset pulse, try this. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> And, as with the decoupling capacitor, keep the components as close to the IC as you can. <A> I have been having problems of a similar nature using a CD4040 being fed ok with a 2Hz clock from a CD4060 set up with a 32,768 Hz xtal. <S> The CD4060 provides a Q14 output. <S> I was trying to get the CD4040 to divide by 120 in order to get a one minute pulse. <S> So I diode ANDED Q4 thru Q7 for all ones with a 22k pullup and fed the ANDED signal to the reset. <S> Q7 NEVER <S> WENT <S> HIGH.I took out this circuit and strapped the reset to ground. <S> Note that on ALL the diagrams I have, Q7 appears on pin 4. <S> So I expected to see, and measured the signal periods on all Qs up to Q7. <S> Q7 was now toggling OK. <S> All was well up to Q6 which showed 32 second period. <S> BUT Q7 on pin 4 measured 128 seconds! <S> I then guessed that probably the circuit diagram data was wrong. <S> And reading your report above today stating NXP data shows Q7 at pin 13, confirms that SOME DATA SHEETS HAVE AT LEAST Q7 WITH THE INCORRECT <S> PIN NUMBER . <S> As stated I believe Q1 to Q6 are OK. <S> I will now search for this NXP data sheet to confirm. <S> I will retest my circuit using pin 13. <S> UPDATE <S> I have checked my old National Semiconductor datasheet, and it seems they have marked Q0 as Q1, Q1 as Q2, Q2 as Q3 etc. <S> So Q7 actually is on pin 13! <S> I have not checked the other pins. <S> This appears to offer some clue as to why YOUR reset circuit didn't work.
| Get rid of them.
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How should a SPDT switch be wired for a digital input? I want to connect a switch to a digital input on a logic board and make it so that the input goes high when the switch is pressed. I'll be using a SPDT switch: Should I connect the digital input to COM, GND to NC, and VCC to NO? Or could I connect VCC to COM and the digital input to NO? (this would leave the digital input "floating" when the switch isn't pressed, but is that necessarily a bad thing?) <Q> You want to use the first option: connect the digital input on the micrcontroller to COM, GND to NC (1T), and VCC to NO (2T), except use an resistor, e.g. 4.7K, between VCC and the NO (2T) terminal. <A> I would connect it thus: simulate this circuit – <S> Schematic created using CircuitLab <S> The resistor will pull the input high when the switch is pressed. <A> Connect a series resistor 4k7 from the digital input pin to the COM 1P. Connect a series resistor 1k from VCC to NO 2T. Connect NC to GND. <S> The first series resistor protects the digital input if it can be configured as an output; it may drive to VCC or GND. <S> Omit if it cannot be reconfigured. <S> Omit if the logic board already has a series resistor on the digital input. <S> The second series resistor protects the supply from a fault that may develop in the switch, a fault which shorts NO to NC. <S> The capacitance of the digital input pin will cause the existing voltage level to remain while the switch is being moved. <S> Obviating debouncing. <S> However, if the logic board digital input has a pullup or pulldown resistor, you must either add capacitance to defeat it, or code for debouncing. <S> All this uncertainty might be cleared up by describing the digital input more thoroughly. <S> ;-)
| Usually not a good idea to connect VCC directly to an input pin on the micrcontroller -- for example, if it became a digital output for some reason and was at GND level, you would have a nasty short.
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Analog signal over long cable I would like to use 50mV FSD signal to sense DC current at a distance of ~70 feet with a twisted pair cable and then perform signal condition there. Does anyone have any experience with this? " Originally asked"Is it advicible to take lower side shunt 50mV FSD signal at a distance of about 70feet with STP cable , and then make signal conditioning there at remote location. I need to enquire does experienced members do this way?. BTW I am trying to sense DC current. <Q> STP I assume means screened/shielded twisted pair and if you want to measure current flowing down one of the conductors then do so at any point. " <S> Condition" the signal where the measurement is made or do the conditioning half locally and half remotely. <S> The minimum amount of conditioning to be done locally is to amplify the shunt signal so that the wires from the amplified shunt conditioning circuit do not pick-up excessive extraneous EMI on route to the final circuit that monitors the shunt current. <A> If the signal can be low-pass filtered (you say it is DC current), and your source impedance is low (presumably a shunt) and your input impedance high and suitably filtered, there's really nothing especially tricky about running a 50mV signal for 20 or 30 meters. <S> We do it all the time with thermocouple signals which are in the same voltage range (sometimes less). <A> Yes, it can be done, but the signal integrity would be bad. <S> You will need to do at least one of two things to the signal to ensure good communication: <S> Amplify the signal. <S> Make the signal a higher voltage. <S> A simple op-amp may do this for you nicely. <S> Optionally turn it into a differential signal. <S> Send the signal twice - once normally and once inverted. <S> You can then subtract one signal from the other to get a nice clean signal out the other end. <S> Again, op-amps can do this for you.
| Given the low level of the signal and the length of the cable I would say this was a bad idea.
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How/Where to boost voltage in a 555 timer to drive piezo transducer I am trying to create a simple and cheap frequency circuit with a 555 timer to drive a piezoelectric transducer. The transducer I am using calls for a 48v input with 1.66Mhz frequency timing. I've used a CMOS 555 timer to create the frequency (using 2 trim pots to do some testing). I don't have an oscilloscope but my multimeter shows I'm right on track. Now, I am trying to use a 5v USB cable and I obviously need to boost the voltage to 48v. I'm not sure how or where I should be putting this boost converter, or even if I'm doing this right? Here's the circuit I'm using for the 555: simulate this circuit – Schematic created using CircuitLab Do I just turn the 5v input into 48v input? or do I have to boost it after the output from the 555? When I measure the voltage with the multimeter (in place of the crystal) I am getting 1.4v. Obviously not enough to drive the piezo element. <Q> Put a capacitor (about 0.1 µF) between the 555 and the transformer to eliminate the DC bias. <S> However, you should note that 1.66 MHz is rather high for a 555 in the first place, and it might not be able to drive much power at that frequency. <S> Also, you'll probably get better results if you can get the duty cycle of the waveform closer to 50%. <A> You obviously need a boost converter for this. <S> You could then apply the output of the 555 to the base of a transistor (or darlington pair). <S> The transistor would then actually drive the piezo transducer. <S> Then refer to the data sheet of the piezo to find out which capacitor and resistors would be adequate. <S> I have done something similar in the past but with 9v buzzer and the output of an MCU (instead of 555) and it worked fine. <S> Another solution, <S> but I do not know how stable it would be, would be to use one of those boosters with a shutdown pin, and let the 555 control the shutdown pin, but this would probably mess up the frequency if it even work. <A> I have seen cheap door alarms use a very small transformer to convert two button cells worth of voltage into a nasty voltage which can drive a low power piezo transducer.
| The simplest solution would be to use a 1:10 transformer to transform the 5V swing of the 555 output to about 50V at the transducer.
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What kind of metal are the legs of resistors made from? Can anyone tell me what kind of metal is used in electronics to make wires going in/out to the elements? Can longer of this metal wire be found anywhere? Actually I am looking for sth like copper wire but in silver colour. What's its name? <Q> You can buy rolls of tinned copper wire, as Andy says in his comment, and many resistors (especially precision types) still use tinned electrolytic copper leadwires. <S> You can tell the difference with a magnet. <S> If it's copper wire, the leadwire itself will be non-magnetic (the end caps may be magnetic). <S> Brass was once used for the end caps, but tin prices spiked.. <S> so they're generally steel as well. <S> Steel wire is worse in every way except cost- higher thermolectric effects, lower thermal conductivity etc. <S> so its use would be confined to low cost products rather than precision resistors. <S> Buying tin-plated copper-clad steel wire in relatively small quantity may prove more difficult, most distributors still carry the tinned copper wire (at insanely high prices compared to what a resistor factor would be willing to pay), and the makers of the commodity wire may want a minimum order of around a metric ton. <A> There is no one specific metal used. <S> It varies from manufacturer to manufacturer and by age. <S> The general principle is the same though. <S> Most are either tinned copper, tinned aluminium, or one of many alloys, using copper, zinc, nickle, tin, etc. <S> Some chips use FeNi42, a form of "invar" according to Wikipedia. <A> These days, most lead wires are made from tinned copper-plated steel, and the single strand tinned copper wire available on spools is generally called buss wire.
| However, in modern times copper prices have spiked, and most leaded parts are cheap commodity items, so tin plated copper-clad steel wire has been used in many through-hole parts.
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How to deal with grounding and the capacitors in this opamp schematic? Here is a schematic I'm planning to make (right click on the image and 'open image in new tab' to see it better'): I have a transformer and AC to DC converter that makes +/-12V. (1) Does 'power supply 1' correspond to the +12V wire and 'power supply 2' to the -12V wire? (2) When I see the symbol 'ground' on the schematic does that mean that I need to connect those to the -12V line? I have no 0 volt line in the circuit so what is the ground line? (3) Why are C3, C5 non-polarized capacitors and why are the other ones polarized? The current flows from negative to positive. Does that mean that the polarized capacitors need to be placed with their positive legs pointing to the left? I already asked the grounding question but I got only answers that were theory. I would like specifically to know how do I deal with grounding in this circuit and why there is no grounding symbol in an arbitrary "powering an LED with a 9V battery" circuit. Thank you very much. This will clarify a lot of things for me. <Q> (1) Does 'input 2' correspond to the +12V wire and 'input 1' to the -12V wire? <S> NO. <S> This is connected to the power supply connector . <S> (Pin 1 is the positive, pin 2 the ground or 0V). <S> The INPUT connector is where the audio signal goes (2) is the live and (1) is the ground. <S> (2) When I see the symbol 'ground' on the schematic <S> does that mean that I need to connect those to the -12V line? <S> I have no 0 volt line in the circuit <S> so what is the ground line? <S> All voltages are relative to each other. <S> If you only have TWO wires from the AC/DC converter then the most positive is taken as the positive and the other one is the OV or ground. <S> A simple check with a voltmeter will determine which is which. <S> (3) Why are C3, C5 non-polarized capacitors and why are the other ones polarized? <S> The current flows from negative to positive. <S> Does that mean that the polarized capacitors need to be placed with their positive legs pointing to the left? <S> C3 and C5 are small value capacitors (0.1uF). <S> These can be easily made as physically small, non-electrolytic types. <S> They have the advantage of being able to decouple (short out) the higher frequencies. <S> The larger values (uFs) are made as electrolytics as these can be made with high values in small physical packages. <S> They are generally much poorer at handling the high frequencies. <S> By combining an electrolytic with a non-electrolytic capacitor in parallel (eg C3, C6) you get a much better response over a wider range of frequencies. <S> In this case they are used for 'smoothing' the supply voltage, preventing hum and hiss. <S> The positive plate of an electrolytic is shown as an open rectangle but left and right (or up, down) have no meaning in terms of connection as this will be determined by board layout. <S> Conventionally current is taken to flow positive to negative. <A> The TDA2003 datasheet indicates that this part can operate from a single supply of 8 - 18 volts. <S> It does not use a negative supply. <S> From your desctiption, your power supply produces +/-12 <S> volts, or 24 volts between your "first" and "third" wires, which is more than the part is designed for, so you should use just the first and second wire - you will need a voltmeter, or the power supply documentation, to determine which wire is positive. <S> All the ground symbols on the schematic must be connected together, and to the middle terminal of the power supply. <S> All the ground symbols are your "zero volt" line. <S> All electrolytic capacitors must be installed with their positive terminal connected to the more positive part of the circuit. <S> C1 and C4 should be installed with their positive terminal towards the TDA2003 <S> (I think C1 is drawn backwards). <A> (1) Does input 2 correspond to the +12V wire and input 1 to the -12V wire? <S> No. <S> On the POWER SUPPLY connector, pin 1 is the +12V input from the supply and pin 2 <S> is the -12V input from the supply. <S> (2) <S> When I see the symbol 'ground' on the schematic does that mean that I need to connect those to the -12V line aka. <S> input 1? <S> I don't know what you mean by "aka. input 1"; the power supply -12V is supposed to be connected to pin 2 of the POWER SUPPLY connector. <S> I have no 0 volt line in the circuit <S> so what is the ground line? <S> Everything connected to pin 2 of the POWER SUPPLY connector. <S> (3) Why are C3, C5 non-polarized capacitors and why is C5 not? <S> C5 is ground-referred, and the signal exiting C4 is real plus-and-minus-about-zero-volts AC, so the signal at the top of C5 will swing both positive and negative with respect to ground. <S> The current flows from negative to positive. <S> It does in the world of physics, but in this one it's called "Conventional current flow" and it's considered to flow from positive to negative, thanks to Benjamin Franklin. <S> Does that mean that the polarized capacitors need to be placed with there positive legs <S> >pointing to the left? <S> In the drawing, the lighter-shaded electrolytic capacitor plates should be marked with a "+" sign. <S> I already asked the grounding question <S> but I got only answers that were theory. <S> I would like specifically to know how do I deal with grounding in this circuit and why there is not grounding symbol in an arbitrary "powering an LED with a 9V battery" circuit. <S> In a stand-alone LED circuit powered with a 9V battery, the ground reference/symbol isn't important because nothing else will be connected to the circuit.
| In this circuit, you deal with grounding by connecting everything with a ground symbol to the negative side of the supply, and that's usually considered to be the "zero-volt" reference for whatever in the outside world is plugged into the circuit.
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How to prevent PA system from interference of 4G signal I apologize if this is not the right place to post this. I am working on a small PA system for a weekly gathering. It catches the signal from wireless mics and output it to the speakers. I am pretty sure that the machine itself is not damaged. Recently, I've found that there are very sharp digital noise coming out from the speakers and it is so annoying that it masks the voices and songs. I've done some investigation and found out that it could be caused by 4G cell phone signal. So I tried some experiments and the observations seem proved the theory. I've learned from my online research that I can either ask everyone to turn off the 4G signal (which is unlikely since the population is around 100), or replace my PA system. Is there another way out? Such as a 4G signal filter or jammer? If so, my concern is, will these filters and jammers also get rid of the signal from the wireless mics? Thanks. Update 1: Thank you all for your advice. Unfortunately, I cannot get access to the equipment until this weekend. I'll mark the models of the gears and post them here then. I'll also try to upload a sample noise for your analysis. <Q> I've worked on small PA systems for weekly gatherings for about 20 years. <S> I've never had any success with filtering, every success was from changing the wiring. <S> Describe the wiring totally; the type of connectors, the number of wires in cables (balanced, or unbalanced), where the cables come from and go to, and the type of amplifier. <S> Photographs of the amplifier would help. <S> You may be able to shorten this effort by testing one channel of the system at a time, with other channels disconnected. <S> If a particular channel is sensitive to the problem, change to balanced cabling and test again. <S> Check the earth of the system exists in only one place. <S> If the wireless receiver unit is separate from the PA, disconnect it and use another device to listen for the noise. <S> If you truly have noise arriving with the wireless signal, you'll probably have to replace the wireless receiver. <A> I would begin by working out for sure whether the interference is being picked up: the wireless microphone receiver (does it go away if you turn the receiver off but leave it plugged it?) <S> the leads between the receiver / other microphones / instruments and the mixer <S> (i.e. does it go away when you unplug all the mixer inputs?) <S> somewhere else in the system (i.e. occurs simply by having the mixer plugged in to the speakers) <S> If it's the wireless receiver picking up 4G phone signals, the receiver may not be 'faulty' so much as 'not designed to cope with that much interference'. <S> Pickup on the mic leads may be solved by a filter such as the circuit given. <S> I found googling "microphone radio interference filter" turned up a number of ready-made commercial products and DIY projects. <S> If the interference is somewhere else in the system, it's possible something (the mixer or the amps) is faulty, or there is bad earthing, or faulty wiring (e.g. screen disconnected). <S> I take it <S> you have balanced connections on all interconnections between receivers, mixers, amps, etc? <A> I am an Audio Technician and spend a lot of time working with PAs. <S> The digital noises that are caused by cell phones typically only appear on unbalanced or unshielded items. <S> For example my computer speakers are very susceptible to the noise. <S> I have used a lot of Shure product from the SLX (low end line) to the UR (high end line) and have not had issues with the receivers being the pickup point for the cell phone noise. <S> (I'm not saying it's impossible). <S> The only time I have had a phone cause problems at the RF end is when the phone was directly between the mic and the receiver and very close to the receiver. <S> And even this only caused dropouts not the noise. <S> This is most likely due to Shure's use of a pilot tone. <S> I have had unbalanced cables pick up the cell phone noise. <S> I have also had balanced mic cables with a damaged shield pick up AM Radio, so make sure your cables are all balanced and good quality. <S> The other pickup point for cell phone noise was the podium mic. <S> There are a lot of them that are not adequately shielded, so if there's one in your system, that may be the problem. <S> I concur with James that 95% of the time you can solve your problem by ensuring your cabling is correct. <S> Filters are great, but you don't normally need to go to that extent. <S> I will look for your post of how the system is setup and hopefully that will point us in the direction of your problem. <S> I understand your frustration, and if I can help, I will. <S> Nothing is more frustrating than having your sound destroyed by an uncontrollable outside force! :-)
| The problem you describe is common, but usually fixable, and usually indicates a design issue with the wiring of the PA system. You might be able to move the antennae closer to the transmitter; if that's not possible you either have to buy a 'better' wireless mic system (or one operating on a different frequency band), or persuade everyone to turn their phones off.
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Simple low voltage disconnect circuit for Arduino Preface: Under impression to need a low voltage cut out, to prevent 2x AA NIMH rechargeable battery from being drained too now, and damaging them. (right??) In deployment this circuit will commonly be exposed to complete discharge event if allowed. Concept: S1 momentary switch is held down, activating Q1 to power up 328p Sketch begins and Setup() calls digitalWrite(A5, LOW) S1 can now be released as Q1 is held LOW by A5 Battery gets low, triggering brown out detect (BOD), 328p resets, and disconnects Q1 Notes: R1 is resistor wheel and i experimented with different values to get it to work. (as i have not done the math) I have this assembled on breadboard, radio and lcd module are not pictured With A5 disconnected it works well (while S1 held down), When off 0ma at Ammeter. With A5 connected it can work with higher value of R1 (47k) to prevent it booting up itself, but when off Ammeter reads about 2ma. This 2ma is still enough to drain the battery significantly if left flat for some time.Im guessing the 2ma and need for 47k resistor are related to what appears as A5 'leaking' while arduino off. I have been able to reduce 2ma current consumption while off by putting pullup resistor between A5 and +V I have limited space so need to keep things simple. Feedback: Any improvements / suggestions, to get around the leaky A5 issue ? Or generally meet my design needs better ? <Q> Your design requires pin A5 to be at the same voltage as the battery in order to shut off the system. <S> The problem is that when you remove power from the Arduino, pin A5 can no longer remain high — there are internal protection diodes that force its voltage to be no higher than V+ plus the diode drop (about 0.7V). <S> The simplest solution is to add a second transistor (NPN) that inverts the logic of pin A5 — when A5 is high the system is on, and you drive A5 low to shut it off. <S> This shuts off both transistors and the quiescent current should be zero. <S> Also, you should consider using a P-channel MOSFET instead of your PNP BJT. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> I've used an N-channel FET instead of an NPN transistor. <S> The Arduino can drive it directly, and the only "wasted" current is the 30 µA or so that flows through R1 when the system is powered up. <A> I have read several forums regarding this topic but with my components I used different approach and created overall schematic of Arduino that measure the battery voltage and turn it off when it reaches the low voltage limit. <S> A0 measures the battery voltage via voltage divider <S> 1:1 <S> D5 is high and when the battery voltage reaches the lover limit, Arduino pin goes to LOW 4N35 OPTOCOUPLER is used to decouple Arduino pin with the battery in off state <S> S1 push button is used to start it <S> I used the simple 2N7000 N-channel MOSFET since I have current less than 100mA. <S> If more current is required I would use STP36NF06L or IRL2203. <A> If you need a low-voltage cutout perhaps a specific voltage detection IC might help. <S> I've seen the Microchip TC54 used in similar applications. <S> It's a pretty small device so it won't take up much space. <A> Try setting pin A5 to high-Z (tri-state) when not using it. <S> Page 77 here . <S> You're going to have to step outside the bounds of the Arduino libaries to do that though - <S> but it's very simple and only two lines of code. <A> The problem with the first circuit at top of page is that the AtMega328 is being powered via its A5 I/O pin. <S> Current passes from battery positive via BC557 emitter/base junction through R1 to A5. <S> From A5 it passes internally via top anode/cathode of internal upper protection diode to final destination of VCC. <S> Change BC557 to a P channel mosfet and problem is solved, because there is no current passed between source and gate as there is with emitter/bass junction.
| It will be more efficient: less voltage drop to the Arduino when on, and no wasted current through the gate terminal.
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Non linear device with two frequencies I got this on an interview question and I still don't know the answer. What output will have a nonlinear device if we will give two random frequencies \$f_1\$ and \$f_2\$? I answered that it will have two carriers but I really unsure about my answer. <Q> The correct answer is that you don't know without more information about this non-linear device. <S> Non-linear only rules out linear, but can otherwise mean anything else, including putting out a fixed signal regardless of the inputs. <S> To show that you understand the concept, you should explain that a linear device can produce at most only the frequencies put into it. <S> It can change the amplitude and phase angle of the input frequencies, but it can't create new ones. <S> It could, for example, attenuate either or both frequencies well past the noise floor, so even for a linear device <S> , the answer is none, one, or both of the input frequencies. <S> What they were probably after with this question is for you to say that the non-linear device can produce additional frequencies beyond those you put in. <S> If the non-linear device happens to include a product term, as Leon is apparently assuming, then you will get the sum and difference of the two frequencies included in the output. <A> In general, you will get: the two original frequencies f1,f2 <S> The sum and difference of the two frequencies f1 + f2, f1 - f2 (aka intermodulation products) <S> The harmonics of each of the two frequencies (f1*2, f1*3, f1*n, f2*2, f2*3, f2*n, ...) <S> Harmonics of the intermodulation products Intermodulation products between any of the above and any other component <S> TL/DR : a mess. <S> In practice while you can get all of these, the amplitudes of many of them may be negligibly low or actually zero, depending on the form of the nonlinearity. <S> One special case is a perfect multiplier or balanced mixer with one carrier on each input. <S> This produces the sum and difference on the output, and nothing else. <S> Another special case is a device with a square law or quadratic transfer function (vacuum tube triodes and some FET amplifiers can approximate this) - you will get the original frequencies, the sum and difference, and almost purely second harmonics only. <S> Makes for a good guitar amp or an overpriced "audiophile" hi-fi system... <S> And so on. <A> Do you understand what happens when you put a single frequency into a nonlinear device? <S> If you put two frequencies into the nonlinear device, you'll not only get harmonic distortion on each of those frequencies, but you will also get intermodulation distortion , which means you'll also see energy at frequencies that are the sums and differences between all of the possible combinations of the two original frequencies and their harmonics. <S> In RF circuits, the latter effect is sometimes used deliberately (in conjuction with bandpass filters) to translate signals from one frequency band to another. <A> if f1 and f2 is fed to a non linear transfer function, the output will be the sum of products of f1 and f2 and its harmonics (f1)(f2)(f1+f2)(f1-f2)(f2-f1)(2f1+2f2)... <S> (3f1+ <S> 3f2).....infinity <S> this how rf mixers work
| In general, putting any single frequency into a nonlinear device will produce harmonic distortion , which means that you'll see additional frequency components at the output that are integer multiplies of the input frequency.
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