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Zero or low-current voltage divider for switch identification Is it possible to design a voltage divider circuit with switches that uses zero current, or very low current? Would it work to use a transistor to connect/disconnect the end of resistor string from ground? Background The circuit will do two things. Each switch in the circuit will wake up a IC (ATMEGA328P) by using a transistor to send an interrupt pin LOW. Once the IC wakes up, the ADC will sample the voltages coming from the circuit, allowing the IC to know which button was pressed. The entire project will be battery operated, and having this voltage divider may significantly impact battery life. The project will 1) read files from an SD card, 2) go to sleep, 3) wake and play sounds when one of 16 buttons is pressed, and 4) go to sleep and repeat process starting at #3 on button press. I anticipate that when it is running it have a significant current draw. Existing Voltage Divider Circuit Possible Transistor Solution Placing a transistor between the end of the resistor string and ground, and having the button press activate the transistor, connects the end of the voltage string to ground. This will result in an initial voltage reading of 5V, and once the transistor is on, the actual read voltage will be the intended voltage. I have no idea if this will work. Other Solutions Per question 28897 I could use high values of resistors such as 10 MOhm. But this will still have a current draw of hundreds of nA. I'd prefer zero. <Q> New answer <S> But, your 2nd schematic has a bug, I think. <S> The ADC will always see V be of the transistor, which is always 0.7V or so. <S> This variation shouldn't have this problem, because there's a resistor R39 between base and ADC. <S> Old answer, which wasn't an answer <S> The divider can be switched with a transistor to save the battery. <S> However, it has to be a high side switch. <S> If you switch at the ground, then the battery voltage will appear on the A/D pin, which could damage the input. <S> (Originally, the schematic was posted in this thread .) <A> You don't even need a transistor to disconnect the resistor string, you can just connect it to an MCU output pin. <S> Set it to the same value as the other end of the string, and it will use near-zero current. <S> I have used this approach and it works fine. <S> (In your diagram, give the transistor its own ground and connect the MCU pin to the bottom of the resistor string). <A> Is it possible to design a voltage divider circuit with switches that uses zero current... <S> This should do the trick and no ground switching required. <S> A divider is connected to the battery only when a switch is closed and the ADC input is pulled to ground when all switches are open. <S> For 5V Vcc, the ADC input is: 5.0V = <S> SW1 closed 3.3V = <S> SW2 closed 1.7V = <S> SW3 closed 0V = <S> all switches open <S> Of course, you can adjust the resistor values to your liking. <A> Given two <S> I/O pins with reasonably-consistent switching thresholds, one could wire a grounded cap to each processor pin via small resistor, tie one cap to each end of the resistor string, and have each switch connect a tap on the string to VDD or ground (whichever is more convenient; I'll assume VDD for this discussion). <S> Have significant resistance between either end and the first switch. <S> Sometime when no switch is pushed, ground both pins long enough to discharge the caps; then float one and set the other to VDD. <S> Time how long it takes for the floating pin to change state. <S> If the inputs' thresholds might differ, repeat the test for the other input. <S> Then ground both pins, and then float them <S> --this is the idle state. <S> Once a pin has been observed to change state, ground both pins long enough to discharge the cap, and float them. <S> Time how long it takes for each pin to change state. <S> The ratio of this time to the baseline measured above will tell you the resistance from each pin to VDD. <S> Ensure that the sum of the two measurements is reasonably close to the total resistance of the string (otherwise the button isn't making good contact, so the reading may be faulty). <S> If the processor may draw excess current when inputs are floating away from the rails, it may be a good idea to periodically discharge the caps even when no button is pushed. <S> If this is done, quiescent currents for the system should be pretty minimal.
| Your approach can work.
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Why do batteries with the same voltage have different currents? Experts say "current depends on voltage". So, if the voltage is high, current would be high. Agreed; (I = V/R) If the voltage is low, the current would also be low. Agreed -> I = V/R But why then do two different batteries available with the same voltage (say 2 V) not deliver the same current? If the voltage is same for two different batteries, current should also be the same, right? Which factor plays the role in delivering a different current for these batteries? If two different batteries (with the same voltage) delivers different currents, how can we say that they are both 2 V batteries? Why do the batteries not obey the rule? Do they have another principle? <Q> Current depends on Voltage". <S> So, if the voltage is high, current would be high. <S> Agreed; <S> (I=V/R) <S> True, if you're asking about resistance . <S> But, you're asking about a (non-ideal) voltage source - a battery . <S> The voltage to current relationship of a battery depends on the chemistry, temperature, etc. <S> Cells and batteries are not resistors . <S> Now, it is the case that a first approximation of a battery is an ideal voltage source in series with an ideal resistor, as another answer points out, but it must be kept in mind that this effective resistance depends on a number of factors. <S> See, for example this . <S> If two different batteries(with same voltage) delivers different currents, how can we say that they are both 2V batteries? <S> You've answered your question yourself. <S> They have the same open-circuit voltage. <S> Clearly, two identical cells connected in parallel provide the same open-circuit voltage as one of the cells but <S> two cells in parallel can provide twice the short-circuit current of the single cell. <A> Battery is not an ideal voltage source <S> **. <S> For the purposes of your question, a battery can be modeled as an ideal voltage source \$V_{ideal}\$ with a series resistance \$R_{series}\$. <S> While voltages of the batteries may be the same, the series resistances may be different. <S> Let's model two of your batteries as identical ideal voltage sources, but with different series resistances. <S> \$V_{batt}=V_{ideal}-I_{load}R_{series}\$ is the observed battery voltage. <S> When there is no load current, the voltages of both of our batteries are the same. <S> Open circuit voltages. <S> Let's gradually increase \$I_{load}\$. <S> The voltage of the battery with higher \$R_{series}\$ will be falling more, and it will fall out of spec at a smaller current than that of a battery with lower \$R_{series}\$. <S> __ _ __ _ __ <S> ** <S> Battery is a nonlinear function of everything. <S> tm <A> The basic concept you need to know is that a battery tries to keep its voltage constant, it is what they are designed for - it is basically a voltage source . <S> When a load is connected to a battery, it is the nature of this potential (voltage) to deliver current to this load. <S> The load is basically a way for charge to travel from high to low potential (one terminal of the battery to the other). <S> Since accumulation of charge is what creates the potential in the first place, if they find a way to go from one terminal to the other externally to the battery, voltage would decrease, so it is up to the battery's chemistry to internally force them through and elevate them back to the source terminal so that voltage can be maintained. <S> If the current being allowed to flow is too much for the battery to handle, voltage cannot be maintained and will drop. <S> Hence the higher current capacity the battery has, the better it is at maintaining that voltage from dropping without getting internally damaged. <S> Resistance is then a measure of how much the load is, well, loading this voltage source (battery), so low resistance means that it will let a lot of charges travel through it per second for a given voltage. <S> High resistance means that few charges will be let through for a given voltage (the battery's voltage in this case), so the battery can relax and not work very hard to keep that voltage up there. <A> The critical parameter to batteries, capacitors, and any power source is the (effective) Series Resistance, Rs or ESR. <S> Generally the bigger the capacity [Ahr] rating, the lower the ESR [Ω]. <S> Here the batteries are graded by ESR. <S> In your case, if you are using the same grade and size of battery the differnce will lie in the aging factor and remaining charge capacity which both affect ESR. <S> So you can estimate each batteries current load capacity based on the ESR which must be tested somehow. <A> The voltage a battery carries depends on the concentration of the chemical compounds (acid or whatever) and the current delivering capacity depends on the amount (quantity) of the chemical compound. <S> Higher the chemical concentration higher the voltage. <S> And more the quantity of the chemical compound, higher the current delivering capacity. <S> A simple example of this would be a motorcycle battery and a car battery. <S> Both carry 12v <S> but the car battery can deliver upto 40-60 amps and the bike battery <S> upto 7-9 amps. <S> And the sizes of both batteries you know!
| You see now that batteries can have the same voltage, but their current is a capacity rating, not something they are outputting all the time.
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What's the difference between linear and audio tapers, and why should I care? I'm stocking up on a variety of potentiometers to keep on hand for my audio projects, and I'm trying to figure out what to buy. The projects I'm working on seen to call for either linear & audio taper pots. I know the difference between linear and log, but where does audio taper fit into the mix? Here's what I've researched so far: No schematic or parts list has BOTH linear and audio. Looking at the Bourns pot spec sheets, they seem to be using linear and audio taper synonymously, but they have different part numbers. Sites like mouser and digikey often have, say, a 10k linear and a 10k log, but no 10k audio pot in a search. However, when I search on the Bourns part number for an audio taper, the part is described by mouser as linear taper. A lot of audio forums I've been trolling talk about pots as either audio orlinear, but they don't appear to be synonymous in the discussions. See why I'm confused? So once and for all...can someone school me on the difference between linear and audio taper? I'm also interested to know how to decide which to choose for a particular application, and if they are different enough to have distinct uses, what types of interesting and unexpected results will I experience if I substitute one for the other? <Q> Some rotary Pot's com in Linear, Audio and reverse Audio (where audio=log taper) <S> Actually it is "quasi"-log scale and not precision logarithm. <S> Notice <S> it has a rapid transition from two different linear slopes. <S> which gives you more sensitivity over a wider range of audio inputs rather than an apparent Off to ON effect if using a linear pot. <S> If one adjusts audio equipment over a 30 dB range, that might occur in the 1st 5% of a Linear pot. <S> Such as Many options may be selected when you order. <S> Precision Log pots would be mcuch more expensive than <A> Very simply, with a linear taper, if you turn the pot halfway, the resistance measured on each side will be equal. <S> With logarithmic (aka "audio") taper, that is not true; the resistance follows a logarithmic progression. <S> Log taper is used because it gives a more natural progression when used for volume control. <A> An audio taper is a volume control or bass or treble control etc. <S> In a fender amp <S> the control that feeds the adjustment for the balance is linear taper since it adjusts the resistance between the screen grids. <S> The idea that you could use two controls is a good idea and I would recommend this process. <S> Just remember if you do any mods check your work carefully and test your voltages to make sure everything is ok
| The difference between audio & taper is if you want to use a pot that is linear taper it is to provide an adjustment for a device or a feed to a device.
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Converting a 2.54mm pitch header to screw terminal blocks I'm trying to avoid soldering or modding of any kind as I'm 7,000 miles away from where the project is based! I'm trying to come up with a simple solution I can get a layman to do on my behalf. I have the following IO board: http://www.phatio.com/ which exposes 2x14 IO ports (2.54mm pitch pin header connection) but need to convert the connections into screw terminal block type. I was thinking of using some angled double pin header connectors and then connecting them to pluggable screw terminal blocks (http://www.metz-connect.com/en/products/311661) but the smallest size I can find is 3.5mm. Anyone have any suggestions or a better/alternative idea? (screw terminal blocks are a must) Thanks <Q> For a similar requirement, converting a 20-pin header strip to screw terminals, we fabricated a tiny adapter PCB with screw terminals like this one: <S> The screw terminals are on 0.197" centers, and they interlock nicely on a 0.2" pattern. <S> The PCB used was thus: The screw terminals are arranged in two rows atop the PCB on opposite sides of the header strip, facing outward for ease of connecting wires. <S> The 0.1" header strip is on the underside of the PCB, facing the board to be connected to - Either make or female header strips are soldered in, depending on requirements. <S> You could as well fabricate this using single-sided stripboard or veroboard, just that the header strip on the underside would then need to be soldered with a bit of finesse. <S> That version's a bit fidgety to use though. <A> You can buy screw terminals with 2.54mm pitch. <S> For example, SparkFun has some . <S> Google yields some more hits, for example Farnell . <A> I just made this little breadboard screw wing from parts I had lying around. <S> The trick is using strip board which takes care of all connections from the pin header to the terminal blocks. <S> Two rows of 5.08 mm pitched terminal blocks back to back with <S> 2.54mm shifted apart allows to access every pin. <S> I pushed the pins of the header all the way through the black plastic so the pins are long enough to plug in the breadboard and still easy to solder. <S> The wing layout has the advantage that it covers as few breadboard holes as possible. <S> I made this one to match an Arduino Mini which has 12 pins (on each side). <S> Pretty much similar idea as in Anindo's answer, but a different layout.
| Another such adapter board of ours uses the screw terminals all on one side of the header strip, with a bit of a gap between the 2 rows of screw terminals.
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How do I implement Lights-Out game using logic gates or flip flops? First, for those unfamiliar with the game, this is how the game works, The goal of the game is to turn off all the lights, hence called "Lights Out", and each press of the button/light inverses its state as well as its north/south/east/west adjacent neighbors, and that's pretty much it. Now, what I could think of is by using SR flip-flops, or JK flip flops. This is due to its ability to act as storage element (the initial state and next state). But I can't seem to think of ways to actually implement them. Another idea is that each set of button and its adjacent (NSEW)button/lights will have its own truth table, like this: but, is it possible to have the input variables be the same as the output variables? Are there any other ways to do this? <Q> The obvious approach would be to use a processor and do all this in firmware. <S> However, if I really needed to do this with stone knives and bear skins for some reason, I'd dedicate a toggling flip-flop to each square. <S> Of course those button presses need to be de-bounced. <S> Again, this would be easier in firmware. <S> A hardware solution wouldn't be all that complex, but everything would be replicated 25 times, making it large and tedius to build. <S> Added: <S> Apparently the description above is not clear enough. <S> Here is a diagram of what is in each cell: <S> The other 4 inputs to the NAND gate are driven from the debounced signals of the 4 surrounding buttons that are also supposed to toggle the state of this square. <S> Likewise, the debounced signal from this button also goes to one of the NAND gate inputs of each of the 4 surrounding cells. <A> I would say that T flip flops would probably be the easiest as you can toggle their output state with a single input. <S> You could use a single flip flop for each LED and with the input tied to your button and the output tied to your LED. <S> Then you could have each button tied to the inputs of the 4 adjacent flip flops in order to toggle their state as well. <S> If you wanted to use JK flip flops, you can make T flip flops out of them by passing your input to both of the inputs (J and K) <A> Shift data through the shifter in groups of 8 bits to drive a multiplexed display and scanned a multiplexed keyboard. <S> Have a seven-bit "flip light" counter which will run any time the bottom six bits are non-zero, or when the state of the top bit matches the state of the presently-decoded button. <S> Flip the state of the current light whenever all of the following apply: <S> 6-bit counter isn't xxx1116-bit counter <S> isn't 111xxx7-bit counter isn't <S> xxxxx007-bit counter isn't <S> xx00xxx7-bit counter is 00xx0xx <S> Note that while a significant amount of logic would be required to decode those counter states, it would be trivial compared to the number of chips required to implement each light separately.
| The flip flop of each square would be toggled by the press of its button or either of the four neighboring buttons. If one wanted to build such a game up to size 7x7 out of discrete logic, the most practical design would probably be to use a circulating shift register to hold the state of the board, and a six-bit counter to keep track of the shift position of the data within the register.
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device for outputting HIGH when input have been HIGH? I am looking for a device or circuit that outputs HIGH when a pushbutton has been pressed (if the button is pressed again it is supposed to still output HIGH), also, if possible, another pushbutton to make the output LOW. Is this possible without an IC? How to do it? <Q> What you want is called a set/reset flipflop. <S> Various flipflops have R and S inputs, but you can also make a primitive one out of two NAND gates. <S> For example: Follow the logic signals thru the circuit to see how it works and how it has memory. <A> One device that may work (High only, no low) is a silicon-controlled rectifier or SCR. <S> Once the SCR gate goes high with the pushbutton, the diode will conduct until a power cycle. <S> Is there any reason why you wouldn't use an RS flip-flop? <A> Olin is right, you are probably looking for a S-R latch. <S> But if you want to do it without an IC, you'll need to search for circuits using its older name : bistable multivibrator. <A> You can put a normally closed contact in series to that circuit to allow the 'reset' function: ( image from this wikipedia article ) <S> This is the oldest fashion you can create a st/reset flip flop :) <A> SRC: <S> http://www.mikroe.com/old/books/keu/06.htm
| This could be done with a thyristor like so: S1 could replaced with a push switch instead of a toggle switch so that the output is reset, when S2 is pressed. If you want achieve that without any semiconductor, you can use a relais and use a normally open contact to feed the coil as soon the relaiy switch on.
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Making a schematic in KiCad; how to represent connections from another circuit board? I'm using KiCad to design a circuit for my latest project. This board includes a number of transistors whose bases are connected to off-board signals. What's the proper way to represent this, so when I build the netlist and PCB, these off-board signals become header pins or pads. Yeah, newbcakes here... EDIT - I found a 12x1 connector which does exactly what you think it does. Now, that happens with this when I convert the schematic to a PCB, I don't know. Stay tuned... EDIT 2 -I have learned nothing happens to the 12x1 connector automatically. Instead, on the next step towards pcb-ing, you have to tell KiCAD what everything is. I did and it worked fine, though there weren't many options so I selected a 7x2 DIP. I'm 95.982% sure I can make what I need, however. This is the least of my worries, lol. <Q> You can split your design into a hierarchy (e.g. multiple pages), then have local nets to each page and also global nets . <S> You can also use specific page to page connectors (so the page symbol is like a component you can drop into another page (over on the right hand icon set you will see an icon labelled "Place hierarchical pin in sheet" and another named "Place a hierarchical label" and a couple of other related icons) Have a play around with the demo projects to see how it works. <S> There are also some good tutorials out there that go through this stuff. <S> Tutorial 1 , Wiki Kicad , Hierarchy Tutorial . <S> Go into wherever Kicad is installed (e.g. Program Files/Kicad/Share/Demos), and look in the Demos folder for plenty of examples of different ways of doing things. <A> To make them actually connect you need to make them the same net name. <S> Described here on step 89. <S> The net labels will be how you show an off page connection (doesn't have to be off the page, just no wire between them). <S> You can see an example of that the the signal "Input" here : <A> <A> The proper way to do this is to create a connector symbol and connect your transistor bases to this connector.
| Use header pins and/or pads as devices in your schematics.
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How do I measure startup/inrush current? I have some small devices that draw 20mA. I'd like to add a resettable (PPTC) fuse to them but I am not sure what value to select. My thought is I need to find a value that will tolerate the startup or inrush current when the circuit first turned on, but I don't know how to measure that. (My meter isn't that fast, but I do have an analog 100MHz oscilloscope.) The devices have a capacitance of about 122μF. <Q> LittleFuse provides a very nice app note with a lot of details for the design of these kind of devices. <S> As for measuring the current, you could use a very small resistor (1 or 0.1 ohm resistor) and use an oscilloscope to measure the current. <S> Be careful not to ground the device. <S> Just get 2 probes, connect the tip to each side of the resistor, and the grounds to the actual ground. <S> use oscilloscope math capability (or just yourself) to do a difference between the two channels and this will be your current. <A> The good thing about PTC fuses, in this context anyway, is that they take a bit of time to heat up enough to "blow" i.e. become an open circuit. <S> If your inrush current is brief, the PTC does not get enough high-current time to react, before current has settled into steady state. <S> If the capacitors are not being charged through any significant resistance, then inrush should be quite brief. <S> Then pick a current between the expected steady state and this estimated overcurrent, and use a PTC of approximately that current rating. <S> If you have (or can request samples of) several PTCs spanning a range up to your power supply rating, this becomes even simpler: <S> Use a fuse two levels higher rated than the highest PTC that trips on start-up. <A> At my last job, our switch mode power supply wasn't working on our USB device. <S> We checked it on the 100Mhz scope and everything looked fine. <S> We then went to the vendor and she had a 1Ghz scope. <S> There was a 3ns 6A spike at power on we didn't see with our scope. <S> Most likely, the spike was from all the bypass caps we had. <S> We changed the switcher design to use an older IC from a different vendor and it worked.
| You need to use two probes for doing this measurement and subtract the value of one from the other - DO NOT CONNECT THE GND OF THE PROBE TO EITHER SIDE OF THE RESISTOR- you will in effect cause a short circuit if you do this. A good strategy to follow in such cases, is to estimate the current at which either the power supply, or some malfunctioning part of your device, is likely to suffer damage.
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Reading datasheets and applying Ohm's Law I have a CPLD that wants +3.3v and has a maximum allowable current per pin of 8ma. I need to supply a clock signal. The oscillator I selected accepts 5v and outputs 5v-ish levels. I'm finding it challenging to internalize the datasheets of the CPLD and the oscillator. Now, the oscillator's datasheet says for my clock speed (1MHz), it typically likes 10ma ( Supply Current, typical ). Which means that as long as 10ma are available at 5v ( Supply Voltage, typical ) we're gonna have a lot of wiggly fun. This is where I get confused. Since the output of the oscillator is up to 4.5v ( Output Voltage Levels ) and this exceeds the 3.3v required by the CPLD, I need to add a resistor. But I don't know what the current should be. I arbitrarily decided that I will allow 4ma to reach the CPLD. This will prevent the smoke from coming out of it. The oscillator will source 16ma ( Output Current ) so I believe this isn't a problem. Ohms: (4.5v-3.3v) = .004R1.2v = .004a * R1.2v/.004a = R300 ohms = R So a 300 Ohm resistor between the output of the oscillator and the CPLD's clock pin will prevent damage to the CPLD by limiting the current and reduce the voltage to the limits required by the CPLD. Q1: Does any of this make any sense? Q2: I assume that when the datasheet says the output current for "1" is -16ma, this means it will source 16ma. Why does the oscillator's datasheet say the oscillator's "0" current is 16ma? I'd have expected 0. EDIT - Wow, thanks for the great feedback. I will look for a 3.3v-compatible clock. While in retrospect it's just a better idea, I am surprised that adding a resistor would be so problematic. They seem to be used frequently in digital circuits. Reference circuit... here's the datasheet for the CPLD . I have not seen a reference circuit as such. However (boy am I dumb!) I do have a working dev board 3' from me that sports an oscillator having "8.000 G MEC AL8GS" stamped onto the tin. (It's my assumption that's the oscillator. Looks like one. The only other thing nearby is a tiny Atmel chip of some sort that I suspect is related to JTAG programming.) Edit 2 - I'm not married to any particular brand or type of oscillator; for my purposes a low frequency clock is ok. I believe anything over 500KHz would be fine. Back to DigiKey and Mouser! <Q> The reason companies provide this is that they want you to get started using their device and unless you have some exotic requirements, it usually does the trick. <S> You didn't provide the datasheet for the CPLD or the name, but I can tell you that you should choose another oscillator that is 3.3V compatible. <S> CPLDs don't typically operate at 5V and this will likely violate the specs in the datasheet. <S> As I said above, selecting the same as in the reference design (or find any other board that uses the same chip online and see what they're using) will be the best. <S> Even just using resistors will introduce issues that can cause the CPLD to malfunction due to all the capacitance and perhaps loading. <S> You must make sure that the power supply you have can provide plenty of current for the worst case of the CPLDs. <S> From my experience with FPGAs, the inrush currents are very very large and quick, and you should plan for this. <S> Again, the recommendation is to use whatever a reference design uses and perhaps tweak as necessary. <A> CTS now make an identical MX045LV <S> that is spec'd for 3.3V but not listed on D-K site at this time. <S> Comparing these specs with the MX045 <S> They appear to be identical except configurable to order from 1~108 MHz vs. 1~50 MHz for the LV series. <S> This is normal as lower voltages also reduce the maximum speed of the inverter, but your choice is the minimum so your std 5V Oscillator may operate OK at 3.3V. <S> They don't spec it this way as it affects their universal datasheet from 1~108 MHz and might complicate it to show a declining freq per voltage scale. <S> But I can tell you standard CMOS logic for 74HCUB04 <S> inverter gates most commonly used in such oscillator designs work from 2~6V at 1MHz. <S> You have 3 choices; <S> Test OSC from 5V down near 2V for wiggles on output and accept for use at 3.3V ( call factory Tech support to confirm if you must) Use a resistor divider for the CMOS output using a total of 1~2 mA load to reduce to 3~3.3V range . <S> 5V/1.5mA=3K3 and 3.3/5.0 <S> = 66% <S> so choose 1K1 out and 2K2 to ground. <S> RC time constant for CPLD clock input <S> Ref Chapter 5 p4 specs indicates input is 8 pF. <S> so rise time will be <S> Rs*Cin ~ <S> 5nS <S> so no risk of slowing wiggle edges. <S> and 4.5Vmax Spec actually will result in 3.0V input for Vmax which is fine. <S> Order the LV part or sample from CTS. <A> Your CPLD uses 3.3V logic signals but you chose a 5V oscillator!? <S> That makes no sense. <S> Your basic problem is you picked the wrong oscillator. <S> Get one that runs on 3.3V. <S> Then you can connect its output to the CPLD directly. <S> As for the current requirements. <S> The datasheet is telling you what the oscillator can drive and still provide a signal within its other specifications. <S> You didn't describe the CPLD, but most likely its clock input is a regular CMOS digital input, which has very high impedance. <S> It won't draw anywhere near what the oscillator can put out. <S> There is no need to try to limit the current into the CPLD input somehow. <S> The CPLD input will only draw a tiny amount in the first place. <S> Don't forget the bypass cap accross the oscillator power and ground pins.
| Get a 3.3 V oscillator and connect its output directly to the CPLD clock input with no additional parts between, and all will be fine. Trying to reduce the signal level of the oscillator is a very bad idea because you'll introduce all kind of issues such as capacitance, frequencies, noise, etc that are bad bad bad. I highly recommend that you find the reference design for the CPLD and if possible, recreate it and even use the same components.
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What are some smallest, cheapest microcontrollers with USB built in? The cheaper, smaller, and fewer external components needed the better. AVR would be preferred, but anything with free SDKs is good. <Q> The best way to find those microcontrollers would be at a site such as digikey. <S> Here is the list of all their microcontrollers. <S> Use the filters to pick Atmel as a manufacturer. <S> Then filter to leave their AVR family on the list. <S> And finally, filter by connectivity. <S> Choose all items on the list that have "USB" or "USB OTG". <S> Hold down CTRL to pick multiple items on the list. <S> Off the top of my head, one of the simplest would be ATmega88U2. <S> The U2 tells you it's a regular ATmega88 with a USB controller built-in. <A> V-USB project provides needed software for bitbanged low-speed USB device and is available under GPLv2 or commercial licenses. <S> As for external hardware, 12 MHz crystal is recommended, but it should work with RC oscillator too. <S> It also needs a voltage divider to set up slow speed and depending on operating voltage of the microcontroller, it may need a couple of zener diodes to limit voltage to D+ and D-. <S> With it software USB works on microcontrollers as small as ATtiny45 for example. <S> Two other popular microcontrollers for use with the library are ATtiny2313 and ATmega8. <S> Another library based on V-USB is USBtiny . <S> It's GLP2 <S> + and should be a bit easier to work with. <S> Both can be made to work with libusb drivers on GNU/Linux and Windows. <A> If you are willing to go the PIC route: the PIC16F1455 has only 14 pins, and can do USB without needing a crystal. <S> It has an internal, calibrated 48MHz oscillator, and does clock recovery from the USB lines to ensure it stays in sync. <A> The PIC18F14K50 is a nice simple chip with USB built in, free IDE and compiler, good examples and low cost. <S> I sucessfully taught a high school class to build and program it use this nice kit <S> https://www.tindie.com/shops/TAUTIC/pic18f14k50-usb-development-kit/ <S> it is a open source design so you can get the components and build for less. <S> But you will need to get a boot loader for USB programming installed on the PIC, so you will need a programmer if you do not buy a chip that is pre-programmed, like Tautic.com 's product. <A> Another alternative for you is an EZ430-F2013 or similar. <S> It's not AVR, but rather a TI MSP430. <S> It's small and cheap, with only the USB-key sized device needed. <S> There a few options for programming, and by far the best is TI's CodeComposer Studio v5 . <S> It's available for Windows and Linux and has two different ways of using it free, including a code-size-limited (16 kB) option specifically for the MSP430. <S> And CCS's Grace tool made developing for the MSP a joy when I used it. <A> Take a look at the LUFA project, it's a software stack that supports many AVR microcontrollers (both low-end and high-end ), with different capabilities depending on each of them. <S> A plus is that it uses the commercial-friendly MIT license. <S> The compatibility table is a good listing of devices with USB support in the AVR family. <A> A lot of information is available at their site: http://www.atmel.com/products/microcontrollers/arm/sam-d.aspx <S> Full disclaimer: I am not affiliated to Atmel by any means. <A> Atmel XMEGA is AVR based, and if you can get past the fact that they are only available in surface mount, they are zero external component for USB and run up to 32MHz. <A> At90usb162 is good. <S> it has a hardware usb in it and runs at 16Mhz (external crystal ) <S> and it is about 4$
| The relatively new ARM Cortex-M0+ based Atmel SAM D also include USB embedded host (D21 family) and USB device capabilities (D11 and D21 families), at very competitive prices.
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What happens to copper pours that are 'too narrow'? What happens to copper pours that are narrower than PCB manufacturer spec? Let's say spec is 10/10 mils (trace/space) and the copper pour in between soic-08 is only 5mils because it can't be wider (soic-08 pad pitch - 2x 10mil space). Should I 'take out' that copper pour or leave it there? Does it get processed? Result? <Q> It depends on the fab. <S> It's possible that a PCB fab would process these thin areas, but they will be fabed wider than you drew them. <S> Could <S> a that cause a short between the copper pour and a pad? <S> Unfortunately, yes. <S> Removing these risky places to avoid spec abuse is a good idea. <S> Fortunately, this is not difficult to do in the layout software. <S> In Eagle, change the width property of the copper pour. <S> Below is an example with 2 copper pours. <S> They have identical geometries. <S> The only difference is the width property. <S> The upper pour has width=0.005 <S> (dimensions in inches). <S> The lower pour has width=0.016 . <S> Notice that the lower pour didn't "go" between the pads. <S> U1 is SOIC. <A> What you are doing is violating the design rules for your process, therefore something won't come out right, at least not reliably. <S> Some PCB houses will flag the errors and put the board on hold until you tell them what to do. <S> Some may run the board and you get what you get <S> and it's your problem since you violated the rules. <S> Some may automatically delete traces that are to thin. <S> In any case, it's your fault if they specified the design rules and you gave them specs for something they can't build. <A> First, most shops in the US prototype business today offer 4/4 design rules as standard. <S> Even if their standard is 6/6, a 4/4 design should result in hardly any premium charge. <S> So if your design ended up with a 5 mil copper feature, you should be able to get this manufactured with no problems, so long as you choose the right shop and specify the correct rules. <S> But, lets imagine you are designing for extremely high volume and want to get the absolute best price, so you are designing for 6/6 or even 8/8 rules. <S> In that case, the best thing to do is make your design tool aware of your rules. <S> I can't help with Eagle specifically, but any worthwhile layout tool should have settings for the space/trace design rules, and it should respect them when laying out polygons. <S> Even if your tool doesn't do this, there should be workarounds. <S> The key thing is that polygons are built by simply serpentining a long trace back and forth to fill an area. <S> As Nick shows, one work around is to change the width of this long trace so that it won't fit into the gaps that are causing the problem. <S> The trade-off, as you can see in Nick's images, is that other features of the polygon, like the corners, becomes more coarse or rounded. <S> Another option is that most layout tools have fairly sophisticated design rule options <S> (unfortunately if your tool is so simple that it can't understand a simple space/trace design rule, this might not be available to you). <S> For example, you should be able to specify a specific clearance rule for polygons on the top layer and SOIC footprints. <S> If you increase this clearance, you should be able to prevent the tool from attempting to fill the space between the pads.
| Traces may be shorted together if the spacing is too thin, or traces may dissappear if the width is too thin.
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Dual AC feed to power a DC unit (Raspberry Pi) with Diode and Polyfuse for protection I wish to place some Raspberry Pi's in remote locations (data centres) for telemetry and other services like serial OOB access. Typically dual AC feeds are presented to customers in a data centre which come from completely separate feeds from the local power grid. Raspberry Pi's take a single DC 5v 1mA input. In the event that one of the power feeds in the data centre goes down I want the Pi to stay up (and in the event they both go, it will have a battery pack). This isn't a natural feature of the Rasp Pi (two power supplies) so I need to design a safe and fault tolerant system that will do this. My knowledge of electronics is limited and I will be consulting others and testing in a lab style environment; As a basic premise though, does the following sound sane or a have I grossly misunderstood electronics and risk exploding all the kit in my rack and potentially others; Dual 16A or 32A feeds per rack (A and B feed) Attach a regular 5v DC Rasp Pi transformer to each feed to present two 5v DC feeds Attach a diode to each feed to stop any current passing back towards the source A or B feed Attach a polyfuse in case of a surge coming in from the source A or B feed Attach A feed to Rasp Pi battery pack, then to Pi 5v inpit Attach B feed as the power supply to a powered USB hub, which is then connected to Rasp Pi USB port (A Rasp Pi can be powered by either it's main USB mini power input, or voltage coming from a powered USB hub, both together as far as I have read, won't harm it) A Feed B Feed | | DC DCTransformer Transformer | | Diode Diode | | Polyfuse Polyfuse | | 5V DC 1mA USB HubBattery Pack Power Input | | Rasp Pi Rasp Pi 5V Input_________________USB Socket I understand this question is somewhat Rasp Pi specific, but in terms of general electronics, would this work? What should I be looking out for on the Rasp Pi that could cause this plan to fail? My main concern is that I will connect the A and B feeds directly together and blow everything attached to both feeds. Also, what if the A and B feeds are out of phase as they are coming from two different sources?; I assume this doesn't matter as they are both being converted to DC? Many thanks and sorry if this seems silly, my knowledge of electronics is very limited. <Q> Given the relatively low cost of a Raspberry Pi, this could be one approach: Power the Pi with an off-the-shelf online charger -- <S> > <S> > battery pack -- <S> > <S> > <S> UBEC such as used in remote control model aircraft. <S> This provides a regulated voltage from the UBEC output, takes care of charging and safeguards for the battery, and also, with some chargers, will provide a visual alert when the battery has aged beyond a specified extent. <S> Power <S> the charger through any failover mechanism of choice from both raw power lines. <S> Set up two such independent Pi + power units, in separate racks, to duplicate each other's function - since the RPi is not an industrial reliability product, failure during continuous operation is not unlikely. <S> Incorporate remote heartbeat code on each RPi, polled from your central Network Operations Center or equivalent, so <S> the failure of any RPi generates an alert on the monitoring panel. <A> That way you do not have to worry about blackouts and you will not have any power from B to A or B to A. <A> I run a hosting company and we have to think about dual power redundancy a lot. <S> If I were to build a solution to your setup I would simply use an ATS (Automatic Transfer Switch). <S> These are AC devices that have two power inputs and one output. <S> Connect one input to circuit A and the other input to circuit B. <S> The output would then power your 5V rectifier. <S> If circuit A dies, the ATS seamlessly switches loads to use circuit B, without any interruption of the loads. <S> APC makes a whole line of ATS devices. <S> You can find several rackmount versions here: http://www.apc.com/products/family/index.cfm?id=14 <A> Examining all common sources of failure to find solutions; I see the following opportunities; power line transients from lightning on both feeds. <S> Power meters protect only >6kV with built-in arc gap, transformers are usually rated at 1.5 to 3kV. Line filter reduces that peak voltage substantially for both common mode and differential mode. <S> Some use MOV for extra surge suppression (not polyfuse) as long as impedance protected. <S> People who don't design their computer systems to 6kV are at risk in exposed power line areas. <S> That was our Corporate Std for all equipment at Unisys. <S> Choice of power supply must have demonstrated reliability (not any old Sparkfun transformer) <S> Battery pack must have capacity over worst case power outage duration and demonstrated reliability over environmental range, otherwise take your chances. <S> Polyfuse is for thermal protection from permanent overvoltage or short-circuit failure. <S> Connectors are a common point of failure and may need additional strain relief from inadvertent damage or pulling on cables. <S> Consider HALT-HASS testing with vibration and low voltage to ensure stress operation of units after high temperature high humidity cycles while operating. <S> if there is a fault do you want to detect it or just have redundant power? <S> Raspberry <S> Pi's may not have demonstrated reliability yet so Halt/HASS helps immprove confidence level. <S> - EMI is a common point of functional failure Consider Ingress from sensors, <S> ESD, industrial noise crosstalk on sensors and communication. <S> Typical tests are 15kV ESD, 1kV radiated pulse dump from sheet metal, pulsed radar, PLT test, HIPOT etc <S> and... use CM chokes on all sensor cable pairs. <S> Ethernet already has that. <S> consider any other single point of failures in your system and risks. <S> added <S> also I might suggest hard wiring AC as you don't have enough wetting current to prevent moisture/dust corrosion with standard consumer grade plugs or get industrial grade plugs with line filters above installed in octagon boxes and tie wraps or suitable restraints for plugs.
| I personally would power the raspberry from a battery which normally would be charged from A and by using a relay I would continue to charge the battery from B in case of a blackout.
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Random number generators using a GPIO pin? I just read through question AVR Random Number Generator and encountered a number of ways to generate random seeds in an AVR: Use a special-purpose "Secure AVR" Use an internal temperature sensor Read unwritten EEPROM Measure time intervals between user input Use a floating ADC pin. Why not just a single digital pin, configured as input without pull-up and floating? In theory that should generate a stream of random bits. Why not use this? Is the state too slowly changing? Does it tend to stuck at 1 or 0? Any other problems? <Q> The self-evident difference between using a digital input, and an analog one, is that there's a greater span of seed values possible in analog. <S> Second, but perhaps more crucial: <S> If the digital pin's floating "median" value is not precisely midway between logic sense levels (which are not the power and ground rail levels, but switching thresholds for the input circuitry) <S> the resultant bit stream will have a strong bias towards 1 or 0. <S> This bias skews seeds much more strongly than a skew in analog levels would. <S> It would take very little coupling, either from nearby traces or signals, or through some resistive pull-up / pull-down, whether designed or accidental, to push a "floating" digital pin to unfloat - and that could as well happen post-deployment. <S> In engineering any design, this form of failure mode is best avoided. <A> A floating input is undefined. <S> It could give you random bits, but it could also tend to get stuck at a 0 or 1. <S> Most of the time it will get stuck. <S> And when it is truly floating, it could cause increased system noise and power consumption. <S> Floating inputs are generally a bad thing, even when the inputs are not being used for anything. <A> Use the bit-stream output to an input on the micro-controller. <S> Sample the incoming bits at intervals. <S> To ensure a roughly even number of 1s and 0s, simply use the change of state as a 1 and no change as a zero. <S> Google for transistor noise generator for more information. <A> "Random" is a tricky word. <S> In some contexts, it simply means "unpredictable", but in other contexts — in particular, the ones associated with signal processing and cryptography — it means "statistically uncorrelated". <S> Even if the value read from a floating pin is unpredictable ("random" in the first sense), it is not likely to be useful in the second sense. <A> One random generator that is available in many microcontrollers is the error between two clock sources (perhaps one internal and another external). <S> When you compare them, they drift slightly due to noise and that can be used. <S> In the end it depends what you need it for? <S> A little randomization and <S> you can even use an ADC pin connected to nothing. <S> Or just use an algorithm like Mersenne twister <S> http://en.wikipedia.org/wiki/Mersenne_twister <S> None if this is cryptographically secure though, but good enough for most purposes. <A> The quality of a random seed and its algorithm is determined by the amount of entropy or uncertainty in each random bit. <S> Then the best generator of a seed uses high entropy with very few steps rather than many steps of low entropy. <S> A good example of high entropy uses the Latin Squares method for a random seed. <S> The floating input is easily biased by flaws, leakage or ingress noise. <A> In a PIC18F microprocessor you need to have a seed for the random number. <S> The seed can be anything from 0 to 32767 (15 bits). <S> I would suggest at the start of the program you make one of the pins and analogue input pin. <S> Read this pin and take the least significant bit and put it into a 16 bit variable. <S> Then shift the variable to the left by 1. <S> Read the pin again and put the LSB of the ADC result into the first bit of the variable you stored the last bit in. <S> Shift left again..... <S> Repeat this until you have 15 bits in your variable. <S> This is the number you need to put into your seed. <S> After you have done this, you could simply make the analogue input pin a digital input pin and enable a weak pull up on it, or make it a digital output pin. <S> Either one will stop this pin from floating.
| The simplest solution is to build a hardware random signal generator circuit out of a misused open-collector transistor circuit.
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How to dim an LED with constant current supply? I have a power RGB led with 350 mA max per color. If I get 3x 300-350 mA constant power supplies will I be able to dim each individual color from 0 to 100% power with Arduino (with NPN transistors)? I am a bit confused since constant current psu's are the best way to power led's but since they provide constant current can they be dimmed with pwm as one would expect with any power supply? I am thinking of this PSU . <Q> Yes, you can use PWM to dim LEDs with constant current power supply. <S> Personally I would go with MOSFETs instead of NPN transistors. <S> If nothing else MOSFET will dissipate less heat. <A> I have used this constant current LED driver with success in the past. <S> May be worth checking out. <S> It can be controlled with an analog voltage or a PWM signal. <A> The name "constant current" is kind of misleading, I prefer "current limited". <S> You can let the LED pull on the supply as hard as it wants, but it will only ever get 300 mA. <S> If you shut off the LED the supply won't push harder than its rated voltage to force 300 mA through it. <S> With the supply you linked that is 13 V, so it would just output 13 V and no current.
| As others have said, yes, it will work with PWM to control brightness.
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How can I power and dim two sets of LED lights where one requires12VAC and one 12VDC I am building a wall unit and I would like to use some LED Tape Light (12 VDC , 48ft Total 1.45W/Foot) to light the shelves. Then I'd like to use a 12 VAC adjustable LED spot light (5W) for the center area. I'd like to have all lights able to be dimmed by using either an ELV or standard in-wall dimmer. I think I can basically use the 12VAC electronic transformer and then use a full wave bridge rectifier (and maybe some caps?) to turn the 12VAC into 12VDC for the LED Tape lights. Is this correct? The 12VAC will be output at 20kHZ. Will frequency play any issue here? I can use a standard magnetic transformer as well if that makes things easier. Any suggested schematics or part numbers would be helpful. The total DC load will be ~75W with AC load will be ~5W. EDIT: Here is the transformer I'm considering using. WAC Lighting: EN-12100-RB-AR http://www.waclighting.com/data/TOOLS_SPEC/en-1260-rb-ar.pdf <Q> For the 12 Volt DC dimmer, this answer provides suggestions like this one <A> A crude but usable solution is to use a standard triac wall-mounted 120VAC dimmer for each line - such as the smoothly variable ones used for ceiling fans. <S> There are multi-channel dimmer fittings available as well, which might better suit your purposes. <S> Some useful information about dimmers is available on this page . <S> The LED strip lights are then driven by a full bridge rectifier with reservoir capacitors. <S> The flicker, if any, on the LEDs will be at twice the line frequency, hence barely perceptible but not distracting. <S> The reason to use a 9 Volt transformer instead of 12 Volts for this, is that after rectification 9 volts would result in 9 <S> * 1.4142 = 12.73 Volts DC , which would be within tolerance for the strip. <S> The "12 Volt AC" lamp would be driven straight off its transformer. <S> Depending on the lamp's internal circuitry, this should work fine: A couple of CFL lamps in my house don't take kindly to triac-dimmed input, but most other fittings work fine. <A> Can you link to the AC LED, as the LEDs themselves surely works with DC. <S> If you can find out how the AC LED spotlight works/how the driver circuit looks, you should be able to power all the LEDs via DC dimmed by PWM.
| Each dimmer's output would then be wired to separate 120V-->9VAC (for strip light) and 120V-->12VAC (for AC LED lamp) step-down transformers.
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Using a voltage regulator to power MSP430 microcontrollers I'm trying to power a MSP430 microcontroller at 3.0V. I have a voltage regulator whose output is 3.6V, but I have a Bluetooth module that can only tolerate 3.0V. I plan to use a voltage divider to convert the 3.6V output of voltage regulator to 3.0V. Are there any recommended values for resistors? Will impedance mismatch be a problem? <Q> Very wide range of output voltages, only 2 capacitors are basically needed. <S> SOT23 package (and even TO-92) is easy to use. <S> Max current is 200 to 250mA which is plenty for MSP430 plus circuitry. <S> And, it's inexpensive. <S> Still, it has short circuit and thermal protection. <S> Do not use voltage dividers. <S> You'll waste too much power. <S> MCP1700 Datasheet http://ww1.microchip.com/downloads/en/DeviceDoc/21826b.pdf <A> MSP430 is happy to operate at 3.6V. <A> Well, you could use such a divider. <S> It's just a matter of order of magnitude of the allowed voltage ripple you can afford. <S> A MCU such as your MSP430 does have variation in its power consumption depending on what it is doing. <S> Theses current spikes generates, due to the power supply line impedance (inductance), voltage drops that are not recommended and you have to put decoupling capacitors close to the MCU pin to reduce the impedance seen by the MCU pin. <S> If you use a voltage divider, you adds deliberately a resistor in the power path on the MCU. <S> Variations of power consumption (A) will leads to voltage drops in the power supply pin. <S> Of course if your voltage divider uses high power resistors or ~1 or 2 ohms (!), the impedance increase on the MCU power pin is negligible but the current drawn by your voltage divider will be of several amps ! <S> I suggest that you use a voltage regulator that is able to provide the right voltage. <S> But if you can't, as Jonny B Good said, add a diode in place of your voltage divider. <A> And simply set the output to 3.0V with the resistor. <S> If this is somehow impossible for you — use the diodes to drop the voltage, as Jonny B Good suggested.
| I've had great success with Microchip's MCP1700 line of regulators. You don't need to drop the voltage at all. If you're already using a voltage regulator (I assume it is a linear one) then just replace it with adjustable regulator, I'm sure you can find a pin-to-pin replacement to you regulator.
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How to decide between a TVS or Zener diode for overvoltage protection? I'll leave my specific circuit out of this for a more broad question: What criteria does one use when deciding whether to use a Zener diode or a transient voltage suppressor, for overvoltage protection on a particular application? Also when would one use a bidirectional TVS? <Q> The characteristics of TVS diodes are similar to those of zener diodes. <S> However, the TVS diodes are specifically designed, characterized and tested for transient voltage suppression. <S> Zener diodes are designed towards voltage regulation. <A> The primary tradeoff is the thermal management or derating features that must be considered in these combined dual roles. <S> If a TVS device is also used as a Zener for continuous voltage regulation, the device would then be operating with sustained power dissipation where internal p-n junction temperatures are driven well above ambient from its own thermal resistance junction to ambient RθJA. <S> That type of operating condition will reduce P(PP) capabilities. <S> " Zeners and TVSs have different design considerations and are also tested differently for their primary operating features. <S> As a result, careful consideration in selection for overall optimum performance and thermal management considerations are required for each. <S> Only under judiciously chosen operating conditions can they be combined with careful consideration and also optimally accompanied by special device testing. <S> " <S> Says Kent Walters in his MicroNote 134 application note. <S> (1) You can have a quick read of his application note if you wish to know more. <S> (1). <S> MicroNote 134 <A> A Zener diode is primarily designed to handlesteady state power. <S> The Zener functions as a voltageregulator (shunt style) or voltage reference in a powersupply system. <S> While electrically similar, the TVS unit has adifferent construction and is designed to absorb largeamounts of energy (joules) in a very short period of time(milliseconds).
| The difference between the TVS and Zener diodes isconstruction.
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Stepper motor troubleshooting with Pololu DRV8825 Driver I salvaged a bipolar stepper motor (QH4-4242), and a power supply from an old printer that I'd like to tinker with before scaling up to larger motors. With the limited knowledge about the stepper (such as current rating), but knowing that the power adapter has a 24V/.55A and 5V/.2A supply, I'd assume the stepper must have a rating equal to or less than .55A. I've found coil wire pairs by measuring resistance (found to be ~8.5ohms). Using a Pololu board built around the DRV8825, I've followed the instructions on their website connecting the correct pins A1,A2,B1,B2, to the corresponding coil pairs. I've supplied the motor with the 24V/.55A output and tuned the current limiting trim pot while measuring the Amps in a single phase while the Step pin was high such that the current was .29A (since the phases only get 70% of the current at full stepping). And I've set the sleep and reset to high as instructed. Using a FEZ Panda II as the microcontroller, I wrote a simple code that turns high the step pin and sleep pin for 1 s, then low for 1 s. The stepper motor makes a high pitch noise (as I've read is fairly normal for a chopper driver) and the shaft becomes difficult to turn, however it does not rotate. While the pins are low, if I slightly move the shaft from the 'cogged' location, when the pins are back to high the shaft rotates slightly as if to snap back into place. However it will not rotate nor step on its own. I would very much appreciate any input or suggestions for debugging this issue. Update: Sorry for the confusion responders, I thought one thing and wrote another last night.My high and low for stepping is 1000ms or 1 second. Pseudo code would look like: while(true){ step(true) sleep(1000) step(false) sleep(1000)} Update 2: Per Richman's suggestion, I've added back to back LED's to monitor the switching of the polarity when a Step pulse is sent this appears to [pun] shed some light [/pun] on the problem. And indeed found the issue. The polarity was not switching when the Step signal was being sent. This was because in addition to sending a pulse to the Step pin, I was also toggling the sleep pin which enables/disables the power to the motor. I read this was done to minimize any high pitched noise, but I think I misunderstood and that enabling/disabling is done after ALL of the steps are carried out. That's sorted out now.Sorry to the responders for the confusion with the timing, I'd give you all the right answer since I explained it wrong if I could. One unknown consequence is that the stepper steps on its own when the Step pin is left at low. I am curious if this is because of noise in the signal which is perhaps causing the Step pin to toggle high/low even though my logic isn't telling it to? Or perhaps there is a grounding/shielding issue because if I touch the motor casing, wiring, or even just put my hand near it, it begins to step faster and faster eventually becoming erratic. Any thoughts on this new conundrum? <Q> I think you're driving the stepper motor too fast. <S> With a 1000uSec high and a 1000uSec low, you're producing a 500 Hz square wave. <S> As such, you're trying to step 500 time a second. <S> Try slowing things down to maybe 50 steps/ <S> second (make your low duration 20 mS, or 20,000 uSec). <A> Start slow and increase durations by a factor of 10x and then experiment with a ramp profile_ _/--- <S> \ _ for speed control. <S> If you don't have enough scope channels, use LEDs to monitor phase voltage and use two LED's back-to-back with a 1K resistor for ~ 20mA <S> Use Green Blue or any different color for ease of reading and slow your program to 1 second per step. <S> Ensure each phase is being driven properly. <S> No need to increase the sleep time I think that is more dependant on the load factor. <S> Generally there are two design interfaces. <S> Direction and Pulse count Like Floppies and old ST506 <S> HDD's and Pulse Fwd/Rev. <S> At the motor interface there are full-steps, half steps and micro-steps. <S> Dead-time is a critical factor in large motors eg 1~10us as the latency and shoot-thru effects from loading can easily fry drivers from both drive-high-low being on with an inductive load. <S> All CMOS works without dead-time <S> but then they don't recommend you drive motors with them. <A> A bit too fast. <S> Try it 5 times slower. <S> If it still doesn't run, look into how you are sequencing the steps; are they in the correct sequence and correct polarity? <S> If there's any doubt about this, sequence a few steps by hand with two wires from the power supply, and check that the driver is generating the same sequence.
| Stepper Motors need to ramp to max speed and down for optimal performance.
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How can I make a plated-through hole in Eagle? I have mounting holes (0.125 in) at the corner of my PCB project in Eagle, made with the standard "hole" tool. However these are not electrically connected to anything. There doesn't appear to be a way to make these plated through by specifying a net (unlike a via). If I want my mounting holes to be connected to the ground plane, do I have to just make an oversize via? Is there some other way? <Q> No, you don't want just a oversized via. <S> Step back and think about the problem a bit. <S> You don't just want a mounting hole, but rather a genuine electrical part. <S> This part should show up in the schematic, be something you place on the board, have a pin connected to a net, and a pad that connection can be routed to, just like other electrical parts. <S> The correct answer is therefore to use such a part. <S> This would have a schematic symbol with one connection. <S> The package would be a thru-hole pad with the hole size matching your mounting screw and the pad diameter a little larger than the screw head diameter. <S> If you want this to be a ground connection, connect it in the schematic to your ground net. <S> I have done this before and have a few such mounting hole parts in my library, including one for a #4-40 machine screw. <A> If I remember correctly, Eagle comes with a library called holes.lbr, with many plated hole sizes that you can place on your sch/brd, or use as template for a non-std size. <S> Those parts are basically a single through hole pad with some layers used to establish proper clearances. <S> Connect in your schematic to whatever you need the plating to be electrically connected, such as ground. <A>
| An oversize via will do what you want. The other thing you can do is to have a premade part with a single plated hole of a specific size and an appropriate keepout area for the screw head.
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Must IO pins each have their own pull-up/down resistor? A very basic question: Must each IO pin of a microcontroller that is serving as an input from a switch or jumper have its own pull-up/down resistor? In this example, each pin is pulled down by a resistor, then pulled up by a 1P3T switch to VCC. It seems straightforward, but as more IO pins are used, are more resistors absolutely required? Is there some clever way to keep the resistor count down? Related question: Sharing a pull-up resistor <Q> As per the schematic provided, if all 3 inputs were sharing a resistor, then any of the lines being pulled high via the switch would raise all 3 lines to high, countering the purpose of the design - the MCU would not know which switch position is selected. <S> A common way of reducing part count , not resistor count, for such designs is to use a common bus resistor network or array: (from here ) <S> These are available as through-hole SIP/DIP as well as SMD, in a variety of resistor counts, depending on your needs. <S> The bus pin is connected to ground, and the other pins are connected to the respective MCU inputs as in your schematic. <S> (from here ) (from here ) <A> So - no resistors necessary. <S> As to why pullup, not down, it's a habit left over from the 1970's TTL logic circuitry, where it took much less current to pull an input up than down - a pulldown resistor would waste more power. <S> This no longer applies with today's CMOS logic, but the tradition of pull-ups has persisted, so that early 5V CMOS chips were compatible with older TTL logic. <A> You never want to leave an input to logic open on the assumption that it would pull itself up or down. <S> if an input is left open, it is a little antenna and also subject to the currents within the logic device. <S> So you pull up or down to ensure you have a clean and predictable input. <S> I learned this rule while I worked at Fairchild Semiconductor in the 1980s.
| Most MCUs have optional pull-up (not down) resistors built in for each pin, so it is normal to pull the pins DOWN with the switch (and deal with the polarity inversion in software).
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Usage of a transistor configured as diode I faced lots of circuits that has transistors connected as diodes (gate connected to the drain). I know some of these circuits employ such transistor for safety reason but I couldn't figure out the reasons of others. My question: Is there any reason behind connecting the transistors as diodes other than the one I mentioned? Some of my colleagues suggest that they may be used to realize high resistance but I think connecting the transistor in such configuration (\$V_{g} = V_{d}\$) will force the transistor to work in saturation region, not in linear region. Am I right? <Q> First, I am assuming that the poster is confusing drain with collector. <S> If he is talking about MOSFETs or JFETs, then ignore the rest of this post. <S> It is a common practice in precision analog electronics to use Bipolar transistors as diodes. <S> The purpose is to obtain a very low leakage diode. <S> For example, a 3904 type transistor will have <1pA of reverse leakage using the Base Emitter junction. <S> However, it turns into a zener diode at around 6.8V. Works great for 5V and lower voltage logic circuits. <S> Higher current and reverse voltage is achieved by using the Base as anode and the collector as the cathode. <S> Still an excellent low leakage diode at around 10pA <S> and now you get the voltage rating of the transistor and improved current. <S> This will not be a high speed diode. <S> Higher speed is achieved by shorting Collector to Base (Anode) and using the emitter as Cathode. <S> However, reverse voltage must be limited to <5V. <S> The other purpose for using MOSFETs and other types of transistors as diode connected is for current mirror circuits, where a diode connected junction will track the active component's junction over temperature. <A> NMOS connected in diode configuration: simulate this circuit – Schematic created using CircuitLab <S> Since Gate and Drain are shorted, the following saturation condition always holds: $$V_{DS}>V_{GS}-V_T$$ <S> This means that once \$V_{DS}>V_T\$ the transistor both begins to conduct and enters saturation. <S> In saturation (after substitution \$V_{GS}=V_{DS}\$ for diode mode): $$I_{DS}=\mu C_{ox} \frac{W}{2L} (V_{DS}-V_T)^2$$ The equivalent resistance of this device is: $$R=\frac {V_{DS}}{I_{DS}}=\frac{2L}{W} <S> \frac{1}{\mu C_{ox} <S> } \frac{V_{DS}}{(V_{DS}-V_T)^2}$$ <S> Now you can see that the equivalent resistance can be controlled by changing the dimensions of the transistor (\$W\$, \$L\$). <S> However, this resistance is not constant - it depends on the applied bias. <S> This is bad, but it is not that you have too many alternatives in integrated circuits <S> (you can implement precision resisters by various techniques, but they are usually costly). <S> On the positive side - there are many application which do not require precision in resistances. <S> Can you implement a big resistor with diode connected transistor? <S> Yes. <S> There are two approaches: Long and narrow transistor <S> Ensure that \$V_{DS}\$ does not rise much over \$V_T\$ <S> However, "big" resistor in integrated circuit is not the same as big resistor as discrete component - in integrated circuit <S> all resistances are relatively low. <A> If you have 5 transistors and one diode in the circuit and there are no particularly exotic signal requirements then replacing the diode with an extra transistor means you have one less item (the diode) in your Bill of Materials and simply increase the quantity of the transistor by 1, which in large scale mass production can yield substantial cost savings (per unit) when buying parts in larger quantities. <S> There are secondary benefits to this also... <S> The manufacturer can use fewer reels on their pick and place machines, which saves time, money and maintenance. <S> Obsolescence is less of an issue with fewer parts in the product. <A> I can't give you technical details but MOSFET's are used as clipping diodes in instrument effects pedals such as guitar distortion pedals. <S> A search for the Mad Professor Fire Red fuzz schematic will show you how this is implemented. <S> However, in the schematic it shows the drain and source tied together. <S> That may just be an error by whomever traced the original circuit. <S> But it should give you an idea of another way a MOSFET transistor is used as a diode. <S> Hope <S> this helps a bit.
| In some cost sensitive products (cheap electronic novelty toys or games for instance) it may save $0.01 per unit to replace a diode with a transistor in the following situation.
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Does PCB artwork affect operation and how is the effect validated? Sometimes PCB designers will draw some art on unused PCB areas. Like here and here . Looks great, but does this affect schema functioning? According to answers to this question an entirely isolated stripe of aluminum foil may act as a stripe antenna. So I'd imagine the following scenario is quite possible: a PCB layout is designed and some area is left unused an artwork is added there the artwork adds capacitance or otherwise affects operations Does PCB artwork affect schema operation? How do designers validate the effects are tolerable? <Q> Artwork on the copper layer could increase parasitic capacitance in neighbouring traces. <S> Since it's rare for modern professionally produced boards to not have a separate silkscreen layer, I suspect the only people doing artwork in the copper layer are people producing their own boards - and they'll validate that it's not a problem simply by empirical testing. <A> I have done artwork, professionally, in the copper layers. <S> Often it is easier to "slip the artwork past the bosses" <S> if it is done in copper, since does not stand out as much as white silkscreen on green boards. <S> Sometimes I would do the artwork in "negative" on the power planes. <S> To view the art you would have to hold the PCB up against the light, and allow the light to shine through the PCB. <S> In theory, any extra (or removed) copper could effect how the circuit operates. <S> But when when I did this I was careful to put the art in areas of the PCB that were basically unused and any side effects were minimal to non-existent. <A> The 1st Mickey Mouse watch had the cartoon character etched in metal on the silicon chip. <S> We saw with our SEM back in the early 80's. <S> Designer Logos are common often hidden from obvious viewers. <S> Fab shops often add their own to the silkscreen. <S> If a logo was added, I would suspect the project designer would have been shown it and approved it. <S> I did not see any interference in the examples I've seen.
| Most artwork - with the exception of homemade boards - is done on the silkscreen layer, not in the copper layer, so should have no electrical effect. It would be silly to add artwork that negatively effected circuit operation.
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What is the easiest way to lower the DC output (reduce voltage)? I have 13.5 VDC output, and i want to power up a small radio that takes 4.5 VDC. I thought about adding 5/6 used 1.5V batteries in series, so it would be about 4.5 VDC for radio and about 8 VDC for batteries. Would that "battery voltage reducer" work? If not, how can i reduce voltage without using any complicated electronical stuff? (I am not an electrician) <Q> Your battery solution will not work and may be dangerous. <S> You need something like this ; (search for "dc dc converter"; anything car orientated will be suitable, given that car battery voltage is somewhere in the 12-15V range) <A> You said "small" radio, but didn't say how much current it draws. <S> Does the radio actually say on it that it needs 4.5V, are you getting this from the fact that it takes 3 1.5V cells, or something else? <S> If this is battery operated, then most likely it will work fine on 5 volts. <S> If you are worried about the exact voltage, use a adjustable regulator to make 4.5V. Keep in mind that a linear regulator dissipates the difference in voltage times the current as heat. <S> If the radio draws 100 mA, for example, then a 5V linear regulator would dissipate 850 mW. <S> That's about the limit for a TO-220 package standing up from the board in free air. <S> Put even a small heat sink on it <S> and it should be fine. <S> If the radio draws only 50 mA, then just a bare 7805 in TO-220 package is all you need. <S> If 200 mA, then you should start to seriously consider a switching regulator. <A> Something like linear voltage regulator 7805 probably will be excellent for your case as it is very cheap, very simple to use, you need only 2 very cheap capacitors (something in the range from 22pF - 100nF will be OK for you) for "just in case" smoothing and optionally but highly recommended one rectifying diode as 1N4007 before 7805. <S> You don't need a heat-sink for 7805 with 13-14 volt input. <S> This setup will provide you with arguably the cheapest solution (bill of materials wise) and most simplistic one. <S> Unused power will be dissipated as heat but it will be negligible and you shouldn't feel it. <S> P.S. <S> I will edit my post little bit later to provide you with schematic to use. <A> The problem with using the batteries is that they will always be charging. <S> Draw it out and notice how current will enter the positive terminal of the battery. <S> That's opposite to the way batteries are usually used and means the battery is charging, instead of supplying. <S> Always charging means the batteries will eventually pop. <S> Unfortunately, to convert voltages you need complicated electronic stuff. <S> The simplest solution is the linear regulator others have mentioned. <S> It's almost as simple at your battery solution (one lead goes to the source and one goes to the battery). <S> The only difference is that there is a third lead which needs to go to ground. <S> Depending on how hungry your radio is, it also may get (too) hot, but that's electronics for you...
| If only a little, a 7805 voltage regulator may be all you need.
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Power management for a 5V 12A load I need to deal with a peak load at 5VDC and about 11.25A but I'm not sure how to manage this. I have 5 individual loads of 2.25A and small loads from an MCU and some LEDs all of which should fit into the 12A. Another requirement that I have is that power supply can be placed in a sealed enclosure so that it can be placed outside and not have water get in. I suppose an aluminum case might help with this. I am not quite confident enough to build my own power supply. Dealing with mains is a bit scary. I've found some a RC car battery charger power supplies that will supply 14VDC @ 16A (Turnigy T-20Pro) . They have fans in them which wouldn't do much being in a sealed enclosure. My though was that I could use simple 5V linear regulators for each of the 2.25A loads and another for the MCU and LEDs. What alternatives do I have and/or what trouble am I going to run into? UPDATE: I wasn't explicit about the power source, sorry. It will be 120v 60Hz AC. UPDATEv2: Most of the load (5x2.25A=11.25A) will be outside the enclosure as the project is a LED Christmas light controller. <Q> If you're not comfortable working with mains voltage, and assuming that your loads are in the same enclosure as your power supply (i.e. you're generating a lot of heat in an enclosure), you might consider a AC->DC power supply mounted remotely from your enclosure, and running 5V into your enclosure. <S> Normally, voltage droop would be a concern under such use cases, as it only takes 0.04 ohm of line resistance to drop 10% of your supply voltage. <S> Many power supplies have remote sense leads, such that the power supply closes its control loop around the supply voltage at your enclosure, compensating for any transmission losses. <S> The benefits of this approach include reduced power dissipation in your outdoor enclosure and only hand-friendly 5V in the enclosure (but watch out for shorts!). <S> The downsides are the need for a separate (hopefully indoor-rated) enclosure for the supply and possibly a slightly higher cost for a AC-DC supply with remote sense capability. <A> Building your own AC-DC power supply will be quite tricky. <S> As others mentioned, you won't be able to do this easily with a linear regulator, and it's just not worth designing a SMPS on your own. <S> After all, computer power supply units regularly provide lots of current on the 5V rail (20 A is typical), and they are very cheap. <S> Since you need to be outdoors, I suggest looking for a fanless ATX power supply, so that it can be sealed in a case. <S> It's even safe to test, since power supplies have an integrated thermal shutoff if they overheat. <S> This datasheet on thermal management suggests you will need a case with an internal surface area of at least 4 square feet to manage a 60 W load without ventilation. <S> In practice, you probably need more. <S> When looking for a computer power supply for this job, keep in mind that many newer ones provide only a +12V rail and use a DC-DC switcher on the mainboard. <S> Look for an older one with a dedicated +5V rail. <S> To make a computer supply work in this way, you need to have a way to toggle its "power button" signal. <S> Have a look at the ATX connector pinout to see what's required. <S> You could make it a switch, or use a pulldown resistor to make it permanently on. <S> I believe you should also put small dummy loads on the other power rails, because some power supplies don't behave well when they are completely unloaded. <S> Usually the +5V rail is the one used for regulating the supply, so you shouldn't have trouble on that front. <S> Also keep in mind that most power supplies have a standby +5V that has a low current limit; you won't want to use that. <A> Exterior power supplies are a big challenge because they need to be sealed (to keep rain / snow / dust / dirt out) which compromises cooling (forced air doesn't really help) plus they need to withstand some extreme temperature swings (-40C to +50C isn't uncommon) which limits you in some nasty ways (most electrolytic capacitors significantly lose their capacitance below 0C). <S> Off-the-shelf exterior LED power supplies are probably your best bet. <S> (Make sure you find one that's safety-approved, unlike the example I cited!)
| You may be able to get away with a regular power supply with a fan if the case can be used as a heat sink, and because the load will be low compared to a computer.
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Any microcontrollers which can be directly programmed on a breadboard? Does anyone know of a microcontroller which can be directly interfaced and programmed through a PC using serial/parallel/usb port? I am looking for a microcontroller which can be plugged into a breadboard and connected to a PC (without the programmer) to upload the bootloader and the program into it. <Q> Most (all?) <S> NXP LPC ARM and Cortex microcontollers have a built-in serial bootloader. <S> You will have to provide the UART signals to these chips at 3v3 (NOT at RS232 levels). <S> The LPC1343 also has a built-in USB bootloader: it can present itself as a drive, you simply copy your firmware to it. <S> Sounds simple, but a bit tedious in practice if you want to do this often. <S> For most microcontrollers bootloader software is available (in most cases multiple versions). <S> You (or someone else) must first download this software to the chip. <S> In most (but not all) cases you will have to configure your application to work with the bootloader (not occupy the same addresses). <S> I once wrote a bootloader called ZPL (zero pin loader) that interfaced to its host only by its reset pin. <S> It timed the interval between the resets, and decoded the bitstream from this timing. <S> The reset signal was provided by a UART signal from the PC. <S> IIRC <S> this idea was not mine, I read (or at least thought I read) somewhere about a bootloader using only the reset pin. <S> Once you know (or think that you know) that it is possible the reset is just work. <A> Look at the Metaboard or Arduino serial . <A> since serial port does have 2 input(DSR/CTS) and 2 output(RTS/DTR) lines,you couldon and off and use them to generate the ICSP signals to program microchip devices. <S> ICSP specification is here: http://ww1.microchip.com/downloads/en/devicedoc/30277d.pdf <S> This my own DIY tutorial about how control RTS line on PC, http://www.expertcore.org/viewtopic.php?f=13&t=3670 <S> You need a level converter, but I've checked on my USB2Serial device voltages, <S> it does not reaching more than 5V in any line [since that's USB].So <S> you could definitelyuse it directly. <S> Microchip have low voltage programmable chips ,but they are littlebit expensive. <S> On the other hand you need to level up using a transistor and external12V supply. <A> Teensy is designed to do exactly that. <S> It is also inexpensive: <S> http://www.pjrc.com/store/teensy3.html
| You can use any ATMega with serial or USB bootloader.
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Finding the right motor I need a motor and I'm not sure what to be looking for. I know the power and torque I need (3 N*m @ 1200 rpm = 1/2 HP). Is finding a motor as simple as matching these two values? Edit: The motor will be used essentially like a fan motor; It will have to maintain a (fairly) constant speed for long periods of time. It's power will come from the wall (120V) so it could probably be AC but doesn't have to be. It'll be sitting still at room temperature. I'm mostly trying to understand how to choose a motor based upon torque/power requirements as I don't know what I'm doing when it comes to this. EDIT: I can incorporate something like a gearbox, I just don't know the process of picking a motor + gearbox that does what I actually want. <Q> Things to ask yourself: <S> What is my input voltage? <S> AC or DC? <S> Are you running this off of a battery or line voltage or a control or something else? <S> What am I doing with the motor? <S> Is it for a fan and will be running at one speed for 8 hours a day? <S> Is it for a servo type application? <S> What's the duty cycle? <S> What is the environment it will be running in? <S> Does it need to be dust proof? <S> Water-proof? <S> Explosion-proof? <S> Will the ambient temperature be 20 C, 65 C or -40 <S> C? <S> How many hours does it need to run before failure? <S> 5000 hours, 20,000 hours, 50,000 hours? <S> Your answers to those questions would give me a pretty good idea of the type of motor you need and what its rating should be. <S> EDIT: <S> In general, sizing a motor is as easy as you suspect. <S> The rated power on the nameplate of a motor is the power that motor can continuously output without overheating. <S> Generally these ratings tests assume that the ambient temperature of the environment is below 40 degrees Celcius. <S> So if you have a motor that is going to be running at a constant torque and speed at a temperature below 40 C, finding a motor is simple. <S> Things get trickier when you have short duty cycles or a duty cycle where the torque changes abruptly or higher ambient temperatures. <A> EricE did a great job summarizing some of the major points of motor selection. <S> This is a really broad topic, so I'll attempt to summarize a few other (non-comprehensive) factors for consideration: <S> What is the peak torque requirement? <S> Is the load torque somewhat proportional to the shaft speed (like a fan), constant regardless of shaft speed (like a winch), or decreasing with shaft speed (like a flywheel). <S> The load characteristics during motor startup (i.e. from stop to running) can be a major design driver. <S> What is the speed requirement? <S> Is some speed variation (slip) acceptable with changes in load, or is tight speed regulation required (synchronous motor or closed loop VFD/servo). <S> Must the motor speed be controllable? <S> Within what range? <S> (If so, a VFD or servo drive may be required. <S> If very low speed, high torque operation is required without a gearbox, an external forced air fan may be required). <S> Are there efficiency considerations? <S> Does form factor matter? <S> Brushless DC (BLDC) motors are available in a wide variety of aspect ratios, while induction machines tend to be of a similar aspect ratio. <S> Does motor mass matter? <S> Will the motor be mounted on a moving assembly? <S> Does the application require a particular acceleration or deceleration profile? <S> If so, the peak torque requirement may increase - <S> it may be driven by acceleration/deceleration torque rather than continuous running torque. <S> Is monitoring of the motor speed or torque required? <S> Are commercially available motors available in the desired speed range (if fixed speed)? <S> If not, a gearbox or other reduction system may be more cost effective than a special purpose motor. <S> I'm sure I've missed a few key ones, but those are the ones that jump out (and are complimentary to EricE's post.) <A> I'm sorry to tell you that this equation is wrong. <S> 3 N*m @ <S> 1200 <S> rpm = 1/2 HP <S> it should be corrected to 3 <S> Nm x 1200 <S> rpm x <S> 2 pi = 1/2HP <S> Just selecting <S> a 1/2 HP won't give you 3Nm and 1200rpm either.[Note <S> that 1200rpm is wronganyway]. <S> DC motor could be easily controlled with filed voltage. <S> Well AC motors too, but notinductor motors or synchronous motors. <S> AC commutative motors could be controlled withregulating voltage. <S> EDIT: I can incorporate something like a gearbox, I just don't know the process of picking a motor + gearbox that does what I actually want. <S> No need to worry that much of. <S> Motor have it's torque-speed curve. <S> So for a applicationlike a printing press[Hamada 700 that I worked with] <S> it uses a mechanism that adjustthe tension of the slip belt to adjust it's operating point and operating speed. <S> In otherwords you could mechanically do it or do it using a feedback system that precisely controlled with a microcontroller with the help of control theory.
| If you know the power you need the motor to output, then you find a motor that is rated for that power.
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What is difference between I2C and TWI? What is difference between I2C and TWI interfaces? Wikipedia says, that TWI is variant of I2C. Some other sources says that only name is differ... <Q> I agree with everything @Wouter said... <S> Namely, TWI is used by manufacturers (like Atmel) to refer to their I2C interface to avoid legal trouble with Philips since I2C is a registered trademark. <S> Some manufacturers (like Atmel) have implemented proprietary features on top of I2C so that their TWI is an I2C superset. <S> To answer your follow-up question, yes. <S> It is enough just to change the name. <S> A trademark only protects the name. <S> If it were patented, the situation would be different. <S> However, it is difficult to defend a patent on a modulation/encoding scheme since it is mostly just math. <S> Often, copyright is used for this, but it is usually pretty easy to circumvent, since the copyright protects only the exact specific instantiation (implementation) of the protocol, not the protocol itself in most cases. <A> The name TWI was introduced by Atmel and other companies to avoid conflicts with trademark issues related to I²C. A description of the capabilities of TWI interfaces can be found in the data sheets of corresponding devices. <S> Expect TWI devices to be compatible to I²C devices except for some particularities like general broadcast or 10 bit addressing. <S> Also, for now, there is also no TWI high speed mode. <S> If you have an I²C analyser it will work with a TWI set-up as well without modifications. <A> TWI devices typically don't conform completely to the I2C specification. <S> Notably, clock stretching is usually not supported.
| From i2c-bus.org : TWI stands for Two Wire Interface and this bus is identical to I²C.
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Repurpose an old lamp with LEDs. Where to start I have an old lamp which I would like to repurpose to use LEDs. I'm not really sure where to start so I'm looking for some pointers if this isn't the place. I also don't know how much of this is practical or sensible. Here's my ideal. I'd like to replace the existing incandescent bulb with LEDs. I'd like (if possible) to have the lamp battery powered. I was wondering if it would be possible to have a "charging" plug to charge batteries. For charging or power I was considering using something plug-able, like a micro-usb charger. I've done basic electrical stuff (replacing ceiling light fittings is about as advanced as I get), I have a very basic knowledge of electronics, but I'm quite a geek so I'm hoping I can make it work. I realise I may be biting off more than I can chew but I wanted to get some feedback, if possible. <Q> I'd break this down in the opposite direction. <S> This isn't a full solution, but the two parts can be addressed with googling or looking elsewhere on this site. <S> 1) <S> Starting from the USB charger: that delivers 5v. <S> 2) <S> Battery: <S> Lithium polymer batteries have a 3.6V-4.2V range, and ICs and circuits designed to charge them from USB 5V are widely available. <S> This sort of thing (not necessarily the cheapest option). <S> Make sure your design checks the battery's thermistor. <S> NiCd and NiMH are the other common battery options, but they also require special circuitry. <S> Lead acid is easiest to charge but not very portable. <S> 3) LED and driver: you could have one big one or lots of small ones. <S> If you have lots of small ones you might be able to drive them directly from the battery (their forward voltage is about the same as the battery voltage), but this is non-optimal. <S> Have a read of this paper from Maxim. <S> Again, there's already lots of stuff on electronics.stackexchange.com about "how to drive LEDs". <S> If you go for one big one it will definitely need a charge pump based controller of suitable current rating. <S> If it's really big then you need to think about cooling it. <A> Making it battery powered and charging from the wall would be too complex to explain, take it one step at a time! <S> (I actually did this last summer!) <S> Get yourself a LED flashlight that takes 3 AAA batteries. <S> Discard the batteries and use the battery holder to hookup a wire to the positive side that fits in the flashlight, and your negative wire to the metal case (you might to have to scratch off some paint.) <S> And mount your LED flashlight in your lamp, and feed the wires thru it. <S> Now cut the wires on an old cell phone charger, they usually are 5 volts DC and will only light the LEDs one way. <S> Feel free to have it plugged in while you play with it, 5v won't hurt you, just don't let it short out for very long. <S> DON'T wire the LEDs to the power cord and plug it in; the LEDs won't last long at 120/240v! ... <S> talk about a flash-bang! <A> Assuming it's a directional light with a 60W bulb, you will want at least 6W <S> and if it's not directional probably 11w which is quite a load for a low voltage battery system. <S> If battery power is not important then buy a mains powered ready made bulb. <S> But if you do want to go battery powered, Your best bet is as John U suggests, buying ready made 12V bulbs; <S> either car bulbs or GU10/ <S> Mr16 12v spotlight bulbs and using a sealed lead acid battery and charger. <S> If you've got the space <S> A 2.3Ah 12v battery should give around 4.5 hours with a 6V bulb. <S> Be warned though: the charger will take the battery Voltage to over 14V <S> so you either need a bulb that will take that (car ones should, but not many cheaper ones state the upper voltage limit), or you will need to regulate the voltage going to the bulbs. <S> You will need a low drop out regulator (LM2940 should do the trick) otherwise the voltage drop across the regulator will take the voltage too low for the bulb when not charging. <S> If you don't have the space for a lead acid, then you will have to use lithium, NiCad or NiMh and decide whether you're making your own LED arrays and drivers, to run at the standard battery pack voltages (so you can use ready made chargers), or making higher voltage battery packs and charging systems to run the 12v bulbs, but you must use a properly designed charger for the specific battery type, to avoid destroying the batteries, or possibly causing a fire/explosion. <S> Again, it may be worth buying a multi-LED or CREE Cycle or Caving head light, and using the parts to make your Lamp. <S> Whatever you do, I don't think USB power is very useful to you
| If you're talking about converting a mains powered incandescent lamp to battery powered LEDs, you will either have a considerable drop in lighting, or will need a substantial battery and charger to get a similar light level for any length of time.
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What was the advantage of a 64-bit processor in the N64? The Nintendo 64 debuted in 1996 and featured a 64-bit MIPS processor. My understanding is that PCs didn't start appearing with 64-bit CPUs until 2003. What was the advantage of using a 64-bit CPU in 1996? Presumably, back then addressing more than 4GB of memory would not have been an issue for a consumer electronics device. <Q> This is mostly content-free marketing hype. <S> Yes, the processor may have been "64 bit" (however that is really defined), but that by itself means little. <S> As you say, it is unlikely that addressing more than 4 Gb would have been of any use in 1996. <S> Presumably, a "64 bit" processor can do arithmetic manipulation on 64 bit quantities as a whole, and that could have speeded up certain operations. <S> However, there are many other architectural choices that result in various cost versus speed tradeoffs that would have had a much larger impact on the result than whether the processor was 64 bit or 32 bit or something else. <S> The fact that at least parts of the processor were 64 bit is therefore pretty meaningless when looked at in isolation. <S> What really matters is the resulting performance and <S> what price you have to pay to get it. <S> This is somewhat like making a big deal about whether the engine in a car has a overhead cam or dual exhaust valves, for example. <S> That by itself tells you very little since the many other tradeoffs in the design of the car and its engine swamp the effects of that one design choice on the resulting performance and price. <S> Just like "64 bit", these are mostly just hype because marketing thinks it sounds cool and that you're dumb enough to think it matters. <S> Unfortunately for most consumers, marketing is right. <A> On many machines of that era, any time the display controller spent fetching data from RAM was time the CPU or other chips couldn't access that same RAM. <S> The less time the display controller spent on RAM access, the more time the RAM would be available for other purposes. <S> If at a particular display resolution and RAM speed, a 16-bit bus would be 90% consumed by the graphics controller's display-data fetches, other chips would only be allowed to access the bus 10% of the time. <S> Moving to a 32-bit bus would drop the display controller's overhead by half (to 45%), meaning other devices could access the bus 55% of the time. <S> Even if the other chips could only perform 16-bit accesses, they'd still see a 5.5x speed improvement. <S> Pushing the bus to 64 bits would cut the display controller's share in half again (to 22.5%), leaving 77.5% available for other stuff--another 40% improvement. <S> Not as huge an improvement as going from 16 bits to 32 bits, but significant nonetheless. <S> Note that if one were using a higher color mode, it's possible that 16-bit bus utilization would have been over 100% (meaning a 16-bit bus simply couldn't handle such a mode), and 32-bit utilization might be over 50%. <S> Under that sort of scenario, the improvements going from a 32-bit bus to a 64-bit bus would be more significant (again, even if no chips other than the display controller can exploit 64-bit accesses). <S> Of course, if some other chips (e.g. ones that are designed for copying large chunks of display data) can exploit 64-bit accesses, those chips would receive even further benefit. <A> http://en.wikipedia.org/wiki/Nintendo_64#Central_processing_unit claims that the 64 bitness of the processor wasn't used that much, which makes me suspect that it was mostly a marketing effort. <S> The graphics co-processor may have had a SIMD-like behaviour (ie do 8 lots of 8 bit operations at once, or 4 16-bit operations), although I can't find a handy reference that confirms this. <S> The nearest is this N64 manual, which says "At maximum speed (usually only attained with rectangles), two 32-bit pixels or four 16-bit pixels are written per cycle." <S> (The memory bus was apparently only 9 bits wide, which may have been a major limitation on polygon fill rate!) <A> From any practical point of view I can imagine it was mainly a marketing trick. <S> From today's point of view it is hard to even define what does "64bit CPU" mean. <S> We can look at standard x86 family of CPU's. <S> They are known as "32bit", although the latest models like Pentium 4 have got 36bit address bus and 128bit data bus. <S> They contain several 128bit wide data registres, 128bit instructions, and can directly access 64 GB of memory. <S> The real main reason why we use x64 platform today is that Microsoft was too lazy to implement correctly the support for 36bit memory addressing into Windows XP. <S> 64bit opearting systems eats a lot more memory, because you actually need to have both 32bit and 64bit versions of the system, also 64bit programs are bigger, because instructions are longer, all address are longer, etc. <S> This means that a faster bus is needed. <S> Another point is that since their beginnings, gaming systems always used many special graphic acceleration chips so they were often successfully implemented with just an 8bit CPU without any significant degradation in quality of graphics etc. <A> If implemented through a single opcode, this could be true regardless of what mode the processor is in. <S> How much of this happened on a regular basis would probably depend on the game being played, though.
| If nothing else, copying portions of memory ought to go twice as fast on a 64-bit processor with a 64-bit bus than on a 32-bit processor with a 32-bit bus given similar clocks speeds. The use of "64 bits" was probably as much a marketing term as anything, but not entirely without meaning.
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Does "50Mhz Oscilloscope" mean that it can plot up to only 50Mhz frequencies or its bandwith is 50Mhz? I couldn't find it in the books. I mean the fixed freq. value on the oscilloscope. <Q> I'm not sure what the distinction is between the two choices in your question, but the real answer is that the gain of the amplifier(s) <S> inside the scope is "flat" up to that frequency. <S> In other words, the voltage measurements you make on the screen will be "accurate" up to that frequency. <S> Above that frequency, the values will appear smaller than they actually are. <S> But the truth is that the specified frequency is usually the "-3 dB" point of the gain curve, which means that measurements at that frequency are already off by 30% — the waveform is only 0.707× the height that it should be. <S> If you want more accurate measurements, you need to look for the -1 dB frequency (10% accuracy) or even the -0.1 dB frequency (1% accuracy). <S> But in most cases, you're not really making precision measurements with a scope; instead, you're looking for qualitative changes in the waveforms, etc. <S> As long as you're generally within the specified bandwidth, you should be good to go. <A> A 50MHz oscilloscope has a bandwidth of 50MHz. <S> What it can plot will depend on its samples/second value (assuming vector plotting, first order interpolation will fill in more than what is actually measured) . <S> This is explained a little further in this related question . <A> For an analog scope it means that a 50MHz sine will be attenuated 3 db (down to 70% in voltage terms). <S> A digital signal of the same frequency contains components of much higher frequencies, so it will be severely distorted. <S> A ballpark estimate I read somewhere is that you should not trust the picture of (non-sine) signals higher than 1/3 of the bandwidth. <S> Hence when you see a signal change from 0 to 1 that change could have happened anywhere between those two sample moments, hence your timing uncertainty is +/- <S> 10 ns. <A> Apart from the accuracy in the waveform shown, you'll notice that triggering on signals with higher frequency will get increasingly difficult. <S> Good/stable triggering is essential to obtain a stable/any image on screen.
| For a digital scope it means 50M samples/second.
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How to search for a component? Whenever I start a new project, one of the first things I must do is search for suitable components to see if the project is feasible. Most of my projects are desperately space constrained, so I don't always have the luxury of just using parts I'm familiar with. Often I have to start from scratch, visiting each manufacturer's web site, and trawling through dozens of parts, using their part filters (which often don't filter by the attributes I'm interested in, like part size). I'll also go to the suppliers (Farnell, Mouser, etc.) to find parts. I often need to read each datasheet to discover the missing details. Finally I record them in a huge spreadsheet, so that I can select the 'optimal' part. Even when I've done this, I can't help feeling that I must have missed some manufacturers. Maybe I should have checked Allegro, or Melexis... This must be a common problem for electronic engineers trying to select the optimum components. It certainly is for me. How do other EEs approach this task? Is there some global electronic component search engine I haven't heard of? <Q> If I can not find the exact part that I need on digikey, I would visit the manufacturer's websites of the parts that the narrowed down digikey search returns. <S> This is the approach I have taken. <S> It is not perfect and can be time consuming at times, but other times you can find a suitable part very quickly. <A> Its always helpful to start the search on digikey , which provides the most accurate category classification, parameter, value and unit. <S> If you only have a few keywords in mind but not sure of part number or which category to start, try descriptive search at ciiva . <A> If you live in Australia, it may be more useful to try Element14 <S> Australia as they have a very easy to use search tool of an extensive list of components as well as serving the Australian audience. <S> You can choose which parts you want by their specific characteristics. <S> For example if you want to choose a voltage regulator, you can choose the output voltage, input voltage range, output current and IC pinout style. <S> Also to note, if you see multiple components that looks exactly the same, well they're not. <S> Scroll further down <S> and you'll se that they are manufactured in different places and have different weights. <A> There are a couple meta-search engines out there, I don't have one of them handy at this computer, but the other is findchips.com which you can use to search for "battery" or "fet" or even just "purple" (if you wanted to see all purple LEDs, cases, or whatever, on the sites they cover.) <S> They search on Digikey, Element14, Mouser, and a number of other sites. <S> I know how frustrating it is to need a specific form factor or size etc.; Also I find that when ordering from one supplier, too often they don't carry at least one part / <S> brand I need, I've had to settle for getting spares when I order from one place and then alternating orders from various places, not ideal <S> but then it's not an ideal world.
| I always start my search for components on digikey ; they have a fairly good interface for narrowing down the search results (for example, you can specify the package and mounting type you need).
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What is "gain" a measurement of in electronics? I'm familiar with the term "gain" used in sound and music, but I'm having difficulty finding the definition in electronics. I'm seeing the term low gain and high gain being used in the description of the Panasonic Grid-Eye infrared sensor. What exactly is gain a measure of in electronics? <Q> Gain is just the ratio of two quantities. <S> When talking about circuits, it's almost always the ratio of some measurement of the output of a circuit or component to its input. <S> As Olin said, the most common case is the voltage gain . <S> That is, the ratio between the voltage amplitude of the output signal to the input signal. <S> Another common specification is power gain, the ratio of the output power to the input power. <S> But some circuits are best characterized by a current gain. <S> Some (transimpedance amplifiers) are best characterized by a transimpedance gain --- the ratio between the output voltage and the input current. <S> Another common usage of the term <S> gain is in antennas, where the gain of an antenna is "the ratio of the power produced by the antenna from a far-field source on the antenna's beam axis to the power produced by a hypothetical lossless isotropic antenna, which is equally sensitive to signals from all directions." <S> (per Wikipedia) <A> "Gain" is usually voltage gain unless otherwise specified. <S> Keep in mind that power gain is proportional to the square of voltage gain. <S> Sometimes gain is specified in dB, which is actually a way of expressing power ratios. <S> A dB is 10*Log10(power1 / power2). <S> This becomes 20*Log10(voltage1 / voltage2) if the two voltages are at the same impedance . <A> In a sensor, "gain" is synonymous with "sensitivity". <A> It's actually typically exactly the same thing as in audio. <S> Or, rather, the audio use of "gain" (for signal voltage amplification) derives exactly from the electronic "voltage gain" used in amplifiers. <S> This means that the dB ratings of gain mean the same thing -- <S> 20 dB means a factor 10 in voltage which means a factor 100 in power.
| A high-gain sensor will have a larger output (voltage, current, or both) than a low-gain sensor, given the same level of the physical input "signal".
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Is there a preferred placement of termination resistor for a differential clock line? Looking at a reference design board from Xilinx, we noticed the placement of the termination resistor for a differential clock line (~300 MHz): The resistor is placed not at the closest possible location, which would be right under the vias which receive the clock (there appears to be a lot of space), rather, it is placed "far away", to the left of the vias. We are wondering if there is a good reason to place a differential termination resistor that way. Any ideas? <Q> Termination resistors sit at one of two places: As close to the driver as possible (for source termination), or as close to the end of the trace as possible (for the kind of termination that you have). <S> Notice that I said "as close to the end of the trace as possible", and not "as close to the receiver as possible"? <S> There is a difference! <S> Most people think of the signal chain for end termination being like this: <S> Driver --> <S> Long Trace -- <S> > Termination --> Receiver. <S> But that is incorrect. <S> The termination should be at the end of the signal. <S> Like this: Driver -- <S> > <S> Long Trace -- <S> > <S> Receiver -- <S> > Termination. <S> If you have several receivers, then you get this: Driver -- <S> > trace -- <S> > <S> receiver1 -- <S> > trace - <S> > <S> receiver2 -- <S> > termination. <S> The reason for this is that the termination prevents the signal from reflecting back to the receiver(s) and corrupting their signal. <S> In theory you could have a super long trace from the last receiver to the terminator and that would work just fine! <S> That is why your termination resistor is not near the driver-- because it doesn't have to be. <S> The resistor is at the end of the line, right where it should be. <S> There are other reasons why the resistor is placed that far away, but that isn't too important. <S> As for the little squiggles, that is because in a differental signal the lengths of the two traces should be identical in length. <S> The Squiggles are done to adjust the length of one trace to make it match the other trace. <A> Like embedded.kyle says, the squiggle is a clue, but I disagree about the meaning. <S> It's absolutely certain that they need to design with controlled impedance for 300 MHz, squiggle or now squiggle. <S> What the squiggle says to me is that they are being very careful about matching the trace length between the two lines in the differential pair. <S> The reason to do this is to minimize common-mode reflections on the line, which could conceivably cause an emi problem. <S> I think they moved the terminating resistor to make room for the squiggle. <S> However, given the frequency (300 MHz) and the size of the squiggle, I believe they are over-designing this. <S> That small a squiggle is unlikely to have any meaningful effect at that frequency. <S> That said, they don't cause any problems with this design. <S> What they've done is just move the termination down the line past the receiving part. <S> This is a perfectly good design, and better than what you usually see, which is to put the termination first and then run a stub for a few more mm to reach the receiving part. <S> In the case you show, the receiving part just causes a slight capacitive shunt at a midpoint on the line. <A> The squiggle in the one trace is a clue. <S> Since they put that squiggle in there, the designers are concerned with matching the impedance of the traces. <S> The following is from an Altera App Note : Differential Pair Termination Differential signal I/ <S> O standards require a termination resistorbetween the signals at the receiving device (see Figure 31). <S> For theLVDS and LVPECL standard, the termination resistor should match thedifferential load impedance of the bus (i.e., typically 100Ω). <S> Xilinx has similar literature <S> but I found the Altera note to be more concise. <S> So the reason the termination resistor is not as close as possible to the source <S> is because the impedance of the traces matters when trying to adhere to the Low-Voltage Differential Signaling (LVDS) or Low-Voltage Positive Emitter-Coupled Logic (LVPECL) standards. <S> Edit: Taking into account your comment and the other answers, the above does not make sense. <S> As the others have said, it is to match the trace length to reduce the effects of common-mode reflections that can occur when using single-resistor termination of a differential signal. <S> Dr. Howard Johnson explains it better than I ever could: <S> Terrible things can happen to these common-mode artifacts if yourtrace delay equals one-quarter of the clock period. <S> In that case, thelittle common-mode artifacts build and superimpose, cycle after cycle,magnifying your common-mode signal-input-range problems at thereceiver and also magnifying the radiated emissions. <S> So, as David Kessner said, they have extended the trace length beyond the receiver and before the termination resistor in order to adjust the trace delay and reduce the possibility of reflections getting out of hand. <A> Regarding terminology, this seems to be an example of a "fly-by" termination. <S> From http://www.interfacebus.com/Design_Termination.html#f : <S> Placing the termination at the far end just at the input pin works well in many situations. <S> For larger chips, such as FPGA's, which can be 1 inch square a technique called Fly-By termination is used. <S> In this case Fly-By termination increases the trace length by an inch, the resistor is still an inch from the input pin, but at the end of the trace and not an inch before the input pin.
| Fly-By termination places the termination past the device which puts the termination at the end of the trace.
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Do I need a current limiting resistor when powering an LED below its forward voltage? I've reviewed the questions: Why can you use an LED below voltage? Is a current-limiting resistor required for LEDs if the forward voltage and supply voltage are equal? But combining them begs another question: If an LED (example: a 3.2 \$V_F\$ white LED) lights satisfactorily when powered using two alkaline cells (3V), is there a need for a current limiting resistor? Assuming the battery source never varies above the \$V_F\$ of the diode, can too much current ever be applied? If it is recommended to always have a current limiting resistor, as in Martin's answer to the second question , does that mean the best option is to use three cells (4.5V) and add the resistor? <Q> Have you tested the two alkaline batteries? <S> I'd bet they are slightly above 1.5V each, bringing your total voltage up to the LED forward voltage. <S> However, when the LED starts conducting, its low impedance pulls the battery voltage down slightly, and the battery itself acts as the current limiting resistor. <S> As to the second part of your question, I can't say whether its always better to add the resistor and use more cells. <S> If you're making an LED throwie, then it definitely isn't. <A> Note that the forward voltage of an LED as specified by the manufacturer is only an average value. <S> Different parts may vary in their actual forward voltage at their specified current. <S> Therefore, it is not a good idea to blindly try to exactly hit the specified forward voltage (3.3V, for instance). <S> To be safe, you could apply the correct forward current and measure the actual voltage for your specific part. <S> This will give a pretty good value for the 'correct' voltage to use. <S> If that's not feasible, some safety margin can be considered to make sure to not overload the LED in even the worst case (specimen with relatively low actual Vf being used in high temperatures). <S> Powering a white LED (Vf typically 3.2-3.3V) with 3.0V should be enough of a safety margin. <S> - On the other hand, your LED ist probably not expected to live those 50-100k hours, so overdriving it a little should not be an actual problem. <A> Martins answer to the second question refers to "using a voltage source" and "using a current source". <S> You are not using a voltage source, and you are not using a current source: you are using a two-cell alkaline battery. <S> Given that you've got a 3.08V circuit, I expect that (1) <S> The battery will discharge rapidly and wastefully initially, then (2) <S> The light will drop out while the battery is still almost fully charged. <S> In between (1) and (2) you should get a brief region where most of the battery discharge is going to power the LEDs (rather than heating the cells or just switched off). <S> If the cell discharge curve exactly matches the LEDs voltage curve, this could be most of the energy in the cells, but I doubt I would be that lucky. <S> At 3.08V with alkaline cells, I would expect the crossover to be fairly short, and up near the fully-charged end of the discharge curve (but it's the kind of thing you could measure) <S> If you use 3 cells and a resistor, you waste energy in the resistor, but I think you will use far more of the discharge curve of the cells, and not leave them almost fully charged when they dip below the LED voltage. <S> So you may pay less for cells, even though you use three at a time instead of 2. <S> Obviously, the main problems with the simple circuit are blowing up the LED or exploding the cells, but clearly that hasn't happened with your experiment, which supports my idea that you are operating up near the top voltage of the cells.
| If you want a consistant runtime that doesn't rely on hitting a bullseye between battery voltage and current draw, then probably.
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Measuring voltage while maintaining isolation I'm building a variable power supply, and want to provide the ability to digitally measure the outputted voltage and current. The digital power supply used to drive the ADC, microcontroller and friends is separate from the unregulated power supply used to derive output voltages. The power supply for the microcontroller is galvanically isolated from the one for the output voltages — the first is an unregulated transformer, while the second one is an SMPS supply. Now, measuring the current while maintaining galvanic isolation is easy, since I'm using a Hall Effect sensor for that, but I don't see a similar method for doing that for voltage measurements. My plan was to use a resistor divider and shoving that voltage into the ADC. However, that completely ruins all the effort I put into galvanic isolation of the two rails, since the downscaled voltage output would get into the ADC, causing all kinds of problems. Is there any kind of solution out there that will let me measure that voltage, while still maintaining galvanic isolation? I don't think optocouplers will work, seeing as they're either on or off, and I'm not familiar with any other kind of isolation, and I'm not really familiar with any other kind of isolation techniques, but I'm thinking that perhaps transformers could be useful in this situation? <Q> These are great if you're looking for a monolithic solution that is easy to integrate, but might not be cost-optimized. <S> Some even provide a little isolated (HV-side) power via an internal DC/DC converter to run scaling op-amps, etc. <S> I've had good luck with the Burr Brown ISO series of isolation amplifiers (acquired by TI a few years back). <S> The ISO122 is a general use product, and the ISO124 has higher accuracy. <S> Analog Devices also makes a few general purpose isolation amps . <S> Just be aware that some sort of modulation is required to move the signal across the isolation barrier within an isolation amplifier. <S> Some do it inductively, some optically. <S> Either way, plan on a little ripple on the output signal. <S> The specs usually do a good job of outlining the limitations. <A> If there is any useful power supply on the variable PSU side (i.e. one that won't get turned down to 0.5V) like an auxiliary 5V or 12V, I would move the voltage measurement across the isolation barrier and transmit its results back to the microcontroller. <S> For example, a voltage-frequency converter driving an opto-isolator; the microcontroller measures the frequency. <A> I would definitely suggest using a transformer if space is not a problem. <S> You will get the isolation you require + you can get or make one that can output a secondary voltage to the full scale of the ADC. <A> There are couple of linear opto couplers from Avago - former HP subsidiary. <S> HCNR200 <S> / HCNR201 will probably fit your needs.
| You're looking for an isolation amplifier .
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Is it fairly common to fry a programmer during the learning process of PIC programming development? Background:Despite having a degree in Computer Engineering, working for 2 years as a vb.net developer, and someone that enjoys messing around with electronics. I've narrowed my buying options down to the PICKIT 3 or the ICD 3. Cost differential aside, I'd like to get something that I can get started with quickly, and will 'grow' with me as a PIC developer. As such, I'm heavily leaning towards the ICD 3. I'd rather invest in something good up front, then only wish I bought the better thing later. I've fried a power supply to my computer trying to build something before, so the option to replace the ICD 3 is lucrative to me. Question:Is it fairly common to fry a programmer during the learning process of PIC programming development? Worded differently, and based on your experience, would you agree that the option to replace the programmer outright is valuable? <Q> I damaged the power supply section of the first generation PICkit 2 when I was getting started. <S> Then I looked at the schematic and fixed it. <S> It is important to understand how your tools work. <S> Troubleshooting broken electronics is not something to fear: If you are getting into electronics, this is exactly what you will be spending most of your time doing. <S> The PICkit 2 cut some corners. <S> More advanced programmers will be harder to damage. <S> Still, you should have some idea what you're doing. <A> Cooking programmers can and does happen, especially if your micro is a part of a bigger system that operates at higher power levels (like an industrial controller or a power supply). <S> If you compare the PICkit3 and the ICD3, there's quite a bit of commonality: <S> To me, it's always worth keeping a spare device on hand in case the primary one breaks. <S> PICkit3s are cheap enough to be a backup, as well as a perfectly good programmer and reasonably good debugger if something happens to your 'better' device. <S> (I'm not sure how strong ICD3s are, but at my previous employer we had a bonepile of fried ICD2s ...) <A> I've used both PICkit 3s and ICD 3s. <S> The ICD 3s of course are more expensive (and much faster). <S> The good thing though is the ICD 3s have a lifetime warranty; if you have a problem with one, they include a little test board to verify whether the problem is in the ICD 3 or your circuit. <S> If the test results in an error message, then you can send the ICD 3 in and they will replace it free of charge. <S> I have done this twice in the last year and a half, no questions asked. <A> I've never blown up a programmer. <S> I've had some forward compatibility issues with the earlier Microchip ICD's, getting stuck w/o current OS drivers. <S> After that, I've been an incredibly faithful user of the CCS MachX programmer . <S> It's very useful to be able to slam a dip into a ZIF and program it, and you still retain your ICP ability. <S> If you're leaning toward the more expensive option, you might want to consider that one instead.
| Never had a problem so far with the PICkits, but have fried a couple of ICD 3s.
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Can I reduce rpm further of a gear motor by using PWM? I am working on a project where I need to move the object from one point to another in a dead slow speed , The object doesnt weight more than 5 pounds .I found a DC gear motor with has a RPM of 0.5 but I thinks its still very fast for my requirement . This is the motor : http://www.servocity.com/html/0_5_rpm_gear_motor.html This is the PWM : http://www.electronickits.com/kit/complete/motor/CKMX033.htm Please suggest a way where I can reduce the speed drastically (Either simply using a potentionmeter or any variable device) . motor needs to be small the rest doesnt matter .Or you can propose me a entirely new solution . <Q> What about a leadscrew and stepper motor? <S> A stepper with microstepping will smoothly move an object down the leadscrew as slow as you'd like, with no discernable backlash, and you can get leadscrews up to 3 meters long pretty easily and cheaply. <S> If a leadscrew is too pricey, what about a bog-standard threaded rod? <S> edit: Since you don't want a leadscrew, how about a stepper and a friction-fit drive wheel? <S> Drive the stepper with <S> a Pololu stepper driver , a pulse generator (555, PIC10F, whatever). <S> For a stepper, any 4-lead bipolar stepper will work. <S> A sample one can be found at Allelectronics . <S> The Pololu drivers are easy to use. <S> A hookup diagram is provided on their site. <S> Its worth noting if you do end up using the leadscrew, the lower torque from the stepper will be amplified oodles. <S> The same mechanism can work for the gearmotor as well, just use a coupler. <S> 1/4-20 requires 20 revolutions to move an inch, 3/8-12 requires 12 revolutions to move an inch. <S> 0.5 RPM on the leadscrew means you'll be moving a whole inch every 6 hours :) <S> ================== parts: =================== <S> As for where to get parts:Mcmaster has 12ft 3/8 <S> " ACME leadscrew (true leadscrew) for $16 ... search for 98935A622. <S> A corresponding nut is $2.28: 94815A106 <S> 1/4-20 threaded rod is as cheap as $8: <S> 98957A029 and the nut is probably in your garage somewhere. <A> Using PWM controller will help you vary the motor speed. <S> Simple potentiometer won't help as it won't be able to handle the power unless it is a heavy wire-wound one. <A> There are two ways one can PWM an ordinary permanent magnet motor: one can switch the motor between being connected to voltage and being open-circuit, or one can switch it between being connected to voltage and being short-circuited. <S> The former approach will almost generally cause the motor to end up using less overall supply current, but will not offer very good speed control; speed control is more precise at lower PWM rates. <S> The latter approach may work well, and offer both better control and better energy efficiency if the PWM rate is high enough relative to motor inductance, but not so high to cause significant switching losses. <S> Energy efficiency of the latter approach, however, may be absolutely abysmal at moderately low-PWM rates.
| You can also try various fixed wire-wound resistors to reduce the speed but it will be fixed.
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Would a 100k ohm resistor or motor emit more heat Would a 100k ohm resistor produce more heat than a motor that had the same resistance? +|---------(100k ohms)---------|- +|---------(100k motor)--------|- Both circuits have the same amount of current running. Which produces more heat? Note One of these circuits is converting electrical energy to mechanical energy as well as heat. The exact same amout of current is running in either circuit. The motor is designed to maximize mechanical output. <Q> 100k is a very high resistance for a motor .. <S> The resistor will get hotter, because as you say the motor is transferring energy somewhere else. <S> If you put the motor plus whatever it's driving in a sealed calorimeter, and don't have it driving an energy store (winding up a spring, lifting a weight etc), then the total heat rise in the calorimeter will equal the energy put into the system. <A> In theory, the resistor would produce more heat, given identical power supplied and consumed. <S> The motor converts some of the energy to rotational motion. <S> Excess energy is lost in heat (sub-100% efficiency). <S> However, given a resistor with sufficient power-handling capacity, it might simply stay cooler than the motor. <S> Final answer: it depends. :) <A> The motor produces more heat. <S> Both will have the same I^2R losses but the motor will have core losses, friction and windage losses and some other losses. <S> Edit: <S> Let me explain a little more. <S> I'm going to change the example to 10 ohms resistance for the resistor and the motor. <S> Let's also say each is drawing 10 <S> A current. <S> I^2R losses in each is 10^ <S> 2*10 = 1000 W. <S> The resistor will require a voltage V= <S> I*R = 10 <S> *10 <S> = 100 V to produce that current. <S> If we say the motor is outputting 10 kW of useful power to the shaft and is 90% efficient, then the total input power required will be 10 kW/.90 = <S> 11.11 <S> kW. <S> Given our 10 A assumption, this would mean the voltage required would be 11.11 kW / 10 <S> A = 1111 V. <A> It would really depend on the nature of the motor and the resistor. <S> If you had a resistor design thats meant to generate heat, it would - for example nichrome wire or a incandesent light bulb used as a resistor. <S> It would also depend on the rating of the resistor. <S> Likewise for the motor - it would depend on the design, and what you intend to use this heat for - without knowing your end goal, this is a unanswerable theoratical exercise. <A> The law of conservation of energy Look here: http://en.wikipedia.org/wiki/Conservation_of_energy <S> If two devices are connected in parallel in a DC circuit, and they have the same resistance (and current), they consume the same amount of power. <S> (energy = power x time) <S> Motor then converts part of its energy to motion, the rest of energy is converted to audible noise and heat. <S> But what can a resistor do? <S> It just heats. <S> So a resistor produces more heat. <A> The answer depends on what the motor is doing. <S> If the motor is simply spinning idly at constant speed, then it's generating only heat due to its internal friction, the noise that it generates, and the direct dissipation of heat in its wiring. <S> (The motor can provide some light, too, if it has brushes that generate sparks. <S> But if we count all radiation as heat, we can lump that together with heat.) <S> If the motor is working against a load, it may be storing energy. <S> For instance, suppose the motor turns a winch, winding a cable onto it, such that a weight is lifted against gravity. <S> There is some loss which turns to heat, but considerable energy is being stored by raising the gravitational potential energy of the weight. <S> An idle motor also stores energy in the form of rotational kinetic energy, during the period of time when it accelerates. <S> Once it is at full speed, it then consumes power only to balance the losses of energy, so that it maintains speed.
| A motor will require more voltage to produce that current because in addition to the core losses and friction and windage losses, it probably also has a load connected to it which will require more input power.
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What resistor (or indeed do I even need one)? I have a device which charges a built-in AA battery. The factory AC/DC adapter outputs 200mA at 3v. This adapter has died one me. I have another one lying around with the correct-size plug that can be set to output at 3v (measured 3.5v with my multimeter), but the current output is 1800mA. The device in question is not intelligent - it just sends all the current to the battery, and predictably, the battery gets insanely hot within a matter of minutes. How can I reduce this current output to a max of ~200mA? I assume (likely wrongly) that I need to splice in a resistor on the positive wire. If so, what resistor do I need? <Q> You talk about a single AA cell - about 1.5V, and a 3V adapter. <S> Power would be V*V/R or slightly over 0.25W. <S> I would use an 0.5 or 1W resistor for the purpose, and expect it to get quite warm. <S> it'd just take a little longer to charge. <A> If the device only uses 200mA then it will only take 200mA. <S> The maximum output current is the total the the supply can give. <S> So there is no need for a resistor. <S> But this is not the case in your situation. <A> No device (battery/AC adaptor/step-down transformer) can give 1800mA(1.8A) at 3.5v DC unless it is SMPS in computer which can give 2.5A at 5V. At its peak voltage (say 15v/18v/24v), your another device may give 1.8A <S> but at 3.5v it will give very lesser current like 500mA/700mA. <S> As your device has built-in re-chargeable batteries, you must limit current otherwise, discharged batteries simply short the charging terminals logically, thus eating up all the current which makes batteries hot in minutes. <S> you must add current limiting resistor in series before connecting the batteries to your new device's terminals. <S> Calculate (Ohm's Law): <S> I = <S> V/R <S> (I=charging (required) current in Amps, V=Supply Voltage, R= Series Resistance in Ohms); convert it into : R = V <S> /I = 3.5V/0.2A(200mA <S> ) = 17.5 Ohms next nearest resister you can get is 19 Ohms, 1W resistor. <S> if your battery capacity is 500mA, approximately it will take (500/150mA due to 19 Ohm resistor)3.5 hours to full-charge. <S> divide cell capacity by charging current, result is charging time in hours.
| A problem only arises when the devices current is higher than that of the supply. So you need a resistor capable of dropping about 1.5V at 200ma - that would be 7.5 ohms. Exact resistor value won't be too critical - if I had a 10 ohm resistor handy
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How to complete 3D Export PCB in Altium I know how to export a STEP file in Altium, but is there a way to export the whole PCB, just as it is displayed in Altium's screen? I mean with the traces, vias, components, etc. <Q> No there isn't a way to export that. <S> You only can export the 3D models included in your PCB. <S> There are some tricks for solve this issue but they are really hard and IMHO <S> they don't solve the problem. <S> You can find them in the links below Here are some topics from Altium Design Forum <S> (you must be registered to see them): <S> http://forum.live.altium.com/posts/193722 <S> http://forum.live.altium.com/posts/193565 <S> http://forum.live.altium.com/posts/192662 <A> The best way that I have found that actually yields good results is to upload the design to oshpark.com, they have a script that automatically converts gerber files into a 2D images (front and back) for preview, with all of the silkscreen, pads, and traces in fairly high resolution. <S> The file is generated with purple mask, I usually use Photoshop's replace color function to change the mask to green. <S> Here is a corner of one of my boards: <S> Then I generate the step file from altium and import it into Solidworks and simply add the aforementioned images to the board surfaces. <S> I assume that this would work with any 3D rendering tool that reads step files. <S> It yields really good results. <S> Here is a part of a fully rendered example: <A> There is such Tools- <S> > Legacy Tools- <S> > Legacy 3D View. <S> Running. <S> Click Tools- <S> > Export. <S> There we disable everything except text and silkscreen, otherwise we will never wait for the render to finish. <S> As a result we get STEP of pure silk-screen printing (by the way, it can also unload the trace there). <S> This STEP is loaded to its motherboard, like the usual 3D_Body, directly on top of the board. <A> Take a look at: Module 20: Interfacing to 3D Mechanical CAD <S> It describes how to do this to export to CAD. <A> CADD EDGE sells a Desktop EDA plugin for Altium that provides full integration with SolidWorks. <S> More information can be found here: http://www.caddedge.com/altium/desktop-eda-solidworks-modeler-for-altium-designer/ <A> In the case of the PCB you can make it a colored fiberglass material <S> and then you can add 'lables' to the PCB material <S> ('label' is just lingo for "flat" 2D art). <S> The label could be a simple .png/.jpg export of the traces and/or silkscreen layers.
| If you need a faithful pcb render you could use rendering software such a Keyshot or 3D Studio Max where you can add material properties to each part including the PCB. There isn't a way to export directly with traces etc...
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Power a device from 12v adapter instead of ac wall adapter Specifically, I have an Alarm Clock iPod dock with speakers that gets its power from ac wall adapter, but is only providing 12v. Is there an adapter to go 12V to 12V that will power this? I want to put this in my camper, but run on 12V so I dont need to be plugged into shore power or have engine/generator/inverter running. I already have inverters and a generator, but this is for when dry camping and not running the engine. I have 200 watts of solar, so I am not too worried about running down my house batteries. I would rather use a standard cigarette adapter than hard wiring it to keep it portable. Thanks <Q> If this will be a permanent installation, wire it directly to a 12V tap. <S> If needed temporarily, wire it to a cigarette lighter adapter. <S> Take your barrel connector to RadioShack and you should be able to find it there . <S> You may also pick up some wire and cigarette lighter adapter . <S> Make sure to watch for polarity when you build the adapter. <A> If it needs 12V it needs 12V. <S> You should be able to replace it with anything with similar rating. <S> Check the back of the wall adapter and check the current rating as well as the 12V rating you already know. <S> Even a low current rating might work, but it should be pretty close. <A> Both answers above miss the fact that 12v systems in vehicles are not 12v <S> - they can vary from <10v under cranking to over 14v when the engine is running & the battery is being charged. <S> Some equipment will not care about this, some stuff will be relying on a well-regulated wall-wart to provide a steady 12v. <S> If you don't know which side your device falls, or it <S> does need regulation, the best thing you can do is buy a decent quality regulated in-car adapter / power supply, they're not expensive and are available with interchangeable tips. <S> Heed the advice above about checking polarity, if in doubt, measure it with a multimeter before plugging in!
| You just need to find the right barrel connector for your clock and wire it to your camper 12V battery. Anything with a higher current rating than the wall adapter should work.
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Can I substitute a mono potentiometer for a stereo one? I have a PCB for a headphone amp that has a 6-hole pad for a single potentiometer. I've only got 3-pin ones in my box o' parts. Can I substitute the mono pot for the stereo pot, tie the sets of pads together with jumpers, and achieve the same results? Or should I order the right part? Based on how I'm visualizing it (below), it seems like I should be able to. Am I missing anything conceptually? or is this a good solution. 1 2 3+-------+---+| O O | O | Jumpers between 1-4, 2-5, 3-6 +-----------+| O O | O | <-- Mono Pot here+-----------+ 4 5 6 <Q> Based on how I'm visualizing it (below), it seems like I should be able to. <S> Am I missing anything conceptually? <S> Perhaps you're visualizing that the pots merely affect the control voltages fed to voltage controlled attenuators, one for each channel. <S> However, it is most likely that the pots actually form two separate voltage dividers feed by the analog signals for each channel. <S> If you combine the L and R signals in a single pot, you will no longer have separate L and R signals; each channel will carry a L + R signal. <S> As Kaz points out, it is also possible that you might damage the stages feeding the volume control. <A> You'll turn it into a mono amplifier by doing that. <S> I doubt that comes within your definition of 'the same results'. <A> You can use two pots. <S> Then you have separate left and right volume controls. <S> Or go to an electronics store and spend the dollar or two to get a dual-ganged pot. <S> It is two potentiometers driven by one shaft.
| Your box o' parts has two or more single potentiometers.
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Eliminate hiss and noise ATTiny is generating PWM based tones with 32kHz sampling rate. Then I've created R-2R DAC and then trying to amplify it for the speaker. I need to make low pass RC filter, to cut PWM hiss during playback and some white noise during silence. I have tried to make use of bypass pin but no difference. So where should I place RC filter and what values in order to filter out PWM hiss and noise during silence? Update(without scope, I've tried to describe the problem as much as I can): This is captured on DAC output using sound card, to describe the signal. Enlarge actual .wav file of the capture download: https://www.dropbox.com/s/pwakovmu9h695ji/beep1.wav Since my sound card is not able to capture the white noise I hear from speaker, I've tried to record voice using my phone. http://www.youtube.com/watch?v=MM2n1Q92hIo&feature=youtu.be Sec 22 I turn off speaker pin. Sec 31 turn on again. <Q> The DAC appears as a 10k resistor. <S> What value of C? <S> That depends on the highest frequency you want to generate. <S> I'm going to assume 3dB attenuation at 4kHz <S> is OK, so <S> R = 1/(2*pi*f*C) where R=10000, f=4000, solve for C. <S> I get 4 nF. <S> (Incidentally the electrolytic cap C6 is shown wrong way round, + is connected to approx 0V) <S> I believe you also need to attenuate the input to the amp, it must be clipping at the moment. <S> That attenuation will also reduce the hiss, and allow a second stage of filtering. <S> For example, 20K in series with C6, 10K and 4nf in parallel from pin 3 to ground. <S> Add this if the first filter isn't enough. <A> You have not addressed the fact that you have far, far too much gain. <S> The LM386 has a fixed minimum gain of 20. <S> Your R-2R DAC has a output range of 0V-3.75V <S> With a gain of 20, your theoretical output voltage range would be 0-75V <S> Basically, you have a 2-bit DAC, with only one-bit of effective resolution. <S> You need to use a better amplifier. <S> An more common op-amp, such as a TL081 or similar, would work a lot better. <S> Just use it as a unity-gain buffer. <A> I would think that placing the filter (assuming you're talking about a simple 1-stage RC filter) between your coupling cap (C6) and the opamp input would be best to get rid of the PWM hiss. <S> The values will depend on the filter characteristics you want (i.e. how much you want to attenuate the 32kHz).
| You can use the DAC itself as the R part of the filter, and put a C from the DAC output to ground.
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Connecting 5 pin 3.5MM Stereo Jack Socket to TDA2822M I'm trying to make a portable amplifier based on TDA2822M IC. I've bought several stereo jack sockets but faced the following issue: I don't know the pin-out of these sockets. I'm interested in how i can discover the pin-out using a multimeter or some other way? <Q> One pin is connected to the outer ring, this is nearly always used as ground. <S> The other four pins connect to the two channels. <S> Let's call them L-SIGNAL, L-LSP, R-SIGNAL and R-LSP. <S> Without a plug in the socket, L-SIGNAL and L-LSP are connected, and the R pins are likewise connected. <S> With a plug inserted, L-SIGNAL and R-SIGNAL connect to the plug, and the <S> *-LSP pins are disconnected. <S> Using a multimeter or your eyes you should be able to find out which pins are the R pair and the L pair. <S> The standard use for such connectors is to connect the *-SIGNAL pins to the amplifier, and the *-LSP pins to the internal speaker. <S> This way the speaker gets the signal when no plug is inserted, and only the headphone (or external speaker) gets the signal when a plug is inserted. <S> When you just want to connect a headphone (no internal speaker) you can simply connect both L-SIGNAL and L-LSP to the L output of your amplifier (and likewise for R). <A> Since you bought these, this is a customer support issue. <S> Whoever sold you those should provide you with access to a datasheet. <S> A stereo jack has a minimum of three connections: one for the tip, one for the ring, and one for the sleeve. <S> If there are only three pins, it is a configuration without switching. <S> With five pins, there are various possibilities. <S> Take a look at this chart, courtesy of Switchcraft, which gives a "zoo" of many possible TRS configurations, but is far from exhaustive! <S> http://www.switchcraft.com/Documents/Jack_Schematics.pdf <S> Look at variant XII for instance. <S> The two extra pins here are extra connections to the tip and sleeve pins, which break when a plug is inserted. <S> This is an example of a plug with non-isolated switching: two switches are provided, but they involve signal lines. <S> This has various uses. <S> For instance, it can be used with input jacks, to pull the inputs to the ground when there is no plug, to reduce noise. <S> Or with output jacks to switch from an internal load/speaker to the external line out or headphones. <S> Then there is variant XX: this has a single, isolated switch between the two extra pins. <S> In the diagram, the double bar connection between the sleeve contact and the switch reed represents a non-conductive spacer, which binds the two so they move together. <S> This is an isolated SPST switch with "make" semantics: it closes when a plug is inserted. <S> To reverse-engineer the behavior, first figure out which pins are the tip, ring and sleeve. <S> Then investigate the connectivity to the remaining two "X" and "Y", pins, writing down a chart like this: PLUG IN PLUG <S> OUT <S> T R S Y T R S YX <S> #Y / # / A # mark means that a connection exists. <S> In the above example, X connects to T, and Y connects to R when the plug is inserted, so this is a pair of non-isolated "break" switches. <S> Y of course connects to itself, which is not interesting, so we put a / there. <S> An isolated "make" would look like this. <S> When the plug is in, X connects to Y. PLUG IN PLUG OUT <S> T R S Y T R S YX <S> # <S> Y / / <A> I used to do a lot of electronics dabbling but it's been many years. <S> I'm currently trying to figure out how to install a 3.5mm headphone jack in a Leapfrog toy. <S> I've done it on a different toy years ago but can't remember how I did it which led me here. <S> I know it's many years later now after these posts <S> but I did make a layout showing the pin outs years ago and still have it. <S> Perhaps it may help someone down the road :)
| This would be useful for generating an event indicating a plug insertion or removal, or simply as a power switch for turning an amplifier on when the headphones are plugged in. I don't know the pinout, but you can deduce it either by looking or by measuring. It looks like the same part so it should be right.
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Difference between negative terminal and copper ground? What's the difference between Negative and Ground? Like a negative terminal and a copper ground? I was reading that they are both the same; one person stated that negative is just more negative than ground. What's the difference? Do I even need ground if I have a negative terminal? <Q> Strictly speaking, ground is a connection to the Earth. <S> When we speak about a "safety ground", it is this kind truly earthed connection. <S> In circuits, the "common return" path to the power supply is informally called "ground", even though it is not actually earthed. <S> Battery-powered devices, and the electronics inside airplanes, still have grounds. <S> All circuits have to have return paths to the power supply. <S> Ground is only a convention. <S> The designer decides that a particular network connected to the power supply is at a 0V potential. <S> Other voltages are measured with respect to this potential. <S> But, of course electrons do not read schematics and do not understand what ground is, and what point is 0V. <S> Moreover, conventional current is opposite to the flow of electrons: Benny Franklin got it wrong. <S> So when the negative terminal is grounded, electrons are not actually returning to ground, but emanating from there. <S> If the power supply is a battery, and the devices in the circuit are designed for a positive supply, then the negative terminal serves as the ground. <S> The positive terminal could also be ground, if all the circuits are designed for a negative supply. <S> "Designed for a positive supply" means that the key voltages in the circuit are positive, measured from the negative return. <S> For instance, many NPN transistor circuits, such as an emitter follower, have a positive sense. <S> The collector is connected to a positive supply, and the emitter branch of the circuit is on a network which is goes to a negative ground. <S> The input signal is understood to be between ground and the transistor base. <S> However, the same circuit could be built using a PNP transistor instead, using a negative power supply, whose positive terminal serves as ground. <S> There also exists, in wide use, circuits that run on dual voltage power supplies. <S> The power supply provides a positive voltage, a negative voltage, and a ground. <S> The two voltages are usually approximately equal (but opposite) so the ground is nearly exactly in between them. <A> Ground is a reference and as such can both source and sink current while maintaining it's reference value. <S> Power supplies often can only either source (for a positive rail) or sink (for a negative rail). <S> You can build power supplies that act as virtual grounds. <S> Additionally your circuit may behave correctly using a sinking negative rail. <S> So the correct answer is - <S> it depends, more information is needed. <A> It is often most negative than the other charge levels and therefore often connected to the negative terminal of a battery, but it doesn't have to be this way. <S> You can easily have circuits that have ground (i.e. 0V), positive and negative voltages. <A> In any circuit there can only ever be one point which is truly at 0V ( the - terminal). <S> Since the ground plane has a finite non-zero resistance, any current flowing through it will develop some voltage across it. <S> For example, consider a ground plane that has a resistance of 10mOhm/inch and points A, B, C, <S> -terminal D in line, with currents Ia (20mA), Ib (10mA), & Ic (10mA). <S> Vc = <S> (Ia + Ib + Ic) <S> * 10mOhm = <S> 400uV <S> Vb = <S> Vc + <S> (Ia + Ib) <S> * 10mOhm <S> = <S> 700uV <S> Va = <S> Vb + <S> (Ia) <S> * 10mOhm <S> = <S> 900uV <S> So it looks like the further you get away from the negative terminal in terms of current paths crossed, the greater the voltage drop at that point on the plane.
| Ground is simply referring to a common reference charge level across the circuit.
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Undervolting PWM fan from 12V to 5V What will happen if I feed a 12 Volts rated PWM fan with 5 V driving voltage? Given all the other variables stay the same (PWM control signals) - will it reduce RPM? By how much? What happens to the RPM range window? Will it shift downwards? Will it shrink? <Q> PWM controlled fans have an IC chip inside and one or more hall sensors. <S> When the applied voltage to the fan gets too low the IC chip will fail to operate correctly. <S> I have found in some experiments that I have done that I was able to get the fans to operate OK at voltages down to about 6.2 volts. <S> Below that the fans would not start reliably and the fan motor lacks torque. <S> I would personally recommend that you operate a 12V fan at 12V and then use the PWM feature to regulate the fan speed to the level you require. <A> OK, just took the plunge and re-wired the fan from 12V to 5V. To my surprise it did start no problem, and is going quite strong with more than sufficient torque - unlike another fan (non-PWM 120mm PSU fan), which doesn't start even at 7V by itself and the torque is very faint when eventually started manually. <S> Here are the actual experimental results: Nominal 12V supply. <S> RPM range was, as observed from the fan sensor wire monitoring: 1000-2500. <S> When modded to 5V, the RPM range got down to: 325-780. <S> Not quite what I was hoping for (I'd be happier with at least 1k RPM at the top end), but still acceptable, at least for the winter time and especially considering the tremendously reduced fan noise level. <S> Note: the fan used is the Intel stock CPU HSF. <A> Some fans may spin, others may spin and stall, others may not start at all. <S> These are the risks one takes when they operate devices outside of their specifications - behaviour is not guaranteed. <S> If you want guaranteed performance, build yourself a simple PWM circuit, run the fan with 12V and use the PWM to control the speed.
| Running a PWM controllable fan designed for operation at +12V may run OK at a lower voltage as long as you do not go too low. It's going to vary by fan manufacturer and by fan.
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How can I drive a 12V automotive relay with a uC running on 5V I have a project where I need to engergize a 40A relay in a car. My uC based circuit is running at 5V via a 7805 and is mainly doing its own thing, but I need a digital out to energise the 12V relay, which in turn is driving an air conditioner clutch (as it happens). My thoughts were (discounted in turn, possbly incorrectly)1) a transistor (worried about 5V on base, 12V across CE, and if there's sufficient current)2) an opto-isolator / opto-darlington (worried if it can drive the current, and at 12V if the power dissipation will be too high)3) a relay (feels a bit silly using a relay to energise a relay, but otherwise seems sensible) but even so I'm likely to need more coil current than the uC can drive (or do I?)4) an opto-isolator driving a relay (handles the 5V/12V disconnect nicely and the currents should be high enough) but now I feel like I'm badly overengineering. I should mention that while I (and everyone) says 12V, in a running car it's usually >14V. What would be my best and easiest approach, bearing in mind I'll be building a fair number, so cost is also a consideration...? <Q> Higher-power transistors generally have lower gain, so you almost always end up using a small transistor (2N3904/3906) to drive the big bugger, which in turn drives the relay. <S> For automotive though it is easy to find relays with less than 100mA of coil current (easy to drive) but with contact ratings you'll be looking for to drive the A/C clutch. <S> But transistors to switch big current are passé; why not use a FET? <S> What are the current requirements for your A/C clutch? <S> I disagree about needing an opto; your micro is already sharing the same common as the rest of the vehicle and the inductive kickback <S> would be handled by flyback diode in the FET or with the diode you place across the relay coil; it's highly unlikely that a failure mode would present 12V at sufficient current to damage an I/O pin. <S> If you were nervous enough you could always use a transistor to switch the FET. <A> Update <S> If you think your relay will draw more than 300mA, or there are temperature concerns, or you just want to be on the safe side, you can use a TIP102 (or TIP101 or TIP100) instead of the BC548. <S> The TIP102 is much beefier, and comes in a TO-220 package. <S> And it is still very cheap, ~$.50 US. <S> This will only end up costing a couple dollars at the most, and is probably the cheapest solution. <S> There are a number of different NPN transistors and diodes you can use, if for some reason you can't find these ones. <S> Most uC's can only source about 20mA at 3.3v to 5v, so you need to use a transistor to switch that much power (kind of like using a relay to switch a relay, like you mentioned.) <S> You need the diode because current flowing through the coil of the relay creates a magnetic field which collapses suddenly when the current is switched off. <S> The sudden collapse of the magnetic field induces a brief high voltage across the coil which is very likely to damage the transistors. <S> This is how it would be wired: See Transistor Circuits for more information. <S> And perhaps this Arduino to Relay thread may help (although I haven't read it.) <S> My thoughts were (discounted in turn, possbly incorrectly <S> ) 1) <S> a transistor (worried about 5V on base, 12V across CE, and if there's sufficient current) 2) <S> an opto-isolator / opto-darlington (worried if it can drive the current, and at 12V <S> if the power dissipation will be too high) 3) <S> a relay (feels a bit silly using a relay to energise a relay, but otherwise seems sensible) but even so I'm likely to need more coil current than the uC can drive (or do I?) <S> 4) an opto-isolator driving a relay (handles the 5V/12V disconnect nicely and the currents should be high enough) but now I feel like I'm badly overengineering. <S> Don't worry, that's what transistors do! <S> This isn't really the correct part for the job. <S> Like you said, it is a bit silly, and most uC's can't drive one on their own anyway. <S> Absolutely unnecessary! <A> First question is the coil current of that relay... <S> My suggestion would be an opto-isolator (protect the CPU!) <S> driving a transistor on the 12V side, rated for that coil current. <S> And don't forget the protection diode to protect the transistor. <A> You just need a BJT transistor with a proper design to have it working in saturation mode ( ie acting as a switch ) <S> Here a couple of article, one explaining why you need a diode to protect the transistor: http://electro-am.blogspot.it/2012/12/why-do-we-need-diode-in-reverse.html and the other one to design the component for the BJT driving a relay: http://electro-am.blogspot.it/2012/12/how-to-use-bjt-as-switch.html
| You don't need an opto isolator to protect the CPU, but a diode in reverse parallel to the coil in order to protect the bjt. Logic level FETs are available and capable of switching dozens if not hundreds of Amps at the voltages you're looking for, and appropriately sized would need a modest (if any) heatsink. All you need is a BC548 NPN transistor, and 1N4004 diode.
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What are the important design considerations when laying out a PCB? After filling a bunch of breadboards with some successful experiments, I'm ready to start making PCBs and I'm wondering if there are any design considerations of which I need to be aware? Some PCBs I've seen are spacious with long traces. On others, components are packed like sardines. Before I go packing a bunch of stuff into the minimum space possible on a PCB, are there any considerations apart from the ones listed below of which I need to be aware? DETAIL: Since I'm using toner & cupric acid to etch the boards, I know I can't have crazy thin traces or spaces between them. HEAT: I need to leave ample space for air circulation so I don't fry my circuit with heat buildup. However, I see this more a factor of enclosure (e.g., big enough to allow circulation or a fan if deemed necessary) and heatsink choices. What else should I know? <Q> I'm going to say "Location, location, location". <S> You can make or break a PCB by selecting where the components go. <S> Some (like switches, or transistors that have to be on a heatsink) may have fixed positions; but you can make a huge difference by moving the rest around until they are in the best place. <S> Actually routing the signals is less difficult! <S> Some components call for placement close together to minimise the length of high current paths. <S> That can also make the difference between a nice quiet board and one that fails EMC emissions tests. <S> Some components call for the right orientation to connect to their neighbours without all the tracks crossing over each other. <S> Or enough distance between them to allow all the tracks to pass. <S> And so on. <S> Spend time placing the components, watching the "rats nest" pattern of unrouted traces, and simplifying it as much as you can. <S> Make several trial placements and save them, and route the best. <S> When you get stuck, move the problem parts around and start routing again. <S> It really does take practice. <S> Minor points : <S> if the boards are to be hand assembled, there will be many fewer mistakes if all the ICs and polarised caps are the same way round! <S> if all the components are on the same side of the board, build is easier you can simplify testing if every signal has at least one accessible pad (test pad or existing IC pin or socket pin) on the opposite side of the board... <A> If you're designing low-freq boards (e.g. some MHz or below) <S> I don't think there's much more to consider. <S> I'm assuming it's not high-voltage also, obviously. <S> Keep the space between traces wide enough that you won't accidentally short when soldering (or if it happens, that it's easy to remove the excess). <S> For the traces themselves, there are many trace width calculators on the Internet, like this one . <S> For simple transistor & TH IC boards, I generally use 0.5mm trace-and-space (20 mil). <S> These are almost always enough unless you need high power (in which case you can calculate specifically) and easy to produce good results with thermal transfer - my preferred method for single-side, basic boards. <S> Given that, the other things I consider are: <S> The placement of external connectors, heat sinks, etc. <S> related to the case <S> How easy it'll be to solder and remove any of them <S> How much spare space there is for the traces - and jumpers, if needed How long are the traces on average, and are there higher speed traces unnecessarily long <S> I generally start with a bigger board than needed, use the autorouter to get a sense of where traces can go and how much of the board is filled up, then remove the traces, adjust the relative position of components, reduce the board to what I believe is the minimum, and autoroute again. <S> When I arrive to the minimum board that's still comfortable to work with, I route manually the traces. <A> This is a response to the O.P.'s comment/request for books: <S> IPC standard IPC-2221 Generic Standard for Printed Circuit BoardDesign. <S> 1998 Must read. <S> The 1998 version of the standardcan be found on the wed free of charge. <S> The 1998 version isadequate for introductory purposes. <S> Kraig Mitzner. <S> Complete <S> PCB Design Using OrCad Capture <S> andLayout. <S> 2007 <S> ISBN-13: 978-0750682145 <S> Even through thetitle emphasizes use of OrCAD tools, the first portion of the bookis a general overview of PCB technology. <S> IPC standard IPC-2152 Standard for Determining Current-CarryingCapacity in Printed Board Design. <S> Donna Dearinger. <S> PCB Fabrication for Designers. <S> 2003 Printed Circuit Design, pp.10-11, UP Media Group Inc Dan Beaulieu. <S> Printer Circuit Board Basics. <S> ISBN: <S> 0-9743561-0-7 <S> (4th ed. <S> 2003) <S> Basic processes used forPCB fabrication in the industry.
| The relative placement of the components to minimize trace lengths
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Does appliance wattage differ by country, since the voltage does? As an example, I have an AC-DC "wall wart" that specifies an input voltage of 100-230 volts and an input current of 0.3 amps. Since my mains supply is 230v: Does that mean that it uses 230*0.3=69 watts? If I were to use it in a 110v country, would it use 110*0.3=33 watts instead? <Q> First, it will only "use" what the load demands. <S> Second, you don't know where it's limitation is. <S> However, the maximum power it can put out may very well be limited by something else. <S> In that case, it will only put out up to that maximum power before it will either shut down, drop the output voltage, violate the input current spec, or catch fire. <S> Look more carefully at the nameplate. <S> It probably says something about its output voltage and current. <S> That is what you can rely on as long as you give it input voltage within its specified range, which in this case seems to be to simply plug it in anywhere in the world. <A> Short answer: no. <S> First of all, the way you compute the wattage works with DC, in AC the calculation is different because you need to use the rms value of the voltage, and the phase between current and voltage might not be zero, <S> that adds another corrective term. <S> Anyway, assuming your power supply is switching, and looking at the input voltages allowed I'm quite sure I'm right <S> , the power consumption is practically the same and depends on the power consumption of the utilizer. <S> If you read carefully the tech specs you'll find the maximum output current, that times the output voltage gives you the maximum output power (that's DC now!). <S> The efficiency of a switching PSU is quite high, so that roughly corresponds to the power required. <S> If you want to deepen the AC power computation topic you can find all what you need on wikipedia: AC Power Switching power factor Long story short, the problem is that inside a switching PSU there is a full bridge rectifier and huge electrolytic capacitors, so the current is absorbed only when the input voltage exceeds the capacitor one. <S> This leads to very high current consumption spikes that make the consumed power computation a bit tricky. <A> To a first approximation, no, appliance wattage does not differ with line voltage because the electrical power (wattage) required to do a task at a given rate is independent of the voltage. <S> However, practical implementations for one voltage or another may be mildly more or less efficient, and it may also be the case that dual-voltage designs are more efficient at one voltage than the other. <S> In some extremely crude cases - let's imagine a transformer, rectifier, filter cap, and zener diode regulator - the power draw may reflect the applied voltage more than it does the load. <S> And of course a simple resistive load like a heater would be an extreme example, <S> though there operation at half or twice the intended voltage is unlikely to produce an acceptable result. <S> Also, distribution at higher voltages incurs lower losses. <A> Those are maximum values. <S> You've not said what's on the output DC side; let's say it's max 30V at 1A <S> = <S> 30W. That accounts for the input of 110V 0.3A, plus 10% internal losses. <S> If you then plug it into a 220V supply it will most likely draw 0.15A, resulting in the same power throughput.
| If the true limitation is its input current, then yes, you get less power at a lower input voltage. Many motors can be wired in their terminal box to configuration appropriate for various line voltages.
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Does lacquer coat protect a PCB from ESD damage? Does lacquer coat protect PCB from ESD? It seems that it actually does, because lacquer is isolation and it's a barrier from my body to the circuit. But I'm not sure. <Q> Yes and no. <S> Yes: <S> No: <S> The protection that lacquer offers is not enough. <S> Not even close, actually. <S> The lacquer would protect the PCB from a direct zap to its circuits. <S> The issue is that direct zaps are not usually the problem. <S> The problems that ESD causes are usually from zaps to the cables and connectors (which cannot be covered in lacquer), or due to induced spikes on the wires from nearby ESD zaps. <S> In the latter case, lets say that you touch a door frame and it causes a zap. <S> That zap will radiate lots of RF energy which can get picked up by cables and traces on the PCB. <S> Those cables and traces act as an antenna, receiving the RF energy that the zap creates. <S> If the zap is strong enough, or close enough, it can cause things to not behave correctly. <S> I have seen badly designed PCB's die from a static zap 8+ feet away. <S> Lacquer won't protect from that. <A> When I learned about ESD conformity testing, the most surprising thing for me was the part where you put a metal plate upright next to your equipment under test. <S> You discharge some high voltage load onto the metal plate, and you look how the device under test, some decimeters away from the metal plate, reacts. <S> Note that there is no, really no, direct discharge from the ESD gun onto your gadget-under-test. <S> This means that a coating will not help at all during this part of the test. <S> It's the field alone that may cause some unexpected behavior, like resets being triggered or memory cells being altered, causing your code to do weird things. <S> Note: <S> Industrial electronics and the fried-up configurations you will encounter are fun. <S> Reality is always worse than any EMC standard. <A> Put in simple terms: Not in any way you could rely on. <S> Coating something with an insulator will of course have some effect, but not in any way you could rely on to give ANY useful ESD/EMC protection. <S> That must come from the design of your circuit / device.
| It is possible to cover a PCB with something like lacquer and this would offer some protection.
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Having trouble soldering headphone wires I was fixing my friend’s headphones today and I’m having trouble soldering the tiny wires. I know they have a tiny coating layer that works as an isolation, I have stripped this layer using a gas burner (not sure if that’s the right word, I’m talking about Dremel Versatip, a kind of a lighter on steroids). There was one wire that wasn’t isolated and the solder sticks to it perfectly, flowing all over the wire and creating a nice top layer. However the other wires don’t seem to like the solder at all, it’s far away from the situation on the first wire. What am I doing wrong? <Q> You should use a wire stripper or a pair of pliers to strip insulation from wire. <S> Small wires can be stripped with your bare hands: just pinch the insulation with a fingernail and yank it off. <S> If you catch it just right, the insulation tears and slides off, but the conductor is left intact. <S> This layer will prevent the solder from flowing onto the metal and bonding with it. <S> Even if the plastic was completely vaporized, it's likely that the metal oxidized. <S> Solder does not like oxidized surfaces. <S> (Other than soldering flux. <S> Speaking of which, you should have some and use it!) <S> One of the functions of soldering flux is that, being mildly acidic, when it gets hot, it etches away the thin layer of oxidation from copper. <S> It will probably not eat through the carbonated remains of burnt insulation, however. <S> Still, it is worth a try. <S> If flux doesn't cut through, then just cut a piece off the end of the wire and try again, and this time strip it mechanically rather than, err, pyrognostically. <S> :) <S> Update : if the wire insulation is just a fine lacquer (similar to what is used on wire used for winding voice coils or transformers), it may be possible to solder with the lacquer in place, with better results than torching the lacquer. <S> Some hobbyists spray lacquer on DIY circuit boards, without covering the solder pads, and solder <S> right through it anyway. <S> It may also be possible to remove the lacquer by using a solvent such as acetone. <A> Usually what I do is to hold the iron tip with the solder on the tip of the wire for a while, until I see that the solder melts on the wire. <S> That means that the protection layer is gone, and then I put some new solder on. <S> It takes a few seconds before the coating melts, be careful to have some headroom to avoid melting the rubber coating. <S> After soldering, remember that you removed the insulating layer, and you have to replace it with heat shrinking pipe or something else. <A> You don't use flame for reaching bare metal - what covered the metal will burn and combustion products will coat the surface. <S> You have to mechanically clean the surface by using something like a knife or a piece of finest sandpaper. <A> Methylene Chloride - a chemical that deserves respect - works for some materials. <S> One source suggested using a Bic lighter to make the wire red hot meaning that the insulation is a goner. <A> Some wires have (in addition to insulation) <S> a strand of high-tensile stringwound with the wires, to reduce breakage. <S> If you do not trim back such astring, it might melt, wet the strands, and prevent solder from adhering. <S> Usually, fine strands aren't individually insulated, but if they WERE, onecould use a small solder pot (dipping the wire ends can burn throughpolyester varnish and the varnish acts as flux). <S> Solder Pot
| To solder, you need clean surfaces free of oxidation or other residue. It sounds like by burning the insulation, you have fouled the metal with the products of combustion of the plastic. Transformer manufacturers use some fluid into which to dip the wire ends.
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Using MOSFET to switch device Will my circuit work for the following goal? When INPUT_N is HIGH (i.e. micro is not driving it and it is pulled up to 3.3V), VCC pin on device will have 24V. When the microcontroller drives INPUT_N to be LOW, VCC pin is at 0V (so device is off). Should circuit below work as intended? <Q> No, that won't work. <S> The Mosfet will still conduct through the body diode so the device will receive a small voltage, not the 0V/24V of an on/off switch. <S> However, even if you did get the mosfet to act like a "short circuit", you're shorting the device VCC to ground which is wasting a good amount of power to turn off the device. <S> A better way would be to put the P-channel mosfet in directly inline with the device closer to the 24V source. <S> This would require the pullup voltage to go up to ~24V (minus the turn on threshold) to turn the mosfet off, so you'll most likely have to add some other circuitry to the MCU pin <S> so it doesn't fry. <S> A potential solution might be to use a second N-Channel mosfet to drive the P-channel mosfet gate pin to ground, then use a pullup resistor to the 24V rail. <S> Something like this <S> : VG1 is directly driven by the microcontroller (high voltage is on, low voltage is off). <S> R2 is the load circuitry. <S> T1 is a P-channel mosfet and T2 is an N-channel mosfet. <S> R1 is a pull-up resistor to the 24V rail. <S> There might be a simpler way to do it, this was just the first thing I thought of. <S> edit: If you can have a low-side switch, the switching circuitry will be much simpler: just a single N-channel mosfet with the drain tied to the load, the source tied to ground, and the gate tied to the microcontroller with high voltage being on and low voltage being off. <A> Usually a P-Channel mosfet is used as what is refered to as a " highside " switch. <S> That is the Source is tied to VCC, and the drain is connected to a resistor to gnd. <S> Your load is connected between the drain and the resistor to gnd. <S> This allows you to pull up VGS to vcc when the unit is off. <S> This allows you to drive VGS negative by appling 0V on the gate to turn the FET on <S> The more negative VGS becomes, the lower the internal resistance of the FET and the more current passes through the FET and into your device! <A> You are using the transistor in a wrong polarity. <S> You could better use ULN2003 series IC to drive this kind of load. <S> You might need a level translator to drive from 3.3v. <S> http://www.ti.com/product/uln2003a
| P-channel mosfets require the Gate voltage to be lower than the source voltage by some threshold in order to "turn on".
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Are there any alternatives for 74LS__ chips? A beginners electronics book tells me to get the following:74LS92 counter chip, 74LS06 open-collector inverter chip, and 74LS27 triple-input NOR chip. Now, this book is from a couple years back, and I see that these chips are quite expensive on eBay (I have to pay $6 for 2x the first one, e.g.) So I'm wondering, are there any alternatives to these that would be cheaper/more readily available? How could I know this? <Q> They have similar performance, the same I/O specs, but are implemented in CMOS instead of bipolar technology. <S> 74HC chips are similar, except that they use CMOS I/ <S> O specs, which means that they are compatible with other CMOS (e.g., 4000-series) chips. <S> CMOS and TTL chips can talk to each other under most circumstances, but you need to pay a little more attention to the details. <A> Then, are you sure that eBay is the best source of these chips? <S> In our local store:74LS92 €1.0074LS06 <S> €0.60 <S> You should specify your location a bit. <S> In Europe we use for example Farnell, but there are other good eshops as well. <A> Checking Octopart.com, large distributors offer 74LS92 for ~$2.50 each. <S> And 74LS27 for less than $1. <S> These are reasonable prices when you are asking them to box and ship onesy-twosy parts. <S> If you want to learn with discrete chips (which isn't a bad idea) these prices are fair. <S> If you're willing to skip building things out of individual gates, you can get a demo board with a CPLD or FPGA and jump straight to designing in Verilog or VHDL. <S> You might pay more than $50 for the board, but you will be able to re-use it for many different types of projects. <S> The main drawback will be you will miss the experience of building things with physical chips representing individual gates. <S> Depending on your learning style, this could end up being no problem, or a major barrier to understanding.
| A more modern replacement for 74LS chips would be 74HCT chips.
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Is there any arduino module to control water flow? I want to build my super cool auto flower water gadget. I'm looking for some arduino module like valve or water gate, which can be controlled by input power. So, it can open or close water flow from some bottle by a signal. Any ideas? <Q> Do you need to control only the on/off flow of water? <S> Do you need control over variable flow? <S> What amount of flow and pressure are you talking about? <S> As you can see, there are a lot of additional questions that would shape an answer. <S> From a bottle suggests fairly low flow and pressure. <S> I am not really familiar with something that size (maybe a drip irrigation system valve); but I have controlled solenoids before with power transistors and/or relays. <S> Solenoids such as what you'd find on a Rainbird irrigation valve . <S> These are often 24VAC and pull somewhere from 125 to 250 mA (depending on model). <S> (I've also used these successfully with small air cannons like for confetti throwers.) <S> If you connect the valve with some PVC pipe to your bottle you should be able to control the on/off state easily. <A> You can use a 5v logic level mosfet + a solenoid(google valves4projects) <S> But you will also need to provide a higher voltage power source for the solenoid. <S> This can come from any dc source, I would just use a higher voltage acdc wall adapter. <A> Due to the date of post this is more for others who come across this question in search of a solution for a similar project. <S> Most sprinkler related valves require a high pressure system similar to what your city water supply has (~60-90 PSI). <S> To open the valve requires both a "high" pressure source of water and the solenoid to be activated. <S> If you are connecting to a garden hose this is fine. <S> If not this is no easy feat. <S> I used a 390 GPM pump and couldn't open the valve. <S> If you are connected to a 2L bottle or rain barrel you'll need something designed for low pressure. <S> Look for a gravity fed solenoid valve instead. <S> Here the maximum pressure is around 7ft of head with water (or 3 PSI) <S> and there is typically no minimum pressure. <S> I ended up using one from valves4projects (suggested by earlier answer) and was happy with the results.
| If you need the ability to control the flow in a variable way, you will probably need to look for some kind of a stepper-motor or servo controlled valve.
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Electricity theft from radar transmissions I recollect reading in a newspaper in england, maybe 20 years ago, that someone had been prosecuted for stealing electricity from a radar. This was detected because the air defense radar had a black spot in its propagation field and apparently he hooked up a load of wires in his roof. I an certain I read this, but it seems physically impossible. Is this false memory syndrome or could it have happened? Could it happen with a microwave link if they pass over head? Thanks – dave <Q> Most radars transmit high power pulses but have low duty cycles so the average power is not very high. <S> They also generally transmit over a narrow beam that is rotated either mechanically or electronically to cover a full circle. <S> Also the power in the beam falls dramatically with range. <S> Based on those three features, the average amount of power that intercepts a single roof of a house would not be very much. <S> In any event, one would then have to efficiently receive this power and convert it to low frequency AC in order to use it for powering one's household. <S> I can't give more quantitative answers without knowing the radar parameters and the distance between the radar and the house, but I seriously doubt the story. <S> Additionally, air defense radars by definition are designed to survey the sky and not the ground. <S> Thus the radar beam would not be directed low enough to intercept a house other than by sidelobes which are much lower in amplitude than the main beam. <A> As Barry said, a house is so low that it shouldn't be in the way of the radar beam in the first place. <S> However, even if it was this would still be nonsense. <S> First, enough radar power is normally transmitted thru the house that harvesting it would be meaningful, it would cause a health threat to those living there. <S> Second, what if the house happened to have a aluminum roof? <S> That would reflect the radar instead of absorbing it, but either way it wouldn't serve its supposed intended purpose. <S> If this myth were true, then aluminum roofs would have to be outlawed too. <A> I asked a friend of mine who served at a USAF radar installation where they used over the horizon ICBM detection radar. <S> We're talking MW output here. <S> He was saying that close to the dish site there were lines on poles etc. <S> that said "do not go higher than this when dish active" and in the offices/barracks, if someone left the lid on a pot/kettle that the pot would buzz every-time the sweep would go by. <S> So that might give you a lead to go look at to verify independently. <S> I'd say it's possible, but the circumstances would need to be particular. <S> Here is a link to WHO site giving some context for energy levels. <S> http://www.who.int/peh-emf/publications/facts/fs226/en/ Aircraft traffic control is ~ 100 KW.and here is a snip from that site. <S> However, because its power is radiated over a large surface area, the power densities associated with these systems vary between 10 and 100 W/m2 within the site boundary. <S> Outside the site boundary RF field levels are usually unmeasurable without using sophisticated equipment. <S> However, small military fire control radars on aircraft can be hazardous to ground personnel. <A> Not very much and not very usable. <S> All you need to be able to get some energy is an antenna. <S> Its much like the old crystal radios, it is certainly possible to get some energy from RADAR. <S> It will, as others have stated, however only be when they transmit in your direction and it will probably only be very little power. <S> I don't see why one would be prosecuted because of this for another reason than breaking some kind of building regulating law. <A> Perhaps you're looking for a slightly different story/rumor. <S> I heard it in the 1970s like so: a farmer in the UK discovered that his wire fences were electrically "hot," so he connected up an electric heater and was using it to heat his outbuilding. <S> It turned out that he was kilometers from a BBC antenna farm, and his long-fences heater was punching a hole in the antenna pattern for the "World Service" AM transmissions being beamed at Europe. <S> In the story this was not illegal, so the BBC cut a deal. <S> They paid to have power lines run out to his remote location. <S> Big BBC transmitter is Rampisham, out in Dorset farmlands. <S> Unless there are other megawatt World Service sites, that's probably the one. <S> (If the story happened at all, I mean.)
| You'd need to be able to get close to a very high output Radar system, which are typically remote or limited access. It is definitely possible to take some of the energy transmitted to the house.
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How can create a simple circuit that will drain a battery in a few hours My son and I are working on his school project. He wants to examine the effect of temperature on battery performance (longevity). He thought he could just put an LED on a battery and watch. We soon discovered that there is more complexity than we would have thought. First we just put a lead from the LED on each side of an AA 1.5 v battery. Nothing happened. Then we tried two AA in series and the LED came on and promptly blew. Then we put each side of a LED on a CR2032 3v lithium battery and all worked well. It worked too well. More than 24 hours later the light was not dim. We then set up a breadboard with a vibrating motor in series with the LED. The LED went out after about 60 seconds and the motor kept on spinning but slower than it had at first. Now I have set up the LED in series with a 1000 ohm resistor (I think). Will the LED light go out any sooner or will it just burn dimmer for longer? I have thought of getting incandescent bulbs or simple old-fashioned flashlights. We want to setup the circuit in three temperatures and replicate the observations 3 times. <Q> It'll just burn dimmer for longer that way. <S> What you need to do is pull more current out of it. <S> First I would get a cheap voltmeter you can probably find one for $15 or less. <S> Then if you look at this datasheet for an energizer AAA battery you'll see they have a discharge chart in there. <S> So if you want to kill a battery in less than 10 hours you'll need to suck out around 100mA of constant current. <S> Starting at 1.5V that would mean you'd need about a 15 Ohm resistor. <S> It'll need to be able to dissipate that power <S> so P = <S> 15Ohms <S> * 100mA^2 is about 150mW of power. <S> So look for a 15 Ohm quater watt resistor. <S> As the voltage goes down so will the current draw <S> but that should be fine for this experiment. <S> Just take a voltage measurement every hour or after 10 hours, it's up to you to decide what voltage will mean it's "dead". <S> Don't just short the battery though that could be dangerous. <S> If you want to do the LED thing because it looks more exciting :) <S> You just need to pick a lower resistor value and maybe a higher power or more LEDs in parallel (with current limiting resistors). <S> The reason you blew the LED the first time was you pushed too much current through it, and the reason it didn't work with one Battery is because there's a voltage called the forward voltage for the LED that must be met before current will flow. <S> You can find it in the LED's datasheet along with the max current the LED can use and use that to calculate the lowest value of current limiting resistor you can use. <S> Here's another helpful chart for you to use from that datasheet. <S> That gives you and idea of what to expect will happen to the battery over temperature. <A> A typical LED needs around 2 volts to light but it also needs a means of limiting the current. <S> When you put two AA batteries in series (3 volts) with the LED, it lit up but there was nothing to limit the current (other than the internal resistance of batteries which is very low) and hence it blew. <S> With a 3 volt lithium battery, there is enough voltage to turn on the LED but since the lithium battery has a higher internal resistance, the LED current did not reach the burn-out level. <S> With a 1000 ohm resistor in series, the current is much less, the LED will glow dimmer, and it will last much longer. <S> Plotting a graph of voltage vs time at different temperatures would make a good presentation. <S> You don't even have to wait for the battery to be depleted to see a difference. <S> Voltmeters can be purchased for only a few dollars <S> so I recommend this approach as it is much more quantitative than just observing the brightness of an LED. <A> Battries capacity is defined by mAH so basicly a 1000mAH (mAH = milliamp hours) battery. <S> To figure out the amount of run time you have you need to divide the amount of current your system draws with the capacity of the battery. <S> For example if your LED with its series resistor draws 10mA, than the battery will run for 1000mAH/10mA <S> = 100Hours. <S> If you increase the current draw by either lessening the series resistance, or putting a resistor in parallal than your run time will adjust accordingly.
| If you had a voltmeter to measure the battery voltage over time (say every hour), you would be able to get a much better picture of battery performance. When you put one AA battery (1.5 volts) in series with the LED, the voltage was below the LED turn on voltage and nothing happened.
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Is this desoldering hack likely to work? I found an interesting YouTube video with a neat de-soldering trick that requires no special equipment. Essentially you bend a reasonably thick piece of copper wire round into a right-angled S shape, add a layer of solder over the bottom edges, then place it over the chip. By heating the copper from the top, it melts the solder and provides a near perfect coverage of the IC's pins. From there you can just lift it off the board with tweezers. Is this likely to work without damaging the IC or the board's pads, as described in the video? <Q> It can work, it's not as easy as it makes it seem. <S> You can have issues with the copper wire getting stuck to a pin here <S> and there, or not being able to get the wire hot enough to actually melt the solder, since it takes a lot of heat. <S> Probably not enough to damage the chip unless you leave it there for too long. <S> I wouldn't recommend using this method. <S> Essentially you're pulling the wire between the pin and the PCB's pad, and just a bit of heat allows the wire to push the pin up and separate it from the solder. <S> Quick easy and harmless! <S> It helps if you have one side of the wire stuck or tied to something, that way you only need to pull on one side (it's a bit easier since one hand is using the soldering iron.) <A> Is it likely to work? <S> cant say for certain untill i have tried it. <S> But as far as i know the Solder wick is made of copper and is rather easier to use. <S> As for Damaging the parts(mico-chips): my suggestion would be, if you are starting out, always have some spare parts and learn how to use the Solder Wick. <S> The picture shown above might prove to be right, the solder will stick to the copper, and the chip will also <S> BUT then you have to figure out how to remove chip from the copper wire (even if its not stuck to the board anymore). <S> Also check out this question/answer: How to tell if chips overheat while soldering <A> Given that it mimics some standard desoldering tools, If you are careful, I see no reason why not. <S> From the Hakko website :
| A much easier/better/safer method to remove surface mount ICs is to take a very thin wire (like magnet wire) and thread it under all the pins on one side, then pull the wire while briefly touching the pin with a soldering iron. Also, this method would probably leave the PCB's pads and the chips pins messy, so you would still need to clean them.
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Why is the output of stateful elements often named Q? In logic circuit diagrams, I've seen various conventions for naming inputs and outputs of logic gates and combinatorial circuits. However, stateful elements like latches and flip-flops often have their "state" called Q. I suspect there is a connection with abstract Finite-State Machines from theoretical computer science, where "state" is often noted Q as well (so I asked them too :-) But why have people picked this particular letter ? <Q> Alan Turing used the letter <S> q to denote states in what came to be known as Turing machines. <S> Presumably the q stood for quanta, emphasizing a state's discrete rather than continuous nature. <S> This happened in the 30s when quantum theory was permeating the scientific æther. <A> So a quick trip through the USPTO database is revealing. <S> This is only a partial answer, and the answer will be much harder to find, simply from observing the differences in what is DOCUMENTED you see a lot disparity in usage. <S> i.e <S> the terminology is not uniformly applied. <S> I traced back through Integrated semiconductor solutions to discreet and even tube systems. <S> Hughes pat # 2903606 issued '59, Filed in '55 discusses a JK FF using the J,K and Q, /Q notation. <S> Computer research corp. <S> pat # 2644887 issued '53 filed in '50 speaks to FF's, and uses A,B, <S> C input terminology for the logic. <S> But does NOT use Q and /Q designing counters. <S> Column <S> 13 lines 1 and 2 talks to a "1" and a "0" for logical states. <S> Monroe calculating machine pat <S> #2603746 issued '52, filed '50 uses tubes and the terminology of a & b as inputs and implements 1bit adder and subtractor. <S> Using carry etc. <S> Column 8 lines 56 ff talks to again logic levels as "1' and "0 <S> " <S> There were lots of later ones in IC's etc. <S> but these are early and implemented using discreet components. <S> It is very clear that the terminology predates IC's. <S> It is also very clear that the terminology is inconsistently used across inventions. <S> The "1" and "0" notation almost certainly comes from earlier work, I would conjecture the work of Boole might reference that. <S> attached is a list of patents I looked at, if someone else wants to look at them and follow the threads into even earlier ones. <S> I only followed one thread through this. <A> Not a definitive answer, but the first flipflops had two inputs, to Set and Reset them respectively; another early type had a single Toggle input. <S> That conveniently allocated the letters R,S,T for input signals, so I guess the choice was between Q and U for the output! <A> George Mealy (1955) used Q to design the "present state" S.C. Kleene (1951) <S> Used q1 ... <S> qn to represent each state, but he use /q/ <S> because /p/ was already used. <S> Previous works related with finite states machines, are based on brain behaviour so the theory of "status quo" is reforced. <A> Maybe Q is used because it looks similar to 'O' (for O utput) but it cannot be confused with digit 0 (Zero). <A> I was under the impression it had to do with the q-point (quiescent point) of a transistor, and that Q started referring to all transistor based components somehow. <A> Because simply if output is designated by O , and output of the flipflop is 0, then it would get confusing , that's why.
| The letter /Q/ is used to design the set of states that the automata can be in a specific moment, so /Q/ is also used to design the "status quo" (Lat: "the state in which").
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Can I draw power for a component while this charger is charging the battery? So I am working on a design for battery-heated wrist warmers. My mother has Reynauld's and arthritis, and as winter approaches it gets much worse. My basic design is that I will take a form-fitting sleeve (made from a shirt or tube sock), weave nichrome wire through the sleeve, wrap thinsulate or space blanket around that, and then fix a nice-looking knitted wrist warmer over it. At the very least I think it would be a good start even without electric heating. For the electric heating, I am thinking of using a 1000mA LiPo battery , a button switch, and a USB charger . Ideally, the charger will charge the battery while also providing heat. My specific question is: can the charger I linked to provide that? In a more general sense, I would appreciate any suggestions you might have! Thanks, and happy holidays! <Q> LiPo batteries require complex charging patterns to charge the battery quickly and properly. <S> The IC used in that package goes through a preconditioning, fast charge, and then constant voltage mode before it considers charging complete. <S> Drawing power from the LiPo while charging will interfere with this charging mechanism. <S> It may work, but it will abuse the LiPo and probably affect its life. <S> It may overheat. <S> Normally the charging circuit will power the system directly while also charging the battery. <S> This can be implemented connecting <S> Vcharger -> <S> diode D1 - <S> > Vin, and Vbatt -> diode <S> D2 -> <S> Vin. <S> When charging Vcharger > Vbatt, so D2 will be reversed biased and the battery effectively disconnected. <S> When there is no charger, D1 prevents the battery from energizing the charging circuit which is obviously not a good idea. <S> If you are concerned about the voltage loss due to the diodes, the concept can be extended such that the diodes control the gates of power MOSFETs instead of directly passing the current. <A> TI and other IC manufacturers offer ICs that can handle both powring the system and charging the battery at the same time (after all, how can your cell phone stay on while being charged if it will affect the battery). <S> Look at the Battery Selection Tool <S> It's important to note that you will need enough current going through the charger IC for both battery charging and powering the system. <A> It can be considered a serious safety concern to be using a LiPo type battery when it is being charged. <S> You need to consider that wires can fray and short when used in an application where there is constant flexing going on. <S> The safety circuitry could also do well to detect temperature and ensure that conditions do not exist that could burn the wearers wrists.
| You should also seriously consider that your warmer should include some safety circuitry to prevent uncontrolled current flow from the battery into the heater.
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Problem with high voltage in a project I have a circuit that should be protected from high voltage. Is there any component other than a fuse that I don't have to change every time? Something like a switch that activates only when high voltage is supplied, diverts the flow and grounds it, and then closes. Sorry, I can't scan the schematic right now but I can give you information. I have a triac connected to a submersible motor via a MCU IC mounted on PCBA. The problem is sometimes there is a short circuit due to water. I want to know how to protect the circuit in this case as I already have fried it twice. Is an optocoupler or solid state relay advisable for isolation? <Q> So in general the way I do this is that I have a circuit which looks something like this: <S> So essentially, the idea is that there is an input power mosfet which controls the flow of power to your circuit. <S> You push a button, which causes the gate voltage to rise, and power flows to your circuit. <S> " <S> VDD" in this diagram is the battery positive rail. <S> Ground is assumed to be available and shared all over this circuit. <S> The "small gap" is a physical one. <S> If water connects ground and the wire coming off of that op-amp, then the op-amp will output low, and you'll suddenly lose power. <S> Now this circuit needs a lot of things, it's only meant to be the beginnings of a circuit that you could use. <S> But it should push you in the right general direction. <S> The robotics team I was on built an underwater vehicle and I can tell you for a fact that the idea behind this circuit is tried and true... <S> the schematic I have for that part of the craft though was a bit more complex because it was dealing with a lot of other parts, so you get the simplified but incomplete version for now! <A> In a comment you say you're worried about shorts between live wires and ground, causing high currents. <S> But in your question you say you maybe want a switch that activates only when high voltage is supplied, diverts the flow and grounds it, and then closes? <S> This is unlikely to work because when there's a short from a power net to ground, it tends to cause the voltage on the power net to fall (due to resistance of the power wires), not rise. <S> In any case, the behavior you ask for is essentially a short from power to ground. <S> It's not clear to me how shorting power to ground would protect you from shorts between power and ground. <S> However if I've misunderstood your requirements and you really do want this behavior, what you're looking for is called a crowbar circuit. <S> other options <S> One thing that might help you is a polyfuse or resettable fuse. <S> These work like fuses, in that they increase their resistance dramatically when current rises too high, typically shutting down the load. <S> Their advantage over a traditional fuse is that they revert to normal operation after cooling down, so they don't have to be physically replaced every time they operate. <S> However like fuses, they may take tenths of a second or seconds to operate, which could be too long to avoid damage for some sensitive equipment. <S> Linear Technology, for one, offers several regulator ICs with an overcurrent protection feature. <A> Unfortunately I would say "you're doing it wrong" - if there are common shorts to ground that blow your kit up, you need to design the drive circuit to sense & shut-down, not just to <S> not explode . <S> This could include feedback, passive current limiting (NTC / inrush limiter), active current limiting/load-sensing, fail-safe principles, and <S> some sort of lockout (is the short when someone is accessing the connections, can the operation cause some feedback that kills the circuit while the user is fiddling?). <S> This is ignoring the safety issues around whatever the heck you're doing with power & water, as I'm guessing this project is not going into commercial production. <S> Some important distinctions: <S> Fuses protect from over-current. <S> Tranzorbs/MOV's throw themselves across the rails when a large voltage comes by, but may or may not survive repeated hits.
| Another option would be to modify your power supply to include overcurrent protection. Fuses are not really meant to protect your equipment, just to stop it catching fire and damaging things around it.
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Ethernet Without a Switch? How do you wire up Ethernet electrically without a switch? An approximately equivalent phrasing of this question would be what is going on electrically inside an Ethernet Hub? To clarify, I'm not trying to connect two computers here. I'm thinking specifically of creating a small network of microcontrollers on a single board. Can it be done without a switch or a hub through just electrical connections. This would be for 10BaseT or at most 100BaseT. <Q> Proper hubs are active devices. <S> They decode the physical layer and listen for incoming frames. <S> When data is detected on one of the ports it's repeated to all of the other ports. <S> When a collision is detected a "jam signal" is output from all of the ports to ensure the whole network sees the collision. <S> There can be a mixture of port types (older hubs often had an AUI port and/or a 10BASE2 port in addition to the 10BASE-T ports. <S> A pure hub can only operate at a single speed at a time. <S> A device that wants to support multiple speeds needs at least some level of bridge functionality. <S> Some hubs had a physical switch to select between 10 and 100 megabit. <S> Some hubs were essentially two independent hubs, one for each speed, possiblly with a two port bridge to connect them. <S> Note that 802.3 doesn't use the term "hub". <S> What the market commonly calls a hub is regarded by the standard as a multiport repeater. <S> Similarly what the market calls a switch is a "mutiport bridge" per the standards. <S> You can read further details of what a repeater is required to do in IEE 802.3 . <S> Bridges are covered in IEEE 802.1D <S> You can get the 802 series standards free of charge from http://standards.ieee.org/about/get/802/802.3.html . <S> Note however there is a 6 month time lag <S> so if you REALLY need the latest version you may have to pay. <S> Someone did hack together a 3 port passive hub for 10base-T and apparently got it to work. <S> http://www.zen22142.zen.co.uk/Circuits/Interface/pethhub.htm <S> It's definitely not a method approved by the standard though <S> and I don't think there is any reasonable way to extend it beyond 3 devices. <S> If your devices have Mii ports it may be possible to program a CPLD to act as a hub. <S> I don't know how much behaviour you would have to simulate to make that work though. <A> Use a cross-cable if it is only two devices and if it doesn't work with a straight cable. <S> It basically swaps transmit and receive pairs, but many modern interfaces are auto-cross. <A> This often also requires disabling cable autodetection and things on the communicating device side for modern cards. <S> Ethernet 100baseT and anything even relatively modern requires an actual switch even for "hubs" - a microcontroller and each port controlled separately. <A> For a small network on a single board, the easiest solution IMO is a switch ic. <S> At 10/100 these are easy to use, and often let one device be the MAC of a microcontroller, with the others being full ethernet devices. <S> Micrel is an obvious choice of manufacturer. <S> You tend to be able to pin-strap these devices to wake up as an ordinary unmanaged switch, which will work exactly like a cheap office switch. <S> If you want more control or diagnostics then you can talk to them from a micro.
| Ethernet 10baseT can work with a pure hub: just connecting all the wires together suitably, with very minimal electronics.
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Best way to control 130 - 140 LEDS with Arduino Uno 32/64 bit led drivers? I am new to electronic prototyping and had a few questions. I need to control 132 LEDs individually. I want to make 3 rows ( 2 rows of 60 and 1 row of 12) I think i can matrix them but i would still need 63 outputs (3 high, 60 long). I don't want to use 8 different shift registers. Are there any 32 bit or 64 bit led drivers(or shift registers, not really sure what there are called)And where can i buy them. Any help is appreciated . <Q> This precisely matches your application requirement at least on the LED count. <S> AMS has a versatile set of LED drivers: For instance their 144 LED driver ( AS1119 ) with its integrated charge pump for driving LEDs from a power rail lower than the forward voltage of your LED, might be of interest for your application. <S> To answer your specific questions: Yes, AMS has both 32 channel and 64 channel constant-current LED drivers, besides the 132 channel and 144 channel ones. <S> You won't even need current limiting resistors with such parts. <S> Other vendors appear to top off at 48-channel LED drivers, e.g. the Linear Tech LT8500 , from a cursory search. <S> You can buy the AMS parts mentioned above from their web site, on the LED drivers page. <S> Also, they have a generous sampling program, so you could try that route for your initial experiments. <S> The challenge you might face is on ready-to-use library support on your microcontroller or platform of choice, for the various LED drivers. <S> For this reason, my go-to part for similar requirements has been the merely 16-channel Texas Instruments TLC5940 , in cascaded arrays, simply because libraries and general public-friendly write-ups exist on the web for it, for almost any microcontroller platform I have wanted to use with them - including for the Arduino platform . <A> If you Charlieplex the LED's, you would only need 12 I/ <S> O pins to control all 132. <S> (\$12^2-12\$) <S> The advantage of a Charlieplex is the reduced pin count, but there are several drawbacks. <S> The main drawback is that you can only light one LED at a time (not quite true, but for practical reasons this is so). <S> Another drawback is that you can't drive the LED with more current than the microcontroller can source. <S> With the Atmel MCU's I use, this tends to be 40mA. <S> If you need to create a persistence of vision illusion of multiple LED's being lit at a time, the inability to drive LED's with more current can be an unacceptable limitation. <S> You can also Multiplex the LED's, as you said in a matrix; however I think you can do this with 24 I/ <S> O pins (12 x 12 grid = 144). <S> Obviously the physical arrangement of LED's would still be 2x60 and 1x12, but the electrical arrangement could take advantage of a 12x12 multiplex. <S> Edit: <S> You could also get by with 23 I/ <S> O pins multiplexing, as 11x12 <S> = 132. <A> As @DaveTweed says, you don't have to match physical to electrical layout. <S> Multiplex/charlieplex is an option, you have enough pins. <S> O expander chips (such as MCP23017). <S> You could also use shift-registers such as 74HC595 to convert a serial stream to a load of directly driven pins.
| You could get a driver chip or several to do it for you (like MAX7219), or just use I2C/SPI I/ There is actually at least one 132-LED driver IC, the AS1130 132 LED driver with PWM ($2.99).
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Can an old pentium III processor still be useful? I have an old PC around the house, and i have no use for it at the moment. I was thinking (in the good old DIY (AKA Destroy It Yourself XD)) about scavaging the computer for whatever parts i could re-use. I have an arduino UNO, and often I find myself needing more sampling rate or faster computation and I thought that using that old pc's Pentium III processor might be a good idea. However, since I'm not that familiar with processors (only with arduino) I was looking for some pointer on how to make use of it. A few ideas I came with are: An oscilloscope A general-purpose meter (think home automation) If you could give me some pointers about the feasibility (or even how to make something) out of those parts, much appreciated. Oh, and if you have any ideas for projects (preferably space-related since it's my major) please share them with me. <Q> It would be very difficult to use. <S> It's a processor and not a microcontroller, thus you will have to provide everything : <S> external ram (is the ram controller inside on PIII? ) <S> DDR are not that easy to implement. <S> Power supplies. <S> I am sure it needs several voltages and you will have to follow a precise power up sequence. <S> Chipset? <S> how do you interface something with the high speed bus of the PIII? <S> There is no connections for FLASH, IOs, USB or something. <S> Theses are generally done by the chipset. <S> You could replace it by a high speed FPGA, but you need to have access to the PIII datasheet and bus specification to know how to interface with it. <S> etc... <S> It is feasible, but it's very complex. <S> In a personal scale from the easiest to the hardest : <S> HERE, EASIEST PROJECT <S> EVER <S> Do something with an Arduino <S> Do something with a Raspberry Pi Design your own circuit and PCB <S> (not a clone of something found on the web) <S> Use a recent (not PIC16F84) microcontroller without libraries. <S> Just bear metal programming using only the MCU datasheet and a good compiler <S> Do some tricky analog design <S> Do some simple FPGA design (including the board) Design a processor platform from scratch (a new kind of raspberry pi) Design into the RF world <S> Do advanced FPGA design <S> Try to use PC world busses from the low level <S> (PCI PCIe BT USB2.0 ...) <S> Design a PC <S> MOST DIFFICULT <S> Trying to do a kind of chipset to interface with a high speed bus of a PIII belongs to the end of my previous list. <A> If you still have the "old computer" in bootable condition use it in its entirety. <S> Learn about MSDOS or FreeDOS and boot it up in that simple environment. <S> Learn how programming used to be done on early PCs from times gone by <S> and you can make strong use of your old computer chip. <S> As others have said...it is hard to try to make your own motherboard for using old microprocessor chips. <S> Even finding some of the necessary components to complement the old processor may be nearly impossible. <S> If your old computer is junk and not bootable then your best bet would be to junk it. <S> Go around to garage sales and check places like Craig's list <S> and you can find plenty of old computers that you may find in still bootable condition that can be had for very low cost. <S> Spending very much money on it is extremely questionable because if you need more performance for some little project than your Arduino UNO can provide <S> then there are plenty of other ready made "development" boards around that can meet that need far better than trying to re-use an old Pentium <S> The old Pentium may be much better used as a historical display like the one I made with magnets glued on it so it can stick to a metal surface... <A> To make the best use of it... <S> Format the hard drive, load a legal OS, browser, etc, and donate it to a charity that can get it into the hands of someone who needs it. <S> Take a $100 tax deduction and use this to feed your electronics addiction. <A> At least you will need to interface to RAM and a chipset that provides peripherals (e.g. I/O). <S> This could be done with an FPGA but would be a major task - at least you will need the datasheet for the CPU. <S> Also, power management and putting it all together on a complex circuit board (you will not be able to build it in a breadboard :). <S> If you really want to use it, get an old motherboard, RAM etc... and run a lightweight Linux OS on it - then you could use it e.g. as a server (which could be the most adequate task for it, since you considered home automation). <S> But then, the price/performance or power/performance ratio might not be so great. <A> I think that a Pentium III is far away from being something useless. <S> I have a few suggestions you can try out. <S> But first, make sure you can boot the system, install some operating system, and run some stuff. <S> This probably have some serial RS232 and parallel ports. <S> You can do lots of good stuff with these: light control/home automatio, several control related stuff (motor, step motor, reading sensors, etc). <S> Google is full of nice projects related. <S> Just search! <S> You have to install a lightweight operating system and fine tune it. <S> The process itself is very fun. <S> Then you can you install emulators and built a nice case around it. <S> You can make this your laboratory for learning stuff: set down and try to install a usable graphical Linux environment. <S> You probably won't get it right the first time, but I assure you you learn a lot in the process. <S> Also you can use this as a testing machine for your software. <S> You will them force yourself to write better (faster and less memory eater) code for your own tools. <S> If you succed in getting the environment to work, try to setup a development environment: compiler, editor, debugger. <S> Try to write some useful code with that. <S> Try to transform this into a full featured internet router with every fancy stuff a router should have (and some it shouldn't).
| This can be a basis for a good old-style arcade PC. Building your own motherboard for a Pentium III will be a difficult, if not downright impossible, task.
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Using a USB device locally on 2 computers I would like to make what I can best describe as a reverse USB switch, where one USB 2.0 device (a dongle) is physically connected to 2 computers (located side by side). The dongle is powered through USB. I.e. what I'd like to make is a Y-split, with one female going to two males. Could this be done simply by wiring together 3 cable ends with the appropriate connectors? Do you think the dongle would be recognized locally and useable on both computers? Should I make sure that power only runs from one of the computers? <Q> USB is a point to point connectivity protocol --- that is a single host end connects to single target end. <S> The connection that you describe would be a multi-host single target type bussed connection and USB is not designed to support that. <S> As a side note you would never want to interconnect the power between two computers either. <S> GND to GND would be OK but not the 5V to 5V. <S> There are available products that perform a function close to what you want. <S> These are two way USB switch devices that allow for multiple hosts to share one target device - but one at a time through either a mechanical switch/relay or through an electronic switch component. <A> The USB protocol is not designed for this. <A> No, you can't do this by simply wiring three cables together. <S> You need an actual switch. <S> If you google for "usb ab switch" you'll get all kinds of hits for both mechanical and electronic USB device-sharing switches. <S> (Which makes this a shopping question, and therefore off-topic.)
| As already suggested, you can use a switch, or you might want to try an USB over IP solution to share the USB dongle over the network. You will not be able to make a Y cable as you describe and get it to work.
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When to use STD_LOGIC over BIT in VHDL Whats the difference between using: ENTITY MyDemo is PORT(X: IN STD_LOGIC; F: OUT STD_LOGIC );END MyDemo; and ENTITY MyDemo is PORT(X: IN BIT; F: OUT BIT );END MyDemo; What are the limitations of using BIT over STD_LOGIC and vice-versa? Are they completely interchange able? I understand that if I've define STD_LOGIC I can't use it with a BIT_Vector to access the elements in the array. But I can't seem to see the difference. <Q> Bit is a predefined type and only can only have the value 0 or 1 . <S> The Bit type is an idealized value. <S> type Bit is ('0', '1'); std_logic is part of the std_logic_1164 package and provides more realistic modeling of signals within a digital system. <S> It is capable of having nine different values. <S> Typically within your code you will only use 0 , 1 , and Z (High-Z). <S> But U (Uninitialized) and X (Unknown) are also very useful when modeling the system in a testbench. <S> ------------------------------------------------------------------- -- logic state system (unresolved) <S> ------------------------------------------------------------------- <S> TYPE std_ulogic IS ( 'U', -- <S> Uninitialized <S> 'X', -- <S> Forcing <S> Unknown '0', -- <S> Forcing 0 '1', -- <S> Forcing 1 'Z', -- <S> High Impedance <S> 'W', -- Weak Unknown 'L', -- <S> Weak 0 'H', -- <S> Weak <S> 1 '-' -- <S> Don't care );-- attribute ENUM_ENCODING of std_ulogic <S> : type is "U D 0 <S> 1 Z D 0 1 D"; ------------------------------------------------------------------- <S> -- <S> *** industry standard logic type <S> *** <S> ------------------------------------------------------------------- <S> SUBTYPE <S> std_logic IS resolved std_ulogic; The std_logic_1164 package also provides conversion functions to convert std_logic to Bit . <A> Most people use std_logic . <S> That allows for <S> u (undefined), x (unknown) and z (high impedance), which bit doesn't. <S> While you may never tristate in a chip and therefore don't need z , u is useful for finding missing resets. <S> x is useful for finding multiple drivers. <A> std_logic has a resolution function <S> Not only does std_logic have more useful states besides 1 and 0 , it also has a resolution function defined. <S> A resolution function is a VHDL language concept. <S> It is a function that is associated to a type, and it determines what happens when multiple values of that type are applied to a single signal. <S> The syntax is: SUBTYPE <S> std_logic IS resolved std_ulogic; where std_ulogic is the unresolved (and thus much less useful) version of std_logic . <S> In particular, this implies nice things like 0 and 1 leads to X : <S> library ieee;use <S> ieee.std_logic_1164.all;entity std_logic_tb isend std_logic_tb;architecture behav of std_logic_tb is signal s0 : <S> std_logic;begin s0 <= '0'; s0 <= '1'; process begin wait for 1 ns; assert s0 = ' <S> X'; wait; end process;end behav; <S> This makes intuitive sense, as we understand X to be the state where multiple incompatible values are applied to a single wire. <S> std_logic also knows how to resolve every other possible pair of input signals according to a table present on the LRM. <S> bit on the other hand, does not have a resolution function, and if we had used it on the above example, it would lead to a simulation error on GHDL 0.34. <S> The possible values of std_logic are a good choice because they are standardized by IEEE 1164 and deal with many common use cases. <S> Related: https://stackoverflow.com/questions/12504884/what-is-the-purpose-of-the-std-logic-enumerated-type-in-vhdl <A> There is also the boolean type, which, like bit , has two values. <S> It is the result type of comparisons, <S> the type expected after an IF [bool] or a WHEN [bool] , often used for selection constants : constant ENABLE_DEBUG_INTERFACE : boolean : <S> = true; One place where bit can be preferred to std_logic is for large arrays, memories. <S> On optimizing simulators, bit occupies less area in the simulator's memory than std_logic . <S> And it may matter if your design instantiates one GB of RAM. <S> It can also be faster for very large designs, for example something automatically generated from post-synthesis gate-level netlist. <S> Of course, this performance aspect is not part of the language, and depends on the implementation of the VHDL simulator.
| std_logic is richer than bit , and should basically be used most of the time.
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Time of flight difference in hardware I have an ultrasonic emitter, and two ultrasonic receivers. When the emitter emits a pulse, the receivers get the signal at different times. I am only interested in the difference between the two reception times. At first, I was thinking of using a microprocessor, measuring individual arrival times, and then taking the difference. Taking two absolute measures when only a difference is required feels somewhat unnecessary, and may add inaccuracies. Is there a way to measure pulse time differences in hardware, i.e. analogically? Would this be advisable? <Q> Use a so called TDC, <S> e.g. the acam-gp21: <S> http://www.acam-usa.com/GP21.html .It <S> even has an example for ultrasonic measurement in the manual. <A> You could do it in hardware with some comparators, a fast clock, a couple of counters and a subtracting circuit... <S> There are probably some other creative ways of doing this but it seems like a race to an ever-more elaborate Rube Goldberg machine . <S> This is the kind of application a small micro excels at, and depending on your implementation, would be far more accurate at than an analog circuit requiring precision components and careful tempco planning. <A> A few MSP430 devices have a peripheral called Timer D that has 4ns time period <S> and I've seen it used for these kind of applications. <S> See MSP430 with Timer D <S> It should simplify development significantly. <A> From your previous post I know that you will going to use a 400ST/R100 ultrasonic transducer. <S> The main problem with these transducers that they are narrowband. <S> Look at received signal (The first measuring results with oscilloscope) <S> http://www.technik.dhbw-ravensburg.de/~lau/ultrasonic-anemometer.html <S> maybe now you will understand the problem. <S> TOF detection methods 1) <S> Reference level crossing, problem <S> - it's depend from received signal level, if you going change distance <S> it's will be a problem. <S> 2) <S> Zero crossing or signal max possition finding requires signal digitization. <S> To find max possition for such signal is hard because it's have a flat top (look at signal envelope). <S> 3) Correlation or L1, L2 norms calculations. <S> To get max possition of correlation accurately use parabolic interpolation. <S> If you calculate correlation between received signal you will get something like this. <S> (It's autocorrelation of received signal after transmiting three 40KHz pulses) <S> As you see the main problem of correlation is side lobes. <S> If you change excitation signal to Chirp side lobes will be lower. <S> Despite these side lobes I was managed to get 0.9ns standar deviation of TOF, with equipment I mentioded in your previous post. <S> Witch method to choose? <S> It's depends of accuracy you need.
| You could also use the comparators with some op-amps configured to integrate, then subtract the difference in voltage level between the two integrators, then use some means to convert that voltage difference into a meaningful time value.
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Solder does not stick to tip of soldering iron? Basic question here, but it's driving me insane! The solder does not seem to stick to the tip of my soldering iron. After the iron heats up, the tip appears black (sometimes on all sides and sometimes on one side), then either it doesn't melt the solder at all, or it I fiddle around with it for a while; it melts the solder, but the solder rolls into itself into a ball and doesn't stick to the tip. I've seen this many times while soldering before and usually it's solved by cleaning the tip on a soldering sponge or cooling it down and scraping the tip off with a blade, but I have scraped this particular one many times and still it's the same problem: it heats up and gets black, then the solder doesn't stick to it. It doesn't help that this soldering iron has a very fine tip and I have to use it to solder very tiny chips and wires under a microscope (I'm just starting to learn soldering under microscopes). Any ideas on what I can do? PS: Is scraping the tip a bad idea? I have been doing it with my other bigger soldering irons and it seemed to work (at least in most cases) but I am not sure whether it is a good solution or not. <Q> First question: How long have you had the tip? <S> You could probably easily and cheaply replace it. <S> If you've not had the tip long, then a few pointers to keep in mind: <S> When the iron is not being used, keep some solder on the tip. <S> This prevents oxidation and corrosion. <S> When using the iron, keep a little solder on the tip when the iron is in the stand. <S> Wipe clean before using. <S> Avoid excessive wiping on a wet sponge as this can cause temperature fluctuations which expand and contract the metal, and stress the tip. <S> Consider using brass shavings instead of a wet sponge. <S> Never use sandpaper or abrasives to clean the tip. <S> Also avoid dipping it into flux to clean. <S> (Except for re-tinning which should be infrequent.) <S> Use a good quality solder. <S> Old solder can have contaminants which leave behind residue and basically make soldering more difficult. <S> (By "old" I mean solder that's been around collecting dust, not necessarily that it has a short shelf life.) <S> Finally, if you've got a cheap iron, it may be too hot or cold and not melting solder at the correct temperature. <S> Consider getting a temperature controlled unit. <S> I could recommend a Hakko FX-888D for ~ $100 US, see also this tutorial . <S> Some additional soldering tip pointers: http://www.inlandcraft.com/uguides/tipcare.htm <A> My first guess would be that's just a low quality tip and no scrapping would help. <S> There are some tips nowadays that arent properly coated. <S> They're some alloy plated or even might be died with something and they're just for show. <S> But in that cases whole iron is low quality also. <S> If it's some quality iron and tip (in its own time) and you scrap it too often <S> it might be <S> you overscrap it <S> or it might even oxidation of inner copper alloy to degrade tip <S> so it's for scrap. <S> If it's HQ tip over years of use it should be replaced. <A> imho, there is no substitute for using tip tinner. <S> You can find it easily on eBay or Amazon, and probably other places. <S> Just heat up the iron and push it into the stuff for a second, then wipe it off on a damp sponge. <A> It is possible you are attempting to melt solder that is a higher temperature grade than your iron <S> is capable of melting. <S> ( I had this problem once ) check it out. <S> A typical iron is 50W <S> you may need 150 or higher.
| This may sound silly, but some solder is designed for higher temperatures. If the iron doesn't have replaceable tips, get rid of it!
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Can I use I2C bus or GPIOs as I2C to connect I2C devices? This question is regarding use of I2C Bus/GPIOs as I2C. My application processor has three I2C controllers. Is it preferable to connect all I2C devices (obviously we can not connect more than 128 devices on I2C) on these buses? Or, is it preferable to use GPIOs as I2C? (My application processor also has GPIO lines which I can use for I2C purposes.) When is it preferable to use I2C controllers and when it is preferable to use GPIO lines for I2C interfacing? <Q> That's what I2C was designed for. <S> To spread the load, using 2 or 3 I2C buses can make things easier (EG keep different systems separate) <S> but it can mean you have to do the same thing 3 times in software. <S> You may do something like use one bus for "high speed" transfer of lots of data (EG to/from a storage device) and another bus running a lot slower to do less critical activities like read a temperature sensor every second or turn an LED on/off. <S> Bit-banging I2C using GPIO pins is a "last-resort" as it's usually much easier to use the built-in I2C device, and most processors have things like interrupt handlers and DMA to make the work much easier, automate the transfers, save processor load / software overhead, etc. <S> It's much nicer to have communications handled by interrupts rather than trying to do it by GPIO <S> + waits/timers etc., especially if you have a lot to do / a lot of data to move. <S> Edited to add: It can also be handy to use one bus for devices which need a level shifter (EG 5v devices interfaced to 3.3v micro on one bus, 3.3v native devices on the other bus) <A> If you are implementing an I2C master for a single-master system, I would suggest that bit-banging may be preferable to trying to use built-in hardware. <S> Among other things: If any devices on the I2C bus get "confused", it can sometimes be difficult to convince hardware master implementations to generate the best waveforms to get them "unstuck" (recommended approach: if SDA isn't floating, or if your device is driving SCK low, assert SDA, assert CLK, release SDA, release CLK, and wait for CLK to go high; repeat as needed, and then assert and release SDA with SCK high). <S> Note that if an EEPROM device thought it was in the middle of a write cycle, this waveform will cause it to abort the write in progress; if the processor doesn't know why the EEPROM would be expecting to write data, aborting the write is probably better than writing data the processor knows nothing about. <S> Easy to do with bit-banging; harder to do (if even possible) on many hardware I2C implementations. <S> Many hardware I2C implementations require that the processor decide, before reading each byte, whether or not it should be positively or negatively acknowledged. <S> In many cases, it would be more convenient to have a "read byte" routine which send a positive acknowledgment <S> acknowledges the previous byte (if any), reads the next byte, and doesn't the acknowledgment until the next byte is needed; once no more data is needed, call an "end-read" routine to send a negative acknowledgment. <S> Such an approach is easy when bit-banging, but I know of no hardware I2C implementations that allow one to read a byte without having to know--in advance--whether it will be the last one. <S> The ease or difficulty of writing a bit-bang I2C master implementation is essentially independent of the number of devices on the bus. <S> I2C slave implementations are generally difficult or impossible without at least some level of hardware support, but big-banging an I2C master is easy. <A> If your application processor has I2C hardware which can operate in master mode, you certainly can use it to communicate with slave devices. <S> Compared with bit-banged GPIO implementations, it can be faster in terms of development time to use the on-board hardware. <S> Most likely there will be some libraries which will simplify mode setting and the general activity that the I2C master will need to perform (reads, writes, ACKs, NACKs, general call, serialization of bytes inand out, etc.) <S> However, if the I2C hardware is buggy, or the libraries don't give you the functionality you need, you may find it better to go with GPIO bit-banged I2C and avoid using the I2C hardware. <S> The downside, of course, is having to write much more code.
| Depending how much data is going along the bus, and if there are "conflicts" of device ID's, or data being transferred at the same time, it's no problem to hang all your devices on one bus.
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How to determine type of through-hole resistor? Everyone should be familiar with standard carbon through-hole resistors, but there are of course other types like wirewound, metal film, metal oxide, etc. The trouble is, how do you identify the type of resistor if you just have a bunch sitting in a parts bin? I inherited a grab-bag of parts and haven't been able to identify some components. Here is a sample of the resistor-like parts: The topmost part is a standard 1/4W carbon resistor. Below that, I believe a metal film or metal oxide. Below that, I am simply not sure. (I think the last part may even be an inductor.) Is there a standard color coding for these parts to identify the component and its composition? <Q> Resistors cases are usually tan, brown, blue, or green, though other colors are occasionally found such as dark red, dark gray, pink and light green. <S> Is there a standard color coding for these parts to identify the component and its composition? <S> Not really however tan tends to be carbon film, whereas a blue/light blue resistor is likely made of metal film. <S> Resistors with 5 or 6 color bands are almost always metal film. <S> Take for example your second resistor, it has 5 bands while the one above it only has 4. <S> Your items from top to bottom, (The power ratings are just a guess, and the links show a similar item.) <S> 1/4W <S> 5% Carbon Film Resistor <S> 1/4W 1% Metal Film Resistor <S> An Inductor* 1/4W 5% Carbon Film Resistor??? <S> 1/2W <S> 5% Carbon Composition Resistor <S> * <S> As @JYelton pointed out, measuring it with an ohmmeter would be definitive in separating resistors from inductors. <S> Inductors wouldn't obey the resistor color code and thus measured resistance wouldn't agree with the markings. <S> How to Tell the Difference Between Carbon & Metal Film Resistors <A> However, it is generally true that cheap, carbon-film resistors are usually brown, and metal film resistors are usually blue. <S> It's also true that it's very rare to find carbon resistors with 1% or better tolerance. <S> About the only definite thing you can say is that carbon-composition resistors are pretty much universally a straight cylinder with flat ends (like the last resistor in your picture). <S> Everything else is a guess. <S> Anyways, for the resistors you posted: Inexpensive 5% tolerance, probably carbon. <S> 1% tol, probably metal film <S> Could be an inductor. <S> That epoxy color is more common for inductors then resistors. <S> Inexpensive 5% tolerance, probably carbon. <S> Old-style carbon-composition (not film) resistor. <S> 5% <A> 4 or 5 rings are hints <S> Some history, long time ago (vacumn tube to early solid state), composite carbon resistors are popular with 10% tolerance and 4 rings (2 significant digit for value, 1 multipler, 1 tolerance). <S> Next was carbon film, popular tolerance is 5% and 4 rings. <S> Next was metal film, popular tolerance is 1% (or even better). <S> 4 rings is not enough and it become 5 rings to add one more significant digit in value. <S> Above works well for resistor used inside ordinary/low cost consumer products. <S> The last resistor in the photo likely a special carbon composite resistor (based on casing style) of low tolerance. <S> Resistsive element is straight cylinder for this casing and low in inductance and good for rafio frequency use. <S> Film (carbon and metal) are coil structure (add inductance). <S> You can see the structure by Cutting into half and PEEL off the cover paint. <S> CARE about work safety!!!
| Unfortunately, there really isn't any genuinely consistent way to identify resistor composition, short of destructive testing.
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Considerations in a bench/lab power supply? I'm looking at adding a low-cost variable power supply to my electronics hobby bench. After years of cobbling together LM317's and LM7805's, it's about time... Considerations: I've found a few options which I am unsure about. Dave at EEVBlog states in various videos ( #8 , #30 , #166 (@04:16), #168 , #272 , #314 ) that you should look for the following: Multi-turn knobs for voltage and current adjustment (as opposed to pushbuttons and separate coarse/fine knobs) A power supply that can range from 0 volts on the low end instead of 1 or 1.2 volts A switch to turn on/off the load The supply shouldn't overshoot the target voltage when switched on It should have a separate mains earth ground terminal (to allow combining supplies) Prefer linear for less ripple and noise (though switching offers smaller size and cost for higher currents) Of course some factors depend on the intended use, some depend on preference, and some depend on budget. (My intended usage is primarily microcontroller projects (3.3V-5V, <1A), automation and control projects (solenoid, stepper motor, etc) (12-24V, <5A), and low voltage lighting projects (12-24V, <3A). Budget ideally < $300.) I asked earlier about the last point, linear vs switching, in a previous question . Olin convinced me that the disadvantages of a switching supply are minor, and certainly for my application, make sense for budget, bench space, and other reasons. Thoughts: To touch on the other points: Multi-turn pots must be preference. I think they are better than single-turn pots, but I would prefer coarse/fine adjustments (or direct-entry via buttons) over turning a knob multiple times to make a larger adjustments. I haven't needed to work on projects below 3.3 volts, but I am mindful of the pico-power AVR micros , which can run at 0.7V. Certainly a supply that can go below 1 volt would be useful for this. A load switch is nice, but not a deal-breaker. How essential is it to have separate earth ground? I don't think I need to combine power supplies. Possible Supplies: I've found a few supplies that I like, but I can't commit because none meet all the criteria. BK Precision 1550 , 1-36V 0-3A, $150. Concerns: Doesn't go below 1 volt. Uses up/down buttons for adjustment. Circuit Specialists CSI3005X5 , 0-30V, 0-5A, $130. Concerns: Unfamiliar brand and low price. (quality issue?) BK Precision 1671A , 0-30V 0-5A, $220. Concerns: Unknown adjustment turns; no earth ground terminal. Is there some other supply or brand that meets all of these requirements? What points (if any) above, are truly important? <Q> BK Precision 1550 <S> This is a switching supply. <S> The up-down adjustments would make this a non-starter for me. <S> CSI3005X5 <S> A whole bunch of companies re-brand this unit. <S> They're actually fairly decent. <S> The voltage pot is a 10 turn, the current limit is button-driven in 0.03A increments. <S> The most common resaler of the power-supply is MPJA. <S> It also comes in a bunch of voltage and current ranges: 0-30V 5A, 0-60V 3A, 0-120V <S> 1A. <S> One thing you can't see in the pictures is that the unit has a set of screw terminals in parallel with the output banana jacks, below the cover plate labeled "EXT OUTPUT". <S> If you need more permanent connections, you can use the screw terminals. <S> The schematic for the whole supply is available. <S> This makes it enormously more repairable (and hackable) <S> then ANY of the others. <S> BK Precision 1671A <S> The funky extra output connections on this make me nervous (speaker terminals? <S> really?). <S> I would guess that the potentiometers are single-turn, both from the artwork on the case near the knobs, and the fact that it does not mention multi-turn knobs, as that's normally a significant selling point at this price range. <S> On the whole, If I had to choose from the supplies listed, I would wholeheartedly recommend the CSI3005X5, more because the alternatives are considerably worse. <S> Anyways, I would say that even if you don't think you need a floating output power supply (what you really mean when you discuss a separate earth terminal), you almost certainly will find it useful in the future, so I think you shouldn't dismiss it. <S> Just being able to string multiple power supplies in series for higher output voltages is tremendously useful. <A> Here is the list of the generic china ones that are sold on Amazon under the names Mastech, Elenco, Sinometer and others. <S> $185 <S> Mastech TRIPLE LINEAR DC POWER SUPPLY <S> 30V 5A HY3005F-3 <S> Dual adjustable outputs: 0-30V and 0-5A Fixed output: 5V and 3A <S> Line regulation: CV <= 0.01% + 2 mV, CC <= 0.2% + 2 mA> <S> Load Regulation: CV <= 0.01% + 3mV, CC <= 0.2% + 3 mA <S> Display reading accuracy: + <S> /-1% for voltage and +/-2% for current <S> Having the extra adjustable power outputs are convenient. <S> They can be connected independently or in series or parallel to allow for higher amperage or voltage if needed. <S> The disadvantage is that the pots are all single turn, so it's hard to get very fine changes. <S> Although you can replace them with 10 turn ones. <S> Also, the voltage and current displays come in LED (as shown in the picture,) or LCD. <S> As I mentioned in my other answer <S> I have this one <S> and it is pretty nice and the extra outputs come in handy. <S> $130 <S> Sinometer HY3005D Variable Linear Lab Power Supply 0 <S> -30V 0-5A Adjustable outputs: 0-30V and 0-5A <S> Input voltage: <S> 110V AC and 220V AC switchable <S> Ripple noise: <S> CV <= <S> 0.5 mV RMS, CC <= <S> 3 mA RMS Coarse and fine control for both current and voltage outputs. <S> Line regulation: CV <= 0.01% + 1 mV, CC <= 0.2% + 1 mA Load Regulation: CV <= 0.01% + 3mV, CC <= 0.2% + 3 mA Ripple <S> noise: <S> CV <= <S> 0.5 mV RMS, CC <= <S> 3 mA <S> RMS LCD reading accuracy: + <S> /-1% for voltage and +/-2% for current <S> Just a very cheap adjustable unit that doesn't take up much bench space. <A> I watched Dave's Review/FAIL video of it and then watched the follow up video and a follow up video by Charles Holtom from TrioSmartCal . <S> For ~$100 I thought it was worth giving it a try. <S> From SRA <S> Features: <S> Compact Design <S> Dual 4 Digit LED Displays Digital Controls <S> Output switch control Switchable between 110v and 220v <S> 4 sets of parameters can be stored inside for fast recall Keyboard Lock to prevent accidental changing of set parameters Circuit protection for over load, short circuit and over temperature. <S> Automatic switching between mA and A display Quiet variable speed fan 2 Year Warranty! <S> Operating Conditions: Input <S> voltage:110V +/- <S> 10% switchable to 220V <S> +/- <S> 10% Frequency range: <S> 50- 60Hz <S> Operating temperature: <S> 32- 104 F (0- 40 ºC) Operating humidity: < 80%RH <S> Storage temperature: <S> 32- 158 F (0- 70 ºC) <S> Storage humidity: <S> < 70%RH <S> Stable Voltage: <S> Output Voltage Range: <S> 0- 30V Continuously adjustable Setup Resolution: 10 mV Setup Accuracy: <0.05% +20 mV Ripple: <S> < 1mVRMS <S> Read back Accuracy: 10 mV Temp. <S> Coefficient: < 0.1% + 10mV <S> Read back Temp. <S> Coefficient: < 100 <S> ppm+10 mV <S> Stable Current: <S> Output Current Range: <S> 0- 5A Continuously adjustable Setup Resolution: 1 mA <S> Setup Accuracy: <0.1% +3 mA Ripple: <S> < 3mARMS <S> Read back Accuracy: 1 mA Temp. <S> Coefficient: < 0.1% + 5mA Read back <S> Temp. <S> Coefficient: < 100 ppm+5 mA <S> This power supply seems like it should be mentioned when considering a bench power supply. <A> If you don't mind something old, and affordable, a Tektronix PS503, or two, or three, or four, in a TM50x mainframe fits the bill perfectly. <S> Two channels, 20VDC each, can be put in series, current limiting, no overshoot on turn on, control down to zero, ground available, regulated, ... <S> Everything you asked for. <S> And you can add a DM50x to monitor it with backplane connections to economise on front wiring.
| $80 Mastech HY1803D variable dc power supply Adjustable outputs: 0-18V and 0-3A I just order a KORAD KA3005P - Programmable Precision Variable Adjustable 30V, 5A DC Power Supply from Amazon.
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powering 12 v lights from an electric fence I have an electric fence running around my farm. I would like to be able use this to power a number of 12 v lights along the road so I have a ready source of light, especially since the electric fence is on continuously.. The specs I can find for the fence is - dc 5000 V, pulse about 500 micro secs every 1 sec, 1.5 Joules (these vary for different energisers). (as you can see I am no electronics or engineering buff. ) Can anyone help me with the electronics to convert this power pulse into a steady 12 v power supply for lights? Many thanks. <Q> That's not a good idea, use a separate power supply for the lights. <S> 1.5 Joule once per second means you have at most 1.5 Watt of power for your lights, which probably is too little for several lights (and then the electric fence wouldn't work any more). <S> And this is far too optimistic, because the huge voltage difference would cause big energy losses. <S> And components for 5000V tend to be expensive. <S> Edit: <S> Best is probably to use those inexpensive solar powered garden lights. <A> Run a separate cable along the fence. <A> I would use a high voltage capacitor with a diode and a regulator on top of that. <S> It would maybe be possible to power a small LED about it. <S> I don't think you can power much from the fence, and even less if you want the fence to actually work as an electrical fence. <A> If your fence is current controlled (does it adjust the power to compensate for weeds, etc. <S> touching it?) <S> then you can get a bit more energy out, but in the end it's going to be limited to avoid killing people who touch it. <S> I can get a pretty bright glow from a white LED + <S> 10k ohm resistor connected to my horse fence, but it's not really usable for anything. <S> Remember that it's pulsing <S> , so you get one flash about every second or so. <S> I only built that circuit because I couldn't find the "real" fence tester and I needed to check a section of fence (gets painful repeatedly shocking yourself just to check the wiring). <S> Much easier to put a few solar powered lights on the fence posts and turn them on when needed. <A> I you're using a multistrand fence isolate two of the earth wires from the fence and use those to power low power led lighting. <S> You can run them off a battery to give lighting when the power is down on the grid.
| The simplest way is probably a small neon bulb such as the fence testers use, but you won't get much light out of it.
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Why do we use bridge rectifier? If we already having a center tap rectifier with just two diodes doing the same work then why is there a need to use four diodes or the so called Bridge rectifier. <Q> I'll assume you meant full wave bridge rectifier circuit for "bridge rectifier". <S> To be clear, here is a full wave bridge: <S> Look at this for a moment and see how it works. <S> It basically performs the absolute value function on a voltage. <S> Actually is looses two diode drops of voltage in the process, but that's not the point right now. <S> If you have a single AC signal, then a full wave bridge is one way to make it all positive. <S> If you already have the AC voltage coming from a center trapped transformer secondary, then you can use the extra connction to your advantage to simplify the rectifying circuit: <S> Look at this a bit and see that you always get a positive voltage from V- to V+. <S> So why wouldn't everyone always do it this way? <S> It should be obvious that this second circuit is only possible in limited situations where you have a center tapped transformer output available. <S> If you do, this can be a useful way to do rectification. <S> One advantage is that there is only one diode drop in series with the absolute value of the AC voltage, not two like with the full wave bridge above. <S> But, think about the cost. <S> Note that only one half of the secondary is conducting at any one time. <S> You are paying for the extra stuff you only use half the time. <S> Diodes are cheap and small compared to transformers, especially at low frequencies like line power. <S> Usually the deciding issue is whether you need a transformer for other reasons anyway, like isolation. <S> In that case the incremental cost of the center tap and winding the secondary with longer but thinner wire is relatively low. <S> There is another reason for using a center tapped secondary, which is if you want both a positive and negative supply: <S> Follow thru what happens over a whole AC cycle, and you should be able to see how you get both the positive absolute value and negative absolute value from this circuit. <A> The biggest disadvantage of the center-tap (CT) transformer is that you only use half of it at each half cycle of the mains. <S> That means your transformer will be twice as heavy and large than one with a full-wave rectifier. <S> The two extra diodes pay for themselves easily this way. <A> The reverse voltage specification of the diodes is double for the 2 diode verses the 4 diode solution as described in the harshly down voted answer, so this may favour 4 diodes if voltage rating affects their price by more than double such as in higher voltage supplies where around 1500V PIV starts to get expensive. <S> There are also other reasons for selecting between the arrangements. <S> It also offers the opportunity to share windings with a ground referenced AC rail (or perhaps even a two phase mains voltage output) that will not be possible as neither of the transformer terminals is at a fixed ground/common voltage when the full bridge method is used for a single rail. <S> A practical reason for having two totally independent output windings is to allow for series or parallel connection of the transformer allowing use of just two diodes with a centre tap or 4 diodes with the windings in parallel for one voltage or the windings in series with a bridge to select double the output voltage like might be desired in a simple car/motorbike battery charger. <S> In a microwave oven HV transformer the inner end of the winding is grounded to the core to reduce the insulation requirements (and capacitive loading) between the HV output and the transformer core and to keep the HV-ground at chassis potential thus no diode is placed on the ground side and rectification (or doubling) has to be done on a single grounded winding even though it might be cheaper to do with full-wave rectification <S> a floating HV supply would be unwise especially under fault conditions. <S> Many practical reasons dictate which arrangement is used and all have up and down sides. <S> EDIT: <S> Another thought came to mind. <S> With very low voltages it is prudent to avoid wasting the second diode drop in the diode if it forms an appreciable portion of the output voltage. <S> This might be most relevant in single cell NiCd NiMH charging circuits.
| The two diode centre tapped solution has value even if one only wanted a positive rail at a specific voltage there might be the desire to have other voltage taps (perhaps selectable) for lower voltages, simultaneously full-wave rectified or perhaps even a simple single diode half-wave rectified rail, these could then share the common ground with the centre tapped rail.
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Best way to remove pins from IC storage tubes? I often buy ICs that come in clear plastic storage tubes like this: There are plastic pins on either end that keep the ICs from sliding out. I'm sure you're all familiar with this.. Sometimes I can easily push the pins out with my fingers, but other times they are extremely difficult to get out. What's the easiest way to get the pins out? Thanks for your help! <Q> Just use side cutters, keep them open (no cutting involved) and rock the handles down. <S> The angle on the cutting surface will provide the leverage needed. <A> I use a tack puller. <S> Its quick and easy. <S> Pins like this are common in automotive paneling as well, and breaking one in those circles means you can't put your car back together. <S> Your local hardware store should carry various sizes. <A> Or you can use a right angle needle nose pliers (or crimpers or klines/cutters, just be sure not to cut or break it,) to grab it from the top (but below the flat top, and pull or use them as leverage to pull the pin out. <A> Ball point pen with a cap. <S> Put it on the underside, and give it a nice tap with the flat end something, or your palm. <A> For the ones that have a 'rounded' mushroom-top I use the wire stripper that I always carry with me, in other words: my teeth. <S> For the more challenging ones I use a pair of scissors to push it out from the bottom side.
| I usually use a pair of needle nose pliers and push them out from the bottom wile twisting it back and forth (pulling them from the round top tends to break the pin.)
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What should attaching another magnet to the back of the speaker magnet do? The dupe speaker I have here 8 \$\Omega\$ 0.5W had it's magnet split from the coil/cone. So I did something thoughtless - attached the speaker magnet to the back of the other speaker (same specs). I don't notice any difference except (this may be my imagination) it seems to sound a little louder. Should attaching another magnet to the back of the speaker magnet make a difference?What should attaching another magnet to the back of the speaker magnet do? <Q> Depends which way round it is. <S> It's almost completely out of the main magnetic circuit, but its fringes will either add, or subtract, a little - maybe 10-15% - to the flux in the field where the voice coil operates. <S> Probably about 1dB louder which is quite difficult to hear. <S> (Some preamps have digitally generated 1dB steps so you can test what 1dB sounds like, <S> and yes it is just audible). <S> If it adds to the flux, the speaker will be quieter - but its bass will be better controlled. <S> I have seen some poor guy flown from Japan to Europe to spend a few weeks supergluing magnets onto the back of 5000 drive units, to increase the flux where the original magnets were weaker than the manufacturer had promised. <A> It will cause the voice coil to not move how it was designed too, since the induced magnetic field will be different. <S> It will most likely effect different frequencies differently, for example higher frequencies will probably not notice a difference, but lower ones may not be as sinusoidal as they originally were (because the cone may have less resistance going backwards, and more going forwards.) <S> However the overall effect will probably be minor. <S> One thing to not is that it could cause premature damage to the speaker as the voice coil may make contact with the side at some points. <A> The back of the magnet is usually flat iron. <S> Any magnet stuck to the back will have very little effect. <S> The magnetic flux lines from an external magnet will not pass through the air gap that the voice coil moves in so they will have no effect. <S> If the speaker should be of a budget type that has a magnet exposed at the rear and no outer iron pole then there might be some reinforcement if the external magnet reinforces the field due to the existing magnet and the lines of magnetic flux pass into the metal gore and then radiate out through the voice coil. <S> Below is a diagram that shows how the typical speaker is constructed and how the add on magnet at the back will have very little effect at the voice coil. <S> Illustrated are a more exotic annular pole magnet and a common top bottom pole magnet.
| If it subtracts from the flux the speaker will be slightly - paradoxically - louder (the same way an electric motor goes faster when you reduce the field strength - more motion is required to generate the same back EMF).
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Anyone know the purpose of these two inductors around a metallic ring? So this link on ebay shows a power board that I have taken out from my microwave. What is the purpose of these two wrapped around a metal ring? Also in general what is the purpose of this power board it takes in an input voltage of 120V AC but what does it output? The circuit diagram shows the inductors without the toroid. The inductance is incorrect, I am not sure that the Henry value would be for them. <Q> The board does two things. <S> It protects the microwave from shorts, electrical noise, and line faults such as voltage spikes. <S> The wires wrapped around the metal toroid form an inductor. <S> This is to prevent electrical noise the microwave generates transferring to the powerline, and electrical noise on the powerline from transferring to the microwave. <S> It's done primarily in order to meet emissions requirements (in the US the FCC regulates this), but also to prevent line noise from affecting microwave operation. <S> You can read more about noise suppression ferrites here: http://audiosystemsgroup.com/SAC0305Ferrites.pdf <A> At the top left, the tan (with color bands) resistor and yellow capacitor form a RC-snubber network. <S> This mainly filters spikes due to instantaneously switching off an inductive load (main transformer); According to your diagram R1 and C1 are in parallel, then it is not an RC-snubber like I expected. <S> R1 is there to discharge the capacitors after disconnecting the device from mains power, so you won't get zapped when you touch the leads when you unplug the device from the wall socket. <S> Top middle thing with the two coper wires is a common mode choke. <S> It is a bit difficult to explain how it works, so I'll leave it to the others to answer that. <S> Until then here is an article on Wikipedia ; The blue things, capacitors to Ground. <S> Together with the common mode choke they form a filter to ground for high frequency noise. <S> ; You probably recognized the fuse at the bottom left. <S> Apart from the fuse all these subcircuits form a filter with specific goal. <S> The overall goal is to prevent the microwave from emitting electric noise onto the power lines and cause other devices (eg. <S> your radio or TV) to malfunction. <S> At the same time prevent power line noise from other devices to disrupt the microwave from properly functioning. <A> Definitely this is an EMI filter. <S> C1 is between line and neutral and is what's referred to as an X-capacitor. <S> Its purpose is to suppress differential-mode noise between line and neutral by showing a low impedance to high frequencies (usually in the high kilohertz to low megahertz range). <S> Because this cap is across line and neutral, it will have its own safety mark on it somewhere. <S> The toroidal inductor is a common-mode choke. <S> The starts and finishes are on the same 'side' of the circuit. <S> It provides a high impedance to common-mode noise (line to earth, neutral to earth) and doesn't affect the normal current drawn by the device (which is differential by nature). <S> The construction of the choke (two bunched windings spread apart) gives you some leakage inductance, which is a high impedance to differential mode noise. <S> The other two caps are Y-capacitors, which provide a low impedance path for high frequency <S> common-mode noise (line to earth, neutral to earth) and will also have their own safety marks. <S> the CM choke and the Y-caps (common mode filter) the leakage inductance of the CM choke and the X-cap (differential filter) <S> The fuse is to isolate the unit from the mains in the event of a catastrophic circuit or isolation failure.
| It also protects the line from noise the microwave generates. From the perspective of EMI, you have two LC filters here:
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Can DAQ devices be used as Oscilloscopes? Recently, I took computer architecture course, and got interested in analog electronics. At present moment, I can't afford good USB oscilloscope such as Agilent or Cleverscope (though, they are not that terribly expensive - $1500) I am curious if USB Digital Acquisition Devices(DAQ) can be used as inexpensive temporary substitution for oscilloscope ? What would be drawbacks (apart from not having probes and having to record the data to PC) ? Thanks ! Edit: some of the DAQ devices I have looked at: http://www.ni.com/products/usb-6008/ (NI make an array of different USB DAQ) http://www.keithley.com/products/data/multifunction/usb/?mn=KUSB-3100 (though it looks like this DAQ is better suited for power electronics) Some USB oscilloscopes I am interested in: http://www.cleverscope.com/products/CS320A http://www.home.agilent.com/en/pc-1418982/usb-modular-oscilloscope?nid=-34492.0&cc=US&lc=eng <Q> Most of them will maybe do ±10V input range. <S> Probably won't support offset subtraction on the inputs, like scopes do. <S> DC-coupled only, unless you supply the series cap. <S> Inputs can be low(ish)-impedance (some may have buffer amps, on cheap ones the input may literally just connect to the ADC pin). <S> Not the 1MΩ standard that scopes have. <S> Most importantly: <S> PC based oscilloscope interfaces suck <S> Also, it's likely a DAQ won't even have a traditional oscilloscope-like software tool . <S> You may have to write your own. <S> Anyways, if you have a situation where you have fixed or low voltages, and don't mind doing a bunch of work on the PC end, a DAQ could be used as rather pokey oscilloscope. <S> They're really different tools, though, and while they do share some characteristics, they have very different intended uses, and this tends to show in their approach and the software design considerations. <S> It's also worth noting that most DAQ systems are designed for continuous, rather then triggered data acquisition. <S> This means you're maximum sample rate is largely limited by the interface the DAQ uses. <S> For example, USB2 only has 480 Mbps(more like 400 Mbps real-world) bandwidth. <S> As such the best sample rate that could ever be achieved would be 50 Msps(million samples per second) at 8 bits resolution, and very few implementations will even approach that. <S> Somewhere in the range of 1-10 Msps at 8 or 16 bits is more realistic. <S> Extracting all the available bandwidth from USB is very challenging. <S> Another consideration is what you're going to <S> do with all the data. <S> 1 Msps is a lot of data. <S> If the 1 Msps stream is 16 bits, that's 2 Megabytes of data per second , or a gigabyte every 8 minutes. <S> I don't know what you're intending to do with this pseudo-oscilloscope, but you can't just take samples willy-nilly, unless you're just displaying them and then immediately discarding them. <S> I've actually written a minimal real-time visualization tool for some IOtech-branded DAQ systems at work. <S> It's kind of an oscilloscope. <S> It works well, but I've also designed all the PCBs that interface with the DAQ system, so I could design them to work to the DAQ system's input specifications. <A> Check ebay for low cost second hand equipment. <S> HP and others brands has very long life and they are generally still very good even second hand. <S> The 2 DAQ models are lower in resolution and "high frequency" bandwidth than PC sound card. <S> They can go down to 0 Hz DC whereas sound card typically cut off at 20Hz. <A> Likely many DAQ ADC cards come with such software. <S> Some (not a lot) software use PC parallel port as very low speed logic analyzer. <S> You can buy adaptor to make "parallel port' out of latest USB-only PC. <S> However, this may limited the speed further. <S> Many software use PC sound card as oscilloscope/spectrum analyzer as well as signal generator. <S> Apparently, these may more suit to EE engineering than computer architecture course. <S> Allow on screen 'experiments' too. <S> Like, http://www.qsl.net/dl4yhf/spectra1.html <A> Personally, I wouldn't go there. <S> You didn't specify a piece of equipment, but you would likely have speed issues. <S> I would look at something like this . <S> It's not very fast, but with a 25 MHz sampling frequency, you can easily look at 5 MHz signals, and theoretically up to 12.5 MHz. <S> The transient recorder, bode plotter, and spectrum analyzer are nice features. <S> The function generator is an added bonus, and most importantly, it doesn't cost $1500. <A> A scope is design to let you quickly and easilly probe a range of signals. <S> This means a few key features. <S> In most cases a high input impedance to minimise disturbence to the device under test (high end scopes often also have a low impedance mode for high frequency work) <S> A systems to let you quickly and easilly change the settings. <S> Physical knobs are much better for this than widgets on a computer screen. <S> An input circuit that is tolerant of incorrect settings. <S> You don't want your scope to blow up because you set it to the 1mV range and then put 10V on the input. <S> Flexible triggering options to maximise your chances of getting a stable display. <S> Data aquisition devices on the other hand are more suited to longer term measurements. <S> They typically have higher precision than scopes and are designed to sample continuously (whereas most scopes can only sample at their headline sample rate in bursts) but have a much simpler input stage that will often require external circuitry to get the signals into the right range. <S> It's up to you to make sure that external circuitry is sufficiently tolerant for your application.
| DAQ systems can make very functional low-speed oscilloscopes, with a number of caveats: You're not going to get a very broad voltage range.
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How to shift differential signal to read with single ended adc? I'm using a current sense resistor (0.02 ohms) for a high current battery monitor for a hobby project. I'd like to shift this to a 0-5 V range to monitor using an ADC input (single ended.) I understand how to set the gain of an opamp using a resistor network. I also understand how to use a differential input of an opamp. But I can't seem to figure out how to do both at the same time. The negative feedback resistor will cause a direct path to ground (this resistor is high side at 12-17V.) In the case of unity gain, there will even be a short! So what am I missing? <Q> If your current sense resistor is in the 12V to 17V line the easiest way to convert the sense voltage to 0V to 5V is with a current sense amplifier. <S> If the current flow is uni-directional, you could use INA139 , or AD8212 <S> There are many others. <S> It is also possible to find some that can sense bi-directional currents, like AD8310 . <S> Using something like <S> these mean all you need are the current sense resistor and a small signal resistor to set the gain. <S> Problems with using a differential amp will be things like being forced to have low gain to keep the input within the common mode range of the Opamp, and having somewhat low input impedance (10 KOhms to 100KOhms) due to the differential resistors that set gain and divide down the input voltage. <S> Then having to add the gain later (and you are going to need an overall gain like \$\frac{5 V}{I_{\text{senseMax}} R_{\text{cs}}}\$). <S> Having control of tolerances can also be a challenge. <S> Getting around these problems become complicated, usually meaning a couple of additional amplifiers and precision resistors. <S> You could also use an instrumentation amp like an AD8223 , or an INA826 could work. <S> The nice thing about instrumentation amps is that the manufacturer has take care of the problems you run into when you make your own diff amp. <S> So, you get high input impedance, easily adjustable gain, and don't have to worry about resistor imbalances or offset voltages. <A> Additionally, if you are lower precision, you can build your own poorer version using op-amps look up <S> I-Amp datasheets to get guidance. <A> This appnote from diodes inc can give you more information about current sensing. <S> Some time ago I was also looking to sense current. <S> Here is the schematic I conceived with some help. <S> You can ignore Q1 and G1. <A> There are stand alone Current Sense Amplifiers in the market. <S> They are internally Gain set to a value, or have additional isolated feedback pins for adjustable gain values. <S> The advantage of the current sense amplifiers are that, they are purposefully designed for Over the top conditions. <S> Say for a supply voltage of 5V, some CSA's can take in upto 76V of Common mode voltage and have an internal gain of 25 to 50. <S> This can be something close to what you are looking for. <S> ThanksAshwin
| One option is to use an I-Amp (Instrumentation Amp) some of which will be way over kill for your application, but there are less expensive ones.
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Connecting to an EEPROM I am trying to download the contents of a ROM chip following this tutorial however the chip is surface mounted and I can't remove it to put it in a breadboard. I don't want to solder wires to the chip but there has to be a way of attaching them other than holding loose wires to the pins. Specifically This chip <Q> Update Here is a good list of micro grabbers/test clips (or whatever you want to call them): <S> http://www.jameco.com/jameco/products/prodds/684481-ds01.pdf <S> As mentioned, it would be helpful if we knew the board and the chip. <S> But without that information here is what I recommend. <S> First off, you need to buy some micro grabbers . <S> These will allow you to connect to individual pins on medium sized surface mount components (usually around .5mm pitch). <S> They look like this: <S> Now if you power to board and listen in on the bus, you can gather needed information like i2c address, etc. <S> and see how active it is. <S> From here you may be able to dump the memory without any ill effects. <S> Otherwise you can use my method for removing surface mount parts to only remove the data communication pins . <S> Then read out the memory like you planned. <S> This will make it much easier for you to resolder the two or so pins you removed back on, and is easier on the chip than completely removing it and trying to power it, etc. <S> If you don't have or want to get micro-grabbers, you can very carefully place a ~22 gauge solid wire under the pin but on top of the non-conductive material, so that the wire is making contact with the pin. <S> This will be hard to keep from disconnecting but with a bit of work you will get the hang of it. <S> If at all possible, I would stay away from soldering wires directly to the pins! <A> The datasheet you linked to shows only big fat packages like DIP and SOIC. <S> It should be easy to solder wires to those pins. <S> You say you don't want to but don't give any reason, so this is still your best option. <S> I would use thin "wire wrap" wire, which is around 30 or 32 guage if I remember right. <S> Such wires won't be strong, but good enough to talk to the chip to dump its contents. <S> Then you unsolder the wires when done. <S> When it's not working, you won't know whether that is due to firmware, wrong connections, or the connections you did make aren't working. <S> By soldering, you at least eliminate one of these three. <A> Best bet is to find the data lines that are on the board and can be tapped to get the right data. <S> Note that depending on how you connect to it, the circuit in which it sits might interfere with reading the data. <S> Note that some of these can be a bit expensive but are worth it. <S> See: <S> 3M Sockets Ebay listing for sockets - Seems to be inexpensive
| If you do finally decide that desoldering is OK for you , then several companies make Sockets for surface mount devices. Place something non-conductive between the pin and the pcb board so it won't make contact. "DIP clips" and the like can work, but in my experience they are also flaky.
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Why do some power adapters output an AC voltage? Out of most of the electronics I use, I have two, both of them speakers, that use a power adapter that simply steps down the voltage without rectifying it. Why did the designers choose to put the rectification circuitry in the speaker instead of the power adapter? <Q> (Like +5, 0, -5 V). <S> Another reason: Some devices run directly on AC, e.g. Christmas light bulbs can do. <S> Yet another reason: Some devices need two different voltages, like 5 V and 12 V. <S> Yet another reason: An external step down transformer provides just safety. <S> Voltages no higher than 50 V are usually accepted as safe (although not very safe with high current). <S> So the transformer makes it safe, the rest is in the device. <A> There could be any number of reasons it was done the way it was, and only the designers have those answers. <S> Some possible scenarios: A suitable transformer may not fit inside the device's enclosure. <S> In testing, the magnetic fields from the transformer may have caused unacceptable audible artifacts. <S> The transformer wall warts were cheaper than a chasis or PCB mount transformer. <A> It can be two things, and since it's audio, probably both. <S> The first, most electronic device companies specialize in one thing, the main product they sell. <S> They tend not to get into the highly annoying area of power supplies. <S> Their engineers figure out what voltage/current/quality power supply they need, and then they buy generics bulk from a Power Supply manufacturer (often paying extra for custom labeling or tighter quality control). <S> It's just way cheaper that way. <S> The second thing would be, especially in audio devices, additional quality control. <S> Might cost 10 cents for a ac to dc power supply compared to 5 cents to put in a rectifying stage inside the speakers while providing better rectifying control (eliminating cheap wall wart rippling or ground loop issues, shorter signal/power/trace lengths to deter emi etc), they will do it. <S> The power supply manufacturer(s) might not have a cost-effective standard part with a quality dc output.
| One reason is that you usually need positive and negative voltage for audio amplifier.
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Unable to find a source for rotary mirrors, how should I search for them? Can someone suggest a source for rotary mirrors? I'm looking for something along the lines of the ones in the image below, but I'd be interested in any alternatives as well. (I'm building a continuous laser harp, so the basic idea is I need to sweep an arc of about 45 to 60 degrees, or so.) From what I understand these are fairly common in some barcode scanners and laser printers, so I'm open to salvaging them if there isn't a good place to buy them independently, though I'm not entirely sure what I should be looking for (in terms of what kinds of barcode scanners / what parts of the laser printer). <Q> The items you've posted are all complicated, specialized integrated assemblies. <S> They're also very purpose specific. <S> As such, I can't think of many places you could just buy them as-is, independently of the device they're intended to work in. <S> Anyways, I do know that most laser printers to indeed use such a device, though I'm ~90% sure that the devices in the picture from the OP are all from bar-code scanners. <S> Basically, in a laser printer, they're used to sweep a modulated laser across the photosensitive, electrostatically-charged drum. <S> The charge on the drum is affected by the light striking it, and then becomes patterned, affecting how toner adheres to it in the next step. <S> Anyways, in such a situation, the laser is only swept in a line, so the aspect-ratio of the mirror is small, so they're typically very short. <S> In the mirror assemblies in the question, the mirror drums are very tall. <S> Since the height is not needed in a laser printer, they're probably from a bar-code scanner, that needs to sweep the laser in an x-y range. <S> Anyways, short of a few OKI printers (which are billed as a "LED printer" anyways), anything sold as a laser printer should have some sort of rotating mirror apparatus inside it, though how much luck you'll have making it work outside of the printer itself may vary. <A> BUILD, NOT BUY <S> Why not make a crude one using a bunch of tiny mirrors attached around the rim of a wood or plastic disk w/DC motor? <S> If too crude, learn from mistakes and make a better one on second attempt. <S> Perhaps make a 3D-printed base which is designed for whatever mirrors you've found. <S> (PS, they're called 'octagons' in the barcode biz, even if they're not 8-sided.) <S> Me, I'd try one of those piezo flapper-fan thingies and stick a tiny front-surface mirror on the oscillating blade. <S> I've seen very tiny f.s. mirrors at Surplus Shed , the online optics junk shop. <S> That, or slice up an old CDROM. <A> A few things you could look for in shopping for surplus parts either to repurpose or build your own: <S> First Surface Mirror <S> : These mirrors are coated on the front and are typically used in optics and laser applications. <S> They're available in "experimental," "commercial," and higher grades. <S> Unless you need specific wavelength specifications, go for the experimental or commercial grades (but even those can be costly).Sources: Edmund Optics , Anchor Optics <S> Galvanometer-driven X-Y Scanner <S> Pair <S> aka "X-Y Scanner Pair" or simply "galvos" or "scanners" in the laser entertainment industry <S> : This is a device commonly used in laser entertainment. <S> It consists of first-surface mirrors mounted on two galvanometers which are used to control the angle of a reflected beam on an X and Y axis. <S> You can find these for sale on eBay from time to time. <S> The price ranges from $50 to many hundreds of dollars depending on the quality and speed. <S> Depending on your application, you may be interested in something like this instead of mirrors on a rotating cylinder. <S> Stepper motor : <S> A motor that can be controlled in fine increments or "steps" for precision speed control. <S> This appears to be what the mirrors in your picture are mounted to. <S> They come in several types including unipolar and bipolar, and vary in the number of steps. <S> You'll have to check various surplus outlets for these. <S> I found a list of surplus optics and equipment at http://amasci.com/surplus/surpopt.html .
| Personally, I would check places like All-Electronics or ebay, if you can't track down some old laser printers locally.
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Can I drastically undervolt/underpower a relay? For instance, if I have a 24VDC/10A relay would running significantly less current (say 12VDC/2A) through it cause it to not operate properly? <Q> For the actual contacts, it's a good idea to provide a minimum wetting current to maintain good contact (the datasheet will sometimes specify a minimum load current). <S> Other than that you can run them at whatever current you like under the maximum value (10A in your example) <S> The coil should be driven at it's specified rating (24V in the example <S> you link to) Reference on switch contacts <S> Reference 2 <S> You should really make sure that your relay has a decent datasheet that provides enough info - <S> the one you link does not appear to have one, so I'd avoid it (unless you can google the part number and get it elsewhere) <S> Here are examples of the contact and coil ratings from a typical datasheet (in the contact section, note the minimum switching load and the different maximum switching ratings for AC/DC): Contact Ratings: Coil Ratings (note the release voltage is much less than the operate voltage - once the relay has switched, it takes less power to keep it in position): <A> Varies by manufacturer and device. <S> By drastically underpowering a relay, you risk not triggering the internal magnetic coil. <S> Normally though, a relay stated as 24v/10A, would mean it is triggered by 24v on it's coil pins, and rated to carry 10A on it's nc/ <S> no pins. <S> The 10A would not be it's relay trigger current, though there are industrial relays that can take that much. <A> If you are referring to the coil specifications, you must provide enough voltage to actuate the coil; this is often refered to as the "pickup voltage" specification.
| You would need to look at the datasheet for the relay to see if it is within range. If you are referring to the ratings for for the load (switching) path in the relay, you may run under the maximum current and voltage ratings with no problem.
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How to safely discharge high voltage (> 42 V) capacitors? I have a 1200 watt PC power supply (yeah that's for a gaming rig!). Unfortunately it's failing on me after about 3 years. As the guarantee is over and I don't want to have shipping costs, I want to open it myself and see what can be wrong. I have low to no experience with power supplies of this wattage and rating. So my question is, how can I make sure when I open the PSU casing, I am not going to kill myself. Is it enough to just leave it off for few days? Maybe not, because I can see huge capacitors inside. Maybe connecting it to PC and trying to turn it on without plugging it into wall socket could discharge the caps? And a silly question, the huge inductors inside are not a threat while it's not connected to wall socket, right? <Q> The best method is using a power resistor (3...5 W, 30...100 k\$\Omega\$), insulated with heat shrink tubes over the body and most of the wires. <S> If you are absolutely sure you will not connect the resistor across the capacitors of a power supply that is still turned on, lower resistor values may be useful for faster discharging. <S> With 330 \$\Omega\$, your discharge current will start at something like 320 <S> V / 330 \$\Omega\$ = 0.97 A, a safe value for most electrolytic capacitors. <S> Just remember that this resistor will dissipate ( <S> 320 V) 2 / 330 \$\Omega\$ = 310 W for a brief duration when used for discharging only ( safe! ) <S> - but the power will be dissipated continuously when your resistor is used over a live power supply's capacitors ( shock hazard from the voltage, burn and fire hazard from the large power dissipated by a relatively smallish resistor! ). <S> Here's a discharge resistor I built for developing and fixing switching power supplies. <S> The wire ends are (hollow) ferrules as used for connecting stranded wires into screw-on terminal blocks. <S> They connect really well to the ends of the PCB-mounted capacitors' wires pointing through the solder connections - and they won't slip off and produce shorts to neighboring components. <S> Y'all, feel free to copy the design and work safely <S> ;-) <S> Also a neon lamp with a series resistor of ca. 100 k\$\Omega\$ is a fairly good indicator of lethal voltages. <S> It will glow above ca. <S> 70 V. Ok, 70 V is more than SELV, but the lamp is still better than nothing. <S> Shorting the electrolytic capacitors with a screwdriver may cause a very high discharge current, possibly damaging them. <S> See here for some details . <S> A plain short is really quite the redneck way to go... <A> The inductors get safely discharged after a short period of time, about a second. <S> It would be safe for you to discharge them using an insulated screwdriver over the terminals the one of the caps at a time. <S> However, first of all you need to get to the terminals which will require you to take the circuit board out of the PSU. <S> the current will go in and out of the same hand, not via the torso. <S> I'm also assuming that you are not grounded while you are doing this. <S> Also, make sure the manufacturer doesn't have some kind of extended warranty before you do this, some corsair models have 5-7 years for example. <A> Short them using a resistor and that's it. <S> Just be sure not to take 1ohm one... Or even more safe solution: connect a small lamp (not a LED).
| You should be safe if you make sure that only one hand is in the PSU, that way IF you get a shock (which you shouldn't, I've taken apart over 20 PSUs and never gotten a shock) The only capacitors you would need to short are the high voltage ones(should be one or two of them).
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Why gate drive transformer is not connected directly to the gates in this App Note? Here’s a circuit from TI App Note SLUA323 , which shows a synchronous rectification in a current doubler circuit. U2 is a gate driver. Transformer T2, which looks very much like a gate drive transformer, provides both signals and power to U2. The output voltage is only +3.3V, which is a bit low for gate driving. Transformer provides the gate drive IC with higher voltage. Obviously, this circuit could be simplified if gates were driven directly by the secondary windings of T2. What was the rationale behind adding U2? <Q> This would help with the overall efficiency of the circuit. <S> Note that when either transistor is switched on, it's carrying the full output current (i.e., the current through both inductors). <A> Two things come to mind: <S> You want to insure that the driving waveform has high enough dV/dt to insure that the transistor is turned on quickly to minimize heating by operating the transistor with lower R DS(ON) . <S> You want to insure that the drivers turn on only when there is sufficient voltage being developed. <A> U2 would be there to snap-up the drive waveforms to the SR FETs. <S> UCC37324s are pretty hot drivers. <S> But, more important here is, it is doubtful that the author of the app note ever built this circuit as drawn because it won't work. <S> Input to the UCC37324 is into a MOS gate, and can't be allowed to float or the output will go high. <S> The schematic shows no way to pull U2 inputs low after being made high through the diode resistor combo.
| Without digging into the application note itself, I would think that the gate driver U2 and its local power supply are used to get higher current (faster edges) for switching the MOSFETs than you could get from the transformer alone.
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How to identify faulty or Fake (Duplicate) ICs? I am developing an application on LPC2138.I prepared prototype using a standard development board. After completion, I shifted to final board.But I am not able to download any program my LPC2138. Connections are correct. I compared with my development board and checked every single pin. Voltage difference at VCC and VSS is 3.3V. Reset pin is high. Crystal is supplying fine square wave. My final doubt is on Chip. I might have purchased a faulty or Fake LPC2138.Is there any way to identify that Chip is good and I am missing something in programming. P.S.: I grounded BSL pin, RESET the circuit. Ideally it should go in Bootloader and if I send "U" it should return "?". This happens on my development board but no response from final board. <Q> It's unlikely your part is counterfeit, especially if you purchased it from a reputable distributor. <S> Have you verified the following from this datasheet : <S> Is your P1.26/RTCK pin pulled low? <S> After reset the LPC2138 will sample the RTCK pin and if this is low the JTAG pins will be used as JTAG, otherwise they will be used as GPIO pins. <S> (Page 12) <S> The P0.31 pin MUST NOT be externally pulled LOW when RESET pin is LOW or the JTAG port will be disabled. <S> (Page 11) Reset pin is high. <S> How is your RESET pin pulled high, what value resistor? <S> Detecting Counterfeits <S> American Counterfeit Detection is one such company. <S> ex. <S> 0657 <S> Date codes that are in the future <S> If parts are moisture sensitive they require a dry pack and a Humidity indicator card. <S> Often the counterfeiters forget some piece of the moisture sensitive parts requirements. <S> This is a great resource for identifying counterfeit parts by visual inspection: Detection of Counterfeit Electronic Components <S> The following pictures show chips that have been sanded. <S> Normally when a chip is molded it will not have any scratch marks. <S> Here you can see a marking that is easily wiped off. <S> Here you can see that they have attempted to re-stamp over the existing stamp. <S> Aside from a external visual inspection, here are some other methods used: Functional Testing X-Ray, X-RF , or SAM Inspection Decapsulation Hermetically Sealed Electronic Component Leak Detection <S> Integra Counterfeit IC Detection Methods : <A> Waitaminit ... when you reset it, it should go into bootloader? <S> So when you bought an LPC2138 for your target board, how did the bootloader get programmed into it? <S> Did you buy it specifically from the same vendor as the development board, as an LPC2138 with <S> the correct bootloader pre-programmed? <S> Or did you buy an un-programmed device and program in the bootloader yourself? <S> EDIT : <S> the bootloader appears to be pre-programmed at manufacture. <S> However the only way to clear security keys on these chips is a full chip erase, and it is unclear to me what that does to the original bootloader... <S> So here's one scenario <S> : the bootloader is not being started after reset. <S> I found a dev board manual with some troubleshooting information on starting the bootloader - at reset. <S> P0.14 must see logic level 0 to neter the bootloader. <S> Another possible scenario : a genuine, hopefully working, but re-used part after a full-chip erase. <S> So some of the following may still be useful - <S> it ought to be possible to read the flash memory via JTAG and verify whether the bootloader is present and correct... <S> Here's an outline : you need programming hardware. <S> The dev board may be able to do that : is there anything in its doc about "programming target devices"? <S> Otherwise, programming hardware can be as simple as an "ISP interface" usually a 6-pin JTAG connector on your board. <S> This connects to a JTAG adapter (probably your dev board) by ribbon cable, and communicates (usually via USB) with programming software on your development computer. <S> Finding, installing and using it is up to you... <S> Add a link to your development board for more detailed help. <A> The easiest way to identify a fake is if the labeling on the chip is substantially sloppy compared to a known good chip ( http://www.bunniestudios.com/blog/?page_id=1022 ). <S> This only helps for obvious fakes. <S> Sometimes a official factory is used to produce fakes off the clock, so you can't even tell by that. <S> It's even a problem for the government ( http://www.bunniestudios.com/blog/?p=2037 ) <S> Aside from the obvious manufacturing issues (not exactly the right product packaging, or ic packaging [you get a 0.65mm pitch ic when it should be a 0.5mm pitch], the labels aren't what they should be), the only other way to identify if it is a fake, is by cutting it open, and again, comparing it to a known good version. <S> Not something the average joe can do <S> (remove the top layers of the protective packaging, take a electron microscope to compare) <S> (unless you are lucky and can plainly see that the chip has no internals).
| There are companies you can send a part to and have it tested, Some of the signs to look for are: Misspellings on the manufacturers labels Scratches or sanding marks Verification that the date code on the label match dates codes on the parts Date codes that are not possible.
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What exactly does a diode do? Disclaimer : I'm a total beginner here. As explained in other answers here , current in a circuit doesn't flow from negative to positive and the direction doesn't matter. So what should I make of diodes, which are supposed to make the current flow only in one direction? <Q> The other question is about the fact that current flows in complete circuits --- there is no current without a complete loop for current to flow through. <S> Some of the answers went off on a tangent discussing the difference between electron current and conventional current. <S> In circuit design, you can safely ignore electron current, and always think in terms of conventional current. <S> But the direction of current flow absolutely does matter. <S> Specifically, above a certain "threshold" voltage, it only requires a very small increase in voltage to increase the current to astronomical levels: (CC image from openwetware.org ) <S> The reverse breakdown behavior (large reverse current when high reverse bias is applied) is normally considered a non-ideality rather than part of the ideal diode behavior. <S> Some diodes have other effects, such as LEDs, which emit light when current flows through them; or zener diodes which are normally used in the reverse breakdown region. <S> Edit <S> what's the purpose of including a garden-variety diode (not an LED) in a circuit? <S> Typically, you use them when you want to be sure current can only flow in one direction. <S> For example, To protect a circuit from a battery installed in reverse. <S> To form a full-bridge rectifier circuit (using 4 diodes) to convert AC power to DC. <S> In a peak detector circuit. <S> In circuits using ac signals, proper manipulation of the dc bias point of a diode allows it to be used as a switch to route those signals. <S> You might also see a diode used in cases where the designer knows that current will be flowing in the proper direction to create a rough-and-ready "fixed" voltage drop of about 0.7 V. <S> Another use is to use a (properly designed) diode's ability to sink large amounts of current (in the forward direction) to protect more sensitive circuits from overload or ESD conditions, or in a snubber circuit to reduce transmission line ring. <S> Another use is that, going beyond the dc properties, a diode in reverse bias has a variable capacitance depending on the magnitude of the bias. <S> This variable capacitance can be used to tune oscillators or filters. <S> Diodes specially designed for this use are called varicaps . <A> For a beginner, the basic answer is that a diode is a one way valve for current. <S> Of course none of them are perfect. <S> There will still be a little bit of pressure (voltage) <S> accross the valve even when current is flowing in the forward direction, and there is some amount of backwards pressure (reverse voltage) at which the thing breaks, but "one way valve for current" is most of what you need to know to get started. <S> After you get into this a bit more, then go back and look at Photon's answer and see all the nasty little details of real diodes. <S> For now though, I think that will confuse more than illuminate. <S> One way valves for current are useful in a variety of circumstances. <S> One obvious one is to make a DC power supply from AC. <A> A diode always allows current through in one way only. <S> It has 2 electrodes , an anode and a cathode. <S> Usually you use them when you want to be sure that the current can only flow in one direction . <S> For example:- to protect a circuit from a battery being installed in reverse .- to form a full-bridge rectifier circuit (using 4 diodes) to convert AC power to D.C.
| As for diodes, ideally, a diode allows current to flow through it in only one direction, from anode to cathode.
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Using a WiFi antenna for UMTS I'm thinking of increasing my 3G USB modem's signal reception using an antenna. I know a UMTS(2.1 GHz) antenna is recommended. I need to know whether a WiFi (2.4 GHz) antenna will do the job (as I can find one easily); will it? (I found that two frequency ranges won't intersect, if the antenna is built properly ! ) <Q> But I personally don't recommend to do this mess. <A> Will it work? <S> Likely a little bit depending on whether the WiFi antenna can support a wide bandwidth. <S> In reality, the results are likely to be horrible and the connection non existent. <S> You will then try and troubleshoot problems due to this. <S> I highly recommend that you get the proper antenna. <A> You can try what I did to improve reception for a 3G cellular wireless modem. <S> Build a reflector for 2100 <S> Mhz. <S> Check out this link. <S> It is for a biquad antenna. <S> http://buildyourownantenna.blogspot.ca/2014/07/double-biquad-sector-antenna-for-1700-mhz-4g-lte.html <S> In place of the biquad element you substitute your wireless modem at the specified spacing. <S> The one with the wings is the one I constructed. <S> The results were fantastic. <S> According to what is said on the plans it gives you 13.87 db gain. <S> My modem that used to crank out 100mW in all directions is effectively cranking out 2.5W in one direction. <S> I get full bars from a cell tower about 5-6km distant with decent upload/download speeds.
| A simple whip antenna may work because it will probably have a wider frequency range, but better antennas can't work because narrowing the frequency range is one of the main ways of improving the antenna gain.
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How to check if my multimeter's fuse is damaged? I recently bought a cheap multimeter. I played around measuring stuff on simple circuits and I think I managed to damage it as I don't seem to be able to measure current anymore (always get zero amperes, although voltage, resistance, and the rest work just fine). I googled around and found out that it is not uncommon for beginners to blow their multimeter's fuse when trying to measure current the wrong way; I think I've fallen in this category. Though I would like to confirm, so I took out the fuse from the multimeter. I'm not sure how a damaged fuse should look like, yet it doesn't look broken, it is nice and clear and a very fine wire is visible. It is around 2cm long and on it is written F200mAL250V which I guess means that it is a fast glass 200mA fuse rated for 250V. Can I use my multimeter to find out if the fuse is broken? What metering mode should I use for this particular fuse and what should I look for? <Q> If it's very low (close to 0 ohms), it's still good. <S> If it's very high (open circuit), it's blown. <S> A 200 mA fuse should have a very fine wire visible inside the glass. <A> You may find there is a second, spare, fuse inside the multimeter. <S> That was the case with a cheap analog multimeter I bought at Radio Shack. <A> You can usually use the resistance measurement function to check -- connect the (usually) red ohms measurement wire to the current range input (also usually red) -- no need to connect the black wire. <S> If it reads < 10 ohm, then the current range (and fuse) is OK; if it reads open, then the fuse is blown.
| Measure the resistance of the fuse. If it's completely clear, the wire is gone (blown).
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What is the purpose of a capacitor to ground? I have here a filtering circuit from a microwave. What is the point of the capacitors to ground. Another answer in a previous question of mine said they were used for filtering however I don't understand why. The inductors by the way are part of a common mode choke. <Q> A capacitor's complex impedance is given by <S> $$\frac{1}{j \omega C} \Omega$$ <S> where \$\omega\$ is the frequency, and \$C\$ the capacitance. <S> For a low-frequency signal, you can see that the impedance will be relatively high. <S> Conversely, you can see that for a high frequency signal, the impedance will be quite low. <S> A high-frequency signal will see the capacitor connected to ground, and travel through it, since it is a low impedance path, but a low frequency signal will not be affected by it. <S> See this question . <A> I expect C1, C2 and C3 in your diagram are filtering capacitors. <S> They filter unwanted high frequencies from power line. <S> Their impedance is low for high frequency signal and high for low frequency signal. <S> This results in acting like a short circuit for high frequency signals. <S> All these capacitors are in dangerous places - in the case of their failure. <S> Because of this, special X and Y capacitors are used in these places. <S> I expect your C1 is X2 rated, while C2 and C3 is Y2 rated. <S> You can find more information why this is used if you search on Google for Y2 capacitors . <S> I think you should draw L1 and L2 differently, because it is a common mode choke on a single core. <A> You see these kind of filters ( common mode choke with capacitors to ground and between the power lines ) also on switching power supplies like computer power supplies. <S> Their purpose is not filtering noise coming in but preventing your device injecting noise into the mains. <S> Switching power supplies are notoriously noisy and they have to add these filters to pass EMI. <S> I don't know <S> but I guess the microwave oven might have a similar problem. <S> After all it is producing 2kW at 2.4 Ghz <S> so some of it might leak out into the power lines. <A> These are simply common mode filter capacitors. <S> In combination to the common-mode choke they filter out common-mode noise (noise present on both lines in respect to ground, or simply (Vline1+Vline2)/2. <S> This is different from Capacitor C1 which filters out differential noise.
| The capacitors to ground form a low-pass filter for the lines they're connected to, as they remove high-frequency signals from the line by giving those signals a low-impedance path to GND.
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How to attach normal Phone handset to jack 3.5 I have this Question : How can I attach the normal phone handset to jack 3.5 to use with my cell phone? For example : there is a product, http://www.cellphoneshop.net/redtel.html Could someone give me DIY How-To ? I want to do it all by myself. I'm Noob in electronics Fields , please explain to me simply. Thank you in advance. <Q> What they sell is no normal handset as it is known from old phones, it is just a normal hands-free equipment wrapped <S> (3$ in every store) in lots of plastic. <A> Take a telephone handset, take it apart. <S> Put the wired headset in the telephone handset. <S> Screw it back together. <S> http://hackaday.com/2009/12/23/bluetooth-handset-hacks/ <S> There really isn't much to it. <S> Maybe at most you need to unsolder some wires and add in a longer one or use some hotglue to hold it in place. <S> Then again, you can get these for like 12 dollars at a big box store like Sears at this point. <A> As @passerby and @Tom have very correctly noted... <S> That product isn't an old telephone. <S> Those would require an audio amplifier and microphone pre-amplifier to get levels that would work with your cell phone. <S> But, if you do have an old phone, do as @passerby suggested: Open it up, remove all the electronics you find in there, replace them with the electronics from a cheap wired headset .
| The simple method: Take a wired headset, take it apart.
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General Purpose Solder for Beginner's Toolbox? I've been asked by a friend to make such a recommendation, to go along with his new Weller WES51. I've told him: 60/40 tin/lead rosin-core good name brand, such as Kester What's a good size to recommend for general purpose use? <Q> In general, you should size the solder with the components. <S> The smaller the components, the smaller the radius of solder (especially if doing SMD use very thin solder). <S> For cases where you're soldering to large amounts of metal, it's best to use larger solder and more powerful soldering iron or it will take too long. <S> A general recommendation is Kester 0.010" solder for SMD, and Kester 0.025" for through hole. <A> I use 0.5mm diameter for normal stuff and 0.35mm for SMD work. <S> Though 1mm would also be OK for through hole, and you would need much less length for the same amount of solder. <A> Here's your standard size list . <S> For general-purpose electronic tinkering/prototyping, I recommend <S> In Kester terminology <S> it's this one: SN6324550.015 <S> 63/37 <S> , 245 FLUX, 50 CORE, .015 DIA, 1 LB
| For small components, pretty much any of the standard sizes will work. 26-28 awg sized no-clean rosin cored tin/lead solder (63%/37%).
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DVD player power supply circuit strange behavior. Need help to understand root cause I have an old (~8 yrs) Cyberhome dirt cheap DVD player that worked well for the time I was using it. I bought it in 2004 from a Best Buy and was using it regularly till 2010. After we got a new DVD player, it had been lying around packed nicely and securely. I recently brought it out just to connect it to another TV and found that it doesn't power on. I opened it up and was able to isolate its power supply (the pic shows it was very easy to do so). It works from 110 V to 240 V AC input 50-60 Hz. I have been using it in India which has 240 v AC @ 50 Hz usually. The DC out of the power supply circuit is like so +5V-GND-GND-+12V. I have assumed that the +5V was for the decoding circuit while the +12V would have been used by the player motor and maybe also the Class 1 laser.The +5V out works fine, but the +12V doesn't. The first time I checked it showed +17V and remained steady there. When I switched off the power, it very slowly (over 10-15 mts) ramped down to nearby 0 V. Based on this, I decided to wire up a voltage divider to bring 17 to 12 on a breadboard. Before I connected the +17V to the voltage divider, I checked again to make sure and now it showed ~+12V! It was slowly ramping up and 5 mts later was at +14.30V and has been holding steady there since last 10-15 mts. Switching off the power supply doesnt slowly ramp it down but goes down to 0 V pretty fast. What is the reason for such behavior? Obviously, something has gone poof in the power supply circuit, and I am interested to know the root cause. Some pointers will definitely help and using those I may try to debug through the board to figure out more. (I plan to wire up the input of the system to +5V-GND-GND-+12V independent of this power supply board to check the player but that's later as I don't have a +12V source handy currently. I tried with +5V and the dvd player powers on fine - but without the +12 doesn't do much apart from that). <Q> Most likely there's nothing severely wrong with the power supply. <S> Your description leads me to believe that the power supply has its feedback on the +5V rail and having zero load on the +12V rail causes it to drift high. <S> In the application (the DVD player) there's always load on the +12V rail which keeps the output closer to +12V. <S> High-pitch acoustic noise often comes from ceramic capacitors or the magnetic components. <S> A faint noise is generally not a cause for concern - it was most likely always there <S> but when it was enclosed you couldn't hear it. <S> That being said, it's possible that the power supply can no longer deliver its rated current, or the rest of the hardware is drawing too much current and forcing the power supply to protect itself. <S> ICs are tricky (especially if the unit is using some custom ASIC). <S> (Realistically, the way consumer electronics are designed these days, eight years is a remarkably-good run for a device that most likely only came with a 1-year warranty.) <A> Is it my imagination that the below shown two capacitors have a slightly bulged top? <S> If so it could be a strong indicator to replace all the caps on the power supply board. <S> The appearance could just be an angle of the lighting. <S> I do have to say that I replaced all the caps on the power supply board of a friends LCD TV. <S> In that case the reported problem was that the TV would not turn on. <S> None of the capacitors showed any sign of leakage or bulging <S> but I replaced them all anyway. <S> He reinstalled it and announced that his TV is now powering up and working!! <S> Power supplies in this consumer equipment are so competitive and built so cheaply. <S> In the past three years I've personally replaced the power supply caps in about 15 LCD monitors that ranged in age from 3 to 6 years and restored them to functionality by doing so. <S> In fact I'm using two of those monitors right now as I type this. <A> Q. if it has a 12V regulator then it might be shorted. <S> 17V <S> in 12V out. <S> The 17V is bleeding down faster because the divider network you installed is providing a lower resistance path at switch off than without it thus discharging the storage capacitor quicker, search <S> (time constants regarding capacitors). <S> Being cheap it is worth while to change at least the capacitors on the primary side of the supply (between the fuse and the blue chopper transformer) <S> If you measured 17V whilst everything was connected and turned on then the 17V rail is on load and if the supply requires a load to regulate down to 12V (poor state of affairs) (but you did say it was a cheapy,as most things are nowadays) <S> But for us curious and courageous ones, us who "need" to know I say press on dear fellow and fix everything that has died before we get buried in this stuff.
| You need to check for short-circuits and dried-out electrolytic capacitors on both the power supply and in the rest of the DVD player itself, as well as probe any diodes and transistors you can find. A common indicator is the one sided phenolic boards that are used and the strange off brands of electrolytic capacitors that are installed.
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Connecting 3.3 and 5 V devices The requirements are different from similar questions I've found here.TXB/TXS are expensive for my needs. Sparkfun's logic shifter is bi-directional, while I need two uni-directional conversions, RX<->TX pairs. As far as I understood from datasheets of both devices, the logic levelsATMega328P (VCC=5V) page 313 \$V_{l}\$ = 0v - 0.3v \$V_{h}\$ = 0.6v - 5v and GPS receiver's page 16 \$V_{l}\$ = 0v - 0.8v \$V_{h}\$ = 2v - 3.3v I've tried to do that with voltage divider, but doesn't look like it's possible or maybe I am missing something. <Q> I think you are using the wrong figures. <S> You must use the output figures for the output, and the input figures for the input, and do that in both directions and for 0 and 1. <S> For the AT->GPS: <S> AT out maximum for 0 <S> : 0.9 V GPS in maximum for 0: 0.8 V Strictly speaking the AT might not pull the line low enough, but in practice (with no load on the line) <S> this will be OK, even without a voltage divider. <S> AT out for 1: min 4V2, can be up to 5V GPS in for 1: min 2V0, max Vcc (3V3) <S> This is a problem. <S> A voltage divider will do fine at this low speed, don't take the values too high. <S> I would take 1k5 / 1k, but <S> probably 10k / 10k will do just as well. <S> For the GPS <S> ->AT: <S> GPS out at 0 <S> : 0 .. 0.4 <S> GPS out at 1 <S> : 2.4 .. 2.8 <S> AT in at 0: 0 .. 1.5V AT <S> in at 1: 3 .. 5.5V <S> One problem here: the output level of the GPS for a 1 is not enough to be guaranteed to be seen as 1 by the AT. <S> You can go for tricky passive solutions here, but a simple HCT gate at 5V will do the trick, their minumum input level for 1 is 2V0. <S> Take for instance a 74HCT00 (2 gates in series), or any other non-inverting set of gates. <A> For 3V3 -> 5V use a simple transistor: just drive its base using 3V3 signal. <A> This is how V-USB project does it for communication between a 5V AT(mega|tiny) and 3.6V USB. <S> The trick here is in what the AT sees as a HIGH (0.6 × Vcc). <S> So if your GPS module will output a HIGH as 3.0V then it should work just fine with the zener setup. <S> Maybe if you carefully design the wires between GPS and AT, the 3V output for GPS won't be a problem. <S> Notice that your project uses 3.3V, whereas USB data lines are at 3.6V.
| For 5V -> 3V3 use a simple resistor divider or a transistor.
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Physics behind switch bounce when opening? I was surprised to learn that a mechanical switch (SPST) bounces when it is opened. What causes the contact to temporarily return to touching? Does this phenomenon only happen with certain types of switches (e.g., with sliding switches but not with most pushbuttons)? <Q> Basically no substance is perfectly smooth, or inelastic. <S> When two objects come together with any force, there is kinetic energy that needs to die down upon contact. <S> So you get some "ringing", plus since the surface isn't smooth the contact makes/breaks a bit randomly until settling (e.g. even with a sliding switch the contact resistance will vary upon movement) <S> It happens with all switches; sliding, pushbutton, relay, etc, though sliding or wetted contacts (e.g. mercury) are better. <S> It also happens upon opening the switch, not just on closing. <S> Some are better than others - you can get some awful ones that will bounce for tens of ms, and some good ones that settle within <5ms. <S> The widely varying resistance this creates appears similar to the bounce during closing. <S> When pressure is released, one part of the contact may disengage, but another part may briefly obtain a lower resistance than it had previously, the more irregular the surface and the more of an angular force between the contacts, <S> the more pronounced this effect is likely to be. <S> Although it looks smooth to our eye, the copper surface looks rather different under a microscope: <A> Think about the reverse. <S> What if you were tasked with designing a switch that did not bounce on opening. <S> How would you guarantee at the microscopic level that the two conductors always moved away monotonically without any wiping action (sliding, which can cause momentary opening and closing)? <S> You have to consider that everything is a spring at that level, <S> no matter how rigid it may seem at a macroscopic level. <S> Even if it were somehow possible to not have any wiping action as the contacts separate, how do you guarantee there will not be any oscillation of the contacts as they move away? <S> You have to consider the velocity starts out at 0, so even a little backlash or ringing due to breaking of the static friction can be significant. <S> Think about it and describe us a dry mechanical switch that won't bounce if you actually think that is possible. <A> In the 1950s and early 1960s I worked as an electrical engineer in a team designing high current (>100 amps) regulated dc power supplies. <S> We had been using magnetic amplifiers for control, but then what we called SCRs (Silicon Controlled Rectifiers, now called Thyristors) began to appear. <S> They were very susceptible to overloads, and would blow faster than the fuses available, so one way to protect them was a "crowbar", that is a switch that short circuited the transformer secondary winding, so that the supply was cut off from the SCRs, and fuses in the supply line would blow -- the hefty transformer and its cabling could take the shock safely. <S> The trouble was that there was an awful lot of dangerous, and frighteningly noisy, sparking as the switch inevitably bounced. <S> The clever engineers in the research lab did design a "no bounce" switch, but it needed very delicate adjustment <S> so was no good for production equipment. <S> In the end improved fuse design and more robust SCRs made it unnecessary. <S> As I recall from a high speed film of the switch, an upper contact descended on to a lower one, and the combination was sprung so that they stayed together as they recoiled and swung up and down. <S> When we tested one there were indeed no sparks from the switch, but a loose connection to the transformer went off with an almighty bang. <S> (before the t dotters and eye crossers complain, <S> yes I know it should be " <S> there was no spark", but what I wrote sounds more like reality)
| To add more on the opening of the switch, it's due to the irregular surfaces opening in a not perfectly perpendicular manner.
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Sensibility and health hazards of reflowing computer PCBs in a domestic oven This is a bit of a fringe question, SE-wise, but I think EE is the best place to ask it. Some acquaintances of mine had a problem with their laptop - the graphics card stopped working. They've received (from someone else) instructions as to how to "reflow" it. Unfortunately I don't have the link to the text (and it's not in English anyway), but the relevant bit instructed to heat the card in a convection oven set at 180 C for 5 minutes. I was asked about my opinion about this fix and answered that while it's conceivable that it might work (my only idea was that it could perhaps remove solder connection defects, since the board is old enough to be pre-RoHS), I advised strongly against performing the fix in an oven still used for food preparation. To my dismay I recently learned that the acquaintances in question went ahead and performed the fix, in an oven used for home cooking no less. Interestingly enough, it worked, at least for now. Therefore, I want to ask the following two-part question: What are the health hazards associated in performing this sort of rework in an oven used for preparing food, and how to deal with those dangers? I would prefer for the answers to concentrate on the possible contaminants left over in the oven, rather than, e.g. increased risk of catastrophic failure of the PCB's components. How can this sort of rework be effective in rendering the board operational again? Preferably, I'd like an answer that addresses both parts, but for obvious reason I'll also be satisfied with a quick response on only the first one. UPDATE: Thank you for the answers so far. Since the most popular ones present opposing opinions, and the number of the respective upvotes is comparable, I think I should wait a day or two before accepting either. This is in the hope that someone would be able to provide concrete data relating to the subject. UPDATE 05.01.2013: I'm still going to leave the question without an accepted answer for some time. Seeing as how none of the answers have hard data to support them, I'm a bit apprehensive with going the either way. Sorry for that. <Q> They are almost certainly not in any danger. <S> Not any more than the fact that they handled the board and possibly ate something before washing their hands. <S> Direct transfer probabilities are much higher than a double outgassing transfer, first to oven and from oven to food. <S> I doubt that there is a lot of this. <S> On top of it, people typically preheat the oven which would spike the temperature up again beyond this "reflow" temp and this would liberate the bad stuff (if it was there) and carry it out of the oven. <S> i.e. it would be baked out. <S> Boards are relatively clean, lead is not very volatile <S> , there are VOC present but once heated they dissipate etc. <S> you wouldn't want to use an oven that is exclusively used for this, but once in a while is most likely very very safe. <S> In fact the oven elements when new outgas on first use, after cleaning the oven there are chemicals left on surfaces that outgas. <S> Those would be the best comparables and more concerning than the PCB. <A> Who knows what kind of junk is released from plastics, metals, solder, etc. <S> I wouldn't do it in my oven, but <S> hey... <S> It's a ghetto way of reflowing the solder balls on BGA chips. <S> Technically, there are "reflow profiles" that should be followed to melt the solder properly and all that jazz. <S> A reflow profile is a description of the way a temperature should ramp up or down over a certain period of time. <S> You can Google around for more info on that, but a regular kitchen oven is not going to be able to follow any sort of reflow profile correctly. <S> It should not be considered a long-term fix. <S> There is still a potential the same exact problem will occur again. <A> I would not do it. <S> Maybe I'm paranoid, but the risk of exposing my dinner to lead and other nasty stuff is too high. <S> Doing it once might be OK, if you are careful, but I deal with electronics all day, every day, so I'm more careful about such things. <S> As for the effectiveness... <S> The process of reflowing a PCB is normally tightly controlled. <S> You don't want it too cold, or to hot, or do it too fast or too slow. <S> Too fast or too cold will result in the PCB not reflowing properly. <S> And ideally you want some flux on the pins/balls that need reflowing. <S> Reflowing in your home oven does not control for either time or temp. <S> It might work. <S> Or it might not. <S> Or you could make things worse. <S> If the choices were to attempt a reflow or throw the PCB away <S> then maybe it is worth it to try to reflow it. <S> Worst case you'd throw it away anyway-- which you would have done anyway. <S> But if I did, here is how I would do it: <S> Get a cookie sheet and cover it in aluminum foil. <S> Cover the insides of your oven too, as much as possible. <S> Preheat the oven, with the cookie sheet in the oven too. <S> Attach metal stand-offs on the PCB. <S> Remove anything from the PCB that you can. <S> CPU, heat sinks, etc. <S> Place the PCB on the cookie sheet. <S> The standoffs will keep the PCB itself from touching anything. <S> When the time is up, turn off the oven and CAREFULLY open the door. <S> Do not shake, bump, or move the PCB. <S> Just let the oven, PCB, and cookie sheet cool with the oven door open. <S> Thrown away the aluminum foil. <S> This method should keep your oven as uncontaminated as possible, while keeping the odds of a successful reflow as high as possible. <S> Without practice, your odds of success are probably less than 50%. <A> Using the home oven is a terrible idea. <S> Health concerns aside, please see documents from IC manufacturers, e.g. https://aerospace.honeywell.com/~/media/Images/Plymouth%20Website%20PDFs/Magnetic%20Sensors/Application%20Notes/AN216_Mounting_Tips_for_LCC_Magnetic_Sensors.ashx <S> Excerpt: <S> Most LCC packages have no special requirements beyond normal procedures for attaching SMT components to printed circuit boards. <S> The exception to this process is Honeywell HMC products that have ceramic or FR4 substrate packages with epoxy top encapsulation. <S> These package designs use two solder types with differing reflow temperatures. <S> Inside these packages, a high-temp reflow solder is used that reflows at 225°C and above to make internal circuit connections. <S> On the package outside, low-temp solder is recommended with a reflow temp range from 180 to 210°C . <S> Three heating zones are defined in SMT reflow soldering process; the preheating zone, the soaking zone, and the reflow zone. <S> The preheating zone includes the soaking zone, and nominally ranges from 2 to 4 minutes depending on temperature rise to arrive in the 160°C to 180°C soaking plateau to active the flux and remove any remaining moisture in the assembly. <S> Preheat rise times must not exceed 3°C per second to avoid moisture and mechanical stresses <S> that result in “popcorning” the package encapsulation . <S> So, unless you're damn sure it's staying uniformly between 200C-220C and not heating faster than the specified rate, you risk damaging the type of IC being referred to. <S> If you look at a reflow oven, they provide a variety of different profiles that are geared towards different IC types + solder (leaded vs lead-free, etc)... <S> do you really want to disregard all of this and throw it in the oven with your pizza, risking health of both you and the silicon?
| With practice, this might work well. In the oven of course you have the potential for outgassing, which then coats out on the inside of the oven. Again, I would not do it in my oven. Doing it too slow or too hot can damage the PCB and chips.
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Is it safe to use a higher value capacitor? I am trying to replace a bad 820 µF, 6.3 V capacitor on my PC's motherboard. Right now I don't have a 820 µF capacitor in my home and also don't have enough time to buy a new capacitor. I have two 470 µF capacitors rated at 16 volts . The damaged capacitor is located near a RAM socket of the motherboard. Can I use a combination of two 470 µF capacitors that is almost 940 µF? Is it safe to use higher value capacitors in computer systems? <Q> Most likely the two 470 µF capacitors in parallel will be fine. <S> At these values, they are almost certainly power supply caps. <S> About the only drawback of more capacitance is higher inrush current when the supply comes up, but the 15% extra capacitance is very very unlikely to cause any problem whatsoever. <A> When it comes to capacitors, apart from capacity and voltage you also need to ensure that the ripple handling capability and ESR (equivalent series resistance) are well-matched. <S> In any application with high-frequency switching (i.e. DC/DC converters on motherboards) <S> the capacitors will be exposed to high frequency (100s of kHz or higher) ripple current. <S> A general-purpose capacitor (one that's only rated at 120Hz) will die a painful death within hours or days in this sort of application. <S> Another factor is operating temperature and rated life. <S> A 105C capacitor can be swapped in where an 85C capacitor was found with little risk. <S> The opposite case is not so safe to do. <S> Also don't substitute a 1000 hour rated capacitor in a circuit where a 5000 hour rated capacitor has failed. <S> Another issue is that if you drastically change the capacitance value and/or ESR by changing capacitors, you're changing the characteristic of the LC filter in the DC/DC converter. <S> This can be a problem if the feedback compensation of the converter is aggressively tuned, as you may render the converter unstable by the capacitor change. <S> If there are some ceramics in the circuit, their very low ESR tend to swamp out the influence of the electrolytics, making the swap somewhat safer. <S> All that being said, paralleling capacitors to match capacity is generally fine, so long as you're not substituting two GP caps (or a GP in parallel with a low-ESR), you're not going with a lower temperature part or drastically cutting the rated hours, and you're not radically changing the overall ESR of the circuit (usually not an issue if there are ceramic capacitors in parallel with the electrolytic caps). <A> Capacitors should be matched for capacity (in farads), esr, and resonance. <S> Changing this on a delicate electronic circuit like a motherboard is asking for trouble. <S> Caps arn't chosen at random by the manufacturer after all. <S> You can blow coils, fry chips, maybe even plague the other caps. <S> Doom and gloom aside, if your not willing to wait for a replacement capacitor, and are willing to buy a new computer if you screw up, you want to wire the two capacitors in parallel. <S> The slight difference in capacitance might not be enough to cause problems. <S> What might cause problems is a high esr rating. <S> Computer circuits arn't big fans of noisy caps and interference. <S> Certainly, it does depend on where in computer the cap comes from, but if it's near the ram, it's for the cpu, or the ram itself. <S> Data corruption is possible. <S> A power supply cap wouldn't be as risky. <A> Just letting you guys know, after seeing this site and real world testing it myself... <S> I'm not an electronics expert, but I have gotten an older (not functioning) ASUS mainboard where 15 caps of the same capacitance and voltage (6.3 V 820 µF) were (visibly) bad. <S> I decided to swap the caps from an even older mainboard with higher capacitance caps ( <S> 6.3 V 1000 µF). <S> Installed OS and software, and <S> after 2 days of using it, I think I can say it did the job.
| The Voltage rating is a safe limit, so increasing that is fine. There are capacitors branded as 'low ESR' and/or 'high ripple current' which are rated for operation at 100kHz - these are the sorts of capacitors you need to use in motherboard applications.
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Documenting Digital Design - Schematics and Figures I'm working on a small digital design using a Xilinx picoblaze softcore processor, and I'm finding that producing schematics of acceptable quality to be frustrating and time consuming. I've attempted to shoehorn previously drawn schematics (From LTSpice, or Eagle) into documentation, but the results look fuzzy or terrible. I've been using LibreOffice Draw, with subpar results - because the drawings do not contain any information about connections, having to propagate a small change through all items is really frustrating. For example, if I want to add a port onto a part, but that part appears multiple times in my schematic, I have to add them to every instance by hand. Similarly, when deciding the format of my wires and bus lines, perhaps I'd like the wires to 'jump' when not connected to another line? What if I'd like to add a bubble to indicate connections instead? These items must all be added by hand, which is ridiculous. I'm sure something like this could be done with LaTeX, but I'd like the feedback of being able to look at my drawing as it is being built, and I need to convince other people who are working on this project to use similar tools, and not many people are willing to undergo the learning curve of LaTeX just for this purpose. I've tried Inkscape and Dia and found them to be lacking, I've used LTSpice, but that's for SPICE and not digital schematics, and the diagrams are not very appealing anyway - a similar argument for EAGLE schematics. Is there a better way, something that I'm missing or not considering? I've just begun examining Electric VLSI - it seems to be a much better fit, but still rough around the edges. Does anyone have a good suggestion?Thanks! <Q> Here is a better way: <S> Don't! <S> The best way to document the design is by using the original code (VHDL, or Verilog). <S> This is the most accurate, and will always be up to date. <S> For simple designs, schematics might be more readable but that is not true for medium to large FPGA designs. <S> At that size, schematics become hard to follow, large, cumbersome, difficult to modify/update, and are impossible to debug. <S> So just don't use them. <S> Doing anything with more detail is a frustrating and fruitless task. <S> I would also argue that VHDL/Verilog are not difficult to follow. <S> If someone can't follow them then the problem is the person, not the code. <S> I have some FPGA designs that if printed out would take up about 800 pages of standard paper. <S> That same design, if shown using schematics, would require 2,000+ large pages. <S> It is easier to get proficient at reading VHDL/Verilog than to transcribe 800 pages of VHDL code into schematics (and then keep it up to date and accurate). <A> While I agree with David that trying to produce detailed schematics to mirror your FPGA design is more trouble than it's worth, sometimes just writing clean code is not an option. <S> I have been in situations where the design must be documented fully and the documentation must be readable by people who can't read code and have no intention of learning. <S> My solution was to adapt a verbose commenting style and use doxygen to produce HTML or PDF documents that fully documented the design in a fashion that was easily understandable by those not inclined to be able to read code. <S> The actual code structure can be automatically extracted and it will group and link the relevant sections which would be of more use to team members trying to figure out how different code sections fit together. <S> Some discipline needs to be maintained so that comments always reflect the code but at least you aren't jumping back and forth between programs trying to keep things in sync. <S> There is a slight learning curve but much less (IMO) than \$\LaTeX\$ . <S> It supports Markdown which is what the Stack Exchange uses. <S> So anyone on your team that uses this site should already be familiar. <S> It supports VHDL out of the box with a plugin available for Verilog. <S> After a slight learning curve and some initial setup, it becomes a one click solution to generating documents that can even be scripted into your build process if you're so inclined. <A> Have you tried using CircuitLab ? <S> It isn't digital-oriented, and we have some room to grow the symbol library, but it does produce what I think are nice-looking schematics and a lot of our users <S> say they're embedding the resulting PDF/PNG/EPS/SVGs into documentation/reports/presentations. <S> You can use the "custom part" tool to define rectangular items, and add some standard digital symbols around that. <S> (Disclaimer: I'm one of the developers and wrote some of the rendering code, but not the UI.)
| When I have to document an FPGA design with schematics, I do it with Visio and make it more of a high level block diagram than schematics.
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Use XBee IO with out microcontroller Im going to be using XBees in my project to wirelessly communicate between things, one of the devices is going to have a set of switches and 1 output - the xbee has enough IO to handle it. Is it possible to program the XBee to do the processing instead of having to interface a microcontroller with it? As I'd like to reduce the need of these extra microcontrollers. The idea of the system is that if a switch that is attached to the xbee is pressed the output is toggled, and then if it is toggled on the remote device, the output is toggled on the device. Is this possible or will I have to use microcontroller as well? <Q> However, integrating a microcontroller isn't always expensive. <S> Take a look at Ti's line of ValueLine MSP430. <S> They should do what you need and for less than $1 (perhaps even less that 50 cents depending on your requirements). <S> They come in PDIP <S> so they're much easier to use than other SMD microcontrollers. <A> You can use the new Programmable XBees. <S> I'm testing them right now <S> and I'm debugging an I2C sensor using nothing but an XBee and its integrated S08 MPU. <A> I would probably not use XBee radios as I am a heavy user of NXP/Jennic JN5148 based modules. <S> Since you stated you are trying to do this without an external microprocessor or software programming, here is a simple IEEE802.15.4 based alternative that you might want to consider: <S> http://www.starmanelectric.com/Products/DataBridge/tabid/70/Default.aspx <S> Note that the SMT RF modules used in the DataBridge product are available without firmware from NXP/ <S> Jennic: http://www.jennic.com and can also support Zigbee Pro development if you desire. <S> You can also do a single-chip level design without a module if you like to do RF work... <S> I've used many other RF chips including those from TI, Freescale and Atmel and have found Jennic to be one of the best. <S> The development environment is free and based on Eclipse IDE, they have good application notes and reference designs making it easy to learn.
| I don't believe there is any xbee with an integrated micrcocontroller.
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What do the two series-connected, forward biased diodes do? This is an example circuit providing a constant current for a laser diode: What does the two diodes in the circuit do? <Q> At a 'moderate' current (set by R1) a single Si diode drops ~ 0.6V, two in series drop <S> 1.2V. <S> Hence the voltage at the top of the potentiometer is ~ <S> 1.2V, more or less independent of the battery voltage. <S> The rest of the circuit serves to get a constant current through the laser diode: the OpAmp will vary its output to get the same voltage on its two inputs, hence the voltage across the resistor R4 will equal the voltage set by the potentiometer. <S> If the voltage across R4 is fixed so is the current, which is the same as the current through the laser (minus the much smaller base current for Q1). <A> They're generating a stable reference voltage. <S> In this case, the designer relied on two diode drops, which would give a reference voltage of 1.2-1.4V. <S> The potentiometer can then be used to adjust the voltage at the non-inverting pin of the op amp, and control the intensity of the laser diode. <S> The use of a dedicated reference diode would guarantee the required reference voltage, with much tighter manufacturing tolerances. <A> The provide a reasonably accurate voltage reference (twice the forward voltage of an 1N4148) that is then divided by the pot and fed into the opamp. <S> There it is compared to the voltage drop over the shunt resistor, which is proportional to the current through the laser diode and shunt, thus a constant current (adjustable with the pot) is maintained. <A> U1 will keep the voltage on its pin 2 the same as pin 3. <S> The voltage at pin 3 when divided by R4 yields a current that will be very close to the current in the laser since the collector current of Q1 will be very close to its emitter current. <S> Q1 provides current gain so that U1 does not have to provide the laser current. <S> Thus the overall circuit acts to provide a constant, stable current to the laser. <A> The Multisim analysis [before someone vandalised the question to delete that part] is wrong; perhaps the supply voltage is about 3V in which case it would be about right for entirely the wrong reason. <S> The two diodes will ensure the voltage across them is two diode drops, regardless of the supply voltage. <S> As to why, there is probably a note that accompanies the schematic. <S> The diodes and current sense resistor R4 are marked with an asterisk; the note presumably explains why. <S> It might explain that the diodes have to be mounted in close thermal contact with the laser diode. <S> Then as the laser heats up, the diode voltage reduces, and so controls the laser current. <A> The power supply provides the current (through R1) to forward bias the two diodes. <S> Each diode, when forward biased, has a voltage drop of approximately 0.7V. <S> In series, they will drop about 1.4 V. Potentiometer R2 functions as a voltage divider that allows adjustment of the voltage at pin 3 of U1 between ground and ~1.4 V. <A> It looks like they are acting like the bottom part of a voltage divider. <S> This post talks about some of this: <S> What are the advantages and disadvantages of using signal diodes instead of resistors?
| The 2 diodes provide a reasonably stable low voltage reference which is picked off by R2 and input to pin 3 of U1.
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microcontroller for custom laptop I'm new to electronics (outside of twiddling with Arduino and a basic college course - I'm more of a software guy) and I'm attempting to build a custom, extremely low-powered, extremely thin, laptop. The ideal is a keyboard, a microcontroller or some chip running command-line Linux distribution (potentially uClinux) and an LCD display. It could potentially run on double A batteries for the time being. It would also need wifi access. My current biggest question is what kind of microcontroller (or other device) I should look for as the brain of the device. I've considered just sticking in a Raspberry Pi, but that's actually much larger than I would like, so I'm willing to attempt a custom-soldered solution. Are there sufficient ones which already have Linux installed? Are there some which have enough input pins / proper inputs for a keyboard / LCD display? I'm also willing to write a driver or two for the keyboard, LCD, etc. as necessary, FYI. <Q> For anything to run linux, it can't be a simple 8 or 16 bit microocntroller. <S> Your best bet is an ARM based system, in particular Cortex A8 or A9. <S> Basically aside from the raspberry pi you've got beagle board, beagle bone, panda board and a few similar other boards. <S> I'm not sure there is a custom apps processor module available in general. <S> If you're an OEM then of course you can find something. <S> If you are willing to build your own solution you can make it very small. <A> Microcontrollers would not be a good choice for a laptop. <S> They are good for embedded control and not for personal computer. <S> To build a laptop, you would need other peripheral controller chips like drive controller, display controller, <S> I/O controller, etc. <S> These other chips may not interface with a "microcontroller" as cleanly. <S> Further, you can buy a bare-bone kit and build your own laptop. <A> AA batteries don't have enough capacity. <S> When you're using the LCD backlight (and you will be) <S> the device will consume probably around 300mA - 500mA, depending on peripherals and other hardware. <S> Using AA batteries of possibly 1500mAh <S> you would be done in a few hours the first time, the second time in like one hour. <S> Use a LiPo accu on 3.7V with a charge pump to 5V, it is flatter and has more capacity. <S> When you're using uClinux, the FAQ and Getting <S> Started pages on their webpage can help you out on what device you should use: an ARM controller. <S> However I would recommend you to build your own OS with only the functions you need - it would be way faster and it's more fun to make :-) <S> I'm building my own laptop-ish device, CText. <S> This is my setup, you might find it interesting when you're going to build it from scratch with your own OS: <S> A PIC microcontroller keys which interfaces the keyboard of an old psion 5mx series A PIC microcontroller graphics which interfaces an LCD screen A PIC microcontroller <S> main which calculates everything and communicates with the other PICs (over I2C), and handles RAM, SD card memory, RS232 communication, etc. <S> Wi-Fi access using a microcontroller is a bit hard, but there will be modules available on the net. <S> However you could also decide to use a chip with a USB peripheral to connect to your android phone (if you have one) and use the internet of the phone. <S> You can build your own app to do this. <S> Updates on this project can be found here: http://camilstaps.nl/topic/electronics/electronics-projects/ctext/
| To build a effective laptop you would be better off using a microprocessor.
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Resistance to Digital converter I am trying to make a digital ohmmeter on a parallel port interface that can measure resistance with at least 95% accuracy. I have tried a 555 timer monostable assembly which generates a pulse with respect to the connected resistor. My program detects the width of pulse and then after calculation gives a value of resistance, but the results are not accurate, especially for small values. Is there any IC avaliable for resistance to digital converter? The output may be serial or parallel. <Q> If you are familiar with microcontrollers then you can use an ADC. <S> Place the unknown resistor as one half of a voltage divider (with known precision resistor), apply known voltage and measure the output at the junction. <S> Do the math and you can work out the unknown resistance. <S> There are a few variations on this (see Wheatstone bridge , pictured below) and you can get quite complex with it the more accurate <S> you need it to be (see LCR meters <S> also - they use a known frequency and measure amplitude and phase angle, so you can obtain a lot more data, e.g. DF, Q, θ, etc) <S> Wheatstone bridge configuration: <S> You can also get ICs and off the shelf modules that can be configured to measure voltage/current/resistance/etc. <S> Obviously a multimeter (e.g. with USB/RS232 link) will do what you are asking for, but I'm guessing there must be a reason you want to avoid this route. <A> Make that a constant current is driven through the resistor. <S> Measure the voltage over the resistor. <S> For example, when the current would be 1A (way too high of course), the voltage would equal the resistance following U = <S> I×R. <S> But when you make a constant current source of 1mA, the voltage equals the resistance in kilo-Ohms. <S> So 1V equals 1k Ohm. <S> Choose a current that gives measurable and reasonable voltages on the reach you want to have. <S> You can also do something with a potentiometer to allow multiple currents. <S> A way to create a constant current source can be found on Wikipedia . <S> Measuring the voltage can be done with a microcontroller's ADC or the 7106 IC if you want to display it on an LCD screen immediately. <S> Got the idea from: http://www.electro-tech-online.com/general-electronics-chat/97919-measure-resistance-ic.html <A> You specify 95% accuracy. <S> This is more commonly expressed as 5% inaccuracy. <S> But you must also specify the range of resistors that you want to measure. <S> The wider the range the more difficult it gets, and very low and very high <S> values are an extra challenge (even if the range is small). <S> I think pulse length measuring using the parallel port will suffer from interference from other activities on your PC. <S> The audio input is probably better suited for this case: make sure the frequency you generate is in the audio range, determine the frequency in software (that's a SW problem, not for discussion here), and calculate back to the resistance. <S> If you really want to use the parallel port get yourself an A/D converter chip with two inputs and measure the reference voltage and te voltage over your resistor, with a known series resistor. <S> A/D converter chips are available in parallel, SPI or I2C fashion (to name a few), which can be interfaced to a parallel port (with some SW effort). <A> As mentioned previously, it would be difficult to make a single circuit to measure a wide range of resistances.
| If you put the resistor into a non inverting amplifier as the input or feedback resister, you could easily convert resistance into voltage, and then convert to resistance after measuring the output voltage via an ADC, which is widely available on cheap ucontrollers.
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