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How to remove enamel from wire? I want to repair my computer mouse cable which has become disconnected from the PCB.The wire is insulated with enamel and I have to remove it before soldering it back onto the board. How should I remove the enamel? <Q> Common ways of removing the insulation on enameled wire include: Scraping the enamel carefully off with a sharp knife or blade, while trying not to nick the underlying metal wire <S> Sanding the enamel off with either fine sandpaper, or a polishing / sanding head on a Dremel-type rotary drill Using a lit matchstick or a cigarette lighter to melt off the enamel, if the wire is thick enough to not get damaged in the process. <S> Really thin "magnet wires" tend to clump up with such treatment Not removing the enamel at all, just connecting through it to the metal wire, using either solder + flux + soldering iron, or a "vampire crimp" or "Insulation Displacement Connector" (IDC) type approach. <A> If you are only doing this for a few wires (assuming USB or PS/2, that's only four to 6), just scrape it off with a knife, or use some sand paper (or even an emery nail board). <A> Get a pill of aspirine. <S> Put your enameled wire on its surface and press it down with a soldering iron's tip heated by operational temperature. <S> Aspirin melts, boils, gets in contact with enamel and desintegrates its coating on wire's copper. <S> So you can strip it with the same soldering iron's tinned tip pretty easily in a couple of seconds (up to 10 sec in fact). <S> Then let it cool down for a several seconds and remove the brittle (like a colophony) <S> melted aspirine's rests away from naked and tinned enameled wire's tip... <S> Do not breath in the aspirine smoke. <A> I had to strip 60 ends of 0.2mm enameled wire for my little project of dead-bugging and dumping a 32-pin EEPROM : <S> By the end I arrived at a method of laying the end of the wire on a piece of smooth wood, and scraping the enamel off with an edge of a thin fine file. <S> It's not as sharp as a blade would be, so you don't damage the wire so much, and you can control the length of stripped area better than with sandpaper. <S> Since I had to strip short pieces, it was hard to hold them against the pull of the file, so I made a little holder by gluing two pieces of fine sandpaper to the inside of a clothes <S> peg: <A> But it is less smell and soldering <S> is much more comfortable. <S> Use it like resin. <S> 1-3 seconds is enough usually. <A> I'm surprised no one has mentioned magnet wire strippers . <S> They're handheld motorized tools that give you a perfect strip with a sharp edge every time. <S> They're adjustable over a fairly wide range, the ones I can find all seem to strip magnet wire between 0.3mm to 2.5mm in diameter. <S> That won't help you if you're using 30 gauge or smaller magnet wire, but 28 gauge and up can be stripped very nicely, quickly, and effortlessly with one of these tools. <S> and you can find them fairly easily. <S> They all seem to be essentially the same product, down to the model number (DF-6). <S> They typically cost $50-$60, so they're certainly not for everyone. <S> I think most hobbyists would find themselves hard-pressed to justify the cost. <S> But if you need to strip a lot of magnet wire, these can save you a ton of time and effort. <S> Also, if you need precision magnet wire stripping capability (you people know who you are), then there is a much higher end tool for stripping magnet wire called a rotary collet stripper. <S> They cost $700 to over $2000 and if you need one, it is probably safe to say you already know about them, but I figured I might as well include them too. <A> Put the enamelled copper wire in 98% sulphuric acid and wait 10 minutes. <S> The coating lifts and floats away. <S> At least, this works for 'Block' brand copper wire whose coating (I think) is polyurethane.
| Melting the enamel off with a hot soldering iron in a quick motion so as not to cause a blob of carbonized enamel to stick to the wire Hydroperitum (Hydrogen peroxide - urea) can be used like aspirin. Just do a google, amazon, or ebay search for magnet wire strippers
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Microcontroller with rotary encoder and bluetooth capabilities I am looking for advice on (what to use/how to make) a rotary encoder that I can connect to through bluetooth. I am writing an app that needs to display on screen in real time to the user the number of rotations as they are turning it. Also, I need to have the micro-controller record the revolutions per day so I can keep track of revolutions if their computer is asleep. Criteria: Must be stand alone Has to be bluetooth compatible with a pc, mac, android and iphone. Has to be able to store up to 7 days of data (this would be a very small amount of data, just the number of rotations per day and the timestamp). Has to be small and cheep. I was thinking a microcontroller with just a rotary encoder and bluetooth capabilities. I have seen microcontroller like the Arduino and think they would be good to start with but I was looking for something that could actually be used in a final product. <Q> Any small micro plus a cheap BT module should be suitable for this. <S> For example, a PIC 12 or 16F , with something like this BT Module <S> (I have used a surface mount version of one of these successfully in a recent prototype) plus the encoder could be thrown together very quickly for proof of concept purposes. <S> On the PC side, Java would be best so you can port to Android, Mac, Linux, etc. <A> TI sells a complete bluetooth + microcontroller solution ( <S> CC256x + MSP430 or Stellaris). <S> Check out their evaluation platforms . <S> The EZ430 solution is very nice and should work for you since it's compact and you can connect a rotary encoder. <A> The primary feature you should probably be looking for in the microcontoller is high speed inputs. <S> Your encoder will be completely useless if your microcontroller can't keep up and it will all look like noise (assuming you're using a quadrature encoder). <S> You should also do some research on noise filtering encoders for the same reason. <S> That being said, if you're looking at creating a prototype, (without having checked on the speed of the inputs) go with the Arduino because most of the work is already done for you and just get a Bluetooth shield. <S> You can always look into making your own Arduinos (they are open source after all) and you can get the price around $3 a board. <S> If you don't go the Arduino route, it's a matter of specking out a Bluetooth module and looking at what interfacing options are available. <S> [Edit] <S> It sounds from the comments like I'm used to using higher resolution encoders than many general applications require, so I may have overstated the necessity of high speed inputs. <S> Either way, you should at least ask the question about what kind of resolution you need and whether your inputs will lose pulses. <A> what is the rate of rotation? <S> How many rotations per day are you designing for? <S> The (old) BC4 has options where you can write you code for a "virtual" chip running inside it. <S> That would get you down to a single chip to do everything. <S> The traditional way of doing this is to use any of the readily-available bluetooth modules that expose themselves as bluetooth serial ports, and then use a small micro to talk to their "real" serial ports. <S> Others have already answered with example modules. <S> Make sure you use a module that your micro can just talk to, otherwise you may have to implement a lower level protocol such as H4, BCSP or somesuch in order to use it. <S> (and if you go that route, try Matthias Ringwald's excellent btstack ; I have successfully pared this down to fit in an LPC2100 and talk to mobile phones.
| You may be able to get away with just the bluetooth chip and writing for its microcontroller to watch your encoder.
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Is there a special way to solder to these LED thermal pads? [I don't know the correct names for these things, so please correct my vocab as needed.] I just tried to solder a wire to the one of the rectangular contact areas on this LED. I'm a beginner hobbyist with a radio shack soldering iron, but usually I can solder wires okay. But in this case the solder just balled up on the surface and refused to bind to it. Is there a trick or a special way to connect these to wires? Actual part is here <Q> The pad is thermally connected to the aluminium plate. <S> That's a byproduct of the LED cooling scheme. <S> As a result, it's harder to heat the pad with a soldering iron. <S> Heat the pad longer before actually trying to solder to it. <S> Perhaps, use a larger tip in the soldering iron. <S> Putting a drop of solder on the tip increases the contact area and improves heat transfer. <S> If possible, pre-heat the whole board using a heat gun or a hot plate. <S> Temperature of the pre-heat should be much lower than melting point of the solder. <S> Use leaded solder. <S> It has a lower melting point. <A> Video on how to do it . <S> You just didn't heat up the pad enough (and 25 W Radioshack iron is good enough for this). <A> Those don't look like thermal pads, they're surface contacts for powering the LED. <S> The actual thermal pad is underneath the LED and already soldered to the aluminum core PCB <S> it's mounted on. <S> However, since the board is metal core, it will conduct heat away faster than normal. <S> What's the wattage of your iron? <S> I know I've done it easily with a 35W Weller iron <S> , it may be more difficult with less power. <S> Usual suggestions apply: be sure the iron is clean and the tip is tinned, use extra flux if the solder doesn't flow properly. <A> I have the same thing, and I have fixed these LED's to a HAMADA printing machine. <S> The easiest way to solder this is take the LED and: turn it other side and past some solder on the terminals. <S> Verylittle, believe me very very very little. <S> [you don't believe me verylittle do you? <S> please very little]. <S> Use a crocodile clip and temporary mount the LED to the heat sink. <S> Like this: Heat up the terminals with your soldering bit, until lead spreadthrough the surface unique. <S> Believe me surface tension is the force which applicable in here. <S> That's whatglues LED to the heat sink. <S> So use very little , very little soldering paste. <S> DISCLAIMER: <S> Make sure that your soldering iron have no leakages. <S> This is important whensoldering LED's. <S> Because LED's have very little avalanche breakdown reverse voltage. <S> Thisis very important. <S> If your LED will lit up while soldering then you already killed it up to some extend and it will not light in <S> it's full light intensity.
| Until the pad is heated sufficiently, the solder will not stick to it.
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Arduino Mega 2560 stuck at uploading I have two Arduino Mega 2560 boards, both gives errors when trying to upload sketches (I bought the second board thinking its a fault with the first one, but same thing happen with the new board). The new board started to fail after a day! The boards worked perfectly in the first few (~10) uploads. I can't imagine how this happen, but it's like the board is dying with time! The same sketch, which could be easily uploaded previously, can't be uploaded now. If you ask me what happens, It starts uploading and gets stucked at about 95% of the progress. (please see the following image) I found this is the exact issue mentioned in this thread . I found several other posts too regarding this matter so I hope I'm not the only one to face this. But unfortunately, still I couldn't find any working solution. Can anyone please suggest me a proper workaround? At least, will downgrading the Arduino IDE will work, permanently ? Arduino IDE version, 1.0.1 <Q> You are using IDE 1.0.1. <S> Most probably the problem is with the stk500v2 protocol used for uploading to Arduino Mega. <S> It's changed to more reliable 'wiring' protocol in v.1.0.2. <S> You shouldn't downgrade but upgrade to a latest version. <S> http://arduino.cc/en/main/software <A> The other answers are not complete. <S> The best thing to do is to burn a new bootloader into the Arduino Mega 2560 board, together with the newest Arduino IDE, and remove all old things that are on your computer. <S> Do you have a programmer ? <S> You should get one. <S> The trouble with the '!!!' is fixed, and also a number of other bugs have been fixed in the bootloader for the Arduino Mega 2560. <S> Remove the Arduino IDE. <S> Delete the C:\Program Files (x86)\Arduino folder, because there could still be files that mess up the new version. <S> Remove all the usb-serial drivers that you have installed. <S> It is not normal to have COM26. <S> I have COM5. <S> Remove Java completely. <S> The Arduino has its own mini version of Java, and an other Java could get in the way. <S> Find the hidden "arduino15" folder(s). <S> Delete those folders. <S> This is important, don't think you can skip this step. <S> After that you can start with a fresh new install. <S> Install the newest Arduino IDE from the official website: https://www.arduino.cc/en/main/software <S> When using a cheap clone board with the CH340G, use the drivers of the official website: http://www.wch.cn/download/CH341SER_EXE.html <S> When you use the Arduino IDE to burn the bootloader, then also the fuses will be set to their proper values. <S> When there is a problem with the reset (the reset is needed to connect to the bootloader to upload a sketch), then get rid of that board. <A> I had the same problem, upload get stuck on my Arduino Mega 2560. <S> In my case the issue was a string with more than one exclamation point!!! <S> Removed and worked again. <S> NOTE: <S> This did not upload: Serial.println("Hello!!!"); This did upload: Serial.println("Hello!"); <S> NOTE 2: "!!!" seems to be an escape sequence to put the bootloader into "monitor" or "command prompt" mode (see: avr-developers.com/bootloaderdocs/index.html ) <A> Have you tried this: https://www.youtube.com/watch?v=tAzjO4v7oF4 <S> This guys says to manually hit the Mega reset pushbutton just as the green progress bar on the IDE from the compilation hits the right end. <S> Another hint was to set the Preferences for Verbose (both compiler and bootloader), which consumes a bit more time with extra messages. <S> Since it worked forme, this would seem to indicate a timing problem with the Reset pulse being too short (not getting recognized) on some systems using the CH340/341 USB to serial chips. <S> The symptoms and his fix seem to say the hardware Reset pulse is too short in some cases, but there is enough of a timing window (in software) for the IDE to accept a manual reset if you get it at just the right time. <S> I don't have a schematic or scope handy to check this for sure, but it may also be that the coupling cap to the Reset line from the CH part to the Mega2560 Reset is too small. <S> Might try bridging/paralleling with a leaded 0.1uF cap across the SMD cap to see if it makes any difference, and solder one in if it fixes. <S> Best Wishes <A> I found another solution: Unplug the USB cable from my monitor / usb hub and plug it DIRECTLY into my motherboard USB port and VIOLA! <S> Sketches are uploading no problemo! :)
| Use a programmer and the Arduino IDE to burn the newest bootloader into the Arduino Mega 2560.
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Why does the ionosphere reflect HF but allow VHF signals to penetrate through? Wikipedia says: ... Unlike high frequencies (HF), the ionosphere does not usually reflect VHF waves (called skywave propagation) so transmissions are restricted to the local radio horizon less than 100 miles... Why does this happen? Is it merely that a VHF signal of say, 100Watts, carries more energy than an HF signal of the same power? Are there any other factors in play here? <Q> Essentially it comes down to the fact that the ionosphere reflects different wavelengths differently. <S> To understand why we'd have to get into plasma physics but a short version is that the plasma characteristics are a function of density, species, ionization rate and ionization energy. <S> Of course this is a coarse simplification. <S> This is similar to how you can have wavelength selective optical filters through interference effects. <S> Metals, which are typically shiny and mirror like, have their conduction band full of electrons which act as a sea of charge carriers which the light can interact with. <S> But even different metals have different Colors, which indicate a similar phenomenon to the ionospheric interaction. <A> The size of ionized plasma particle is relative to the wavelength of the HF signal and when these signals hit the plasma particles since size is more less eqaul to wavelength these HF signals will be reflected as in case of vhf wavelenths are very small and can penetrate. <S> One more reason is effect of the refractive index of the medium mostly depends on the wavelength of the signal wchich is passing through that media. <S> At HF frequency this effect is more compared to VHF. <A> The analogy I use to explain, in the most generic and elementary way, is the RF is like a drill bit. <S> The wavelength is the groves in the drill bit and the speed of the drill bit is the frequency and the material is the ionosphere, like wood or metal. <S> If you have ever tried to drill a hole with a very slow speed and large drill bit, the drill has a very difficult time penetrating the metal or wood. <S> This would represent the ability of HF waves being reflected off the ionosphere. <S> If you use a smaller size drill bit, and drill it at a very high speed, the drill bit goes thru the material very easily, just like VHF or UHF waves penetrate the ionosphere. <S> Don't laugh, my kids got it, and <S> many non- hams also! <S> For all you Amateur Radio academics, hope this doesn't upset you. <S> Lol.
| For certain wavelengths the ionosphere can interact with the wavelengths of the RF energy and acts like a mirror, at other wavelengths the frequency is such that the ions don't interact as well and thus don't reflect the energy back.
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Picking out correct MOSFET for solenoid PWM control? I'm attempting to control an automotive solenoid with PWM at 1250hz. At peak, the solenoid draws 3.75 Amps at 14.4 Volts. The specific problem is that any MOSFET that I've tried gets too hot and destroys itself as I increase the PWM frequency. The fastest that I can successfully control the solenoid is 40hz with the BUK9275-100A. I designed a test PCB to isolate the solenoid, CMS04 freewheel diode, MOSFET and MCP1402 MOSFET driver. Here are the MOSFETs I've tried: BUK9275-100A NTD4855NT4G IPD040N03L G NTD4804NT4G IRLR2705PBF QUESTIONS: What MOSFET characteristics should I be looking at besides RDS(on) to successfully control this solenoid at 1250hz? Will I need to move away from the DPAK form factor to control the heat? I have up to 11 sqin of PCB surface to dissipate the heat. Is there anything potentially incorrect with my PCB layout for the driver? I plan to use the same driver layout for the MCP1402 in the production board. I've added the schematic: http://www.firepunk.com/albums/About-Me/mosfetbreakout.png <Q> Let's first rule out static losses as the cause for your troubles: Your MOSFETs have an on-resistance of approx. <S> 100 <S> m\$\Omega\$ (or something much lower). <S> With a load current of not more than 4 A, the power dissipation for a full (100 %) duty cycle should not be more than P V, max = R DS, on \$\cdot\$ <S> I 2 P V, max = 100 m\$\Omega\$ \$\cdot\$ <S> (4 A) 2 P V, <S> max = 1.6 W To adress your question #1: Don't try to use a MOSFET with a super low R DS, on when you don't have to. <S> The low on-resistance comes with the price of a larger gate charge, making it harder for your MOSFET driver to switch it fast. <S> Also, a DPAK should be able to handle the static losses with a PCB like yours (your question #2). <S> Having checked this, and reading your note on not being able to use more than 40 Hz as a PWM frequency, I suspect something is wrong about getting a clean signal from your µC board to the power PCB (question #3). <S> How long is the connection between the microcontroller and the MOSFET driver's input? <S> How does the overall supply wiring look? <S> Edit: <S> Now that things are a bit clearer after you have added your schematic, I feel that your input side (driver IC and MOSFET gate) is in danger. <S> The flyback energy released by the solenoid after switching off needs a place to go. <S> Your paralleled 1 µF and 100 nF capacitors may not be enough, and the voltage may rise beyond the max. <S> voltage allowed as VDD for the IC or as V GS for the MOSFET. <S> It is not clear how long the wire from the next stiff source (read: good capacitor) to your board's input is, and I strongly recommend a large, local electrolytic capacitor (1000 µF, 35 V). <A> Wait... <S> what? <S> I think of solenoids as on/off devices which operate relatively slowly - 40Hz is already fast for a solenoid, and actually operating at 1250Hz is well into voice-coil actuator territory. <S> And since they are usually on/off, PWM would seem unnecessary unless you are controlling it proportionally? <S> Are you really moving a solenoid at 40/1250Hz, or moving it much more slowly with a PWM frequency of 1250Hz? <S> I suspect the answer will become clearer once the problem is clearer. <S> For example, for proportional control you probably need a low pass filter between PWM and solenoid, to prevent the MOSFET driving the inductance directly. <S> Perhaps a link to the device would be useful. <S> (Sorry to post this as an answer <S> but it's a bit big for a comment) <A> Your MOSFET may be oscillating on the switching transitions. <S> It is often recommended to put a small-value resistor (e.g., 100Ω) or a ferrite bead in series with the MOSFET gate, placed as close to the gate terminal as physically possible, in order to suppress such oscillations. <A> If your main problem seems to be running out of current, you can put multiple MOSFETs in parallel. <S> Since a MOSFET usually has a positive temperature coefficient, should one of them have a smaller ON resistance than the rest, more current will pass through it. <S> Because of the larger current, it will get hotter, thus increasing its resistance. <S> And so, your parallel MOSFETS will be intrinsically balanced.
| It could happen that every time you switch on the MOSFET, the ground voltages of your power circuit and your small-signal circuit bounce with regard to each other, causing your MOSFET to switch quite a number of times whenever it should just switch once.
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Receiving input from RS232 and reading it with Arduino board Let me begin by stating that I don't have much experience in hardware-related projects. I need to obtain info from a machine that outputs its data from an RS232 cable. http://i.imgur.com/fkqnf.png My goal is to then take this data and send it to a server. I thereby plan on using an Arduino Uno Board, attached with the Arduino Ethernet board, and write a looping program that will send the data every few seconds. What device should I use to collect the information from the machine? Can I simply use an RS232 to USB converter and plug it directly on the UNO board, or are there RS232 boards for this? And finally, are there better alternatives to my current problem? I'll gladly clarify anything that is not clear enough, thanks. <Q> A ready-made RS232 shield for the UNO such as the one I linked would be the quickest way to get going collecting data <S> (it was the first search hit; I have no experience with the particular product or vendor.) <A> Having been trained as an industrial electronics technician, specializing in computer systems of the mid 80's (I have personally upgraded over 200 IBM PC to IBM XT with 20 MB (yes, that's Mega-Bytes)) and have worked with serial port modems (200 baud and up!). <S> I suggest an old laptop running windows and Hyperterminal software. <S> I'd try a straight thru cable and a breakout box with leds. <S> The leds on the terminal unit will blink as it tramsmits. <S> Once you select the serial port com number and applicable baud rate, start/stop and data bits ( 9600,n,8,1 for example). <S> You can then "read" the output data and save to a data file. <A> This is quite straight forward. <S> I have done this when I had to receive uart data from 8051 to TI LM3S3748. <S> However for this TTL levels was sufficient. <S> Between <S> , I know I am bit late to answer this question, however, taught I'd add a little more details to it. <S> Now coming to your question: 1st - Since Tx Machine is giving data output in RS232 format, you will need a Rs232 converter at your Arduino end to receive this data successfully. <S> We need this RS232 converter at the receiver (Arduino) end, so that it can convert the RS232 levels to TTL levels, that can be interpreted by the microcontrollers. <S> The hardware connections for this is simple: |Machine| <S> Tx pin----(Data-Rs232)--->Rx pin <S> |MAX232 <S> /RS232 <S> Board| <S> Tx pin----(Data-TTL)---> <S> UART <S> Rx <S> pin <S> |Arduino| <S> The above shows the connection sequence. <S> While receiving the data from MAX232, make sure that the ground of the Arduino and MAX232 is common reference at the receiving end. <S> 2nd - For the above set, another important thing to keep in mind is the baud rate. <S> Make sure that your receiving device i.e. Arduino's baud rate matches with that of the transmitting machine. <S> For e.g. If the baud rate of your Tx machine is 9600 bps, then your Arduino should be programmed to receive the data with a baud rate of 9600 bps. <S> 3rd - For you question about, If you can connect directly to your Arduino USB, yes you can, provided you have and on board USB chip such as a FTDI. <S> Hope <S> this helps. <S> Good Day! <S> ~VD
| If learning and doing are more important, you could make up some RS232 <-> 5v logic level-shifters and write or find a software UART to read the bit-stream.
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1 x male micro USB to 2 x female I have an Android tablet and have it fixed up onto an external keyboard via a micro USB cable. But whilst using the tablet I would also like to be charging it. So would like to know if there is such a cable that splits from one male to two female micro USBs. I can't seem to find one anywhere. Thanks in advance for any help. <Q> That's an excessively non-standard cable. <S> You need a cable for OTG Host mode, and charging. <S> That one will give you a micro male to plug into the tablet, a micro mini for charging, and a usb A female. <S> You might need a usb a female to micro female adaptor. <S> Well, it might not work though. <S> http://mehrvarz.github.com/usb-otg-power-nexus7/ points out that the Nexus 7, the plain vanilla, pure android and no manufacturer bs tablet, needs two driver/kernel patches for USB Host Mode AND Charging to work at the same time. <A> And a USB compliant host device won't be charged even if you provide power through an homebrew cable. <S> The otg Y cables you will find are meant to power the device you attach and not the host, so they won't help much. <S> You can use them in order to lighten the load on the host though. <A> USB connections are NOT designed to be used as a "Y" connection. <S> You will not likely find the type of device that you are looking for. <S> Certainly do not try to cobble up something on your own unless you 101% understand the internal circuitry in both the existing device and its charger. <S> Is such a device feasible? <S> It may be if it was incorporated as a feature into the charging unit itself. <S> Short that stay away from a hacked solution.
| But "Micro USB Host OTG Cable w/USB power" is what you want. The cable you are looking for does not exist.
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Diodes in parallel or series I am wondering what the effects are of putting diodes in parallel or putting them in series. (like current capabilities, voltage capabilities etc.) Let's say I have a datasheet of a diode. What characteristics would change ? My estimation is that putting parallel would increase the current capabilities, but may have a negative effect in reverse leakage. I have no idea if I am right or how to test it, so any info on diodes in parallel or series would be great. <Q> Putting diodes in series will add the diode drops together. <S> Reverse leakage (and capacitance) should reduce in this configuration. <S> In parallel, the drop will stay the same (reverse leakage and capacitance will add), but the current capability may not be much higher, due to the possibility of thermal runaway (since as a diode gets hotter <S> it's Vf drops, then it draws more current relative to the rest, gets hotter still, and so on). <S> You can avoid this somewhat by placing the diodes in thermal contact with each other, and/or using a small resistor in series with each. <A> Connecting diodes in series ( AK-AK --|<--|<-- ) will increase the forward voltage of the resultant diode. <S> Connecting diodes in series ( AK-KA --|<-->|-- ) will cause an open circuit until peak inverse voltage (smallest diode) is applied on total resultant. <S> Connecting diodes in parallel ( AK/AK --|<-- + --|<-- ) will increase the current carrying capacity of the diode. <S> See "Current Sharing" document below. <S> Connecting diodes in parallel ( AK/KA --|<-- + -->|-- ) will not get you a resultant diode conduction in both sides. <S> Diodes in parallel: Diodes are frequently connected in parallel in switching power supplies in order to share the current. <S> Here is a document on "Current sharing in parallel diodes" . <S> Thermal Runaway really depends on the diode package and the heat-sink (dissipation) that they are mounted on. <S> The diode in my hand right now has a maximum Tj of 150 C ( Vishay STPS30L60CW-N3 ). <S> Provided enough dissipation in the design, the design can deliver higher current in "diodes in parallel". <A> What you say is partially true, but putting two 600v diodes in series does not result in an effective 1200v working, unless you put current balancing resistors across each diode. <A> Diodes in series with the same polarity each behave no differently than a single diode. <S> The voltage drop and current capabilities of each diode remains the same. <S> The overall voltage drop of the series combination of the diodes will be equal to the total of all of the diode voltage drops. <S> The current capability of the diodes does not change. <S> Diodes in parallel with the same polarity each behave no differently than a single diode. <S> However, due to the fact that the current into each diode is lower because of the current divider rule, each diode will have less current flowing through it, and therefore its voltage drop will be lower, as this is a characteristic of diodes. <S> Although each individual diode's current capability does not change, the parallel combination of diodes can handle more current overall , once again because of the current divider rule. <S> Great illustration here: http://youtu.be/ZH4fs6xkWbk <A> https://www.daenotes.com/electronics/basic-electronics/diode-in-parallel In a parallel connection, diode with the lowest forward biased voltage drop will try to conduct more current causing Thermal runaway .
| Therefore, assuming the diodes are all very similar in Vf, the overall voltage drop of the parallel combination of diodes will be lower than it would be for a single diode.
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What is the green/red drag line in Altium? When I drag a component in Altium there is always a line (sometimes it's red, sometimes it's green) connected in the center of the part and pointing to some place in the PCB. I always wonder what is this line for anyways? <Q> I asked the Altium support if there is either a way do disable this feature or to get a detailed description to understand what it does. <S> This is loosely translated from the German response: If you are seeing green/red lines when placing components in the PCB document, this can be interpreted as follows <S> : It is a kind of feedback from Altium Designer regarding the length and position of the connecting lines. <S> Red lines indicate, that there is still a better position available with respect to intersections and connection length. <S> Should the line show green, then this is an indication that the length of the connecting lines is well matched. <S> I am not really satisfied with that response <S> but I could just disable .... <S> oh wait. <S> However, there currently is no way to turn off this mode during placement. <S> Well, at least its a great example of implementing a feature that no-one needs because its not even documented how it works :) <A> It turns out that this feature (Optimal Placement Vector, or OPC) is documented in the tutorial ( http://techdocs.altium.com/display/ADOH/Tutorial+-+Getting+Started+with+PCB+Design ) <S> Here's what it says. <S> As you move a component around in the workspace, a thick green or red line will be displayed, traveling from a point within the component, to a location on the board. <S> The vector has two distinct properties: its proposed target location; and its color. <S> To determine the locations for each end of the vector, the feature uses the centroid of the polygonal shape defined by the locations of the end points of the connection lines. <S> There are 2 centroids of interest, one defined by the ends of the connection lines terminating on the component you are moving (the component centroid), the second defined by the other ends of that set of connection lines (the target location centroid). <S> The Optimal Placement Vector is drawn between these 2 centroids, with the component end highlighted by a dot. <S> Because it is a relative indicator, when you first click to start moving a component the vector is always drawn in green. <S> The 2 centroids are continuously re-calculated as you move the component, because the connection lines can move from one pad to another as they are automatically re-optimized to maintain the applicable net topology for the moving component. <S> Because of this net re-optimization, the target end of the OPV can jump around as the component is moved. <S> If the centroids move apart and the OPV becomes longer, it may change to red. <S> If the centroids move closer together and the OPV becomes shorter, it may change to green. <S> The length of the vector is not the only condition used to set the color, the color of the OPV is also affected by the overall length of the connection lines attached to the moving component. <S> If moving the component results in the overall length of the connection lines increasing, then the OPV becomes red. <S> Alternatively, if moving the component results in the overall length of the connection lines decreasing, then it becomes green. <A> Ok, the line seems to be showing the mid-point between each end of the two net-lines, at least when dragging a two-pin device. <S> It seems to be part of the "Dynamic Reconnector", a term which does not appear at all even in the altium docs. <S> In any event, if you press N while actively dragging a component, it stops following the component around. <S> However, you also loose the live net-lines. <A> It is some feature that is supposed to show you optimal routing based on component position. <S> Like, when you put rotate a component one way, and it turns green, that's supposed to be the best position for that component. <S> It is mostly BS, IMHO.
| This line is called the Optimal Placement Vector, its function is to give an indication of whether the new location is better (green) or worse (red) than the previous location. I think the only way you're going to get more information is either wait for altium to document this feature, or contact them directly.
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How to join 3 traces is a PCB? Usually I route my PCBs never making a 90º turn as best practices recommends, but sometimes there are points were 3 traces must intercept, if I always do 45º turns, in this intersection will be a 90º turn, so is it ok, or there is a better way of doing so? <Q> The 90º intersection will not be a issue for most traces. <S> If your trace is very thin (< 15 mill) and/or the PCB boards aren't being professionally manufactured you can use mitered traces (chamfer) as helloworld922 pointed out. <S> For high frequency traces (> 1GHz) it also helps reduce signal reflections. <S> The lines have constant impedance along their length regardless of routing which is a highly desirable property. <S> Note: this is less true at the highest frequencies where abrupt corners in a track also cause reflections. <S> For this reason most PCB designs employ mitred (45°) or chamfered (curved) corners. <S> Best practice in circuit board design <S> There are other methods to help prevent signal reflections, like the following image shows: <S> The image above is from Microstrip, Stripline, and CPW Design . <S> For more reading material, take a look at 90 Degree Corners: <S> The Final Turn . <A> You can add some internal chamfers to the PCB trace. <A> Either straight 90 degrees or added mitres is fine. <S> One thing to avoid is angles of less than 90 degrees, like the internal angles in a 'Z', or the "increased inductance" example in Garrett's answer. <S> These are sometimes called "etchant traps" and cause manufacturing problems with some board manufacturers. <S> For a hobbyist PCB I wouldn't worry too much, but if you were building a million I would re-design to avoid them.
| Adding chamfers will eliminate the 90º intersection and help to strengthen the trace.
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PIC: UART transmit corrupted after watchdog sleep I have a PIC16 (datasheet here ) for which UART transmit works, except right after a watchdog sleep, where the first few characters are corrupted. I have posted my reduced code below. What could cause corruption of the first few UART transmit characters right after a watchdog sleep? // Baud rates#define BAUD_115200 0#define BAUD_9600 1// Watchdog sleep times#define WATCHDOG_SLEEP_4S 0b01100/* Note: The watchdog operating mode configuration is done in configuration word 1 */void watchdog_configure(char interval) { // Configure the watchdog interval (e.g. 10010 for 256 seconds) WDTCONbits.WDTPS = interval & 0b11111;}void watchdog_sleep(void) { WDTCONbits.SWDTEN = 0b1; SLEEP(); WDTCONbits.SWDTEN = 0b0;}void UART_setBaudRate(const unsigned char type) { // See table 25-5 if(type == BAUD_115200) { SPBRG = 68; } else { SPBRGH = (832 >> 8); SPBRGL = 832 & 0b11111111; }}/* Configure the UART for full-duplex asynchronous transmission */void UART_initialise(void) { UART_setBaudRate(BAUD_115200); // Configure the RX/DT pin as an input TRISCbits.TRISC4 = 0b1; // Enable the asynchronous transceiver TXSTAbits.TXEN = 0b1; TXSTAbits.SYNC = 0b0; RCSTAbits.SPEN = 0b1; RCSTAbits.CREN = 0b1; TXSTAbits.BRGH = 0b1; // Use 16 bits for the baud rate BAUDCONbits.BRG16 = 0b1;}void UART_write(const unsigned char character) { // Wait while the transmit shift register empties while(!PIR1bits.TXIF) {} // Indicate the character to be written to the Transmit Shift Register TXREG = character; // Add a one cycle delay to ensure that the TXIF flag is valid _delay(1);}void UART_writeString(const unsigned char *string) { unsigned char i = 0; while(*(string + i) != '\0') { UART_write(*(string + i)); i += 1; }}void main(void) { // Perform initialisations watchdog_configure(WATCHDOG_SLEEP_4S); UART_initialise(); watchdog_sleep(); UART_writeString("UART TESTING");} <Q> I would guess that the problem is not actually occurring when the PIC wakes up, but rather when it sleeps. <S> If the PIC goes to sleep between the time the code enqueues a character for transmission and the time the UART finishes sending it, some bits of that character will get sent before the PIC sleeps and some will get sent after it wakes up. <S> The first falling edge that occurs after the PIC wakes up will likely be interpreted by the receiver as a start bit rather than a data bit. <S> If a start bit occurs within 8 bit times of that, the receiver will not detect that as a start bit, but will instead detect the next falling edge (which may well be some other data bit). <S> This condition is referred to as a "framing error"; note that nearly all framing errors will cause incorrect data to be received, but only some will cause the "FE" bit to be set. <S> It's worthwhile to note that when sending a byte value of 00, 80, C0, E0, F0, F8, FC, FE, or FF, it's possible for a misdetected start bit to cause an incorrect value to be received for that byte, but the byte following one of the above values should be received correctly in any case <S> (since will be no falling edge between the start bit of any of the indicated values and the start bit of the next byte). <S> By contrast, if a framing error occurs on byte value in the range 40-7F, it's likely that the next byte will be corrupted as well, since there's a falling edge between bits 6 and 7. <A> According to the datasheet, as you quote the wake up must be all zeros, although this looks to be for the RX line. <S> It says any number of bit times for normal RS232 devices. <S> Try just reinitialising the peripheral, clearing any interrupts/flags and/or sending a couple of zero characters on wakeup, then proceed to send your real data. <A> A clock oscillator requires some time to stabilize after starting up, such as after a watchdog timer / SLEEP reset. <S> Until this stabilization happens, the clocking of the UART will remain unstable. <S> Follow this up with a byte or two of value 0, before sending your first useful data packet.
| A simple fix would be to add a short delay after the SLEEP ends, before beginning to transmit your bit stream.
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Can we attach LED to VGA port and make it light up, the same way we do in Parallel Port? I just connected an LED to a parallel port in my PC and controlled it successfully through inpout32.dll I was wondering, if it was possible to do the same with the VGA port. Where can I find information about the functions and layout of the pins of a VGA port and how to use it to light up an LED? Tried to search myself but was really tough - any Google search using the terms "VGA" and "LED" is only resulting in the TV and Monitor related websites. <Q> This is probably a better pinout <S> The video and sync signals are 0-1V and not under software control; they're connected to the video hardware output of your card. <S> You might be able to put the card into "power saving" mode and turn them off altogether, but you don't have the fine-grained control that is possible with the parallel port. <S> There's also a 5V signal, and the I2C signalling used for "EDID" monitor detection (this allows the computer to know which resolutions are acceptable). <S> It might be possible to access that more directly from software, and likewise it might be possible to turn the 5V off by disabling the card or putting it into power saving mode. <S> Anything that you do discover will depend on the model and drivers for your video card. <S> Note that, given the ability to toggle a pin at a high enough rate, you can do the reverse: generate video from something that is not a video output. <S> It's also possible to use the video card to output an RF signal: http://bellard.org/dvbt/ <A> If I remember correctly, the VGA sync outputs switch polarity to instruct antiquated monitors to switch their vertical refresh rates. <S> Thus, each sync wire will be high most of the time in some modes and low most of the time in others. <S> Alternatively, one could probably construct a circuit using a 556 or similar dual timer, a fast analog comparator, and a 74HC74 or similar dual flip-flop, to latch the state of a particular region of the screen. <S> One of the timers should be set for a delay of around 2-12ms triggered by vertical sync, and one should be set for a delay of 20us or so triggered by the horizontal sync. <S> The second timer should trigger one latch to capture the state of the first timer, and output of that latch should trigger the second latch, which should capture the state of the comparator, which should check the voltage on the R, G, or B pins. <S> The net effect of the circuit would be to have the state of the LED indicate whether the brightness of a particular "pixel" was higher or lower than a certain threshold. <S> By adjusting the timers, one could control which pixel was output. <S> For reliability, one would probably have to display a rather large rectangle rather than a single pixel, but this approach could easily be extended to work with dozens or hundreds of LEDs. <A> Here is information on VGA: http://en.wikipedia.org/wiki/VGA_connector <S> But you cannot control the pins of the connector directly. <S> The only thing you can do is write to the screen and the VGA will output something different. <A> I can light up an led by putting it to pins 6/7 (-) and 12 (+). <S> It worked constantly for over a year, still continues to work and supplies the led with 2-3 volts with relativly low current. <S> I don't know is it making any damage, but no errors has been recorded yet.
| When using a newer monitor with a vintage VGA card, it would probably be possible to change the sync polarity as a means of controlling an LED; newer monitors probably wouldn't care about the sync polarity, though switching it might cause a momentary display glitch.
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Can I connect vcc to vdd? I have an USB-UART cable . The connectors listed are Red VCC, Black GND, Green TXD, White cable RXD. I want to connect it to a Alfa Hornet-UB board (http://www.youtube.com/watch?v=i58XOF4Rfc4) with pins for GND, TX, RX, VDD. If I connect: Black (GND) - GNDGreen (TXD) - TXWhite (RXD) - RXRed (VCC) - VDD Will this damage the board? I want to be cautious because I have previously connected a 3v to the VDD pin, which broke the board. I assume TX is the same as TXD (Transmit), and RX = RXD? <Q> Based on the images used on the manufacturer's page, found two sites WifiPineapple Wiki (very detailed) and Opening The Alfa AP121U (AP121U is the commercial product version of the Hornet-UB board) that use the same image and reference the same board. <S> Both state DO NOT CONNECT THE VDD PIN . <S> So that's the main source of your problems. <S> The WifiPineapple Wiki mentions that any 3.3v uart adaptor will work. <S> Oh, and importantly, you want to crossover the uart cable. <S> The Uart Cable TX would connect to the board's RX, and Uart Cable RX connects to the board's TX. <A> And VEE and VSS are the same (-), just refer to different technologies . <S> Will this damage the board? <S> Nope, it is exactly how it is meant to be used. <S> But it's always best to be cautious! <S> Hang on - <S> what about the voltages? <S> Do you want to power the board or just communicate to it? <S> If just the communicate, you can leave the positive voltage disconnected and as long as they share a ground and the logic level voltage is the same (3.3v or 5v) <S> you will be able to communicate without an issue. <S> If the logic voltage is different you need to use something like a logic level converter to correct for this. <S> (There are many other options, including voltage drivers, etc.) <S> The board is powered separately by an external 12v power supply, but I don't know what the UART Interface wants <S> Ok, then like I said above, don't bother connecting the VCC/VDD. <S> Just stick with ground. <S> As for the uart voltage level, according to the not so helpful links I have found for the USB PL-2303HX device, PL-2303HX to connect directly to 3.3V ~ 1.8V devices <S> So to be on the safe side, I would test the the TX line of both the USB and the board. <S> If you only have a multimeter, will probably have to set both devices up to be flooding the transmit line with data and at a slow datarate. <S> Don't have them connected together, <S> just one at a time put the positive probe of the meter on TX and the negative on ground. <S> If the voltage reads above 3.3v then that device is 5v data, if under 3.3v (never exceeding 3.3 or 3.4 if you have a cheap meter,) then it is most likely 3.3v, same goes for the 1.8v <S> but this is fairly rare, and I highly doubt either is. <S> The reason you need to set a very slow data speed and a lot of data is that the line is at this voltage for such a short period of time, you wouldn't notice it on a multimeter. <S> If you have an osiliscope, then this test will be easier. <A> As long as the voltages are the same there is no problem. <S> Vcc and Vdd are just other terms, just like Vee and Vss. <S> USB voltage is 5V, so if your board uses 5V as well this will work fine. <S> However be aware of little differences between the voltages, as the voltage difference will be used to produce heat.
| Yes, VCC and VDD are the same (+). Without the voltages I can't say for sure. This board is based on the AR9331, an Atheros SoC. Atheros never releases datasheets for the chips, but they have been reversed engineered by OpenWRT users , and these chips are known to use 3.3v.
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How can I detect my vehicle's shifter position? Will a Hall Effect Sensor work? I drive all Manual Transmission vehicles, and after a long battle of wills with my girlfriend, she has conceded that she would learn if there was a display that showed what gear was selected, thinking this might deter me. To her future dismay, I have an Arduino and I'm not afraid to use it. So here's what I want to build: a small bracket that will fit nicely under my shift boot and relay enough information to the Arduino to determine the position of the shifter. This is different than some methods that try to infer the relationship by RPM and speed, as they cannot work when stopped or with the clutch in. Ideally, I do not want anything touching the shifter. I could use two (or four) spring potentiometers, but they are $122/each or more! Plus I don't want any tension or touching of the shifter. I could use microswitches, but they may eventually get weak or be very sensitive to the bracket position, and would still touch the shifter. Hall Effect What I'd like to use is either Hall Effect Sensors or Proximity Sensors. Hall Effects are cheap (~$0.25 on eBay), and I'd prefer to use them. Would they be sensitive enough to pick up the shifter moving close to them? Would they continue to sense the presence, or do they only sense movement? Would I have to magnetize the shifter? Do I want Latching ones? Could someone explain how to use these in a manner that the Arduino can sense, and/or provide a diagram for this? Proximity Sensors The other option would be Proximity sensors, which look to be about $22/each at the cheapest, so buying six would be pricey and not fit well under the shift boot. Would these work without magnetizing the shifter? Are there other options? Edit I'm now between using microswitches and the cheap hall effect sensors pictured above. Could someone familiar with HE sensors draw me up a simple wiring diagram for how best to wire one to an arduino? They say they'll sense 5-8mm in distance, so this should be enough as I'll be mounting them near the base of the shifter shaft. Also, what magnet(s) should I use on the shifter? And help on how many / what strength I'll need? <Q> This is actually harder than it looks to get repeatable. <S> I would not recommend hall effect sensors for this application. <S> Their sense distances are generally small, meaning that the mounting of the sensors and magnet need to be done with some precision. <S> You will find it very challenging to align all of the halls to work consistently and under all operating conditions. <S> The transmission will move on its compliant mounts as a function of engine torque. <S> If you're dealing with a top-loader transmission, the shift lever will also move relative to the body as a function of engine torque, making it almost impossible to obtain repeatable measurements. <S> Edit: <S> The OP has a top-loader style transmission, where there is no external linkage to take advantage of. <S> For any other transmission, there is external linkage between the shifter and transmission housing - usually rods or cables. <S> For a top loader: I'd recommend X and Y axis potentiometers on the shifter, with the pots mounted to the transmission (not the body), which should be accessible through the floorpan opening that the shifter comes up through. <S> You'll need to decode the analog readings from the two pots to match gear selection. <S> For any other transmission: I'd recommend using a sealed potentiometer used as a voltage divider connected to each shift linkage (under the body, not inside the transmission). <S> You'd then do the calibration + windowing in your firmware, where it will be easier to tune. <S> If your transmission has multiple linkages, it essentially de-muxes some of the multidimensionality out of the shifter for you, again making it easier to map discrete gears to analog voltage ranges. <S> Reverse is easy--just tap the back up light circuit. <A> Edit: <S> Well, if HikeOnPast is right, this won't work for you either (as you have no shifter linkages to deal with mechanically)... <S> The mechanical design would be a huge PITA, though. <S> I'm not deleting this answer though, since I think this is a decent solution to someone else who might want to do the same in a different car (like me! <S> I've been thinking about something like this for a while). <S> Another idea would be to use optosensors/magnetic sensors along the shifter cables (assuming your car uses shifter cables, and not some other sort of system, like hydraulics). <S> I think optosensors would be cheaper. <S> You'd need 2-4 of them, two for each cable. <S> You put the optosensor around each cable (one end on one side of the cable, one end on the other side), then attach some sort of mask to the cable that will block the sensor when the shifter is in a certain position. <S> Attaching the mask to the cable wouldn't affect the shifter travel/tension at all, and would be a lot more repeatable than using sensors directly on the shifter itself. <S> The biggest problems I would foresee would be the availability of space around the shifter cables under the dash (before they cross into the firewall), and the fact that you'll have to take apart a good amount of your dash to get a good amount of room to work with. <S> In this diagram, it's assumed you have a 5spd car that uses two linkages. <S> /O for the uC. <S> The red bars/arrows signify the optosensors, the blue bars the mask attached directly to the moving part of the linkage, and the black rectangle the linkage itself. <S> Extrapolating the position of the shifter is real simple with this. <S> If both left sensors on the left cable are unblocked, you know that the transmission is in neutral (regardless of the right cable). <S> Otherwise, the transmission is in gear, and which gear is determined by the sensors that are blocked. <A> Mounting something to the shifter itself, hidden under the boot, is probably the best way to go. <S> A two-dimensional linear encoder strip, perhaps made of plastic, could be made as a collar that is mounted around the shifter and attached to it. <S> The collar is white, with black patterns printed on it. <S> Or perhaps transparent. <S> You might be able to get away with a laser-printed transparency. <S> These patterns are scanned by optical sensors. <S> The arrangement is such that if the stick is moved back and forth, only the back-and-forth sensors are driven with a quadrature signal which can be decoded to give the direction of movement and position. <S> The side-to-side sensors do not see a change in the pattern because it is wide compared to conventional linear encoders. <S> And vice versa. <S> The picture gives a hint at the basic idea.
| The best bet for this to work for you would be to create some sort of mask that sits on top of the shifter unit under the boot with 6 optosensors, and then attach a mask to the shifter to obstruct the sensors as necessary to determine which position it's in. The simplest way would be to use 4 optosensors to act as digital I
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Is it possible to solder or otherwise connect to this camera chip by hand? [Disclaimer: I'm a beginner.] I ordered this OV7740 camera chip from Digikey without really researching it, thinking maybe I could connect to it with a microcontroller. (I have only Arduino at this point). Digikey part page "Datasheet" link I just took these photos, next to a penny. How do you connect to that? Is there a name for that type of connection? Can it only be handled by robots, or is there something I can learn to do by hand that could connect to those 32 contact points? The sides of that chip are about 2-3 mm. <Q> That looks like a BGA surface mount component. <S> It is possible to solder these parts to a PCB by hand using a variety of different methods. <S> Use a hot-air rework station <S> Use a reflow oven. <S> This can be a professional one, or more commmon for hobbyists is to build your own from a toaster oven or a hot plate. <S> For really cheap you can try using the toaster oven/hot plate without any extra controlling circuitry, just carefully watch the process. <S> Note: If you use the second option, be sure NOT to use the same toaster oven/hot plate for cooking food you're planning on eating or feeding to someone else. <S> Whichever method you choose you'll want to get a pair of very fine point tweezers, or even better a vacuum pickup tool. <S> You'll also want some kind of magnifying system. <S> This allows you to precisely align components. <S> Some people use a headband magnifier, others use a full-on desk microscope. <S> Soldering BGA packages also usually requires solder paste and flux. <S> You can get breakout prototyping boards for BGA packages, these are a basic PCB board which route the BGA's pinout to some through-hole pins you can solder headers or wires to. <A> This is a BGA (ball-grid array) package. <S> After soldering, it's usually X-rayed to check for defects. <S> There's a particular type of SMT prototyping board called Schmartboard. <S> Guys who make it claim ( in this video ) that it can be used for hand-soldering BGA packages. <S> There are sockets for BGA chips (somewhat similar to ZIF sockets for DIP), but they can be very expensive. <S> Also, you could look for a breakout board (or a development board) for your sensor. <S> The sensor would be professionally soldered onto the board. <A> This could be described as a Chip Scale Package (CSP), as the "package" is just larger than the chip itself. <S> It is also a Ball Grid Array (BGA) package, as it is designed to be attached to a printed circuit board (PCB) by reflowing the solder balls you see on the bottom of the package to pads on a PCB (and the solder balls balls are arranged in a grid pattern). <S> A "robot" is not required to place or reflow the device, but professional quality "pick and place" and reflow equipment make the job easier. <S> These "pick and place" machines can be manually actuated or automated (robotic), as can the reflow equipment. <S> I would not try to approach this task with a hand-held soldering iron. <A> It is possible but likely difficult. <S> Best approach I would consider is to create the PCB as it should be and then apply a little paste manually to the device, place it carefully on the board and reflow using a skillet. <S> Using a skillet will make the heat come from below, avoiding possible damage to the top of the package where the imager is. <S> Also, it will allow it to center itself due to surface tension. <S> Make sure you have plenty of flux since it'll help avoid shorts.
| It's soldered by automated pick&place equipment or special BGA rework station. It's not meant to be soldered by hand.
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Why JTAG connectors are available in 10pins/14pins/20pins when JTAG is of 5pins AFAIK, JTAG is requires 5 pins ( Wikipedia article ): TDI (Test Data In) TDO (Test Data Out) TCK (Test Clock) TMS (Test Mode Select) TRST (Test Reset) optional. Let's add two more Power supply Pins (Vcc and GND). If it is of 7 pins then why most JTAGs connectors come with 10 pin or 14 pin or 20 pin variants. Most of the pins are either NC or GND. Why so many GND pins are provided? Is there any special reason? <Q> I don't know the exact reasoning for JTAG, but when high speed signals are used and according to the best practices, you should put a GND in between every signal of a flat cable. <S> JTAG can be considered as a High Speed Signal. <S> Multiple GND wires are used to avoid crosstalk between the signal lines. <S> And they also provide a separate return path for every signal. <S> Indeed, in high speed signals, the return current "prefers" the path of least impedance. <S> That path is, for high speed signals, the closest GND. <S> Thus the different signals will have different return paths and that avoids crosstalk of the return path. <S> The final goal is to guarantee good signal integrity, reduced emissions and a better immunity to external disturbances. <A> One of popular existing options was Everex (10-pin IDC connector, formerly used for connecting DB-9 serial ports to the PC motherboard). <S> 14- and 20-pin IDC connectors are popular options as well, but JTAG isn't limited to these. <S> One important property of IDC connectors is the strict 1-to-1 pin to wire relationship, so if you need more GND lines, you have to add more pins. <S> You could make a specialized 7-pin connector with several GND wires connected to a single pin and it would work just as well, but such a multipoint connection would mean you cannot simply crimp your connector to a flat cable and be done with it. <A> 20 year ago during the transition from serial and ISP programmers to JTAGs. <S> The 10 or 20 pin headers supossedly concentrated not just one JTAG bus <S> but as many busses as microprocessors had a unit. <S> not sure if someone could confirm this true?
| They avoid capacitive coupling between adjacent lines. There are no official standards for physical JTAG connectors, existing standard connectors are used instead.
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Looking for a quick and easy way to boost +5VDC to +24VDC I am looking for a simple IC solution how to boost +5VDC to +24VDC 1A. Requirements: THT (the project is on a 2.54mm(0.1") experimental breadboard) low-cost (this is a hobby project, so arround $5 for the whole converter system) simple (fewest external elements) has been around for a while (most recent chips are delivered to my country mainly by the Farnell, charging high transport prices). Is this attainable? What are a part number or two, that I can toss in the local electronics shop and get the device working by the afternoon? <Q> Note : This solution can not provide 1 Ampere output while boosting from 5 Volts, as pointed out by @markrages. <S> Also, the product mentioned is not suggested to be operated beyond a step-up ratio of 1:3, i.e. up to 15 Volts from a 5 Volt source. <S> Hence, this answer does not meet the criteria specified in the original question. <S> With the amended 12 Volt source, it would work . <S> As has been pointed out by @Wouter van Ooijen, one is unlikely to find a simple single-IC solution for boosting a 5 Volt DC source to 24 Volts at 1 Ampere. <S> An alternative that fits the specified budget but not the full requirements is a pre-built adjustable DC-DC boost module such as this one on eBay for $4.49 with free international shipping. <S> Input voltage ranges from 3.5 to 30 Volts, and output can be adjusted from 4 to 30 Volts, as long as the output is not set to more than 3 times the input voltage . <S> Searching on eBay and other sites may yield lower prices for these modules. <A> I'm pretty impressed at Roman Black's simple +5V to +13V voltage boost circuit <S> that uses only extremely common low-cost components. <S> It also appears to me that all of the components in this circuit are available in through-hole packages (THT) which some people seem to think is easier to assemble. <S> A few people can get a (low frequency) switching regulator working on a solderless breadboard. <S> ( Emre , Roman Black, J. B. Calvert , Bob Pease , etc.).Alas, solderless breadboards are notorious for causing problems with switching voltage regulators. <S> ( Bob Pease , again. ) <S> A full 1 <S> A output would require swapping out most of the components with higher-current components, and increasing the current limit by reducing the resistance of R2. <S> It might also need a few heat sinks, since it's "only" 72% efficient. <S> There are several ways to improve the efficiency, but I all the ones I can think of off the top of my head would make it more complicated. <A> TI's selector gives a variety of choices with those requirements including with a LM3478/88/81. <S> They cost about $5 in parts, but that isn't in small quantities and with shipping.
| It appears to me that replacing the 13V zener with a 24V zener and some relatively minor part value tweaking would produce a +5VDC to +24VDC booster. The module uses the Texas Instruments LM2577 step-up voltage regulator , and incorporates the required inductor and adjustment preset on the board.
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Trying to understand range and frequency of an RFID reader I am trying to understand how range of communication between RFID reader and tag is related to frequency and power. So if I increase frequency, the range will be shorter or larger? why? (I need to understand basic mechanism) I have an RFID reader . This one is having a range of only 5-10 cm. I want to increase its range to say 1 meter. I understand that if I can increase output power I can transfer energy to larger distance, so I can energize the tag kept at larger distance. So can I increase the range by amplifying the signals and possibly connecting it with a different antenna which can send the signals to a larger distance? If yes how? <Q> You cannot change the frequency because it has been tuned for specific frequency, not to mention the legalč stuff even if you could. <S> For given frequency (or better said wave length) there is a relation which determine how far the so called 'near field' extends. <S> And only within this near field passive RFID system could possibly work, because it works similar to the transformer /the energy 'taken' from the primary side indicates the presence of the tag in the field/. <S> The antenna near field extends to 'wavelength / (2 <S> * Pi)' <S> So if you made the frequency higher, you would actually shortened this reactive detection range. <S> Long range systems (10 m or so) works based on the back scattering principle, much like the radar. <A> Neither. <S> Frequency is just how long the wave is, how long one period takes. <S> Also, the system is set to a specific frequency and both the transmitter as the receiver have to be on that frequency. <S> When you change the frequency of one, it won't work anymore. <S> You won't get it as far as 1 meter, the module can't handle such a dramatic power increase. <S> You however might use an extern amplifier. <S> Dismount the antenna, implement the amplifier, and use a (new) antenna. <S> I'm sorry, but I do not know what you should use for this. <A>
| If you change the frequency, the system will stop working; it's tuned to a particular frequency.
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Ground loop problem with Power over Ethernet I am trying to build a weather station, powered using Power over Ethernet.Originally I planned to use a Nanode as the microcontroller, but since the Raspberry Pi is so cheap, I decided it was much easier to use that instead. I bought a TP-Link power over Ethernet splitter, made up some custom cables and after checking the voltages, connected it up to the Pi. All worked fine, until I tried plugging in the HDMI cable to my TV (for debugging). At which point the board kept turning off and then the switch cut out. At that point I discovered that there was a 40 volt difference between -ve on the PoE output and the ground on the HDMI. I tried a couple of other PoE boards and found there was the same issue on them too. Is there a solution? Although I don't really care about plugging it into my TV, I am a bit worried about plugging sensors in, that might be grounded and end up putting big voltages across them. There is a photo here (click for full-size): <Q> The -VE being (around) -40 <S> V relative to ground is as expected. <S> PoE provides -48 V DC. <A> I just checked the standard and PoE is supposed to be isolated at both ends (switch and device, section 33.4.1, since at least 2008). <S> So it looks like TP-Link are cutting corners with their PoE devices and omitting isolation circuitry to reduce cost, violating the standards. <S> I purchased an isolated DC-DC converter from an electronics supplier (in my case a Tracopower TEL 5-1211) and soldered some connectors onto it. <S> I installed it between the TP-Link splitter and the device I wanted to power, and it solved the problem. <A> TP-link power splitter <S> I managed to find <S> seems to be non-isolated. <S> Bad thing for you is that PoE usually have diode bridges on input (to be polarity-independent and work on straight or X-cables) <S> meaning that if your 48VDC supply is not galvanically isolated and for instance it's minus terminal is grounded, sparks can fly if you miss to crimp ethernet cable right way. <S> Try to swap 4,5 and 7,8 pairs, measure ground potential difference again. <S> Of course, for careless PoE usage best solution is to use isolated splitter . <A> PoE does not necessarily have to be isolated, so it may be the case that your TP-Link isn't. <S> (Although it seems to me they'd want to sell isolated versions so they would be more foolproof/robust...) <S> Since you are connecting HDMI to your TV though, that counts as a user connection and you really should be using an isolated splitter.
| If your PoE device has no outside world/user connections, then you can safely use a non-isolated scheme.
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Why would this wire be shielded? I'm trying to work out how I might emulate this assembly to replace the seat weight sensors below. I'd guess the sensors are just strain gauges, so a variable resistance or maybe capacitance. Unfortunately the control module is a black box. I can't see why the red/blu wire would be shielded. Any ideas what signal it might be carrying? <Q> The Weight Sensor Control Module provides amplification and filtering for individual seat sensors. <S> (sorry - automotive electronics is tough if you don't have support from an OEM (Original Equipment Manufacturer) or supplier. <S> A typical arrangement looks like this <S> (your sub-circuit in green box -- note that in for the vehicle shown, the line is not shielded). <A> With the limited information you've given, I can only guess that it's a low level analog signal that needs to be isolated from potential interference. <S> Perhaps it's being routed near a noise source, or perhaps it's a longer run that has more potential for pickup. <A> @BobT <S> 's nailed it. <S> Automobiles are a horribly noisy electrical environment coupled with high reliability requirements and cost sensitivity. <S> That line has to run some distance back to the main body/cabin control module (typically near the passenger foot well). <S> Long-run plus sensitive signal = <S> shielding required for reliability. <S> Without more detail it will be difficult to say precisely why. <S> (signal is the agressor ).
| Either it is to protect the signal from interference (signal is the victim ) or it is to prevent the signal from generating interference to nearby wiring It interfaces to the airbag control module using a proprietary protocol that you will need to reverse engineer with a scope and/or protocol analyzer First note that you are messing with systems directly related to the control of your vehicle's airbags - a safety critical system that can really hurt you or kill you: If you need it to work and it doesn't, and If you expect it to not work and it does Work on this system at your own risk.
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Is there a PIC development tool like Texas Instruments's EZ430‑F2013 to help introduce me to PIC? I would like to familiarize myself with Microchip's PIC product line and IDE, I don't want anything too basic or cheesy, etc. I can't seem to find a device like Texas Instruments's EZ430‑F2013 USB development kit: The features I like about TI's board, and what I'm looking for in a PIC version, are the following: (In order of 1 = most desired to least) Works with their IDE Quality The detachable PCB with chip pinout, and the size of this PCB (because it can easily be used in a basic project...) Size and surface mounted desing USB connectivity Price Can easily be used to program other MSP430's I also want to learn with a processor line that I will be able to use in production units. I don't know the PIC processor line differences well, but a processor in the price range of $1.80~$6 per 100 quantity, is my goal. From a bit of research and answer's like this one I'm thinking that to best meet my desirements, I may need to get a separate programmer and dev board (like is needed/recommended with AVR's.) If so, it seems like the PICkit2 or maybe PICkit3 would be the way to go? And I don't know what dev board/processor line etc. to start with. Price isn't too big of a deal, especially with regards to the programmer/debugger, so if there is a better option I may rather buy it. However, a cheap device like TI's is fine too, as long as it meats my processor goal, etc. <Q> Once you get comfy with the 8-bit parts, you can move up to an Explorer 16 board and play with PIC24 and dsPIC devices. <S> The board supports add-on cards called PICTail boards, which provide extra functionality (Ethernet, power supplies, etc.) <S> There's also a standalone <S> PIC32 <S> starter kit that doesn't need a programmer. <A> Id'd go for something like the PICKit3 Starter kit , or the PICkit3 Debug Express, it has all you need. <S> If you want USB connectivity from the dev board itself, then something like the MPLAB Starter Kit for PIC18F MCU <S> (DM180021 - shown below) might be an idea (it has an on board programmer) <S> All the starter kits are on this page . <S> There are also thrid part vendors with soem good options, like Olimex ( dev boards , proto boards ). <A> Most of Microchips dev boards only meet the first two of your features, but the ez430 doesn't meet much of it either. <S> The EZ430 is a mostly abandoned dev system (it can't program some of the newer <S> msp430's released in the last year and a half.), has no firmware updates available, it's only replacement boards are other MSP430F2012 with minimal flash and ram. <S> The target boards themselves are just basic breakouts for the device. <S> They are as simple as the msp430 line gets. <S> The Msp430 Launchpad is a much better dev tool, based on price, included mcus, and range of use. <S> For PIC development, there really isn't any official one that matches the form factor of the ez430. <S> There was the dlp-flash2 that had a debugger/programmer witha detachable target board, but that hasn't been sold in quite a while. <S> Honestly, the best thing for PIC development is a PICKIT2 or 3. <S> Any board with a ICSP connection will work great with it. <S> As for target boards, you can make them from a piece of protoboard for dip sized chips, or with a standard soic breakout, not hard at all. <S> Update: Actually, found the DM330013 - Microstick for dsPIC33F and PIC24H. A 25 dollar development stick with built in programmer <S> (Supports only 4 parts though), able to plug into a breadboard, supports Microchips standard IDE, and socketed, so you can swap mcus out. <S> Every bit as good as the ez430, minus the smd boards.
| Consider the PICkit3 Debug Express kit which includes the PICkit3 (a tremendously useful little programmer), MPLAB IDE on a CD and a 44-pin evaluation board with a PIC18F45K20 populated on it.
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Patching stereo audio into an old car's tape deck I drive an older car before the days of 3.5mm line-in jacks. A couple years ago, I found an interesting idea online where some owners of the same car patched in a line-in directly into the L/R channels of the cassette player deck (since no one really uses cassette players anymore). http://www.matthewsvolvosite.com/free-ipod-input-for-your-volvo-radio.html I did the same thing, and it seemed to work great for a couple Nokia phones that I had and also an old iPod shuffle. Recently, however, I got a new phone - the Samsung Galaxy S III and it seems that it doesn't like the line-in at all. When the tape deck is powered off, the phone detects the line-in properly (e.g. it shows that headphones are connected, and any audio is routed to the headphone jack; of course there is no sound played in the car because the tape deck is powered off). However, once I power on the tape deck, the phone detects that there is nothing connected to the headphone jack. Oddly though, if I start playing a song before powering on the tape deck, then the phone maintains the connection when powering on the system and the music plays fine through the car's audio system. However, the moment the song ends or play back is stopped/paused, the phone thinks its no longer plugged into anything again. Curious about this, I assumed that perhaps there's something odd with the grounding of how I hooked up the cable's wires the first time. The PCB I patched into had a labeled terminal for Left and Right, but none for Ground - so I had just grounded the cable to the chassis (for better or worse). In opening up the stereo system again and checking with a volt meter, I noticed the following: When the tape deck is powered on, there is a 2.8 V potential between the signal lines (e.g. L or R) and ground When the tape deck is powered off, there is a 0 V potential between the signal lines and ground (I suppose that would make sense; but what about the first bullet point?) Interestingly: The potential between the signal lines and ground of the 12 V outlets in the car also show the same 2.8 V potential There is a 10 V motor in the tape deck - its ground also shows a 2.8 V potential from the audio signal lines When I play music through the system using the method described above (where I "trick" the phone into keep the connection), the potential between ground and signal is something less than 1 V (close to 0 V when measured with a basic multi-meter). I thought perhaps I needed to add some pull-down resistors to the signal lines, but I'm not sure if this is true? I tried this with a couple different resistor values (680 Ohms and 10 kOhms) but neither seemed to have any noticeable benefit (the difference was still greater than 2 V). I was wondering if anyone had any ideas about this? <Q> I would start with a value of 10uF with a voltage rating of 16V or more. <S> It would be best to utilize non-polarized capacitors but if you cannot find those then consider <S> one of two options: a) Use two 22uF capacitors in series wires + to <S> + <S> and then one - lead to the jack and the other - lead to the cassette input. <S> b) Try to see how a polarized cap would work. <S> Put the + lead toward the cassette player amplifier and the - to the jack. <S> These capacitor ideas block the DC bias that is apparently present at the connection point where you attached the signal wires into the cassette amplifier board. <S> Some of the earlier devices that you connected to the cassette jack may very well have had capacitor coupled outputs. <S> On the other hand your newest device may not have an output configured like this. <A> the Galaxy s3, and really, any smartphone post iphone2, have finicky 4 conductor (TRRS) <S> 3.5 <S> mm <S> jack (L/R/G/Mic, or a variation of that). <S> When you use a regular 3 conductor (TRS) <S> 3.5mm plug, the 3rd and 4th conductor are shorted together. <S> This causes problems, sometimes. <S> Especially with a higher than 0 potential. <S> You are essentially triggering the mic/remote sensor. <S> You will most likely need to get a TRRS to TRS adapter. <S> These have a 4 conductor plug connected properly to a 3 conductor jack, without shorting any of the conductors. <S> Also, Did you connect the ground to the actual chassis ground? <S> That might be the problem. <S> Most radios are going to isolate audio ground, digital ground, and chassis ground as best they can. <S> You would want to look for a ground point close to the cassette audio points. <S> Just as a proper setup. <A> I cannot add a comment but make sure to use a capacitor at least on the ground/common line between the phone and the car. <S> A lot of internal audio input stages are floating at half the supply voltage (so it might be 6V or more).You will not see any problems until you connect a charger which closes the car ground to the audio input middle point, through your phone. <S> Source: own experience <S> , I fried an iPod doing this.
| Sounds like you should consider placing some good quality capacitors in series with the audio signal lines from the jack that you added and to where you connected the inputs to the guts of the old cassette player's amplifier input.
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Can I replace the power adapter of 12V/0.8A with one of 12V/2A? Possible Duplicate: Choosing power supply, how to get the voltage and current ratings? Can I replace the power adapter of 12V/0.8A with one of 12V/2A? I have a Digital device that uses a power adapter to 12V/0.8A, and it stopped working.Can I replace it with other adapter that is 2A? Would this work? And what effect would it have? <Q> Yes. <S> Just make sure that the polarity on the connector matches. <A> Yes, with caution. <S> 2A is quite a bit more (as opposed to a -LOT- more, like, say, 20A would be) than 0.8A. <S> You device might not even have bothered with a proper fuse, thinking ' <S> oh well, the power supply will limit to 0.8 A'. <A> This question will probably have been closed next time you blink. <S> It is asked often and not deemed a proper question by those who prognosticate on such things. <S> As Ignacio says, no problem. <S> If the voltage is correct the target determines how much current to draw. <S> In a very very few cases the equipment may rely on the power supply sagging under load to work correctly BUT such equipment is rare and should be shunned. <S> One case where it may matter is cheap power supplies for cheap appliances using NiCd or NimH batteries. <S> Drills of say 12V to 24V rating may be like this. <S> Such systems rely on the transformer open circuit voltage not exceeding the battery fully charged voltage "by much". <S> They usually use unregulated supplies. <S> Such systems cook their batteries if left connected when charged at the best of times. <S> They will do so more happily if a higher current capacity supply is used. <S> This is a very nasty method of battery charging and best avoided if possible.
| The device will only draw as much current as it requires.
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Methods or any innovative ways to check the correct layer order independent of the customer given layer stack data As a PCB manufacturer do we have any fool proof method to check correct layer order?I tried to check the impedance with standard Er, H. It works, but I'm not satisfied with the methodology I followed. I want to ask the community if you have any measures! I want the methodology check to be independent of the customer provided stack up drawing or information. <Q> Such markings, when done correctly, can be used to very quickly determine that all layers are in the correct place. <S> Here is an example of how these could look for an 8 layer board. <S> On layers with internal plane fills the plane is cut back away from the layer indicators and the outer solder mask contains an opening on each side over the marker area. <S> Once the board is all fabricated the layer markers appear as follows and can be easily seen when the blank board is held up in front of a light source. <S> This scheme also is useful for sorting layers of Gerber films and the best part of it is that it is totally under the design control of the customer. <A> Taking Michael's feedback a step further, in the area that you use for alignment structures, have your customers place those numbering "tags" beside each board in the waste area. <S> Enforce that they place YOUR structures on the proper layers upon submission. <S> Have them sign off that the count sequence is correct. <S> If you produce boards that count properly and it's still built wrong <S> it's their fault. <S> This is what's done on the semiconductor side. <S> No one ever gets it wrong and still keeps their job. <A> In general the layers should be labeled in the gerber files. <S> And the assembly drawing should show the stackup. <S> If the gerber layers aren't clear (e.g. GND instead of LAYER1), then look at the stackup. <A> http://www.speedingedge.com/PDF-Files/Test%20Structures.pdf <S> Another neat trick you can do is build vertical copper lines on perpendicular sides of your board. <S> This will confirm none of the layers were shifted in the X-Y plane before being laminated. <S> Of course if you're using a high end fab, this is almost always a non-issue. <S> But its something useful for those using cheaper fab solutions.
| I am a strong proponent of putting readable layer indicators on each layer that can be seen through the board after it is fabricated.
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Servo Current Draw When Moving This may seem a very silly question, but does a servo draw more current when it's moving? I need to be able to log when a servo flips, and being able to log the current drawn from the battery pack seems to be the most obvious and simple solution <Q> Assumption : <S> Servo motor in question is a hobby-type servo, not an industrial high torque or heavy duty servo. <S> The bulk of the current through a (hobby-type) servo motor is the current through the coils of the DC motor typically used within the servo housing. <S> Some current is required by the servo controller circuitry, but that is negligible in comparison. <S> Current through the servo will thus be high when it is changing position , as the DC motor must turn till target angle is achieved. <S> Current is then low after it reaches the new position and stops, as long as there is no opposing torque that the servo must fight to retain the new position. <S> Current will be high again if a torque is applied to the servo arm ... <S> this last can actually be higher than current while moving, up to the stall current of the internal DC motor. <S> This is because the servo controller drives the motor hard to force it to hold the target position. <S> Thus, depending on current as a means to detect servo traversal is unreliable at best. <S> Instead, a mechanical solution such as a homemade rotary limit switch made using a conventional limit switch levered by one of the servo arms, would be a recommended approach. <A> Like a motor, if it isn't moving, there's no current in the windings (or insufficient current). <S> (Though technically a motor can be under too heavy a load <S> and there's a lot of current despite its inability to move, hastening its failure, but that's another subject.) <A> The amount of current a servo draws will be related to the load on the servo (and consequently the servo's torque. <S> There also may be additional torque required (/current drawn) when moving a load from a stand still.
| Simply put, a servo that isn't moving is not powered, and therefore is not drawing current.
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A separate MCU to cutoff LiPo on low voltage? I'm designing a project that is going to work on a flying quadcopter. It features an ATmega328P MCU in the main 5V circuit, and is powered by a small (<500 mAh) 1-cell LiPo battery. I want to provide low-voltage cutoff mechanism to protect the battery. Currently I'm considering the three options: Just let the main MCU to monitor the battery voltage , using VCC as reference. VCC is usually 5.1 V, but to be sure I will also need to measure own VCC. Pros: no additional circuitry. Can signal cutoff reason using a LED. Configurable cutoff voltage. Soft cutoff. Cons: in case of malfunctioning of the main MCU (software bug) the battery voltage may be left unobserved under a high load. Complicates software. Requires powering the 5V network. Use a Reset Monitor like MAX809 . Power-down current is 0.5 μA. Pros: highly reliable. Cons: no signalling. Not easily configurable. Hard cutoff only. Additional circuitry (voltage divider). Use an intermediate MCU like ATtiny25/45/85. It is tiny but can measure own VCC. Power-down current is 2 μA. Pros: signalling. Configurable. Easy to test. Measures voltage before powering other things. Soft cutoff (can signal the main MCU that it's a time to gather stones). Cons: requires additional (but very simple) software. Weight (1g?) Option #3 is illustrated here: I'm thinking of going with option #3. Does it make sense? <Q> I'd do #1 or #1 + #2 . <S> The main controller would be the primary battery monitor. <S> If something goes awry with the main controller, the analog reset monitor would be the backup. <S> #3 makes sense if you think that later you might add more functionality to the dedicated battery monitoring controller. <A> We do #1 and #2, working thus: Board is powered from 24v, so our 3v3 supply is regulated down from that. <S> If the 24v goes down we have a long time (in CPU cycles) to see that happen using a voltage-divider and ADC pin on the micro, and make a note of it by setting a bit or whatever you want to do. <S> Reset monitor catches the "hard cutoff" of the 3v3 rail having a wobble, going down, spiking, etc. <S> and forces a hard reset (grabs the micro /RESET pin and holds it low until it's really happy with the 3v3 line) which prevents the micro from missing a problem due to high load (although that's what hardware watchdogs are for). <S> A point worth thinking about is how, using #1, you can actually store the reason in a way that survives the reset/power-loss and doesn't risk corrupting something by loss of power half-way through a flash-write cycle or something. <S> We have plenty of time as our power source has a long way to drop (20.7v) before the 3v3 line dies, but in your application you may not. <A> Try this circuit instead, for complete independence on any MCU (I can't upload an image, so I'll describe the circuit): 1 Zener diode, 5 resistors, 1 NPN transistor and one PMOS. <S> They can be SMD or TH. <S> The Zener diode acts as the voltage setter, the NPN keeps the PMOS conducting in normal operation and the resistors act as current limiting and bias providers. <S> If the input voltage falls under the Zener threshold (a little higher actually), the NPN stops conducting and the PMOS stops supplying power to the load. <S> Connect the components like this: <S> Supply line to zener, arrow pointing to the positive voltage, then a resistor in series (e.g. 1k) to ground. <S> Take the voltage from the mid-point and feed a voltage divider (e.g. 5k-50k). <S> Take the voltage from the mid-point of the voltage divider and connect it to NPN (e.g. BC817-40) base pin. <S> NPN "arrow" goes to GND, collector goes to a resistor (e.g. 1k). <S> Connect the end of this resistor to the gate of the PMOS <S> (e.g. IRF9130). <S> Connect the source of the PMOS to the positive rail and the drain to the load positive terminal. <S> And you have a simple UVL circuit. <S> A zener diode with a voltage close to where you would like the Under voltage Lockout circuit to start working is required and this is what acts as the "switch". <S> Simulating or building the circuit might help in understanding the advantages and disadvantages: it draws a very small current (zener diode leakage + other leakages) when off. <S> I got 0.6mA in simulations. <S> The above values were chosen for a 10V battery pack (LiFe) and they work great for cutting off the battery under 8.6V. <S> For different voltages, use different Zeners and maybe also smaller/bigger resistors in the first 3 places. <S> The schematic can be doubled and reversed for a +/- <S> supply configuration, if required. <S> I hope this helps.
| A 50k resistor should be connected between the gate and the positive rail, to make sure the PMOS does not conduct when not needed (and discharges the gate capacitance, avoiding linear turn-off).
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How do I make a Polypropylene box static resistant? I am beginning to organize my collection of electronic components and parts but have run into a roadblock of not knowing how to properly store anything that is ESD sensitive. I have been using ESD shielding bags that I have acquired over the last few years to store those components and parts, but now that I have implemented a drawer system I would like to keep those parts accessible in the same manner as my passive components. By perhaps lining the the drawers with the ESD shielding bag material would the components be safe from ESD, or will another approach be necessary? For reference: Oh and as a side note, if anyone has a good schema for keeping resistors quickly accessible I would love to hear it! <Q> [This is more of a comment. <S> I'm posting this as an answer, because I'd like to add a picture.] <S> Storing SMT resistors and capacitors in a binder provides the quickest access. <S> I've tried several methods for storing SMT component kits, until I started to use binders. <A> I store my ESD sensitive parts by sticking their leads in ESD foam, then putting them in an ordinary bin. <S> The foam looks like this: <S> I've never purchased it; I've just acquired it over time from ordering parts. <S> This scheme probably isn't up to spec for a manufacturing operation, but for storing cheap parts for tinkering at home, quite sufficient. <S> There are also ESD plastic tubes molded for bigger components. <S> They look like this: I've collected quite a few for DIP and TO-220 packages, and store some in a pencil cup, or rubber-banded together. <S> For resistors, see if you can find one of these gems: Ohmite made them years ago <S> and I've never seen anything better. <S> No longer in production, but you can find them on ebay with "ohmite resistor drawer". <A> You can get a bunch of small ones fairly cheap. <S> Even better, most parts come packaged in one with the part number printed right on it. <S> Why expend more effort and subject the parts to unnecessary handling? <S> There are treatments to create a static dissipative surface on plastics, but those treatments only reduce the static generated by the plastic. <S> They don't provide shielding from discharge, and the chemical has to be reapplied every month. <S> If you really want to make your bins conductive, you could apply HVAC tape (the heavy duty aluminum foil+glue kind) to the inside surfaces. <S> The parts will still be less protected than in a bag because they aren't completely enclosed. <A> Old post but my $0.02 as an EE... <S> I prefer Black conductive foam. <S> As long as you keep the parts away from the sides, it should not be an issue. <S> The metallic bags would give you more protection, but I find it is a hassle to store them. <S> As a note there are several misconceptions about ESD storage materials. <S> You will find two terms " ESD Dissipative " or shielding, and " ESD Safe ". <S> ESD Dissipative materials conduct and prevent static from damaging a component. <S> Black ESD foam, back bags, silver bags, and ESD mats fall under the "ESD dissipative" term. <S> ESD Safe simply means the packaging won't create static, but provides no protection to the components. <S> This would be the parts tubes, pink bags and foam. <S> They will allow a static charge to pass right through them. <S> The challenging thing about ESD, is it often won't brick an IC. <S> But will cause damage that will result in a shortened life. <S> For a company its a big issue, for home projects not so much.
| Keeping the sensitive components in an ESD shielding bag is still a good idea.
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Is a Schottky diode appropriate for reverse polarity protection? To narrow the focus a bit from an earlier question : Is a Schottky diode appropriate for reverse polarity protection? I'd like to prevent mishaps from a user connecting DC power in reverse, but I'd also like as low a voltage drop as possible. Can you explain what the reverse leakage current is and whether it would be a concern or not in this scenario? The application is a small device that operates on 9-12 volts DC at less than 100 mA. Edit: Just as an example, I am expecting users to be able to use 6 AA cell batteries in series, either alkaline or NiMH. In the latter case, the batteries are 1.2V, so the total voltage is only 7.2V. I am using a 5V voltage regulator with a dropout voltage of 1.3V, so therefore my minimum operating voltage is 6.3V. A bias protection diode with 0.7V drop is going to raise that minimum to 7.0V. As the batteries are drained, I expect to dip below the 7.0V requirement very quickly, and therefore not use the full capacity of the batteries efficiently. If a 0.3V diode is used, the minimum requirement is lowered to 6.6V, which I feel is a better fit for use with NiMH batteries. <Q> The bigger problem you're likely to run into is operation under forward bias conditions. <S> Schottky diodes still have a voltage drop under forward bias, say 0.25V. <S> That means at 100mA, you're dissipating 25mW of power. <S> Better than a standard silicon diode, but not great especially for a battery constrained device. <S> A better way to get reverse bias protection is to use a P-Channel MOSFET. <S> MOSFET's act more like a resistor when saturated, and it's possible to get MOSFETs with low on resistances. <S> Let's assume we have a 1 ohm on resistance. <S> At 100mA, that's a 0.1 V drop across the MOSFET and 10mW dissipation. <S> 1 ohm on resistance is kind of lousy for a MOSFET, you can get some which have significantly less on resistance. <S> I'm not entirely sure about the leakage current through MOSFET's, but I seem to remember it being quite small. <S> To hook up the mosfet: Connect the drain to the positive battery terminal, connect the gate to the negative terminal, and connect your load to the source. <S> For added protection you can add a zener diode and a resistor across the source/gate. <S> A more complete explanation can be found here. <A> Reverse leakage current would only apply in the opposite polarity case. <S> This is current that is leaking through the semiconductor material when the diode is in the off condition because it is incorrectly biased. <S> Thus in your application, it is not an issue. <S> However, you are right to be wary of the forward voltage drop. <S> Only you can decide what an acceptable voltage drop is based on what your power source is and the voltage needs of your regulator or power supply are. <S> Since you are buffering your power supply using a voltage regulator, I would guess that a small voltage drop (less than 0.7 V) would be acceptable. <S> As an example, such a voltage drop might be important if the circuit protection was for a analog voltage sensing application instead. <A> I'd say that a Schottky is an excellent choice as it's forward voltage is less than what the ESD (Si) diodes on your semiconductors probably have. <S> You are then diverting this mistake, and it's current away form more delicate parts into something that is beefier/ more robust. <S> This spec will be in the datasheet and increases with temperature. <S> Reverse leakage current is simply the deviation away from the ideal case in which there would be zero current flow. <S> Having reverse leakage means it will eat some of your battery power. <S> At 100 mA it will be unnoticeable, but it is the offstate leakage that you care about. <S> Just make sure that the Schottky is AFTER the power switch <S> and it's leakage <S> will be moot at that point.
| Reverse leakage can be an issue with battery powered devices a Schottky's do have higher reverse leakage current (due in part to their lower forward voltage).
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Why can't resistors be used to connect cathode pins to GND instead of segment pins to VDD with a 7-segment display? When using a common cathode 7-segment display, is there a reason that two resistors can't be used to connect the two cathode pins to GND instead of using seven resistors to connect the seven anode pins to VDD? I'm assuming there's a good reason, because I haven't been able to find any schematics that do it this way, but I'm unsure of why this would not work? <Q> The purpose of the resistor is to limit current to a segment LED, by dropping the surplus voltage (over the LED Vf) across itself. <S> Thus, the voltage dropped would change as well. <S> This will lead to intensity of digits changing with number of lit segments. <A> You can, but you get uneven lighting in good situations, and in worst case, since LEDS (or any given diode) are not perfect systems, variations in them will cause one led to conduct more current then the rest. <S> Eventually it will burn out before the rest, which would lead to the same current to be divided by a smaller amount of imperfect diodes, which would lead to (rinse, lather, repeat). <S> See <S> Why exactly can't a single resistor be used for many parallel LEDs? <S> for more info. <A> Each resistor should only be used to control the current through one LED at a time. <S> If on each display one only wanted to drive one segment at a time, one could simply use one resistor to limit current to the display's common wire. <S> This could be a reasonable approach if e.g. one had a ten-digit display and wished to drive it with 1/7 duty cycle. <S> Most of the time, though, one will wish to energize one entire digit at a time, and would thus need one resistor per segment wire.
| When a resistor is used on a common cathode, the current through this resistor would vary by the number of segments lit at any time.
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What does the abbreviation IP in "System-on-Chip (SoC) infrastructure IP applications" stand for? I'm looking at a Texas Instruments datasheet, and noticed the phrase: " System-on-Chip (SoC) infrastructure IP applications " and was just wondering what IP stands for. <Q> As already mentioned by Chetan, it stands for "Intellectual Property". <S> This Wikipedia article should make the context clearer: <S> In electronic design a semiconductor intellectual property core, IP core, or IP block is a reusable unit of logic, cell, or chip layout design that is the intellectual property of one party. <S> IP cores may be licensed to another party or can be owned and used by a single party alone. <A> In this particular context IP stands for Intellectual Property <A> Many, many, many other things. <S> In this case, it's probably Internet Protocol. <S> Basically, they're saying their device is good for Internet-enabled applications. <S> Saying a device is "IP-enabled" generally means it has the processing power/integrated peripherals to handle communicating over the internet easily. <A> My best guess here is Intellectual Property as others have already mentioned. <S> If you want to know for sure, please provide a reference to the datasheet. <S> In ASIC design terms : modular, license-able, on-chip component designs can be referred to as "IP cores" that other companies can purchase and integrate into their own ASIC. <S> Since ASICs are described with HDL languages, ASIC components can be surprisingly portable across several different designs. <S> The reference to "SoC infrastructure" seems to suggest this, but again, please provide more context for a better answer. <S> Some examples of IP cores that you can buy and integrate into your ASIC design may be memory controllers, entire CPU's (Xilinx microblaze and others), peripheral blocks, etc. <S> There are companies that make entire businesses out of selling licenses for IP cores.
| IP can stand for: Intellectual Property Internet Protocol
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How do I pan in EagleCAD with an Apple MagicMouse I'm using a MacBook Pro with OS X Mountain Lion, but can't figure out how to pan the schematics? <Q> The center click on my Thinkpad doesn't work either because it uses some nonstandard scroll thing with the Synaptic driver where it's treated like scrolling on the touchpad instead of a true third mouse button. <S> You might want to give Kicad a shot instead <S> , it lacks a normal middle mouse pan function in favor of more zooming in and out that normally bugs me, but it could work out to be more intuitive with the MagicMouse -- the nonstandard middle mouse on my Thinkpad ends up working as a pan+zoom function. <A> This seems to be solving the problem okay for me: http://magicprefs.com/ <A> I run Eagle on OSX. <S> Eagle 5.x has no way to pan without a real middle button, but 6.x pans with the normal scrolling gestures. <S> I expect that a magic mouse would work the same way as a trackpad in this regard.
| You need a mouse with a standard no-frills middle click button for Eagle.
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Cellphone charger has no transformer? I opened up a really tiny mobile phone charger I have, to see how it is designed. The entire "charger" is integrated into a small 2-pin mains plug, 1 x 1.25 x 0.5 inch, that has an USB socket for the phone's USB charging cable. I could not find anything that seems to be a transformer anywhere in the circuit, and yet I have tested that it is an isolated 5 volt well-regulated output. The tiny flexible PCB has merely a dozen or so SMD parts, ranging from 0402 to 4516 (metric), plus the connectors at the two ends, for mains and for USB. The SMD parts all have part numbers sanded off. How do they manage the isolation in these chargers? Responses to comments: This is a no-name "Hi-Standard USB phone charger, Extra Powerful!" I have just bought in Korea, that is supposed to work with any USB-charged cellphone. They have pictures of a half dozen different cellphones on the box, and a hydra USB cable inside that has mini-USB, micro-USB and some other types of connectors. I bought it just to see whether it is safe or not. That is why I tested for isolation first. <Q> Although the question has provided limited details, this answer presents a somewhat different hypothesis from the standard assumption that there's an inductive coil hidden in there somewhere. <S> Does the charger looks somewhat like this? <S> If yes, the designers have used a Piezo transformer instead of a conventional one. <S> Interestingly, the source of this image is a paper in a Korean academic publication. <S> This makes the hypothesis even more apt. <S> A piezoelectric transformer designed for Mhz operation, 500 mA secondary current with 5 Volt signals, using Polyvinylidene Fluoride (PVDF) as the piezoelectric medium, could be fabricated as a thick 1210 SMD part. <S> Since the question mentions SMD parts up to 4516 metric i.e. 1806 imperial, one of the largest of those components is probably the piezoelectric transformer providing the isolation as per the question. <S> Some interesting information gleaned while investigating this mystery <S> charger: <S> Piezoelectric transformers deliver 80% to (recent experimental versions) 90% efficiency, impressive in transformer terms <S> These transformers can provide galvanic isolation at multi-kV levels - of course, not in a SMD 1210 size, where the contacts would be too close together. <S> PVDF exhibits piezoelectricity several times greater than quartz. <S> Hence it is ideal for making Piezo transformers. <S> Many LCD display CCFL backlights are made using Piezoelectric transformers instead of the inductive coil ballast used in earlier versions. <S> So it isn't really new technology. <S> Equipment used in magnetism-sensitive areas (e.g. MRI labs) are expected to transition to non-magnetic electronics, hence Piezo transformers where a transformer is needed. <S> ( n.b. <S> Any current flow, however, will still generate some magnetism courtesy H fields ) <S> Some articles of interest: Piezoelectric Transformer With Very High Power Density <S> Panasonic's Piezo transformers for LCD backlights - Of course, much bigger than the dimensions usable in USB chargers. <S> A comparison of magnetic and Piezo transformers by Texas Instruments <S> Full disclosure <S> : I have never worked with, or even seen Piezoelectric transformers before today - the above information was a new learning for me, in the process of investigating the mystery charger . <A> The isolation is managed by a transformer. <S> However, it's just very small; it's probably one of the bigger surface-mount components you see in there. <S> See How do these new-fangled phone chargers achieve step-down? <S> for a more detailed description of the technique involved. <S> Note the transformer in the middle of the circuit. <A> How do they manage the isolation in these chargers? <S> A high-frequency transformer can be very small, especially when triple-insulated wire for one of the windings. <S> so the 2mm clearance needed for a 150VAC circuit rated at pollution degree 1 (since the charger is essentially a sealed box) allows for very small construction. <S> I bought it just to see whether it is safe or not. <S> That is why I tested for isolation first. <S> If you want to judge safety, the first step is to check for any regulatory agency marks on the package (UL, TUV, CSA, etc.) and cross-reference the file number with the respective agency to make sure it's not fraudulently applied. <S> If the device doesn't have any safety marks, or has a fraudulent mark on it, the onus is on you to establish if it's safe to use in your jurisdiction (by getting a special inspection certificate for the device, for example) which means the prudent thing to do is <S> throw it away and buy a safe device given the price of the device vs. the price of a special inspection. <A> It comes on in the dark and goes out in daylight. <S> It contains a 47muF capacitor in series on the active wire followed by two diodes and then a NPN transistor controlled by the light sensitive resistor. <S> The LED is across the transistor and that's about it. <S> There is a electrolytic capacitor across the diodes for smoothing and a 70K resistor also in series but in parallel with the 47muF cap. <S> It works. <S> There is a detectable 50hz blinking but if you don't look away, the LED appears steady enough. <S> So there's a DC couple of volts from 240V AC definitely with no transformer.
| The charger in question possibly uses a Piezoelectric Transformer instead of the magnetic (inductive) transformers usually seen for isolation. The device is most likely only designed to work at 120VAC If there is a valid safety approval, the device would have been subjected to isolation and other safety tests before release to market. I just bought a "night light" which is an LED that is in a 240V plug.
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What thickness of wire should be used to power Arduino on 9 meters? I have built a standalone Arduino device which has a few buttons, LED's and one servo. I also have a power source which on the casing says 4.9VDC ~ 700mA. (Actual reading is 5.7V). This power source worked for a whole year non-stop powering my Arduino UNO at 9 meter length with two solid core wires. [See below.] Now I built this device to be standalone and also to use a more neat wire. If I use the dual-wire the device gets powered, but malfunctions. If I connect the device directly to the source and skip the extension, it works fine - this brings me to the conclusion that is in fact the wire. The voltage on the other side of the extension seems normal, so I guess it is the current that drops to below required? What type or sort of wire can I use to prevent to much power loss and to power my device at 9 meters? - obviously as thin and neat as possible. UPDATE I investigated the "malfunctioning" and it seems as if the board resets the every time when the servo moves. As if the servo is draining the current or if the servo causes a major voltage drop. Voltage on both wires are the same on the end as at the source without anything connected and with the device running (without the servo moving). The Power source: The old but working solid core wire: The new but not working dual wire: <Q> Your servo is pulling too much power from your Arduino every time it moves, causing your Arduino to reset. <S> You can use some smoothing capacitors to store some extra charge for quick use (although I've never tried this myself), or you can hook up an extra set of batteries/power adaptor for the servo. <S> Connect the +ve and -ve power from the extra batteries to your servo, and the -ve of the extra power to the -ve of your original power supply <S> (so that the servo positioning signal from the Arduino can ground back to the original supply). <S> I have a photo of how I have this set up on my site , along with a description of how I set it all up. <A> I wouldn't be so quick to conclude the problem <S> is the wire. <S> A fatter wire may have a lower resistance and mask more fundamental design issues. <S> Have you included any capacitors to filter your input power? <S> Have you inspected your power rails with an oscilloscope? <S> A thinner or longer wire, with its higher impedance, will degrade the voltage regulation and transient response of your power supply (since it can't know what's happening at the far end of the wire), but with sufficient filtering, it's unlikely to cause the serious issues you describe unless you really use the wrong wire. <A> So thicker cross-section (i.e. AWG) wire should give you less resistance. <S> A good reference for wire gauge recommendations for current requirements is http://www.powerstream.com/Wire_Size.htm . <S> So based on that, I would recommend no thinner than 22 AWG wire for power transmission. <A> As vicatcu mentions, a wire with a smaller cross section will have higher resistance. <S> The resistance of a wire (with a given cross section) increases with length. <S> The voltage drop across the cable is given by Ohm's law as V = IR. <S> So, the voltage drop is current dependent. <S> To see the voltage drop effects at the far end of the wire, you must measure the voltage under load. <S> Perhaps you did not see a voltage drop because your Arduino was unconnected or was not running under full load when you measured the voltage. <S> It is also possible that your device presents a time varying load (pulses for the servos?). <S> Load current spikes could cause quick voltage drops that your meter may not be able to capture. <A> Regardless of whether the wire registance is a problem or not, I think it is worth mentioning, that given equivalent wire size (thickness), stranded wire (paradoxically, perhaps) conducts better than solid wire. <S> Therefore, in conductance calculations, whether the wire is solid or stranded needs to be taken into account as well. <A> You have a classic "ground loop". <S> The high current to the motor is running on the same wire as the Arduino supply, and when the motor causes a big voltage drop the Arduino resets. <S> Solve it like this: Make two runs of wire to the adapter. <S> Power the Arduino with one, and the servo with the other. <S> Then, the wire to the servo can drop several volts without causing the Arduino to brownout. <S> This gets your circuit functioning without requiring giant wire gauges or similar heroics. <A> You would be better off running a higher voltage over the long cable run, and using separate local regulators for the servo and arduino (the latter already has one on board of course). <S> A modification of this idea would be to run the AC mains to the device (with code compliant wiring of course!), and place the power supply there. <S> And perhaps choose a more appropriate power supply while you are at it. <S> If you really want to run the arduino off that supply, you might consider switching to a 3.3v regulator and lowering the clock speed accordingly.
| Thicker wire has lower resistance per unit length. Unless you are bypassing the regulator, ~5v is well below the designed input range of the Arduino.
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What does this schematic symbol mean? I'm trying to understand what the following symbol is for. It's found in this schematic . <Q> It protects the device from static voltages that may develop when inserting or removing USB devices to connectors. <S> The four diodes (on side) act as clamping diodes that conduct to suppress ESD. <S> This or similar device can be found in this Semtech catalog <A> Looks like electrostatic discharge (ESD) protection on a USB port. <S> Any port that a user can touch should have protection against finger zaps frying the chip connected to it. <S> Have a look at the datasheet for TI TPD2E001 to get an idea what the specs might be. <S> http://www.ti.com/product/TPD2E001 <A> it is 5 symbols enclosed in one device. <S> 4 normal diodes in a bridge configuration with a zener diode in the middle for simple regulation.
| This device is a transient voltage suppressor for USB D+ and D- signals.
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How to connect multidigit 7seg common anode/cathode display to cascading shift register? How can I connect this type of 7 segment display to cascading register like 74HC595? I know I can use 7219, but at the moment I have those cheap registers and would like to see if it's possible. Not looking for MCU scanning solution! The problem for me is that segments are interconnected... What is the simple way to connect? UPDATE: I've found the following in google. Is it close to what I am trying to achieve? There is only 1 shift register instead of 4 though. UPDATE 3: Enlarge <Q> It can be done with the 74HC595, but you will need an extra chip or ic. <S> You will need to drive both the common anode for each display, and the OE (OUTPUT ENABLE) pin on each shift register at a separate time. <S> The OE pin allows you to disconnect the output pins without clearing the shift register. <S> Creative use of this allows pwm, or in this case, by cycling through through a digit/shift register pair, allows you to implement scanning. <S> The easiest way to do this would be a 555 timer driving a decade counter (or ring counter). <S> The 555 would provide a clock signal, while the decade counter loops through the four shift register OE pins. <S> Can't be done without scanning. <S> Especially not with a simple shift register. <S> Not without cutting open the display and somehow wiring individual cathodes for each digit. <S> Those multidigit displays are designed for scanning in order to save on pins. <S> It sacrifices software efficiency for hardware resource reduction. <S> What you need is a LED Display Driver ic, like the MAX6965 LED Driver with PWM Intensity Control (I2C) or <S> ICM7211, <S> ICM7212 Four Digit Display Decoder/Drivers (Shift Register like) <S> (Of course, check datasheets first before you buy. <S> You need a common-anode driver) <S> Siemens provides a detailed appnote on how to interface to 7segment displays Drivers For Light <S> Emitting Displays Appnote 24 including a list of drivers suited for this. <S> The other option is to take a second microcontroller like a attiny and make your own driver, implementing it's own scanning so the main arduino doesn't need to do it. <S> Inverting Transistor setup. <S> Uses a weak pull-up to disable OE pin when the base has no current, pulls to ground when the base has a current. <A> They are assuming you would only have one of the 7-segments sections powered at once. <S> You will need to sequence through powering the 3 sections fast enough they all appear to be on at once. <S> Your circuit for shifting appears good. <A> I've written a tutorial on the theory behind cascading the 74HC595 for a clock. <S> I am not entirely clear how you are driving the 7-seg displays, but it should be helpful in explaining how to get a desired output from the chip. <S> From this you can wire it appropriately to your driving mechanism. <S> :edit: Looking at your circuit, since you are required to multiplex <S> , perhaps you can use 3 more bits on the registers to drive the anode of each digit. <S> Seems like you could just use two registers cascaded, one for the eight cathodes and one for 3 anodes. <A> May I point out that if there is a common anode/cathode for each digit and all segments have the same pins like almost every multi-digit display, there is no way to avoid scanning. <S> A long time ago I remember a special BCD encoded LED display that you sent it the actual BCD of the digit <S> and it took care of the rest. <S> But I could not tell you if they are even still manufactured or even if they have multi-digit versions. <S> I think your best bet is either have <S> an ATTiny receive the data and take care of the display with either single digit/shift registers pair per digit (no scanning) or just use 2 shift registers with the multi-digit display (scanning which you cannot avoid).
| If each digit is separate digit and has its own set of segments and the segments are not tied together, then you could use a shift register for each digit and cascade them without scanning as you want.
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How can I drive 12v LEDs from my Arduino Nano? Background I am building a scale RC helicopter and ordered a bunch of 3mm LEDs to give it some nice navigation and strobe lights which I plan to drive with an Arduino Nano. Problem is that I goofed and didn't notice that the LEDs are 12v until they arrived. I know that the Arduino Nano can accept 12V on pin 30, but I think the voltage output on any of the "D" or "A" pins is always 5v (correct??). Question If the voltage output of the Arduino pins is 5v, then how can I drive the 12v LEDs using the Arduino, or is it impossible? <Q> If driving only one LED, use a transistor. <S> Otherwise use a ULN2003 IC to drive multiple LEDs. <S> Better off, order the normal LEDs. <S> You will save area on your RC device. <S> Here is a video <S> that will hep you using ULN2003. <S> In the video ULN2003 is used to drive relays, but you can replace them with LEDs. <S> Using a transistor: <S> Using <S> ULN2003: <A> I'm not aware of any individual LED which is actually a 12V device. <S> Usually, a "12V LED" is an LED plus some circuitry (most usually just a resistor) designed to make it trivial to drive by applying <S> only 12V. Can you locate this resistor, remove it, and replace it with a wire, or maybe a \$0\Omega\$ surface mount "resistor" designed to do exactly this? <S> Once that's done, it will be an ordinary LED, and you can drive it like any other. <S> I'd explain in more detail, but seeing that driving LEDs is an especially common application for an Arduino <S> , I doubt it's necessary. <A> Using OHM's Law. <S> Assuming the leds are set for I = 20ma current and use Vf = 2.4 to 3.2 volts (you didn't provide a color, but the voltage difference is minor) at Vs = 12v <S> the Leds would have a resistor close to 450~480ohms. <S> R = <S> (Vs - Vf) / <S> I. <S> If you use that same resistor with a lower input voltage (5v as you mentioned), you need to calculate for I. I = <S> (Vs - Vf) <S> / R. <S> (Using average numbers) I = (5 - 3) / 450, I = 0.0044A or 4.4ma. <S> Using those leds with the 5v Arduino output, the leds would work at ~4.4ma. <S> They won't be very bright, but they should work. <S> The other option is to replace the resistor if you can. <A> It might be too complicated for use in this situation, but in theory you could use this circuit I came up with a while back: <S> It was intended to boost 3.3 volt supplies to run white LEDs, but could also boost 5 volts to run a "12 volt" LED. <S> A1, A2, A3 are sections of a hex inverting schmitt trigger like the 74HC14. <S> Drive the end of R2 from a microcontroller pin to turn the LED on and off.
| You cannot directly drive 12V LEDs from arduino.
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How to connect mic on arduino, with opamp? I'm trying to connect a contact mic to an arduino.My knowledge in electronics is sparse, so please bear with me. I got a couple of lm833n audio dual opamps, which I'd like to put to use. I'm still struggling to understand opamps, though. I've been reading a instructable on audio input on the arduino. It uses the following circuit: However, this is for the TL072 opamp ; I'm not really sure if I can use the lm833n in its place? And is this circuit suitable for a contact mic? Also, it would be great if anyone could point me to a good (easy) introduction to opamps - the wikibooks page is a bit dense for me <Q> I know it's old, but may still interest someone. <S> With a piezo contact mic, you may want to also connect two diodes allowing current to go from ground to mic output and from mic output to vcc. <S> Under normal conditions there is not voltage over them, but a piezo can create some serious voltage spikes if hit, and this will protect your amp / arduino, so excess voltage can go to power supply rather than amp. <A> It will work, and Oli's suggestion of a capacitor is a good one. <S> I'd point out a couple more things: <S> You don't have any capacitor to block any DC offset on the input. <S> This is fine, if you are going directly to your microphone <S> and you know there won't be any. <S> However, if this is going to a connector on the box, you never know what people will plug in there, and it may work fine, until it doesn't, and your amplifier is fried. <S> It's probably best to go for a non-polarized type here, since you don't know what people will plug into it. <S> Another issue: "Contact microphone" usually means a piezoelectric microphone . <S> These are different from most other microphone types in that they have a very high output impedance, on the order of \$10M\Omega\$. <S> The input impedance of your amplifier is an order of magnitude less, which will result in significant attenuation of the signal and alter the frequency response of the microphone-amplifier system. <S> It will work, but it may not sound good . <S> This is of course, subjective and depending on the timbre you desire. <S> The solution is buffering , converting a high impedance output into a low impedance output, or equivalently, amplifying the current. <S> An op-amp can do this in a circuit called a voltage follower : <S> Put this between your microphone and the input of the circuit you already have (on the microphone's side of the cable, if you can), and your amplifier's input impedance will be that of whatever op-amp you use, without otherwise changing the operation of your circuit. <S> Oddly, I don't see an input impedance listed in the LM833N datasheet, but since it has a BJT input stage, it's probably some megaohms. <S> This is "high", but not higher than the piezo. <A> Yes, the LM833N should work fine in it's place. <S> The circuit looks okay for a basic mic amplifier - I'd probably add a capacitor across the 100k resistor though, to roll off the gain at higher frequencies (i.e. > 20kHz) <S> 50-100pF should do, but don't worry if you don't have one in this range, it will likely work <S> okay. <S> EDIT - note Phil's point about the piezo issue - if your mic is a simple passive contact mic (with no battery powered preamp) then use the buffer he suggests. <S> Looking at the link I notice the original circuit is intended for a dynamic mic, which have a far lower impedance <S> The formula for the -3dB point (0.707 of the initial voltage) for the filter formed with the capacitor added is: \$\dfrac{1}{2 \pi R C}\$ <S> so: <S> \$\dfrac{1}{2 \pi \cdot <S> 100\cdot 10^3 <S> \cdot 100 <S> \cdot 10^{-12}} = <S> 15915Hz \$, which is fine for most audio purposes. <S> A good intro to opamps is "Opamps for Everyone". <S> Edit: TI appears to no longer be hosting the PDF version of "Op Amps for Everyone" but Google finds a number of versions that are still out there. <S> I'm not sure about the propriety of linking to them, so I haven't included a direct link, but this search should get you started.
| You'd want to look for an op-amp with a MOSFET input stage with a very high input impedance: TL072 is a common such type with the datasheet listing input impedance at \$10T\Omega\$.
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What is this component, and how can I use it? Please see the attached photograph of the component: My guess is that it is a solenoid, but I am not sure. If it is a solenoid, I would really like to use it. The problem, however, is that I do not know anything else about the component. There are no part/model numbers or any identification of any kind on the component. In a situation like this, how do you go about putting a mysterious component to use. I generally don't throw away things simply because I don't know what they are, and am sure I could use this for something. <Q> It looks like an electromagnet. <S> Start with a DMM - check the coil winding resistance. <S> From there, you can probably make some educated guesses about where to start testing using a benchtop DC supply. <S> Slowly increase the supply voltage. <S> Keep tabs on when it starts to get hot, and you've likely found its continuous current limit. <A> After searching, I am quite certain it is a holding solenoid, similar to those used to hold doors open, or to lift piles of metal/cars at a junk yard. <S> Compare to the pictures below. <S> Near identical in operation as a linear solenoid (The push/pull rod type). <S> You will need some ferromagnetic metal (like carbon steel - stainless steel will not do) to interact with it, or use a compass to verify it is working. <S> Practical uses are to hold a door open. <S> Impractical would be to make a miniature magnetic crane game, or scale sized scrap yard crane remote controlled car, or rig a dart board. <A> An LCR meter (Inductance Capacitance Resistance) is a great tool for testing unknown components such as this. <S> I have a Mastech MS5308 (can be found for £133 on eBay ) and can highly recommend it. <S> For the price it is comparable with more expensive meters (from e.g. Agilent, BK Precision, etc) comes with proper 4 wire kelvin clips, 4 wire <S> SMD tweezers, an isolated USB link and case. <S> It tests at frequencies of 100Hz, 120Hz, 1kHZ, 10kHz, and 100kHz (the more test frequencies the better - many lower quality meters only have a couple and do not test at 100kHz, which is the frequency most capacitors are tested at for the datasheet specs) <S> It has a minimum resolution of 0.01pF, 1nH and 1mΩ which is very impressive indeed, tests ESR (Equivalent Series Resistance), Q, DF, θ and has a sorting/tolerance mode you can use to quickly find parts that are out of rated tolerance. <S> For comparison the Agilent U1731C only tests at 120Hz and 1kHz, and has a minimum resolution of 0.1pF and is over £200. <S> Anyway, you might guess I'm extremely happy with mine, but have a browse around before deciding just in case <S> I'm talking rubbish :-) ) <S> Also, you could carefully peel a little of the covering away to reveal the windings (assuming it is a coil) <S> When you do apply power, preferably use a current limited bench supply and ramp up slowly. <S> Try swapping leads around to see if this makes any difference (just in case there's a series diode hidden in there somewhere) <S> A function generator may also be of use, possibly combined with an amplifier in case it requires AC drive. <S> Having said all the above, it does look like either an electromagnet or part of a solenoid/relay assembly (but it could be a buzzer, or...)
| Depending on what test equipment you have available, before you apply any power, you can test it's resistance, inductance or capacitance (probably the first 2 are most useful in this case since it looks to be a coil of some sort)
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What is the minimal set of parts for a circut with this AVR microcontroller? My Arduino Uno has a ATMEGA328P-PU microcontroller, and bunch of other stuff on the board. I'd like to program the chip without the Arduino software, and only the minimum of other components. I want to eventually create things without the cost of Arduino, and I want to learn about the other parts on the board and add them back as as needed. I'm comfortable with C and gcc, so I can probably figure out the software part. But what, if anything, besides these two products below, do I need to have in the breadboard with the microcontroller? Atmel Programmer Microcontroller <Q> All you need apart from the programmer and the chip is a couple of decoupling caps, and some way of connecting the programming signals to the breadboard. <S> So: 1 largish electrolytic cap (e.g. >100uF) <S> You can maybe do without this if your source is nice and quiet (e.g. battery) <S> 1 100nF ceramic across the power pins of the micro 1 10kΩ resistor to connect from the reset pin to Vcc to hold the micro out of reset. <S> (you can buy ready made jumper wires, but I use a 22AWG roll and cut my own - much cheaper if you do this a lot) <S> A header to plug your programmer cable into. <S> According to the user guide <S> it looks like you need a 2x3 pin 2.54mm pitch header. <S> Optional <S> A button to connect your reset line to ground if you wish to physically reset the chip (I don't use AVRs, but I'm 99.9% certain the programmer can do this from the IDE) <S> A crystal to use instead of the internal oscillator <S> 2 * 22pF capacitors for the crystal (place from either side to ground) Passive components, LEDs, sensors, etc in order to do something useful with the code you write <S> ;-) <S> Apart from the header (which is 10-pin with some unused pins - you can use the labels as a guide for your connections, the names are the same) <S> this schematic is about the simplest I could find with a quick Google: <S> Funnily enough <S> I just wrote a blog on doing pretty much the same thing with a PIC microcontroller (may be worth a look, the two are very similar) <A> I'm only posting this as an answer to hide the ugly URLs. <S> You should certainly have a look at Atmel Appnote AVR042: AVR Hardware Design Considerations . <A> I also found this helpful link from the Arduino folks: https://www.arduino.cc/en/Main/Standalone <S> It explains how to go from a chip you just got from DigiKey (or where ever) to loading the bootloader and the bare minimum support components.
| Some jumper wire to connect up nodes on your breadboard
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Is telephone and LAN wiring symmetrical? Is telephone and 10/100 MBps LAN wiring electrically symmetrical? I mean is it allowed to swap wires in the pairs, or is one of them like GND so they cannot be swapped? It's important to know when I plan longer wires if I can use just two wires of the same color. I am unable to find some information on how telephone and 10/100Mbps LAN communication works on this physical layer because I don't know the correct technical terms I should search for. (I am not talking about symmetrical speed of digital lines.) <Q> You cannot swap wires in the pairs for either, but for different reasons. <S> In telephony, the AC voice signal is overlayed on a DC supply, which is used to power telephones. <S> Research starting point would be hybrid . <S> If you filtered out the DC part, and only transmit the AC part, you could actually swap the wires. <S> In 10/100 Mbit <S> /s <S> LAN <S> you cannot swap the wires, because the signal on the wires is differential, the signal of the second wire gets subtracted from the signal on the first wire. <S> Start out here: physical layer <A> I am a telco repairman. <S> ALways go with the convention. <S> Standard is Blw = first line Orw = second line Grw = third line Brw = fourth lineEthernet = <S> Blw and Grw = standard for data and Orw and Brw are spares, unless using Cat 6 or something exotic. <S> You can lay wire on the ground and carry voice a LONG way when necessary (miles). <S> The pairs contain different values of twist. <S> VDSL is preferrred to be wired from the NID (protector) to the modem location on the GrW pair. <S> The twist has been determined to be the best for VDSL applications. <S> From experience whatever you need to repair a circuit is what you do for service. <S> Changing pairs is not an issue. <S> But keeping continuity isolated and straight ahead is the key. <S> You cannot combine circuits. <S> That is a crossed line. <S> Buzz and humm will be your problems as well as shorts. <S> As you should know ethernet has a 350 foot limit. <S> However it can be longer, it just isn't shown in the specs and you could have a problem tommorrow after it works today. <A> The wiring is not symmetrical. <S> However most phone systems are indifferent to the polarity of the line. <S> Some older PABX and enhanced PSTN systems did require the correct polarity before enhanced functions would operate. <S> The bulk of modern electronic phone line termination circuits incorporate a bridge rectifier before the DC clamping/hook/flash circuits <S> so do not know which polarity is connected. <S> 10 and 100 Mbps UTP <S> LAN cabling makes use of a RJ45 8pin connection. <S> Only 4 of the circuits are used even if all 8 are connected. <S> The pinout requires the transmit pins of one device to connect to the receive pins of the other device. <S> This requires proper pin selection. <S> The transmission is a differential drive so the transmitter has a positive and a negative pin and these must match the appropriate positive and negative on the receive side. <S> The LAN interfaces were made with the transmit either on pins 1 and 2 and receive on pins 3 and 6 but on devices like network switches and hubs the transmit was on 3 and 6 and receive on 1 and 2. <S> This allowed straight through cabling in this standard configuration. <S> The cable pairs had to be on 1 and 2 and the other on 3 and 6. <S> Now when there were more unusual connections cross over cables were required. <S> All this was a bit confusing so automatic detection of which pairs were connected to which pins was implemented. <S> Modern devices appear to also tolerate crossing of positive and negative as well as swapping of pairs. <S> Modern Gigabit LAN uses 4 pairs and does not need cross over cables as it establishes the wiring in the hardware and communicates appropriately. <S> I am not certain if the standard requires them to ignore pair crossing but this may be possible. <S> So the answer to the LAN question is modern systems also seem to be increasingly indifferent.
| If you switch the wire, you reverse the polarity of the signal, effectively munging the bits.
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Prevent overcurrent while filter capacitor is charged I have this circuit: On start-up, the power supply experiences an overcurrent condition, because when the capacitor is charged, it looks like a short circuit as seen from the power supply. How can I prevent this? Adding a current-limiting resistor before the capacitor is not suitable for me, because of the load-dependent voltage drop - the load, of course, will look smaller once the capacitor has charged at the maximum current of about 2 A. Maybe there is a simple circuit to temporarily enable a current-limiting resistor? Or another simple solution? ADDED. I have these ideas. Please feel free to comment. <Q> Several manufacturers make what are called hot-swap controllers which essentially act like programmable resistors (using MOSFETs) to apply a load to a power supply output in a controlled manner, independent of the power supply itself. <S> This allows for gradual charging of capacitors without overwhelming the power supply, plus once the caps are charged, the low \$R_{ds(on)}\$ of the MOSFET makes the losses manageable. <S> Most of the time, there's no need for a series current limiting resistor when using a filter capacitor. <S> That being said, some types of capacitors (tantalum and some organic semiconductor types) do need charge/discharge current limiting. <S> You haven't mentioned the wattage of your power supply or how much capacitor you're trying to charge - this information is obviously needed to judge whether or not the cost and complexity of a hot-swap controller is justified. <A> I think what you are looking for is called an "inrush current limiter". <S> Here's a search on Digi-key that you might find what you are looking for. <S> And here is a Wikipedia article on the subject . <A> If your ciruit has a "reset chip", and if it won't draw much current while in reset, I'd suggest perhaps using a MOSFET to turn input current on and off, but with a moderate-value resistor to allow some current through while the MOSFET is off. <S> Until the capacitor has charged up to the reset threshold and been above that point long enough for the reset chip to be happy (meaning it will have had time to charger further above the threshold), the current into the device will be limited but the device itself won't use much current. <S> Once the reset is happy, the MOSFET will provide an excellent conduction path, thus allowing the device to make good use of the supply current. <A> A solution from the old days was to put a resistor in series, and then use a relay to short-circuit the resistor. <S> The relay coil goes across the capacitor, so that the relay pulls in after the voltage has risen to a suitable level. <A> Another solution is to specify a DC power supply with current limiting (say at 3A, if your load is 2A) instead of tripping when it senses overcurrent. <A> Usually, you try to make sure that your capacitor is smaller than the maximum output capacitance allowed for your supply. <S> Some DC-DC converters allow as little as 100 µF. <S> If the output characteristic of your supply is of the fold-back type, this is a very important piece of information - and you would need a supply that doesn't shut down or hiccup during over-current events if you want to use any size of capacitor. <S> If there is no such rating in the data sheet of your supply, you are relying on pure luck, even if you are experimenting with series resistors. <S> The questions it always boils down to: <S> Will your power supply go into shut-down, and how fast will it go there? <S> If it shuts down faster than it takes you to charge the capacitor and obtain a low current, it won't work.
| Well-designed power supplies get around capacitor load charging in a few ways: a pre-charge, mate-first contact that has a resistor or NTC thermistor in series with the contact; this path pre-charges the cap through the resistance and gets shorted out when the other power pins mate a soft-start feature where the output voltage ramps up over tens or hundreds of milliseconds, decreasing the capacitor charging current a brick-wall or constant-current mode of operation; when there's heavy load, the power supply goes into current regulation mode (it keeps the output current constant and lets the voltage drop out) which will charge any capacitors in a controlled manner Since your power supply doesn't appear to have any of these features, you need an external solution.
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How do I know I'm not drawing too much current for my 3.3V supply? Continuing my learning expedition, I've decided to use the LM3S5R36-IQR80-C3 ARM Cortex-M3 MPU, to try and learn to build my own little experiment PCB, almost like the Teensy 3.0. The idea is to learn the entire process of designing the PCB from scratch. I've come up with the following circuit so far, working on ESD protection for the USB port, as well as the 3.3V supply for the Vdd pins on the ARM chip. I'm a hobbyist, so I have no electronics degree or background, so I'm using the datasheets for the ESD array I selected, as well as the voltage regulator , to hopefully get a steady 3.3V voltage supply from the 5V USB Vbus. Here's my circuit so far (if it's broken, it's coz I had no help, except Google) : I'm not sure if I can ask here whether the circuit looks right or not, I am looking for some constructive criticism, however. So feel free to comment, or refer me to a site where I can have my circuit scrutinized by the community. More, importantly, let me state my actual question... My next step is to start connecting the 3.3V supply from the regulator, to the ARM chip's Vdd and Vddc pins to the 3.3V supply. The thing I want to know is, how I can be sure I am not pulling too much current through the regulator or the USB 5V Vbus . The last thing I want is to sit with a bunch of scorched spares and a broken USB port on my Mac. There are 4 Vdd and 4Vddc pins, so I understand I can get the 3.3V to each pin, by simply connecting them to the supply in parallel. This however, depending on the resistance the chip causes between the 3.3V and GND, makes it tricky, and I'm not sure how to go about making sure I don't draw too much current. I guess it comes down to how much current the ARM chip sinks, but I'm not sure how to effectively work this out. I hope my question makes, sense, feel free to edit to clarify it. <Q> The largest value I can find in the datasheet for the ARM is 126ma when running flat out and peripherals on. <S> This is unlikely to cause problems. <S> If you're very worried you can buy desktop power supplies with current limiters. <S> At the very least, when powering it up for the first time use a phone charger not a computer and have a suitably-ranged current meter in the positive line. <S> Are you intending to connect the ARM to the USB? <S> There will be a reference circuit for that - use it. <S> That bit also requires careful PCB design. <S> I would start with a reference design for the Stellaris - there's a cheap dev board available for it - and work through understanding why everything is there. <S> If you decide you don't need it, remove it. <S> (These parts aren't as forgiving and easy to use as PICs / AVRs, and therefore less likely to produce a working circuit) <A> If the USB port follows the USB standard, it is current limited. <S> There is a default minimum current limit, and USB device need to negotiate for more as part of the enumeration process. <S> Even with the extra current negotiation, there is a reasonable maximum. <S> The USB port should be unable to output more current than that. <S> My windows OS warns me when something is trying for extra, but I'm not sure if you'll see the message on a mac. <A> According to the spec, you can enumerate yourself and request up to 500mA in USB 2.0 spec. <S> However, these limitations are rarely enforced physically, meaning that usually you will be able to draw until some circuitry in the PC limits you. <S> So, in general, less than 500mA is safe. <S> The voltage regulator you chose is an older regulator that has very little short circuit protection, reverse voltage protection, overheating protection, etc. <S> I suggest you select a more modern part that can do what you need and includes all of these things. <S> Note that the microcontroller will likely need direct connection to the USB VCC.Selecting the 3.3V regulator, you need to take into account <S> primarily output current and dissipation (Aside from the obvious fixed voltage output and a Vin that includes 5V), although for 500mA it's not that big of a concern. <S> Your circuit needs a fuse as well. <S> Select a fast acting fuse. <S> Also, I didn't see the typical ferrite bead and decoupling capacitors that are usually added. <S> Your microcontroller likely needs direct connection to 5V bus to be able to operate with USB, so check the reference circuit for the part you're using. <S> Finally, if you're that worried about your Mac at first, power the board using some AC to 5V converter like the ones used for charging phones. <S> Measure the current and make sure that the it's ok. <S> Also, always measure whether there's a short circuit between the USB Bus VCC and GND before connecting it to anything. <S> The ARM processor you're using will be much less than the limit. <S> Look at its datasheet on page 1013, it tells you that at 80MHz fully running it is 90mA. Add the extra stuff and GPIOs to it and USB is plenty.
| Technically For USB, you should typically draw no more than 100mA, unless you're enumerating the device and specifically asking to draw higher current.
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Is it valid to consider back-EMF in a DC motor equivalent to increased inductance? I know that back-EMF can be considered as a voltage source in series with the motor which is proportional to speed. This is the common understanding, and I totally get it. Before I understood this, I developed an alternate explanation on my own, and I wonder if it has any validity. Think of this: an inductor resists change in current. A bigger inductor resists it more. A stalled motor resists change in current. A spinning motor resists it more. A small inductor at a given current has some stored energy. A bigger inductor at the same current has more stored energy. A stalled motor at a given current has some stored energy. A spinning motor at the same current has more stored energy. Hopefully you can see what a student might intuitively hypothesize: a motor's windings exhibit an inductance that increases with the motor's speed. Not because it's magically growing more turns of wire of course, but perhaps it's a sort of mechanical inductor, storing energy in the motor's momentum, rather than in a magnetic field. My intuitive understanding of an inductor is, after all, a flywheel. Maybe this is an inductor that actually is a flywheel. Can this analogy be stretched further? In a resistive and inductive load, AC current lags behind AC voltage. Add more inductance, and current lags more. In a motor, current lags behind voltage. If the motor is spinning faster, does it lag more? And if that much is true, can it be shown that back-EMF is equivalent to an inductance that increases with motor speed? And if not, why? Intuitive examples would be appreciated first, then the math. I never seem to understand when presented in the opposite order. <Q> Interesting. <S> The back-emf (modeled as a voltage source proportional to speed) is not equivalent to an inductance that depends on speed. <S> Furthermore, there is no possible L(w) <S> you can come up with that will make that assertion true. <S> I will describe a simple experiment, but in essence I'll be saying that they can't be equivalent because upon a motor load change, an inductor dependent on speed L(w) will not affect the stationary state current (torque after all transients have died down, becoming a contradiction), while a voltage source dependent on speed v(w) will (which makes sense). <S> Assuming a DC motor, a simple proof is to imagine that the load on the motor gets reduced. <S> Because there is less load, the motor speeds up. <S> Also imagine we wait for some time so that all transients go away <S> (t=inf.). <S> Now let's see what happens with both models: <S> With the back-emf modeled as a voltage source, its voltage increases because speed increased. <S> This means that the current decreases, because the difference between the power voltage source and the back-emf voltage got smaller. <S> This means torque decreased, which makes sense because we reduced the load on the motor. <S> On the other hand, no matter what inductance value you give to the "back-emf inductor", the current on the motor would remain the same, because inductors are short-circuits in dc. <S> But this does not make sense, because torque is proportional to current and if the current remains the same, torque remains the same, but we started this analysis saying that we reduced the load on the motor. <A> An ideal motor may be modeled as a "transmission" between the electrical and mechanical sides, with a "gear ratio" of "k volt seconds per revolution" for some constant k. <S> Just as a mechanical transmission bidirectionally commutes changes in one side's torque or rotational speed to changes in the other side's torque and rotational speed, so too with the motor. <S> A normal transmission scales by a dimensionless quantity, but that doesn't pose a problem. <S> I can't figure out how to make Google's dimensional analysis to work with torque, but one assumes that a motor drives something some particular distance from its shaft, one can then change the formula to use meters instead of revolutions. <S> If one assumes k equals pi, then applying one amp to the motor will yield (1 amp * (1 volt second per meter)), which is to say one newton of force. <S> Applying one volt to the motor will cause motor's output to move at a rate of (1 amp / (1 volt second per meter)), which is to say one meter per second. <S> Moving the output at a rate of one revolution per second will cause the voltage to be one volt; applying one newton of force will make the motor draw one amp. <S> Just as with an ideal mechanical transmission, the motor establishes an instantaneous correspondence between what's happening on both sides. <S> Of course, real motors don't behave quite like ideal motors, but most real motors may be modeled as an ideal motor with a series inductor and resistor on the electrical side, and with an attached mass and some friction on the mechanical side. <S> Commutation issues may cause behaviors to vary somewhat from that simplified model, but in many cases it works well enough to be useful. <S> Because of commutation issues, a motor's inductance may vary slightly depending upon its exact mechanical position. <S> Nonetheless, a motor's inductance is relatively independent of speed--the faster a motor is turning, the faster the inductance will vary between the values it has at different positions, but for the most part it will behave like a relatively-constant inductance. <A> No, they're not at all equivalent. <S> The voltage depends on the speed of the motor and nothing else. <S> Any current that flows as a result of that voltage depends only the external impedance connected to the motor. <S> On the other hand, the energy stored in an inductor is essentially a current source, and it will (attempt to) produce whatever voltage is required to get that current to flow in the external circuit, which is what gives rise to the "inductive kick" effect. <S> Of course, the magnitude of the current in question is modified over time by the terminal voltage of the inductor. <A> OK. <S> Back to "Back EMF." <S> As for the original question: "Is it valid to consider back EMF in a motor equivalent to increased inductance?" <S> The answer is NO. <S> An inductor gives you back the energy you apply against the Back EMF-to build the magnetic field-as electrical energy. <S> A motor CONVERTS the energy you apply against the Back EMF-into mechanical energy.
| Back EMF is, as you say, a voltage source.
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Why not SRAM for FPGA in image processing? I'm beginning with VHDL coding and I've done some basic image processing on my development board. I've noticed that most FPGA development boards often use DRAM (SDRAM, DDRAM) as RAM.For example, I'm using a FPGA dev board from TERASIC and it uses DRAM.Although DRAM memory seems to be cheaper that SRAM (Static RAM), from my point of view it is not optimized for image processing, let me explain:DRAM data is accessed by block, SRAM data can be accessed (addressed) byte by byte (or word by word, or 36 bit at a time, whatever). This feature is more matched to image processing, for example, a simple mean filter where a pixel is the mean of its neighbour, say Pix(x,y) = (Pix(x-1,y-1) + Pix(x,y-1) + Pix(x+1,y-1) + Pix(x-1,y) + Pix(x,y) + Pix(x+1,y) + Pix(x-1,y+1) + Pix(x,y+1) + Pix(x+1,y+1))/9 is difficult to do with DRAM, you need to have an internal buffer of at least two lines to memorize lines y-1 and y (doing the processing on the y+1 line). Also it seems to take a huge amount of time to Quartus to do this simple processing (I'm using 2ports FIFO). If the board had used SRAM like IS61LPS25632A from ISSI for example, the processing would be straightforward since the address of a pixel can be for example ByteAdress <= x + (LINEWIDTH*y); From a hardware point of view, with 9 read/write memory clock cycles the process is done for one pix and with some tricks in parallel processing this can be improved I suppose.Then, is my understanding true? If not, where is the mistake? EDIT: Using internal FPGA RAM is not an option for me since there is not enough memory inside the FPGA device I use (Cyclone III) and I don't know if paying for a more expensive development kit just to have some more RAM inside FPGA is judicious. My question is more related to code efficiency. Say for example that one has 8MB of SRAM. From my point of view, the code produced with this kind of RAM will be more efficient (speed and buffering) in the case of 2D processing (like image processing) because true random access is well suited for accessing data with (big) offset between then, like on this statement: ByteAdress <= x + (LINEWIDTH*y); Reorganising information is time consuming. I use a camera that outputs data pixels sequentially, line by line. I can only store these data on two port FIFO RAM component built with DRAM. I can't do even a simple transposition before storing the data with this. Surprisingly I didn't find any code on the web that deals with this 2D problem. <Q> No fundamental reason why not. <S> Synchronous SRAM is truly random access, fairly inexpensive, and easy to interface to. <S> Its downside in that it occupies a fairly narrow niche between the on-chip BlockRam (not much smaller, free until it forces you to select a larger chip, massively parallel and more flexible) and external DRAM (massive storage capacity at a price SSRAM can't match). <S> So up to 0.5 or 1MB, external SSRAM is unnecessary, and above 8MB or 16MB <S> (numbers may vary according to your budget and current prices!), SSRAM becomes expensive enough that DRAM takes over despite its limits. <S> Then - if you need random access - you have to massively reorganise the computation to read chunks (bursts or pages) from DRAM into BlockRam where you can process it fast before writing back bursts etc.... <S> But if you have a role for SSRAM within that window, go for it. <S> I have added simple home-made SSRAM boards to augment commercial FPGA platforms where necessary. <A> Coupled with using the FPGA's on-chip SRAM as line buffers, it allows pretty much any whole-image computation (including the 2D FIR you give as an example) to be pipelined to the point where you can process one pixel per clock. <S> If you're working with HD video in real time, being able to do this is essential. <S> Don't discount the importance of capacity. <S> Modern image sensors are in the range of multiple megapixels for video and 10s of megapixels for still images. <S> In many applications, multiple frame buffers that have 4-6 bytes per pixel are required, for things like per-pixel correction data and ping-pong image processing. <S> The total memory required quickly exceeds what can be economically implemented with SRAM. <S> Even if you're only interested in still images, the development kits are designed to support a broader range of application development. <S> BTW, I have a Terasic board that has both DDR SDRAM and ZBT SSRAM. <A> Both SRAM and DRAM I think is necessary. <S> for exertion mask on picture SRAM is useful because of convolution. <S> but to store the result of picture Dram is efficient.
| SDRAM is actually a very good choice for image processing, because both the capacity and the bandwidth (especially with DDR) are improved relative to SRAM.
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Common ground and voltages Further to a question I asked earlier, I realised that common ground confuses me. I have a USB powered Arduino and a separate Battery supply. The battery supply negative is connected to the Arduino ground. The voltage of each supply is the difference in potential between its two sides and my understanding was that you didn't need to worry what those actual potentials were, just the difference. But when you connect negatives surely the potential is important as the 2 potentials will now "average out" thus potentially (just noticed pun on proof read) changing the original voltages of both supplies. The fact that it is okay to connect the two supplies means my understanding is flawed. Could someone please straighten me out ? Section 7 of this document is basically what I'm doing <Q> One key concept that might help in clarifying any misunderstandings is this: <S> A voltage is the potential difference between two points in a circuit, it is not an absolute value of any physical characteristic at a single point in a circuit. <S> Thus, there is no absolute potential involved, it is relative value, a difference. <S> How this applies: <S> The "ground" of the Arduino is the point (or PCB trace, to simplify the concept) relative to which, the potential on the "Vcc" trace of the Arduino is measured. <S> Thus, when powered by the USB cable, the Arduino's "ground" is the measurement basis, and is incidentally at the same potential as the "ground" of the computer whose USB port is being used. <S> The "negative" of the battery is merely one of two points across the battery, between which the potential difference <S> i.e. the voltage of the battery is measured. <S> Relative to the rest of the (electrical) universe, the "negative" terminal of the battery is floating , i.e. has no specific relative value until this so-called negative terminal of the battery is connected to a circuit. <S> Thus, only while such a connection is made, the "positive" terminal of the battery is at a potential of "battery voltage" compared to the Arduino's zero level or ground. <S> If, instead, you connected the battery "positive" to the Arduino ground line, then the other terminal of the battery would be at a negative value compared to the Arduino's zero level or ground. <S> The battery, or for that matter any arbitrary voltage, is relative to the reference you provide. <S> It is a difference , from your defined base point in the circuit, not an absolute value , to reiterate the basic principle mentioned at the start here. <A> You are right that connecting a battery generates a difference in potential. <S> There is no universal "0 V"; voltage is a relative measurement. <S> When we talk about "5 V", we probably mean "5 V from the ground symbol on the schematic". <S> If you connect one end of a battery to some other voltage, it will add its difference: <S> You only run into problems when you have batteries in parallel: <S> This situation can't actually occur. <S> In reality, wires are just very small resistors, and what happens is that wire on the top gets very hot as a huge current flows through it, and it vaporizes. <S> Or, the batteries aren't able to supply enough current to maintain their voltage and they go dead. <S> In any case, something gets hot and breaks. <A> If the negative connections of both supplies are connected to a common conductor (ground) and the positive connections are connected to a common conductor ( V+) then any difference in Voltage will cause a current to flow. <S> Assuming the USB is at it's correct 5V then if the battery pack is only at 4.8V the usb will try to charge it. <S> If the difference is much greater then damage may be caused to the circuitry from high current flow, or the batteries from excessive charge rate. <S> If the battery pack is over 5V, Without checking the specs for USB <S> I can't be sure, but I would assume it was pretty well protected from minor overloads, so will have a diode stopping any reverse current flow. <S> If both supplies were designed for the device, and can be plugged in at the same time, I would assume (provided the batteries aren't flat) <S> it was safe to do so, but why would you want to.
| When you connect the negative terminal of your battery to the ground of the Arduino, you are providing a reference value for the battery, relative to the Arduino.
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Why are the pins on the official arduino motor shield so long? Here is the Official Motor sheild website. The shield has pin connectors at the bottom that are 2x longer than on any other shield I've seen. Why is that? <Q> These are the normal stackable headers used on OFFICIAL Arduino brand shields <S> (they specifically state they are used on the Proto Shields, on the header item page ), and on most Adafruit shields. <S> They even sell them as standard components (Arduino 14.5mm pin length, <S> Adafruit 10.5mm <S> pin length headers). <S> Because they are meant to stack, you have to account for board length, plus imperfect solder joints, and component height, on the solder side. <S> Pricing for headers is based more on plating and pin count than it is on pin length. <S> They would get a price break due to volume as well. <S> I'm guessing the others you seen <S> haven't been official shields. <A> One reason could be due to high voltages present on the motor board. <S> Slightest short could turn into flames if the shield touches any high component (like an electrolytic capacitor) on the board below. <S> Also it could destroy the arduino in case of short. <S> As you can see in the above stackup that the screw terminals are only a few mm apart from the DC connector. <S> The board is also close to the USB connector shield. <A> My guess is it may be to keep any high current switching as far away from the Arduino board as possible, to reduce the risk of EMI problems. <S> I have never used an Arduino before though, so I may be completely wrong in my understanding of how these "shields" plug together.
| This are used to stack, because normal length headers are meant to be soldered on one side, and connected on the other, flush on the board.
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Does CW or SSB consume more power? CW pumps out the full carrier but has a better SNR. On the other hand SSB suppresses the carrier, and one side-band but this suppression demands additional circuitry. There are probably additional considerations with SSB - poor SNR, operator dialect but these are not part of the question here. When the available options to transmit are CW, and SSB in ARES - which of the two will conserve the battery more? <Q> Your question suggests an actual use (that is, "pass a message on ARES"), so you've got to consider bandwidth as well. <S> Not so much how much band space you will occupy, but how long you'll be on the air. <S> How long does it take to send a message on CW vs. reading it on SSB? <S> Transmit time will affect the battery as much as the indicated power level. <A> It should be noted that this question assumes HAM (amateur) radio jargon/conventions and in that context it is strange to compare CW (a "digital" communication modality used with morse code) to SSB (an "analog" strategy used with voice communication). <S> If you transmit a monotone signal (like morse code) using SSB modulation the effective signal in the air consists of only one frequency as well (assuming SSB-SC -- carrier suppression). <S> That signal is offset from the carrier by its tone frequency, but there is still only one tone in the air. <S> Therefore, at a theoretical level this question is silly since you are talking about a single tone and whether you On-Off key (OOK) the carrier (CW), or offset a single frequency from this magical carrier (SSB-SC) and OOK that, is completely irrelevant. <S> The same energy is in the air. <S> That is to say, if your CW carrier is 1MHz, the equivalent carrier for SSB-SC transmission of a 3kHz tone is 1.003 MHz assuming you use the lower side band. <S> Both strategies result in an airborne transmission at 1MHz. <S> So from physics, energy is proportional to frequency (given constant amplitude), the signal energy is exactly the same. <S> If you now expand the question to include the radio hardware that is generating/receiving the signals, OOK (CW) hardware is typically simpler and, therefore obviously, requires less quiescent and operational power. <A> You can find here that CW uses up to 5W and SSB up to 10W, most of the time. <S> This however does not mean that CW uses less battery power. <S> How much battery power you need (or how long the battery will last) depends on more than the output power. <S> You cannot say which one conserves more power for you battery. <S> We can only say that SSB may transmit more power in QRP operations, but that doesn't answer your question. <S> If you want an answer to your question, I'd recommend you to do some testing. <S> You can simulate a battery with a regular power supply and measure the current in both modes, or you can test both modes in the field and measure how long a fresh battery lasts in that operation. <A> CW communication carried out by highly trained operators is well known to be more power efficient than SSB voice for transmitting readily transcribable information. <S> That's mostly because voice is not very efficient as a representation of information, requiring a relatively high signal to noise ratio for the amount of data carried, unless you include more nuanced things such as expressive tone in the total of information that is being transmitted. <S> However, CW does have the downside of needing more highly skilled operators with the benefit of a lot of experience. <S> The real answer today would be a digital mode, combining the spectral and power efficiency of CW (though probably not OOK modulation) with a fully automatic implementation - leveraging modern low power DSP instead of trained ears. <A> To answer your battery conservation question, CW will conserve battery more than SSB. <S> That's why CW is QRP (Low Power). <S> Hope this answers your question. <S> Some more information about QRP from wikipedia.
| There might be circuits in CW generation that draw much current, so that it needs more power than SSB, while SSB has a higher Peak Envelope Power.
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Is it practically possible to transfer electrical power around the World by wires? Assume that we installed plenty of solar cells in all continents; mainly in Asia, Africa and America. Can we transfer the electrical power harvested from them around the World through wires? So that the continent(s) in daytime will feed the one(s) in nighttime. Assume that we use a higher technology then we use today in practice. For example, we will use higher insulation technologies to transfer with higher voltages to decrease copper losses. Do the practical limits make this scheme feasible? <Q> No. <S> It is completely impractical. <S> I would even go so far as to say that it will never be practical given that the cost of transmission system construction will never outpace that of local generation. <S> The copper capacity tables are here <S> and I could do a bunch of math for you, but it is pointless. <S> Generating electricity is a relatively easy thing to do. <S> There are so many different strategies: coal, wind, hydro, solar, nuclear, CNG, mixed-fuel, biofuel, et. c. ad. <S> infin. <S> At least one of them will be practical in any given locale -- at least more so than running greater than 20,000km of cable would be. <S> I'm not even sure it's absolutely possible since the tremendous voltages you would need to overcome the staggering losses might make insulation requirements beyond the reach of today's materials. <S> Actual Power Lines <S> The longest power line on the planet is in China and only runs <S> 2,059 km using a special extremely high voltage DC configuration (avoids the phase synchronization issue). <S> It doesn't run through the ocean as that would dramatically increase the losses. <S> It isn't cost effective (read: practical), which is why China is the only country with these types of lines in mandated use. <S> As a case study it cost 3.5 billion to put in and can transport 7.2e6 kw. <S> For perspective, consider that 3.5 billion is the cost of building a local plant that can generate almost half that much power (and deliver it). <S> In the Chinese case all they got for their money was the delivery -- they still have to finance the generation. <S> The longest conventional AC lines in the US are <S> ~240 km . <S> Even Space is More Practical <S> It is even much more efficient and practical to beam power from space using Wireless Power Transmission (WPT) technology (~160km), than to move it terrestrially half-way around the world (~20,000km). <S> It is also more politically palatable since you don't cross any other foreign territory and "energy security" is a sensitive topic. <S> Google "Solar Power Satellites Wireless Power Transmission" and you will find a broad base of literature on this idea. <S> Here are a few references to papers to get you started: 1 <S> 2 <S> 3 <S> It appears to be gaining traction. <S> Japan and others are seriously considering deploying such an energy architecture. <A> No, it is not practical with today's technology or what we can reasonably forsee well today. <S> However, what you ask about is theoretically possible in that no laws of physics are being violated. <S> Therefore it is impossible to say whether it could become practical in the future. <S> Saying catagorically today that it will always be impractical is just wrong. <S> Think of what would have seemed totally impractical not that long ago. <S> Right after the Wright brother's first flight, how many would have thought something like a large commercial jetliner would ever be practical, let alone in just a few 10s of years? <S> What about a man on the moon just a half century later? <S> The list goes on. <S> To make this practical, long distance transmission would have to become much cheaper relative to power generation than it is today. <S> I really can't see how this can happen with today's copper conductors, but the future may not be limited to that. <S> Who is to say what advances will have been made in the next 100 years in superconductors, or carbon nanotube fibers, or something else we haven't even heard of yet? <S> The cost of generation may also go up in the future relative to the average of the economy. <S> That's quite likely even from today's viewpoint for any generation relying on fossil fuels. <S> So, definitely not today, almost certainly not for the next 10s of year, but in the range of 100 years or more, so much unforseeable technological advance is possible that it would be irresponsible to say it would be impractical then. <S> Think about how relatively little back in time you need to go to where trans-ocean communications cables and communications satellites seemed completely impractical. <A> It will only become practical when we have reasonably priced superconducting cables. <S> That way losses would be eliminated. <S> Ideally that would use room-temperature superconductors , which unfortunately are not available at the present time - and may not be possible in practice. <S> However, we already have materials that are superconducting at liquid hydrogen temperatures. <S> One proposal is the Supergrid , which places the superconducting cable inside a liquid hydrogen pipeline. <S> That would keep the cable cool, and at the same time allow us to use hydrogen as a fuel for vehicles. <S> Practicality of the system is a different matter though. <S> For starters, it assumes that we will have an efficient and safe way to pump liquid hydrogen at ordinary petrol (gas) stations. <S> Hydrogen is not a safe liquid! <S> It also overlooks the energy needed to keep the hydrogen liquid. <S> Over long distances that would become prohibitive. <S> This site has a discussionof the system.
| It just isn't practical to go planet-scale with transmission lines.
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Dangerous AC or DC Is alternating current more dangerous than direct current? And give me explanation for that... Assume that 230 V, 50 Hz AC voltage and 230 V DC voltage. Which is more hazardous to human life? Which has more probability of risk? <Q> AC has some time period - frequency. <S> It repeats itself and periodically touches zero. <S> However DC remains at a constant voltage level. <S> Let me compare on different points:- Due to such nature of AC, there is a "Let go" thing with AC <S> , At a specific point it may let you go and you could be safe. <S> However, in case of DC, there isn't any chance. <S> It also depends on factors like Skin Moisture. <S> Moisture conducts!! <S> When you are shocked with AC, it generates ripple in your body of 50Hz (In this case) which contracts your muscle and you get heart-attack. <S> When you are shocked with DC, it electroculates your body i.e <S> It separates chemicals from your cells. <S> (In both cases, You Die!) <S> For DC, Path Matters! <S> If DC passes from say one finger to other, you might feel burns <S> that's it... <S> But AC spreads across body. <S> At High frequency, AC is comparatively safe... <S> Yes.. <S> Due to Skin effect, it flows from your skin! :) <S> No such case for DC (0 Hz) <S> So both are dangerous under appropriate circumstances! <S> This is refered as "War of Currents" between Thomas Edison and George Westinghouse! <S> P.S. <S> IN ANY CASE, DON'T PLAY WITH ANY.... <S> BOTH ARE DANGEROUS!! <A> This is like the question, "Which is more dangerous, a pit full of spikes, or a giant swinging blade?" <S> You can argue about the different ways in which they may kill you, but it doesn't much matter when you are dead, does it? <A> The common 50Hz may causes the heart to go in fibrillation. <S> Without a quick medical attention you'll die in minutes. <S> It is true that it is the current that will cause the damage, but in AC only 30mA is sufficient to put you in fibrillation if endured for several second. <S> It is also sufficient to block your lung muscle. <S> Given the body resistance, it only need ~50V AC to have that much of potential.
| AC is more dangerous for a human being.
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Simplest circuit that flashes and fades color LEDs synchronized to music? This is about designing a circuit that can control multiple LEDs synchronized to music or sound, as simply as possible. I am looking for some conceptual ideas to start out with, not a complete circuit design or anything. My question is inspired by this YouTube video (now those are heels to kill for!) I have read the question Flash an LED to music but it does not begin to address a requirement like those shoes. I realize that there are many different and increasingly complex ways of achieving the desired result, but I am looking for the simplest approach, even if it is not the cheapest, and not the most precise. Also, I am not looking for a ready-made pair of shoes that has this (though if there are any, would be interesting to know about them). <Q> Depends on your meaning of simple. <S> @Tcrosely answer gives a good example, a simple passive color organ, all parts that can be found through hole. <S> Using resistor/capacitor pairs as bandpass filters. <S> Or you might think a microcontroller solution is simpler. <S> Which there are three ways of doing. <S> Completely in software <S> (FFT libraries for Arduinos/MSP430/Pics), using another chip to capture the audio and turn it into digital information you can use (The MSGEQ7 + microcontroller is a popular version of this), or dedicated chips (All do three band, bass/mid/treble, with adjustable cross over points, or randomization of the LEDs) <S> TI's/ <S> NatSemi's LM4970 (most ""hobbyiest"" ""friendly"" package) or other's in their Boomer family (really small SMD). <S> The LM4970 needs to only be enabled by i2c from a microcontroller, while some of the Boomer chips have an auto on mode. <S> AMS's AS3665 and AS3668, both which have auto audio sync mode. <S> ISSI's IS31FL3193,IS31FL3196, and IS31FL3199. <S> Same as the LM4970, needs a simple i2c on command. <S> Simple really depends on your experience and how easy you can get parts. <S> A dedicated chip method would need a lot less parts or board space than a passive component color organ, and can be powered by a lower voltage. <A> To address the conceptual ideas: You need a very simple and small microphone, an electret mic would do for this. <S> A small power source, like a coin cell or two. <S> A couple of transistors, and LEDs <S> Some basic filtering that directs e.g. bass frequencies to one colour LED and higher frequencies to the other (you can have many filters if you want more LEDs) <S> This could be done entirely in a simple analog fashion, but you could also consider a small microcontroller like a PIC12F - this would give you much more flexibility with stuff like LED blinking patterns, possibly changing between timed routines, reacting to overall amplitude levels/gaps in the music, etc. <S> It would be a fun beginners project for a microntroller. <S> If you need less "concept" and more circuitry, let us know <S> and I'm sure you will get a few good ideas you can pick and choose from. <S> I watched the video, and to be honest <S> I wasn't too impressed with the synchronisation part <S> (I think this could be improved significantly - the purple LED didn't seem to be doing much) , although the lights did undoubtedly look cool. <A> What you are looking for is called a "color organ". <S> I did some looking around on the net, but most circuits used a combination of op-amps and drivers and were much too complicated (IMO) for your needs. <S> But I found the following one which appears simple enough: It is designed to run off a 9v battery, but perhaps you could use three 3v coin cells in series. <S> It is also designed to take its input from a MP3 player such as an iPod; for use with an electret microphone the author recommends one with a pre-amplifier like this one : <S> It runs off of 2.7v to 5.5v, so you could tap off of the first 3v coin cell to power it. <A> Anything I have ever seen that is "synchronized to music", be it a flashing LED, dancing flower, or the visualization plugins in Winamp, seem no better than random to me. <A> By looking at the video, it seems like different LEDS are blinking depending on the different sound frequencies. <S> One way, not sure simple enough, is to use a MSGEQ7 IC along with a small microcontroller (8pin - 14 pin). <S> The MSGEQ7 splits the sound input in 7 frequency bands. <S> This information can be passed on to the microcontroller which takes the relevant band information and lights up designated LEDs.
| So, if "simple" means low component count, then I'd hook a couple LEDs to a microcontroller, implement a simple pseudo-random number generator, and flash the LEDs on and off, randomly.
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Can a charge-pump be 100% efficient, given ideal components? A recent question about cyclically charging a capacitor reminded me of something I read once. As I remember, it demonstrated that it's impossible to construct a charge pump that is 100% efficient with ideal components, but it is possible to build a 100% efficient boost converter with an inductor if components are ideal. Does this resonate (no pun intended) with anyone else? Any way to demonstrate or refute the truth of this? To be clear: we are assuming we have ideal components . I realize no real circuit will be 100% efficient with real components. Diodes may have zero voltage drop. Transistors may be ideal switches that take no energy to change state. Wires may have zero resistance. <Q> It is all about dualism. <S> With ideal components, you can make an ideal SMPS type voltage converter <S> (= <S> using an inductor to do the work). <S> You can't make an ideal voltage converter using switched (flying) capacitors. <S> That is not the universe being unfair to capacitors: you can make an ideal current converter using switched capacitors, which is not possible using inductors. <S> I can't do the math out of my head, but the problem with capacitors and a voltage source is like this: take a voltage source with a certain source impedance (= series resistor). <S> Connect a capacitor to it and load it for an infinite time (any finite time will do too). <S> Calculate the amount of energy lost in the series resistor as a function of its resistance. <S> Now mathematically take the limit of that formula twoards zero resistance. <S> You will find that the energy loss will stay the same. <S> Intuitively this is because a smaller resistor causes a higher initial loading current, and hence a higher RI 2 loss. <S> management summary:You can't connect an ideal voltage source to a capacitor, because that would result in an infinite current which is impossible in itself and would cause an infinite magnetic field which would destroy the universe (just kidding, remember this is the management summary). <S> But you can approach this ideal as closely as you like, and the result will still be the same: a fixed amount of energy is lost while charging the capacitor. <S> Hence: sorry boss, no ideal flying capacitor voltage converter. <A> An inductor-less charge pump made with ideal components may be 100% efficient if the source current and voltage waveforms have the proper relationship with the load current and voltage waveforms. <S> It is possible for either the source or the load voltage to be constant DC, but not both (except in the trivial case where both voltages are the same and the charge pump doesn't have to do anything). <S> Note: <S> a charge-pump which contained an internal current source could be 100% efficient at converting input power from a constant-voltage source to an external constant-voltage load, with any energy that was drawn from the internal current source in one cycle being replaced on the next. <S> On the other hand, such a current source would simply be taking the place of an inductor. <A> For a boost converter you can design one with idealised components and all the equations still make sense, voltages and currents remain finite. <S> From these voltages and currents you get an efficiency of 100%. <S> A charge pump with zero stray resistance simply cannot be analysed in this way Trying to do so results in absurd answers. <S> What happens when you connect a perfect capacitor to a perfect voltage source through a perfect switch? <S> Trying to calculate the current results in a divison by zero. <S> The same problem applies to connecting two perfect capacitors. <S> Lets say we have a capacitor charged to a given voltage and connect it to a voltage source of a higher voltage via a resistor. <S> Lets assume for now that we let it charge fully (ignoring for a moment that doing so would take infinite time). <S> We find that changing the value of the resistor does not change the efficiency, the total energy drawn from the voltage source remains the same. <S> The efficiency is however dependent on the ratio between the starting voltage of the capacitor and the voltage of the voltage source. <S> A smaller voltage difference leads to a higher efficiency tending towards 100% as the voltage difference tends to zero. <S> In our charge pump there is not infinite charging/discharging time so resistance does affect efficiency but as resistance tends to zero efficiency (for a finite voltage difference) tends towards a finite number less than 100%. <S> The charge transferred on each switching cycle is related to the change in voltage on the capacitor by the capacitance. <S> To transfer a finite average current to the load we either need to transfer a finite charge per cycle or we need to have an infinite number of cycles. <S> So making your 100% efficient charge pump would require either an infinitely big capacitor or an infinitely high switching frequency. <A> Well it really depends upon how far we go with "ideal components". <S> If diodes had a forward voltage drop of 0 volts, BJTs had a base threshold of 0 volts a saturation of 0volts and infinite current gain, and FETs have a gate threshold of 0volts and a Rds of 0 ohms <S> then it may very well be likely be possible to realize a 100% efficient change pump. <S> Even in the case of the boost converter it will not be 100% efficient unless the switch FET and flyback diode are ideal in the sense that I described above. <S> Likewise the inductor has to have a DC R that is equal to 0.
| An inductor-less charge pump cannot be 100% efficient when powering a constant-voltage load from a constant voltage source.
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Which connection(s) to a decoupling capacitor should be shortest? Circuit Cellar-Issue 210- has this question "Which connection(s) to a decoupling capacitor should be shortest?why?" in its "Test your EQ section". I couldn't understand what they meant 'Which connection(s)'. Can somebody explain this? <Q> First let's look at what a bypass capacitor is good for. <S> Let's talk about digital circuits, like discrete logic, or an FPGA or a microcontroller. <S> Contrary to what's been said in comments, in many many cases it is a pretty good assumption that the output drivers are the dominant source of power supply switching noise. <S> Exceptions could be very high-end microprocessors or very large FPGAs. <S> For traditional logic designs it generally also is reasonable to assume capacitive loads. <S> If your outputs are driving transmission lines, you've got some other issues to worry about, but it won't actually change the results of the analysis much. <S> Let's say an output driver is switching from low to high. <S> Then the high frequency components of current flow basically as shown in this diagram: <S> When the output is switching from high to low, the high frequency components of current flow basically like this: I've put the reference ground down at the end of the line because that's where the receiver will be trying to decide between a '1' and a '0', so that's where the noise margins will matter. <S> First thing to notice is the bypass capacitor doesn't have much to do with what happens for 1-0 transitions. <S> We'll leave the issue of ground bounce for another question. <S> Second thing is that for the 0-1 transitions, the current path goes from the capacitor to the driver, and it comes back to the capacitor from the signal's return path. <S> So in this case, what's critical is the path from the capacitor to the driving chip's Vcc pin, and from the ground plane (you are using a ground plane, right?) to the capacitor. <S> Of course if you're doing analog filters, or power, or something else, you should look for the current loops and do an analysis of your specific situation to know how to optimize your bypassing layout. <A> For reference, here's the original answer to the question: <S> The most important connection is between the supply side of the capacitor and the supply pin of the chip. <S> The biggest transient currents in most integrated circuits are related to the switching of the output pins. <S> When a pin switches from low to high, a pulse of current is drawn from the chip's external supply to charge the output node, which includes the capacitance of the PCB trace and the input gate of the next chip. <S> This current is primarily supplied by the decoupling capacitor. <S> Note that this current "returns" to the ground end of the capacitor <S> NOT by way of the same chip's ground pin, but rather by way of the PCB ground plane and the ground pins of the chip(s) <S> that it is driving. <S> For this reason, it is not particularly important that there be a short ground connection between the driving chip and its capacitor, <S> but it is important that the capacitor be bonded as directly as possible to the PCB ground plane. <S> When an output switches from high to low, the capacitance of the output node must be discharged, which causes another pulse of current. <S> However, this pulse does not involve the decoupling capacitor at all. <S> Instead, it flows from the capacitance of the output node, through the pulldown transistor and into the ground plane, where it is "returned" to the other end of the distributed node capacitance. <S> This is why the ground pins of all chips need to be be bonded as directly as possible to the PCB ground plane. <S> If any part of this is not clear, I'd be happy to expand on it as needed. <A> In an ideal world both of the leads of the decoupling capacitors should be as short as possible and be as close to the IC supply pins as possible. <S> Now that said, with many logic chips using threshold specifications following the old TTL standard where the noise margins near ground are less than those near the VCC or VDD <S> then it is desirable use extra care to keep down noise on the GND bus. <S> Thus if you have a non-ideal condition and it is necessary that the bypass capacitor has a tradeoff of where one connection has to be longer than the other then it will be appropriate to favor the shorter connection at the GND side of the capacitor. <S> If a circuit used all CMOS type logic thresholds where the noise margins at GND and VDD are the same <S> then it may be best to try to even out the bypass capacitor connection length on both sides. <A> This puts the inductance of the trace in series with the power trace , where it can act to suppress noise, instead of in series with the capacitor, where it just acts to render the capacitor a less effective filter.
| You want to make both connections as short as possible; if you can't make them short , an acceptable alternative is to "zigzag" the power and ground routes to the capacitor instead of using T-traces to "stub" the connection in.
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Why are 10-pin DIP integrated circuits so uncommon? This is a tangential thought from this answer , specifically the line " There are 10 and 12 pin DIPs ". After some searching around, I finally located only one 10-pin DIP part: NTE1450 , 5 Watt audio power amplifier. The part is evidently obsolete, and the one datasheet I found, I cannot locate again. I found many other components in 10-pin DIP though: LED displays, DIP switches, resistor packs and of course DIP sockets. There are many 8 pin PDIPs 14 pin PDIPs flood the logic gate listings, but 10 and 12 are apparently pariahs. I haven't actually searched for a 12-pin DIP IC yet, but I suspect that too will be a rarity. Why are 10-pin DIPs so uncommon? <Q> As chip production increased, manufacturers may have had to produce new tools simply to keep up with demand <S> (if one has a four punches to produce DIP14 lead frames, one can only produce four such lead frames at a time), but they would have had more inclination to add tools for sizes that were slightly larger than existing ones (so they could offer chips that could do things previously not possible) than for sizes that were smaller. <S> The only real use for a 10-12 pin DIP would have been for an application where an 8-pin chip would have been insufficient, but a 14-pin part would have been too big. <S> Even before the invention of surface-mount technology, there weren't many such applications, and once surface-mount technology came on scene there really weren't any: someone who needed a chip with 9-12 useful connections but couldn't afford the size of a 14-pin DIP wouldn't use a 10-12 pin DIP--they'd use a surface-mount part instead. <A> You don't see very many 10-pin DIPs for historical cost reasons. <S> Extra pins are expensive. <S> Especially back when these were extremely common. <S> I don't recall a lot of analog parts in DIP packages. <S> Transistors would come in 3-leaded packages. <S> Anyway, most logic gates (OR, AND, NAND, etc) are going to have two input and one output pin. <S> Plus you need to add a power and ground pin. <S> So, if you have two gates in a package, then it ends up being 8 pins. <S> If you have 4 gates in a package, you end up with 14 pins. <S> Any other combination give you extra pins, and extra pins are expensive, especially for large volume commodity parts like they were back then. <S> That was the reason back then, anyway. <S> The cost of the extra pins isn't as big of a deal now as it was back then, but now these parts have been in catalogs for decades, so it doesn't make sense to change. <A> You need power and ground. <S> That leaves 8 pins. <S> Two standard two-input gates requires 6 pins. <S> That leaves 2 pins extra. <S> On the other hand an 8 pin part fits that perfectly. <S> Similarly 12-pins would be 3 gates with 1 pin extra. <A> You need power and ground, that leaves 8 or 10 'used' pins. <S> This is just a wild guess, but I'd say packages with a 'nice' number of 'used' pins are most common. <S> These packages would be more common because there are multiple things of the same thing stored in one package, most of the time. <S> Look at the MAX232 device, which randomly pops into my head. <S> It is a 16-leader, of which 12 'used', since there are two power pins and VS+ and VS-. <S> All pins are in pairs: C1 <S> + and C1-, <S> C2+ and C2-, R1IN and R1OUT, R2IN and R2OUT, T1IN and T1OUT, T2IN and T2OUT. <S> Additionally, these pairs are in pairs: C1 and C2, R1 and T1, R2 and T2. <S> Since we have pairs of pairs, the number of 'used' pins is dividable by 4. <S> 14-leaders are quite common. <S> That gives 12 used pins, which can be divided by 1, 2, 3, 4, 6 and 12. <S> 12-leaders <S> with 10 used pins give only 1, 2, 5 and 10. <S> 10-leaders would be more used, since 8 used pins is dividable by 1, 2, 4 and 8 - powers of 2. <S> Disclaimer: it's just a wild guess.
| While today's computer-assisted manufacturing would make it relatively inexpensive to create and manufacture new styles of lead frames, tooling used to be much more complicated and expensive, and there were relatively few discrete sizes of chips. Back in the 90's, DIP packages were mostly for digital logic (TTL back then). A number would be 'nice' when it has lots of divisors.
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Common ground and voltages, Part 2 To further my understanding of what happens when 2 circuits have a common ground could someone please explain what happens when the following 2 circuits (charged capacitor, high resistance load) : share a common ground What happens to the 2 voltages and capacitor plate potentials ? <Q> For the (somewhat theoretical) case you show nothing different from the not-connected case will happen, in fact nothing CAN happen because there is no (additional) closed current path. <A> What Wouter van Ooijen said, but also the bottom halves of the two circuits are guaranteed to be at the same potential. <S> You seem to be wondering why anyone would ever care about this. <S> To demonstrate why this important, go find a stereo or something with an audio plug coming out of it like you'd plug into an iPod or such. <S> Turn up the volume just a bit, and touch just the tip of the connector. <S> Hear that buzz? <S> It's because your body is picking up all sorts of electrical fields from the electronics in your house and nature. <S> These are at different potentials from the stereo, and it amplifies them as sound and you can hear it. <S> Now, touch all the parts on the plug. <S> Squeeze <S> so you get a good connection. <S> The buzz is gone! <S> Why? <S> Because now you have a common ground. <S> All those fields are still there, but now your body and the stereo are at the same potential, so there's no difference between you and the radio, so you don't hear anything. <S> (Or not as much, anyway. <S> The connection isn't infinitely conductive, so you will still hear something, if you turn the volume up more.) <S> So, your stereo amplifies the voltage difference between its idea of "ground" and the left and right channels on the connector. <S> But if something (your body, an iPod, whatever) isn't connected to the stereo's ground, then the stereo will be amplifying the voltage output by the iPod and also the difference in ground between the iPod and the stereo, which is likely some crazy random noise and 60/50 Hz hum from powerlines. <S> Giving the two a common ground reduces this difference to nearly zero, so that the two circuits can have a common reference, so they can agree what "4V" means. <A> Enough electrons will flow between newly-connected plates to bring them to the same potential. <S> From a practical perspective, though, unless something else not shown on the schematic behaves as a significant capacitance connection between the two halves of the circuit, however, the number of electrons that will have to flow to equalize the potentials will be really, really small. <S> Really, really, really, really small. <S> If one point has a higher voltage, electrons will prefer moving in that direction to moving in the reverse. <S> The higher the difference in voltages, the greater the preference. <S> If an electron arrives on either one of a capacitor's plate, it will make every part of the capacitor be much less accepting of electrons (i.e. decrease its voltage). <S> Conversely, every time an electron leaves one of the plates, it will make every part be much more accepting of electrons (increase the voltage). <S> Note that if many more capacitors enter an electron than leave, both plates will have a very large negative voltage, such that electrons will jump at the opportunity to leave. <S> What is significant about a capacitor is that when an electron enters one plate, it causes that opposite plate's voltage to decrease slightly less than that of the plate receiving the electron. <S> Consequently, if the balance of electrons entering one plate exceeds that of the other, the plate which receives excess electrons will start to have a more negative voltage than the plate which doesn't, but the amount of imbalance between the two plates that is required to achieve that vastly exceeds the imbalance of electrons entering or leaving the device as a whole which would be required to change both plates' voltage by that amount. <S> Thus, unless there is some effective capacitance or other connection between the left and right sides, the number of electrons that might have to flow between the two sides to balance out their voltages would be vanishingly insignificant. <A> Next to nothing happens because the connection you have made does not create a circuit, and so current will not flow through it, except to equalize any differences caused by static electricity buildup. <S> Static electricity buildup will appear on both plates of the capacitor: it's an excess or deficiency of electrons in the circuit as a whole. <S> That exchange will not change the charges on the capacitors. <S> Neither capacitor can discharge via the connection to the opposite circuit because that connection isn't a complete circuit. <S> What the connection means is that the two circuits now share a common reference voltage. <S> Thus now it is meaningful to answer questions like "what is the potential difference between the top of the left resistor and the top of the right resistor". <S> If you add one more connection bridging the two, then you have a complete circuit. <S> Some current can then flow through one connection and, of course, an opposite and equal current will return through the other. <S> A "common ground" serves this purpose: to return the currents exchanged by the other connections. <S> It is not a true ground in the earth sense; only a common return that is called "ground" out of hubris. <S> Which connection is "ground" is just a convention (but of course there are design reasons for choosing a particular connection, rather than choosing arbitrarily!). <S> We assert that the voltage on that network is 0V, and then other voltages are referenced to that.
| Think of the voltage at any given point as relating to its willingness to accept or donate electrons; if two points have the same voltage, electrons will have no particular desire to favor one direction or the other.
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Testing graphic LCD without code and wiring I am using a KS0108 compatible graphic LCD (JHD12864E). I was wondering if there is a method to see if the lcd is working or not with a simple test. I tried sticking 5 volt between vcc and ground but nothing lights up. Is there a way I can test this module without messing around with the wires on the breadboard? <Q> It is not possible to test the module without wiring it to a microcontroller. <S> When power is applied to the module, the display remains off and uninitialized. <S> In order to turn on the display, you need to initialize the display with certain commands and parameters. <S> After which you can send characters to the display. <S> This requires it to be wired to an intelligent chip like a microcontroller that can carry out the commands to initialize the module. <S> Edit:There are also other impractical ways to drive these modules as suggested in other posts. <A> Unfortunately all of the LCD modules that I have worked with, whether they be graphic or character mode types, have required a "smart source" to check out whether they work. <S> For graphic mode displays it is highly unlikely that you can get it to display something unless you cobble up an interface and program the sequences needed to "talk" to the display module's controller chip. <S> Part of the difficulty comes into play because there are many varieties of graphic displays using many different display controller chips that all work a bit differently from one another. <S> To make matters even more difficult for the graphic displays is that the interface connection to the display will be different pinning and connector type for each model of display. <A> There is no self test mode with the JHD12864E lcd, nor with the KS0108 compatible controllers. <S> These lcds are always manufactured for production use, so it's expected that an appropriate test system (microcontroller) is setup by the end user. <S> Since you say you don't want to mess around with your breadboard or wires <S> , that's pretty much the answer, no. <S> That said, the protocol for the KS0108 lcd controller is pretty simple. <S> It's essentially a variation on the HD44780 character lcd controller (The difference being that the HD44780 has a built in character generator map [i.e. Fonts and Alphabet]). <S> The pinout and protocol is roughly the same. <S> You can test the lcd with Tedious flipping of switches. <S> And you can combine that with a 555 timer, to rapidly fill or clear the lcd with a specific pattern. <S> (Set the initialization bytes by hand, then set up the 555 timer to fill in the rest of the lcd by toggling the E pin. <S> Or just rapidly hit the E switch). <S> There are ways of doing this without a microcontroller/code, but you are way better off with one, instead of the frustrating dipswitch method. <A> No, that is not possible. <S> The display may show some pixels when you connect Vcc, ground, Vout and Vo, but you cannot tell if everything's working. <S> For correct connections: (forget about the PIC)
| Sometimes you can be lucky with character mode types and connect them in place of an existing display on another piece of equipment to see if that can drive some content on the display for test purposes.
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Drive Fight on 7400 Series I'm currently designing a 7400 series circuit to drive LEDs from a set of shift registers (74595). Each shift register's tri-state input is wired to the output of a 2:4 decoder (74139), and the decoder's input is driven by a 4 bit counter (74191). The counter has a 200Hz clock input. In steady state, this circuit runs fine. My concern is during the transition. According to the 74LS139 and 74HC595 datasheets, the decoder will transition from one output state to another faster than the tri-state output of the shift registers. This means that there is a brief period of time (about 10ns) where both shift registers will drive their outputs before they complete their transition. My question is, will this brief transitional period damage and eventually destroy the shift registers? I've attached a simplified schematic illustrating my question. Do note that this is a simplified schematic. The real circuit has way more shift registers. As such, multiplexing each shift register is prohibitive (I would need a 96:32 multiplexer, a.k.a. lots and lots of 7400 series parts). <Q> Possible solution: instead of having one R1 per driving transistor, put R1 on the output of each shift register. <S> Then the maximum current that can flow between two shift registers is limited by 2x 47k resistors, but the drive path between shift register and transistor is the same. <S> (I'm not sure if the 10ns shoot-through would actually be a problem, but given the amount of discrete logic it's worth considering <S> what happens if part of it fails and two shift registers get stuck on) <A> I don't think it's a problem provided you aren't switching between the shift registers at silly speeds (MHz instead of kHz). <S> I base this on: <S> the 10ns overlap is between typical \$t(en)=15ns\$ and \$t(dis)=23ns\$ on the 595 datasheet. <S> So it's less than half the turn-off times. <S> I doubt that the drivers will both be fully on during this overlap. <S> Drive strength (to the TI datasheet) is guaranteed 6ma, quite low. <S> Worst case drive current will be several times this figure - maybe 25ma at 5V, 0.125W <S> * 8 (no. of outputs). <S> A permanent short circuit may eventually destroy the drivers through overheating, but: The duty cycle and thus the excess heating is quite low: 10ns per (insert your switching period) - <S> multiply the power above by this (rather small) number. <S> Finally, if you are still worried about this, you could use OEn on the 74139 decoder to disable both 595s during such transitions. <A> See CMOS (or CMOS compatible) <S> shift register with latched open-drain outputs
| You could use a shift register with open drain/collector outputs so they simply can't fight each other.
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SPICE Simulator at Linux What is the better choice to get opensource SPICE simulator on Linux?I am fan of Proteus (Windows) is there any alternative like it? <Q> I'd say either ngspice with gspiceui (part of gEDA <S> I believe) or LTSpice with wine as Renan has already mentioned. <S> Here's a screenshot of ngspice on KDE (with one of the graphical addon packages like nutmeg): <S> I have a linux box <S> and I use the second option (rarely, since I'm mostly on my Windows laptop), simply because I'm used to LTSpice. <S> There are also many others, including some interesting variants like eispice , which claims to be the only open source SPICE that provides native IBIS model support (this may be pretty useful for many high speed digital designers on a limited budget - I will certainly be checking it out). <S> It was aimed at PCB signal integrity simulation initially, but has expanded to include more general purpose features. <A> SPICE was developed under, is and always has been UNIX based with most instances of the software also being actively run under Unix. <S> There are very few windows variants in comparison. <S> As a result there is a many different flavors of spice out there and many that are FOSS. <S> gEDA <S> , XSpice etc. <S> A quick search under EDA on freecode (the old Fresh meat repository) shows 16 instances, but not all are free. <S> Another search under source forge shows GNUspice and ngSpice. <S> There is even a FEL spin (Fedora Electronics Lab) that is slightly out of date and has many, many EDA tools bundled into one focused version. <A> If you don't mind using Wine to run Windows applications in Linux and using a closed-source application, LTspice runs perfectly there <S> (this is what I use, usually) <S> Otherwise, there's ngspice , for which Oli already has pointed to an ngspice GUI. <A> There is another interresting project you could take a look at which is QUCS .
| As far as I know, there isn't something like Proteus for Linux.
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Are both of my SSRs broken or am I doing something wrong? I have one side of a 125VAC panel-mount indicator connected to the AC load points of a solid state relay. The control points are hooked to a microcontroller with a basic program that switches the indicator on and off depending on characters that arrive on the UART. The problem is that even when no voltage is applied to the control points, the indicator light still glows dimly. When I engage the control points, the indicator lights up brightly. At first I thought the SSR had failed, so I ordered another one from a completely different company, but it's doing the exact same thing. Is it just a coincidence that I got two broken SSRs in a row from different manufacturers or (more likely) am I doing something wrong? EDIT: Here are the SSRs I've tried. ECE ESR5102401000 24-240VAC/10A 3-32VDC SHZHE SSR-25DA 24-380VAC/25A 3-32VDC <Q> You don't have enough of a load on the output of the SSR. <S> Most SSRs contain a snubber network that consists of a resistor and capacitor in series and then connected in parallel with the output terminals of the SSR. <S> This leakage current is enough to illuminate neon lamps and will even cause some 120 Vac strobe lights to flash periodically. <S> Connecting a small incandescent lamp (5 Watt Christmas bulb) across your load terminals is usually enough to swamp out that leakage current. <S> Note: <S> the snubber network is required for the triac inside the SSR to not false-trigger when certain conditions are just right. <S> But the snubber does introduce its own set of problems. <A> You need to check the input current on the DC side, the led on those relay will blink even if the circuit is not closed. <S> I have 25A relay and that one requires 50mA on the DC side to activate the relay. <S> i think the 10A relay needs like 25mA. <A> Are you driving the relay high (with 5 Volts), or are you sinking (taking the pin low)? <S> Some MCUs sink more current than they can source. <S> I have similar setup with 80 relays, 3 different kinds and current abilities, all running from an Arduino Leonardo/UNO with no issues. <S> Try this, if you don't already have it this way:1) put 5 volts on the + side of relay input. <S> 2) Take the common (-) side to a pin on your MCU, Arduino or PIC3) <S> To drive it, enter a 0 (or LOW) instead of a 1 (or HIGH) in your code, eg digitalWrite (relay, LOW). <S> This drives the pin low and will trigger the relay.
| The capacitor allows some leakage current to flow.
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connecting Wheatstone bridge to pin 3 of LM324 changes voltage? Built a Wheatstone bridge with two 10k resistors, one 5k rheostat, and a thermistor.Adjusted rheostat so that voltage across bridge reads < 0.013v, close enough. Powered LM324N, pin 4 to 6.07v, pin 11 to 0v (just one 6v battery supply) When I connect pin 3 of the LM324 (Vin1+) the voltage of that leg of the Wheatstone bridge changes to 5.4v. I assumed I had blown the opamp previously, and tried a fresh one. Same result. So I simplified the circuit. The only connection to the Wheatstone bridge is a single wire that goes to pin 3 of the op amp. Nothing else is attached, and yet connecting that wire changes the voltage between the two resistors to 5.4v. When I disconnect, they go back to 3.03v as they should be. Here's a picture of the breadboard in case I am just out of my mind. I can't believe power is leaking out of the input of an LM324, I must be doing something boneheaded but can't see it.Picture attached. The perspective isn't 100% ideal, but the wire from the Wheatstone bridge is connected to pin 3, and pin 4 is Vcc. This was my attempt to simplify the original problem. I built a differential amplifier to amplify the output of the Wheatstone bridge, and current was leaking out of both positive inputs back into the bridge. So I wanted to isolate the problem. I don't have a picture of the board implemented that way, but here's the schematic: R1 was 10k, Rgain was 1k, R2 and R3 were all 1k <Q> Here is a 5 second answer to start. <S> First, put a decoupling cap (0.1uF or something like that) between LM324 pins 4 and 11 (its just good practice). <S> Just for testing connect LM324-1 to LM324-2 (unity gain), because an OpAmp is meant to be used with feedback. <S> Of course you know that you will need a differential amp sensing both sides of the bridge to get any advantage out of using the bridge. <S> Edit <S> Thanks for adding the schematic. <S> The output differential stage, as drawn is positive feedback. <S> So, you will need to exchange pins 9 and 10 of the LM324 connection to have negative feedback. <S> Also, I didn't think about this much earlier, but the 4th channel would be best to be connected as a voltage follower with the (+) input connected to return. <S> I know that each channel is supposed to be independent, but they are not quite really. <S> The die will have parasitic structures that can affect all the channels. <S> For example, parasitic SCRs are common in many monolithic ICs. <S> So, it is always best to have all channels connected. <A> Connect all inputs to some proper configuration, for example - voltage follower. <S> Inputs of other opamps in a package left floating can affect the performance of the whole package. <S> Also, connect opamp 1 in voltage follower. <S> Also, put a 100nF decoupling cap between pin 11 and pin 4. <A> The answer appears to have been none of the above. <S> The op amp was blown. <S> I replaced the op amp, and the circuit simply works.
| While placing a capacitor across Vcc+ and Vcc- is probably a good idea, as is connecting all negative inputs to outputs, it appears not to have any effect.
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Can I safely substitute a 4.7uF cap for a 5uF cap? I'm working on a circuit that calls for a couple 5uF caps. I have 50v and 25v 4.7uF at my workbench. No cap voltage is specified in a schematic or parts list. The circuit has 12VDC coming in, one negistor that'll be connected to the path of the caps (though downstream), so I think the 25v will be sufficient. But is it safe to substitute 4.7uF for 5uF? What's the worst that can happen by having less capacitance in the circuit by 0.3uF? <Q> Without knowing the details of the circuit, it is impossible to tell. <S> Specifically we don't know exactly how this cap will be used, and so we do not know how critical the value is or what tolerance is required. <S> In many applications that substitution is acceptable, but in some it might not be. <S> But, wait, there is more! <S> 5.0 <S> uF is not a common value of cap. <S> In fact, I have <S> NEVER seen a 5.0 uF cap. <S> I'm sure there is one out there, but it is probably some weird cap intended for some weird application with an equally weird price. <S> Basically, not something that 99.99% of the people will ever use in real life. <S> Even then, 5.1 uF cap would be a more likely value (because it is a more "standard number"). <S> Here is a web page showing standard cap values: http://www.rfcafe.com/references/electrical/capacitor-values.htm <A> Electrolytic caps have a tolerance of minimum 10% (the good ones), so 5 uF may be anything between 4.5 uF and 5.5 uF. <S> For the 4.7 uF <S> that's between 4.23 <S> uF and 5.17 uF, so both ranges overlap for the most part. <S> If the operation in the extremes would be important a capacitor for that value would have been chosen. <S> So it's safe to say that 4.7 uF is a valid alternative for a 5 uF. <S> But 25 V on a 12 V supply leaves very little headroom, so I would use 35 V at least. <A> It's likely well with in the tolerance of the speciaiction. <S> and the voltage rating is safe too. <A> For the oscillator you're referring to because you've exceeded voltage rating required the only effect will be a change in frequency. <S> However a typical aluminum electrolytic capacitor has a tolerance of 20% so a hyphothetical 5uF capacitor could be anywhere from 4-6 <S> uF <S> so your 4.7uF subsitution will be fine, although technically a 5uF would be closer to the original design mark. <S> For future reference keep in mind if you did need an odd capacitance and were using parts with a low enough tolerance to make sense (some types can be closer to 1%) <S> then you can place standard values in parallel to get a close match. <S> For example a 4.7uF plus a 0.33uF in parallel would yield 5.03uF
| For the voltage 25 V may be OK, since you hardly see more than twice the power supply voltage, unless you're actually building a voltage multiplier (step-up converter of Greinacher circuit, for instance).
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How do I select the correct inductor value for the following buck regulator? First of all, I suck a bit at math, and I'm no electronics genius, so the stuff I do is for fun and for learning purposes... I'm working on a buck converter circuit to convert my USB Vbus 5V to 3.3V. I've selected the AP5100 and finding it quite challenging to figure out the correct values on some of the components. The datasheet neatly specifies the values for R1(49.9kΩ) and R2(16.2kΩ) in Table 1 on page 6, to establish an output voltage of 3.3V, but I'm finding it a bit of a train smash understanding how to calculate the inductance value for the L1 inductor. The datasheet indicates 3.3µH on page 2, Figure 3: I'm wanting to understand better how the 3.3µH was calculated, and if this is in fact the correct value for my application. Now back to the datasheet, the formula for calculating L is stated as: $$L = \frac{Vout \times (Vin - Vout)}{Vin \times \Delta IL \times fSW}$$ Where ΔIL is the inductor ripple current, and fSW is the buck converter switching frequency. The datasheet states: Choose the inductor ripple current to be 30% of the maximum load current. The maximum inductor peak current is calculated from: $$IL(MAX) = ILOAD + \frac{\Delta IL}{2}$$ Alright, this is where I'm horribly lost, and trying my best to wrap my tiny brain around the value. I know the following: Vin = 5V (USB Vbus) Vout = 3.3V fSW = 1.4MHz I = 2.4A (I think) How does one determine the ΔIL (ripple current) in order to get to the inductor value? My formula should look something like this in the end, right? $$L = \frac{3.3V \times (5V - 3.3V)}{5V \times \Delta IL \times 1.4MHz}$$ But what is ΔIL? Also I thought the buck converter was supposed to allow a range of inputs for Vin, in the case of this one, 4.75V to 24V? Here's my schematic I'm drawing in Eagle CAD: <Q> Choosing an inductor value for a buck regulator comes directly from V = \$\frac{\text{L di} }{\text{dt}}\$ . <S> Where V is the voltage across the inductor, and i is the current through it. <S> First, you want to design for the case where the inductor is in continuous conduction mode (CCM). <S> This means that energy in the inductor doesn't run out during the switching cycle. <S> So, there are two states, one where the switch is on, and another where the switch is off (and the rectifier is on). <S> Voltage across the inductor during each state is essentially a constant (although it is a different value for each state). <S> Anyway since the voltage is a constant, the inductor equation can be linearized (and rearranged to give L). <S> L = <S> \$\frac{V \text{$\Delta <S> $t}}{\text{$\Delta <S> $I}}\$ <S> this is the basis for the equation you saw in the app-note. <S> \$\text {$\Delta <S> $I}\$ is something you define, not determine. <S> You will want to maintain CCM operation, so define \$\text {$\Delta $I}\$ as some small fraction of inductor current (I). <S> A good choice is 10% of I. <S> So, for your case \$\text {$\Delta $I}\$ would be 0.24A. <S> This will also define the ripple current in the output capacitors, and less ripple current means less ripple voltage on the output. <S> Now you can choose an optimal value of L using \$V_{\text{in}}\$ and \$V_o\$ <S> (and hence the duty cycle D = <S> \$\frac <S> {V_o} <S> {V_ {\text {in}}}\$). <S> But you can also make a quick over estimate for the inductance where you don't consider \$V_{\text{in}}\$ using L <S> ~ <S> \$\frac{10 <S> V_o}{I_o <S> F_{\text{sw}}}\$ (for more on this look here How to choose a inductor for a buck regulator circuit? ). <S> An over estimate can be worthwhile, especially if you are early in development or uncertain exactly how much the output current will be (output current tends to end up higher than expected usually). <S> Since you are looking at Linear Tech you should (as Anindo Ghosh pointed out) also look at using their CAD support. <A> For designing a buck regulation circuit, it might be better to start with one of the various free online power design tools on the web sites of manufacturers, such as: <S> Fairchild Semiconductor: Power Supply WebDesigner <S> Linear Technology: <S> LTPowerCAD <S> Texas Instruments: <S> WeBench Power Designer and SwitcherPro Design Tool <S> By providing your requirements (including acceptable ripple for instance) as parameters, the tool would typically shortlist a set of controllers that meet the purpose. <S> This is usually a safer approach than starting with a controller already decided upon, then attempting to deviate from the datasheet specified values for support components. <S> Many of the mentioned free "power designer" tools provide a complete bill of materials as an output - including the inductor(s) needed, typically with part numbers. <S> Some (e.g. TI WeBench) also provide recommended layout and required board space estimates. <S> Some tools further allow desired board space as a design parameter, as also component count, cost, and other preferences. <A> It might help to understand what happens if you select an inductor of the wrong value. <S> If you select an inductor with too low a value, the current through it will change too much in each switching period. <S> The current might grow so much in a switching period that it exceeds the current capability of the circuitry driving the inductor. <S> This high ripple current also isn't nice to the capacitor on the output side. <S> ESR losses in the capacitor will be high, or ripple current will exceed the capacitor's rating <S> and it will fail. <S> If you select an inductor with too large a value, you will be paying for a lot of inductor you don't need. <S> Inductors with a core have a saturation current. <S> This is the current at which the core can not take any more magnetic flux, and the inductor stops being an inductor with a core, and starts being almost a wire. <S> For a given core of a given size and material, you can make an inductor with higher inductance simply by putting more turns of wire around it. <S> But, each of these turns contributes more magnetic flux, so by adding more turns, you are also decreasing the saturation current of the inductor, since your current will be multiplied by the number of turns of wire to arrive at the magnetic flux through the inductor. <S> Thus, if you want a higher inductance at the same saturation current, you need a physically larger core. <S> I'll leave an explanation of the math to another answer. <S> I'm not the best at such things. <A> How does one determine the ΔIL (ripple current) in order to get to the inductor value? <S> Use the rule of thumb provided by the chip maker (select a value that's equal to 30% of your expected maximum DC output current). <S> Also, there's a note on page 8 that says "A 1μH to 10μH inductor with a DC current rating of at least 25% percent higher than the maximum load current is recommended for most applications. <S> " <S> I = 2.4A (I think) <S> However, it sounds like you aren't sure what your expected maximum DC output current is. <S> Take a look at this waveform shamelessly borrowed from Wikipedia: <S> The inductor current is shown at the bottom. <S> The magnitude of the ramps are defined by \$V_L = <S> L \cdot <S> \dfrac{\Delta I}{\Delta <S> t}\$ <S> This is why it's important to know your maximum expected DC output current before trying to choose the inductor. <S> Also bear in mind that this part has internal error amplifier compensation, which will put constraints on the output LC filter (don't deviate from their inductance range unless you have equipment to measure the closed-loop frequency response of the converter).
| You define your inductor by the size of the ramps: that is to say, by choosing a value that's 30% of the maximum expected DC output current (\$I_{av}\$ in the waveform).
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How do you simulate voltage noise with LTSpice? Is there a way to setup a voltage supply with voltage jitter/noise? I want to experiment with filtering out noise on various voltages etc. but not sure how to configure LTSpice to create a noisy voltage supply. <Q> Yes, you can inject noise using the arbitrary voltage (or current) source, then use things like the random or white function to create some noise. <S> Here is an example circuit (I separated the noise from the signal just to make things clearer - <S> obviously you can combine them together in one function if you wish): <S> Simulation: <S> All the functions are detailed in the help under circuit elements -> arbitrary behavioral voltage or current sources . <S> Noise simulation mode <S> Also, just in case you were not aware, SPICE has a noise simulation mode, to quote from the help <S> files: .NOISE -- <S> Perform a Noise AnalysisThis is a frequency domain analysis that computes the noise due toJohnson, shot and flicker noise. <S> The output data is noise spectral density per unit square root bandwidth. <S> Syntax: .noise V(<out>[,<ref>]) <S> <src> <S> <oct, dec, lin> <S> <Nsteps> <StartFreq <S> > <S> <EndFreq> <S> Basic example: <S> Simulation: <S> The above is rather boring as it only models the resistor noise (I stepped the resistor through various values to show how the Johnson noise increases with resistance). <S> But it can be very useful with more complex circuits containing diodes/transistors/opamps/etc. <A> (Not enough rep. yet to create a comment on Oli's post, so this goes in a post of it own). <S> Oli's post above is very useful, but for the LTSpice beginner, it is perhaps worth explaining how to actually create of one of those "arbitrary behavioral voltage source" : I was naively expecting to be able to modify the value of a normal voltage source to enter the white(...) formula, but of course, it does not work. <S> Instead, you have to press the "component" button in the toolbar, and in the window that opens, pick a component of type "bv". <A> SPICE <S> ( I can't tell you if LTSPice is a subset of normal SPICE or not) <S> normally has the ability to model the noise that each device generates. <S> I think your question is more about how to you measure how effective your filtering is and how much an external interfering signal may affect each node. <S> To do that what you need to do that is .AC <S> analysis of the circuit. <S> To do a noise analysis you need to use both <S> .ac <S> and .noise. <S> So noise analysis is a subset of ac analysis. <A> <A> To create a voltage noise source in LTSpice, select the component "bv" For the function, enter "V=white(time) <S> " <S> This gives you white source of +-0.5V with upper band around 1Hz. <S> Multiply time*BW to move the upper band of the noise to BW. <A> In series with your voltage source add an E element whose inputs are attached to a 60.34e18 ohm resistor and whose gain is the desired V/rt-Hz. <S> This resistor value produces 1V/rt-Hz at 300K. <S> (Warning: you must ground one side of the input.) <S> You can use the same resistor value as the input to a voltage-controlled current source whose gain is the desired Amps/rt-Hz. <S> (Again, one input lead must be grounded.) <S> The noise current source connects in parallel with your noiseless current source. <S> You must use a separate resistor for each noise source.
| As you want to inject noise from the power supply, I think easiest is to put a small amplitude AC voltage source in series with the DC voltage source you already have and sweep its frequency through the range you are interested in.
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How to make traces on an universal PCB? I hope this question won't be closed as too subjective. I'd like to know best practice - how to make traces on universal PCB's with individual holes without traces (like the following image). My idea is to bend the ends of discrete components and use them to make traces to other components. Is this approach acceptable, or nasty? Although this is called "prototype" PCB, I'd like to use it for my simple final circuits (low frequency, low current applications), because I hope it will save time. <Q> Protoboards are up to you how to use, if it works, it works. <S> The three common methods of using them is using jumper wire, solder bridges, or using leads. <S> Or all three, depending on your needs. <S> Solder bridges take a lot of solder, especially long ones. <S> They are good for two or three adjacent points. <S> And jumper wire is best for when you can't/don't want to go around an existing solder joint. <S> It all really boils down to what you need. <S> Of course, solder bridges can be very sloppy if you don't have practice at them. <S> And using too many jumper wires or bare leads looks ugly and not well thought out. <S> But this is your project, you can figure out how to mount them, and what's acceptable or not. <S> If you need much of either, what you really want to do is think about the placement of your components, move stuff around to minimize the need to use jumpers or long bridges. <A> Typically a board like this has components connected using wire from pad to pad. <S> I usually use 30-AWG wire-wrap wire . <S> I often place components in such a way that leads that need to be connected are placed in adjacent holes. <S> It's quite easy to create a solder bridge across two pads. <S> Sometimes I'll need to connect three or four leads. <S> If possible I place these along a horizontal or vertical line of pads and create a solder bridge across the length of them. <S> It's somewhat more difficult than a 2-pad bridge, because the solder will have a tendency to bridge only at the position of the iron, where the heat is. <S> With a little practice, it's entirely possible to create lengthy solder bridges that span several pads. <S> However, it can be ugly, waste solder, and is difficult to change. <S> Multiple bridged pads is something of a hack, so <S> while I do it sometimes, I don't recommend it for best results. <S> You can improve the bridged pad idea slightly, by adding stripped 24-gauge or smaller wire to the pads and using it to connect them. <S> This image and caption/tutorial from <S> brewpi.com : <S> ...use some naked wire for long tracks and just solder it to the board. <S> For very short connections, you can do it without a piece of wire: Set your soldering iron to a lower temperature, and first put some solder on both pads. <S> Then put some solder on your iron and tip in between the pads. <S> When you are lucky, you’ll connect them. <S> It takes some practice but it is definitely easier at lower temperatures. <S> I personally find bending the lead of a component to connect to another component on the underside (solder side) of the board less desirable than creating solder bridges. <S> The reasons are: the leads are thicker than the wire I would use instead the leads are equivalent to bare wire and could easily make contact with an adjacent pad that should not be connected difficult to remove/replace a component <S> In conclusion, if you need some bus lines, I would highly recommend spending a few dollars to get a solderable perf board that already has them as part of its design. <S> This 6.3" x 3.94" board is $5 at All Electronics . <A> I know this isn't exactly an answer to your question <S> but I thought it might help others. <S> I still use perfboard a lot because it is more commonly available <S> and I have a bunch of it laying around, but since I discovered stripboard, it is my preference. <S> If you workout your layout, you can minimize the number of traces you have to create with soldier or wire. <S> This is very similar to what JYelton posted at the end of his answer <S> but you can find boards like this all over ebay for dirt cheap. <A> It is possible and I have to admit I'm doing the same thing.. <S> but I cannot recommend it because it's very hard to take the parts out later, when something's broken. <S> The holes stick not that good to the board most of the time, so with too much soldering you would get the hole of your board. <S> That's what happens sometimes when you're taking out parts again. <S> So it's better to connect wires on the backside instead of using the leads of the components.
| Using the leads, or bare wire, is great for straight lines and buses, using less solder than solder bridges. Instead, you use a drillbit to break the strips of copper where you need in order to create your traces.
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Powering a 12V and a 9V device from single power supply with LM7809 voltage regulator I have a little electronics project where it would be nice if I could power two devices from a single power supply (12V switching adaptor). One of the devices is a high power LED with 350mA, 12V and the other is a 9V lower amperage device. Using a 12V, 1A power supply, would the circuit below work to regulate down the voltage?Will the LM7809 will have problems with the load if I switch off the LED? Sorry if the questions seem silly, it is the first time I am dealing with that kind of stuff. EDIT: I'm aware that the LED is not added correctly. This was more for illustration really. It is a high power 10W RGB LED that I was hoping to connect with this constant current circuit <Q> The LED does not affect the load of the voltage regulator. <S> It affects the supply. <S> The voltage regulator's load is the 9V device. <S> The only way it will cause a problem for the regulator is if its current draw triggers a voltage drop below the point that the regulator is able to maintain 9V. <S> The main question about your circuit is: is the incoming 12 volts regulated? <S> Or is it just the nominal voltage on a transformer device such as a wall adapter? <S> If 12V is the maximum rating of the LED device, but the supply is unregulated, you may want an additional 7812 regulator there. <S> (What kind of LED is it; and does it have a built-in resistance for current limiting?) <A> You should be fine regardless of the state of the LED. <S> As a side note the more current you draw from the 9V regulator though the hotter the regulator will become. <S> At 300mA I would expect it to get warm without a heat sink. <A> Also, you will need a heatsink if you want to draw up to <S> 750mA. <S> The regulators <S> thermal resistance θj-a is 65°C/W, so for every watt dissipated it rises 65 <S> °C above ambient temperature. <S> Your input voltage is 12V, and the output 9V, so at 1A, the dissipation equals: (Vin - Vout) <S> * Iout -> <S> (12V - 9V <S> ) * 750mA <S> = 2.25W <S> so the temperature rise above ambient without a heatsink would be: 2.25W * 65 <S> = 146.25 <S> °C - too much, the ICs thermal protection will kick in and shut it down. <S> So you need a heatsink, to bring the θj-a down to at least, say 35°C/W, preferably much lower. <S> Here's a reference on Heatsink Basics . <S> This is an example of a heatsink that would help keep your IC cool. <S> It's thermal resistance is 24°C/W, so added to the 5°C/W thermal resistance of the junction to case (which we can't do anything about) <S> we get 24°C + 5 <S> °C = 29 <S> °C/W. <S> So the temp rise above ambient is now 2.25W <S> * 29 = 65.25. <S> Not great, but you have some margin there. <S> For your maximum ambient working temp, subtract this value from 125°C (maximum working temp for the regulator) and you get 125 - 65.25 <S> = 59.75°C. <S> This can easily be reached in an enclosed circuit, so usually you would want a bit more "headroom" than this, <S> e.g. the ability to work up to 70°C ambient. <S> There are loads of heatsinks to choose from out there though, many with very low thermal resistances that will handle this dissipation easily.
| The 7809 will have no problems if you switch off the LED, no. As long as you keep the input voltage above 11V (regulator output voltage + regulator dropout voltage), it will regulate okay.
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Mobile AC Power Source for Projector? I need to use a projector for a presentation (about 30 mins, but I'll need to make it about 3 times), but there will not be any wall outlets available. I thought about using an inverter and a 12/24v battery (auto?), but I'd like to know if there are any easier options, or if I'm missing something in my original idea.The projector is a BenQ MS612ST, it has/needs a power supply of 100-240 VAC and consumes 275W.Can anyone help? Thanks so much in advance! <Q> A 280W load will drain a car battery rather quickly. <S> That's something like having seven headlights on. <S> Three repeats of 30 minutes? <S> You're lucky if you get through one. <S> The capacity you will see on a fast discharge rate will fall short of the theoretical Ampere-hours capacity of the battery. <S> I would borrow or rent a portable generator that runs on fossil fuels. <A> An inverter is not 100% efficient, and then you would have some more loss through the mains adapter too... <S> but if by some bit of luck your projector ran directly from the 12v battery (you might still want to use a DC regulator but that will be more efficient) <S> you might just swing it. <A> What you are describing is essentially a UPS. <S> You can go to APC's website and browse their products to get an idea of how feasible this is. <S> It looks like the smallest unit they have which would have a chance of running your load for 90 minutes is the APC Smart-UPS 2200VA (about $1000 list price), but given manufacturer's tendency to overstate performance, you probably want to go bigger. <S> As you can see, not cheap or small. <S> Even if you build it yourself, you can't overcome the physical limitation of battery energy density.
| It depends if the projector uses mains voltage directly (EG to power a 240v lamp) or if it actually uses a mains adapter to drop the supply to low-voltage DC.
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Is it ok to connect the output of buck regulator in parallel? I am using MCP16322 buck regulator powered from 12V and outputs 5V and 2A. Is it ok to connect the output of two of these in parallel? Does connecting the outputs in parallel mess up the maximum capacitance values on the output of the regulators? Is it better to connect the outputs in parallel via diodes? The diodes will cause a .7v drop though which I rather avoid. Here is the application circuit. <Q> Directly connecting the outputs of multiple regulators, switched or linear, is inadvisable for the following reasons: A marginal difference in output voltage would cause high currents to flow between the regulator output pins, potentially damaging one of the regulators. <S> The MCP16322 is rated for 2% precision, hence for a 5 Volt nominal output, one regulator could be at 4.9 Volts, the other at 5.1 Volts. <S> The 0.2 Volt gap would cause current flow between outputs limited only by the rail impedance of the regulators. <S> Any delay in powering up or powering down of either regulator would cause a back-feed from the powered regulator to the non-powered one. <S> By design, the approach stated in the question will have one of the regulators operating while the other may not be - if one of the power sources is off at a given time. <S> This is a failure mode with strong likelihood of device damage <S> Even if the two regulators were powered by a common source, there will be mismatches in power-up timing while the two oscillators are starting up. <S> This is why sequencing of power supplies is required, and there are special-purpose parts for this sequencing. <S> A buck controller that supports synchronization and sequencing would be required, instead of the selected device. <S> If the design proposed in the question is used as-is, even if there is no immediate failure, component deterioration would reduce the expected longevity of the device due to repeated exposure to stresses not designed for. <S> The design can then use a single buck regulator instead of multiple. <S> The datasheet indicates that the regulator will not have any trouble using a 11.3 Volt input instead of 12 Volts, to produce a regulated 5 Volt output as desired. <S> This article about sequencing of multiple voltage rails might be useful reading, it discusses the sequencing and component degeneration issues. <A> It is generally not a good idea to parallel the output of two power supplies. <S> Both power supplies are unlikely to be at the same exact output voltage. <S> As a result one will tend to try to supply all of the load whilst the other one will tend to idle along at low load. <S> Depending upon the filtering characteristics used in the feedback networks on the two power supplies it is possible that oscillation could also happen. <S> Now all that said there are power supplies designed that are specifically designed to be able to be paralleled. <S> These often have a special sense line that connects between all the power supply outputs that is used to support a balanced current sharing between the supplies. <S> Current sharing supplies also have to add additional levels of fault detection to ensure safe operation/shutdown in the event that the common current sharing scheme fails or some component in a particular power supply fails. <S> It is not uncommon to see this type of parallel usage power supply used in server computers where power delivery is added in modular manner to the server as additional CPUs, memory and I/ <S> O boards are added to the server. <S> Many of these power supplies contain internal microcontrollers that run sophisticated fault detection algorithms to make them safe in failure modes. <A> To overcome the problems due to parallel connection (mentioned in earlier answers) you can make some circuit modifications as listed in this article at Electronic Design or this application note by Texas Instruments . <S> Of course it all depends from circumstances but this recommendation may help you to enlarge current capability of yours design. <A> No it is not recommended. <S> As already pointed out there will be tolerances in the circuits, leading to slightly unequal output voltages. <S> The buck with the highest output voltage will 'win' and supply all of the load current at first. <S> When it reaches its current limit, its output voltage will collapse. <S> At this point, the DC-DC with the lower output voltage will begin contributing. <S> It is quite common in industry to parallel DC-DC converters using a technique called voltage droop. <S> This is basically creating an artificial resistor in series with the output of the DC-DC converter to aid current sharing. <S> You can read more in this article . <A> Yes, this is possible but with a lot of complications involved using transformer based solutions such as managing current sharing. <S> There are capacitor based solutions as well, which are recently being invented that don't have to worry about current sharing as much, and are detached by frequency of operation, though there are probably other issues. <S> A lot of the industry uses POL to manage the conversion right at the CPU or point of load, but there are some companies looking for alternatives. <S> Though I imagine multi-phase solves this for certain applications. <S> Though it's quite a bit expensive in terms of board area and cooling solutions. <S> Cuk converter may be a good alternative as well for reliability and long lasting solutions.
| There will be higher peak voltage / peak current demands on output stage capacitors of the regulators, due to additive effects of the (non-synchronized) ripple voltages of the two. Designs of this type are more expensive and do add additional components to the circuit board. The solution : Instead of a diode-OR of the outputs of the two buck regulators, use diodes to merge the 12 Volt input sources .
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Convolution perfomed by an analog circuit As a Electronic Engineering student I have a fair knowledge about convolution and DSP. But, I was wondering if it is possible to perform a convolution only using analog circuit (without memory)? And if it is possible, what would be the restrictions? In short, I would like to project this using only an analog circuit: $$y(t) = (x * h)(t) = \int^{b}_{a}x(\tau)h(t-\tau)d\tau$$ Clarifications: Both signals would be an arbitrary input (x and h in the above formula). I am willing to make simplifications of all kinds, since it is doing what I am asking. <Q> Before digital processing got fast and cheap enough to do convolutions, various ways were developed to do it in analog electronics. <S> If you want to convolve two arbitrary signals, then you're out of luck unless you are willing to make a lot of compromises and/or spend a lot of money. <S> Either way, some storage is required for each signal, but with one signal fixed it can be implemented by a "permanent" memory, which allows for a lot more possibilities than doing it on the fly. <S> You still have the problem of storing some portion of the live signal, since some interval of that needs to be multiplied by the kernel as the signal passes by. <S> Systems have been developed that do this with delay lines, traveling electron beams, bucket-brigage charges on a CCD, and accoustic waves. <S> There are probably others that I am not aware of or forgot about. <S> Once you can somehow store a snapshot of the live signal wide enough to match the filter kernel, you will have to then multiply it by that kernel and sum up the products. <S> In delay line systems, this would be done with "taps" at regular intervals. <S> The signal at each tap would be multiplied by a fixed gain (the filter kernel value at that tap), then all these resulting signals summed. <S> CCDs had split pickups over each charge bucket so that the gain for each bucket was set by where the split was located. <S> This would be set when the chip was made, so there were CCD filter chips with certain pre-determined filters. <S> The most common use was for a sync filter, which is a low pass filter with a sharp frequency cutoff. <S> Surface accoustic wave devices had the signal propagate accross the chip accoustically, which is much slower than light <S> so a large enough time snapshot would be on the chip at any one time. <S> Like with CCD, the pickups were arranged on the chip with pre-determined gains. <S> These parts were typically used for IF and RF notch filters at a well tuned frequency. <A> For a Linear Time-Invariant system convolution is equivalent to filtering. <S> When you pass a signal through an LTI system you are simply convolving it with the impulse response of the system. <S> If you want to convolve two signals though then that's a lot trickier to do in the analogue domain. <S> It would certainly need "memory" in some form, e.g. a delay line. <A> If you look at your equation, you will need to replay X & H at many values of Tau as you integrate on the fixed interval from a to b. <S> This means that you will need storage/memory of some sort. <S> But what a good question. <S> On one end of the spectrum you have a sampled and digitized sequence (commonly called "digital") <S> on the other you have a purely analog signal. <S> Intermediate between the two is a sampled analog system. <S> The act of sampling and storing (whether analog or digital) allows operations such as convolution, and non-causal filtering which is what your equation is one form of. <S> The very first CCD's (Charge Coupled Devices) where developed for similar signal processing tasks as you describe. <S> Although those early signal processing chains were notably less complex than your choice, being simple delay lines and feedback/feedforward systems. <S> For example, guitar effects like a flanger and echo were done using CCD's. <S> (I may have the terms guitar effects wrong - please correct me). <S> In your case you would need to run the sampled system many many time faster than the sampling rate of the incoming signal. <S> If your sample depth is say 16 then it would need to be run \${16}^2\$ = <S> 256X faster. <S> I do know that these devices exist still, in some signl processing applications like support chips for image processing in analog signal chains. <S> And they would be called analog delay lines or sampled analog delay lines. <S> But in purely analog sense without sampling you would still need an analog memory of some sort which is replayable.
| Historically, analog convolutions were limited to convolving one real time signal by a pre-determined fixed signal, called the "filter kernel".
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Ideal temperature and humidity for preventing problematic ESD conditions? I know manufacturers often keep components sensitive to ESD problems in climate controlled rooms and I'm just wondering what the ideal temperature and relative humidity is to prevent ESD problems? <Q> Nobody relies on temperature and humidity to prevent ESD. <S> Some people might use it to minimize ESD, but they don't rely on it to eliminate the problem. <S> If you want to eliminate ESD as a potential problem then there is no replacement for proper equipment and proper handling of parts. <S> Ground straps, conductive/dissapative tools, floor, and furniture. <S> All the usual stuff. <S> Temperature does not matter much, but humidity does. <S> Ideally, the whole factory would be under water. <S> Clearly that is not practical, as it prevents the soldering irons from heating up. <S> So anything less than "underwater" is a compromise of ESD safety. <S> Ironically, it is becoming common to store sensitive electronics in low-humidity environments. <S> Some chip packages, like BGAs, are humidity sensitive. <S> When stored in a humid environment they absorb water from the air. <S> Before soldering, the parts have to be baked to get that water out. <S> After soldering water absorption from the air isn't a factor (for reasons I don't completely understand). <A> From Henry Ott re: static discharge (ch 15.3) in "Electromagnetic Compatibility Engineering" Charge accumulated on an object leaves the object by one of two ways, leakage or arcing. <S> Because it is better to avoid arcing, leakage is the preferred way to discharge an object. <S> Charge can leak off an object through the air, because of humidity. <S> The higher humidity, the faster the charge will leak off the object. <S> Temperature affects humidity, so that is where the temperature sensitivity arises. <S> Temperature per se does not directly affect ESD conditions. <S> One techniques that is used , aside from ground straps etc. <S> is the use of ionizers on benches when there is extreme ESD sensitivity. <S> MIL-STD-1695 addresses relative humidity levels between 30 - 70 percent in areas where electronic parts and hybrid microcircuits are handled or processed. <A> Actually, ionizing the air — rather than humidifying it — is much more effective at encouraging the leakage of static charges, and it doesn't incur all the other problems associated with high humidity. <S> However, ionization systems need to be carefully designed so that they don't end up depositing charges where you don't want them.
| Temperature matters only in that it can effect humidity.
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Why does the CAN bus use a 120 ohm resistor as the terminating resistor and not any other value? I know the reasons for using terminating resistors on a CAN bus and how important it is. But why 120 ohm? How did this value come up? Is there any specific reason to use 120 ohm? <Q> You need to be familiar with Transmission Line Theory to understand the deeper physics in play here. <S> That said, here's the high-level overview: <S> Here length is determined in terms of wavelengths. <S> If your bus is shorter than one wavelength over 10, the termination is irrelevant (practically) since there is plenty of time for the reflections introduced from an impedance mismatch to die out. <S> Length defined in wavelengths is a strange unit on first encounter. <S> To convert to standard units you need to know the velocity of the wave <S> and it's frequency. <S> Velocity is a function of the medium it travels through and the environment surrounding the medium. <S> Usually this can be estimated fairly well through the dielectric constant of the material and assuming free-space surrounding that medium. <S> Frequency is a little more interesting. <S> For digital signals (such as those in CAN), you are concerned with the maximum frequency in the digital signal. <S> That is well approximated by f,max = 1/(2*Tr) where Tr is the rise time (defined 30%-60% of the final voltage level, conservatively). <S> It isn't specifically important which value they picked within a broad range (for example, they could have gone with 300 Ohms). <S> However, all devices in the network have to conform to the bus impedance, so once the CAN standard was published there can be no more debate. <S> Here's a reference to the publication (Thanks @MartinThompson). <A> That type of CAN bus is intended to implemented by a twisted pair of wires. <S> The transmission line impedance of unspecified twisted pair isn't exact, but 120 Ω is going to be close most of the time for the relatively large wires commonly used for CAN. <S> The resistors also have another function in CAN. <S> You can think of CAN as a open collector bus implemented as a differential pair. <S> The total of 60 Ω is the passive pull-together of the CAN bus. <S> When nothing is driving the bus, the two lines are at the same voltage due to the 60 Ω between them. <S> To drive the bus to the dominiant state, a node pulls the lines apart, about 900 mV each, for a total of 1.8 V differential signal. <S> The bus is never actively driven to the recessive state, just let go. <S> That means the resistance between the lines needs to be low enough so that the lines go back to the idle state in a fraction of a bit time. <S> Note that the actual CAN standard says nothing about the physical layer other than it must have these dominant and recessive states. <S> You could implement a CAN bus as a single ended open collector line, for example. <S> The differential bus you are thinking of is very commonly used with CAN, and is embodied in bus driver chips from various manufacturers, like the common Microchip MCP2551. <A> CAN Bus is a differential bus. <S> Each differential pair of wire is a transmission line. <S> Basically, the terminating resistor should match with the Characteristic Impedance of the transmission line to avoid reflection. <S> CAN bus have a nominal characteristic line impedance of 120Ω. <S> Due to that we are using typical terminating resistor value of 120Ω at each end of the bus.
| How important termination is to your system is almost exclusively determined by how long the bus wires are. Why it's 120 is simply a function of the design limited by physical size.
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Detecting dry-running of water pump What are some reasonably reliable ways of detecting that a water pump (electrical motor based) is "dry running", i.e. no water is passing through the pump either because there is no water available at the inlet, or due to "air-block" ? I'd prefer mechanisms that are least intrusive, i.e. ideally - do not require re-plumbing. However, if all practical mechanisms involves some plumbing work, then it might as well. An indication regarding possible cost of sensors, installation skill required, would be very useful. <Q> Given this additional detail: <S> Not familiar with the typical plumbing terminology, but it is certainly not submerged. <S> Pump inlet is fed by an underground sump (about 10 feet deep), pump itself is at ground level, and it pumps up water into an overhead tank, which is about 40 feet above ground. <S> Typically it is a 1.5HP 2000W 220VAC monoblock pump. <S> I am just guessing that the configuration is "in-line", I could be wrong. <S> I've got a very simple, cheap, reliable solution for you: remote monitoring. <S> Nothing says your sensor has to be at the pump's inlet. <S> You could place your sensor at the system's inlet. <S> Submerge a cheap float sensor at the level of the inlet to the feeder pipe in your well (or whatever water source). <S> Run wires back to the pump's power controller. <S> When the float sensor says the water level is below the feeder pipe, turn off the pump. <S> Ten bucks , totally reliable. <S> Done. <S> =) <S> This figure shows multiple float switches at increasing depths in a tank to illustrate operation. <S> You just need one at the bottom of your feed pipe. <S> You do not need to disturb any existing plumbing to do this. <A> Note that AC motors all have a current surge at the very start, so <S> your circuit should probably look for the current level for a few seconds before making its determination. <S> As far as how to wire it up -- a current transformer would be slipped over ONE of the motor leads -- preferably the hot lead. <S> You might want to also sense whether the motor is being asked to pump or not so that you could also detect a loss of power or a broken motor lead. <S> The circuit could also easily detect a jammed or stalled motor by a current draw above the normal pumping current level, but this too would have to be tested to ensure that the "pumping water" vs "stalled" conditions were different enough to reliably detect. <S> Alternately, you could just count the time that the motor is on... if the motor has been running for more than x minutes, it's likely that it's not actually pumping water since the water level hasn't fallen enough to turn the motor off. <S> That's of course assuming that the pump duty cycle has bounds that are acceptable to your "dry" condition test. <A> Sensing motor current may not distinguish "spinning freely" as the motor ages and lubrication starts to fail. <S> A flow switch is a minor plumbing job but will definitely tell you if water is running.
| If you can sense motor current this might be the least intrusive method, but you must test it first to make sure that there is a reasonable difference between "pumping water" vs "spinning freely".
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Does a self-powered USB device have to connect to VCC from the host port? A self-powered USB peripheral provides its own power, so is it okay to connect just D+, D- and GND to the host port, or do you always have to connect +5V as well, then wait for the negotiation phase for the device to tell the host that it is self-powered and therefore doesn't want any power, thank you very much? The specific peripheral I'm looking at is based on an FTDI FT4232H USB-to-serial adapter, but I'm really hoping for a generic answer to the question. <Q> With a self powered device <S> don't connect it's 5V to the host's 5V <S> , you may blow the either power supply. <S> GND and D+ , D- will do fine. <S> Mind you <S> that levels for D+ and D- are rated for +3.6V max, not to 5V as you might expect! <A> The generic answer to this question is yes, the VBUS (+5V from cable) must be connected to the device even if it is self-powered. <S> The reason is as follows: <S> To start the connect process on host side, the device must pull up D+ (in case of FS/HS mode), or D- (in case of LS device). <S> However, USB specifications have a mandatory requirement that no USB device should source any current on any interface pin unless it is connected to a cable, see section 7.1.5.1, which reads, The voltage source on the pull-up resistor must be derived from or controlled by the power supplied on the USB cable such that when VBUS is removed, the pull-up resistor does not supply current on the data line to which it is attached. <S> If a USB device doesn't have this control, one of data lines will be a source of current. <S> Premature assertion of pull-ups were a source of problems for some legacy USB hosts. <S> That's why this rule was instituted, and there is a special test for this in USB-IF certification program. <S> Therefore, the USB VBUS is an important "side-band" signal in USB connect protocol. <S> As such, normal USB device ICs do have a separate input pin to sense the presence of USB host. <S> Some IC manufacturers (e.g. FT232H, MCP2221, etc.) <S> skip on this requirement, assuming that their chip will be solely used in bus-powered configuration, where the pull-up control requirement is automatically satisfied. <S> However, when designing these chips into self-powered designs, some extra circuit efforts are needed to link the enabling of pull-ups with presence of VBUS on the USB port. <S> Regarding the USB connect "handshake" protocol, USB doesn't rely on current drawn from VBUS. <S> The protocol is this: Host port must have VBUS active; VBUS is connected to device; device sees the VBUS and pulls-up 1.5k on one of D+/D- wires; host sees this connect, and after a 100ms delay asserts USB_RESET signaling (SE0 etc.). <A> Consequently a self powered device downstream need only connect it's local +5V to its VBUS in one of two ways: directly, and cut the VBUS line in the cable; alternately, wire OR local 5V + host VBUS, preferably with ideal diodes like the LTC4357 <S> * that regulate an NFET like the FDB3632 that regulate a 25mV drop. <S> * https://www.analog.com/media/en/technical-documentation/data-sheets/4357fd.pdf
| As @GustavoLitovsky observed, there is no requirement for a host to monitor VBUS current except for an over current condition.
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Calculate power supply on distance I need to provide 12V DC to a DC brushless motors that need 12V 30A at distance of 30m from the source. The cables I can us are 20 Gauge , I know that the cables are loosing power, how can I calculate the power that I need to provide from the source to the motors cable in V DC and current that I need to provide at distance of 30m from the motors? <Q> 20 gauge wire has a resistance of about \$ 33.31m\Omega / m \$. For a 20 meter cable, it's effectively a resistor \$33.31\frac{\Omega}{m} \cdot <S> 20m \approx 1 \Omega \$. <S> Your circuit is this: <S> You want 30A through the motor, and since this is a series circuit, that means 30A through R1 and R2, also. <S> The voltage drop over a resistor is given by Ohm's law: voltage equals current times resistance: \$ <S> V_{R1} = 30A <S> \cdot 1\Omega = <S> 30V \$ <S> So, you lose 30V in R1, and another 30V in R2, so you will need 60V plus whatever voltage is required at the motor to get <S> 30A. Worse, power equals the product of voltage and current. <S> We know the voltage across R1 is 30V, and current is 30A, so power lost in the wire is: \$ P_{R1} = <S> 30A <S> \cdot 30 V = 900 W \$ <S> For comparison, a typical electric heater is around 1000W. <S> You have two wires, so <S> your total losses are 1800W. <S> It's quite likely you will trip a circuit breaker if you can magically prevent the wires from bursting into flames, and we haven't even powered the motor yet. <S> There are two obvious solutions: make the wires shorter <S> make the wires fatter <S> If you can't do either of those, there's a less obvious solution: keep the power the same by raising the voltage and reducing the current <S> This is the solution the electric company uses to avoid huge losses in power transmission. <S> If you combine the two previous equations, you can see that the power through a fixed resistance can be calculated through the current alone: \$ <S> P = <S> I^2 R \$ <S> But, the losses in your wires will be much less, by minimizing the \$I^2\$ term above. <S> To do this, look for a motor that operates at a higher voltage, or put a mechanism for converting the voltage near the motor. <A> This is not possible. <S> The problem is that there will be a voltage drop over the wires. <S> This calculation shows how much volt you'll need: You'll have to calculate the resistance of the cables. <S> When you know how much resistance 1m cable has, you can calculate the total resistance by multiplying with 2*30. <S> You can find here that 20Gauge has a resistance of 33.31Ohm per kilometer, so 0.03331 per meter. <S> Thus: $$R_{cables <S> } = 0.03331 <S> \cdot 60 = 1.9986\Omega$$ After that: $$R_{motor} = \frac{U}{I} = \frac{12}{30} = 0.4\Omega$$ <S> $$R_{total} = <S> R_{motor} + R_{cables} <S> = <S> 0.4 + <S> 1.9986 = <S> 2.3986\Omega$$ <S> $$U <S> = <S> I <S> \cdot R_{total} = <S> 30 <S> \cdot 2.3986 = <S> 71.958V$$ <S> You see this is pretty much. <S> I'd recommend you to place the power supply near the motor or to use bigger wires. <A> All answers so far are forgetting the fact that the motor uses 30A (probably) nominal. <S> But with varying mechanical load, the voltage at motor end with these thin wires will vary greatly. <S> Both answers so far are in the same range 60-70V for the power supply, so for the sake of argument let's say both calculations are correct. <S> This means that when mechanical load (and with that current) decreases, the voltage on the motor can rise from its rated 12V up to 70V worst case. <S> This will dramatically impact expected lifetime of the motor.
| You'll have to use a power supply with a higher output voltage. 12V at 30A is 360W. 360V at 1A is also 360W, and in theory, capable of producing the same mechanical power at your motor.
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What do you call a switch/relay which switches on an appliance for a given set of time? I need a relay/switch, which would have a push-button switch upon which the button is pressed, it would power an appliance on. I have a water pump, and its water pressure switch is defective. I would like to make it such that I can switch it on for 10 minutes, and it automatically switches off after the 10 minutes have elapsed. What is this type of relay/switch called? This is not to be confused with a 'timer', which would say switch on an appliance at a given time every day, like those which are sometimes used for Christmas lighting, where you can switch it on at 5 pm and automatically turn off at 11 pm. <Q> Those switches are called delay timer relays. <S> For instance, see the Omron H3Y , sold on eBay for $8 or less: <S> Note that the sample image above is rated for 5 Amperes 250 Volts AC <S> resistive load, not the rating applicable to your requirement. <A> It is a timer in any case, and you can find it as an electromechanical part. <S> From a more electronic point of view such a circuit will be identified as a monostable, and it is easy to create with a 555 chip. <S> You can use the monostable example, but being accurate with a 10 minute time will require a tantalum capacitor. <A> Try looking for a "timer delay" circuit. <S> These can be implemented using transistors, 555 timer integrated-circuits, or in the most general case microcomputers. <S> For example, this timer delay circuit lights up an LED when a push button is pressed and stays on for an undefined amount of time depending on the value of the capacitor in the circuit. <S> For a time-delayed switching of a relay, however, you will need to be careful. <S> For one, relays require a lot more current than LEDs, and you will need to choose a timer circuit that accounts for that requirement. <S> Additionally, make sure that you include a protection diode when switching an inductive load like a coil relay or risk damaging your relay and your circuit.
| Your application would require a relay with 10 minutes or more on the setting range, and you would need to ensure that the specific relay / relay base you buy are rated for inductive loads sufficient to cover the pump's starting / stall current rating.
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Quiescent power of BJT vs MOS transistors This is a question about the power necessary to control a transistor circuit, i.e. the base current in BJTs vs. the gate current in MOSFETs. Why is there quiescent power dissipated by base currents of BJT's? Why doesn't it happen with MOSFETS? <Q> That isn't exactly a true statement " <S> i.e. why doesn't it happen with MOSFET's ?". <S> It is True for logic type circuits, but for analog circuits <S> both types of transistors consume power, and indeed in BiCMOS (the combination of the two) <S> BJT's can be used to get performance at lower power, in some cases. <S> By restricting the answer to only logic based circuits we can come up with the following: BJT's essentially operate on currents and the transistor (in a simple model) is a current amplifier. <S> CMOS's equivalent simple model is that of a voltage controlled current source, with the control gate being a capacitor. <S> In CMOS circuits, the transistor sources or sinks current onto the control nodes of other transistors "downstream" <S> so the transistors's "current " aspect only sees capacitor loads (other transistors gates). <S> With a capacitor you only need to flow current to change the voltage on it, either by injecting current onto the node to bring it high , or draining charge off to bring it low. <S> Once the capacitor is at a given state, the current need not and indeed does stop flowing. <S> In BJT circuits, you must provide a current (albeit very tiny) to the next down stream transistor in order for the transistor to amplify that current to set it's new state. <S> It is simply the fact that you always have to supply some current in all states that BJT's consume more power. <S> To summarize (using simple models): CMOS flows current on change, BJT flows less current continuously. <A> The key to the answer is buried in the long name of the MOSFET, which is Metal Oxide Semiconductor Field Effect Transistor. <S> The oxide layer isolates the gate from the channel (drain-source). <S> For the MOSFET to be controlled, you apply a voltage between gate and source, but because of the isolation, the voltage cannot cause a current to flow from gate to source. <S> Gate currents become significant <S> only when you change the voltage, i.e. when you put more charge into the gate or you withdraw charge from the gate. <S> The resistace being formed between drain and gate depends on the field created by the nearby gate, i.e. the gate voltage. <S> Thus, MOSFETs are essentially voltage-controlled devices, not current controlled devices. <S> A BJT will allow (and require) current to flow into the base when you apply a voltage between the base and the emitter. <S> This current controls the current from the collector to the emitter. <S> Even if you don't care about the "transistor action", the base-to-emitter part will act like a diode, i.e. it will allow current to flow if biased the right way. <S> Now... Current alone still doesn't explain the power (current * voltage) dissipated by the BJT. <S> Just like with diodes, there is a voltage drop (approx. <S> 0.7 V) across the B-E part of the BJT. <S> Multiply your base current with this "diode drop", and you have a power dissipated by the base current. <S> This explanation is quite short and by far doesn't cover everything, but it gives you at least some ideas where to start digging further. <S> However, your question makes sense only when looking at either transistor in the context of a whole circuit. <S> One npn BJT alone can be off with no base and <S> no collector current, so there is no quiescent current. <S> Consider a larger logic circuit in RTL (resistor-transistor logic) vs. CMOS (complementary MOS) logic. <S> With RTL, there will always be some transistors that conduct and some that don't, so some will dissipate power and some won't, even if the circuit "does nothing", i.e. doesn't change its state. <S> CMOS will only require some gate (dis)charge current whenever it changes its state. <S> Something similar is true for MOSFET-based LDOs (voltage regulators) vs. BJT-based LDOs. <A> If by "quiescent" you mean "off", as in "no drain or collector current desired", then I think you are wrong to think that MOSFETs leak less current than BJTs. <S> BC547's (BJT) datasheet lists the collector current at \$15nA\$. <S> 2N7000 (MOSFET) lists the zero gate voltage drain current as \$1 \mu A\$, about 66 times more. <A> A BJT amplifies current (\$I_c <S> =I_b \times H_{fe}\$). <S> Both currents flow out of the emitter. <S> \$H_{fe}\$ is the gain (see datasheets). <S> A BJT also has a diode drop between base and emitter, <S> typically \$V_{be} \gt 0.7V\$. <S> As a result, the driver circuit for a BJT has to supply power \$V_{be} \times I_{b}\$, if \$I_c \gt 0 A\$, even if Ic is constant. <S> In particular for high current BJTs, \$H_{fe}\$ can be low, often less than 10, so they have high power losses. <S> A darlington (two BJTs stacked) will have a higher \$H_{fe}\$, but \$V_{be}\$ at least doubles. <S> A FET (not just MOSFET) controls resistance in the drain-source path with an electric field generated by voltage \$V_{gs}\$ between gate and source. <S> There is a very large resistance between gate and source, but also a very large capacitance. <S> If \$V_{gs}\$ is constant, the capacitance does not matter much. <S> You charge it once <S> and then no more current flows. <S> If you only care about quiescent power and have the necessary drive voltage available, a FET is a good choice for switching. <S> However, if at any time you have to switch high frequency signals, the capacitive current can get large. <S> You asked about quiescent current, but to give you an idea, in projects that I worked on, the cross over where the BJT drive power became less than FET drive power was around 60kHz. <S> If you are concerned about drive power near the off-state, consider depletion mode FETs. <S> Those conduct with \$V_{gs} = 0\$ and are turned off by applying a voltage. <S> To get the best of both worlds, one can use IGBTs, which combine FET and BJT, but they cost a lot more and drivers can be tricky once you get into the MW range.
| One can say that BJTs are current-controlled devices more than they are voltage controlled devices, although you first need a voltage as the cause for the current to flow.
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why power supplies won't work with laptops? laptops need a supply and it is a DC voltage isn't it? so why my lab power supply can't turn my laptop on. it is tune on the desired voltage and polarity, when i press the on button, the laptop starts, leds on, tries to boot and suddenly dead with a beep!! the power supply can provide 3A and it shows on its front Ammeter that the laptop is consuming 1.5A. what technologies makes my laptop adapter different? <Q> This depends on the type of laptop and power supply, some have additional logic signals, etc. <S> In all likelihood, your lab power supply is not able to provide enough current to power your laptop. <S> Laptops use a good amount of power, if you have the original power supply it came with, you would see a current rating in mA or A. <S> This is the total amperage the power supply can provide, and my guess is that it is at the least, 2 Amps or more. <S> You may be able to charge the battery with your lab supply though. <S> There are other possibilities too, but without knowing the laptop model and the specs on your lab power supply, I couldn't say. <S> Edit <S> the power supply is able to provide 3 Amps and the Current measurement shows that consume doesn't exceed 1.5 Amps <S> There are a lot of possibilities here, perhaps your lab supply isn't accurate or there were power fluctuations. <S> Maybe your adapter plug was not the correct size, so you had a poor or intermittent connection. <S> The laptop may have had an unrelated issue, that caused this problem, I've seen some that turn off as soon as you plug in external power. <S> Perhaps the battery was bad, problem with the motherboard, or maybe a fan wasn't working. <S> There are just a ton of possibilities. <S> the laptop is an old lenovo... <S> "the beep and dead" seemed to be a power security act... <S> HOW can the laptop distinguish between the its adapter <S> Usually a beep like that is a bios problem or warning. <S> Also, according to this , it seems that Lenovo has different power supplies and depending on what one is used, it will throttle it's performance. <S> A power supply for a laptop doesn't just have to provide power, sometimes there is some data communication/feed back, perhaps even just a way to verify that the power supply is a real brand name one. <A> Many laptops have a hidden connection that digitally communicates with the power supply to make sure it's an "approved" one before charging enables. <S> Quite likely the system sort of starts up but the communications check fails. <S> This is often done with a 1-wire EEPROM. <S> If there isn't such protection with your Lenovo setup, most likely the power supply cannot handle the load imposed by the computer, or as others have suggested your laptop has suffered some sort of hardware failure. <A> Some devices may have a protection circuit that checks if the supply is providing a limited current. <S> This is particularly true for Nokia cellphones. <S> Try setting your laboratory power supply to provide both a constant voltage and current into the laptop. <S> You can get the required constant current limit by measuring the current when the laptop is on and charging the battery from its dedicated supply. <S> If that cannot be done, try using the indicated output current of the dedicated supply. <A> Some laptops such as Dell's have a third pin on the power supply that sends data (Dallas 1 wire) to validate that the correct supply is connec <A> Does your power supply have a current limit feature? <S> If it does then that might explain it. <S> When the laptop tries to draw more current than is allowed the voltage is drawn down to accommodate the increased current. <S> This also might happen if the laptop demands a large amount of current quickly - the power supply voltage will drop before the supply can react to increase the current and the laptop dies as a result. <S> Your current measurement is probably wrong - you need more power.
| Your lab supply may very well be able to power the laptop, however when the laptop is first turned on, your power supply may not be able to quickly adapt to the load.
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Two wire vs. three wire serial interface for RTC I'm looking to choose an RTC for a project I have in mind. I have the following options (linked to datasheet): DS1307 DS1302 The main difference seems to be the communication protocol. The DS1307 seems to use something called "two-wire serial interface" whereas the DS1302 uses something called the "3-wire interface". I would like to use one of these chips to keep time on my ATMega microcontroller (probably ATmega8). So how are these two different, and how would I hook them up? <Q> The main difference between SPI (3-wire interface) and I2C (2-wire interface) is, that you have to send an address when using I2C. <S> So it won't be as fast as SPI because you have to send more data. <S> When using SPI you will need an extra pin for the chip enable signal. <S> This will be required for each slave. <S> So it's faster, but you need more output pins. <S> I would prefer I2C because speed doesn't matter in this scenario. <S> The ATMega microcontroller has hardware peripherals for both methods <S> so it's up to you <A> The first runs off I2C . <S> Timing information is in Figure 5 on page 7 of the datasheet. <S> The second runs off SPI . <S> Timing information is in Figure 4 on page 8 of the datasheet. <S> It looks like the ATmega8 has TWI and SPI just like all the other ATmega parts, you can use both with it, simultaneously if you choose. <S> My preference would be SPI. <S> It's a bit easier to work with, and as a protocol tends to be a bit less temperamental. <A> It depends on a few factors: <S> Hardware design: If you are using only this chip with your microcontroller, anyone of them will do. <S> But if your design consists of two or more devices that use I2C interface then it makes sense to use I2C RTC so that you don't have to do additional interfacing, just add the RTC to existing bus. <S> Same applies to SPI but keep in mind when you add more devices to the SPI bus, you need one extra SS signal per device. <S> It also depends on the hardware peripherals that you have in your microcontroller. <S> If your micro has hardware SPI peripheral but not I2C <S> then it make sense to use SPI device over an I2C device. <S> Same goes with I2C peripheral. <S> Software design: <S> I2C involves a bit more work than SPI. <S> You can also find some code implementations over the web. <S> If you already have a I2C code with high confidence, use it. <S> Same applies if you are pretty confident with a piece of code for SPI that you found over the web. <S> Your understanding: The choice also depends upon your understanding of either protocols and how well-versed you are with these. <S> If you have used anyone of them before, it makes sense to use prior experience. <A> The DS1302 does not truly use standard SPI because the single "I/O" data signal is time-multiplexed during communication (driven by the chip during a read between the falling-edge of "CLK" to just after the rising-edge of "CLK", high-z at all other times). <S> Standard SPI has separate MOSI (Master-Out/Slave-In) and MISO (Master-In/Slave-Out) pins for each peripheral (although the MISO can be three-stated by the peripheral if it is multiplexed with multiple slaves). <S> The DS1302 "CE" (previously referred to as "/RST") is absolutely essential in communication to start/stop each transmission, plus store the RTC data after a "Burst" (you can NOT simply connect it to a rail). <S> Also, the first bit sent after CE is asserted indicates if you are doing a Read (1) or a Write (0).That being said, there's still less protocol overhead than using I2C, because you don't need to worry about I2C-specific addressing and timing. <S> Even at the 2V specification (the datasheet gives timing for 2V and 5V), <S> the DS1302 is still faster than using I2C, and WAY faster when using 5V.Be aware that each of the 3 communications pins for the DS1302 have a (nominal) 40k pull-down resistor, to avoid allowing the pins to float.
| Another deciding factor is whether you are implementing either protocol in software.
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What do I do with the extra "ground chassis" tab on a DC power jack? Product Link Datasheet I have about fifteen 9v wall warts within specs for the circuit; I'm going to use the one that fits the jack. There are three solder tabs on the jack. One for the positive tip, one for the negative ring/sleeve, and one for the chassis ground. My schematic has Vin and ground specified. I'm relying on the wall wart to appropriately ground the AC end of things. My pos/neg tabs are used for source voltage and ground, respectively, and that leave the third chassis ground tab. Do I ignore it? Connect it to the negative side ground? To the chassis? What should I do when there's no specification about what to do with the extra tab? <Q> I don't think you are correct about a "chassis ground". <S> Every barrel-connector jack like the image you provide has three pins - Tip, sleeve, and connector-present. <S> It's often used to disconnect internal batteries when the device is powered off of an external supply. <S> Short of a few rather unusual laptop power supplies, every barrel connector I have ever dealt with has been two-conductor only. <A> Fake Name's answer is correct, but I want to add an image from the datasheet: <A> Take a continuity meter to the two 'normally connected' pins out of circuit <S> and I'm pretty sure notice that they ring out... <S> I would say connect them together in your circuit
| Basically, there is an internal switch that is opened by the presence of a plug in the connector. There is no "chassis ground" anywhere in that datasheet. - I have certainly done this in a couple designs (with a center positive jacks).
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Smallest PIC which the TCP/IP stack can be implemented What is the smallest (size) PIC with internal oscillator available which the TCP/IP stack can be implemented (and used) successfully? Also, what is the smallest (size) PIC with internal oscillator and Ethernet stack implemented? <Q> On their Ethernet page , Microchip recommends the PIC18F97J60 . <S> It has MAC and 10BASE-T PHY integrated. <S> IEEE 802.3 compatible Ethernet Controller <S> Fully Compatible with 10/100/1000Base-T networks Integrated MAC and 10BASE-T PHY <S> Up to 128KB <S> Flash <S> 8 KB Ethernet Buffer Support for Unicast, Multicast and Broadcast <S> packets <S> Programmable wake-up on multiple packet formats,including Magic Packet®, Unicast, Multicast, Broadcast,specific packet match or any packet 64/80/100-Pin TQFP Package <A> You may want to explore the ENC28J60 . <S> It's an external chip from the PIC that handles all the Ethernet. <S> It interfaces to the PIC via SPI bus. <S> Just choose a PIC that has enough flash and RAM for the TCP/IP Stack. <S> You can download and compile to a simulator to see how much space things take. <A> On the link provided by m. <S> Alin you can see in the right bottom corner <S> a button named See all Ethernet PIC MCUs which lead to here where you can see different characteristics, including packaging in the most right column.
| If you're going to use Microchip's TCP/IP Stack , the PIC18 is the 'lowest-end' family of PICs that the software stack supports. By the look of it, the smallest PIC32 with Ethernet are 64/QFN 64/TQFP
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Why does solder not conduct sometimes? I was reflowing and a 2K 603 resistor partially tombstoned. There was solder connecting the lead to the pad, yet my DMM showed an open. When I heated both ends and the resistor came down and contacted the pad, the DMM read 2k ohms like it should. Why didn't the solder conduct between the pad and lead? Also, sometimes when I touch a probe to a solder blob on top of pin, it shows an open. Yet when I touch the pin above the solder joint, it shows conductivity. Is this the same case when there is a cold solder joint? The solder is touching the pin and pad, yet there is no conductivity. Isn't solder a metal and therefore should conduct? How come it doesn't? <Q> Sounds like a classic cold solder joint. <S> This happens when the solder and pad are insufficiently heated (time or temp), or the surface is not clean, and the wetting action does not occur properly. <S> It's not that solder isn't conducting, but that it hasn't made a bond to the copper and therefore contaminants or flux <S> are actually creating a non-conductive barrier. <S> As for the pin-to-pad conducting but not solder-to-pad, the pin is resting on top of the pad and likely making a physical connection (held in place by a brittle solder blob). <S> The solder, however, may have a thin air gap left over from flux or contaminants, and isn't actually making contact with either. <A> It's amazing how easily solder (or other metals) can look clean but due to a thin layer of contaminant or oxidisation when you probe them <S> you don't get an easy contact (you have to wiggle and press harder). <S> This is why sharp probe tips are so important. <S> Flux cleaner/Ethanol/Cotton buds are good to have handy also. <A> What you're most likely experiencing is the flux or similar chemicals which are preventing you from properly touching the conducting solder. <S> Using alcohol will help you remove them. <S> The solder itself conducts (unless it doesn't make proper contact), it's just that your instrument isn't connecting to it directly as it should.
| If you use a meter to measure from one point to another on the solder itself, it should conduct, but between the solder and pad, there is not actually a good electrical connection, hence the open.
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How do I change the properties of several objects at once? I have several lines and arcs in an EAGLE footprint library I need to modify to make thicker. In Altium it's straightforward to hold Ctrl, click on a bunch of objects, open up their properties and change them all at once. In EAGLE I have no idea. EAGLE's "group" paradigm seems a little far-flung from the standard, so my attempts to use that may be a dead end, but is there some better way? Is there some magic command I could run to give every single line and arc in a library part on layer X a thickness of Y? How about in a PCB, smashing all the parts and changing all their labels to font Z with size W? <Q> You probably need the change ("wrench") tool available from the GUI. <S> But it might also worth considering using some Eagle commands for that purpose. <S> Assuming for example you want to change the thickness of your wires to 30 mil,first you have to select those tracks as usual, then you might issue the following commands: <S> change width 30 mil(> 0 0) <S> The (> 0 0) part simulate the right-click required to apply changes. <S> If you need to change the thickness of all the wires, this might be scripted even further: display none topgroup allchange width 30 mil( <S> > <S> 0 0)display last <A> Yes, the "group" paradigm is what you're looking for most of the time. <S> Yes, it is as clunky as you think it is. <S> There's no premade magic command. <S> One thing you can try is to turn off all layers except the one you want to modify and then box select everything and use the group tool that way. <S> Other than that, you may want to look into Eagle's ULP/scripting functionality to make yourself a command. <A> Enter the command: cha wid 0.234 <S> Where you replace "0.234" with the width you want in whatever your current units are. <S> This is a short way to enter the "change width" command. <S> Most Eagle command names and other keywords can be abbreviated to three letters. <S> For more details on the "change" command, enter HELP CHANGE. <S> Then just click on any wires you want changed. <S> Yes, it really is that easy. <A> Don't know if it is too late for this, but you could use a ULP(User Language Program) to change the width of all the traces on the board. <S> Go to File - <S> > <S> Run <S> ULP.. -> Type "cmd-change-brd-width.ulp" -> <S> Open <S> It will open up a dialog box <S> and you can use that to change the width of multiple wires at the same time.
| You can shift+drag (pretty sure it's shift, maybe control, been a while) selection boxes to add more objects to an existing group selection.
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Choosing an MCU in a time-effective way I'm spending a lot time figuring out which microcontroller I should choose for my projects. For example , for one project I have the following requirements, in order of importance: I 2 S interface (for connecting at least 2 digital microphones, 4 preferred) At least 32kB RAM At least 40MHz clock Low power. Has to work for at least 24 hours on an 900mAh battery Should come in some kind of a board that I can work with by hand The board should be small, not much larger than a mobile phone This is a one-off project, so cost is not a major issue How does one go about finding a short-list of candidates that meet or exceed a given set of requirements (such as the above), and be sure that they haven't missed anything? Currently, I'm browsing through the element-14 catalogue and opening up datasheets at random. This is a very slow process. Ideally, there would be some table where I could select what columns to display (such as RAM, clock speed, power consumption) and sort MCUs that match my criteria. Things may not be ideal, but surely people have smarter ways to browse the MCU market? <Q> For manufacturers that support various broad categories of MCUs they segment their selection matrrix pages according to these categories. <S> Atmel for example does this according to AVR, ARM and 8051. <S> They quickly assume that you can easily make your selection choice based on these categories before trying to drill down further. <S> The following shows some of the choices available in Atmel's ARM selection matrix. <S> You can quickly find links to the selection guides from various manufacturers by doing a Google search with microcontroller selection guide as the search phrase. <S> This will point to information from the likes of TI, NEC, NXP, Freescale, Analog Devices and Renesas to name but a few. <A> Using websites like this can be very tedious work. <S> I find that deciding what set of microcontrollers I'm interested in first and then using the manufacturers website to reduce to a shortlist helps (but only if that website is any good). <S> For example. <S> If I'm interested in the Microchip PIC range, you can use their Advanced Part Selector to make selection across 8/16/32 bit product ranges easier. <S> Using a few of your parameters I reduce to a few parts, all in the dsPic, PIC24 or PIC32 range. <S> I'm sure some of those have development boards. <S> (PS: I2S is selective under the + submenu in Digital Comm. <S> Features . <S> No, I don't know why they do this!) <S> Other things to think about is the availability of development tools and ease of coding for the part family you are interested in. <S> NXP have a range of ARM microcontrollers and there are many products like the mbed that might suit you. <S> I'm afraid I haven't used I2S before so <S> can't help you out there. <A> I'd say the biggest factor is the support available, not the choice of micro. <S> These days we are spoilt for choice for mid-spec micros which meet or exceed your requirements, the bigger issue is having a good dev/eval board available for good money ($50 rather than $500) and a toolchain/dev environment/debugger with supporting libraries etc. <S> without spending huge sums on pro software (unless you have the budget for that too). <S> Unless your time is free, those factors will have a far greater impact than saving $0.20 per device in production. <S> Based on my current experience: Stay the hell away from Texas Instruments, unless you're buying a million devices a year they won't give you the time of day. <S> I've had reasonable experience with Freescale (relatively!) and heard good stuff about NXP.
| Many manufacturers of microcontrollers have selection matrix pages on their web site to support just the type of thing you are needing.
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Capacitor and ISP issue If I remove C5 I am able to use ISP. However, with C5 on board ISP programming fails... Capacitor is for button debouncing. What is the quick workaround to have debouncing routine and use ISP? I can't change the board because it's already on PCB, just can make a patch. <Q> A simple solution could be to connect the junction of C5 and R1 via a 10k resistor to PB2, and connect the ISP connector directly to PB2. <S> This is a general solution that you can use whenever you want to override a signal with a stronger signal that will not be present (or is in tri-state) when the weaker signal must be read. <S> Of course, the series resistor put 'before' the weaker signal to make it weak must not cause a problem. <S> Generally, digital microcontroller input pins have a very high impedance, so a 10k extar won't be noticed at anything but very high frequencies. <S> (but remember: analog inputs might might have a much lower impedance, so in that case this trick likely won't work.) <A> That capacitor breaks ISP because it limits the rise and fall times on PB2. <S> The programmer can only source or sink so much current, and so only charge/discharge this capacitor so fast. <S> Alternately, make R1 bigger, and C5 <S> much smaller. <S> If C5 is small enough that the edges are still sharp enough, you won't have this problem. <S> Programming at a lower speed will also help. <A> You haven't specified which micro you're using, but in many(most) cases a 1uF cap is going to wreak havoc with the ISP programming clock. <S> I'd nopop the cap and go for a software debounce instead of brute-forcing it.
| If you can't change the board, you will have to remove the capacitor and do debouncing in software.
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Choosing relay to operate water pump Based on my research so far, I've concluded that a relatively low-cost, yet reasonably reliable way to control operation (on/off) a motorized water pump, is a well spec'd EM-relay with flyback diode. I might also put in an opto-coupler for isolation, if that improves safety. The whole thing would be assembled on a general-purpose phenolic veroboard, with point-to-point connectors. Given the pump motor's specs, I wanted help checking if the selected relay would fit the bill, or should I look for another. The pump's specs (the relevant parts): Self-priming 1.0HP mono-block pumpOperating voltage: 220VAC +/- 6%, single-phase 50HzkW / HP: 0.75 / 1.00Current: 5ACap. run: 20uF 440V The operating characteristics are about 100 on/off cycles per month, with each "on"-cycle being about 60mins long. AC power quality is not that great, with voltage fluctuating 10-12% around 220V, and line frequency fluctuating by 6-7% (as told by a local friend, who is knowledgeable in power electronics). The EM-relay specs: General-Purpose 1Pole Heavy Load SPST-NO Power RelayCoil Voltage: 12VCoil Resistance: 185 OhmContact current: 25ASwitching voltage: 250VACLife: Resistive-Load: 100 x 10^3 operations Motor-Load: Min. 200 x 10^3 operations, given (250VAC inrush 80A, cosφ=0.7, cut off 20A, cosφ=0.9) Does that sound good enough ? The above mentioned relay has 3 variants -- Standard, High Isolation, High Current. For the moment, only 'Standard' one is readily available. Would that suffice ? Just curious, what might be the other 2 variants good for. A surprising observations was that the EM-relay has longer life with motor-load, than resistive-load. Is that normal ? Also, the readily available part is a "PCB mount" relay. Is that acceptable, or should I really try to hunt for the socket-mount / tab-terminal kinds ? Finally, is it fair to assume that I do not need any kind of "snubber circuitry". PS, I understand very little of "snubber" though, but it has appeared few times in the material I've read so far. Edit: Signs of aging, and some carelessness. Had asked this question exactly a year back, when I had just began working with relays, and forgot all about Russell's excellent answer. It does cover many of the questions I re-asked here. Thought of linking this together. <Q> An electromechanical relay will work. <S> Hey, they have been used for a looooong time. <S> You don't have any contacts to corrode, arc, etc., and degrade. <S> Using an electro-mechanical solution is not optimal... <S> You seldom see an electro-mechanical relay in modern electronics. <A> Your proposed relay looks fine to me. <S> The "high isolation" variants can be useful for applications where a specific level of galvanic isolation is required (normative requirements for example). <S> That the "high current" relays are designed for higher currents should be obvious. <S> From what i can tell, those are just slightly more expensive but have no drawbacks in regard to the "standard" variant. <S> As far as i know a motor load can cause a high inrush current, while the circuit breaking behavior is much the same as a resistive load. <S> So i don't know why the relays should be able to perform more switching operations on motor load than on resistive. <S> This was based on superficial knowledge, and after giving it some more thought it probably isn't correct. <S> Google did not help me to understand it in detail, since there are many different scenarios. <S> Maybe someone can improve the answer. <S> One should pay attention when breaking an highly inductive load, since that will generally produce heavy arcing. <S> I can't help you regarding the snubber circuit. <A> I am using EM relay ( 10 Amp. 250 ohm. <S> SPST ) for the last 2 years. <S> It is working well. <S> I use it daily about 1 hour to run 1 HP Pump.
| However, a solid state relay would be my choice for a quick solution.
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What is "offset null" in IC 741? What is Offset null in 1st and 5th pin in IC 741 (Op-Amp)? Why it is used, though it is not used in many circuits? Give me explanation regarding the offset null!Why offset voltage was formed in IC 741? <Q> The datasheet gives an example. <S> By adjusting the pot we can null any offset error . <S> An offset error is when the inputs are exactly equal but the output isn't exactly zero. <S> This error is also characterized by the datasheet: <S> It can be safely ignored in AC applications, where this offset will be ignored by the AC coupling. <S> It becomes more important in DC applications, especially amplifiers, since this DC error will be amplified by the next stage. <S> This offset voltage exists because a real omp-amp can't be ideal. <S> There will always be some unintended asymmetries between due to random variation in manufacturing. <S> In all cases, there are op-amp designs that can minimize these errors, but usually at the expense of some other parameter, like cost. <A> The best example I can give you is the old ohmmeter. <S> You would short the leads and adjust the needle to read zero, then you knew your resistance readings would be as accurate as possible. <S> With the 741, you provide the same voltage (signal) to the input pins and adjust the offset null to make sure the output is zero. <A> as ics are manufactured with cascade amplifiers,a smal voltage is enough for its workng n as semiconductors are used the room temperature is enough to get extra voltage so while operating using opamp,offset nul is connected so as to make these voltages zero
| For circuits that accuracy and/or symmetry is critical, the offset null pins provide a means to cancel "internal" discrepancies.
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What is the correct way to indicate a NC (no connect) pin in Eagle CAD I looked around for NC symbols etc. and off course, leaving pins hanging will throw errors when doing a design rules check. So what would be the best practice for indicating NC pins in Eagle CAD? <Q> I asked the same question of Cadsoft support only last week. <S> I mentioned that I prefer to have a symbol on the schematic so that I know that I've properly reviewed that the pin in question can be left unconnected, the response I received was: <S> We don't have an equivalent 'X' symbol in EAGLE. <S> You could create your own no-DRC symbol without a footprint, if you define it as an external device it won't raise an error for not having a footprint. <S> So that's a feasible solution. <S> After finding that I couldn't add a schematic symbol without a package <S> their follow-up was: <S> See the HELP pages for the ATTRIBUTE and PACKAGE. <S> The key to defining a device as an external device is to create an attribute called EXTERNAL in the device. <S> This will allow you to create a part that doesn't need a package. <S> This is handled in the library. <A> Don't connect anything to it, and nothing will be connected to it, and that will be obvious in the schematic. <S> There really is no need to do anything more. <S> No, the DRC (Design Rule Check) doesn't care at all whether pins are connected. <S> The ERC (Electrial Rules Check) only complains about floating input pins, but then again those should be connected to something. <A> NC symbol can be very useful when Net classes are used. <S> It is way how to create net with different net class (e.g. clearance matrix) for not-connected pin. <S> General NC pin has default net class.
| There is no need for a explicit not-connected symbol.
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Why does water short out PCBs? (i.e. Why doesn't electricity follow the path of least resistance?) If copper is more conductive than water (at any reasonable PH), submerging copper electronic circuits in should have no effect, as the electricity should continue to follow the path of least resistance (the highly conductive copper PCB paths, for example) rather than shorting into the mildly conductive water. However, dumping water on electronics clearly shorts them out, even though copper is the superior conductor. Why does this occur when the path of least resistance should be the copper rather than the water? <Q> Fallacy correction: <S> "Electricity" NEVER 'follows the path of least resistance'. <S> Electricity is NOT like a single large Zorb bounding down a hillside . <S> Electricity would be better modelled in this context as a large reservoir of water being poured down a rough hillside . <S> Most of the water will follow major gullies and channels but some will find smaller side channels. <S> The question is not "which path will the electricity follow?" <S> but "how much current will flow along this given path?" <S> Low resistance paths will conduct higher currents. <S> Higher resistance paths will conduct less current. <S> Only infinite resistance paths will conduct zero current. <S> There are NO infinite resistance paths. <S> Plus: <S> When voltage is applied to water above a certain critical voltage, the water will decompose into Hydrogen and Oxygen - - known as "electrolysis" . <S> Any components in the water are liable to also decompose to produce eg Chlorine gas from chlorine products in the water. <S> Even wholly pure deionised water can be decomposed in this manner. <S> Once you get even a small amount of current flow you get ions formed which promotes more current flow, lower resistance, more breakdown ... <S> zap. <S> I have dropped a "pager" into sea water and a portable phone into concentrated chlorine solution* and "saved" both with no long term damage by immediately taking out the batteries and washing them in copious quantities of fresh water and then leaving them to dry completely (a very important step). <S> And I have seen equipment destroyed by exposure to water with the batteries then left in. <S> Note to self <S> : Remove portable phone from top pocket before stirring large bucket of Sodium Hypochlorite solution. <S> When splashing along sea edge and generally having fun your pager should be in a salt-spray protected bag or, better still, left at home. <S> Note to the young: <S> Pager's were things we used to have before they invented cell phones. <S> Effectively a receive only SMS system with no voice or any other features. <S> Doctor's still seem to use them. <A> The problem is that although the traces have a very low resistance, the traces are not a fully meshed network. <S> The PCB is a collection of networks. <S> There isn't a low resistance path between any two nodes, but only between specific nodes which are understood to be on the same network, as dictated by the circuit design which was transferred to the PCB artwork. <S> Between some other nodes that are not on the same network, the resistance or impedance may be, in fact, very high. <S> Close to infinite, in some cases. <S> Moreover, there may be a voltage between these points. <S> A prime example of this are power rail networks. <S> Ion-rich water will create conductive paths between nodes which are not supposed to be thus connected, and the voltage difference will cause a current to flow, creating short circuits or near short circuits. <A> A circuit board has traces on it. <S> These traces are meant to be insulated from one another. <S> Proper circuit operation requires it. <S> If you dump water on the board, the traces are no longer insulated. <S> If the "circuit" was just a solid plane of copper, the water would have no effect. <A> It is following the path of least resistance. <S> In this case, that path is the copper and the ions in the water that allow some more electrons. <S> You need to look at water as a very high resistance. <S> It isn't infinitely high, so some current will flow there, usually with bad results. <S> Just as in the case with two resistors in parallel, the copper is the low resistance, and most current flows through that path, but some will flow in the other high resistor (ions in the water). <S> Note that deionized water cannot conduct due to the lack of ions.
| Another thing usually ignored is that ICs are not completely hermetically sealed, so water and other elements can go in and wreak havoc inside, especially with time when you take into account oxidation and ions that can short the IC.
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Difference between MISO/MOSI and TxD/ RxD What is the difference between USART and SPI ? There are [TXD/RXD] for USART and [MISO/MOSI/SCK/SS] for SPI.I know that SPI is a synchronous protocol but USART is hardware. But, what's difference between Synchronous part of USART and the hardware ? <Q> USART is a device (or peripheral). <S> SPI is a standard method of connecting things. <S> USART stands for Universal Synchronous/Asynchronous Receiver/Transmitter , and is the basic thing you need if you want to transmit using RS-232.422/485/etc. <S> The Synchronous part of a USART is not used very often, and is sometimes that functionality is left out of the device-- <S> and then it's called a UART (pronounced You-Art). <S> USARTs (with an appropriate RS-232/etc driver/receiver) are mainly used to talk with devices over a cable. <S> Sometimes they are used to talk between devices on the same PCB, or within the same box, but it is much more common to talk with another device over a cable. <S> SPI is mainly used to talk with devices on the same PCB or in the same box. <S> For example, an MCU talking with a digital temperature sensor. <S> It is almost never used to talk over a cable, from box to box. <S> The nice thing about SPI is that it is super simple, and the devices using SPI do not have to be MCUs. <S> USARTs almost always require that MCUs of some sort are on both ends of the communication link. <S> But USARTs can be connected using less wires over longer distances. <A> There is a major difference between USART and SPI; Synchronization <S> UART is Asynchronous - <S> That means that the communication doesn't accompany a clock signal. <S> Only data signals are connected along with sometimes optional flow control signals (RTS, CTS, DTR, DSR, etc.). <S> UART can be full duplex or half duplex. <S> RxD carries data from DCE to DTE. <S> TxD carries data from DTE to DCE. <S> UART is also point-to-point connection, there is one transmitter and one receiver on the whole link. <S> SPI is Synchronous - <S> That means that the clock signal accompanies the data signal. <S> MOSI - Master Out <S> Slave In; MISO <S> - Master In Slave <S> Out; <S> SCK - Clock signal from master to slave; <S> SS - Slave Select signal selects slave devices. <S> SPI is a bus and can have multiple devices on the bus. <S> Because of a clock signal present, SPI can be operated faster than UART. <A> The Synchronous (U* S *ART) is hardly ever used. <S> When it is, its actually quite similar to SPI. <S> In fact, many microcontrollers call (one of) the synchronous mode(s), the SPI mode. <S> Some examples are here and here .
| SPI, the Serial Peripheral Interface Bus , is a completely different thing than a USART.
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Using signal in different modules vhdl I am trying to connect a microcontroller(cortex m3) and a fpga(actel a3p060). I am able to read/write successfully with 16 bit databus. My modules in vhdl is structured as follows: Top module(interface with controller) sub module where I am having a frequency divider. The idea is to generate different frequencies fed by the controller to the fpga. In the main module, I'm passing a signal called as the Load_Divider which will be checked in the frequency divider module as follows: process(Load_Division)if(Load_Division='1') Data_Buffer<=DataBus; Load_Divider <='0';end if;end process;if(Counter = X"0000") then Counter <= Data_Buffer; //reloading counter And in the main module : case Address is "00100"=>Load_Divider <='1'; // generating load signal when proper address is found But what happens is my Load_Divider is not going '0' when driven from the sub module. Is it not possible to drive a signal from two modules? <Q> It is possible to drive a signal from two modules. <S> However, it is not straightforward; the two modules have to cooperate - if one drives '1' and the other '0' at the same time, you have created a short circuit across the power supply, and you can potentially damage the device. <S> In simulation, thankfully, the results are less dramatic! <S> Two ways to do so; and the reason not to use either of them inside an FPGA. <S> The first way is the traditional "wired AND" where there is a weak '1' permanently connected to the signal. <S> Traditionally this was a resistor connected to +5V; in VHDL you might say Load_Divider <S> <='H'; in the toplevel design. <S> H and L are weak versions of '1' and '0'. <S> Now every driver can either pull it low, or turn off if condition then Load_Divider <S> <= <S> '0';else <S> Load_Divider <S> <='Z';end if; ('Z' means the high-Z or high impedance state, i.e. turn off the driver). <S> If ALL the modules drive Z, the output is 'H' (equal to '1'). <S> The second way is called a tri-state bus. <S> Here either module may drive 0, 1 or Z onto the bus. <S> However, you the designer MUST make sure that when one module is driving '0' or '1', ALL the other modules are driving 'Z', to prevent the short circuit and damage mentioned above. <S> That requires the cooperation of all the modules in some way: if my_turn then Load_Divider <S> <= my_signal; -- 0 or 1else <S> Load_Divider <S> < <S> ='Z';end <S> if; Either of these techniques are valid ways to communicate between separate FPGAs and other chips (memories, CPUs) on your board. <S> But inside an FPGA there are no tri-state signals (only on its I/O pins) <S> so if you use either technique above, the synthesis tools translate it into an equivalent signal without the tri-states. <S> For example the first technique (wired-AND) might translate to <S> Load_Divider_1 <= '0' <S> ; <S> -- module 1Load_Divider_2 <= '1'; <S> -- module 2Load_Divider <= <S> Load_Divider_1 and Load_Divider_2; where you can see that each signal has only one driver. <S> So why not write that in the first place? <A> When I need to do this I have each module drive its own separate signal, then do whatever I want in the toplevel module. <S> Something like signal sig_a, <S> sig_b <S> , sig_final: <S> std_logic;foo_a: entity foo PORT MAP ( clk = <S> > <S> clk, rst = <S> > <S> reset, out <S> => out_a );foo_b: entity foo PORT MAP ( clk = <S> > <S> clk, rst = <S> > <S> reset, out <S> => out_b );-- for active-high, use or <S> :sig_final <= sig_a or sig_b;-- for active-low, use and;sig_final <= <S> sig_a and sig_b;-- if you have to override bad entity logic: <S> sig_final <= sig_a or sig_b when reset = '0' else '0';-- <S> if you have something really complex and that has to be registered <S> :determine_sig: process(clk, <S> reset)begin if reset = '1' <S> then sig_final <= '0' <S> ; else if rising_edge(clk) then -- do something wild and wooly based on sig_a, sig_b, state variables, phase of the moon, etc. <S> end if;end process; signals are cheap and allow you to be explicit with your code. <S> There is almost never any need to muck about with strong and weak drivers, and NEVER use the high-impedance state ('Z') for internal signals. <A> However, you must provide a resolution function for the signal that determines what the output is if more than one source drives the same signal at the same time. <S> You would typically define a resolution function in a package and then include the package in both modules. <S> The most commonly cited example is that of std_logic in the std_logic_1164 package which can be seen here . <S> That example is probably a little more complex than you need. <S> A simpler example would be a wired AND: package my_package is function wired_and (input : std_logic_vector) <S> return std_logic;end my_package;package body my_package is function wired_and (input : <S> std_logic_vector <S> ) return std_logic is --setting this to '1' ensures that even if there is only one --driver, the AND will output the correct value variable output : <S> std_logic := '1'; begin --AND <S> each input driver with all of the others for i in input'range loop output := <S> output and input(i); end loop; return output; end wired_and;end <S> my_package; The function takes in an array, the std_logic_vector <S> called input , that contains all of the multiple drivers. <S> And it outputs a single bit, <S> the std_logic called output . <S> Instead of using this to assign your signal a value: Load_Divider <S> <= '0'; You would use this: Load_Divider <S> <= <S> wired_and('0'); Though that can get somewhat annoying and can be error prone. <S> It is generally easier to declare a subtype that automatically uses the resolution function like this: subtype my_type is wired_and std_logic; And declare your signal like so: <S> signal Load_Divider : <S> my_type; <S> That way, whenever there are multiple drivers of that signal (even if there are more than two in the example above), the resolution function will automatically be called and the appropriate value output.
| You can have multiple drivers of a signal.
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Is there an open source USB to JTAG circuit I can include in my PCB design? Im trying to find out if it would be possible to build/include a USB to JTAG circuit on my experiment circuit board I'm designing. My aim is not to reinvent the wheel, but use something that is already out there, and simple/small enough to add to my PCB design. I want to be able to program and debug my Atmel SAM3S chip using JTAG, via USB. I've seen devices like the USB to JTAG programmer from Modular Circuits, and it looks like this can be fairly easily integrated into my design, however, I'm not too sure if their license allows this. I've tried to Google around a bit for alternative more open sourced designs, but couldn't really find something. Can anyone refer me to a Open Source project for USB > JTAG circuits, that are available online? <Q> There are many examples on the web of using an FTDI FT232R chip in its "synchronous bit-bang mode" to drive a JTAG interface to configure a microcontroller or CPLD/FPGA, including the necessary host software. <S> One advantage of this approach is that you can subsequently switch the chip to its regular UART mode and use it to communicate with the target application. <A> Note that I haven't used this system myself, but the following project called opendous-jtag is compatible with OpenOCD and may be worth a look. <S> The software license is the fairly liberal MIT license and it's based on open source hardware you should be able to replicate: http://code.google.com/p/opendous-jtag/ <A>
| There's a Bus Blaster of Dangerous Prototypes - it's based on FT2232H - this chip has two serial interfaces - one is used for JTAG and another - to reprogram onboard CPLD to emulate different JTAG debuggers.
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Why pull base of BJT switch? I often notice usage pull resistor for base of bipolar transistor, e.g. \$R2\$ here: Why it's used? I understand pull resistors for FETs, because of gate's high impedance an EMI can easy switch it. But BJT need base's current to open and, I think, EMI has too high internal impedance to give enough current. Is it safe to leave floating base in BJT switch? <Q> A bit longer: When the current ceases to be provided through R1, all there is left to stop the base from being forward biased is the current that flows through the base itself. <S> This works for slow circuits. <S> For fast turn-off, R2 helps to get the charge out of the base. <S> Note that there's a parasitic capacitance from B to E and worse, from C to B. <S> The latter is worse because while B goes down, C goes up, and the C-B-capacitor pushes charge into B while you want to lose charge in B (Miller Effect). <S> Also, with B floating, a tiny current (interference / parasitic creepage path) may be enough to turn the BJT on. <S> R2 helps to prevent this. <A> In addition to what zebonaut said, another reason could be so that the transistor turns on at a higher voltage than its inherent B-E drop. <S> R1 and R2 form a voltage divider. <S> By setting the divider appropriately, you can get a higher effective threshold a the left of R1 for turning on the transistor. <A> Yet another reason for adding a pull-down resistor is that many BJTs have a certain amount of collector-base leakage. <S> If R1 is large, or if the pin that's driving it is not actively pulled down, <S> that leakage pay pull the base up to 0.7 volts. <S> If that happens, the transistor itself will amplify the leakage. <S> If the transistor has a base leakage of 0.2uA and a beta of 50, then in the absence of any sort of base pull-down, the collector would end up leaking 10uA. <S> Adding a pull-down to the base reduces that leakage to 0.2uA. <S> In some situations, 10uA of leakage may be no big deal, but if everything else in the circuit is powered down to draw only 5uA total, having a transistor leaking 10uA could triple one's power consumption.
| The very short answer: R2 helps to turn the BJT off fast .
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Are there existing electronic sensors that can sense if underwater or not? My goal is to to be able to read the water level percent in a small water tank. I was thinking wiring sensors going up the side that read whether they are currently touching water. My plan is to wire the sensors into my Arduino and then calculate the percent full by the sensors that are touching water and the sensors that are not. Are there existing sensors with this functionality? (are they cost effective?) Is there an alternative solution? Thanks so much! <Q> There are a few types of water sensors, with varying levels of applicability to the purpose stated in the question. <S> At the simplest level are electro-conductive sensors, essentially a pair of zig-zagging exposed conductive traces along a PCB, with a small gap between the traces. <S> If the device is submerged in water, there is conduction between the traces, which would indicate presence of water. <S> Of course, the traces must not be covered by solder mask. <S> From this site <S> Since the conduction of electricity through the water is dependent on electrolyte / ion content in the water, results vary depending on the water quality. <S> Further, due to corrosion and oxide build-up, the longevity of these sensors is very limited. <S> If the water tank in question contains somewhat impure water, then a long strip, or several short strips, of such sensors made of standard PCB may serve the purpose. <S> Since the "sensor" is inexpensive to make, regular replacement is an easy recommendation. <S> A variant on this theme is the use of microstrip antennas, essentially the same concept of PCB traces, but designed to be energized via a radio frequency signal. <S> As the dielectric between the two conductors varies with humidity, moisture and of course submersion, the resultant change in antenna load can be used for detection. <S> See this publication for additional information. <S> The hygroscopic material expands in moisture / water, tripping the limit switch. <S> Such sensors may falsely trigger in high humidity, such as in an enclosed water tank. <S> Although this is not the question asked, an alternative approach is to use an ultrasonic or infrared proximity sensor to sense the water level from above , and thereby determine how full a tank is. <S> Similarly, there is the classic float sensor, which consists of a floating ball attached to a lever, that acts upon a flex sensor or absolute rotation sensor, to determine the level of water. <S> These approaches may well be more reliable to implement, than the submersible sensor approach. <S> See also : All the answers to This question , and this other one . <S> Suggestion : <S> Research Rain Sensors for other options. <A> VDO sell water-tank sensors & gauges for boats, or you could possibly get away with using a standard resistive float sensor from a car fuel tank as long as you protect it from corrosion. <S> These will give a varying resistance (typically 0-240 Ohms) in proportion (NOT always linear, note!) to the float position. <S> You can bend/extend/shorten the arm to suit your tank. <S> Something like this: http://www.brit-car.co.uk/product.php?xProd=90015 <A> Another option is a floating magnet (for example embedded in a ping pong ball) rigged up to the edge of the tank and a series of hall effect sensors (or reed switches) up the outside of the tank. <S> You can detect the water level by reading the state of the reed switches. <A> Surprisingly I stumbled on the image of our water sensor (I'm the owner of TempGP). <S> Just to add to comments about our sensor plate, their galvanization (transfer of ions from one side to another) to avoid that variable/ alternative current can be used there. <S> Also in our V3.0-B version we planing to use an ultrasonic sensor to detect distance to the water source, this will allow corrosion resistance and the sensor can be adjusted to tank specs.
| More sophisticated sensors use a stack of hygroscopic material, and a limit micro-switch, to detect water. Even more advanced sensors exist for the type of function described, but they are typically for industrial use, and are very expensive, if at all sold in small numbers.
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How to light a fluorescent lamp? There are a lot of circuits on the internet for lighting fluorescent lamps. But they don't reveal the working mechanism of the lamp itself. I'm more concerned about the theory rather than the practical circuits. Please consider the lamp as a 4-pin device. And state what signals to apply these four pins. I'm looking for an explanation like: For the first 3-5 seconds, connect the B and D pins to the ground, and apply 9-15V DC to the A and C pins to heat up the lamp. After that, disconnect the C and D pins and apply 200-300V 50Hz AC voltage between A and C pins. I understand that the voltage levels and time periods will differ between different products, but at least an example on a commonly used one will make the things clear. <Q> The lamp is simple to operate in principle. <S> It doesn't need four terminals: simply supply a regulated current from one side to the other. <S> In practice, that's not trivial. <S> Like any gas discharge lamp , it exhibits a negative resistance and so requires a ballast to limit the current in the lamp. <S> This is not unlike the series resistor used to drive an LED, but it must operate under different conditions. <S> The tricky part is starting the arc. <S> Once the gas inside is ionized, the resistance between each side decreases tremendously, so the applied voltage must immediately be decreased to maintain a low current that won't destroy the bulb. <S> Cold cathode lamps work this way, starting the arc with nothing more than high voltage. <S> This is not a cold cathode lamp. <S> Between each pair of terminals on each end is an electric heater designed to facilitate thermionic emission . <S> Essentially, the high temperature encourages charge carriers to wander off the cathode, reducing the voltage necessary to strike an arc. <S> What signals you apply to these pins is entirely up to you. <S> There are countless ballast designs, each with a different method of lighting the tube. <S> Some are little more that transformers. <S> Some have microcontrollers. <S> Some are made with cheap integrated circuitry from China. <A> Phil's answer is good on the need for a ballast, and the basics of starting - cold cathode, via high voltage breakdown; and heated cathode via thermal emission. <S> The old fashioned starter had two pins : it was a tiny cold cathode lamp with a bimetallic (thermal) switch that shorted itself out after a brief time. <S> Connect AC via ballast to pins A,C and connect the starter to pins B,D. <S> Apply AC and things happen in the following sequence... <S> There isn't enough voltage to strike the main light, but the starter will light, and heat up. <S> The starter shorts itself out, connecting pins B,D together, powering the heaters. <S> As the starter is no longer lit, it cools down. <S> The starter switch opens, applying AC to both itself and the heated gas in the tube. <S> Now either of two things can happen : <S> (3a) <S> the main light starts. <S> It conducts well enough that there isn't enough voltage to light the starter again. <S> Or (3b) <S> the main light doesn't start : goto (1). <S> You've got to admit <S> , that's amazingly sophisticated behaviour from a little glass bulb with two terminals - sheer genius! <S> I once had a starter that failed short circuit; the switch had welded itself shut. <S> There was a glow from the heater at each end of the lamp. <S> I put the starter in the freezer for half an hour; the extra pull from the low temperature popped the switch open, and it worked reliably for a couple more years! <A> Yes it is the inductive surge that is created by the ballast when the switch suddenly opens. <S> This surge creates a voltage spike that now appears at the cathode and anode and hopefully should create an arc wthin the tube. <S> If no arc the cycle repeats itself as stated previously. <S> The ballasts acts as transformer and choke. <S> The choke part needed to create the arc and limit the current flow of the arc. <S> The transformer portion needed to provide the proper voltage for the lamp and choke used. <S> The longer the lamp the larger the choke?
| You can even put one in a microwave and light it without using the terminals at all. One way is to apply a very high voltage (many kilovolts) from one side to the other until you get an arc.
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DIY power extension with relay My project involves hooking up relays to individual plug points so I can control them with a microcontroller. Unfortunately, I am not sure how to approach splitting my mains 240V into 4 different inputs. All I can think now is to use a metal plate, solder the mains to the metal plate, and solder to the live connection of the 4 inputs to that plate. Then do the same for neutral and ground. Is there a better approach? (EDIT) PS I have two possible methods. Can I know which is safer. (Drawn here at this link) One way option 1 is to solder the main mains wire to a metal strip. Then solder the live wires of my extensions onto the strip. The relay will divide this wire to the extension (I am using an arduino all in one relay module http://dx.com/p/arduino-5v-relay-module-blue-black-121354 ). Option 2 is just to use those mains wire caps to join up all the wires. I feel this is safer as I do not have to expose a metal strip inside my project box. Ideas? <Q> Metal strips are used in swtichboards in my country because they can handle more power and have less risk of catching fire (cheaper maintenance, less human error risk). <S> They are also cheaper to manufacture, since the strips are connected and held in place by screws and not soldering. <S> If you're planning on making a PCB, you could use the PCB itself to distribute power with ultra wide traces (not a good idea if you want to control heavy duty electronics like ACs or room heaters). <S> The PCB would give you the freedom to leave large pins for the thicker mains wire to be soldered properly. <S> The optimal solution without a PCB or buying new components like thimbles, in my opinion, would be your option <S> two where you use the caps as thimbles. <S> I would advise caution, though - the outer surface of the the caps are also conducting and therefore these would have to be well secured and insulated. <S> Remember to not skimp on the wire thickness to accomodate more wires per cap - that could lead to burning plastic, broken wires, and worst case, electrical fires. <S> A fuse may be a good idea to include in your box, to be safe. <A> First off, be extremely careful when working with mains power. <S> You can really hurt yourself (possibly even die,) and your device may start a fire, etc. <S> You should also make sure your relay(s) can handle the power (current and voltage,) that the end devices may require. <S> As already pointed out, your question is somewhat unclear. <S> To explain this better, you would leave one wire going to the light, and the other wire is going through the relay. <S> So when the relay closes the circuit, your light turns on. <S> When it relay opens the circuit, the light turns off. <S> This is the same as how a light switch works. <S> You can easily extend this to 4 devices by having all the devices share the same AC line, and a relay to open/close the second line for each device. <S> You didn't explain how the relay is turned on or off, but just for the visual, here is a schematic of a DC relay. <S> Note that the diode is across the coil so that the relay turning on or off doesn't induce voltage that may damage the logic. <A> By far the easiest and safest way to do this is to use a solid-state relay. <S> You can drive them directly from a microcontroller with 3-5V and <S> perhaps 5-10mA. <S> I would use a powerstrip that already has 4 sockets. <S> If the sockets aren't too close together, you will generally find enough space between them add in the solid-state relay.
| Soldering a wire to a metal strip is not especially easy, though, and there's a risk of not doing a great job with one wire and it coming off and touching something else. However, lets say you have a mains powered light, to turn it on and off with a relay, all you need to due is interrupt power to one of the two wires. Then you merely need to run a cable into the powerstrip with 5 conductors (4 signals plus ground), and you're done. The ideal way, in my opinion, would be to use some sort of thimbles which you can hold together with nut and bolt.
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Input impedance of inverting amplifier clarification Take a standard inverting amplifier: I'm unclear on what the input impedance is. At first I thought it was the equivalent resistance from the inverting input to the ground, which would be Rin || Rf, because there's a ground on the other side of Vin and Vout as well as inside the output of the opamp. However, most sources on the internet seem to claim that the input impedance is Vi/Ii, thus making it Rin. This appears to ignore the fact that the feedback wire connects to other grounds. Then there are answers like this that mention that input impedance is infinity. I'm hoping to figure out a clear definition of what the input impedance represents and a [brief] general approach to calculating it in a more complicated circuit. Thank you! I read tons of pages like this answer and others, but at my beginner level it was hard to extract a clear answer. <Q> When we talk about the input resistance of a circuit, we're describing how it affect the other circuit that is providing the input signal. <S> Specifically, we want to know, in order to change the input current by i <S> amps, how much do we need to change the input voltage? <S> That's why the input resistance is, by definition, <S> \$ \dfrac{\mathrm{d}v_i}{\mathrm{d}i_i}\$. <S> So what's the input resistance of this circuit? <S> The key point is that in this configuration, as long as we avoid saturating the op-amp output, the inverting input of the op-amp is a virtual ground . <S> The feedback in the circuit operates to keep that node at 0 V. <S> So whatever input current we want, by Ohm's Law, the required input voltage is \$\mathrm{R_{in}}\times{}i_{i}\$. <S> Therefore the input resistance is R in . <S> Then there are answers like this that mention that input impedance is infinity. <S> That answer was talking about the input resistance of the op-amp, which was assumed to be ideal, not the input resistance of the whole circuit. <A> The Photon's answer is absolutely right: ideal op-amp, etc, input impedance is Rin. <S> It's important because it can tell the designer about loading effects between the output impedance of the prior stage and the input impedance of the next one. <S> The general approach to calculate input impedance (or output impedance) is to inject a small current into the input node (di) and look at the resulting change in voltage of the input node (dv). <S> Or, equivalently, to apply a small voltage (dv) and look at the resulting current (di) from your test voltage source. <S> Then compute (dv/di). <S> To make it completely clear that this is what's going on, take a look at my answer to How to compute input impedance , where I demonstrated how to use a circuit simulation program to compute and plot the input impedance by literally adding a test voltage source at the input and plotting a custom expression. <S> Hopefully, seeing the voltage source V1 (or in your case Vin) drawn literally as a voltage source will make it clear how to go about setting up the calculations for doing it by hand! <A> The definition of Ri is Vi/Ii. <S> Rf is connected between the output node and virtual ground, and is basically a circuit on its own. <S> It is not involved with the input. <S> What you said is correct; Ri is the equivalent resistance of the input, but that is equal to Rin, not Rin || Rf. <S> As for “other grounds”, there is only one ground. <S> Ground just means V=0. <S> You can treat “real” ground and “virtual” ground as one and the same when it comes to circuit analysis. <S> I would like to add an intuition on what the input resistance actually is and why it matters. <S> In other words, if you apply vi to the amplifier, or apply vi to Ri connected to ground, you get the same current flowing through vi in both cases. <S> Hence Ri=vi/ii <S> As for why it’s important, real signal sources have not only vi, but a signal resistance Rs attached to it. <S> Once you attach the signal source to the inverting amplifier, the input voltage vi would be the node voltage between Rs and Rin. <S> Generally, if you look at an equivalent circuit, the input resistance is the total equivalent resistance between vi and ground. <S> So if you look at the voltage divider rule,Vi=Vs•Ri/(Ri+Rs)Which means the higher the input resistance, the more signal you get at the input. <S> Keep in mind <S> Ri= <S> vi/ii, NOT Vs/ii. <S> In the second case, Ri would depend on the signal source resistance, which is not the case.
| Input resistance Ri is the equivalent resistance between vi and ground. In a more general circuit, even one with non-linear components like transistors, input impedance is a small-signal (linearized), frequency-dependent quantity.
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What properties vary with the number of blades on a BLDC fan? In general, what properties vary with the number of blades on a axial BLDC fan when physical and electrical dimensions are the same (e.g.: rated at approximately 12V and 0.15A and 60mm Ø)? So far I've seen fans with 7, 9 and 13 blades and unfortunately not all fans come with a datasheet. Things that I can think of myself include, but don't know how they would be influenced by blade count: Fan speed (rpm) Volume (\${\text{m}^\text{3}}\cdot{\text{min}}^\text{-1}\$) Application (mounted directly on a heatsink or mounted in equipment housing) Artistic mood of the designer 13 blades: 9 blades and 7 blades: <Q> Hence the USA-type (water pumping?) <S> windmills with lots of blades are slow-runners, contrast them to modern electricity-generating windturbines that have 2 or 3 blades. <S> Applying this to fans means that a higher number of blades means that it can rotate more slowly. <A> More blades allow a higher delta-P (pressure) to be developed across the fan. <S> Thus if your application created higher than average backpressure, a fan with more blades might be better. <S> There are design tradeoffs in everything, and I'd suspect that flow is compromised in order to achieve higher pressure. <S> Just anecdotally, fans with more blades tend to be quieter, but this is highly subjective and really depends more on the shape of the blades and RPM than just on the blade cardinality. <S> Compare the flow curve of a fan with many blades to a similar-sized, similar RPM fan with fewer blades. <A> Fan designs can come in a huge number of variations. <S> Various manufactures will offer a large range of sizes of fans with varying numbers of blades and blade designs. <S> If you get into the mode of system design where you need to select a fan to meet a particular cooling requirement then you really need to look at some manufacturer catalogs. <S> It is rather surprising at the total number of models of fans made. <S> I did a quick survey of just one product family of fans from manufacturer in a specific size and found a large range available for selection. <S> The parameters you select on do not necessarily focus in number of fan blades. <S> Instead you would be focusing on size, voltage, rated current, speed, max air flow, max air pressure and noise level. <S> An example of the fans from Delta in their AFB series in the 60mm x 60mm size yield tremendous variety of selection. <S> ( click here for larger image size ) <S> The above set of fan models can be seen in the Delta Fan Catalogue at these links: http://www.delta.com.tw/product/cp/dcfans/download/pdf/AFB/AFB60x60x13mm.pdf <S> http://www.delta.com.tw/product/cp/dcfans/download/pdf/AFB/AFB60x60x15mm.pdf <S> http://www.delta.com.tw/product/cp/dcfans/download/pdf/AFB/AFB60x60x20mm.pdf <S> http://www.delta.com.tw/product/cp/dcfans/download/pdf/AFB/AFB60x60x25.4mm.pdf <S> http://www.delta.com.tw/product/cp/dcfans/download/pdf/AFB/AFB60x60x32mm.pdf <S> http://www.delta.com.tw/product/cp/dcfans/download/pdf/AFB/AFB60x60x38mm.pdf <S> Just this single AFB series of fans from delta come in all these size selections: And to put even more emphasis on the total number <S> Delta is offering all these series of DC powered fans and blowers: <S> Note that I have no affiliation with Delta other than I've used their fans in some product designs.
| I am not sure whether this applies equally to fans, but for wind turbines (the things that doe the opposite, turn moving air into a rotation) a higher number of blades causes the optimum working point to be at a lower RPM.
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Using any/all Ethernet wires for power transfer? I was looking to get a small amount of power to be transmitted ~100ft, and noticed there is a standard way to move power over Ethernet cables . We actually plan to splice the cables ourselves, and not use a standard Power over Ethernet, but wanted to confirm how much power I can dependably pass through a single pair? There will be no data transmission, no Ethernet devices. Only DC current. Further, I wanted to know if it makes a difference which wires I use ? I am not sure whether all 4 pairs (8 wires) are the same, or if only certain ones are built specially for power, while the others might be inappropriate for power, and only dependable for data? Edit: Additional details: I have a transformer plugs into the 120V wall plug, and it says it outputs 5V 1.2A. So that is what I want the cable to support. I plan to cut the small wire coming out of the transformer and extend it using ethernet wires. If ethernet 1 pair is enough, I might do that. Or I might use 2 pairs just to keep the two transformers and end devices completely separate. <Q> If you don't need compatibility with ethernet, it doesn't matter which pair you use. <S> It doesn't even really matter much that you use pairs. <S> Your main concern in this endeavor will be the resistance of the cables. <S> You can find a chart of wire resistance and from that, calculate the total resistance of your cable. <S> For an example, cat-5 cable is typically 24-26 AWG. <S> From a table, I see that 100 feet of 23 AWG wire has a resistance of about \$3.2\Omega\$. <S> Say you want to move 500mA on this wire. <S> We can calculate the power lost in the cable: \$ <S> P = I^2 R = <S> (0.5A)^2 <S> 3.2\Omega = 0.8 W \$ or the voltage <S> you will lose on this cable: \$ <S> V = <S> IR = 0.5A <S> \cdot <S> 3.2\Omega = <S> 1.6 V \$ <S> Keep in mind, this is for 100 feet of wire, but since you need a supply and a return, the total length of the wire is twice the length of the cable. <S> This 0.8W is not only power not doing useful work, and less power available at your load, but also heat generated in the cable. <S> A longer cable has more area to radiate heat, so can dissapate more power. <S> This is why most tables also include a maximum current. \$500mA\$ is just a bit too much for 23 AWG cable. <S> Also, the voltage drop will reduce the ability of a voltage regulator to regulate the voltage on the other end. <S> The voltage regulator has no way to know what the voltage on the other end is, so it can't regulate it. <S> Another concern is the maximum voltage the insulators in the cable can withstand. <S> Power over ethernet provides 48V, so anything below this should be safe. <S> Anything above that, consult the datasheet for your cable. <A> My past investigations and other people's experiences showed that the resistance of the various pairs varies in cables which comply with the relevant standards and they can be significantly worse than what the standard stipulates in cable from 'less reputable suppliers'. <S> In aquick look now <S> I found no mention of different resistance per piar so that MAY have been CAT5 only. <S> Maybe. <S> Measurement with an Ohmmeter will rapidly tell you what the situation is with the cable that you have but regardless, connecting the wires as I suggested above will give you the best result for a given cable. <S> Pacific Custom Cable say re Cat6 - 26 Ohms max per 100 metres per pair. <S> This article suggests Beldens Cat6 24AWG cable, part 1872A has a resistance of 2 Ohm per 40 feet or about 16 Ohms per 100 metres or in your case 5 Ohms loop resistance over 100 feet. <S> IF all pairs were the same then 4 pairs in parallel will give about 1.25 Ohm loop resistance. <S> Here's some "military tactical Cat6M - Kevlar armour etc. <S> Pretty picture of makeup <S> Specs <S> - they say 23/7 AWG (23 AWG 7 strand), 75 Ohm/km - probably one leg, so 15 Ohm/100 <S> m loop - much as above for 24 AWG. <S> They specify 1.5A max for the RK45 connectors so 4 pair may allow 6A. <A> Yes, a bit late, but just to complete a minor issue. <S> One way to correct for the voltage drop is to use a power supply with voltage sense terminals and use one pair of the cable to as a voltage sensing pair to the distant end. <S> Within reason this allows the power supply to boost the voltage at its end to compensate for the IR losses in the cable. <S> The sense pair carries very little current, so there is no significant voltage drop. <S> The downsides are: <S> The power supply is typically more expensive than your run-of-mill supply. <S> It leaves one less wire pair for carrying current for the end load. <S> Because of the compensation the voltage at the power supply is higher by a few volts under normal circumstances, so equipment plugged in closer to the supply will see that higher voltage. <S> You have to make sure the power supply has enough headroom to be able to correct for the voltage drop. <S> You have to make sure that the load is never powered via the sense lines, it destroys the current sense resistor(s) in the supply. <S> So to be safe switch the AC side of the supply only, or switch both sides of the load on/off at the same time, not just the high or low side. <S> For more details you can search for 'remote sense power supplies EDN', there's a nice EDN article that goes into more detail on this topic including what -not- to do with this technique.
| If you are using the cable simply as a power cable and effectively need two conductors then using the 4 x "white striped" wires together as one conductor and the 4 solid coloured wires together as the other conductor will give you a two conductor cable with equal resistance in each conductor and lowest possible overall resistance.
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Alkaline "% full" from voltage? Is there an equation or even a graph I can look at that appropriates the % full of an alkaline battery? <Q> There's usually a V vs. <S> Ah curve in the datasheet for the model of the battery, which you chose. <S> If the battery comes from a reputable manufacturer, the datasheets may be available on the web. <S> Example: from here . <S> In addition to the useful link , which Brian had posted. <S> You can also make your own measurements and generate/verify the curves. <S> You would characterize the battery yourself. <S> Keep in mind that the curve is dependent on temperature and rate of discharge. <S> Essentially, there is a family of these nonlinear curves. <S> It's usually not practical to try to account for these parameters. <S> As a result, the error can easily be as large as 20%. <S> Battery voltage measurement could be an adequate approach if all that's needed is to make a threshold comparison and blink a "battery low" indicator. <A> Just monitoring the voltage to estimate the condition of a battery is an approach that just does not work very well. <S> I have found that there are some other ways to assess battey life that may be a bit better than by trying to just monitor voltage behavior. <S> One system I actually designed into a product years ago used a method where the measurement process used a transistor to switch on a temporary load on the battery. <S> The load was more than the normal discharge rate of the battery by two to five times. <S> Voltage under load was monitored, then the load was released and the voltage change curve after that was monitored for a period of time. <S> The total time of the load being applied was short, no more than a millisecond or so. <S> The repetition time of the test was made slow enough so that the net loss of battery capacity had a very small impact on battery longevity. <S> Could even be only once per day. <S> It was fairly easy to characterize this methodology for a particular battery family so that a battery life meter in the product display was surprisingly accurate. <S> Another system I designed into a shrink wrapped battery pack used one of the many battery "gas gauge" chips that are on the market. <S> These chips monitor current flow out of the battery by measuring the small voltage drop across a series shunt resistor that is in the battery lead. <S> Note that these "gas gauge" chips are most often meant to be used for re-chargeable batteries but can also be used for a discharge only application as well. <S> This measurement scheme does a pretty good job of accumulating the the total energy drawn from a battery pack. <S> When used in conjunction with a thermistor that be used to monitor and compensate for temperature effects you can end up with a system that is much better than simply measuring the battery voltage. <S> Both of these schemes require some engineering characterization and testing work to fine tune the algorithms used and produce the final implementation. <S> That work can be kind of fun to do. <A> Short answer: <S> No, unless you know the current draw and the manufacturer. <S> There are such curves . <S> However they depend on battery manufacturer since some are more cheaply made. <S> Also higher currents result in less power being provided due to resistance and chemical effects.
| The alkalyne battery voltage versus voltage curves are anything but linear and vary according to a number of factors like temperature, pre deployment shelf life, manufacturer, chemistry and discharge rate.
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What to look for in an oscilloscope? I'm getting back into basic electronics with an Arduino after many years away from the hobby, and I've forgotten a lot of my knowledge. One thing I'm finding is that I really need to get an oscilloscope to assist in designing filtering circuits to give me smooth output from PWM. I'm not after anything fancy and my budget it small, so what sort of things are on a an oscilloscope must have list? <Q> This is somewhat dependent on the work you do, but I think it's mostly an artifact of what you are used to. <S> I came up working with digital scopes (and scope/logic analyzer hybrids), so that's what I like. <S> Bandwidth: Just make sure it's high enough for what you want to look at <S> (I do old arcade gear, which tops out in the < 20 <S> Mhz range <S> , so I don't need a lot of bandwidth). <S> Remember that this is the ANALOG bandwidth - not the sampling rate of a digital scope. <S> Sampling rate: Digital scopes only, of course, but with digital scopes remember that the sampling rate is NOT inherently related to the bandwidth. <S> It's entirely possible to have a scope with a high bandwidth, but a sampling rate barely able to meet Nyquist on that bandwidth. <S> Memory depth: <S> Again, digital scope only, but important. <S> Especially important is to look for shady specs like only being able to sample at full rate with a part of the memory buffer. <S> This trick is kind of common and quite annoying, because it makes comparing the sampling rates of different scopes quite difficult. <S> Specialty items: There are a lot of 'USB scope' things out there these days. <S> BE CAREFUL! <S> Especially if they say 'unlimited buffer', you want to look closely - those scopes turn out to be sample rate limited to however fast the USB bus is on your computer. <S> In an optimal situation they should be able to meet their stated sample rate, but if something (driver issue, competing device, whatever) slows your computer's USB bus down, then your sample rate can drop. <S> On the other hand, they are usually quite cheap. <A> If you want a good scope for not a lot of money, pick up just about any analog scope from ebay. <S> I've had a Beckman Industrial 9022 and never found it lacking for hobby use. <S> It's a 20 MHz scope, and there are few things that require anything faster. <S> The fastest thing I have on my desk is a micocontroller clock, but even if for some odd reason I wanted to run my AVR at maximum speed <S> and I wanted to scope the clock, at 16 MHz it's still visible. <S> Maybe not super sharp, but fully resolved at least. <S> Unless you plan to build any RF circuits soon, I doubt speed will be a concern. <S> Despite the tech gadget factor of a digital scope, there are some things that show up better on an analog scope. <S> The cheap digital scopes I've had the displeasure of using have the problem that they can't update the display very fast, while an analog scope updates the display instantly each time it triggers. <S> This is great when you are looking at noise superimposed on another signal; with an analog scope you can see the trace brightest at the most frequent values, and dimmer regions of the trace for the rarer excursions. <S> On a digital scope all you get is a sharp trace that wobbles, to whatever the noise happens to be each time the display updates. <S> I'm sure there are digital scopes that address these shortcomings, but they will be out of your price range. <S> You might miss the storage capabilities that every digital scope has, but I don't. <S> For things I need to store, a logic analyzer is usually the right tool, and for the other times, dimming the lights, turning up the brightness, and using the single trigger mode works well enough usually. <S> Analog storage scopes do exist; maybe you can find one for a good price. <S> In all, I think what you lose in high-tech gadgetry you will gain in quality construction and robust engineering. <S> New, inexpensive digital scopes are made in China and will break in a couple years. <S> A used analog scope for the same price was made in the same country that designed it with parts made to last 200 years. <A> For the uses you mentioned, at least 25MHz and at least 500Msps. <S> But of course the more, the better. <S> Since oscilloscope manufactures arrange their product lines along a price range, the other parameters will have correspondingly adequate values, usually. <S> Also I'm biased against both pocket and USB models, the former because they are "cheap" (!) <S> and USB because they are more limited and less practical. <S> Sparkfun has a model with quite better features, but at a reasonable price for them. <A> For your given use, just buy anything of a known make 2nd-hand for peanuts, low-speed (<50MHz) <S> analogue scopes go for a pittance these days. <S> You don't need fast or fancy. <S> Anything with a Tektronix badge will do. <S> Used but known stuff is preferable to cheap new chinese stuff as you know there'll be parts & service out there if you need it. <S> DSO's are nice <S> but you have to remember they will lie to you as the signal is by definition always processed. <S> The Seeed studio DSO-Quad / DSO-nano are nice open-source pocket scopes and perform well enough for whatever you want (and can also play tetris), but can be fiddly and the screen is of course quite small. <S> For the money I'd rather have a cheap used analogue or CRT-DSO scope with knobs and buttons as my main or only scope.
| Digital vs. Analog: I've a preference for digital, because I'm often looking for transient stuff, and I like being able to stop on a trigger and really look at the signal.
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What are those wire terminals that mate with standard header pins? I have purchased "dupont" connectors on Ebay that crimp onto a wire. They look like this: http://www.hansenhobbies.com/products/connectors/pt1inconnectors/pt1in_ft/ I'm tired of waiting weeks for these things to ship from China so I would like to purchase these from Mouser but I don't know what they're called or how to identify them on mouser's website. They are the standard rectangular wire terminal that is normally used to slip into a 2.54mm housing but I just use them without the housing because then slip snugly only standard header pins. Can someone tell me how to find these on Mouser? EDIT I just wanted to follow up on this because, as stated by The Photon, there are different varieties of contacts that are IMO quite significant. (source: ioplex.com ) Regarding the above image: On the left is a "dupont" connector from China via Ebay. Its contacts are slim and form a sleeve that holds header pin squarely. This limits wobble to one axis. On the right is TE Connectivity 1-104480-2. Its contacts pinch the pin with two fingers. This allows wobble on both axis. And I can guess that the surface area of contact is less. It also has a spur that useless and potentially annoying when used without a housing. So unfortunately I still have not successfully identified the correct part and as such I have reset this question as unanswered. <Q> Those will be under the "Rectangular Connectors" heading of the connectors section of their catalog . <S> This is one instance where thumbing through the catalog is easier than the online version. <S> Check out the Molex SL Connectors on page 1623 . <S> More generically known as .100 inch headers. <A> http://tech.mattmillman.com/info/crimpconnectors/#minipv <S> Those 0.1" (2.54mm) crimp terminals with the triangular insulation crimp "wings" that are found basically everywhere are called Mini PV basics and are made by FCI Amphenol. <S> There are countless knockoffs which is what you will usually find on ebay. <S> They are NOT designed to be crimped with a generic double D crimp die, if you crimp a lot of them you should track down an HT-95 crimp tool as this will do the job correctly (and much more easily as it has a pin locating mechanism which frees you to just line up the wire and close the handle). <S> You can find old Berg and Dupont branded ones on Ebay for under $100, new they cost over $1000. <S> You would not believe how long it took me to find this information. <A> Unfortunately when it comes to connectors every manufacturer seems to have a different name for the same thing. <S> Once you find the right housing the datasheet should indicate the corresponding pins. <A> The Harwin female crimp contacts ( M20-1160042 ) are a good match for the original question's picture and seem to me one of the better choices for the hobbyist as they can be used in both single and 2 row housings that can be found inexpensively all over the place (Jameco 100812 for instance). <S> I can also vouch for the usefulness of the Japanese crimping pliers made by a company called "engineer" <S> I've used my PA-21 on many projects and types of connector pins, and while manually crimping wire terminals is never going to be fun or fast, it is far less frustrating when the results are actually of consistently high quality. <A> I know this is an old post, however it popped on my screen <S> and I get this question often when helping with arduino and other projects. <S> If it is just for connecting to a header and only 1 position I will suggest using a housing such at Amp FCI part <S> 65039-036LF <S> This uses many contact one of them being 47212-000LF which is very similar to the one you have pictured. <A> Search for m20-1060x00, where x is the numbers of required pins. <S> Unfortunately for me, only from US or large quantity from china. <A> I believe that Harwin make a compatible crimp terminal http://www.digikey.co.uk/product-search/en?keywords=M20-1180042 <A> FYI Molex SL connectors will work just as well (better even) and have some nicer wire to wire connector options. <S> They also have more reasonably priced official tooling, $140 for a service grade tool that has a locator and wire stop and they work better with generic crimpers as well. <S> Relevant parts, the terminals here are the 30 micron gold plated ones, there are cheaper ones with less gold or tin plating as well as a "high contact force" version (which is less forgiving of exact pin dimensions,so I don't use those) <S> If you are using wire smaller than 22-24 AWG then you want the smaller gauge terminals for those wire sizes. <S> https://www.digikey.com/product-detail/en/molex-llc/0016020104/WM2564-ND/1118774 https://www.digikey.com/product-detail/en/molex-llc/0016020117/WM2567-ND/1118775 <S> https://www.digikey.com/product-detail/en/molex-llc/0050579003/WM2801-ND/115005 https://www.digikey.com/product-detail/en/molex-llc/50-57-9403/WM2901-ND/115034 <S> https://www.digikey.com/product-detail/en/molex-llc/0701070002/WM2534-ND/210969 <S> And the Molex service grade tool <S> , these can be found used but make sure it has the terminal locator as I've seen units described as "working" that are missing this important part. <S> Yeah the crimper "works" but it's worthless without the locator. <S> https://www.digikey.com/products/en?mpart=0640160201&v=900 <S> The terminals have a spur that catches a hole in the housing which works opposite of how a dupont terminal works. <S> This makes it easier to get the terminals out of the housing <S> and they don't wiggle around as much either.
| What I tend to do is go into the wire to board connectors section of the suppliers site, select pin count, row count and pin pitch (stnadard header pins are 0.1 inch or 2.54mm) then look through the pictures for ones that look right.
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