source stringlengths 620 29.3k | target stringlengths 12 1.24k |
|---|---|
ESR meter: Are in-circuit readings accurate? I recently bought one of these ESR meters, a 'MESR-100': http://www.amazon.com/MESR100-AutoRanging-Circuit-Capacitor-Tester/dp/B00G7OPBP2 It claims to be able to test capacitor ESR with the caps left in-circuit. It appears to work well from my tests so far. However, how much can I trust the in-circuit readings? Are they going to be less accurate? Are there certain circuit configurations where it wouldn't be possible to measure ESR in-circuit accurately? [EDIT: Added the model as requested] : <Q> It's quite useful for repair purposes. <S> The voltages are low enough that semiconductors don't come into play. <S> When there are parallel e-caps, you should be comparing a total ESR of the parallel combination, but you're really looking for gross differences not even the last 2:1. <S> The typical situation is an older piece of electronics where the electrolyte dries out, causing the capacitance to change a bit, but mostly the ESR to increase, so you get mains ripple or SMPS ripple, which eventually causes the thing to stop working adequately. <S> E-caps do eventually wear out, and replacing them can often restore an older item. <S> Especially the caps in the power supply- because they usually have less margin and because they tend to be in a hotter area and have more self-heating (due to ripple current and ESR-- so the higher ESR gets, the more self-heating occurs). <S> You may have to look up the part number or guess from the schematic or circuit configuration (something like 1000uF/6.3V on a motherboard is just about guaranteed to be low-Z). <S> I wish I had one of those <S> , there's a few times it would come in handy, but I don't do enough repair work to justify it. <S> Usually I just tack a known-good cap across the suspect one and see if it makes a difference! <A> I'm talking about power rail circuits because they are a common usage for large electrolytics where you might be particularly relying on ESR being low. <S> In places other than power rails, electrolytics are of course used but, is the ESR that important in these places? <S> I'm thinking time delay circuits like 555s. <A> I have used a relatively cheap Atlas ESR meter to test capacitors in a faulty Jamo E7 Subwoofer. <S> The meter accurately identified the faulty 10uf capacitor in circuit. <S> Measurement generated significantly higher uf readings as well as leakage. <S> I am happy to report that the sub is still working, 6 months after the cap replacement. <S> (I was expecting the repair to be short lived.) <A> It depends on the design (excitation voltage) and the dut surrounding circuitry. <S> If the excitation voltage is low, it may not turn on a transistor or diode, sand <S> there isn't other resistors or inductors around, it can be fairly accurate.
| The other factor affecting usability is that a lot of recent electronics uses low-Z caps, so what's excellent for a standard e-cap is a bulging mess on a low-Z cap. You can't rely on readings when the capacitor-under-test is in circuit because other components (such as other similarly sized capacitors) may be in parallel.
|
12V 120V inverter and battery for off grid oxygen I have an oxygen concentrator that is 120V. The data plate says 3.0 amps and the specs say 280 watts. My question is this. I want to use it in an RV off the grid but without running my generator all night. I have two Interstate 12V AGM batteries I think they are rated at 75 amp hours each. If I use an inverter will these batteries be enough to power this unit for a full night? If so, what size inverter do I need? Sorry about the basic nature of this question but I am an electronic dummy. <Q> Your question is "what capacity of batteries do I need to power a 280W 120V load from 12VDC lead-acid Absorbent Glass Mat (AGM) batteries for a given length of time?" <S> The rated power draw is 280W. <S> The batteries will discharge to something like 10.5V at the end of discharge, so let's say that the average voltage is 11.2V. Further, let's assume the inverter is 75% efficient. <S> That means the average current will be \$280W \over 11.2V <S> \cdot <S> 0.75\$ = <S> 33A. <S> That means that if the batteries can actually safely (without damage) deliver 75Ah each, the operation time will be about 4.5 hours. <S> You really need to look at the detailed data for the batteries to see what they recommend for discharge "depth" to see if that assumption is true, but it's unlikely it will operate for more than 4-5 hours. <S> For 8-10 hours of operation you'd need about double that capacity of batteries. <S> AGM batteries generally allow significantly more discharge depth than conventional lead-acid batteries, but it still might be only 80% of the rated amount. <S> Note that this assumes that the oxygen concentrator presents a steady 280W load. <S> If the load is intermittent then 280W likely represents the maximum load, and you may get considerably more operating time out of a pair of batteries. <S> There is an inexpensive gadget called a Kill-A-Watt (see Amazon etc.) <S> that can be used to measure the consumption, and it may give you a more reliable number than relying on the nameplate watts. <S> I would definitely make that determination before buying the batteries. <S> Here 's a portable oxygen concentrator that claims two hours at 2l/min continuous flow (the mode I presume you'd be using at night), using a 98Wh lithium battery, so the average consumption of that product is only about 50W. <A> If the thing draws 3 amps at 120 volts, the inverter will draw somewhat more than 30 amps at 12 volts, depending on the inverter efficiency - say 35 amps as a rough estimate. <S> If the unit draws 280 watts continuously, you're looking at 350 ampere-hours to run it for 10 hours. <S> It is commonly recommended that you not discharge a lead-acid battery more than about 50%, so you'd need about 700 AH of battery - a bank of six "golf cart" deep cycle batteries might be adequate. <S> Then you have to think about recharging the batteries, and running the oxygen concentrator during the day.... <A> As others have indicated about 30 amps on the 12 volt side so takes a lot of battery power for the night. <S> though while driving it could easily run off the alternator at 30 amps. <S> you would be best off getting a small quite generator like the 1000w inverter ones. <S> Wouldn't cost much more than 1 set of batteries to run it for the night.
| You will need an inverter rated at more than 280 watts - 400 - 500 watts would be good, and should provide some headroom for possible motor starting surges.
|
Just making sure I won't fry my motherboard? I'm building this circuit , which converts a toggle switch to a momentary one, which sends a key to a USB keyboard PCB by connecting 2 contacts together: I just wanted to check, how do I know the keyboard PCB won't get fried when I flip the switch to 12v DC? Or worse my motherboard? <Q> If nothing but the contacts goes to the PCB you won't hurt the computer. <S> If you want to be super-cautious, add a 0.1uF capacitor across the relay coil. <S> That will ensure that when the switch opens, no serious spikes get coupled to the keyboard scan lines. <A> The momentary buttons make a physical connection when pressed. <S> The relay's actuating part, does the same. <S> The relay provides physical isolation between the coil and the actuating part. <A> It seems a resettable polyfuse would do the trick: USB Overcurrent Protection <S> The Arduino Mega2560 has a resettable polyfuse that protects your computer's USB ports from shorts and overcurrent. <S> Although most computers provide their own internal protection, the fuse provides an extra layer of protection. <S> If more than 500 mA is applied to the USB port, the fuse will automatically break the connection until the short or overload is removed.
| And unless you short a wire out or physically connect them wrong, you won't fry the computer or keyboard.
|
Amplification without transistors or tubes? Are there any means or technologies that can be used to linearly (or close to linear) amplify an audio or LF RF electrical signal that do not require transistors or vacuum tube technologies? (Or equivalent tech such as ICs, photo-transistors, klystrons, etc.) Specifically linear (or "close enough" as are the usual slightly non-linear simple tube or transistor amp circuits), so disqualifying a two state relay as a possible solution. Were any electromechanical gadgets tried before Fleming and De Forest? <Q> Negative-resistance devices such as tunnel diodes can be used as amplifiers/oscillators. <S> There are also magnetic amplifiers , most often used for electromechanical control. <S> A relay has gain, but it's highly nonlinear. <S> Are these the sorts of things you're looking for? <A> A generator without permanent magnets can be used as a amplifier. <S> The extra power comes from the power provided by mechanically rotating the shaft. <S> The magnitude of the output signal is then proportional to the shaft rotation speed and the field winding current. <S> Since extra power is being added to the system via the shaft, you can get more power out than you put into the field windings. <S> If you keep the shaft rotating at a fixed speed, the generator can be thought of as a amplifier from the field current to the output. <S> One problem with this is that these generators usually put out AC. <S> That can be rectified, but will still have some ripple. <S> If you wind the generator with enough poles and spin it fast enough, the AC output frequency can be higher than your highest frequency of interest. <S> In that case, it can be AM demodulated to yield the final amplified signal. <S> Rigging up something like this to work at audio frequencies will be tricky. <S> However, you can demonstrate the concept without too much trouble at lower frequencies. <S> Connect a LEB (light emitting bulb) to the output of the generator and spin it to get typical line frequency out, like 60 Hz. <S> Now you can control the lightbulb, or a bank of lightbulbs, with much less power than it takes to drive them directly. <S> The alternator in your car actually works on this principle to keep the rectified output of the alternator at the right voltage. <S> The alternator spins at whatever speed it spins at. <S> The control system modulates the field current so that the alternator output produces the right voltage to keep the battery charged but to not overcharge it. <S> In this case the "amplifier" is being used as a voltage regulator, but the overall concept is the same. <S> Less power is required to run the alternator field windings, which comes from the 12 V DC, than the alternator puts back onto the 12 V DC line. <S> The extra power comes from adding a mechanical load on the engine. <A> A pure mechanical audio amplifier called an auxetophone <S> uses an air compressor and a gate that is mechanically modulated. <S> This video shows a recreation of an auxetophone where a solenoid is used to move the gate, but it can be mechanically moved. <S> In a similar manner, the gate could be replaced with an adjustable power resistor, so you can build a mechanical amplifier that converts the small signal to mechanical movement, which then modulates a much larger power source. <A> A transformer can provide gain i.e. a "step-up" transformer can magnify the input voltage by the turns ratio <S> but it isn't a power amplifier. <S> They can be fairly linear. <S> A tuned circuit (L & C) can also amplify a voltage signal by many times but it is limited to a small band of frequencies close to the resonant point. <S> Beyond that frequency signal output levels fall off. <A> In the early days of radio, a device called the coherer was used.. <S> It could be described as a detector with gain. <S> Not linear at all. <S> There were various devices in the early history of instrumentation that provided something like gain without actual power gain. <S> For example, the optical lever, the galvanometer, the thermopile.
| Parametric amplifiers can be made with nonlinear capacitors .
|
what do the nm or µm for a process technology refer to e.g 22nm process technology for device fabrication There are many "process technologies" that exist to fabricate circuits on silicon. They are mainly differentiated based on size of some feature. Here is the list of them: 10 µm – 19713 µm – 19751.5 µm – 19821 µm – 1985800 nm – 1989600 nm – 1994350 nm – 1995250 nm – 1997180 nm – 1999130 nm – 200290 nm – 200465 nm – 200645 nm – 200832 nm – 201022 nm – 201214 nm – 201410 nm – est. 20157 nm – est. 20175 nm – est. 2019 What I wish to know is, what precisely do these numbers refer to e.g the 90nm that was reached in 2004, what does it precisely mean? I think it has to do with some dimension of the transistor that is fabricated but do not know the details. So what does it mean? <Q> The nanometer number on memory chips generally refers to the smallest "half-pitch" between identical features on the chip, as in this illustration from an IEEE article . <S> By this definition, a fairly fine-pitch PCB with 0.004" traces and spaces would be defined as 100 micron (100,000nm). <S> A CPU with a 32nm node might have a 50~56nm half-pitch, and a 22nm chip might have a 34nm half-pitch. <A> It refers to half the distance between neighbouring details on the die (ie between two CMOS) <A> Those are standard units, not specific to electronics or semiconductor manufacturing. <S> The "m" stands for meter, and the letter before it is one of the standard prefixes denoting a power of 1000. <S> Here are some common prefixes: Prefix Name <S> Multiplier------ <S> ---------- <S> ---------- <S> G <S> Giga <S> 10 9 <S> M Mega 10 6 k Kilo 10 3 <S> m milli 10 -3 <S> µ <S> micro <S> 10 <S> -6 <S> n <S> nano <S> 10 -9 <S> p <S> pico 10 12 <S> So "22 nm", for example, means "22 nano-meters". <S> This is the minimum feature size of that technology.
| On microprocessors , it refers to the "node number" (the smallest feature).
|
Making my own 8x2 pin connector cable I am currently using two 8-wire ribbon cable to connect 2x8 pins on the Arduino to 2x8 pins on a board. This makes plugging/unplugging very tedious are the ends are not glued together like in a connector Question: How can I create my own ribbon cables with connetors for 2x8 pins? After some researched I found IDC connectors somewhat suitable, but the connectors appear to have additional plastic like the clips which will obscure the 2 pins to each side of the 2x8 pins I'm plugging the cable into. What other options do I have? I'm hoping for a 8x2 version of the following cable, what are such cables called? <Q> Wires . <S> 2x8 housing . <A> One possible solution is to use the 0.1" (2.54mm) double row male headers on one side. <S> and a female 0.1" (2.54mm) <S> double row header in the other side Get some multicolor flat ribbon cable <S> and you can make your own cable at any length and pin count. <S> If you want you can reinforce the wire/plug connection using a hot glue gun, something that looks like <S> Another possible solution may be to use your current jumper wires, just use glue or hot glue to hold them together as one piece and should be fine. <A> The connectors in your last picture appear to me to be "AMPModu" connectors, made by AMP. <S> The housing are available in both single and double row. <S> The contacts are crimped on the wires, then inserted in the housing. <A> Make a PCB bridge Something like this: <S> Either use generic prototyping board or have a custom PCB made just for you. <S> Then solder both male and female 8x2 header to it. <S> Some manufacturers can also do flexible PCB boards. <S> You can design the bridge using Eagle Cad and then you can shop for a manufacturer here .
| Pololu.com sells connectors like this, and they have precrimped wires as well.
|
What will happen if we excite a Synchronous Machine with AC? Normally we excite Synchronous Machines with DC. But what would happen if we provide AC in place of DC in field winding of Synchronous Machine? Will Synchronous Motor start in with AC excitation? <Q> You'll see smoke I reckon. <S> Think of a transformer with an AC voltage <S> applied to both primary and secondary windings - unless this is totally matching the turns ratio there'll be a lot of smoke. <S> Imagine a synchronous motor run like an induction motor - you can easily do this by shorting the rotor slip rings. <S> When first powered, the motor generates tons of torque because the induction into the rotor is 50Hz and this generates a field that causes the motor to start to turn. <S> As it turns the "slip frequency" gets less and less until it is a few Hz <S> - it reaches equilibrium somewhere close to (but below) synchronous speed. <S> If you held the rotor fixed mechanically so it couldn't turn, the rotor induction would be massive and at 50Hz - this is what your question is asking - it would burn. <A> If the AC frequency applied to rotor and stator windings is the same, the motor will consume some energy and turn it into heat. <S> The result depends on the windings' cooling, resistance and applied voltage. <S> If the frequencies are different, the syncronous motor will turn into a double-feed synchronous machine, and it will be stable rotating at two frequencies: f1-f2 and f1+f2 (where f1 and f2 are stator and rotor AC frequency). <A> A doubly-fed machine requires a poly-phase rotor, while the rotor on a synchronous machine is typically only wound on one axis (they are effectively single-phase). <S> In order to achieve a frequency difference, it is necessary to have a revolving magnetic flux on both the rotor and stator, but the flux can only decrease to 0 and reverse on the rotor unless it is wound in both the q and d axis (meaning that it is poly-phase). <S> In short, a normal synchronous machine can't be used as a doubly-fed induction machine. <S> Providing an AC current to the field would cause it to reverse poles at the exciting frequency, potentially causing instability during ordinary operation, but providing no benefit or aid for starting. <S> That said, it would be very difficult to apply 50 or 60Hz AC to the rotor on a synchronous machine. <S> The inductance of the rotor on such a machine is typically on the order of 1 H or greater (depending on the size of the motor), so the impedance to drive at 50Hz would be about 300 Ohms while the impedance at 60Hz would be about 377 Ohms. <S> This means that driving at 377 VAC at 60Hz would result in about 1A of AC current on the rotor. <S> Rotor currents for a synchronous machine with a 1H rotor inductance would tend to run closer to 10ADC, so the impressed AC would make very little difference. <A> About 2009, TECO informed me they could use VFD start on synchronous motor, but had to change excitation to ac. <S> The project didn't get funded so it never went any further.
| To answer the original question, it is entirely possible to apply AC on both the rotor and the stator of a synchronous machine, as long as the current flowing in the coils doesn't exceed the coil rating, but there is no benefit to doing so.
|
Will this relay handle my current and voltage requirements? I have this relay: SONGLE SRA-24vdc-cl I want to use it to run 240V 11A device. I can see in the datasheet it can be up to 20A (section 7, line 4-5), but the relay is so small and its legs so thin, so I am asking here before connecting it. The 240V device should operate for 1 - 1.5 hours with 11 - 12A. The coil is 24VDC, which I will use MOSFET to give it power (I have 300mA 24VDC adapter), and the gate will be connected to AVR MCU. <Q> This relay is inappropriate for what you want to do. <S> You need to actually read the datasheet . <S> In the middle of the first page for #4, RATING, it clearly says 125 VAC. <S> Therefore <S> 240 V is well out of spec. <S> What part of this is so confusing? <S> 11 A would be fine, if you weren't violating other specs . <A> The relay is rated (by the Chinese manufacturer ) for 20A @ <S> maximum 125VAC. <S> However, for some reason, UL does not believe that <S> , they think it might just burst into flames at more than 10A/125VAC <S> (they don't care about functionality or life in the ratings, just about safety). <S> A hint is that they only rate it for 10A at 14VDC, so it's probably getting too hot for their liking due to internal resistance. <S> I tend to pay serious attention to safety agency ratings, particularly when the mains is involved. <S> Think about defending your choice in a court of law after someone dies or their house burns down, and how such a choice is going to sound. <S> " <S> Well, I just picked the highest numbers I found. <S> YOLO". <S> This is engineering <S> and we have a legal as well as a moral requirement to provide for the public safety. <A> Others have already commented along the lines of RTFM <S> so I won't re-answer that. <S> Doubly so when the device is a cheap no-brand one from eBay China. <S> I would normally over-spec things like relays, switches, cable, etc. <S> by a factor of 10-50% in current handling - so if I know I need 20A current handling I would spec at least a 25A relay, and if there was no significant cost penalty I would spec up to 40A. I usually do this with car stuff, the highest spec "standard" car relays are 40A but 30A and lower spec are available. <S> I just keep a box of 40A type as they cost pennies extra and you only need to keep one type. <S> A relay rated to 125v AC will probably not explode when used with higher voltages, but it may be electrically or thermally unsafe, unstable, unreliable, and have a much shorter life. <S> Where mains is concerned, do not cut corners and ideally don't use the cheapest junk you can find. <S> It only takes some brittle piece of low-spec plastic to crack on the relay body etc. <S> to expose lethal voltages. <S> Good quality safety-approved stuff costs more for a whole herd of good reasons. <S> Running stuff near the limit is rarely reliable as the marketing department applies quite a bit of optimism to most of the stuff in the data sheet, and the max rating in one column is not achievable at the same time as the max rating in another column, in fact usually the max ratings are extracted with all other conditions being engineered as close to "ideal" as possible. <S> As stated above, the 20A rating will probably be at low voltage into an ideal load, and could be MUCH lower in non-ideal situations. <S> Good manufacturers will at least give you this information in their data sheet, even if they don't make a big deal about it.
| However, for the sake of reliability you should usually take all data-sheet numbers (especially "headline" specs) with a pinch of salt. So I really don't think it's good for 11A or 250VAC, and certainly not both together. As for the current rating, if it says it's good for 20 A, then you can run 20 A thru it.
|
Can anyone explain this balanced driver circuit to me? I'm looking to generate a differential signal to control a laser projector's galvos, and as I understand it needs to be +5V/-5V (10Vpp). I've found this circuit for a laser harp, but I'm confused about what this specific dual-opamp design does. It looks as if it's a pair of inverting and non-inverting amplifiers with a gain of 1, but they're being fed into each other. Here's a picture: The original can be found here . I'm curious if anyone could tell me what it's called, or how it works, because I've lookat at plenty of 'example circuits' and couldn't find anything that resembles it. <Q> Looking at the topmost <S> op-amp and ignoring the \$100 \Omega\$ resistors, write by inspection: $$v_{X+} = <S> v_{OUTX} <S> + v_{X-}$$ <S> For the bottommost op-amp, write $$v_{X-} = v_{X+} - <S> v_{OUTX}$$ <S> Thus, $$v_{X+} - v_{X-} = <S> v_{OUTX}$$ <S> So, this circuit converts a single-ended input signal, <S> \$v_{OUTX}\$ <S> to a balanced output signal; it's an active 1:1 'transformer'. <S> An interesting 'feature' of this circuit is that, while the differential output voltage, \$v_{OD} = (v_{X+} - v_{X-}) = v_{OUTX}\$ , is well defined, the singled-ended voltages \$v_{X+}\$ and <S> \$v_{X-}\$ aren't . <S> For example, substituting the 2nd equation into the 1st yields $$v_{X+} = <S> v_{X+}$$ and similarly $$v_{X-} = <S> v_{X-}$$ <S> So, in fact, the common mode output voltage output voltage $$v_{OCM} = \frac{v_{X+} + <S> v_{X-}}{2} = ? <S> $$ is not determined without an additional equation (circuit constraint). <S> Update: <S> I know <S> I've seen and analyzed this type of circuit before <S> but I haven't yet found my notes on it. <S> However, I did find this article at the Elliot Sound Products site for a " Balanced Line Driver with Floating Output " which appears to be essentially the same circuit except with a balanced input rather than single-ended input. <S> The whole amplifier, as it is dimensioned here, has a gain of 1. <S> The same amount of voltage across the input terminals appears across the output terminals. <S> This remains true if any output terminal is supplied with any voltage - like transformer coupled outputs <S> do (provided both output voltages stay within the supply voltage area of course). <A> At first I thought the circuit was a Differential Howland Current Pump. <S> Similar to this one here . <S> I thought maybe the cross-coupling makes the current sources share the available voltage. <S> But I did a simulation since the analysis didn't indicate that was possible.. <S> With no load, the (-) output is a virtual ground and <S> the (+) output equals the input voltage, which is not very exciting. <S> With a 1000 ohm load, the differential voltage is 90% of the input voltage (implying about a 100 ohm output impedance) but the (-) output is following the input by about +4%. <S> With a 100 ohm load, the waveforms look like this: <S> Green: input voltage <S> Purple: <S> output + <S> Red: output - Yellow: differential output voltage <S> I'm at a bit of a loss to understand the usefulness of this functionality if it's feeding coils directly. <S> Edit: <S> As Alfred has pointed out, the circuit should have a high output impedance with respect to common, and as I said, the differential output impedance is low and matched to a twisted pair. <S> So it would be a suitable driver for a balanced output feeding a twisted pair, going to receiver that might have a different (by as much as a few volts) ground potential from the transmitter. <S> Very nice. <S> Here is a plot of the common mode impedance measured by applying a 1VAC signal to the center of a split load resistance of 100 ohms, and sweeping from 0.1Hz to 10MHz. <S> As you can see, it's 10K for low frequencies, crossing over at around 2.2kHz and dropping to 150 ohms or so at high frequencies. <S> Perfect for situations where there is mains-frequency voltage between grounds, not so great for higher frequencies. <A> The easyest way should be to ask me directly through my laserharp website ;)i'm the designer of this schematic. <S> It's explained in the user manual of the laser harp. " <S> ILDA wiring" <A> Looking at the schematic you linked, evidently this op amp configuration is used to drive outputs that are part of the standard ILDA interface to laser projectors (as you alluded). <S> http://www.laserist.org/StandardsDocs/ISP05-finaldraft.pdf <S> So the primary task is to create a differential signal from a single signal. <S> A differential signal is usually used to deliver an analog signal in an environment susceptible to noise, as laser shows might well be. <S> Any noise will affect both the positive and negative copy of the signal approximately equally, and when the receiver recovers the signal by subtracting one from the other, the noise gets subtracted out. <S> Resistors R45 and R52 create some protection for the op amps in case the outputs are shorted, and possibly some impedance matching to the cable, though I'm not sure the need for that in this application (don't know the freqs involved). <S> But what about R48 and R49, and the apparent feedback they provide to the "opposite" amplifier? <S> I think they might implement compensation for the attenuation introduced by R45 and R52, useful if the receiver input impedances are not balanced.
| It's an output stage with a balanced/unbalanced output driver.if not used as balanced, you must connect the negative output to the ground to get full unbalanced signal.
|
What does FCY / F_CY stand for? I've come across many equations in microchip datasheets that are based on this value and am looking for a central place to lookup definitions. F_OSC = oscillator frequency T_CY = 1 / F_CY = time to execute one assembly instruction? <Q> This looks like a convention used by a particular microncontroller manufacturer. <S> "F" means frequency, and "cy" means cycle, or instruction cycle. <S> Tcy likewise refers to the period of each instruction cycle, with "T" meaning time. <S> In this case Tcy = 1 / Fcy. <S> Fosc refers to the raw oscillator frequency driving the micro. <S> This is often divided by some integer, like 2 or 4 before becoming the instruction clock frequency. <S> On a old PIC 16 like the 16F877, for example, the maximum clock frequency was 20 MHz, and it took 4 clock cycles to make one instruction cycle. <S> If you connected a 16F877 to a 20 MHz crystal, Fosc = 20 MHz, Fcy = 5 MHz, and Tcy = 200 ns. <A> Microchip has a glossary here . <S> " <S> FOSC" is listed, as well as "TCY" ("The time for an instruction to complete."). <A> Different manufacturers use different conventions in their datasheets. <S> It makes it difficult to make a generalized answer, but here it goes! <S> Frequency (in Hz) and Time (in seconds) are related to each other as you mentioned: <S> T=1/F <S> In a microcontroller, the basic oscillator frequency is the basis of all the other possible clocks. <S> This frequency is often given as F_osc or F_clk . <S> The duration of each clock cycle is given as T_osc or T_clk . <S> Depending on the microcontroller, sometimes an instruction takes multiple clock cycles. <S> For example, Microchip's PIC18 series (and possibly others, but I haven't used them) use four clock cycles for every instruction cycle. <S> The instruction cycle time/freq is referred to as T_inst and <S> F_inst . <S> In this case, if F_osc = 1MHz, then F_inst would be 250kHz (a quarter of the speed). <S> T_osc <S> = 1 microsecond, and T_inst <S> = 4 us. <S> T_inst <S> and F_inst can also be referred to as T_cy and F_cy . <S> Then, the rest of the clocks are derived from these. <S> If you are configuring a timer, for example, the datasheet will state if it is based on F_osc , F_inst , or some other timebase. <S> Some uC's perform one instruction per clock cycle, so here is no need to differentiate between the oscillator cycle time and the instruction cycle time. <S> In this case, F_cy would be also used. <S> Hope this helps! <S> edit: just to confuse matters, let me add: An instruction cycle is not necessarily the amount of time it takes for an instruction to complete! <S> Again, using the PIC18 as an example: most instructions take a single instruction cycle, although some can take two.
| Fcy or variants thereof refer to the instruction clock frequency.
|
What's a loop current? I just started mesh analysis, but don't really understand the concept of a loop current. The current in different parts of a loop is not necessarily constant - two parts of the same loop can have different currents. So how can you define one current for the whole loop? <Q> Any leg (edge) of the mesh may be involved in more than one loop, in which case the actual current in that leg is the sum of the loop currents of those multiple loops. <S> It's like the flip side of the more intuitive observation that currents flowing into and out of a single node add to zero. <S> (All current flowing in must flow out -- current can't just appear or disappear.) <A> The actual loop current is the summation of currents, and you can also have influence of currents from a nearby mesh. <S> So If there is a shared branch between two meshes, say a resistor, then I_R = I1 + I2 , where I1 and I2 are the individual loop currents, which may be positive or negative, and probably produced by a current source or generated by a voltage source over a resistance. <A> In a circuit, with only linear elements in it, the current in any branch can be written as the linear combination\$^{\dagger}\$ of L currents. <S> Where 'L' is the no. of loops in the circuit. <S> These L currents are called loop currents. <S> It doesn't mean that all branches in a loop should have same current. <S> It just means that the current through any branch in a loop can be expressed as the sum or difference of the corresponding loop current and neighboring loop current. <S> OR as you said, any branch current can be 'broken down' into loop currents . <S> \$\small{ ^{\dagger <S> } \text{linear combination with coefficients}\in \{0,1,-1\} }\$ <A> Bottom line - loop "current" is a mathematical construct used in KVL equations. <S> It is only directly measurable in a loop with an unshared branch.
| The idea of a loop current is that for each loop you can identify in the mesh, you can find one value of current "attributable" just to that loop, and which passes through every leg of that loop.
|
How can I slow down the switch time of a MOSFET? I have an NMOS that is switching too fast for my application. Into the gate I am sending a logic-level square wave (PWM). Unfortunately for me, as expected, the output is also a near square wave. How can I get the Vout to be more trapezoidal? Or said another way, what is the simplest modification I can make to decrease the slew rate at the output? Note: (Vin) is the voltage applied at the gate of the NMOS & (Vout) is the voltage seen at the drain of the NMOS. <Q> Not enough Miller time? <S> Just extend it. <S> Spehro has the right approach here. <S> I am going to ride his coat tails and expand on the idea a little, because it is such a good idea for this kind of thing. <S> \$C_{\text{dg}}\$ is special in a FET because it provides negative feedback to the gate. <S> Part of what that means is that it also gets multiplied by the transconductance (\$g_{\text{fs}}\$) of the FET. <S> So, it has a larger effect than it's size would lead you to believe. <S> But, let's forget about \$C_{\text{dg}}\$ for now and instead add an external capacitor from drain to gate (\$C_{\text{fb}}\$), because if you really want to slow down the rise and fall times of the FET that's what you'll do. <S> Here is a schematic to help illustrate: As \$V_{\text{drv}}\$ rises and \$V_{\text{ds}}\$ falls you can probably see how \$R_g\$, \$R_L\$, \$g_{\text{fs}}\$, and \$C_{\text{fb}}\$ <S> all play a part in limiting the value of \$V_{\text{gs}}\$. Small signal transfer function of \$V_{\text{ds}}\$ relative to \$V_{\text{drv}}\$ is: \$-\frac{R_L}{s <S> C_{\text{fb}} \left(g_{\text{fs}} <S> R_g R_L+R_g+R_L\right)+1}\$ <S> And, \$R_g\$, \$R_L\$, \$g_{\text{fs}}\$, and \$C_{\text{fb}}\$ are all involved in forming the pole. <S> (Note, all the FET capacitances are left out here for clarity.) <S> To show approximately how this works out, put in some values into a very simplified model. <S> \$R_g\$ = 1000 Ohms, \$R_L\$ = 2 <S> Ohms, \$V_{\text{drv-pk}}\$ <S> = 5V, \$V_{\text{cc}}\$ = 10V, <S> \$g_{\text{fs}}\$ = <S> 5 S. <S> Here is a plot of \$V_{\text{ds}}\$ on application of \$V_{\text{drv-pk}}\$. <S> The blue curve is \$C_{\text{fb}}\$ = 100pF, and the purple curve is \$C_{\text{fb}}\$ = 1000pF. <S> Of course, switching loss will be huge and huger. <S> It should also be mentioned that adding a Miller feedback capacitor like this will make the circuit more sensitive to dV/dt turn on. <A> The only control you have over the resistance of the FET is the gate-source voltage. <S> You need to slow down the change of that voltage. <S> The most common way of doing that is an RC filter at the gate. <S> The bigger the resistor, the slower the turn-on and turn-off. <S> As a general rule, I always put an RC filter with a pulldown resistor on all FETs. <S> This allows control of the rise-time, and provides improved noise immunity. <S> simulate this circuit – <S> Schematic created using CircuitLab Keep in mind that any time your FET spends not fully "on" or "off", it sees increased losses. <S> If it's on, the device has very low voltage across it. <S> If it's off, the device has no current through it. <S> Either way, low loss. <S> But if you're in between, the device sees both voltage and current, meaning its power dissipation is far greater during that period. <S> The slower you switch, the greater that loss becomes. <S> At what point it becomes a problem depends on the FET, the source, and the switching frequency. <A> You can add a series resistor to the gate. <S> That's often done to slow rise-fall times in order to reduce EMI or prevent excessive overshoot. <S> Obviously this increases switching losses (but not conduction losses), so there is a trade-off. <S> As well as causing the switching to slow, it will also add a delay time, so keep that in mind if there is a chance of cross-conduction or similar problems. <S> The slope you get for a given value of gate resistor will depend on the capacitances from gate to source and gate to drain, as well as the value of Vcc. <S> While the MOSFET is switching, the resistor supplies the current to charge \$C_{GS}\$ as well as the current to charge \$C_{DG}\$ between Vcc and 0. <S> The total amount of charge is often specified in the datasheet (under given conditions) as the gate charge (measured in nanocoulombs). <S> Because of the Miller capacitance (\$C_{DG}\$) <S> the nature of the load comes into play as well. <A> What are the operating condition of your MOSFET? <S> When used as a switch, the MOSFET is most of the time in two states: <S> Blocked: <S> High \$V_{\text{ds}}\$ voltage, <S> no current -> no dissipated power Conduction: <S> Very low \$V_{\text{ds}}\$ voltage (\$I_{\text{d}} \times R_{\text{ds_on}}\$), high current (\$I_{\text{d}}\$) -> small dissipated power (\$R_{\text{ds_on}} \times I_{\text{d}}^2\$) <S> The MOSFET is in a third state, during a very small amount of time. <S> And this third state is when it is conducting a little: - <S> Non negligible \$V_{\text{ds}}\$ voltage, non negligible current. <S> \$I_{\text{d}} \times V_{\text{ds}}\$ may be high! - <S> > possibly big dissipated power. <S> If you plan, by design, to put your MOSFET longer into this third state, you have to ensure that the increase of the temperature of its junction won't let it pass above the maximum allowed temperature for that junction. <S> (found in the datasheet)Reducing the slew rate of a MOSFET has to be carefully studied. <S> I don't know what you are driving with it. <S> If it is a LED and you want to have it becoming brighter and brighter, but slowly, you would better use a PWM on the gate of your MOSFET and still use it as a switch. <S> If the PWM is very fast, it won't be noticeable to a human eye. <S> The same approach is also valid for driving a motor.
| Put a resistor between your drive source and the device gate, and the gate's parasitic capacitance will form an RC filter. If the resistor gets too big, you can have noise immunity issues (false gate triggers and such), so past a certain resistor value (maybe in the 10k-100k range) you're better off adding capacitance gate-source to slow the switching down further.
|
How to Implement this special selector? Is it possible to write a module with 3 wires a,b,c that would output either : z (disconnected) if a=b=c=z a if a=(0 or 1) and b=c=z b if b=(0 or 1) and a=c=z c if c=(0 or 1) and a=b=z x (dont care) otherwise In verilog? <Q> No, it isn't, because z isn't a condition that can be tested for in synthesizable logic. <A> To elaborate on why high impedance isn't a condition that can be checked for: When a signal is high impedance, the FPGA just leaves that signal floating. <S> Usually if you drive a signal to 0 , it will connect it to GND , if you assign 1 it will connect it to VCC . <S> By assigning Z , you're telling the FPGA to leave it floating and not pulled up or down. <S> In this state, the signal is in an indeterminable value, because the CMOS/TTL circuitry that determines if a signal is a 1 or a 0 is not designed to work with floating signals. <S> So what's the point? <S> Why have high impedance signals in the first place? <S> Say for example that on your board the signal is connected to one of the output pins, and the external circuitry connects it to GND . <S> By leaving the signal at high impedance, you're allowing the external circuitry to do the pulling down. <S> In this case, if you were to set the signal as Z , you would still be able to determine its value by checking if it's a 1 or a 0 , because even though you on the FPGA haven't pulled it down, it's been pulled down elsewhere in the circuitry. <S> So when you're doing signal assignments, assigning to z is something that's done regularly. <S> However, when you're checking signals, you shouldn't be checking for z , but rather for 1 or 0 depending on the expected behaviour of whoever really is driving them. <A> always @(in1,in2,in3) begin case ({in1,in2,in3}) 3'bzzz: <S> data_out = <S> 1'bz; <S> 3'b0zz: <S> data_out = 1'b0; 3'bz0z: <S> data_out = 1'b0; <S> 3'bzz0: <S> data_out = 1'b0; <S> 3'b1zz: <S> data_out = 1'b1; 3'bz1z: data_out = 1'b1; <S> 3'bzz1: data_out = 1'b1; <S> endcaseend <S> this seem to work...
| You'd need to find a way to convert the concept of " z or not- z " into a logic level.
|
Properties of inductors with a magnetic core If I take an Iron core,for example a simple nail, and coil wire around it, I create an inductor. Now if I strongly magnetize the iron core, will any properties of the inductor change? <Q> All magnetic materials have a B-H curve as shown below where B is the flux density and H the magnetic field strength. <S> The effect of permanently magnetising the core is to shift this graph along the H axis. <S> Note <S> \$ <S> H \$ is the sum of \$ \frac{N \cdot <S> I}{l}\$ <S> where \$ <S> N \$ is the turns of your coil, \$ <S> I \$ <S> current, \$ l\$ magnetic path length and the magnetic field strength from the permanent magnetisation. <S> The end result being that the inductor will start to saturate at a lower current in one direction and an higher current in the other. <S> (source: electronics-tutorials.ws ) <A> It sounds like you are moving into saturable reactor designs: - With a saturable reactor (aka magnetic amplifier) <S> you can control the AC power delivered to a load (see the lamp above) with a varying amount of DC put through the control circuit. <S> When there is no DC on the control input, the power winding (on the right) is just like a large value inductor and at 50 Hz or 60 Hz it would be designed to have an inductance that largely blocks the AC current to the lamp. <S> The design shown above isn't very good because the output waveform of current through the lamp is asymmetrical but there are better designs that involve two cores so that the AC waveform presented to the load is symmetrical and this type of design also reduces the power AC coupling to the control circuit: - <S> Details got from here <A> Yes, in that case you will saturate the core sooner and the inductance will fall off quicker when current flows in the same direction as the pre-biased magnetic field.
| As the control DC current increases, the iron-based core will start to saturate magnetically and this reduces the power coil inductance and more current flows to the lamp and it gets brighter.
|
PIC32 only works when shorted to ground We're using the PIC32MX795F512L on a number of custom boards. I'm responsible for one design, for which we currently have two prototypes. The first board appears to work perfectly. On the second, the PIC is mostly unresponsive. Pressing the Reset button doesn't help. Occasionally it will report a Device ID to the PICkit but for the most part it appears dead... until I briefly short the 3.3V supply to ground (don't ask how I discovered this). After that it seems to work just fine - every time, without fail. If I power it down for more than a split second, it begins to fail again. I think the successful functioning of the first board rules out most design issues, but there's gotta be something wrong somewhere. These were hand-assembled; I've checked all of the connections but I'm wondering if a particular component might be suspect. Any ideas on where to look? EDIT: thanks for all the replies so far. As requested, a schematic of what's currently connected to the MCLR' line: To clarify, this is our first time designing boards for these chips so there are bound to be some less-than-ideal conditions present (rules we forgot, or thought we were implementing but didn't, etc.); it's clearly serviceable somehow, though, since my first board works. <Q> I agree with Erik that there could be an issue with the hand assembly of the second board. <S> If the first and second boards are exactly the same design, but one works and <S> the other does not, then it's a component or assembly or freak damage/accident issue. <S> I have had a recent experience with hand loading 20 small bat tracking boards I designed, which each have small power circuits, memory, gps, and a microcontroller (Atmel ATMEGA328P). <S> of the 20 boards, 16 worked immediately after assembly, but 4 were unresponsive to programming. <S> I could not work out exactly what was wrong, as all solder joints were fine after inspection with microscope, but after SMD re-work replacing the MCU ICs with some spare ones, (TQPF-32 packages, easy stuff) they immediately worked. <S> I suggest after you try flux + re-heat each solder joint to see if it's a bad join and therefore bad grounding issue, you should attempt to replace the microcontroller IC. <S> Make sure you establish that it is not a solder short circuit or other external fault before doing this! <S> You may just fry/permanently damage another one if you immediately replace before elimination of other options. <A> , wherein some boards don't work this way (it turns out mine wasn't the only one with this problem). <S> Running all Vdd <S> /AVdd pins from the same regulator solved the issue. <A> Since they are hand assembled, look REAL close at the pins, with magnification. <S> Especially if they are the .4mm/PT package. <S> You WILL NOT see shorts on these chips until you use something with high enough power. <S> I personally use a plossl 25m focal length eyepiece. <S> I have had many funny issues through the years with the 460f512l and 795 that are caused by my own bad hand assembly. <S> Sometimes I end up flood and sucking all pins to fix something I know is assembly but can't see. <S> Basically this is running a heavy bead of solder across all the pins, and using solder braid to remove excess solder. <S> If its not assembly, its likely design, or perhaps the chip was static damaged during assembly. <S> Did you follow design rule #1 about putting resistors on all pins, or at least those facing anything like the real world? <A> Another possibility is that, for some reason, the power supply voltage is ramping up too slowly for a proper power-on reset to occur. <S> I don't know how this could be related to the problems with the reset button, but it might have to do with what PRG_MCLR is connected to beyond what is shown in the schematic. <S> I would suggest trying a smaller resistance value for the pullup to improve the rising edge transition on this signal.
| Shorting the supply after it's already up may give the microcontroller a nice fast ramp up on the supply voltage, allowing it to reset properly. So it turns out the problem was our power supply scheme - we had the analog pins on a separate plane, but there's some issue we haven't completely identified with our regulators
|
Use transistor to connect terminating resistor for differential signal? Is it okay to use a transistor to connect a terminating resistor between the A and B lines of a differential signal? I'm not sure because I know that at the source/emitter of an NPN transistor must be at ground potential for the base/gate to turn the device on. My instinct says no but please help. If this won't work is there of something other than a relay that would work to do this with a GPIO pin of a processor? <Q> I'm not going to explain why is this a bad practice, but rather mention why a single transistor won't do the job. <S> Say you have a setup like the following: simulate this circuit – Schematic created using CircuitLab <S> Obviously, you need two FETs simply to tackle the fact, that reverse-biasing the Source-Drain junction triggers the device's intrinsic diode. <S> Second, you'd have to make sure, that at all times, your enable would be higher/lower than your maximum/minimum single-ended voltage + \$V_{th}\$ of the FETs. <S> Say if your signals swing from 1-2 Volts single-ended, then your Enable signal would have to go from 1V to at least \$2V+V_{th}\$ (in respect to ground). <A> Analog switches seem to be suitable for this task. <S> There are also over the rail products from various manufacturers. <A> Many differential cables are terminated in two resistors; one on each leg to ground\$^1\$. <S> There is a good reason that they do this - if the cable is shielded or screened and the signal is not truly differential there can be reflections on each wire to ground that upset things if only terminated with a single differential resistor. <S> If the differential resistor is 120 ohms then the more practical solution is to use 2 x 60 ohms to signal common. <S> This: - Terminates the differential pair at 120 ohms and Partially terminates each wire individually to ground to reduce the problem mentioned above. <S> How does this help? <S> It should work but <S> to be sure I'd need to see the voltage waveforms to 0V. <S> \$^1\$ For ground also read screen, shield, 0V, earth, common, return etc.. <A> Assuming you need to preserve the typical common-mode range of RS485 signals: -7V to +12V and have only a single supply available, this is not all that easy. <S> It's also connected directly to the outside world, so ESD immunity has to be superb, and the resistance of the switch should be perhaps 10 ohms or less or its variability will affect the accuracy of the termination. <S> @gwideman's suggestion LTC2854/55 looks like a good way, if you don't mind using a "boutique" component. <S> There are well proven small telecom-style latching or non-latching relays, but they're fairly large. <S> Finally, there are analog switches with built-in charge pumps, such as the MAX14759 , but availability may be an issue. <S> Of course there are lower tech non-software-programmable solutions such as jumpers and switches, and those ones will withstand ESD that will kill the receiver dead. <A> Short answer is: no you can't use a single bipolar (NPN or PNP) transistor in this role. <S> Just one reason is that the transistor only functions a particular way with current in a particular direction from collector to emitter, whereas a differential signal reverses polarity. <S> As to a reasonable way to do this... head scratching in progress.
| If the signals on the differential pair are logic levels (or biased above ground) then using two transistors simultaneously to connect each wire to ground should work. The characteristics of a transistor are wrong for this role for a bunch of reasons.
|
Is there any difference between the assembly instructions of ARM MCUs from two different corporations? I'm curious to know, are there any differences between the assembly instructions of ARM MCUs from two different corporation? For example between an Cortex-M3/4 of NXP and TI or ST or other corporations. Some of my friends say me that they have no difference. Is that correct? <Q> I think the correct thing to say is that for a given architecture, such as the ARMv7-M architecture of the Cortex-M3 core, the instruction set is the same for all processors. <S> However, the behavior of some instructions may vary because of implementation-defined (i.e. optional) functionality in the processor. <S> Instructions that try to access optional capabilities that are not implemented in a particular processor may cause exceptions. <S> To find the features that may be implementation defined, search the appropriate ARM Architecture Reference Manual for IMPLEMENTATION, in all capitals. <A> Processors within the same family (e.g. Cortex M3) should have the same instructions, but different families have different instructions. <S> The original ARM used a set of 32-bit instructions, then a version appeared which could switch between "ARM" mode and "Thumb" mode, with the latter implementing a smaller set of 16-bit instructions. <S> A job which takes half again as many Thumb instructions as it would take ARM instructions will take roughly half again as long to execute in Thumb mode as ARM mode, but will fit in 3/4 of the space. <S> Many newer processors do not have any 32-bit mode, but some can combine two consecutive instruction words in such fashion as to yield most of the instructions from the 32-bit ARM instruction set, plus a few more. <S> Note that some 32-bit ARM instructions are not implemented. <A> There a number of different variations on the ARM instruction set (see http://en.wikipedia.org/wiki/ARM_architecture for the details), and different vendors' parts might support different subsets. <S> Just as an example, there's no integer division instruction in ARMv6; it's optional in some versions of ARMv7, mandatory in others; and present in ARMv8. <S> Furthermore, a vendor making their own ARM-licensed CPU can in principle add or remove any instructions they care to.
| The net effect is that there is no processor which can perform every ARM instruction; different ARM families have different sets of instructions available to them.
|
Maintaining a constant current with varying loads I've designed a constant current source using a current mirror, we need an output of about 50mA. The problem is that we build it and heat has become a problem.The transistors and the R_load are both heating up which decreases beta stabilization (our amperage through our load resistor increases constantly). I think that by adding an emitter resistor we should be able to beta stabilize but I'm not sure if I'm doing it right. <Q> There are a few other constant current type regulators to consider: - <S> This biases a transistor to have a certain voltage across the emitter resistor and this means the current in the collector to the load is pre-defined. <S> Can be used with a LED as load - it works by the BJT cutting the voltage supply to the gate of the MOSFET when the current thru R2 reaches "the constant current level". <S> Classic voltage regulator turned constant current source <S> My favourite - an op-amp used with a BJT sets (very accurately) <S> the voltage across the emitter resistor and hence the current in the collector is very well controlled too. <S> Pretty much the same as (4) Like (2) but with the load at the top Like (4) and (5) but <S> with the load at the top Also to consider are Wilson current mirrors (a definite improvement over your standard current mirror) and the Howland (and improved H) constant current source. <S> I don't really understand what you are trying to achieve <S> so I can't recommend one over the other. <S> AD have a good paper on current sources too. <A> Given the very large head-room you have available from your primary supply, you should be able to just use the 7805 in current regulator mode: simulate this circuit – <S> Schematic created using CircuitLab <S> This would accomodate a wider range of load resistors if you could swap the 7805 out for a chip with a lower reference voltage and lower drop-out voltage. <A> All you need is a resistor between output and ground to limit your current. <S> $$I_{out} = <S> \frac{V_{out}}{R_{out}}$$ <S> I will refer you to <S> this Maxim application note for a more thorough explanation.
| Linear regulators actually make for good constant current sources because of fairly high accuracy between the output voltage and ground.
|
Easy Soldering to Aluminum? I am building an antenna with aluminum elements. Is there any easy way to solder copper wire to the aluminum? I do not need a very strong connection, just a good electrical connection so I can then apply glue etc. I tried sanding the aluminum and applying cooking oil, but that didn't seem to work. I will try motor oil later, but I don't think that will be any different. The aluminum is a 1/4 inch rod. I got it to work on aluminum foil, but it won't work on the rod for some reason? Would something like this work? <Q> One thing I've seen several times on both DIY and smaller commercial antennas is using a combination of screws and washers or solder lugs to attach the feed to the antenna elements. <S> Here's one idea from Homemade 14 element Yagi antenna for PMR446 that would be pretty easy to put together: <A> It's better to solder with a slightly overheated soldering iron this way. <S> Then resolder the same connection with some neutral flux. <S> I don't know the exact chemistry of this method, but it worked for me to solder aluminium wires (but not thin foil). <S> It's better to do it in a well-ventilated place. <A> Between the time you sand it and apply the oil, it oxidizes again. <S> If you're going to have a chance, you would have to apply the oil and then sand with the oil in place. <S> This still does not make it easy.
| One well-known solution is to use a piece of polyvinylchloride tube as flux when soldering.
|
Why do I have to solder header pins? Coming to electronics as a hobbyist I'm not sure I understand why I have to solder header pins? On more than one occasion when dealing with Arduino / breadboard projects the circuit will not work until the header pins are soldered. Holding them in place will not solve this problem. Why? Low voltage? Need for persistent connection? Holding them in place doesn't work nearly as well as it looks like it should? Why? <Q> Think about it mechanically - you have a straight row of pins and you insert them in slightly loose fitting holes (all in a line). <S> Even if you hold them in place - can you be sure that one of the pins isn't fractionally bent in one direction different to the others. <S> Think about this for a 3 pin header: <S> - Clearly the pin in the middle isn't touching the inside of the yellow hole until you put unreasonable amount of pressure on the connector. <S> Please, no complaints about the colours. <A> The holes in the PC board are somewhat larger than the pins on the header so that you can easily insert the header. <S> The normal headers are intended to be soldered into the board. <S> There are "press fit" headers available that will make a secure connection when inserted into holes of the correct size - but there is no guarantee that the common boards (Arduino shields, etc.) have the correct size holes to accept the press-fit headers. <A> Electrons are really small. <S> What looks like a solid connection to the naked eye at a macro scale could have tiny gaps which look massive to an electron at the tiny scale that they care about.
| Because of the loose fit of the pins in the holes, simply placing the header in the board without soldering will not ensure a good connection.
|
What is the white wire for on this voltmeter? I found this voltmeter on eBay , and I know what the red and black wires are for (live and ground respectively, or alternative names), but I have no idea what the white wire is for. As its top range is 99 volts it can't be high voltage. The listing does not say what it is for, and it has a really complicated circuit diagram which I cannot understand at all. Please could somebody tell me what this is for? <Q> My guess would be that black is ground, red is power supply (5 - 28 volts) and white is voltage to be measured (0 - 99 volts) - but that is only a guess, and I would examine the unit carefully, comparing it to the schematic provided, before applying any voltage. <S> Personally, I would not buy anything that did not come with the necessary documentation. <A> Two of the wires (2 and 1 in the middle of the schematic) provide power to the device. <S> The third (3) is for sensing voltage along with the ground at 1. <A> How I came to this conclusion, was I asked my wife, who has absolutely no electrical background and is deathly afraid of electricity. <S> lol. <S> So I decided to test this for myself. <S> I put the black on ground and the hot on red - the meter lights came on and displayed 0.0 volts. <S> So then I added the white wire to the red one that was on hot - <S> it displayed the voltage. <S> so the last test just to be sure? <S> Add a ground from another DC source and a hot from that other source to just the white wire and leave the red on the hot from the other battery? <S> Can you guess what happened? <S> Yes! <S> Exactly, it showed the voltage of the new source on the white wire and not use the power from white to light the device. <S> Instead lighting its self from the red and measuring from the white. :) <A> Connector P1 is a three wire connector for the red, black and white wires. <S> Nobody is able to categorically tell you which colours relate to the three pins but the three pins are: - <S> Pin 1 on P1 is ground Pin 2 on P1 is battery or dc supply to power the LED display and chips - it needs to be 4.5V to 30V. Pin 3 on P1 is the pin that you use to measure a voltage so that its value can be displayed on the LED display.
| The answer is: black is - (negative) and red is display + (positive), and white is the wire that the voltage is measured off of.
|
Converting a circuit from Multisim into a PCB I drew this metal detector circuit in Multisim. How can I make a PCB board from it? Is there an option in Multisim to convert it to PCB or do I need another program? <Q> NI Ultiboard is the PCB layout counterpart to Multisim and allows you to layout PCBs from a Multisim schematic. <A> Transfering to Other PCB Layout Packages <S> If you are using a PCB layout package other than Ultiboard, you can create files in the necessary formats for transfer to the following third party layout packages: <S> • <S> OrCAD <S> • <S> PADS <S> Layout 2005 <S> • <S> P-CAD • Protel Complete the following steps to transfer the circuit design to a third party layout package: 1. <S> Select Transfer»Export to PCB Layout. <S> A standard Windows Save <S> As dialog box appears. <S> 2. <S> Navigate to the desired folder, enter a file name, choose the desired manufacturer from the drop-down list and click Save. <S> Multisim creates a file of the appropriate format that can then be loaded into the layout package of your choice. <A> You will need a specific tool such as Eagle (free or at least trial for small PCBs) or Altium Designer (kind of costly). <A> I don't think multisim supports this feature. <S> I recommend using Eagle. <S> It's a tool which allows you to draw schematics and layout pcbs. <S> You can download it from here: http://www.cadsoftusa.com/download-eagle/?language=en <A> Of course multism supports to design your own PCB. <S> It has Ultiboard package which can enable us do our PCB. <S> But, be careful while transferring your schematic to ultiboard, since some components will not be transferred directly. <S> Because Multism considers some parts (like resistors, capacitors..) as virtual components.
| Another alternative would be to use RS-Component's design tool: http://www.rs-online.com/designspark/electronics/deu/page/designspark-pcb-home-page
|
What is this component in the picture? I am building a "DIY Cellphone" project ( http://2loo.com/ax ). I have ordered all the components in the Bill Of Materials ( http://2loo.com/ay ). I put together almost everything when I stumbled upon a component that I don't know what it is. It is the component in the figure below (Taken from: http://2loo.com/aw ). I think it says "D" or "0" on it. Is that a capacitor or a jumper? According to the cellphone schematics ( http://2loo.com/az ), one capacitor and two coils are supposed to be there. However, those are missing from the bill of materials and are also missing from the figure above. <Q> Looks like a "0 ohm resistor" (a jumper in some standard size such as 0603). <S> From Yageo's catalog: <A> It is resistor. <S> It's value is 0 ohm, so neutral to the trace. <S> It is used in place of a jumper or wire to bridge two traces while allowing a few features. <S> It can easily replace (or be replaced by) an actual resistor, if it was added then decided that the resistor was not needed. <S> It allows a second signal to pass underneath. <S> And it can be used for bootstrapping, as in only added in when needed, on a single board that can have multiple configurations. <S> Furthermore, due to the nature of production, smd pick and place machines can deal with a smd resistor a lot easier than it can a wire or jumper. <S> And allows keeping the through hole part down, also making it cheaper for production. <A> Such filters are often added to designs at the antenna in order to deal with problems after the design is built, but are not always needed. <S> So the 0 ohm resistor effectively removes this filter from the circuit acting as a jumper. <S> If the final design is shown to have emissions that fail FCC, CE, or other regulatory testing, it may be that adding the correct parts in place of the jumper will resolve the problem quickly without having to redesign the PCB. <S> If testing is satisfactory without a filter in place at that location, then a 0 ohm resistor is very, very cheap. <S> So there's little risk in adding it, and if things don't work as well as planned (perhaps due to a mismatch between the antenna and transceiver) <S> then a few parts can be added in a matter of hours or days rather than waiting days or weeks to re-spin the PCB. <A> Adam's and Passerby's answers make a lot of sense, but let me try an alternative interpretation: BU-SMA-V component is not populated, and that's an antenna jack. <S> Perhaps 0Ohm resistor is there only "tell" the chip that the external antenna is missing?
| The purpose of the capacitor and inductors in that part of the circuit is to act as a filter.
|
Control resistances with voltage Let's say I have an audio oscillator, which a resistor varies the tone. I have 5 different voltage sources, each representing a tone. How can I control the tone (vary the resistance) depending on which voltage source is on? <Q> A microcontroller taking inputs based on the voltage, controlling a Digital Potentiometer. <A> Definately not the best way to make a VCO. <S> But, the question deserves an answer, because it's an interesting and possibly educational way to do a VCO, and voltage controlled resistors are used for other things. <S> One way, that actually was used a long time ago in audio equipment was to shine a light at a light depedant resistor Photoresistors aren't used much on account of RoHS these days though. <S> The other way is to use a FET. <S> A FET pretty much is a voltage controlled resistor, but the resistance varies a bit with the voltage across it and not just with the input voltage, so you have to add two extra resistors to correct for this otherwise it won't quite act like a resistor. <S> http://graffiti.virgin.net/ljmayes.mal/comp/vcr.htm <A> http://en.wikipedia.org/wiki/MOSFET#Modes_of_operation <S> A mosfet in the triode/linear region should be what you're after. <S> Send the different voltages to the gate of the mosfet and you'll get a varying resistance between the drain and the source depending upon the input voltage. <A> This page <S> http://graffiti.virgin.net/ljmayes.mal/comp/vcr.htm has the most direct explanation I have found. <S> See also MOSFET as a voltage controlled resistor <A> If you have a logical signal 1 of 5 (represented by 3 bits), you could very simply use a set of resistors and an analog multiplexer to select one of 5 different resistances. <S> The multiplexer will have some internal resistance (which changes with voltage and with temperature) so this is most suitable if your desired resistances are in the K ohms. <S> And, if you don't have it encoded, a 74HC148 would do that for you. <S> If the supply voltage is 5V, you could use a 74HC4051 , or if it is 15V or +/-5V <S> you could use a CD4051. <S> The latter has an internal resistance of about 1K, the former around 100 ohms. <S> There are better (and much more expensive) multiplexers available that will work at higher voltages and have lower internal resistance.
| To answer the question directly: Yes, it is possible for a simple MOSFET circuit to approximate a voltage-controlled-resistance within certain limits.
|
Override digital out with pushbutton I have an Arduino Digital Out pin connected to a 1k resistor, then connected the base of a 2N3904 transitor. In fact, that same digital out is connected to several resistor->base-of-2n3904. Each transistor is in turn driving a p-channel mosfet. I would like to add a pushbutton to override the digital out. By that, I mean that the digital out is most of the time OFF, but sometimes ON from a brief moment. Pressing the pushbutton will have the same effect as turning ON the digital out. I don't think I can simply add a momentary pushbutton between the digital out and +5V, can I ? This will probably feed 5V directly into the output pin. Should I maybe add a resistor in series with the pushbutton ? do I risk frying the pin by adding the pushbutton like so: simulate this circuit – Schematic created using CircuitLab <Q> You could switch this around and make the output of the Arduino open-drain, tied to +5 through a 10K resistor (R4). <S> Get rid of the three transistors, and drive the gates of the MOSFETs directly from the output pin. <S> (By the way, I'm not sure why you have three transistors in parallel; it seems you could of run the output of one transistor to all thee MOSFETs.) <S> Now, setting the output pin to ground will be the same as turning on the transistors in your earlier circuit (collectors connected to ground). <S> Setting the pin high (tri-stated for open-drain outputs) will leave the outputs at +5 due to R4. <S> Meanwhile, if the switch is now connected to ground, and tied directly to the outputs (no R5), pushing the switch will also drive the output low without interfering with the output pin of the Arduino whether it is tro-stated or at at ground. <S> If you want to use your existing circuit, making as few changes as possible, you should be able to isolate the switch and the output of Arduino using just a 1N4148 diode and adding a resistor to ground. <S> (The resistor in series with the switch is removed.) <A> This is what I was thinking out loud in the comments. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> The push button can be in parallel to the BJT without a problem. <A> The digital output presumably drives actively both in the high and low directions. <S> 10 kΩ to any voltage from 0-5 volts won't have any effect on it. <S> There is a way to do this. <S> Connect the bottom of R5 directly to the base of Q1, which means on the right side or R1. <S> I'd also make R1 higher. <S> The collector load is only 10 kΩ, <S> so 10 kΩ <S> on the base is plenty low enough. <S> R5 of 10 kΩ as you have it now should then work. <S> Remember that the digital output will be actively driving the left side of R1 low during the pushbutton override. <S> With R1 and R5 equal, most of the current thru R5 will go to turn on the resistor since the base will be at about 700 mV. <S> If R1 were much lower than R5, as you have it now, then most of the current thru R5 during override will actually go thru R1 and into the digital input instead of into the base of Q1. <S> However, a better way to do this overall is to connect the button to another input of the processor, then perform the override action in firmware. <S> That way the firmware knows it's happening, the switch will be debounced, and you don't have to worry about current sharing. <S> The only real drawback is that it takes another input pin. <S> As long as you have one available, this is really the better way to solve the problem. <A> I'm not sure if jippie had something like this in mind, but you can use a "master" transistor that will control the rest of the transistors. <S> The transistor I have used is in an emitter follower configuration (to avoid inversion). <S> When the AVR output is high then the emitter will output about 4.3v and will turn the output transistors on. <S> If the button is pressed during that state then it will just feed the output transistors with 5v which will not make a difference for the way they are used (as switches). <S> When the AVR output is low then the transistor will be off, but the button will be able to override and feed the output transistors with 5v. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Schematic created using CircuitLab <S> Probably the simplest way for an existing system. <S> Or if you must use a momentary SPST, one added diode and resistor will make it happen. <S> (But @tcrosley already posted the second solution, so look at his answer!).
| You could use a SPDT momentary pushbutton (or momentary toggle) simulate this circuit –
|
How to cheaply switch between two sets of 7-input-wires? I am trying to make a water level sensor . I want to use a single circuit for two water tanks. For this I need to be able to switch between two sets of 7-wire-inputs to the circuit (the 7 wires that are in the water tank in the diagram below. What is an simple/elegant/cheap/easy way to do this without having to use 7 switches? Circuit diagram: UPDATE: Thanks for all the answers!Some clarfications: I will using a 9V battery source, not mains. I thought of one more solution: the circuit inputs will be connected to a CAT5 socket, and the input wires (CAT5 cables) from the two tanks will have connectors. So I can simply switch cables and use the one circuit. <Q> Truly horrifying circuit, by the way. <S> The transistors are drawn backwards (he managed to get them right on the one he built for himself). <S> It will also cause electrolysis of the probes if you leave it on all the time since he's using DC to detect conductivity. <S> But it is simple. <S> Frankly, I think it would be easier to build two of these than it would be to switch the probe lines. <S> Or you could wire the probes to something that fits a switch box . <S> If you insist, it can be done with three <S> 74HC4053 triple analog multiplexers (that will handle up to 9 lines). <S> Increase the resistors to 1K from 33R and put one on each probe input to give the chips a chance of surviving when bad things happen at the probes, and put the chips in sockets so you can replace them when they inevitably get fried. <S> Ground E (/Enable) and \$V_{EE}\$ (and GND, of course), \$V_{CC}\$ is your 5V supply, connect all the \$S_i\$ together with a 10K pullup to +5, and a switch to ground to control the analog switches simultaneously. <S> Common from each switch goes to a transistor base, and the other two inputs go through resistors to the probes. <S> If you run this crap outside, not only could you kill the muxes at the first electrical storm but you might invite Mr. Lightning in for a visit. <A> Amazing that everyone hates your circuit but no-one seems to have suggested the obvious re-design of using a simple float connected to a potentiometer - these are widely available as they are used in boat / camper van water tanks and very similar parts are used in vehicle fuel tanks (and would likely be ideal for the job as long as you keep the resistive element out of the water). <S> You could make one using a potentiometer and a toilet/water tank ballcock float & arm. <S> Then you only need TWO wires to measure the resistance, and only need to switch ONE wire to measure the other tank. <S> Display could be done with am LM3914 10-segment bar-graph driver IC, or comparators, or read by the ADC input on a microcontroller / Raspberry Pi. <S> I second / third <S> the numerous comments above about safety & reliability too <S> ; DO NOT design your own mains PSU, 5v wall-warts (AKA any old phone charger) are throwaway and likely to be a better option. <S> Be aware of the risks around electricity + water, as well as less obvious factors like corrosion. <S> I realise this does not answer your question but hopefully it goes further towards solving (one of) your problems. <A> What is you water tank here ? <S> A Bath? <S> If the answer is yes: Don't use a home made power supply! <S> Done correctly, it would be ok, but you stated that you are new on this field. <S> A failure in the power supply, may cause the main voltage to be applied to your water tank. <S> And I am never very confident to rely on my circuit breaker to stay alive. <S> What I suggest, as an addition to what has already been told, is to use a battery to power everything that can be in contact with the water. <S> This includes the probes and the sensors. <S> The latter being powered by the main. <S> Or, use use new digital isolation techniques, such as the iCoupler from Analog Devices. <S> Very low power consumption, and still providing true galvanic insulation. <A> <A> Use a digital CMOS multiplexer or build one of logic gates. <S> It should be easy to find an 8-bit 2-channel multiplexer. <S> Connect the sensor wires directly to the multiplexer inputs and pull them down with resistors. <S> It should be better to light LEDs with a low-power MOSFET than with a BJT since most CMOS chips can't source that much current to drive a BJT.
| The easiest and cheapest way to switch is to include a diode in series with each 33 Ohm resistor, connect the probes/resistors/diodes of the two tanks to the one transistor, and to switch the 'power' lead to the two tanks with a single pole dual throw switch. I second the opinions stated in other answers: this is a horrible circuit, and don't even think of powering this from the mains (no, not even with a transformer). You may use opto-couplers, but the LED inside would be powered by the battery, this is not good for your battery life. Then you may use galvanic isolation between this sensor part and the LEDs/buzzer part.
|
What happens when supplying an LED with too low voltage? What happens when supplying an LED with too low voltage? Will it just emit less light? I'm making an octocopter, and need to cut away as much weight as possible. I need to run the LED without cooling if possible. <Q> Look at a typical LED forward bias graph: - Y axis is current and x axis is voltage. <S> At about 2V the LED takes 20mA and illuminates nicely. <S> At 2.2V the current has shot-up to nearly 40mA and the LED is bright. <S> At 1.8V the current is about 5mA and the LED is a bit dim. <S> Below this there is no business until you start going negative then at about -5V <S> applied the LED usually dies, never to recover. <S> This is a typical old-fashioned red LED and, may not 100% apply to more modern LEDs <S> exactly/verbatim <S> but the general idea is the same. <A> Since you have mentioned the word cooling <S> I assume you must be talking about a power LED. <S> Standard LEDs generally do not require cooling. <S> A power LED dissipates roughly 1/3 of the energy supplied to it as heat. <S> Therefore it is possible to reduce the heat dissipation of the LED and hence the need for a heatsink by reducing the power supplied to the LED. <A> As others say, it will depend on that LED datasheet. <S> 50W is pretty big, I doubt you'd be able to run that LED without cooling. <S> Also, you'll want limit the brightness (and power usage / heating) of the power LED with current limiter (eg. <S> driving it with constant current source) or with PWM, not by lowering voltage (LEDs do not behave the same as your typical lightbulb!) <S> And/or you could always use lower specd LED (with less power drawn, less heating, and of course less light produced). <S> Alternatively, as it is for octocopter, I'd guess you have 8 pretty good active cooling parts already, maybe you can use them to cool the LED instead of the big passive heatsink on it (just a small one + wind generated by propellers through it).
| At 1.6V or below the LED hardly takes any current and probably will barely be visible in a dark room.
|
VHDL - how to use inout as inout and as normal out? I wanted to ask if it is possible to use an inout pin as inout and normal out? The two behaviours should be switched through a MUX. The reason for this weird looking implementation is that I have two boards and I want to use the same bitstream. On one board, the same pin is connected to a LED through GPIO and on the other it goes to my I2C bus connection. The software tries to detect the I2C and if successful it sets a register. If not, it clears it. LED_or_SDA : inout std_logic; -- port definitionprocess (register)begin if ( register = '1') then -- software sets this register LED_or_SDA <= I2C_SDA; -- here I want to use it as inout else LED_or_SDA <= gpio_reg; -- here I want to use it as normal out end if;end process; This implementation throws the error "bidirect pad net is driving non-buffer primitives" during translate. Is there a solution for this? <Q> There is no fundamental reason why an inout pin cannot be used as a simple output...just ignore the input signal. <S> I suspect your problem is in the actual VHDL code (rather than the version you posted) or in the details of how you are implementing the design on an FPGA. <A> Most FPGAs do not have internal tri-state buffers except at the IOB (I use Xilinx terms). <S> In fact, given an inout port "DataBus", I create signals "DataBus_in" and "DataBus_out". <S> Everywhere I read from the data bus, I use "<= DataBus_in", and any time I want to write something to the data bus, I use "DataBus_out <=". <S> Other than this snippet, the actual inout port DataBus is used nowhere else, and the _in and _out versions are used exclusively. <S> DataBus_in <= <S> DataBus <S> ; -- always read the value to xyz_intsb_bus_driver: process(OE, DataBus_out) <S> isbegin <S> if (OE= '1') <S> then DataBus <= DataBus_out; else DataBus <= <S> (others => 'Z'); end if;end process tsb_bus_driver; Elsewhere... <S> DataBus_out <S> <= foo; -- example of writing to xyz_outbar <= <S> DataBus_in <S> ; -- example of reading from xyz_in <A> You're unlikely to be able to synthesise connecting your tristate I2C line via a process like that. <S> I would do this <S> (I'm assuming you already have SDA_IN and SDA_OUT signals which you are currently using to create your SDA)... <S> LED_or_SDA <= gpio_reg when register = '0' else '0' when register = '1' and I2C_SDA_OUT = '0 <S> ' else <S> 'Z';I2C_SDA_IN <= <S> LED_or_SDA <S> when register = '0' else '1'; -- idle high
| Therefore it is recommended to put all inout signals at the top-level (with the associated 'Z' driving logic), and use plain old in and out ports throughout your design.
|
My DIY solenoid does not work DESCRIPTION: Hi there, I don't know lot about electricity and I`m having problems with my first solenoid-prototype, I tried to build it watching you tube videos. I have attached the pictures of the wire that I used (diam=1.32mm), and the solenoid that I wired. The core is iron pipe (inner diameter= 1.5cm, outer= 2cm), you could see the size from the picture. There are 500 turns around the core, each turn is about 6.75cm + 50cm additional wire from each side, so the total length of the wire should be 6.75*500+50*2=34.75 meters. I used 1.5v batteries (1st test with 1 battery, 2nd test with 2 of them, so 3v) . I also found from the internet that the resistance for this diameter of wire is 12.597 milli-ohms per meter. PROBLEM: I want it to be a strong electro-magnet. But it doesn't stick any metal at all. Please tell me if the design is wrong, or i wired the circuit wrong, or volts are too little. How much volts do I need to make it work best? <Q> There are a few things I can see that might be the problem: The magnetic field is strongest "inside" the solenoid, and it quickly fall off outside the solenoid. <S> hyperphysics image of solenoid magnetic field: <S> However, this image ignores the fact that your steel core extends beyond the wire turns. <S> Good conductors tend to prevent the "diffusion" of magnetic fields into them, thus the theoretical strongest magnetic field locations outside the solenoid is at the two ends of the steel cylinder. <S> The second issue is with how weak this electromagnet is. <S> Using this solenoid calculator , I found that for ~3.43A through your solenoid (theoretical current for a perfect 1.5V source and your given dimensions) the magnetic field strength was 35.9 mT. <S> If I took into account the internal resistance of the battery ( ~0.15 ohms ), the current reduces to 2.55A, with a corresponding magnetic field of 26.7 mT. <S> This is only a few times stronger than a fridge magnet. <S> Note that this is the magnetic field inside the solenoid. <S> The magnetic field outside will be weaker. <S> Another caveat is I calculated these magnetic fields assuming the steel core was the same size as the portion wrapped with wires. <S> It's not. <S> It isn't immediately clear to me which length to use for calculating the magnetic field strength, So I simply used the length of the wrapped portion. <S> The true length will probably be somewhere in between, so the magnetic field strength will be lower still. <A> That wire is capable of carrying 3A or so, depending on cooling. <S> Given the length, and your resistance per meter, the total coil is probably around 0.4 ohms. <S> Plugging that into Ohm's law, I=V/R, which is V=IR <S> , you find the voltage required is V=3*0.4=1.2. <S> The battery you have is a Panasonic Hyper Manganese D size cell . <S> Sadly they don't provide detailed information on discharge rates, but looking at a similar competitor's datasheet <S> you'll find the battery isn't rated above 500mA of discharge current. <S> However, a careful check of the datasheet will show that the battery has an internal resistance of 150 to 300milliohms. <S> This is pretty close to your coil's resistance, meaning that you're wasting a third your energy inside the battery. <S> If the battery were capable of delivering 2-3A you might be able to get a decently strong magnetic field from the solenoid. <S> But it's not. <S> You'll need to consider using a power supply rated for the current and voltage you need, or several D cells in parallel to supply the current you need. <S> If you put 7 D cells in parallel, each would still be discharging at a fast rate, and wouldn't last long, but you will see a much more noticeable magnetic field from your solenoid. <S> Lastly, note that the magnetic field will appear to emanate from the ends, not the side, of the pipe, and is strongest closest to the coil. <A> The magnetic attraction force from an electromagnet is calculated as: - Force = <S> \$(N\cdot <S> I)^2\cdot\dfrac{\mu_0\cdot <S> A}{2\cdot g^2}\$ where N is number of turns <S> I is current <S> \$\mu_0\$ is permeability of free-space = <S> 4\$\pi\times <S> 10^{-7}\$ <S> A is cross sectional area <S> g is gap from end of solenoid to piece you want to attract with force With 2A, 500 turns, a 1" diameter solenoid and a 0.5" gap, the force is 0.08 newtons. <S> See this calculator
| So the open end of the pipe closest to the coil will appear to have a stronger magnetic force than the other end, or the sides of the pipe. You need a power supply capable of 1.2V at 3A to get as much power from that solenoid as you can reasonably expect.
|
Powering 28 LEDs in series by 220VAC - simple stabilizer I have a home-made thingy that I would like to use as a lightning in my living room. It contains 7*4=28 LEDs (red, 1.86V , 20mA ), originally in 7 groups by 4, each switched on seperately, powered by 12VDC . Now, I would like to power the whole thing by 220VAC . I thought of using a simple stabilizer as on the following schematic: simulate this circuit – Schematic created using CircuitLab If the stabilizer were optimal, I would get (220-28*1.86) / 8.2k = 20.4mA , which is fine for 20mA diodes I think. However, I have some issues: Is my idea of ACDC converter correct? I have a bad feeling that the capacitor is somehow the 1st thing in the circuit and that it's not correct. How well the signal is stabilized by such layout? Especially at the beginning, there might be high currents through the bridge, can't there? Can I rely on the LEDs having moreorless correct voltage drop? It seems that 0.1V per LED would make a little difference in the current so I think it should be safe, but I'm not sure. <Q> That's the warning done. <S> Instead of wasting over 3W of heat in a 8.2kohm resistor, use a dropper capacitor before the bridge on the AC side. <S> At 20mA LED current and with 220V ac, 50 Hz I'd estimate a capacitive reactance of about 11k8 ohms would be about right. <S> \$X_C = <S> \dfrac{1}{2\pi f C}\$ <S> Capacitance will be about 270nF. <S> This should produce a peak current from 220V AC into a load that looks like 60V of 20mA. <S> Here's a simulation: - <S> Input AC voltage of 50Hz and peak of 311 volts (220VRMS) drives via 270nF capacitor a bridge rectifier. <S> A capacitor is on the output but this can be virtually any value and in fact I've just rechecked and it is not needed. <S> A zener diode of 62 volts rating is simulating the 28 LEDs in series. <S> An insignificant 1 ohm resistor tells me what the current is. <S> The response shows a peak current of about 26mA and an average current of about 13mA. <S> The circuit linked to in the comments is unsuitable for driving strings of LEDs because of reverse bias diode breakdown problems. <S> MAKE SURE <S> YOUR 270nF CAPACITOR IS X <S> OR <S> Y RATED AT THE APPROPRIATE SUPPLY VOLTAGE. <S> The diodes in the bridge I've shown as 1N4007 because I know they are rated easily in excess of the potential voltages that might be present. <S> 500V minimum rating for these diodes I'd say. <A> Wiring up a string of LEDs directly to 220VAC presents a possible shock hazard if it isn't completely sealed away. <S> You should probably consider, for safety's sake, a transformer for galvanic isolation and voltage reduction. <S> Or else select a suitable wall wart with 9VDC@250mA, perhaps, and use 7 parallel strings of 4 series-LEDs and resistors. <S> DC voltages higher than about 50V are also considered to be a safety issue. <S> So try and choose a DC voltage below that figure. <A> This doesn’t look good practice to me. <S> You simply do not connect bridge rectifiers to the mains. <S> You need to use a transformer for isolation <S> so you do not have your led’s directly connected to the mains. <S> Rather than connect the led’s in series which requires quite a high voltage with that number <S> usually 3.3V each LED <S> I would redesign your matrix to wire banks of smaller numbers in parallel. <S> You don’t necessarily need full bridge ratification or smoothing, if you use half wave rectification that might be sufficient depending on your transformer selection.
| Safety is a big issue and testing this directly from 220V AC is a hazard you should avoid.
|
Fighting overshoots/undershoots using ferrite beads does not help? I posted relative problem a while ago here but we came to conclusion that the overshoots were just artifacts. Later, when I extended the signal lines, the overshoots increased and were about 2V of magnitude. First thing I tried were unknown to me before ferrite beads. I tried several different types of bandwidths but only one had shown some positive effects. I connected one and then two FB in series making a total of 940 Ohms (MPZ1608B471A), the overshoots were smoothed out, but the undershoots remained: Then I found a similar discussion about my problem here and a simple 100 Ohm series resistor fixed the whole deal: Finally, questions: why a simple resistor (that kills DC as well) fixed the problem, whereas the ferrite bead that should only attenuate high frequencies was not as effective? if I run two wires (one is signal the other is GND), should the ferrite beads go on both wires or just the signal wire? do I need a pull up/pull down resistors if I connect two DI/DO interfaces? could someone, please, add "overshoot" and "undershoot" tags. Thank you so much. simulate this circuit – Schematic created using CircuitLab <Q> Overshoot (and undershoot as used by the TS, which is essentially negative overshoot) are created by a quick transition of the signal (arriving through a long wire) arriving at the end of the line with no place to go (read: arriving at a high impedance). <S> Hence there are three basic ways to reduce overshoot: <S> make the wire short <S> make the edge (transition) <S> less abrubt (a series resistor at the source will do this) give the signal edge somewhere to go (a parallel resistor at the destination will do this, a series resistor will have some effect, but less). <S> A ferrite bead will soften the edge a little bit, but it is mainly used to prevent HF signals (like from radio stations) from entering a device, and vice versa to prevent HF signals from the device from entering the cable which would work as an antenna for these (highly unwanted) signals. <S> HF in this sense means higher than the frequencies you are likely to look at. <S> edit/added: A series resistor will be more effective at the source, but it will flatten the edge, which might be undesirable. <S> Note that to be effective the resistors must match the cable impedance. <S> Something in the order of 30 <S> .. 100 Ohms is a good guess. <A> That's not a very high frequency circuit. <S> Ferrite beads tend to be pretty high Q inductors up into the MHz. <S> They get much more lossy at high frequencies. <S> As you can see from the datasheet , it doesn't cross over into looking more resistive until you hit about 5MHz. <A> It sounds like your issue is as Wouter suggests, too abrupt of a switching action. <S> From your waveform, it would imply to me that you have a very small inductance action happening. <S> Current starts flowing and then has nowhere to go once it hits the receiver resulting in a transient voltage spike. <S> I'm confused as to why no one has suggested a very small capacitor on the receiver side to absorb this transient energy. <S> If a cap is put in place it should be to ground or if it's a differential signal, between the two signals. <S> This should also help relive your original issue of having cross-talk or transient power issues by smoothing out this signal. <S> And to answer your original q's: a ferrite acts on higher frequencies as others have stated. <S> The resistor helps your issue because it's adding a load to the output of the transmitter so part of its output current gets sent to ground rather than it all getting sent only into the capacitance of the wire and reciever.
| A parallel resistor (to ground or power, or a combination) will be more effective at the receiving end, and will not harm the edge, but it will attenuate (lower) the received voltage, and increase power consumption.
|
What do the abbreviations "SPF" and "SPH" mean with respect to fuses? (Single line drawing for 480 V system.) There are two lines connected between the transformer low voltage side and the 480 V bus, one with a fuse labeled "SPF" that connects to a GE power quality meter, another with a fuse labeled "SPH" that connects to a single winding of a transformer. In the context of fuses, what does SPF stand for? It looks like could potentially be single-pole fuse or single pole fixed. I have no idea what SPH stands for. Any ideas? <Q> Maybe SPF for single pole fuse (individual fuse for one of three phases). <S> SPH may stand for "Single Phase", which would fit with your single-winding transformer. <S> SPF fuses seem to be "Solar Protection" which I doubt applies in this case. <A> The SPF Solar Protection Fuse series has been specifically designed for the protection of photovoltaic (PV) systems. <S> This family of Midget style fuses (10x38 mm) can safely protect PV (photovoltaic) modules and conductors from reverse overcurrent conditions. <S> As PV systems have grown in size, so have the corresponding voltage requirements. <S> This increase in system voltage has typically been intended to minimize power loss associated with long conductor runs. <S> Standard circuit protection devices are not designed to completely protect photovoltaic panels. <S> However, the SPF series is UL Listed to safely interrupt faulted circuits up to this demanding voltage level. <S> Littelfuse offers various ampere ratings to match specific requirements in a variety of applications. <S> Features and Benefits 1. <S> Designed to both UL and IEC photovoltaic specifications 2. <S> UL 2579 Listed 1000 VDC maximum 3. <S> 1-30 <S> A ratings available 4. <S> 20,000 A Interrupting Rating Not just LittleFuse, but <S> Keystone and other fuse producers also use SPF for Solar PV system fuses. <A> SPF and SPH are two types of "Failsafe Wirewound Resistors". <S> These are resistors that are designed to act as fuses, and are still common in some power supply and utility applications. <S> Please see this link for a description of why they are used. <S> The specific datasheet for the SPF/SPH parts can be found here (pdf) . <S> Good luck!
| All jokes aside, SPF does stand for Solar Protective Fuse
|
What determines how much current can flow through a 2N2222 A? I've been fiddling around with a brushed DC motor, a 2N2222 and an arduino's PWM to get different speeds out of my motor. Now, based on a video tutorial I watched on youtube, It was recommended to put a 1k Ohm resitor between the PWM pin and the base, apparently to protect the arduino in case the transistor screwed up. I did what I was told to do, following this schematic (My power supply is 1.5V and R1 is 1K, not 220) But the motor would not turn when the microcontroller pin would output 5V ( full duty). So I figured, ights, let me not mess around with my arduino in case that was the problem, so I connected the base, throught the 1k resistor, to the same 1.5V powering the motor...still nothing...put a new 1k resistor...still nothing...so I romoved the 1k Ohm resistor and plugged my base straight to the 1.5v source and the motor started turning. Can someone please explain to me why that is? From what I've been reading, isn't the voltage at the base what determines the current flowing from the collector to the emmiter? -_- <-- Confused face <Q> The motor needs a certain amount of current in order to turn. <S> How much current is allowed to pass through the transistor from collector to emitter, and hence through the motor, is controlled by the current passing through the transistor from base to emitter times the transistor's current gain, known as "h fe ". <S> The base resistor reduced this current to too low a value to allow the motor to turn even when amplified by the transistor. <S> E.g.: <S> 300mA <S> (I motor ) / <S> 70A/A <S> (h fe ) <S> = <S> 4.2mA <S> (5V (V MCU ) - 0.7V <S> (V BE(SAT) ) ) <S> /4.2mA = <S> 1.024 kohm Note <S> that the motor voltage supply is not involved in these calculations, but it still must be high enough after subtracting the voltage from collector to emitter as per motor specifications. <A> In addition to Ignacio's answer, I would like to recommand this web-site to learn "How to use a transistor as a switch": Using Transistor as a Switch myself learnt that how to use a transistor as a switch by this web-page. <A> I'm going to go out on a limb and assume that when you plugged in the base of the transistor to the 1.5V power source, you actually plugged it in across the motor. <S> This would explain why the behavior was as you described. <S> Assuming that your implementation of the circuit was correct, and the transistor isn't backwards (it gets me all the time; it's worth double checking), the problem with this circuit it almost certainly the voltage of your power source. <S> The transistor will chop off about 0.6V in the base to emitter junction, and probably closer to 0.8V between the collector and the emitter, depending on how saturated it is. <S> 5V through 1K will give 5 mA, which with an h fe of about 100 (fairly standard) gives a maximum 500 mA of current through the collector. <S> This should be enough for a small toy motor, but you won't get 500 mA through a motor with a 1.5V-0.8V = 0.7V power supply, especially if it's just a battery or two. <S> Like markrages suggested in the comments, you ought to be considering a power source more like your diagram, 9V. 5V would be a minimum, I'd say, or maybe 3V. <S> The 1K resistor is important, not because it protects your Arduino from feedback (though there's that too), but because it prevents it from shorting its 5V through the transistor, which could burn one thing or the other out (though the current supplied by an Arduino is limited, I think less than 100 mA). <S> You should definitely use it always . <S> Finally, depending on how you power the Arduino, you could share its power source with the motor. <S> If you're powering it from the barrel jack, I'm 80% certain that you can connect directly to that power supply using the V in socket that is next to the 5V and 3.3V and all them <S> (I'm imagining an Uno). <S> This line won't be going through any regulating (again, 80% sure), so you wouldn't have to worry about overtaxing the Arduino's components. <S> Conversely, you could get a decent power supply and power the motor and the Arduino with it by doing the reverse, and connecting the Arduino's V in pin to the supply's voltage like it's intended to be... <S> Good luck <S> ; hope that helps.
| Use what you know about the motor's required current, the voltage across the transistor from base to emitter during saturation, the voltage from MCU pin to emitter, and the transistor's h fe to calculate the correct maximum value of resistor to use.
|
IC to count number of inputs that are on? I have 8 wires, and I want to count how many of these are in a high state and convert this to a 3-bit binary signal. If the count of inputs that are high is greater than 4, I don't care what exact amount of inputs are on beyond that point, I just need to know whether more than 4 are high. So I only need 3 bits of output, if that makes sense. i.e. if 1 wire is on, output 001, if 2 wires on on, output 010, etc. and if 7 or 8 wires are on output 111. Is there an IC that can help me do this easily? <Q> A microcontroller could do it very easily. <S> Since there are only 256 possible combinations, a simple lookup table would be the fastest way. <S> (I did the below outputting 8 for all high for consistency, you can change the last entry to 7 if you want). <S> The code could look something like this:- unsigned char lut[256] = {0x0, 0x1, 0x1, 0x2, 0x1, 0x2, 0x2, 0x3, 0x1, 0x2, 0x2, 0x3, 0x2, 0x3, 0x3, 0x4, 0x1, 0x2, 0x2, 0x3, 0x2, 0x3, 0x3, 0x4, 0x2, 0x3, 0x3, 0x4, 0x3, 0x4, 0x4, 0x5, 0x1, 0x2, 0x2, 0x3, 0x2, 0x3, 0x3, 0x4, 0x2, 0x3, 0x3, 0x4, 0x3, 0x4, 0x4, 0x5, 0x2, 0x3, 0x3, 0x4, 0x3, 0x4, 0x4, 0x5, 0x3, 0x4, 0x4, 0x5, 0x4, 0x5, 0x5, 0x6, 0x1, 0x2, 0x2, 0x3, 0x2, 0x3, 0x3, 0x4, 0x2, 0x3, 0x3, 0x4, 0x3, 0x4, 0x4, 0x5, 0x2, 0x3, 0x3, 0x4, 0x3, 0x4, 0x4, 0x5, 0x3, 0x4, 0x4, 0x5, 0x4, 0x5, 0x5, 0x6, 0x2, 0x3, 0x3, 0x4, 0x3, 0x4, 0x4, 0x5, 0x3, 0x4, 0x4, 0x5, 0x4, 0x5, 0x5, 0x6, 0x3, 0x4, 0x4, 0x5, 0x4, 0x5, 0x5, 0x6, 0x4, 0x5, 0x5, 0x6, 0x5, 0x6, 0x6, 0x7, 0x1, 0x2, 0x2, 0x3, 0x2, 0x3, 0x3, 0x4, 0x2, 0x3, 0x3, 0x4, 0x3, 0x4, 0x4, 0x5, 0x2, 0x3, 0x3, 0x4, 0x3, 0x4, 0x4, 0x5, 0x3, 0x4, 0x4, 0x5, 0x4, 0x5, 0x5, 0x6, 0x2, 0x3, 0x3, 0x4, 0x3, 0x4, 0x4, 0x5, 0x3, 0x4, 0x4, 0x5, 0x4, 0x5, 0x5, 0x6, 0x3, 0x4, 0x4, 0x5, 0x4, 0x5, 0x5, 0x6, 0x4, 0x5, 0x5, 0x6, 0x5, 0x6, 0x6, 0x7, 0x2, 0x3, 0x3, 0x4, 0x3, 0x4, 0x4, 0x5, 0x3, 0x4, 0x4, 0x5, 0x4, 0x5, 0x5, 0x6, 0x3, 0x4, 0x4, 0x5, 0x4, 0x5, 0x5, 0x6, 0x4, 0x5, 0x5, 0x6, 0x5, 0x6, 0x6, 0x7, 0x3, 0x4, 0x4, 0x5, 0x4, 0x5, 0x5, 0x6, 0x4, 0x5, 0x5, 0x6, 0x5, 0x6, 0x6, 0x7, 0x4, 0x5, 0x5, 0x6, 0x5, 0x6, 0x6, 0x7, 0x5, 0x6, 0x6, 0x7, 0x6, 0x7, 0x7, 0x8}; <some initialization stuff <S> > for( <S> ;;) out_port= lut[in_port]; <S> // do this forever <S> I don't know of any logic chip that has this exact functionality. <S> Priority encoders do something a bit different. <A> This is a simple job for a microcontroller. <S> Spehro correctly points out that it can be accomplished with a simple lookup. <S> If you can tolerate lower speed then counting the bits in a loop will do it too, but will take less code space. <S> Another way is to do this in analog. <S> Put a 100 kΩ resistor in series with each digital output, then feed that into a comparator. <S> If 4 out of 8 are high, then you get 0.50 of the supply voltage. <S> If 3 out of 8 are high, then you get 0.38 of the supply voltage. <S> Set the comparator threshold to 0.44 of the supply and it will distinguish between 4 or more, and 3 or less. <A> Shouldn't a 4-bit full adder do the trick? <S> http://nl.farnell.com/texas-instruments/cd74hc283e/logic-4bit-binary-full-adder-16dip/dp/1750337
| If you needed really fast response time, you could program a nonvolatile memory with the lookup table, or use programmable logic.
|
Regulated +5V drops when relays are switched on A regulated power supply RAC02-055C/277 provides 5V to an Arduino driving 2 relays . The power supply is rated to provide 2W of power. Connecting a multimeter between one of the leads of the power supply to the +5V bus (which connects to Arduino 5V pin and the +5V pins of the relays), and another multimeter across the +5V and GND rails, the current reads 98.6 mA and the voltage is 4.969 V . When one of the relay is turned on, the readings are 209.4 mA and 4.911 V . When both relays are turned on, the readings are 301.9 mA and 4.869 V . How can we prevent the voltage output from the supply from dropping/changing whenever the relays are turned on and off? The +5V rail is used to bias a input signal, so a constantly changing +Vcc will give inaccurate readings. <Q> According to the datasheet, the voltage drop is within spec: "Load Voltage Regulation 10% to 100% full load ±6% max. <S> "If <S> "full" load is 400mA for a 2W, 5V output version, then you are getting up to 75% load by then. <S> (4.869V / 5.0V ) <S> * <S> 100 gives us 97.38% of load voltage regulation, which is still within reason. <S> I suggest you use a more constant reference.. <S> Some people use a reference (zener) diode or maybe another small linear regulator that goes from 5V to 3.3V <S> but its more of a reference than a 'power supply' so to speak. <S> Basically you need something to act as a stable reference, and not ever-changing due to such large changes in load as one or both relays turn on. <S> Edit: To use a resistor and zener diode as a voltage reference, see this question and <S> it's answer by Anindo for how to do it, and calculate the required resistance to make it work. <S> Using a Zener diode as voltage reference <A> The biggest things for load regulation are distribution loss (conductive or copper loss), and specified load regulation. <S> In this case the output is dropping about 2% over your test load range. <S> This is within spec for the part (+/-6% from 10% to 100% load). <S> For conductive loss supplies frequently will include remote sense lines, but this model doesn't have any remote sense. <S> It looks like the only thing you can do for a DC drop if you plan to continue to use this supply would be enhanced copper thickness or shorter distribution (and even then it sound like it could still fall short just by performing to spec). <A> Your power supply has an output voltage accuracy of 5%, or +- 0.25V <S> for a 5V supply. <S> It is behaving within its specified limits! <S> You should use a separate power supply for the bias voltage.
| If the relais stay on for only short periods of time, it might be possible to decouple the bias voltage using capacitors & a resistor in the line charging it.
|
Meaning of “with respect to rotor and stator” In the following example, when it says “with respect to stator” and “with respect to rotor”, did they mean the rotor part of the synchronous generator in both cases? Example: The excitation field in the synchronous generator is stationary with respect to the rotor. When the rotor rotates at synchronous speed, the excitation field rotates with respect to stator. <Q> From what I understood from your question you want to know what is "with respect" to refers to in a synchronous motor. <S> This means relative velocity. <S> If two object are moving at same speed it appears like one object is stationary with respect to other. <S> The phrase "excitation field in the synchronous generator is stationary with respect to the rotor" means that both field and rotor have same speed. <S> For example let angular velocity of electric field is 40 rps. <S> for synchronous motor, rotor is synchronous to electric field. <S> Therefore speed of rotor is also 40 rps. <S> Here speed electric field with respect to stator is 40 rps. <S> But speed of electric field with respect to rotor is 0 rps. <A> With respect to stator: does not rotate. <S> is stationary with respect to the rotor. <S> This says that the electric field rotates around with the rotor. <S> In fact leading the rotor into it's advancing position. <S> When rotor rotates at synchronous speed the excitation field rotates with respect to stator. <S> It's likely referring to a brush-less DC motor (BLDC), that the excitation field from the magnets that are physically attached to the rotor. <S> The electric drive field also rotates synchronous to the rotor; it must lead the rotor magnets by a fixed angle ahead of them. <S> As long as the angle is fixed, the excitation in the stator is synchronous to the rotor. <S> If the stator is not driven correctly, the rotor and electric drive can slip out of synchronization; essentially the field is not properly positioned to pull the rotor around to it's target position and speed. <A> Lets clarify this with example of a induction motor(Referred as IM) for easy understanding. <S> Things to know before- <S> 1.Synchronous speed (Ns) is 120*f/ <S> P. 2.Relative speed concept-Speed of object when seen from other reference. <S> 3.IM has a stator(static) and rotor(rotating or static). <S> 4.Stator magnetic field - Field generated by stator. <S> 5.Rotor magnetic field -Field generated by induced currents in rotor. <S> In case of induction motor during normal operation. <S> Stator magnetic field speed = <S> Ns. <S> Rotor magnetic field <S> speed =Ns. <S> Speed of Rotor magnetic field w.r.t Stator M.F or vice versa = <S> Ns-Ns=0. <S> Speed of stator = 0 <S> (static). <S> speed of Rotor = <S> N (actual speed). <S> speed of stator field <S> w.r.t rotor= <S> Ns-N. <S> In general speed of X w.r.t Y , we just compute by = <S> Vx-Vy
| With respect to rotor: rotates with the rotor excitation field in the synchronous generator
|
EL wire brightness, per m not per m\$^2\$ The datasheets and catalog pages for commercial electroluminescent (EL) wire in hobby quantities, e.g., adafruit's , list brightness in candelas per square meter. That's ok for two-dimensional EL sheet , but not for one-dimensional wire. (I understand that frequency and voltage affect brightness, as well as half-life, power consumption, color, etc.) How do figures like 46 cd/m\$^2\$ convert to cd/m, which can be meaningfully compared to LED strips or neon tubes? Just consider the circumference of the cladding? Or of the phosphor? If there's multiple layers of phosphor, which one? Also, the candela measures luminosity only in a particular direction, while EL wire has nontrivial off-axis brightness. Are there specs for lumens per meter of such wire? <Q> Wire does not produce light one-dimensionally - it produces light in all directions and the intensity of the light is maximum right at the surface of the wire. <S> Imagine the light is emitted in all directions along the length of the wire - there is real power entering into the space between the wire and your eyes and, at any particular distance from the wire, the sum total of that power remains the same (just like it is on a radio antenna). <S> What happens at a greater distance is that the power per sq metre reduces because the surface area that the power is passing through gets bigger proportional to distance squared. <S> This is why the lumen per metre does not really mean anything. <S> It's lumens per sq metre. <A> The \$\frac{\text{cd}}{\text{m}^2}\$ units you are looking at are probably intended to give a measure of the Luminance . <S> It doesn't matter that the strips are, in some abstract sense, one-dimensional. <S> They are in fact 3D objects which create a surface that emits light, and luminance is the brightness of that surface independent of its size and shape. <S> Let's make an analogy between luminance and, say, density. <S> We wouldn't use a different measure of density for a steel wire compared to a steel plate or steel ball. <S> That is, not if we had in mind the property of the material itself. <S> There exists the concept of "linear mass": mass per unit length of a wire. <S> But that isn't a property of the material, since it depends on the cross-sectional area of the wire. <S> Fact is that if these strips are made thicker (or doubled or tripled up, etc), they do give off more light, right? <S> If luminance were given for the strip as \$\frac{\text{cd}}{\text{m}}\$ <S> then it would mean that the thickness of the strip is being ignored: for such a measure to make sense as a constant, a given length of EL wire would have to give off the same amount of light no matter how thick or thin it is. <A> I regretfully propose to answer that an EL wire's datasheet is useless for comparing its brightness to that of other illuminants; one must buy samples and measure them oneself. <S> (I still hope to be proven wrong.) <A> that should be a good starting figure for your project. <S> Take losses into account when you don't use optics in situations where the light emits only on one side (built-in wires in furniture etc) <S> Good luck and enjoy the light! <A> What you care about the is the surface, you don't need to care about which phosphor layer/ <S> It is about the exitance of that surface. <S> With the OD (outer diameter) <S> = <S> D of the wire: <S> \$ 46 [\dfrac{cd}{m^2}] <S> * \pi <S> *D [m]\$ i.e. using the circumferential distance around the wire. <S> The next step is a little more complex because the emission from the surface in the direction along the wire is lambertian whilst around the wire it is isotropic. <S> So your steradian calculation will be a volume that is a toroid, with eth wire running up <S> the Z -axis (wire not shown). <S> I calculate this as \$ \dfrac{3}{4} \pi^2\$ [steradians} combining all this... \$ <S> 46 <S> * <S> D <S> * \dfrac{3}{4 <S> } * pi^3\$ <S> [\$\dfrac{lm}{m}\$]
| the numbers per square meter indicate perceived brightness of the surface, which does not say much about how brightly your EL wire is gonna light an object or surface.the luminous flux per meter varies depending on voltage and frequency of the driver and can reach around 16 lumens for 3 mm high-brightness EL wire.
|
Thick vs Thin Sponge for Soldering After purchasing a TC205 sponge and comparing it with the default sponge that came with my soldering station, I noticed that the TC205 sponge is very thin in comparison! Is the thin sponge used for a different application? What are the slits on the sponges for? <Q> There is no difference between thicknesses of sponge as all they are doing is abrading off the solder without damaging the tip. <S> The grooves provide a larger abrasion surface without increasing the volume of the sponge. <A> Brass wool works a lot better than sponges imo, and doesn't require water. <A> Try pouring water on it and watch the magic of science LOL. <S> Seriously it should just expand outwards every which way and look just like (or similar to) <S> the one on the right. <S> Hey I could be wrong but <S> one I did recently expanded beautifully. <A> Others have already noted that you just need to add water to expand your new sponge. <S> I wanted to address your second question about the slits. <S> Line them up parallel with the iron when you put the sponge in its tray. <S> Then when you drag the iron tip through the sponge to clean it, it can do a better job cleaning the sides of the tip. <S> Some model sponges have the slits, others don't. <S> I've never found it to make much difference either way. <A> Brass wool is a correct accessories for cleaning soldering iron. <S> Wet sponges should not be used due to worn out of soldering iron due to temperature difference between iron and wet sponge. <S> Soldering iron will degrade faster while using wet sponges. <S> So it is very clerar whether it is thick or thin, sponges should not be used.
| The sponge comes thin as purchased because the cells have been flattened, but once it is wet the cells relax and the sponge will take its final form.
|
Inverting Amplifier Not Giving Gain of -2.2 A LM358N op amp is used with R_f = 220k and R_i = 100k , which should give a gain of -2.2 in a inverting amplifier. It is powered by a regulated +5V supply. However while the input voltage is 1.000V as read by a multimeter, the output is 0.032V . Why is the gain ~1 and the output not inverted? <Q> It needs a negative rail. <S> Without that it can't go below ground. <S> Also the LM358 isn't really rail to rail. <S> It can work near ground, but not at it, and it needs about 1.5V from the positive supply. <A> As pointed out by Gsills, there is no negative power rail for it to go negative. <S> The LM358 is a pretty old, and for general purposes where you would 'expect' rail-to-rail ( 0V -> VCC) outputs it doesn't cut it. <S> Your comment of using an MCP6002 is a good idea <S> , it will definitely get the job done. <S> Just make sure it's slew rate and <S> in general it's bandwidth ratings can do the frequency of expected signals you need. <S> If you are doing things mostly at DC then it's fine. <S> If you expect the output to increase with the input, then use non-inverting arrangement. <S> If you want your output to go towards 0V as the input increases (negative feedback) then you would want the inverting arrangement. <S> to give it a negative supply, try a simple 7905 series linear regulator. <S> http://www.jaycar.com.au/images_uploaded/LM7905.PDF Also do not forget to place local supply supply decoupling capacitors for supply rails, as fast changes in the opamp output will sag/not behave correctly if there is a lack of available current/energy storage nearby. <A> A DC-DC converter or a 7660 charge pump will do it. <S> Some use a MAX232 to generate +/-9V with the help of a few capacitors.
| If you wish this circuit to function with only a single 5V supply available you are going to have create a negative rail somehow.
|
How does a potentiometer work? A lot of places where I've read explanations about potentiometers either go into so much detail or they're so superficial that I don't understand the actual concept behind it. How would you easily explain to me what a potentiometer is and how it works? <Q> Here's a simple experiment you can do: Find a pencil with a soft lead (2B or something). <S> Make a dark, thick rectangle on a piece of paper about 5cm <S> x 5mm or so (a long strip). <S> Using a multimeter set to measure resistance, put one test lead at the very edge of the rectangle you drew, and measure different points along the strip using the other test lead. <S> You should read lower resistances as the two leads are closer together. <S> A potentiometer uses a similar strip of resistive material (carbon, for example) to provide varying amounts of resistance, based on the distance between two connections to the material. <A> This answer was originally intended for the physics website, but seeing as I was almost done my answer <S> I'll post it here: <S> (B being the ground), and attach some other circuit to W. By turning W clockwise we decrease the voltage supplied to W because we have increased the resistance between A and W. <S> If we align W with B, W is now at 0 volts because B is the ground. <S> We can calculate $$V_W=V_A-R_{AW}I$$ <A> A fixed-resistor potential divider reduces an input voltage to a lower value at the output: - If you imagine that \$R_{top}\$ and \$R_{bottom}\$ are combined into one resistor then the wiper position is where top and bottom resistors meet mathematically ; the resistance above the wiper (\$R_{top}\$) is exactly as it was with two separate resistors and the the resistance below the wiper (\$R_{bottom}\$ is also the same. <S> So, with a potentiometer you vary the tap-point and <S> this means your output voltage can be varied between 0V and Vin. <A> This question comes from over in Physics. <S> I suspect the OP is asking about the measuring device called a Potentiometer that was used extensively until the VTVM was developed. <S> In Juliens's diagram, consider a reference voltage is connected between A and B. Connect a sensitive galvanometer between W and the voltage to be measured. <S> Adjust the variable resistor until the galvanometer reads zero. <S> Now the Vout is equal to the the voltage under test and there is no current flow, so there is no load on the test point or the variable resistor. <S> Now the Vout can be measured with a meter that has lower impedance or the variable resistor can be calibrated so the dial or slide position reads in appropriate units. <S> (There are variations on this that eliminate some error sources). <S> If you ever wondered how they made those great accurate measurements in the 1800's like chemical half-cell potentials and thermopile voltages and the work functions of metals with light shining on them, this was it, along with the ballistic galvanometer and electroplating to measure charge and current.
| When you adjust the knob on a (analog) potentiometer you change the amount of resistance between W and A and W and B. (W stands for Wiper here).Let's say we apply 5 Volts between A and B
|
Voltage regulator for DC motor I'm trying to regulate the voltage of 9 volts battery to DC motor but didn't work. This is my scheme: The original scheme is for led and just wondering why it didn't work for DC motor. Here is the image of DC motor with 104 capacitor. Please give me an idea why it is not working. <Q> The motor needs a lot more current than the LED does. <S> You need to reduce the value of R1, or remove it altogether, connecting the motor directly to the regulator (assuming the motor can take the full 5V). <S> Also, the 78L05 is only good for 100 mA in any case; if the motor requires more than this, bad things will happen. <S> Even at 100 mA, the 78L05 will probably be dissipating more power than is good for it, and it will (hopefully) shut down to preserve itself. <S> Try a value of 47Ω for R1; see if the motor moves at all. <A> You have a huge resistance on regulator <S> so you have (not counting motor resistance) <S> 5V/470Ohms = 10.6mA which is hardly any current to be driving a motor. <S> It's enough to drive an LED tho. <S> Either make that resistor smaller or remove it entirely. <S> I would think that the motor could handle a full 5V. <A> As others have stated, the resistor R1 is not necessary - the old schematic used as reference is for an LED, which usually (in poor efficiency circuits) use ballast resistors for current limiting. <S> The 7805 regulator must be checked for power ratings - the motor will only draw large currents when starting, and when stalling (you stop the motor from moving with your fingers, or something gets jammed in it). <S> Please note this is not actually required for operation - things will work fine without it. <S> the diode 1N4004 will also explode if you do not check <S> it's current rating as well. <S> For the LED circuit it would only be passing 6-7mA in order to give the 5V rail ~10mA for the LED. <S> With this circuit, the diode may have to pass much higher (~100mA, or in start/stall conditions much more). <S> I suggest you completely remove the diode, as the 7805 most likely has a diode-like pass-transistor or MOSFET with diode anyway. <S> If you really wanted, place the diode from the 5V output to ground ("pointing" at the 5V rail) <S> in order to catch reverse voltage spikes from the motor when it turns on/off due to the inherent inductive characteristics of DC motors. <S> The idea is that it's in parallel with the motor pins. <S> Place it in parallel to that little capacitor if you want. <S> If you do actually replace the R1 with a 47Ohm resistor, this will limit the current to ~100mA, but be careful that your resistor does not get too hot. <S> a 1/8 or 1/4 Watt resistor might not be good enough. <S> Try for a 1/2 -> 1 Watt power resistor if you go down that path. <S> Please not that the motor will work fine without the resistor - it has it's own internal resistance and when operating will only draw the current that it "needs".
| If you have like a switch or something to allow the 7805 to regulate before turning on the motor, I suggest you should put a 100uF or higher electrolytic capacitor from the 7805 output to ground, to allow it to store energy for the start current of the motor.
|
Why is L1 cache faster than L2 cache? I'm trying to understand why certain CPU cache memories are faster than others. When comparing cache memory to something like main memory, there are differences in memory type (SRAM vs DRAM), and locality issues (on-chip vs having to traverse a memory bus) that can affect access speeds. But L1 and L2 are generally on the same chip, or at least on the same die, and I think they are the same type of memory. So why is L1 faster? <Q> No, they're not the same type of RAM, even though they're on the same chip that uses the same manufacturing process. <S> Therefore, it is built using larger transistors and wider metal tracks, trading off space and power for speed. <S> The higher-level caches need to have higher capacities, but can afford to be slower, so they use smaller transistors that are packed more tightly. <A> L1 is usually used as a storage for decoded instructions, while L2 is a general cache for a single core. <S> The lower the cache the smaller size <S> it is and faster it usually is. <S> As a rough rule of thumb for PC processors: L1 Cache: 2-3 clock cycle access L2 Cache: ~10 clock cycle access L3 Cache: ~20-30 clock cycle access <S> The design of the L1 cache should be to maximize the hit rate (the probability of the desired instruction address or data address being in the cache) while keeping the cache latency as low as possible. <S> Intel uses an L1 cache with a latency of 3 cycles. <S> The L2 cache is shared between one or more L1 caches and is often much, much larger. <S> Whereas the L1 cache is designed to maximize the hit rate, the L2 cache is designed to minimize the miss penalty (the delay incurred when an L1 miss happens). <S> For chips that have L3 caches, the purpose is specific to the design of the chip. <S> For Intel, L3 caches first made their appearance in 4 way multi-processor systems (Pentium 4 Xeon MP processors) in 2002. <S> L3 caches in this sense greatly reduced delays in multi-threaded environments and took a load off the FSB. <S> At the time, L3 caches were still dedicated to each single core processor until Intel Dual-Core Xeon processors became available in 2006. <S> In 2009, L3 caches became a mainstay of the Nehalem microprocessors on desktop and multi-socket server systems. <S> Quote sourced here from "Pinhedd's" response. <A> There are several reasons why speed is inversely proportional to size. <S> The first that comes to mind is the physical domination of conductors, where signal propagation i limited to some factor from speed of light. <S> An operation may take as long as it will take an electrical signal to travel the longest distance inside the memory tile and back. <S> Another related reason is the separation of clock domains. <S> Each CPU runs off its own clock generator, which allows the CPU to run on multi-GHz clocks. <S> Level-1 cache runs at and is synced with the CPU clock, which is the fastest in the system. <S> Level-2 cache on the other hand has to serve many CPUs, and is running in a different (slower) clock domain. <S> Not only the L2 clock slower (larger tile) but to cross a clock domain boundary adds another delay. <S> Then of course there are the fan-out issues (already mentioned). <A> Apart from inherent performance characteristics, locality also has a role (L1 is closer to the CPU). <S> According to What every programmer should know about memory : <S> It is interesting to note that for the on-die L2 cache a large part (probably even the majority) of the access time is caused by wire delays. <S> This is a physical limitation which can only get worse with increasing cache sizes. <S> Only process shrinking (for instance, going from 60nm for Merom to 45nm for Penryn in Intel’s lineup) can improve those numbers.
| Of all the caches, the L1 cache needs to have the fastest possible access time (lowest latency), versus how much capacity it needs to have in order to provide an adequate "hit" rate.
|
Why Direct Broadcast satellite is interrupted by climatic conditions? Television signals received through direct satellite service are mostly interrupted by the climatic conditions. But other communication systems that uses satellites are not that much affected. Why is this so ? Or is it just because television signals are hard real time? <Q> Normally the higher the frequency is, the more is affected by the rain. <S> On RF links rain fade start to occur at some GHz and it becomes very important if you go above 10GHz. <S> You can check the ITU-R model for the rain attenuation on this link: <S> http://www.itu.int/rec/R-REC-P.838/en Another thing that might make the difference is how fault tolerant the service is (they could have redundancy, diversity…). <A> If one had two antennas located a mile apart pointing at the same satellite, it's likely that during adverse meteorological conditions both antennas would have a significant fraction of data packets disrupted by weather, but the sets of disrupted packets observed by the two antennas would be different and largely unrelated. <S> Likewise if one had two adjacent antennas pointed at different satellites. <S> A typical DBS receiver setup, however, will just have one antenna receiving a signal from one satellite. <S> Further, some forms of satellite communication can use after-the-fact error recovery: a receiver that can tell that a packet got corrupted can request a retransmission. <S> The need for retransmission will increase latency, but can be made mostly transparent to the user. <S> By contrast, if a receiver can't capture a broadcast video packet when it's transmitted, there's no mechanism for it to find out what it missed. <A> Satellite reception degradation is determined by several factors. <S> The lower the frequency the less affected by rain fading. <S> C band less effected, Ku and Ka band more affected. <S> The ground based receiver position in the satellites footprint. <S> The further away from the maximum signal zone the greater the effect. <S> The precision of the aiming of the antenna itself lessens the effects. <S> Rain and snow absorption and refraction of the signal. <S> The altitude of satellite above the horizon. <S> Transmission mode and bandwidth would also be a consideration as well.
| The difference is probably because of the different Radio Frequency Bands used by the different Satellite Services.
|
Are there any disadvantages using magnetic couplers instead of opto-couplers? I'm trying to find the disadvantages of using magnetic couplers versus opto couplers (1). I'm deciding which of the two I should use to create a galvanic isolation between two PCB's that are going to communicate via SPI in a normal environment e.g house living room. (1) http://www.wisegeek.com/what-is-an-opto-isolator.htm I have found several websites and documents that show the advantages of magnetic couplers vs opto couplers. Here is one that summarizes most advantages(2) (2) http://www.nve.com/isopdf/WP1201.pdf (Page 3). But I'm having a hard time believing that there is not a single disadvantage to using magnetic couplers versus opto couplers for galvanic isolation between two PCB's. So my question is: Are there any disadvantages using magnetic couplers instead of opto couplers? <Q> Although that page you linked says that they work to DC, that's only because there is a latch on the other side that maintains the last known state. <S> Consider the NVE IL710 <S> In some cases, this could be a deal killer. <S> Secondly, another disadvantage is cost. <S> You'll never see a magnetic type in a cheap SMPS controller because they're more expensive, even if the other quirk didn't exist. <S> Thirdly, if you're designing things that need to live near high currents or other things that cause large magnetic fields (not uncommon in applications that require isolation), you'd better check out the magnetic field immunity: <A> Disadvantages of magnetic signal coupling versus optical signal coupling: Magnetics are usually bulkier and more expensive. <S> Magnetics don't pass DC. <S> If you need to pass a low frequency signal, then you have to modulate it onto a carrier somehow. <A> Generally speaking, it's a lot easier to provide a very high level of optical shielding to an opto coupler than it is to provide a very high level of magnetic shielding to a magnetic coupler. <S> Optocouplers can be made very small and in huge quantities with modern semiconductor processing. <S> Magnetic couplers can transfer much more power given specific space and cost constraints. <S> So for small signals, optically coupled circuits can be smaller, cheaper, and more immune to the environment.
| For power transfer, magnetic couplers of a given efficiency are generally less expensive. The magnetic ones (and some other non-optical types) transfer only changes to the state.
|
How to detect non magnetic material? I want to design a paintball counter. I have a metallic or plastic tube through which a paint ball passes. How can I detect that the ball passed? What kind of sensors do I need? I cannot see the ball from both ends of the tube, thus I need a sort of non mettalic detector. I dont want holes, and not microphone because of the tube will shake. And no obstacles in the tube. <Q> Use a 1 MHz oscillator with the capacitor as part of the tuning - <S> probably an LC resonant tank type oscillator would work just fine - count the oscillation cycles over a short duration - as the ball is passing, the count should drop measurably. <A> Some possible alternatives: pressure sensor (if the tube is not too much larger); some sort of mechanic switch (not good if you don't want to obstacle the ball though); photo detector (may require a hole, like Matt suggested); microphone (record vibrations from the passage, need to define a pattern). <S> To me the microphone sound the least invasive, I'd go for that one. <S> But it takes some engineering to figure out the sound pattern. <S> There are more for sure, but some more specifications may narrow down the list. <A> What is the nature of the tube shaking? <S> If it shakes because a ball passed through, there's your answer. <S> If it is due to something else, you may still be able to distinguish a ball passing from the shaking--the frequencies should be different. <S> For many plastic tubes (white PVC included), you can shine a bright light through it and see a shadow as the ball passes.
| If the tube is plastic you could arrange plates of a capacitor around the periphery of the tube - under normal circumstances the dielectric of the cap is air but as the tennis ball passes, it is likely that the permittivity of the material of the ball will increase the capacitance - it might increase the capacitance by 30% if care is taken.
|
Played with some lead solder without washing my hands for awhile, should I be concerned? I was using some lead solder at work, and wasn't aware it was lead solder nor had anyone told me that it can be dangerous. For the most part I was just handling it. This was last week and I don't think I washed my hands afterward. I know this may be silly, but is that anything I should be concerned over? I can't imagine my low exposure would be worrisome, it's probably something that multiple hours every day for years would cause issues over, but I just want to make sure. <Q> Lead solder is not all that bad- <S> it's cumulative exposure that can cause problems. <S> Here is the MSDS (Material Safety Data Sheet) for a typical tin-lead solder. <S> It says, in part, Keep away from foodstuffs, beverages and feed. <S> Immediately remove all soiled and contaminated clothing. <S> Wash hands before breaks and at the end of work. <S> Note also that, as @MattYoung says, the "no-clean" flux is quite nasty- <S> LD50 on rats is 4mg/kg. <S> From this occupational health and safety site: <S> What does LD50 mean? <S> LD stands for "Lethal Dose". <S> LD50 is the amount of a material, given all at once, which causes the death of 50% (one half) of a group of test animals. <S> The LD50 is one way to measure the short-term poisoning potential (acute toxicity) of a material. <S> Toxicologists can use many kinds of animals but most often testing is done with rats and mice. <S> It is usually expressed as the amount of chemical administered (e.g., milligrams) per 100 grams (for smaller animals) or per kilogram (for bigger test subjects) of the body weight of the test animal. <S> The LD50 can be found for any route of entry or administration but dermal (applied to the skin) and oral (given by mouth) <S> administration methods are the most common. <S> The reasons for only washing hands before breaks and after work is so that you don't ingest lead from handling food etc. <S> or other contact with mucous membranes. <S> Solder <S> paste is considerably easier to ingest, so more care is called for. <S> Also, if you are pregnant or could become pregnant , much more care should be taken. <A> It has only been more recently agreed that there is NO acceptable level do lead exposure and thus those big warnings on all of the California bound products. <S> Open pit mining <S> and it's affects on ground water lead levels <S> is what has fueled some of the concern and research. <S> The main points of concern have already been pointed out in the various answers, but as a family doctor I will summarize my concerns with lead exposes: 1) length of exposure and cumulative exposure which for you is very little. <S> 2) type of exposure: handlig lead is nothing compared to ingesting it or breathing it in. <S> 3) <S> Age: Children and pregnancy are of greater concern <S> In other words: don't worry about it! <S> This is likely from the flux. <A> You're safe. <S> Although I'm statistically insignificant <S> my exposure might give you some comfort: 10 years old, chewed on the lead plate included in the electronics lab kit. <S> Chewed on lead pellets for air gun (I like to chew). <S> My co worker accidentally dropped 1dl of solder lead in the dishwasher at work, the lead lay in the machine for a couple of washes and probably contaminated the coffee cups... <S> Lots of soldering without gloves or fume extractor. <S> I'm still alive. <S> Although I do get rashes sometimes from flux fumes and residues.
| By the way, long term exposure to breathing in the fumes that are made when solder is melted has been linked to asthma. To be safe: don't do it again!
|
RC filter infront of Analog port pins in pic microcontroller What is the use of RC filter in front of Analog port pin while using port pin for analog to digital conversion in pic 16f877A and in most pic controller the RC filter is advised to be used? <Q> The idea is to filter out noise and bandlimit the signal to avoid aliasing. <S> You therefore want to set the corner frequency of the filter above the maximum signal frequency, but make sure you have no significant signal or noise above half of the sampling rate. <S> With a single-pole filter you will have 3dB of attenuation at the corner frequency and 20dB per decade of frequency rolloff above that. <S> The corner frequency for an RC filter is f = 1/(2*pi*R*C), so choose your components so as not to interfere with your signal. <S> Of course if you're just taking a sample of some slowly varying signal like a temp sensor or something like that a single pole filter will be fine. <A> A lot of MCUs take a pulse of current to "charge" the internal sample and hold circuitry (part of the integrated ADC). <S> If the series resistance of the signal source is too high, it won't charge the internal cap sufficiently and there will be an error in the conversion. <S> Applying a capacitor from the pin to ground means that there is a much greater chance of this working because the external cap can supply the current needed. <S> See this SE post (relating to PIC input impedance) and check out the answer. <S> Of course this makes a low pass filter that is also useful for anti-aliasing. <A> Two reasons: Impedance anti-aliasing Impedance:as said by @Andy aka, you ADC has a small capacitor inside to provide the sample and hold functionality of the ADC. <S> You have to quickly provides the charges to that capacitor. <S> You may use a low impedance buffer in front of the ADC. <S> Or use a capacitor that is several times bigger than the one into the ADC. <S> Anti-aliasing:You will often see that you have to ensure that the frequency content of the signal at the input of your ADC is at least half the sampling rate (fin < fs/2). <S> But why?Because of the signal processing theory. <S> The Nyquist–Shannon sampling theorem said that if you want to be able to sample your signal without any information loss, then the sampling frequency should at least twice that frequency. <S> Ok, fine. <S> But do I need this? <S> Good question! <S> Two answers: <S> No: <S> If you use the signal that enters the ADC to compare against a threshold (for instance when a tank reaches a certain level). <S> Fine. <S> You got a sample that said that the tank was at that level, this is fine. <S> The sample reflects the reality. <S> At least if the noise level on your signal is low. <S> Yes: <S> If you plan to use any signal processing computations. <S> Then your samples are not just numbers, they represents a signal. <S> And this representation matches the reality only if the Nyquist–Shannon sampling theorem is verified. <S> Then it's up to you. <S> You want to know the voltage at certain time: the ADC will tell you this with or without an anti-aliasing filter. <S> You want to consider your input as a signal and perform things such as filtering: You need the anti-aliasing filter.
| You may need more than a single pole filter to avoid aliasing depending on your signal and noise frequency content.
|
Atmel Xmega ADC input voltage range - Can the ADC be damaged by exceeding it's reference voltage, if the input is still less then VCC? I've been working with Atmels ATxmega parts a lot recently, and there is one question I still have that I have not been able to find a answer to, even after scouring the datasheets. Basically, The ADC in the xmega series are somewhat limited by the fact that the \$V_{REF}\$ cannot exceed \$V_{CC}-0.6V\$. This is easily enough accommodated, I am using a 2.048V reference for the ADC with 3.3V rails. However, I cannot determine if exceeding the ADC \$V_{REF}\$ on the ADC inputs would be actually harmful to the device. I'm aware exceeding the \$V_{REF}\$ will saturate the ADC, this is fine in this application. However, having to clamp the input to \$V_{REF}\$ would be challenging in my application, as the \$V_{REF}\$ I'm using cannot sink any current, so if I simply used a simple schottky diode to clamp \$V_{in}\$ to \$V_{REF}\$, any excursions of \$V_{in}\$ would simply cause \$V_{REF}\$ to rise as well. I'd really prefer not to have to add a whole additional buffer just to accommodate potential occasional \$V_{in}\$ excursions. Looking through the datasheet, I'm not really much enlightened: The ADC IO lines are "Pin"s, so I would assume that exceeding the reference, as long as the voltage is kept < \$V_{CC}\$, while the conversion results may be invalid, it won't cause actual damage to the MCU. On the other hand: Note that this isn't specified as an "Absolute Maximum" value. I'm not sure if this implies that exceeding these bounds will lead to invalid conversion results, or actual damage. I've been through both the part-specific datasheet (ATxmega32e5 in this case), and the E series databook, and neither have clarified the issue much for me. <Q> Exceeding Vref isn't a problem providing you don't exceed absolute max ratings. <S> One table is the "don't ever do" table and the other is "don't expect it to function if you do" table. <S> EDIT - <S> The Vref operational rating of AVcc-0.6V relates more to the DAC outputs on the XMega. <S> DACs share the same voltage reference as the ADCs and because they use the same AVcc power rails, I guess the outputs are restricted to AVcc-0.6V in order to produce non-clipping signals without resorting to extravagant rail-to-rail output amplifiers. <A> Atmel explains this in Application note: AVR1300: Using the Atmel AVR XMEGA ADC . <S> To quote from 1.4.2: <S> The voltage on any of the two inputs can be between GND and VREF, but the difference between them must not be larger than VREF/GAIN because this will saturate the ADC and the converted value will only equal the top value of the ADC. <A> Im not sure this counts as an answer:I found your question after i extensively searched the net for the same problem!After <S> I could find no answer anywhere (datasheet, application notes, etc.) <S> and I didn't want to screw up one of my current design, I took a ATMEGA168 on a breadboard an run some tests. <S> @AREF <S> = 2.5V <S> I could ramp up the voltage up to VCC without any damage (I even kept ADC0 for about 30 minutes on VCC). <S> After I played a bit with different voltages I went with the voltage on ADC0 higher than VCC. <S> @About 6V permanent damage occured.
| Don't exceed absolute maximum ratings (in fact stay within them as much as you reasonably can).
|
Multiply clock frequency by three or more times? Frequency of a digital clock signal can be doubled by using an EXOR gate (clock at one input pin and delayed clock at another). Can we use any similar circuit which can multiply frequency by three times ? <Q> You can use a third harmonic filter (inductor and capacitor) to resonate at clk*3 and then a schmitt trigger inverter (or other gate) to turn the sine wave into a decent square wave at clk*3. <S> This works because a typical square wave has fundamental and odd harmonics in its spectrum: <S> - The blue waveform is the output from the tuned circuit when excited by a 1 MHz square wave. <S> This can be magnified to logic levels with a comparator with a little bit of hysteresis. <S> Can we use any similar circuit which can multiply frequency by three times ? <S> I cannot tell you to believe it's similar <S> but in my mind it is. <A> The most common way it to use a PLL based frequency multiplier. <S> Source (www.ee.ucl.ac.uk/~pbrennan/E771/PLL.ppt) <S> The Phase Locked Loop works for frequency of a signal (or more correctly -phase), like an op-amp works for Voltage. <S> It has a high enough gain to keep the two inputs of the phase detector equal in frequency (and usually phase). <S> Applying a reference signal to the input, the other phase detector frequency becomes equal to the reference by virtue of the loop gain changing the frequency of the Voltage Controlled Oscillator (VCO) until the frequency error is zero and therefore the output frequency is Finput <S> * N. <S> I've taken some liberties with the details like lock range, loop bandwidth etc.. <S> but I hope you get the idea. <S> You may find it interesting that an XOR gate can be used as the phase detector in a Digital PLL frequency Multiplier: <A> If one wants a clean 3x waveform with uniformly-spaced rising edges and a known timing relationship to the original, a PLL is the best way to go. <S> As a possible implementation, have the 3x clock operate a three-bit counter a flop which captures the state of the reference, and a flop which captures the state of that flop. <S> Have the counter jump to 000 any time the latter two flops read "01", and otherwise advance once per count whenever its value isn't 101. <S> The LSB of that counter will pulse three times for each reference clock edge, provided that the input clock is fast enough for it to do so.
| If one wants something with 3x as many rising edges as the original waveform, but doesn't particularly care how even they are, a simple approach is to use an oscillator that runs at least 6x the speed of the reference along with a counting circuit.
|
What is the common mode in common mode input voltage range? The common mode voltage at the input of a difference amplifier is defined as $$\dfrac{V_1+V_2}{2}$$ and common mode input voltage range is the range of voltages at the input for which the opamp works as it's supposed to.But what is common mode about that? <Q> Common mode voltage is distinguised from differential mode voltage. <S> The differential mode voltage for two inputs is how different they are. <S> That is \$V_1-V_2\$. <S> As you say, the formula is \$\dfrac{V_1+V_2}{2}\$. <S> We can make this more mathematical by noticing that with these definitions \$V_1 = <S> V_c + V_d / 2\$ and <S> \$V_2 = V_c - V_d / 2\$. <S> You can see that the common-mode term is common between these two equations. <A> For an op-amp in normal operation, V1 ~= V2, so it collapses to just the input voltage. <S> The range of acceptable input voltages is the common mode input voltage range. <A> Fundamentally, the term common mode implies that the signal at the two input terminals of a differential amplifier is identical in both magnitude and phase. <S> When signals V1 and V2 are applied as input we can spilt them into a combination of common mode and differential mode signals in the following manner V1 = <S> (V1 + V2)/2 <S> + (V1 - V2)/2 V2 = <S> (V1 + V2)/2 + { <S> -(V1 - V2)}/2 <S> thus, (V1 + V2)/2 is the common part and <S> (V1 - V2)/2 is the differential part of the signal. <S> One of the signals that is always applied in the common mode is the DC Bias. <S> It is this common mode signal that the Input Common Mode Range is associated with. <S> Therefore, the ICMR is the range of input DC (or Common Mode) voltage range for which the circuit is in saturation (or as intended to be). <S> From the previous computation if we consider V1 and V2 to have both ac and dc components, such as <S> (V0 +/- <S> A sin wt) <S> then V0 would be the DC common mode signal applied to the amplifier.
| The common mode voltage is the part of the voltage that is the same for both, that is, the part that they have in common.
|
Are there stencils for thermal paste/grease? I recently came across a reference to the use of stencils to consistently apply thermal paste/grease between device and heatsink. I'm having trouble finding much better explanation of that concept. Are such stencils commonly used? If so, how does one use, construct, and/or obtain them? <Q> I put "silk" in scare quotes because most screens these days are polyester mesh. <S> I have not used these for this particular purpose, but I have done more than enough silk screen printing to know it would be no problem to print arbitrary patterns of thermal grease on the flat surface of a heat sink (for example, 12 TO-220 package outlines could be printed at once). <S> The problem comes in if the surface has any protrusions (such as fins) which could interfere with the screen or the screen frame. <S> In the case of stencils, the viscosity is similar to solder paste, so printing should be similar. <S> In the case of a screen, they normally work from graphic arts files such as Adobe Illustrator or PDF. <A> I haven't heard of them, so I'll answer with respect to what I'm sure of, and then move off into SWAG territory. <S> You'd make them the same way you make a stencil for solder paste. <S> Pick a layer, and use the gerb processor to put that layer in a gerber file, and send that file to a stencil maker. <S> Use a tool like gerbv to make sure you have it right before you waste money. <S> Now the part I don't know -- <S> It would be very difficult to use such a stencil when there's been any populating of the board at all. <S> It can't be done before the board is populated, or you'll screw it up during solder stencil printing. <S> It can't be done after, because there are components in the way. <S> Maybe its doable as part of a wave solder assembly, or with a small hand-held stencil. <S> I can't think of any particular case where thermal paste is a real precision job, though. <S> I can picture some situations where you might want a metered grease delivery, but there are better tools for that than a stencil. <A> Yes they are. <S> it is a method which not only ensure a reasonable repeatability on the application but also is the fastest and less "messy" method during manufacture. <S> Ideally a pad is used (phase change, graphite etc...) <S> as this is the cleanest method for manufacturing to use (it's typically a good idea to not use fluids in manufacture (glue, paste etc) <S> BUT if paste is to be used a stencil is idea. <S> https://www.infineon.com/dgdl/Infineon-AN2006_02_Application_of_screen_print_templates-AN-v1.0-en.pdf?fileId=db3a304412b407950112b40ed3f71297 <S> https://www.semikron.com/dl/service-support/downloads/download/semikron-application-note-thermal-paste-application-en-2018-01-19-rev-00/
| A stencil could be used as Scott mentions, or a "silk" screen with adequately large mesh for the viscosity of the thermal grease. You could create a gerber file for the stencil maker as you would for a PCB layer.
|
Movement of Electricity in a Simple Circuit: A Few Simple Questions and a Diagram The diagram above is of two batteries a light bulb and some cables. A book that I am reading shows that electricity would not flow through the blue cable, it would flow through the red cable. This left me with the following questions: Is this even true? why wouldn't some of the electrons from the battery on the right still try to flow to the battery on the left? (this kind of relates to question 3). Also suppose that the cable extending from the negative terminal (the anode) of the battery on the left was many times longer than the length of the cable extending from anode of the battery on the right, in such a case would electricity be transferred from the anode of the battery on the right to the positive terminal (cathode) of the battery on the left? In other words do the electrons take the path of least resistance? Pardon the analogy that follows; would a sort of traffic jam be created where the blue and the red cable connect? I mean once the electron from one battery started flowing to the cathode, would they impede other electrons from joining the flow? <Q> This question is about steady-state dc analysis, and my answer applies to the steady state condition. <S> Is this even true? <S> why wouldn't some of the electrons from the battery on the right still try to flow to the battery on the left? <S> (this kind of relates to question 3). <S> The battery on the right is not part of a complete circuit. <S> Therefore no current will flow through it. <S> To form a complete circuit there must be a path for current to flow from the + terminal back to the - terminal. <S> Also suppose that the cable extending from the negative terminal (the anode) of the battery on the left was many times longer than the length of the cable extending from anode of the battery on the right, in such a case would electricity be transferred from the anode of the battery on the right to the positive terminal (cathode) of the battery on the left? <S> In other words do the electrons take the path of least resistance? <S> The battery on the right is still not part of a complete circuit, so no current will flow through it. <S> Pardon the analogy that follows; <S> would a sort of traffic jam be created where the blue and the red cable connect? <S> I mean once the electron from one battery started flowing to the cathode, would they impede other electrons from joining the flow? <S> Without a complete circuit through the battery on the right, it doesn't matter. <S> Even if both batteries were connected in parallel, the connection of two wires would not cause a traffic jam and impede the flow of current. <S> However, the light bulb acts as a resistor. <S> As such, if you apply a given voltage (12 V, say), it will only draw current according to its resistance. <S> If the voltage is applied by two batteries in parallel, part of that current will come from one battery and part from another, so that Ohm's law is satisfied for the resistor (light bulb). <S> In this way, you might consider the light bulb to be causing a kind of traffic jam, in that it resists the flow of current. <A> This kind of question is difficult to answer because you are including too many myths in your question that need to be dispelled before you are in a position to understand a real answer. <S> In other words, you need a basic introduction to electricity, which is too much material to teach in a answer on this Q&A site. <S> I'll therefore just mention a few things: <S> Yes, the book is correct. <S> Current must flow in a loop. <S> The battery at right is not part of any loop. <S> Talking about "electricity" flowing is meaningless and you will continue to be confused until you clear up the basic conceptual problem behind this wrong use of terms. <S> There is EMF (usually measured in volts and often referred to as "voltage") and there is "current". <S> There is no such physical quantity as "electricity". <S> Go learn voltage and current . <S> Until then, trying to talk to you will be too cumbersome and probably wouldn't work anyway. <A> Consider electricity as having two co-existing components (this is not strictly correct <S> but it's good for beginners). <S> The first component is "voltage". <S> Voltage is an electric field which propagates at a significant fraction of the speed of light through a conductor (I think roughly 200 million meters per second). <S> When you first connect the red wire to the + on the battery, the field propagates very quickly throughout the red wire, and even down the blue wire. <S> The second component is "current". <S> Current is the actual motion of the electrons, and current is "caused" by the voltage "pushing" electrons. <S> Compared to the electric field that is the voltage, current moves at a glacial pace of 1 meter per hour . <S> One of the requirements for current is that the voltage can travel in a closed loop from one side of the voltage source to the other. <S> That is why Yes, this is true. <S> The voltage field propagates to the battery on the right, but it cannot cause electrons to move without a closed loop. <S> Electrons take the path of least resistance, but for the blue wire, the resistance of the path is infinite , because it is not a closed loop, therefore no current flows. <S> There is no traffic jam because no current flows down the right wire. <S> However, if you are using the voltage of the signal to transfer information, you may see distortion that is caused by the field propagating down the blue wire and then slamming into the infinite resistance. <S> (this is why The Photon prefaced with the comment about "steady-state analysis")
| No electrons from the battery on the right move because there is no closed loop for them to return to the battery.
|
What is the usage of the negative voltages on a PC motherboard? What is the usage of theses negative voltages? Are they there only for backward compatibility? In nowadays PC power supplies, we have: +12V +5V +3.3V but also: -12V -5V But the current rating of the negative rails are much smaller than the positive ones. If we were back to the 80' where op-amps were always powered symmetrically at +12V -12V: Okay.. But nowadays, almost everything you may find on a motherboard is digital logic only powered by positive voltages. Except for the RS232, which is an almost obsolete bus, I don't see any reason for having negative rails distributed by the power supply. Because it's very high volume, I suppose that cost drives everything here. Thus, why each PSU has to deliver those voltages if they are barely used ? (the very low current rating of the negative rails of PSUs let me suppose this). Wouldn't it be less expensive to let every hardware provider to add their own embedded SMPS when a negative voltage is required? <Q> PCs are stuffed with requirements which relate to backwards compatibility - and -Ve rails are part of that. <S> I'm not sure about -5V, <S> but there's a -12V line on the original PCI bus, so <S> if you want to provide proper PCI sockets, then you need a -12V rail, even if the last person making a PCI card which needed -12V died in 2002. <S> So now you have a -12V rail on your power connector even after people have stopped fitting PCI connectors. <S> Some of these things are remarkably difficult to get away from — <S> the 'legacy free' PC with no PS/2-style keyboard/mouse connections was being talked about as imminent 15 years ago, but desktop machines still tend to have those connectors. <S> It just turns out to be cheaper/easier to keep supporting all the old cruft than it does to drop it and clean-up the design. <S> Or perhaps it doesn't, and PCs have sunk under the accumulated weight of all this baggage and people have moved on to other form-factors... <A> Although –12 V and –5 V are supplied to the motherboard via the power supply connectors, the motherboard normally uses only the +3.3 V, +5 V, and +12 V. <S> If present, the –5 V is simply routed to the ISA bus on pin B5 so any ISA cards can use it, even though very few ever have. <S> However, as an example, the analog data separator circuits found in older floppy controllers did use –5 V. <S> The motherboard logic typically doesn’t use –12 V either; however, it might be used in some board designs for serial port or local area network (LAN) circuits. <S> In fact, –5 V was removed from the ATX12V 1.3 and later specifications. <S> Because modern PCs no longer include ISA slots, the –5 V signal was deemed as no longer necessary. <S> However, if you are installing a new power supply in a system with an older motherboard that incorporates ISA bus slots, you want a supply that does include the –5 V signal. <A> In short, because the ATX spec says so. <S> ATX is an improvement on the old standard of AT, and does go through revisions. <S> -5V <S> rails became optional in ATX12V 1.2. <S> The original ATX specifications were released by Intel in 1995 and have been revised multiple times. <S> Currently at 2.3. <S> But it is slow to update, like any standard. <S> -12V will be phased out, eventually. <A> -5V are needed for dynamic RAM. <S> With early PCs based on 8 bit processors, 64kB was a typical "maximum" memory size, implemented using 32 16kBit RAMs (4116). <S> Only with the advent of 64kBit RAMs was the out-of-rail voltage (which was +12V and -5V for the 4116) generated on-chip with charge pumps. <S> Similar requirements for early EPROMS. <S> So there is little surprise in the bus system for the first IBM PC having those voltages. <S> +12V also was popular for the motors of disk drives, both because of the larger power than on the +5V rail as well because of fewer consequences to the computing from power surges. <S> -12V, in contrast, was almost only used for RS232 circuitry.
| Then if you want to design a standard power connector pin-out which can be used by people building motherboards with PCI connectors on it, then it needs a -12V rail, or else the motherboard manufacturer needs to start adding power supplies to his motherboard. The only reason it remained in most power supply designs for many years is that –5 V was required on the ISA bus for full backward compatibility.
|
Why have two NOT gates in series? I have recently been looking at the datasheets for the 74HC139 IC in order to see if it was suitable for my project, and have come across the following logic diagram which strikes me as a little bit odd: simulate this circuit – Schematic created using CircuitLab For each of the inputs Yn, there are two NOT gates after the triple-input NAND gate; I don't understand why this is necessary as simple boolean logic tells us: $$\overline{\overline{A}}\equiv A\qquad \forall A \in \{\text{TRUE}, \text{FALSE}\}$$ Therefore I am assuming there is some electronic based reason why there are two inverters before the output? I have heard not gates called Inverting buffers before, and these supposedly isolate the circuit before and after, however, I cannot claim to understand the use of this so I'd appreciate any enlightenment! <Q> Possible reasons: <S> Load Balancing <S> The driver of A has an unknown number of fan-out to drive. <S> Fan-out within the circuit and the parasitic it induces can be calculated for the specific circuits, but we do not know the other circuits that are connected the driver. <S> Essentially the inverters are being used as buffer equivalent. <S> and help manage the parasitic. <S> Timing and total current <S> To reduce the transition glitch, the second state inverters can be sized for a faster transition switch. <S> Doing so makes the NAND gates input update near the same time. <S> With the inputs changing less periodically, power can be saved and transition glitches can be reduced. <S> Signal boosting and power <S> Lets say VDD = 1.2V but the input is 0.9V. <S> The input is still a logical 1, but considered weak which causes slower switching and burns more power. <S> The first inverters can be sized to handle transitions better, making the voltage more predictable for the rest of the design. <S> There is also a possibility of the change in the voltage domain. <S> In this case the inverters in the first state can act as a step down, e.g. a 5V input domain to 2V domain. <S> Any combination of the above <A> The time required for a gate to switch is dependent upon the amount of capacitive load it must drive, the size of the transistors, and the number of transistors in series. <S> An inverter consists of one NFET (N-channel Field Effect Transistor) and one PFET (P-channel FET); <S> a three-input NAND gate has three PFETs in parallel and three NFETs in series. <S> In order for a 3-input NAND gate to switch an output low as quickly as could an inverter, each of the three NFETs would have to be three times as big as would be the single NFET of an inverter. <S> For a small chip such as this one, the only transistors which have to drive any significant load are those connected to output pins. <S> Using four outputs driven by inverters, it will be necessary to have four big PFETs and four big NFETs, plus a bunch of little ones. <S> If one assigns the NFETs an area of "1", the PFETs would probably have an area of about 1.5 (P-channel material doesn't work quite as well as N-channel), for a total area of about 10. <S> If the outputs were driven directly by NAND gates, it would be necessary to use twelve big PFETs (total area 18) and twelve huge NFETs (total area 36, for a total area of about 54. <S> Adding 20 little NFETs and 20 little PFETs [12 each for the NAND, and 8 each for inverters <S> ] the circuit will reduce the area consumed by big transistors by 44 units-more than 80%! <S> Although there are some occasions when an output pin will be driven directly by a "logic gate" other than an inverter, driving outputs in such a fashion increases greatly the area required for output transistors; it's generally only worthwhile in cases where e.g. a device has two power-supply inputs and it must be able to drive its output low even when only one supply is working. <A> If the NAND gate is made in the obvious way (three parallel transistors to GND and three series transistors to Vdd) <S> then it will have low source capability, the transitions will not be sharp, and the delay time will be load capacitance dependent. <S> Adding a buffer (or two to restore the logic) cleans up all those problems. <S> Here is what a typical unbuffered inverter (schematic like this)... .. <S> transfer function (output vs. input shown on line (1)) looks like: With a buffer, the line (1) will be much closer to a square shape. <S> (the second line is the current that is drawn). <A> This is silly if you are just trying to communicate the logic of a chip. <S> The internal gates are probably very small with little drive capability. <S> Signals that go outside need to go thru a buffer that can source and sink much more current. <S> Somehow this implementation detail seems to have made it into the logical description, where it doesn't belong. <S> The logic would be the same if the two inverters in series were replaced by a wire. <S> Then there should be a overall speed and current drive spec for the outputs. <S> You could just as well envision slower and more powerful NAND gates. <A> While this may seem like a pointless thing to do, it does have practical application. <S> This will boost the weak output signal. <S> The level is unchanged, but the full current sourcing or sinking capabilities of final inverter are available to drive a load resistance if needed <A> In the past, such arrangement was used for a delay.
| Probably it is drawn this way because internally there are some buffering stages.
|
Why aren't headphone jack shaped plugs used for data? With Apple's lightning cable, and USB 3.0, reversible cables are taking off, and I personally think this is very convenient. But we have had better than reversible for a long time, in the form of the headphone jack, which can be inserted in any direction, not just 2 directions. Why isn't a headphone jack shaped connector used for data more often? All I ever see that shape used for is audio and power supplies (I've seen it used once for data, in the iPod shuffle, but thats it). <Q> Digital signals are highly susceptible to the noise generated by rotating the plug. <S> For audio, these noises (cracks) are rarely audible unless they last longer than 50us (simply because of the fact that we're unable to hear frequencies over 20kHz). <S> So, the cracks becomes audible only when the surface of the connector has deteriorated enough that the period of lack of connection is substantially longer. <S> It might be acceptable for low frequency digital data, as well as power supplies. <S> Finally, most digital standards require quick detection of the disconnect - even though the above issue could be worked around with proper ECC (Error Correcting Codes), USB assumes that any loss of connectivity for over 2ms is considered a disconnect. <S> (USB 3.0 SS is even more strict). <A> A reason why the round barrel connector with multiple ways may be not preferred is that to insert it, you'll push it through one or two "wrong connections" before it finally comes to rest. <S> Ditto when removing it. <S> This means that it could short a power supply out (momentarily or indefinitely if not inserted properly). <S> It could also reverse a supply to a chip and this would be an obvious problem. <A> It's difficult to get more than a few connections in a coaxial form. <S> I have a hard time imagining an equivalent-to-HDMI plug/socket system (19 pins) made coaxially (image from here ). <A> In fact, theye are. <S> One example that comes to my mind is are the Texax Instruments graphical calculators, like the TI-83+ http://education.ti.com/en/us/products/computer_software/connectivity-software/silver-usb-cable-for-windows-mac/features/features-summary <A> There's another reason which none of the current answers have touched on, but which has a very clear user experience impact. <S> Simply enough, lots of devices use more than a single connector. <S> My MP3 player sitting at the desk as I am typing this has both USB and headphone receptables. <S> The backup HDD at home needs USB and external power. <S> Lots of common devices, particularly those that might benefit from something like this, use two or three connections: one for power and one or more for something else. <S> Same connector for different purposes has been tried; look at the IBM PS/2 and its (dubious choice of the) use of identical 6-pin Mini-DINs for keyboard and mouse that has caused, I am certain, untold frustration with devices being connected to the wrong port . <S> Now, that's not exactly the type of connector you're asking about, but it presents the same sort of user experience problems. <S> Had IBM used different style connectors for connecting hardware that needed to be connected to the correct port to work (even just different number of pins with one or more unconnected would have prevented the user from accidentally plugging one into the socket for the other), subsequently color coding the mouse and keyboard connectors would not have been needed. <S> This is largely a non-issue for headphones, but it can be a very serious concern if one of those pairs carry voltage at the ability of delivering significant current. <S> While that could conceivably be worked around at the protocol level using something like a Power Good signal , that's another layer of complexity added for at best marginal gain. <S> Such connectors are also only available in a few different sizes (2.5 mm, 3.5 mm, 6.3 mm, and I think that's about it as far as consumer-appropriate sizes are concerned; I'm not sure if there are other sizes for specialist applications). <S> While this does not necessarily have to be a showstopper, it is a limiting factor particularly if you want to use more than one of them on a single device. <S> Of course, USB too is only available in a few select sizes, but very rarely is more than one USB input needed. <A> They are used, but just in a product that doesn't see a lot of daylight. <S> Check it out: https://support.apple.com/en-us/HT201893 <A> One additional area to consider is the ability to effectively route the conductors within the connector, with a coaxial connection like a headphone jack <S> the center connector is surrounded by a insulator which is surrounded by another conductor, and then another insulator and another conductor meaning for a simple 3 point connection they had to layer 5 times in a circular fashion effectively making the connector (1+(n*2))width of the conductors/insulators. <S> wouldn't this make anything larger than 3-4 conductors impractical? <S> also i can't even count the number of headphone cords i have gone through due to their lack of good protection from bumps/wiggles. <S> even the connectors soldered to most boards are prone to disconnect.
| Standard headphone style jacks also momentarily make connection with wrong cables during insertion or removal. As a rule of thumb, any connector where there are moving parts while the connection is established, is a terrible idea for high frequency digital data. But as explained by qdot, only for low-speed connections. The reasons they aren't prolific have been answered above, but it is interesting to note that it HAS been done, even if it isn't done often.
|
Is it good practice to parallel relay contacts for increased current capacity? Say I have a DPDT relay, like T92S7D12-24 . The contacts of this relay are rated for 30A, but there are two sets of contacts. Can I parallel the contacts to get an effective 60A relay? Further, could I parallel two (or more) relays and get even more current capacity? I see two possible problems. Current may not be shared equally between sets of contacts and between relays. One set of contacts could take more than their share of the current and overheat. Switching times may vary between contacts and between relays. When breaking with current through the contacts, the last set of contacts to open may be carrying far more than their recommended current at time of break. This could cause damage. Are these problems? Are there other problems? If so, can they be quantified and worked around? Or is paralleling contacts and relays always bad design practice? In my specific application, I'm using these relays as part of precharge for a capacitor bank. They are not expected to switch current. They make once the caps are charged, then hold. They should never open under current flow. Under these specific circumstances, should I still expect problems? <Q> No you should not do this. <S> Sometimes it is explicitly allowed on the data sheet (but not that I can see on this data sheet), and when it is, in my experience you never get as much as double the capacity. <S> Paralleling physically separate relays is worse again because they're not physically moving together- <S> expect welded contacts etc. <S> if you tried that. <S> If you can split the load (for example, instead of a 40A heater use two 20A heaters) then you can get an equivalent functionality. <S> You could think about ballasting the loads (wasting power to roughly equalize the currents) and fusing each contact separately, but I don't think that's a good idea at all. <S> Note that using the relay at the maximum rated current will lead to a pretty short life (only 100,000 operations for a resistive load), which might be only weeks or months if it's switching continuously. <S> At 3HP (motor load), the life is only 1,000 operations, so at once per minute it won't last a single day. <S> Edit: <S> With the added information that you're using the relay to switch effectively at a relatively low DC voltage and you're mostly concerned about carrying current.. <S> I can't say categorically this is really a horrible idea with a single relay, but I think I'd get on the horn to the manufacturer and see if it's possible to get any buy-in. <S> When one of the contacts inevitably fails first, I think I would prefer the relay to not emit excessive amounts of smoke or flames). <S> I think you're okay at 40A (with AgCdO contacts) given the UL508 rating, but beyond that is in question. <S> If you really need such a high carrying current, the Omron G7Z appears to explicitly allow paralleling the 40A contacts without derating, for 160A total capacity, but perhaps not with the blessing of safety agencies. <A> Paralleling contacts to pass a higher rating, than the individual contact rating, is a bad practice. <S> Sometimes paralleling of contacts is done for redundancy. <S> Should one set of contacts degrade current can flow through the other set of contacts. <S> When this is done the normal current load does not exceed the rating of the individual contacts. <S> I've seen this practice recommended by NFPA in Electrical trade magazines where the system is critical and a failure of the relay can lead to another hazard. <A> Relay current-handling ability is limited by two factors: the ability of the contacts to pass current when continuously "on", and the ability of the contacts to handle the stresses associated with switching. <S> In general, if two relays are connected in parallel, the continuous current-handling ability of the combination will be almost equal to the sum of the individual relays' abilities, but the switching ability of the combination may be that of the worse relay, and in many cases won't be much better than that of the better one. <S> If one's application never entails opening or closing relays under loaded conditions, it may be reasonable to use paralleled relays to boost steady-state current-handling ability. <S> In general, however, one should only use parallel contacts to boost "live" current-switching abilities if a relay manufacturer specifically allows it. <A> If this happens the whole of the remaining current will end up being passed by the one remaining good contact. <S> This will over load the contact and it will fail as well. <S> Best if you can divide the load and switch them all separately.
| No its not good practice to parallel contacts because in the worst case, one of the paralleled contacts could fail. It comes down to variability in contact resistance vs. the resistance of the connections (plus whatever, hopefully balanced, resistance you add externally).
|
How do you arrange six 6-ohm resistors to have a total resistance of 6-ohm? Is there a mathematical way to know the answer? (or you can do it only by trial and error) Could you prove that it is possible or impossible mathematically? <Q> simulate this circuit – Schematic created using CircuitLab <S> here <S> R5//R1 series to R3 <S> => <S> 3 + 6 = 9 in one branch R4 + R6 + R2 = <S> > <S> 6 + 6 + 6 = 18 in 2nd branch <S> 18 <S> // 9 gives 6 <A> Arrange 5 in your pocket, connect up one. <A> What about these. <S> Are they eligible or just cheats? <S> : - simulate this circuit – Schematic created using CircuitLab <A> It is possible to arrange all possible topologies and calculate the resistance of each. <S> Nice idea for programming homework. <S> Proving that something is possible requires only one example. <S> In your case: one resistor between the two poles, all other resistors unconnected (or connected to one pole, etc). <A> Another possibility would be: (6//6//6) + <S> 6//(6+6) <S> = <S> 2 + <S> 6//12 <S> = 2 + <S> 4 = 6 simulate this circuit – <S> Schematic created using CircuitLab BTW, I did note that you're after a mathematical solution, but since I couldn't think of one, I offered this. <S> It would certainly be possible to solve it algorithmically, with iterations, but a single mathematical solution may not be possible? <S> Very interesting question. <A> This problem is under constrained.. <S> what does 'arranged' mean? <S> Can you use one or four in series-parallel and short the left-over resistors? <S> It's not possible to have them share power equally, however it is possible to actively use all the resistors. <S> Hint: calculate 1/( 1/9 + 1/18 ) <S> If there is a straightforward mathematical way, I'm not aware of it. <A> This appears to be related to: https://mathoverflow.net/questions/66853/number-of-graphs-with-n-edges which leads to just twelve graphs for six edges - quite a suprise to me. <S> You will then need to measure n! node pairs. <S> Oh - I quickly came up with the 'leave 5 unconnected' (a definite cheat) and bridge (not a cheat) circuits. <S> Kudos to the answers where all resistors carry current.
| Proving that something is impossible requires an ad-hoc proof or enumerating all possible topologies.
|
Microcontroller programming: do I need vendor's IDE? I am very interested in the following ARM Cortex-M0 microcontroller: http://www.nxp.com/products/microcontrollers/cortex_m0_m0/lpc1100/LPC11U24FBD48.html#overview However I'm turned off at the thought of having to license their development software, which I found here: http://www.lpcware.com/lpcxpresso Do I have options? My goal is to PCB design a custom development board based off of this microcontroller or similar (Cortex-M0/M3), and be able to program it via (micro) USB. Frankly I'd rather write ARM assembly - and likely be able to find good open source stuff to get me started - than have to use a proprietary IDE. Thanks for any help; also open to alternatives based on my goals. <Q> Pre-compiled versions can be found all over the web, or you can build your own. <S> You could do assembler if you want, but I would suggest at least C, personally I like C++ even better because it allows very efficient libraries. <S> The things that 'hang around' the compiler can be a bit trickier. <S> I wrote my own make scripts, linker scripts, startup code, and some support libraries. <S> That requires some in-depth knowledge, but it is not that much work (at least for the first few chips). <S> I mainly use the LPC DIP chips, LPC1114FN28 and LPC810M021FN8. <S> I am not a fan of the 'heavy' IDEs. <S> I use mostly PSPpad, but the make-script can be used with any editor that can call a shell script, catch the output, and parse a (GCC) error message. <S> I am not a fan of debuggers, I prefer to insert print statements. <S> I use lpc21isp for hands-off serial downloading + terminal emulation. <S> Works OK, except that I had to patch lpc21isp it to reset the chip after downloading (instead of using the ISP GO command, which is broken on Cortex. <S> Blame on you NXP for not fixing this!). <S> An article about how I use C++ can be found here . <S> In about two weeks I'll have my environment up-to-date for my C-on-LPC1114 course. <S> The last-year version can be found here . <A> Yes, you have many options. <S> I've been programming Cortex-M0 and Cortex-M3 processors from NXP and ST using open-source tools for years. <S> ARM maintains a version of gcc that cross-compiles at launchpad.com . <S> You can use Eclipse as an IDE. <S> To do it right <S> you need a Single-Wire Debugging interface. <S> I use the Segger JLink but there are other options for that as well. <S> Some time ago I wrote a paper about putting all of this together for teaching a microcontroller course. <S> There is also a great deal of useful information at <S> Yet Another GNU ARM Toolchain . <A> Most, if not all IDEs are presentation devices for editors, makefiles, compilers, linkers and debuggers that conveniently build-in but hide the command-line details of operation <S> so you don't have to think about those. <S> They usually expose the details for you to modify or replace, so you can almost use your choice of IDE with any toolchain you have access to, such as the gcc set Wouter referred to. <A> LPCXpresso is an Eclipse-based front end for a build of the arm-non-eabi-gcc toolchain, plus some other glue, libraries, demos, etc. <S> A free license with moderate code limit is available by registration. <S> But NXP does not own Eclipse or GCC apart from their own modifications, and for GCC the license even compels them to provide you source code for the exact version they use. <S> Nor can they place legal usage restrictions on GCC (ie, they could add a code size restriction in their customized source, but you can also comment that back out - though the restriction may well not be in the compiler at all) <S> Using LPCXpresso appears to be entirely optional - the included GCC build is usable from the command line, and the IDE even appears to generate makefiles which can be used to rebuild your project without running the IDE again, provided you specify the location of the compiler. <S> You should also be able to learn enough from these to create projects from scratch without running the IDE at all. <S> It's probable that upstream arm-gcc would be able to target the chips; if not, a determined person could figure out whatever is changed in NXP's GCC sources and upstream that.
| The free GCC toolchain supports ARM/Cortex just fine.
|
How to create a 13.56MHz PCB antenna in Eagle? How can I create a 13.56MHz PCB antenna in eagle since I can't find any suitable antenna library in Eagle, even failed to download 13.56MHz antenna .lbr in the internet. Kindly help <Q> SparkFun have this RFID Evaluation Shield that works on 13.56MHz. <S> They provide the Eagle files which contain the antenna portion. <S> You could either use this for ideas or take the antenna portion from the .brd file. <S> Copying this sort of antenna is not a simple case of copy and paste. <S> If you look at the schematic you will see this section with C2-C7: <S> This relates to the following portion of the board: <S> These capacitors and two resistors (used as jumpers) tune the PCB antenna to work well on the chosen frequency. <S> If you choose to copy the antenna then make sure it's as close to the original as possible. <S> This means trace width, spacing, the feed line from your driver etc, even the PCB manufacturing process might make a difference. <S> If you keep it virtually identical to the original then you might get away with the capacitor values used by Sparkfun. <S> You might like to check their licensing model if your application is commercial. <S> Images in this post from Sparkfun, licensed CC BY-NC-SA 3.0 . <A> You are not thinking about this correctly. <S> The antenna itself is not 13.56MHz - <S> it's just an inductor made from copper on the circuit board that requires tuning with a capacitor. <S> When tuned with exactly the right capacitor, the inductance of the copper track/loop and the capacitance of the capacitor form a tuned circuit corresponding to the following formula: - \$F_C = \dfrac{1}{2\pi\sqrt{LC}}\$ <S> Given the above formula there are a wide set of possibilities to choose from so, roughly copy what @David is saying and then fiddle with the tuning capacitors to get peak tuning and optimum Q (quality factor for the tuning). <S> Read the data sheet on the device you are interfacing it to because that will give you the info you need to get this right. <A> I have designed a NFC antenna before (assuming you are using NFC due to this particular frequency :P). <S> On its webpage is also a tool for antenna calculations and it's application note. <S> App note: http://www.st.com/st-web-ui/static/active/en/resource/technical/document/application_note/CD00232630.pdf <S> Software: <S> http://www.st.com/web/en/catalog/tools/PF257518 <S> Note that antenna design is quite tricky and your first antenna might not work. <S> Although using this tool, my first antenna design did work right away. <S> You can include a footprint of a capacitor parallel on your antenna so it can be fine tuned later. <S> Hope this helps.
| I experimented with a ST NFC antenna, ANT1-M24LR16E.
|
Consensus on microstrip trace impedance calculations? I'm designing a PCB that has a 2.4GHz signal that I'm routing to an SMA jack. I'm trying to figure out the geometry of the necessary microstrip trace. This is the first rf PCB I've worked on. The problem is that if I feed the values into different online calcuators, I get considerably different values! I don't know which ones to design to. I'm using a 4-layer PCB, 32-mil thick, FR4. The ground plane is directly under the microstrip, with 5.6-mil of prepreg between them. The microstrip thickness is 1.35-mil (1oz copper). I'm assuming a dielectric of 4.2. The trace is running from a bandpass filter to the antenna jack, and is only 135-long. Say that I make a 10-mil wide trace. Here are the results: eeweb.com = 37.1 Ohm. chemandy.com = 52.2 Ohm using one set of formulas , or 46.85 Ohm using IPC-2121 . If I use formulas from this textbook , I get 42.55 Ohm. This is a surprising spread. What is the best practice here? Thanks! <Q> I ran the calculation for you in Cadence SigXplorer (my favorite tool for this and a lot more): <S> This is a 10.25mil wide trace (sorry units on the image are in metric) to give 50R pretty closely. <S> Always use a 2D field solver for this (as you noticed, formulas are not enough). <S> Be aware that the SMA footprint may not be a smooth 50R without some great care as well. <S> For this you can often get help from the connector manufacturer if you send them your stackup (and is deemed a worthy customer :-). <S> Disclaimer: I provide training in signal integrity often using tools kindly provided by Cadence. <S> Other than that, I am not affiliated. <S> Other tools can do the same thing. <S> The only free one I have tried is called TNT. <A> My suggestion is to take information from your PCB manufacturer as primary, and the online calculators as a double check. <S> If you specify controlled impedance, with a tolerance, then you'll normally have to pay something of a premium, depending on the tolerance required. <S> 10% or 5% are common tolerances, though most high-end makers say they can do better (with exponentially increasing cost, I suspect). <S> You'll want to be able to verify that the manufacturer has done what they say. <S> One method is with a test coupon located in an otherwise unused are of the PCB panel (so it is essentially a second design of PCB included on the same panel with your PCBs). <S> Here's an 8-layer design from <S> this application note: <S> And the cross-section from the same source: If the PCB manufacturer does the panelization, they'll probably do this themselves. <S> You can add pads that are not used in production versions, but that will cause reflections. <A> Best practice is to specify controlled impedance, but that costs money. <S> Most of the online calculators aren't so transparent on that front. <S> As a cross-check, and for unconventional geometries, I use MDTLC , which is a 2D field solver. <S> The site shows some comparisons to commercial calculators.
| For my home projects, I use the Wcalc calculator, since it is open source and tells you where it got its formulas from. Of course another method is to directly measure the traces you care about, but that might not be easy if the traces are buried and there are not connectors available to attach to them.
|
replace fan with resistor I have always had a huge fascination with electronics, but never the time to devote to learning it beyond a couple of quick tutorials here and there (and the basics you learn in school). So, sorry if this is too dumb a question for this forum, but here goes anyway. I have a (SATA) harddrive caddy which I use for backups and moving these from one computer to another. It works well but is notorious for using bad (ie, loud and pretty useless) fans. The device is currently connected so I cannot check, but I believe the fan is 5V (I doubt it is 12V, but it is possible). I can disconnect this fan, and the device works fine, but it causes one of the LED indicators at the front to remain on.. which is also pretty annoying (it's the same LED used to signal when the disc is being accessed, so if it is always on, that renders this function useless too). But this got me thinking, "what if I just connect an equivalent resistor there instead of the fan? This might mean I can remove the fan, and the noise, and not have that LED permanently lit up." So, am I thinking correctly here? I have a multimeter, so I can check the resistance of the fan, and find a resistor of the same resistance. Would this work? And, what should I have in mind when trying to do this? Thanks in advance. Edit: A few of the answers have been asking this, so I thought I would just clarify that this is a 2-wire fan . I don't want to replace the fan, I just want to know (also out of curiosity and to learn more about electronics) if this is possible (which, given some of the answers, it looks like it is) and how to go about doing this (safely). In case anyone is curious, this is the item: KingWin SATA Aluminum Mobile Rack with Single Fan, KF-811-BK <Q> If the fault detection depends on current variation with rotation, the resistor isn't going to fool it at all. <S> After all, current will not drop if the fan stalls with dog hair, but air flow will, so simply detecting the current is about useless for detecting fan failure (and is seldom used). <S> Tachometer fans have an extra lead for this purpose, but it can be achieved without extra bits. <S> You could probably make an oscillator circuit to fool it that would be close enough, but it might be easier to search out the controller and disable it or just replace the fan with a quiet one. <S> For an example of a current detecting controller, the Microchip TC64x series uses a current sense resistor on the low side: <S> the rule of thumb is every ten degrees C of heating halves the life of the product. <A> Do not simply measure the resistance of the fan. <S> This will be incorrect, because the motor changes resistance as it rotates. <S> Instead, measure the current draw of the running fan, and use that to calculate the necessary resistor. <A> A 5vdc Fan 40mm has .29FLA(Full <S> load amps)(random fan <S> I looked up) <S> 5v <S> /.29amps=17.24ohms. <S> That gives you the resistance you need. <S> Then you need to do 5v*0.29Amps=1.45Watts (the resistor has to be rated at that wattage). <S> I suggest looking up the fans FLA if your going to fool a current sensor. <S> Look up ohm's law and power dissipation here: http://en.wikipedia.org/wiki/Electrical_resistance_and_conductance <A> I'm not sure what kind of fan you're using. <S> If it's a 3 or 4 wire PC fan, you might have issues due to the tacho line not returning any info to the computer. <S> See this: <S> http://www.pcbheaven.com/wikipages/How_PC_Fans_Work/ <S> Otherwise, if you have a 2 wire one you could replace it with a resistor, but if there's no tacho line, then you don't even need a resistor since no process should care about whether or not it's sucking power.
| I would suggest replacing the fan..
|
How do I drive a MOSFET working under high voltage while keeping it insulated from the rest of the circuit? I want to drive a MOSFET with a 50Hz, 50% duty cycle clock pulse. The voltage to be switched is as high as 300V DC. The section inside the dashed lines will stay electrically isolated from the rest of the circuit. I am planning to use an optocoupler for isolation (I'm open for any other suggestion). Best solution I can find is using a voltage divider network (as seen in the second schematics) and control it with the optocoupler. But I don't think that this is an optimal solution, because there are two major problems. I have to use a bulky power resistor for R1. Switching losses. Gate capacitance of a typical MOSFET is about 5nF. Time constant for a 5nF capacitor charging over a 20k\$\Omega\$ resistor is 100\$\mu\$s which is 0.5% of the switching period. The maximum current through RL will be 10A, and the R ds,on resistance of the MOSFET will be 50m\$\Omega\$ at most. So the switching loss will approximately be \$(10A)^2 \times 50m\Omega \times 0.5\% = 25mW\$. I can increase the values of R1 and R2, but this time the switching loss will increase. There is a trade off. Either way there will be a heat source in the circuit, and some energy will be lost. What is the efficient way of driving a MOSFET like this? EDIT: What about this circuit? With enough dead-time between on states of the two optocouplers, the required rated power of the R1 resistor may stay below 1W. Do you see any problem in this new circuit model? EDIT2: I decided to put external darlington transistors for supplying more current to the MOSFET gate. The gate voltage will drop down to 1.4V at low state, at which most MOSFETs are completely turned off. Is this new circuit feasible? <Q> Further down I'm suggesting something that should work in a lot of applications <S> but maybe not yours (due the the high gate capacitance of the MOSFET). <S> Because of that I'd consider using a DC-DC isolation module to provide isolated power to gate circuits at the MOSFET. <S> Here's a circuit that is 90% indicative: - B2 would be supplied by the DC-DC converter and U1 can be isolated as well although in this particular diagram they show both sides of the coupler grounded (directly below U1). <S> DC converter like this would be OK: - It can continuously supply 250mA and this can be used in conjunction with a reasonably sized electrolytic capacitor to provide the high current surge needed to charge the gate. <S> I'd also consider the simplicity of using a photovoltaic opto isolator such as the Vishay <S> VOM1271 . <S> It can switch on in 53 us into a 200pF load and produce a drive voltage of about 8V making it suitable for a lot of MOSFETs. <S> Of course if the MOSFET gate capacitance is 2nF then it will take about 0.5 milli seconds to turn on. <A> In your diagram you show a optocoupler-triac, which is not appropriate here (usefull with AC only). <S> Note that whatever you have in your optocoupler, the secondary must be able to handle the full 300V! <S> For your charging calculation, you must assume that the 5nF is charged from a (1/21)*300 = 14V source with a 1//21 ~= 1k resistance. <S> You seem to assume a 20k resistance. <S> Note that you have to take the discharge into account too! <S> You don't mention the frequency you want to achieve, which might be an important factor in the design. <A> Have you considered using a transformer? <S> At high freqeuncies, you could use a small pulse transformer, but at 50 Hz, you'd want a good 1:1 transformer designed for audio frequencies. <S> Connect the scondary between the source and gate of the MOSFET. <S> Drive the primary with the signal you want to apply to the gate. <S> Also connect a diode from source to gate to limit the negative swing on the gate.
| A 'discrete' solution could be to create a 14V supply at in the HV area, with a big capacitor, and use a conventional optocoupler and gate driver.
|
What do we mean by "ground" in an electrical circuit? What do we mean by ground in an electrical circuit? When we want to solve electrical circuits problems, we set a "ground" level which is mean that the potential on it equal to $0$. I wonder how the current will flow if the potential at a point equal to zero. <Q> Ground is an arbitrary definition for the (typically) lower side of a voltage differential. <S> It's normally considered 0V, but since a voltage source is simply a difference in voltage, it's in relation to the other voltage potential. <S> For example, two 9V batteries in series could be 18V to 0V, or can be 9V to 0V to -9V. <S> It's all relative to what you decide <S> is the reference voltage, aka ground. <S> Current flows from one voltage potential to another. <S> If there is no difference of voltage between two points, current will not flow. <A> The electricity flows from. <S> High potential to low potential. <S> So it can flow. <S> Also, ground usually connect to the - of the battery. <S> So that usually means that the ground, connect to the - of the battery. <A> I agree with Passerby, but Passerby's answer describes the term neutral. <S> There is one more use for the term ground, which differs from the term neutral -
| The ground should not change it's potential when a a (large) current is passed to it, the potential of the ground must be stable.
|
How to drive LED when sinking current to MCU? Hopefully a simple one. Should I always control by writing to the GPIO digital output? Or should I always leave it low and then tri-state the output? LED turns on when MCU pin sinks current: In code, I mean the difference between the following for the given schematic: // Control entirely via GPIO TRISBbits.TRISB5 = 0; // configure as digital outputLATBbits.LATB5 = 0; // turn on the LEDLATBbits.LATB5 = 1; // turn off the LED// Tri-state output to always just enable/disableLATBbits.LATB5 = 0; // drive output lowTRISBbits.TRISB5 = 0; // turn on by configuring as digital outputTRISBbits.TRISB5 = 1; // turn off by configuring as digital input <Q> In the case you've shown, it doesn't really matter too much. <S> I'd leave the output as an output and control the LED from the LAT bit. <S> If the input is not a Schmitt Trigger input, you could get unnecessary power consumption from having an input floating around, and having inputs floating around is just not good practice in any case. <S> Where it could matter more is if you're using an older PIC that does not have a separate LAT register (or are directly writing to the port register on a PIC that has a LAT register, in which case, don't do that). <S> In such a case, you could run into a read-modify-write issue where an instruction to modify a different (unrelated) bit in the same port register could flip the LED port bit into a high state, so the next time you set the TRIS bit low the LED does not turn on. <S> It might not happen reliably one way or the other since the pin is floating around. <A> Effectively, the tri-state solution is giving you the options of "small resistor(on)/big resistor(off)". <S> But it's very un-intuitive to someone else reading your code (and perhaps even to you, after a couple weeks away from it). <S> I would only suggest using the tri-state method <S> if it's the only way, i.e. your battery voltage is higher than the µC's VCC and taking the µC <S> pin high only dims the LED. <S> Using the GPIO as an output always makes it more clear what the code is doing. <A> (Because of the LED acting as a diode, this would be very small anyway.) <S> However it is a good practice to use, in those cases where there is no diode, and you are essentially bucking the supply rail via perhaps resistor and <S> whatever load you are driving. <S> Making the output pin OD is similar to switching back and forth between an input and output pin, but is more intuitive. <S> // <S> Control entirely via GPIO TRISBbits. <S> TRISB5 = 0; / configure as digital outputODCBbits. <S> ODCB5 = 1; // ADDED to make pin open-drain . . <S> .LATBbits. <S> LATB5 = 0 <S> ; // turn on the LEDLATBbits. <S> LATB5 = 1; // turn off the LED <A> I agree with others that I'd just switch the output high or low to control the LED because this documents the intent better and avoids the floating input problem. <S> However, there is a case where you need to switch between low and high-Z as apposed to between low and high. <S> Some microcontrollers have special open drain outputs that can withstand higher voltages than Vdd (like RA4 on some older PIC 16). <S> If the LED is being driven from a higher voltage than Vdd but within what the open drain output can handle, then the output must be switched between active low and high impedance. <S> As Wouter pointed out in a comment, this may still allow for the same firmware. <S> Old PIC 16s <S> with open drain RA4 can't ever drive the output actively high, so controlling the PORT bit still works. <S> Keep in mind that this is a unusual case, especially today. <S> Newer micros don't seem to have the high voltage open drain outputs anymore.
| I agree with those that suggest making the pin low to turn on the LED, and setting it high to turn it off, but I think you should also configure the output as open drain so you are not sourcing any current.
|
Strange amplitude fluctuation in video signal I am reading a video signal from a CCD array. For some reason I get this extreme vertical jitter. I use decoupling and bypass caps everywhere possible. But it does not seem to be the noise issue. Any one has an idea of what could be the reason of such jitter? The yellow clock signal is just for reference. Here are superimposed frames of the video: TRIED: Different illumination intensities - the fluctuations present in the dark too. Tried different light sources: flashlight, diode lamp, luminescent lamp, laser. Tried running the microcontroller with the internal clock (no external oscillator). Tried running the setup from a 12V battery (no mains). Tried using electronic switches DG642 and EL7156 instead of direct connection to the microcontroller. Tried switching/interchanging the probes (as you can see, the TTL signal looks just fine). The oscilloscope and the power supply seem to be grounded since the power sockets have the third 'ground' pin. Tried three different kinds of CCD arrays. Do not seem to have any high power electromagentic devices around me. When I measure a signal from a waveform generator, no fluctuations are present. I use bypass and decoupling capacitors everywhere I should. I am seriously stuck. Please, help!!! Perkin Elmer RL1024PQ sensor Vishay DG642 video switch (pin driver) Intersil EL7156 pin driver <Q> Try syncing your scope on the power mains. <S> If the voltage fluctuation becomes stable, then likely you have interference from an outside electrical source coupling into your circuit. <S> Fluorescent lighting would be a good candidate. <S> Try shutting down or moving away from any high power electric motors or other equipment. <S> Also cover the CCD lens and make sure that you have an electrical problem even with a black scene. <A> This is what normal linear Pixel array signals look like without using HSync on scope to HSync on display chip. <S> Clock speed affects gain from light integration time and thus amplitude sensitivity. <S> Variation between pixels shows as your jitter between pixel integration values. <S> Since there are black pixels at each end included with fringe pixels, most of the amplitude variation will be at both ends. <S> Since you are displaying all the pixels overlapped, it will appear as amplitude jitter. <S> I might expect the outer 10 pixels to be most affected by aperture effects. <S> Custom devices without black window apertures will improve on this feature but lack the black level. <S> I suppose that is why they have 10 black (blocked) pixels at each end. <S> I have not found any application notes, but the buffered frame rate is 200KHz out, but the imager input is 70KHz max. <S> Frame rate. <S> - What steady light source are you using and what are the dominant wavelengths. <S> There will be more ripple in the blue range for light sensitivity "non-uniformity" . <S> The older products have specs like +- 6% @50% , 635nm, which is red. <S> What frame scan rate? <S> Where is your schematic and test conditions? <S> Are your scope probes calibrated? <S> The TTL has excessive overshoot, as does the video. <S> Use a very short gnd lead. <S> is there any switched capacitance signal injected by the switches? <A> First off, you are not seeing jitter, jitter is a lateral (time) variation in a vertical edge <S> this is a vertical variation (voltage) in a flat part of the curve. <S> So this sensor is designed to be operated as a CDS type signal processing scheme, sample the reset level sometime after Phi_rg is high and before Phi_1 negative edge and then sample again once the floating gate settles (Q_2 dumps charge onto the floating node). <S> This removes the common signal and the KTC noise signal. <S> In general these types of devices should not be driven from a micro-processor. <S> The voltage variation on the waveform (both jitter wise and in amplitude) couple directly into the sense node and inject a signal into the most sensitive part of the whole sensor. <S> When I designed with these devices we would derive signals from an FPGA and then re-clock them externally with SGL (single gate logic) <S> D-FF to remove Jitter and these would be powered from a separate power supply with lots of filtering. <S> Clean up that signal chain and do CDS. <S> You also have to be careful with edge rates on the CCD transport registers and return current flow on the PCB (signal bounce). <S> Vrd (see page 4) also has to be very very clean. <S> i.e. a separate power supply/regulator. <A> From a quick look at your scope shot, it seems this jitter you are referring to is actually your signal. <S> It seems we are looking at the value of individual pixels being clocked out. <S> Since only a few can be shown on the scope at a time, snapshots of many different short runs of pixels are displayed in rapid succession on the scope. <S> The different brightness values from the different pixels therefore show up as different heights of the output waveform.
| It's possible that you are seeing the optical effects of room lighting variations as captured by the CCD. Your power supply rails of your micro-processor are bouncing your signal that then is capacitively coupled to the sense node through the RG transistor of the sensor.
|
Help my off grid solar system is driving me mad Please can someone help me with this problem I'm having with my solar system? Ok I have 7 × 190w 24v panels wired in parallel,Coming down to a 80A MPPT charge controler which automatically selects whether it's a 12/24/36V and so on. From that I have 16 100A 12V battery bank wired in series, then from the batteries a 12V to 240V AC inverter which is 8000W. Now my problem is they are not fully charging 11.8 to sometimes 12.3v it goes to only and my water pump need at least 13.5 for it to come on. Hopefully I haven't confused you guys, if u need to no any more info please ask. I'm not a electrician but I have a bit of an idea. <Q> I suspect that your 16 100 AH batteries are wired in parallel rather than series, otherwise you'd have ~200 volts instead of 12. <S> - I had this problem on my boat. <S> The voltage would not get over 12 volts, even using a high power charger for a long time. <S> I eventually noticed that one battery required lots of water in all but one cell - <S> once I disconnected that battery, the other battery quickly came up to ~14 volts under charge. <S> If you have normal flooded cell batteries, check the water levels,looking for batteries where one cell doesn't need water, but the rest do - that (or those) batteries will be bad, and removing them from the bank should allow the rest to charge normally. <S> If you have "maintenance-free" flooded batteries, you may be able to pry off the cell covers to do this check. <S> With gell-cell or AGM batteries, you'll probably have to disconnect batteries one at a time to find the bad one. <S> With the batteries disconnected, measure the voltage of each one to find the bad one - its voltage should be about 2 volts less than a good battery. <S> It is generally recommended that all batteries in a bank like yours should be the same vintage, and have the same usage history, so you may want to replace all the batteries at once, rather than one at a time as they fail. <S> Many boaters use 6 volt "golf cart" batteries connected as two 6 volt batteries in series, with several pair in parallel as needed for the desired capacity, for deep-cycle applications. <A> I will agree with Peter Bennett on the likelihood that one or more of your batteries may have gone bad. <S> Being an electrician first and an engineer second, I would like to add a few things. <S> It is a terrible idea to pry open a maintenance-free battery. <S> Take it to a shop and get it tested, or buy (or build!) <S> a loadtester and learn how to use it. <S> They're pretty cheap. <S> The voltage on a bad battery - sometimes even a very bad battery - can still fall within ranges that look acceptable until you try to put a load on it. <S> I've had this get me in trouble a few times, personally. <S> See (1). <S> You may want to check your MPPT charger to see if there is a ripple voltage. <S> Its not likely, but you could be burning up batteries with a faint AC signal from the conversion. <S> Just select 'ac volts' on a multimeter and put it across the terminals. <S> Anything more than maybe half a volt needs to be investigated. <S> Do this while the charger is running, with near full sun available - otherwise, you didn't really find out anything. <S> Other than that <S> , I definitely second Peter's recommendation to keep all batteries in the bank at around the same age. <S> Over time, the internal resistance will continually increase, leading to irregular charging and discharging cycles. <S> Over years, you'll lose the capacity anyway, but having one completely die in the bank can cause reverse-charging, excessive gassing, and eventually, an explosion. <S> Also, you are definitely going to want to invest in a different pump. <S> I don't know what the application is, but if it absolutely must have 13.5 volts, that means its marine/automotive, and built assuming you have an alternator running. <S> There are other pumps that would handle this better. <S> Furthermore, if its as small as I suspect it is, you may want to see if you can get an AC one and hook it up to that beefy-sounding inverter. <S> You'll adding the overhead of the inverter to it, but AC machines are capable of being really quite efficient, so it might pay off. <A> A little late to the game here, <S> and I believe that the above answers are the correct ones. <S> I would like to ask you to look at your panel "strength" and your battery capacity and how much energy your pump would use. <S> I don't know where in the world you are so <S> how much sunlight you get or if your panels facing the "ideal" direction, but at 8 hours of FULL sunlight, no clouds, no lines loss etc, I.E under perfect conditions, it looks like it would take almost 2 days to fully charge your bank of batteries! <S> That is without the pump running. <S> If you are not getting that much sunlight <S> or there are a lot of clouds <S> etc <S> it could take a lot longer. <S> And if you are using the batteries each day, then they may never get up to full charge. <S> Just a thought - unless my math is wrong! <S> Good luck
| It is possible that one (or more) of your batteries has developed a shorted cell You should be using "deep cycle" batteries (often called "marine" or "RV") for this application.
|
3D accurate position sensing of a point I want to sense the 3D position of an object with respect to some fixed reference. The position of the point will be within a meter of the reference. The accuracy should be very high, possibly within a cm. The point will be almost static or exhibiting very less movements. Please tell me which sensors can be used to achieve this task. The sensor unit will be connected to a microcontroller unit. It would be better if the entire unit can be accommodated in a single PCB. The output should not change appreciably with temperature. The measurement will be done only once on system startup. Based on the position of the object , some other actuators will be controlled. As it is a one time measurement so power wont be a big constraint. The system is supported by 6Ah battery <Q> According to your current specifications, a possible solution would be to use two small cameras and stereoscopic computer vision to locate the object. <S> It will require a line of sight to the object, knowledge of how the object looks like, an adequate resolution to the size and distance of the object, quite some processing (OpenCV may help) and accurate knowledge of the position of the detector. <A> Your ability to resolve the exact location of said freely chosen object in three dimensional space is going to be directly related to the number of data points you can get on that object, and how fast you can perform the computations. <S> Not that this is impossible, but you may be looking at some space program stuff. <S> Clabacchio is spot on with computer vision as probably the best choice for budget and function. <S> The right process will give you angle of inclination/declination, x and y coordinates and rotation, but that will require fairly advanced pattern recognition, assuming it doesn't have a very simple silhouette and you can't tag the mystery item with a white ball. <S> Another workable alternative would be radio frequency or ultrasonic tagging. <S> The sensor net would have to be distributed around the environment, but this will give you a pretty accurate reading on x, y and z coordinates of the tag in the space, but without two tags, it can't tell you how the object is rotated or if it has been turned sideways. <S> If the object's position must be known or controlled in some way, typical applications would have an operator put the object in a known position, maybe in a tray or locked into a jig, and then move that tray or jig with servomechanisms, whose exact position would be known electrically. <S> Dangerous items such as reactor control rods get attached to big lead screws with encoders on them that give the position with very high accuracy. <S> These are just some rough drafts. <S> To formulate a better answer, we must know more. <S> The only effective solution for 100% universal applications is a trained human operator, and that can get a little spotty. <S> Is it in a enclosed space? <S> Outdoors? <S> Is it hazardous? <S> If the point is fixed, why is the measurement needed? <S> If the work you are doing is very sensitive or confidential, you should immediately retain the service of a professional engineer or engineering group, to whom which you can more fully communicate your needs. <A> Magnetics is the best - see Flock of Birds by Ascension Technology <S> It is a fixed cube about the size of an AppleTV anyplace in your environment, and a sensor about 1/4 the size of a cigarette on the object to be tracked, and you get highly accurate 6DoF postion (X, Y, Z, Pitch, Roll, Yaw).
| Depending on the complexity of the object, a microcontroller may not be sufficient, and a digital signal processor or microprocessor will likely be necessary.
|
AVRDUDE with buring settings of AVR Studio I make my own electronic boards often with an Atmel ATmega328P. I used AVR Studio for programming. Now I tried to make the same settings with avrdude and it looks, that they are not really compatibale. Is there a way to set the programming speed in avrdude? I want to program my fuses with 2.1592kHz and program my firmware with 500kHz. Does anyone know a command? It seems that AVR Studio decodes the values for the fuses different than avrdude. Especially the efuse is different. I think these two are the same: AVR Studio: efuse:0xFD, hfuse:0xDA, lfuse:0xFF avrdude: efuse:w:0x05:m hfuse:w:0xDA:m lfuse:w:0xFF:m Thank you for your hints. Felix <Q> Is there a way to set the programming speed in avrdude? <S> I want to program my fuses with 2.1592kHz and program my firmware with 500kHz. <S> Does anyone know a command? <S> Use parameter <S> -B <bitrate> <S> : Specify the bit clock period for the JTAG interface or the ISP clock (JTAG ICE only). <S> The value is a floating-point number in microseconds. <S> The default value of the JTAG ICE results in about 1 microsecond bit clock period, suitable for target MCUs running at 4 MHz clock and above. <S> Unlike certain parameters in the STK500, the JTAG ICE resets all its parameters to default values when the programming software signs off from the ICE, so for MCUs running at lower clock speeds, this parameter must be specified on the command-line. <S> It can also be set in the configuration file by using the ’default_bitclock’ keyword. <S> Especially the efuse is different. <S> Normally, the raw fuse byte values should be identical as those are just read out values and are not interpreted in any way. <S> See the answer from microtherion for a possible explanation why the values are different in this case. <A> To answer your second question, for an ATmega328p, only the 3 LSBs of efuse matter, so the two values you show are identical, as far as the MCU is concerned. <S> It's possible that avrdude and AVR Studio treat the "don't care" bits differently when reading the value back for verification. <A> Did you know that Atmel Studio also comes with a command line tool named atprogram? <S> Actually works pretty nicely once you figure out where to put each value :-)
| It seems that AVR Studio decodes the values for the fuses different than avrdude.
|
What's the difference between a FP, gain guided and DFB laser diode? What is the difference between a Fabry-Perot (FP), a gain-guided and a Distributed Feedback (DFB) laser diode? I think Fabry-Perot is the default configuration, gain-guided has something to do with a strip electrode (but I'm not clear on how it works) and DFB uses a Bragg-grating to limit the output of the laser to a single wavelength. However, I'm still a little blurry on the details. Help clearing up the differences would be appreciated. <Q> Fabry-Perot laser diodes are lasers whose mirrors are simply the flat cleaved surfaces at the ends of the laser chip. <S> Distributed feedback (DFB) laser diodes are lasers that have a grating structure in the cavity that produces multiple reflections throughout the cavity. <S> This leads to narrower linewidths than are produced by FP lasers. <S> (Image source: Laser Focus World ) <S> Another type you didn't ask about is the Distributed Bragg reflector (DBR) laser. <S> A DBR typically has two separate grating regions, on either side of the gain region. <S> They're usually distinguished from the DFB because the grating doesn't overlap the gain region, although there is some grey area and the terms are not always used consistently. <S> (Image source: <S> US Patent #6638773 ) <S> In a gain-guided laser there is no waveguide patterned between the reflective structures (end facets or DBR regions). <S> Instead, we rely on a narrow region where current is injected to maintain the transverse mode. <S> This works mainly because any transvese modes that don't overlap the injection region will see much lower gain and so won't lase. <A> The Fabry-Perot laser is the default configuration for a laser (the usual population inversion, stimulated emission and positive feedback), except of mirrors at either end for creating positive feedback are replaces with cleaved surfaces. <S> The gain-guided laser diode, as described here , uses a strip to direct current to the center of the active region to reduce the chance of edge emissions. <S> This is usually combined with index-guiding, which is bordering the active region of a laser diode with materials with a high refractive index, this also reduces edge emissions. <S> The DFB as you said, has a Bragg-grating in the active region to limit the emission to a single wavelength, since a laser actually emits multiple wavelengths called longitudinal modes. <A> A Fabry-Perot laser is a resonator that lases at wavelength selected by the gain bandwidth of the lasing medium. <S> These are called longitudinal modes. <S> It will usually lase with multiple longitudinal modes that fall within the gain bandwidth. <S> The cavity modes are determined by the optical dimensions of the resonator. <S> Depending on the dimensions of the cavity, the laser may also oscillate on multiple spatial modes. <S> To select a single spatial mode, a laser diode is designed with a single mode waveguide, to confine the spatial mode. <S> To select a single longitudinal mode, the laser diode is fabricated with a Bragg grating. <S> A distributed feedback laser, or DFB, is designed with the grating distributed along the entire length of the resonator. <S> A distributed Bragg reflector, or DBR laser is designed with a grating outside of the gain region. <S> DBRs generally have lower linewidth and higher power than DFBs and are used for metrology and spectroscopy. <S> DFBs generally have a wider tuning range and are used in telecom applications. <S> A more detailed discussion about the differences between DFBs and DBRs can be found here ( http://photodigm.com/difference-between-dbr-and-dfb-lasers/ ).
| Where FP, DFB, and DBR types describe how the longitudinal reflections that provide laser feedback are produced, the designation gain-guided describes how the mode is confined in the transverse dimensions.
|
Distortion sine waveform from LT1227 amplifier Schematic for the amplifier: Before connect it to amplifier circuit: Why the sine wave will distorted? any possible reason? is this due to noise?blue is the input signal of an amplifier and green is the output signal <Q> The circuit is oscillating. <S> The reason is as follows: <S> The opamp is a current-feedback amplifier which needs a feedback resistor that must exceed a certain limit (specified by the manufacturer). <S> You should consult the data sheet to see the lower limit of this feedback resistor - typically, some hundreds of ohms. <S> (By the way: Don´t worry about capacitors in parallel to the power supply; they are useless in simulation as long as the dc source is ideal). <A> Check this out from the datasheet : <S> If your feedback resistor is much less than about 1K with +/-15V <S> supplies, it's going to oscillate like a banshee with only a tiny amount of load capacitance (such as a scope probe). <A> On a quick look, I see one problem is that you have relatively large power supply decoupling capacitors. <S> These are likely to have a large-ish internal resistance/inductance. <S> This might permit the op-amp to oscillate. <S> Try adding 0.1uF caps from V+ and V- to ground, near to the op amp. <S> Almost certainly, the center of the problem is that the op amp is oscillating at some high frequency, and that oscillation is added to your intended signal. <S> That's why my suggestion to add fast-responding bypass capacitors. <S> Or it could be caused by some capacitance of the load (possibly the scope probe) causing the feedback loop to have some delay (phase shift), resulting in the feedback being positive for some frequencies.
| This could be caused by power supply connections whose voltage varies when the op amp tries to draw rapidly changing current.
|
How to control three LEDs with two uC pins using just passive components? I only have 2 (3-state) pins available on a microcontroller and need to control 3 red LEDs, uC is running on 5V. I can only use passive components in addition. <Q> To control four LEDs: simulate this circuit – Schematic created using CircuitLab <S> To get both LEDs on a particular pin lit, toggle the pin at a few hundred Hz. <S> Note that this setup requires that the forward voltage of the LEDs be less than Vcc/2. <S> Note also that the resistors consume power all of the time, not just when the LEDs are on. <A> This solution depends on the fact that 5V won't light the three LEDs in series. <S> If necessary, you can add a silicon diode in series with one or more of the LEDs in order to increase the total forward voltage drop. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> If both pin A and pin B are tristate, all LEDs are off. <S> If pin A is driven low, D1 is on. <S> If pin A is driven high and pin B is driven low D2 is on. <S> if pin B is driven high, D3 is on. <A> With two pins you can actually control four LEDs. <S> The free anode end of the first LED gets connected to the VCC of the MCU board through another resistor. <S> The free cathode end of the second LED gets connected to GND (through another resistor). <S> Now the LEDs will light alternately when the MCU pin is set high or low. <S> Wire up the second MCU pin the same way <S> and you now have control over a total of four LEDs. <S> To give the appearance of independent control of the each LED in the pair off the one of the port pins requires a little software work as follow. <S> If LED1 is the one with the anode to the port pin and LED2 is the one with the cathode to the port pin then do the following to establish the four states for two LEDs. <S> LED2 LED1 <S> ActionOFF <S> OFF <S> Set the port pin to tristate level. <S> OFF <S> ON <S> Set the port pin low. <S> ON OFF <S> Set the port <S> pin high. <S> ON ON Toggle the port pin high and low at a frequency over about 120 Hz. <S> Repeat the same actions for the other port pin <S> and you will be looking like you have four independent LEDs off two port pins. <S> This scheme works well for LEDs that have a forward voltage drop that is over half the VCC level. <S> Red LEDs with a 2.1 VF will not work so great if the VCC is 5V for example. <S> On a otherhand a green LED with a 2.5V VF will work great on a system with a VCC of 3.3V. <A> So like this no current flows when they are off, and you can drive as much/little forward voltage as you need to <A> EDN published some related Ideas for Design here <S> Here's one of them:- For typical values with D1 a yellow LED (2.2V on), D2 a red LED (1.9V on), and off voltages of 1.2V and 1.1V respectively, and on-currents of 8mA each, <S> Vcc = 5.0V, the optimum values are R1 = <S> 300 <S> ohmsR2 = 330 <S> ohmsR3 = 1.2K ohms <S> Quiescent current is 2.7mA. <S> To have both LEDs appear to be illuminated, toggle the output pin at 100Hz or greater. <S> I used the Excel solver in the original article, the code may be still available from EDN. <S> The additional degree of freedom afforded by the resistor R3 can avoid the limitations of both Dave Tweed and Michael Karas' circuits, though for the specific case of 2 red LEDs operating from a 5V supply, Dave Tweed's circuit is probably acceptable, but check the Vf carefully, <S> it's not fine for some red LEDs, and may be marginal for others if the port pin doesn't pull all the way down or up.
| To have more than one LED lit at a time, you'll have to multiplex: turn them on one at a time, rapidly enough so that they appear to be continuously lit. To control two LEDs put a resistor from the MCU pin to limit the current and tie it to two LEDs - the cathode of one LED and the anode of the other LED.
|
Replacing adapter with batteries? I have some device with ac/dc adapter [9v, 500mA], and I want to replace the adapter with batteries. If I used [6 * 1.5v] battries, is that acceptable? do I have to worry about the max current of the batteries? could the current of the batteries exceeds 500mA? whenever I try to measure the current of the device connected to the battries, it gives me "1" "the upperlimit of the Multimeter", why? how? And if so, what should I do? HOW CAN I MEASURE THE MAX CURRENT OF THE BATTERIES OR SOME POWERSUPPLY??? <Q> If you wire your batteries up in series to get the voltage you require, the current will be the same as from a single battery, if you do need to change this you need to wire another set of 6 in parallel. <S> That having been said, a very quick search showed that most AA batteries are rated to give a max of about 2400mA, which is ample. <S> HOWEVER 500mA is a fairly large current draw assuming that the device actually requires this much, (which it won't) <S> this means you will run the batteries down quickly, although you do have 6 of them... <S> To use your multi meter to measure current you must place it in series with the supply, so, positive end of supply - positive multi meter probe, (goes through multimeter) <S> - negative end of probe to positive input of your device - negative end of device to negative end of supply (as normal). <S> MAKE SURE YOU HAVE THE MULTI METER IN CURRENT MODE!! <A> As @Tim Mottram said: Amp meter has to be connected in series. <S> But: DON'T SHORT YOUR POWER SUPPLY WITH YOUR AMP METER! <S> A multimeter in amp meter mode is a device that measure a voltage accross a very small resistor that is connected internally between the two probes of the meter. <S> If you connect thoses probes to the power supply you will actuall load your power supply with a very small resistor. <S> Because I = U / R, with R very small, I will be very big. <S> For sure it will be big enough to blow the fuse included into you multimeter. <S> (or blow your multimeter if it's a very very low cost model). <S> A single cell battery would have an internal resistance that is big enough to limit that current and save your meter. <S> But a power supply would for sure have enough power to blow your meter. <A> You don't have to worry about providing too much current. <S> The current is determined by the voltage and the load. <S> To need to make sure that you can provide enough current. <S> An AA battery (alkaline) can provide at least one amp, so your 500mA is within spec. <S> However, AA's have around a 2.5 Amp-hour capacity, so you won't get more than 5 hours (or so) of run time. <S> Of course, the circuit likely doesn't take the entire 500mA, so it'll probably last longer than that. <S> Good luck! <A> Ohms Law, Ohms Law, Ohms Law. <S> If the voltage is 9V and the load is 30 ohms (for example), <S> the current taken by the load is:- <S> I = <S> \$\dfrac{V}{R}\$ <S> = <S> 9/30 <S> = 300mA. <S> If your power supply (or battery pack) is capable of supplying 1000 amps it still has to obey ohms law and that means I = <S> V/R <S> and if the load is 30 ohms it'll take 300mA from a 9V power source. <S> You don't need to measure this to know this. <S> HOW CAN I MEASURE THE MAX CURRENT OF THE BATTERIES OR <S> SOME POWER SUPPLY?? <S> ? <S> Put your meter back in its case and make sure the leads are plugged back into measuring voltage because you might forget next time you use it across the AC in your home <S> and then there will be smoke and <S> it'll be your meter burning possibly. <S> If you want to know how much current a battery can produce, look at its data sheet and don't try measuring it.
| Don't connect your meter (measuring current) across your battery because you'll measure the full output current from the battery and this might damage your meter or burn small wires. You current charger is rated at 500mA and is adequate - this means your load takes 500mA or less (maybe 300mA as per my math above). In your case, it should work out fine.
|
Calculating time spent by a function in Microcontroller Applications I would like to write a simple profiler for my STM32F100VE microcontroller. I have seen that I can include time.h and based on the c standard functions I can use it, but I was confused about _CLOCKS_PER_SEC_ 1000. I'm not sure if that approach will work or not. How would one try measure the time a function took in micro-controller application? <Q> The most basic maneuver for complicated functions is to utilize the onboard timers. <S> Depending on how you set them up, you can get them to count exactly the number of clock cycles it takes for a certain function to execute. <S> Simply read the timer before the function call and then again after the function call. <S> You will have to look through the assembly code to know how much time to take off the timer for the read calls <S> but if it is a lengthy function or if you don't mind being by a couple of clock cycles then it is ok. <S> The only error in the conversion from clock cycles back to time is the possible error in the system clock's oscillating frequency. <S> Again, this will be negligible for functions which take significant time to execute. <S> It sounds like in your case <S> you want something around 30Hz (30 FPS) which is a very long time compared time cycle of the processor clock. <A> <A> Do you need to use that information in your software as well, or do you only need it for debugging/development purposes? <S> If it's the second one (I assume it is, as your main goal is profiling), just toggling an output at the beginning and the end of the function, and measuring the time with an oscilloscope might be the quickest, easiest and most accurate measurement.
| If it is a small function you could just look at the disassembly and count the instructions.
|
Safety of this circuit (involving hand-to-hand current): Note1: The voltage source in diagram is 3 lantern cells in series.Note2: Hands are placed on acrylic/plastic mesh so not directly in contact with metal trays. The hands would be submerged for 20-30 minutes at a time and process would be repeated (say, every day). Given the hand-to-hand contact and submerged hands, I'd like to know what level of current would be expected to flow (in particular through the heart) and if there's any (non-negligible) risk associated with this circuit in both the short- and long-term. I've looked at previous, more general threads about safe levels of current/voltage ( How much voltage is "dangerous"? , Safe current limit for human contact? ) and realise that there are too many variables to give exact answers to those but I'm hoping the specifics of this setup will allow for a more definitive answer! <Q> So the idea is to deliberately plug sweat glands with minerals in tap water? <S> And polarity doesn't matter at all? <S> I see at least one commercial unit has adjustable range 0.1-4mA which certainly has the potential to have some kind of effect on human tissue. <S> It's well above the lethal current if injected directly into the heart, BTW, so I don't know if anyone can guarantee complete safety. <S> The other metric in that Powerpoint is the current density of 0.1-0.5 mA/cm2. <S> I don't know how many cm2 my hand is, but if I guesstimate 500 (that estimate is in line with the 2.5% of body surface area (BSA) from this paper , and the Wikipedia estimate of 1.9m^2 for the average male BSA), that would be 50-250mA, which is well within the range that can cause venticular fibrillation . <S> I suppose that your aim is to increase the exposure time and use a lower current density. <S> If there is an area not requiring "treatment", then exposing it to the water is just putting unnecessary current through the body of the subject. <S> It may well be perfectly safe all or most of the time, but it's pretty hard to predict. <S> Wouter/JoeH's idea of two series (redundant) current limits, keeping the overall limit to less than a few mA (and the voltage to less than 20V and definitely battery-powered with no mains connection at all) may be the best you can do, but I must repeat, I don't think anyone can guarantee safety of this circuit- <S> some individuals may have more sensitivity, less internal resistance or something like a pacemaker that causes them to react differently. <S> It's possible that the high price of the commercial units is at least partially related to the cost of insurance against such claims. <S> As far as I understand it, this device exceeds the 1975 AHA recommended limit for long term DC patient leakage (eg. <S> for external ECG patient electrodes) by a factor of 400:1 . <S> AFAIUI, the IEC-60601-1 limit is also 10uA for long term DC current. <S> There may be some justification for an exception based on the really large contact area, but.. <S> Apparently long-term DC current can cause necrosis or ulceration. <A> I would suggest consulting IEC 60479-1:2005, Effects of current on human beings and livestock, Part 1: General Aspects. <S> This is the standard that informs other electrical standards where safe touch and step voltages are concerned , i.e. for design of substation earthing systems. <S> Here is a snippet from the table of contents: <S> Note, hand-to-hand current is distinguished between other currents (i.e. hand to foot, chest to hand, foot to foot) by a Heart-current Factor F - see Table 12. <A> We're hearing a lot about IEC and NFPA99 codes and such, but those don't necessarily apply. <S> It is CURRENT DENSITY, not CURRENT, that causes problems -- that is, current travelling through a small cross section near the heart can cause fibrillation. <S> In the hospital setting, codes recognize that the sickest patients can have catheters with conductive fluid right in the heart, so requirements are strict. <S> If you look at the graphs, by the way, DC current is much safer than 60Hz AC current from the fibrillation point of view. <S> If the suggested system can be held down to about 4 mA, which is less than let-go current (i.e., the current at which your muscles will contract, making it impossible to "let go" of the source of the shock), this should be relatively safe, as the current will cross the entirety of the chest. <S> AVOID THIS <S> IF YOU HAVE A PACEMAKER, OR <S> ANY IMPLANTED ELECTRICAL DEVICE!! <S> The safer way to do this is to put both hands in the same bath, and use a large surface grounding electrode (like those used on TENS units -- http://www.tenspros.com/Electrodes_c_11.html?gclid=CPKXp_T4370CFUVp7AodEi4A1g ) on both forearms tied to the other battery lead (still with a current limiting circuit of some kind). <S> Consider LIGHTLY abrading the skin under the ground pad with a plastic scrunchy, but this would make a current limiting circuit even more important (A resistor to limit to 4 mA). <S> Perhaps one hand at a time might be better. <S> I think the issue may be more important with implanted electrodes, with smaller surface areas and more electrochemistry going on. <S> Use the largest ground pads you can find. <S> If it hurts, STOP!
| As has been pointed out, watch out for burns, or lesions that might be caused by long periods of DC current. Your circuit depends on some unknown and variable resistances to limit the current.
|
Securing a pcb to be used in a keyboard to its case I'm looking for a way to secure a pcb I will be making for a cherry mx (homemade) keyboard to a case which I'll also be making. Obviously, I don't want the pcb to move. I could use standoffs/screws, but wouldn't the added height between the pcb and case cause the pcb to flex when keys are pressed? Glue/tape seems mega sketch and may mess with conductivity. How is this done on normal (production) keyboards? Advice is much appreciated. <Q> Obviously, you need to use enough standoffs to do the job! <S> Make sure that components and their connections won't interfere with where the stiffener needs to touch the PCB. <A> I would put mounting holes and/or supports in the back (for the PCB to press against) on a square grid with 80mm centers, or triangular grid with 100mm centers. <S> Obviously, the numbers are approximate and the grid doesn't have to be dead-on square. <S> P.S. <S> If you post some mechanical drawings, I could critique your design in more detail. <A> In a previous company, we used VHB 4905 tape by 3M. <S> You can use it to secure a PCB to an enclosure, and then use it to seal the enclosure, too :) <S> It has a body thickness of 20mil, so it's pretty compliant. <S> It's <S> non-conductive (10^16 Ohms/square), and has a breakdown voltage of 630V per mil. <S> We never saw any issues with using it to stick PCB's to aluminum housings. <S> If your PCB has RF components, the dielectric constant of any material may detune your rf circuitry... <S> The datasheet for the entire line of VHB tapes is here . <S> Page 7 shows the electrical characteristics. <S> It's not cheap, but you can get a short roll on Amazon for about $17. <S> As an aside, once I took some home and taped a pewter key hook to my wall. <S> A year layer, when I tried to remove it, the paint and sheetrock tore away before the VHB! <S> Nothing like a little quick spackling as you leave :) <S> Good luck :)
| In addition, if you can't put standoffs every place you think you need them, you might need to use one or more "stiffeners" (usually a metal bar with an L-shaped cross-section) that spans between pairs of standoffs and provides extra mechanical support to the PCB.
|
Why isn't there a potential difference across a disconnected diode? I know this question sounds silly, as if there was a potential difference a current would be created when the terminals are connected together and this would mean energy has come from somewhere. The reason I ask this though is that from my understanding of the depletion region and built in potential of a diode it seems like if you connected a voltmeter across the whole diode it would show the value of the built in potential. This is explained in the image below: At first, electrons flow from the n type to the p type because there are a higher concentration in the n type, and holes do vise versa. This is called the diffusion current. The first electrons and holes to cross the pn boundary are the ones which are closest to it; these carriers recombine when they meet each other and are then no longer a carrier. This means there is a depletion region of no carriers near the pn boundary. because electrons have left the n type material, and holes have left the p type material, there is a surplus of positive and negative charge on the n and p side of the pn boundary respectively. This causes an electric field that opposes the diffusion current, and so no more electrons or holes cross the boundary and combine. In short, only the electrons and holes near the boundary combine, because after they have done that an electric field is formed that prevents any more carriers from crossing. The current due to this electric field is called drift current, and when in equilibrium this will equal the diffusion current. Because there is an electric field at the boundary (pointing from the positive charge to the negative charge) there is an associated voltage. This is called the built in potential. If you sample the electric field at each point along the diode from left to right, you would start with 0 in the p region because there are an equal number of protons and electrons. As you approach the depletion region you would see a small electric field pointing back towards the p region, caused by acceptor impurities which now have an extra electron (due to recombination) and therefore now have a net negative charge. This electric field would increase in strength as you get closer to the boundary, and then die away as you get further away. This electric field means there is a voltage, as shown in graph (d). The p side is at an arbitrary potential, and the n side is at a potential higher than this because there is an electric field between them. This means there is a potential difference across the depletion region; this is known as the built-in potential. But why when I connect a volt-meter across the whole diode will I not see this built in potential? <Q> I think, the answer is relatively simple. <S> Do you know the working principle of a "Schottky diode", which is based on a semiconductor-metal junction?Now - what happens if you connect a voltmeter (or any other load) across the diode? <S> You create two Schottky junctions which exactly compensate the diffusion voltage inside the pn diode. <S> Thus, no voltage can be measured. <S> With other words: You cannot use the diffusion voltage to drive any current through an external load. <A> Err, the rest of the answers seem a little dodgy <S> and I just stumbled upon this question so I'll take a shot at it. <S> I think it's because of the fact that the Fermi level becomes discontinuous under bias. <S> I'm sure you can visualise that what the voltmeter is really measuring is how badly electrons and holes want to cross the junction. <S> At thermal equilibrium, the electrons and holes have no intention of moving across the junction, so the voltage is 0V. <S> IN other words, the voltmeter really only measures the difference in the Fermi levels between the 2 sides. <S> To understand why it does this, you have to know how a voltmeter works. <S> Rather than literally measuring the difference in the energy level of an electron at both ends of the diode (which would be awesome), it just measures the current flowing through its high resistance. <S> In a diode at thermal equilibrium, there's no net movement of any charge carriers and so there's no current. <S> No current means no voltmeter reading. <A> It is a very nice curiosity question! <S> Same question came up to me when I was in my second year. <S> But until I came across the Threshold voltages in Transistors and PN junction voltage drops, the picture became little clear. <S> That is why, to allow the current to flow through diode (PN junction) you would need higher potential from P-type and n-Type such that their difference is larger than the intrinsic potential difference which is in opposite direction to applied voltage across diode. <S> This is what we call forward biased diode! <S> I am sure you know this basics. <S> Now lets go to the real question - <S> > <S> If you were to probe your virtual Digital voltmeter exactly at the two depletion boundaries <S> then I am sure you would see the voltage difference there, but its quite impossible to do with the regular multimeter. <S> I am sure there are ways that semiconductor companies have special probes to sense these voltage differences. <S> But if you were to measure the disconnected diode from your regular multimeter (same this is taken in consideration when you simulate it in LTSPICE that the probing is done at the ends of the diode not internally). <S> Basically, your Graph (D) has this answer it self. <S> Graph shows that both ends of diode have no electric field present. <S> since the Electric field is conservative, and two diode ends (ends of P and N type materials) have no charge and electric fields at the ends are cancelled due to diffusion, as a results there is no electric field present at after the diffusion region ends, that means their difference is also 0 and measured voltage difference is 0 V as well. <S> Hope this helps! <A> If you had an electrostatic voltmeter with a resistance much higher than your D.U.T. Series Resistance, which is possible, but the diode leakage would have to be equally high to prevent discharging the Static Potential. <A> The answer is quite simple. <S> The barrier potential exist across the depletion region not across the diode, so the region of existence of electric field lines is limited to the depletion region only. <S> The multi-meter used is connected across the terminals of the diode. <S> And there exist n and p regions between multi-meter probe and depletion region. <S> The unbiased n and p region acts as an insulator so as result no field lines is received at the probes so no voltage is shown in multi-meter. <A> Giving this question a shot. <S> There are two types of currents at a PN junction. <S> Diffusion currents are caused by carriers moving down a carrier density gradient. <S> Drift currents are caused by carriers moving down an electric field. <S> When no bias is applied to an isolated pn junction, the diffusion current moves carriers across the depletion region, building up charges on each side of the depletion region. <S> The accumulated charges create an electric field across the depletion region, and this electric field induces a current in the opposite direction. <S> The process naturally tends toward an equilibrium in which the diffusion current is exactly canceled by the drift current. <S> One could model this as two equal valued current sources connected in an anti-parallel fashion. <S> If one were to connect a volt-meter across such an anti-parallel connected pair of current sources, one would measure 0 volts. <A> The answer is quiet simple: You confuse the electrostatic potential with the electric potential. <S> What you measure with a voltmeter is a difference in the electric potential. <S> The electric potential however does include the chemical potential of charge carriers. <S> Note: <S> The chemical potential µ, or more precisely the gradient -grad(µ) of the chemical potential, is the "driving force" behind diffusion. <S> In the case of a PN junction, net diffusion of carriers occurs until the difference of the electrostatic potential between the two conductors equals the difference of the chemical potential between the two conductors in magnitude. <S> Since both potential differences have opposite signs, their sum is zero - <S> > <S> there is no electric potential difference to measure, despite a non-vanishing difference in the electrostatic potential!
| You are absolutely right (last paragraph), because there is a change in potential due to electric field in the depletion region, there is higher potential from n-type side and negative potential from p-type side, making the intrinsic potential difference built up.
|
Four-contact push-button switch; what's the pictured switch called? I have single phase submersible pump, around 400 foot underground in a Bore-well and its controlled by the controller shown below. The green button is a push-button switch which if pushed and held for 5 seconds, starts the pump. I would like to know what the switch is called, it has 4 contacts shown below (top right). Here is the complete album of the controller: http://imgur.com/a/U01qs I am planning to replicate the switch function using Arduino and high load relay. The button reads MAX 600V . <Q> Does the button light up to indicate that the pump is running? <S> If so, one pair of leads will be for the lamp, while the other pair are an SPST switch. <A> It's hard to tell exactly, but I'd guess with 4 connections <S> , it is a DPST (double pole, single throw) switch. <S> That is to say, two separate switches operated with a single button. <A> It appears to be a momentary push button. <S> Most likely a dual pole. <S> One pair of leads are normally closed, and the other pair are normally open. <S> You can disconnect from the unit, set multimeter to continuity, then test each pair.
| My advise is to, with power off , disconnect the four wires (marking them so you can properly reconnect) and then use an ohmmeter to confirm the operation of the button.
|
Shunt Resistor on output of Voltage Regulator? I was looking through a popular circuit on Sparkfun and trying to lower the power consumption of the device. I found that the outputs of their regulators are shunted to ground with a 300-ohm resistor. Any idea why they shunted the output to ground with a resistor? Does it have to be such a low resistance? I'm trying to reduce the current consumption of this small device as much as possible. Thanks! <Q> You should find some of the better LDO IC's for example those offered by Micrel (like the MIC5205) which do not need a load resistor. <S> For as low power as possible, make sure the input voltage is also as low as possible to what you eventually need it to be. <S> Like, if you want 2.7V as your system power rail, then use a 3.2V Li-ion coin cell or something, until a 2.7V fixed LDO regulator like the one I suggested earlier. <S> But yes, for the load resistor you see there on Sparkfun, that is because the linear regulator IC cannot maintain the voltage level on the output unless there is >= 10mA or so. <S> This is terrible for power efficiency, and you should avoid that device. <S> The internal pass element (probably a MOSFET) and it's feedback loop (Opamp to the Gate of MOSFET) obviously cannot handle very low duty cycle (no-load condition) properly. <S> This could be due to cheap components or perhaps bad bandwidth of the feedback circuit. <S> If it has no load, it may oscillate/overshoot or just cut off entirely until there is enough load/demand to continue normally. <A> It's a little hard to say without complete information <S> but I would guess that this is a shunt voltage regulator used to provide a precise reference voltage for A/D or D <S> /A converters. <S> The regulator may require a certain amount of current to be drawn in order to maintain precise regulation. <A> Without a part number, we're just guessing, but my guess is that's a cheapie LDO regulator (perhaps an AP7115) to generate an analog supply voltage of 2.8V from the 3.3V. <S> Probably not very accurate.
| The ~9.3mA shunt may be to prevent the 2.8V from rising if they allow current from a higher voltage output into the inputs of a 2.8V chip, but that's just a guess .
|
How many low-power batteries to power up 12V LEDs? I bought some strips of 12V LEDs and it comes with a pack of 10 x 1.5V batteries. The whole pack gives ±14.88V, which I thought to be quite high. Is this normal? Does this reduce the life of the LED strip? Also, if I'm using rechargeable Ni-MH (1.2V) batteries instead of the pack that comes with the LED stip, how many do I need? Thank you. <Q> Y'all feel free to tidy. <S> I may come back to it. <S> Content should be OK. <S> Rushing. <S> If they are White LEDs Vled ~= <S> 3.2V -3.5V <S> typically. <S> 3 are placed in series with a resistor <S> so VLEds = <S> 9.6V - 10.5V typically. <S> VLED varies with current and batch and device. <S> Balance of voltage is dropped across Rser = <S> series R. <S> If they spec at say 13V and if say Vled avg <S> = 3.3V <S> then VLEDS = 9.9V <S> so <S> Vr = <S> (13-9.9) = <S> 3.1V. <S> Nominal rated current flows when <S> Vr = 3.1V. <S> So I_wrt_nominal <S> ~~~~~= (Vin-9.9V)/3.1 <S> x Irated. <S> So at say 15V I = (15-9.9)/3.1 Irated = <S> 5.1/3 <S> Ir ~= 1.6 <S> x Irated. <S> As I rises VLED also rises <S> so actual I will be lower than shown. <S> eg <S> If VLED rises to 3.5V then VLEDS = <S> 10.5V <S> and I = (15-10.5)/3.1 = 4.5/3.1 <S> x <S> Ir ~= <S> 1.45 <S> x Irated. <S> And at say Vin = 11V <S> I = <S> (11-9.9)/3.1 <S> x Irated <S> = 1.1/3.1 <S> x <S> Ir <S> ~= <S> 0.35 <S> x I rated. <S> Again, VLED changes as I changes so here Vled may be say 3V <S> so Vleds = 9V <S> so I is now <S> (11-9)/3.1 Ir = <S> 2/3.1 <S> Ir = <S> 0.65 Irated. <S> SO IF strip is rated at say <S> 13V as above current MAY be about 2/3 rated value at 11V in and ABOUT 50% above Irated at 15V in. <S> For a modern phosphor LED 150% is too high - <S> about 120% is typical allowed abs max. <S> A responsible manufacturer will arrange current so about 120% of rated current flows when Vin is at abs max value liable to be experienced. <S> A typical manufacturer won't. <S> Alas. <S> Manufacturers of lower end products know that buyers equate bright with good and push I up. <S> If you care you can add a constant current source. <S> An LM317 does this well but needs 2V+ drop to operate. <S> The two transistor cct that is often suggested is usually "not nice" but here may be OK. <S> Vdrop min is probably around 1V <S> so it makes the brightness noticeably lower at Vin low but saves your LEDs at Vin high. <S> For max range at constant current a cheap transistor (bipolar or FET) + an opamp section can do well. <S> An eg <S> LM324 gives 4 x opamps in a cheap pkg (< $1 in 1 in US retail, about 5c in volume in Asia). <S> If interested ask about opamp and constant current circuits. <A> Cells in series sum their voltages, thus 10 x 1.5V would equal 15V. Ten Ni-MH cells at 1.2V <S> would be 12V. <S> With time, the voltage will drop more, and at some point the LEDs will fail to light (then it's time to replace or recharge the batteries). <S> Without knowing what LED strips you have specifically, there's no way to know what the tolerances are for supply voltage, whether it has any voltage regulation, or overvoltage protection. <S> The Ni-MH cells at 12V will likely power the strip just fine, but as they discharge, the LEDs will fail to light at the same point where the supply voltage is no longer sufficient. <A> If the strips are made for automotive applications, as it appears from your link, they are probably designed to run of of the car's 12V (nominal) power rail. <S> This rail is actually around 13.5V when the engine is running. <S> It can surge higher when the engine is cranking, or even when you rev the engine. <S> So, if you want a non-automobile power source, I would target <S> anywhere between 12V and 13.5V. Call it 9 alkalines or 11 NiMHs. <S> I expect you can get away with 15V, but I would stay safe unless I needed the extra brightness :) <S> Good luck!
| If the LED strip it not designed to be supplied with >12V, then yes, providing anything over the rated voltage may be driving them with too much current, and they will exhibit failures earlier. The reason you measure 14.88V rather than 15V is because batteries (cells) are not ideal voltage sources, and have internal resistance as well as chemistry breakdown.
|
What simple IC can I use to extract 500mA from a computer USB port? I have very small and simple electronic projects that I power up using the USB power connectors plus a resistor. For one project I require to use more than 100mA, however the USB port requires an enumeration process to be done before giving more than that (up to 500mA). I've browsed TI.com looking for some IC that can help me with this task, however I'm not sure I'm on the right track (I've pre-selected LM3526 and BQ2402x ICs, but I don't fully understand how to use them... I'm still learning....). Is there any simple example circuit design that I can use to solve this? Ideally, it should be something that I can connect to an USB port and that will just give me an output of 500mA and more than 4.5V. Thanks for the help, <Q> Consider Atmel's ATtiny85 with V-USB . <S> It's an 8 pin AVR chip that you would have to program with V-USB, which is a software-level USB implementation that would enable the Enumeration Phase, which would allow you to use the entire 500mA available. <S> It's about a $1.50 in SOIC packaging, pictured below, which saves both space and cost: Easily programmable and inexpensive, whereas the FTDI chip above (FT232R) is about $6 for one. <S> If you want to communicate with the chip, using V-USB also gives you the ability to act as a CDC-class USB device, which is akin to a serial port (UART), just like the FTDI chip. <A> You'll connect your load on the other side of the p-channel MOSFET power switch. <S> They're easily available through distribution , and in not too intimidating a package. <S> The default power setting is 100mA, so you'll have to use a utility to program the 500mA you want into the USB power setting. <S> FTDI has MPROG , which can be used for this purpose. <A> While USB does specify no more than 100 mA are to be taken by the device, there is no USB host that actually implements such restriction. <S> You can easily pull even slightly more than 500 mA (before polyfuse or something similar kicks in) from any computer built in last 5 years. <S> Yes, OS will be unaware of such pull but current going out will be perfectly fresh. <S> :) <S> Based on your project description, you are not actually interested in USB device but just using it as a power source. <S> While ignoring any specification might not be best approach, I am yet to see any computer that limits current under 500 mA.
| You can use a FT232R USB-UART chip, as so:-
|
Is circuit analysis really used? I'm currently in a first year circuit analysis course. While the concepts are relatively simple they don't seem like something you could use to design circuits. I mean it's hard enough to analyze circuits that you already have. But I feel like it would be impossible to work backwards to figure out the circuit you need given the values you want. So, as an electrical engineer working on designing circuits, how much of the circuit analysis techniques taught in a first year course (Nodal/Loop Analysis, Thevenin/Norton Theorems, Steady State Analyis of RC and RL circuits, AC analysis etc.) are actually used in the design of circuits? <Q> Actually, there are usages to circuit analysis, even beyond pure EE fundamentals: <S> Review a schematic from your colleague in a final design review; Reverse engineer an older circuit produced by a retired engineer which was sloppy at documenting its designs (True story during an internship...); Reverse engineer open-hardware design to learn new patterns/circuits; <S> Understanding side-effects experienced by your design when you have parasitic resistors/capacitors affecting your circuit; Troubleshooting mistakes in digital circuits: a colleague had an internal pull-down that he inadvertently activated. <S> Understanding circuits explained the cause of the remnant ~0.4V signal that we saw on oscilloscope without wasting hour debugging code and PCB; <S> Understanding what a circuit that you designed really does. <S> It is not always possible to implement a transfer function literally to components, but with experience, you know what will happen if you add a resistor/capacitor at a given location and if not, you can analyze it quite fast (to guess behavior) prior performing extensive calculus. <A> I find myself using filter design stuff I learned in college quite a bit. <S> The other thing that getting good at circuit analysis does is give you intuition. <S> After you work enough problems, you can look at a real circuit and spot potential problems. <S> I think that's more true in analog design than digital, but it still helps in being able to point to a spot in the schematic and tell if the current is going to be really high or really low <S> and if that's bad or not, etc. <S> RC stuff also comes in handy when figuring out how much decoupling capacitance you need on a chip. <S> None of the stuff you'll see in real life is as nicely set up as textbook problems of course, but first year EE circuit analysis is still worthwhile to master! <A> Analysis is a fundamental skill. <S> Circuit design is synthesis <S> which is pretty much the opposite of analysis. <S> Design (except in the simplest cases where you're simply finding parameters for a known configuration) is inexact, and consists of constituting a workable solution from parts. <S> Analysis is (literally, lysis means "to separate") breaking down a circuit into constituent parts. <S> You can test analysis by comparing with reality, whereas I don't think it's possible to prove a non-trivial design is optimal, only that it's sub-optimal if a better one is presented. <S> When you can break down previously encountered circuits into parts, and reassemble the parts to make new designs, you can apply other reasoning methods such as analogy . <S> This SQUID control circuit you need to design this month may resemble in some way an LF receiver you designed years ago, and you can apply tricks or caveats learned from that experience. <S> A purely mechanical technique might have some analog at GHz frequencies. <S> Plenty of innovation has come from re-applying old ideas in a new way, and in fact that is considered novel and patentable in many cases. <S> But if you don't really understand the old idea (which requires analysis), you're not going to be able to abstract it and re-use it in a new domain. <A> Every analysis tool is like a Chess move you can master. <S> The "best tools" are the ones you know how to use. <S> Once the theory is solid, shortcut tools can be trusted. <S> Designs start with awareness of requirements and how to convert them into functional specs,. <S> These need no design implementation at first, Just specs for inputs and outputs , environmental stress factors , cost, time and go to it. <S> Researching prior methods comes quickly when you can recognize what works, needs changes and then improve to exceed requirements. <S> Analog defines the world of all activity between a one and a zero, the latency, transmission line effects, and all physical effects of insulators, conductors, static charges, DC power, AC transmission, RF communication, Medical sensors. <S> Then you can add smarts with digital processing power, sensors based on physics, Systems on a Chip and create better tools. <S> Commodity products will end up in China or India, but there is a huge world of advanced technology emerging in many fields, from more efficient electric transport , more advanced medical imaging of live cancer cells, safer smart grid technology, more sophisticated communication devices, smart detectors for predicting earthquakes, Carrington Effects, measuring condition based monitoring to prevent transformer arc failures ..from 10kV to 1.5 gigavolt. <S> Each of these requires the foundations of experiment and Physics discovered by Faraday, Ampere, Gauss, Weber, Ohm, Siemens, and Maxwell. <S> From there you can either go into, Research, Applied Research, (design), Test Engineering, Software Development, System Design, or become a Business Developer, raise millions and build teams form a company to do the same. <S> ( as many of my colleagues did.. BNR, Symbol, ImRIS, Cubresa, ... )
| You need analysis to understand and to hone and tweak circuit designs.
|
Is the CAN bus protocol a master and slave protocol? I am investigating if the CAN bus will work for the projects I am building at the moment. I want to create a bunch of input/output controllers that will accept/buffer commands, execute them and then report back progress. My feeling is that a master and slave setup will not work best for project, but I am not sure if the CAN bus works on that principle. <Q> A CAN bus is multi-master and automatically arbitration free. <S> The whole point is that you don't need a single master or main controller to take care of everything. <S> Each message that is sent has a priority, and higher priority messages trump lower priority ones (a lower priority message will wait until there isn't a high priority message being sent). <S> So there are no slaves. <S> It's kind of similar to a peer to peer network, in that it's decentralized. <S> The reason this is done is to reduce failure points. <S> If a single main controller failed, so would the entire network. <S> If there is no central controller, a single component can fail and the rest of the system will keep going. <S> For your network, you should assign priorities to each command sent depending on what's most important. <S> The Wikipedia article is actually quite good. <A> CAN is a low-level protocol, somewhere around the data-link layer of the OSI model (layer 2). <S> Thus it is only concerned with transferring generic data frames. <S> CAN does not specify a network topology, nor how data gets passed around, nor the nature of the data etc etc. <S> So CAN is not a master/slave protocol for the same reason as UART isn't one. <S> On top of the standard CAN protocol frames, you need higher layers that define the communication. <S> The far most common ones for CAN are: CANopen (most common, generic) <S> DeviceNet (generic, mainly used in the automation industry) <S> J1939 (trucks, tractors, heavy vehicles) <S> These work in entirely different ways. <S> CANopen for example, doesn't define a master as far as the data traffic is concerned. <S> But CANopen does define a "network manager" master, who is in charge of supervising the nodes on the bus: making sure they start correctly and remain alive and functional etc. <A> CAN only has masters. <S> One node transmits at a time - <S> if multiple nodes being a transmission at approximately the same time, they will collide, and the node transmitting the lowest ID will win arbitration. <S> The other nodes will listen to the winning node's message and then after some backoff time re-try their own transmissions. <S> And of course, there's nothing to stop you building some kind of master/slave protocol on top of the low-level message-moving that CAN provides you with.
| Data communication between nodes on the bus is only a concern of the nodes involved: there is no master directing data traffic and every node can be configured to speak/listen to any other node.
|
Wind turbine generator: brushed or brushless DC motor? I am trying to use a DC motor as a generator for a wind turbine but then there are two possibilities for the DC motor which are brushed and brushless. Which one is more efficient to use? I know that if I use a brushless motor then I need to use a 3 phase rectifier which includes 6 diodes. Will the voltage drop in these 6 diodes affect efficiency? But if I use the brushed motor then I don't need to use the rectifier. Can this be the advantage over the brushless motor? <Q> A brushed motor has the rectifier built in via the commutator. <S> So you don't need the 3 phase rectifier. <S> However, the brushes and commutator contacts do wear out and the brushes need to be replaced from time to time. <S> There are still losses associated with the commutator, even though not as much as with a diode bridge. <S> In a wind turbine I would think that maintenance could be painful <S> so you would want to minimize it as much as possible. <S> The brushless motor is typically much more reliable and doesn't require periodic replacement of anything. <S> Depending on the output voltage (Speed and back EMF constant) the rectifier losses could be significant. <S> You could use a 3 phase FET bridge instead of a diode rectifier, but that becomes much more complicated. <A> Brushless motor is much better for few reasons: efficiency <S> reliability <S> it can work in harder conditions (brushed motors have much worse tolerance for humidity, dust etc.) <S> no brushes voltage drop <S> smaller rotor inertia <S> (same power brushless motor is smaller) generally better by design <S> Small motor inertia is very important. <S> When wind stops or slows down, and turbine is still turning - you are wasting energy (kinetic) into... working as fan. <S> Thats why turbines have 3 blades instead of 50. <S> Voltage drop on diodes can be reduced to 0.3V per diode (by using diodes with low drop, like Shottky diodes). <S> Typically (at least in my country) - generators for wind turbine have nominal voltage like <S> 48V. Voltage drop on 2 diodes is no more than 0,6V. <S> Thats just 1.25%. <S> In addition - low current diode voltage drop is smaller than nominal. <A> I'm guessing the machine in conjunction with a typical speed control (Darlington Inverter) will operate well as a DC generator even with commutation turned off, whereby the diodes provide natural rectification of 3-phase "DC" output, perhaps smoothed with an output capacitor before current is sent to the battery. <S> Confession: <S> I'm a mech. <S> engineer. <S> Elec. <S> guys please verify or correct my assessment
| Efficiency of brushless motor is better because: no brushes friction
|
Motor Discharges Boost Converter Too Quickly I am trying to step 3V up to 9V via a boost converter, the idea being I have the current but not the voltage to drive a DC motor. I used this this website to calculate the parts values I would need, and assembled the boost converter on my breadboard. I noticed two things: Increasing the duty cycle with which I was PWMing the transistor barely increased the voltage. I was led to believe it would do so more greatly. The capacitor would charge until I provided a load across it, at which point it would reach an equilibrium. This makes sense to me (the capacitor charges when the switch/transistor is off, and discharges if there is a load when the switch is on), but placing the H-Bridge that drives my motor in parallel with the load resistor results in an immediate discharge of the capacitor and the voltage drops from 9V to 4.5V, not enough to power the motor. Placing the H-Bridge in series with the load resistor does not result in it being powered but as if the load was disconnected. I unfortunately do not remember the model number of the diode off the top of my head and won't be able to get to the lab to check it until later in the day. It is some type of fast switching Schottky diode. simulate this circuit – Schematic created using CircuitLab Am I doing something wrong in particular, or is this just something a boost converter cannot do? Should I be using a different circuit of some type? <Q> If you have the current but not the voltage to drive your dc motor you might be disappointed. <S> You might have the ability (in your battery) to supply 1 amp at 3 volts but, to power a 1 amp motor with 9 volts means your battery will need to supply maybe 3.5 amps into the boost regulator. <S> Power in equals power out plus losses and to obtain 9 volts at 1 amp requires a power in of 9 watts plus about 15% extra for the boost converter losses. <S> That takes the input current up to about the 3.5 amp level. <S> Can your battery sustain this level <S> I wonder? <A> There are two parameters that matter in the boost converter: Duty cycle, and switching frequency. <S> (And a third: whether it's in continuous or discontinuous conduction mode) <S> You are saying that the capacitor "drains" when you supply the load -- <S> this means that you're not switching the converter fast enough to re-"fill" the capacitor. <S> The reason for this may be that there's too high resistance in your inductor, or your transistor can't sink enough current (typically, power N-channel MOSFETs are used,) or your battery source has too high internal resistance. <S> A 2N2222 is not a power transistor. <S> Also, you don't indicate how much current your oscillator will source. <S> Given that you're using a BJT, the amount of current through the transistor is directly proportional to the amount of current out of the oscillator (up to the saturation point of the BJT.) <S> Also, the schematic shows a sine generator. <S> You want to use a square wave for a boost converter. <S> Hook up a scope to various points in your circuit, to measure how it's doing, and this will tell you where it goes wrong. <A> This is not kind of question that can be answered precisely, but I will try to give some suggestions. <S> If circuit parts are calculated correctly, and you still have such voltage drop <S> - it may be caused by: too small coil (in size) - you may need thicker wire <S> power source can;t keep up with motor power requirements wrong capacitor type, it should be low-ESR <S> (normal capacitor can't charge quick enough due to its internal resistance) <S> damaged or worn out capacitor <S> If you add more information about your power source, motor, real schematics, information about coil size - maybe someone will answer more precisely.
| Without knowing more, I would suspect using a higher-performance transistor with a higer switching rate (say, 20 kHz) and a lower-resistance inductor would improve the performance you see.
|
Why don't people tend to use voltage dividers or zeners in front of linear regulators After seeing some students yesterday that tried to use a voltage divider instead of a regulator to provide a sensor with a lower power supply, with predictable results, I started wondering about this question. When picking a regulator, it seems many look at the required voltage drop and the power dissipation required. Efficiency aside for the moment, if a linear regulator can drop that power within thermal limits, linear regulators are an option, and if they can't, move on to switching regulators. If one could figure out the range of current draws, and calculate out a voltage divider that would simultaneously keep the input to a linear regulator high enough to maintain regulation and low enough such that the regulator doesn't burn away too much power across the current draw range, is this a viable approach? I can think of a number of reasons why this might not be the best approach: power supply rejection ratio may not be good enough on the regulator; the range of current draws that make this approach feasible might be very small, unless you use small resistors that are likely to exceed their own power ratings; its just more efficient to use a switching regulator; etc. Also, it might be that people do this all the time, and I just haven't noticed it, or maybe a zener is used instead of the divider. It just seems that when the power drop is too big, people mostly run to switching regulators. Anything I'm missing? <Q> This is certainly a technique I have used a few times to overcome the limited power dissipation abilities of the diminutive 78L05. <S> I've known the range of currents that the load is taking and placed a dropper resistor in series with the power feed to the device. <S> Why didn't I use a switching regulator? <S> I couldn't - I was sending power and data down a 50 m cable (phantom power) and the extreme complication of filtering out the switching regulator's current surges meant it just wasn't feasible. <A> Voltage dividers are terrible for efficiency (if you think of output impedance vis-a-vis power consumption). <S> I'd be hard put to think of a good place to put them in front of a regulator. <S> Series zener diode- <S> if you put a 24V zener diode in to knock a 35V input down to 11V for a 9V regulator, you've increased the sensitivity to input variations- <S> a 10% drop in the input means there's only 7.5V left and your regulator drops out. <S> With capacitive droppers you don't suffer much loss. <S> Many of us will also put a shunt TVS that effectively acts as a regulator under unusual circumstances, so I'd count that too. <S> Series or shunt resistors around a linear regulator- <S> I think I used the latter once, the former not so far. <S> (downside is that some power would be wasted if the required current drops below the mean). <A> If one needs to convert 12V to 5V for a load which may vary from 0 to 1 amp, and the regulator needs a minimum of 6 volts on the input, connecting the supply directly to the regulator will cause it to dissipate 7 watts with a one amp load. <S> Adding a 6-ohm resistor in series with the input would cut worst-case power dissipation in the regulator to about two watts over a wide range of load conditions (as current goes up, the amount of voltage dropped by the regulator [as opposed to the resistor] would go down). <S> Series resistors don't help overall efficiency, but they can shift heat dissipation away from the regulator. <S> A key point to note, though, is that the bottom half of a resistor divider wouldn't really help much of anything, since its purpose would be to waste power when the load isn't drawing current, but in that scenario regulator current will be low so there's no need to share it.
| The shunt resistor would be more attractive if the linear regulator was capable of sinking current (some are, but most are not), then you could just set the resistor to handle the mean current and the regulator would tend to run very cool I have used a shunt zener with a capacitive dropper in series with a linear regulator to get power from the mains, and I think that's fairly common.
|
Why is pin protection needed? I have read this article about protecting input pins . One thing is still unclear to me — why is the protection needed at all with such a high resistance on the input pin inside of the microcontroller? The current would be really small and the heat dissipation will be little as well. <Q> The high voltage will destroy the gate oxide of the MOS transistors connected to an input pin. <S> The real threat is from static electricity that builds up during handling and assembly of circuit boards. <A> As others have said, the protection is primarily from ESD , which is usually more of an issue before the component is fitted to the PCB, although stray voltages finding their way in to (badly designed) equipment can also cause damage. <S> ESD damage can happen without being noticed, and the component may fail days, weeks, months later. <S> Some designers, either being very brave or by knowing very well (probably in consultation with the device manufacturer) <S> what they can get away with, will rely on the internal protection / current limiting features to save on adding external components <S> (automotive designs do this sort of thing) <S> but you really never should. <S> Edited to add, there's some good info on pin protection & abuse of it here . <A> Electrons are not only useful as carriers of electricity. <S> They are also responsible for binding atoms together in useful configurations. <S> Some materials have large numbers of "loose" electrons which can move around relatively freely without particularly affecting the material through which they are flowing. <S> Some other materials don't; given enough voltage, the electrons locked into the latter materials' structures can be knocked loose, but the act of knocking the electrons loose may permanently alter the structures in question. <S> The purpose of protection circuitry is to absorb energy over a much wider area, thus increasing the amount of energy that would need to be applied to cause damage. <A> It is the voltage, not the current, that could damage the pins. <S> If you are using short wires, or traces, you wouldn't need to worry about this. <S> But, if your wires are very long (multiple yards/meters), then the inductance of the wires can cause the spikes... <S> Good luck! <A> Because any electrical overstress is overstressing the component. <S> Here's another short article (with pictures!) <S> that talks about different kinds of electrical overstress. <S> I quickly skimmed that digikey article and it looks amazing. <S> This is great advice to live by. <A> Don't think of it as there being a resistor ahead of the input circuit, thus protecting it. <S> Think of the resistor as being part of the input circuit. <S> The resistance is between two sensitive parts of the input, and ohm's law tells you how the voltage will behave. <S> Put a mere 10 microamps through a 1 megohm resistor and E=I/R <S> tells you that you've got 10 volts. <S> This already would make a 3.3 volt input very unhappy. <S> It's not a real resistor in there, but for simplicity and for modeling purposes, it's good enough for illustration.
| That article is talking about protecting against voltage spikes when using long wires to connect to an input pin. A small amount of damage, concentrated in one place, can make a chip useless, and--in the absence of protection--it wouldn't take much energy to cause such damage.
|
How frequently must Microchip's USB HID USBDeviceTask() Task run? I'm integrating the example USB module into my existing app and can only get it to work when I disable interrupts. I'm working off the mla example installed at the path: C:/microchip/mla/v2013_12_20/apps/usb/device/bootloaders/firmware/pic18fxxjxx Details I'm implementing each of USB callback functions exactly as done in the example. USB clock is 48MHz CPU (f_osc) clock is 16MHz pic18f25k50 one high-priority interrupt for Timer1 ticks every 250us, everything else happens about every 1s never go to sleep, yet this still seems to reset Is microchip's USB stack not thread-safe? Am I not running the USB task fast enough? while (1){ // Clear the watchdog timer ClrWdt(); // Run the USB task faster than xxxHz? USBDeviceTasks(); // Do stuff}// high-priority foreground loop takes ~5usvoid high_priority interrupt HighPriorityTasks(void){ // Do stuff quickly, clear sources of interrupts}// low-priority foreground loop takes ~5usvoid low_priority interrupt LowPriorityTasks(void){ // Do stuff quickly, clear sources of interrupts} When it fails, I see device manager continue to refresh accompanied by one of these popups about every 5s in Windows 7: <Q> Try using USB in interrupt mode. <S> USB devices must respond to host requests within a certain time frame. <S> You haven't detailed how long your timer tasks routine takes to execute but from this error it seems that it is taking too long <S> and there is no way for the USB to respond in a timely fashion. <S> This way when the host requests a response from the PIC when it is executing the timer tasks it will pause executing that routine, answer the host, and then return to that routine. <A> USB HID class can poll as fast as every 1ms. <S> So, ideally, you want to call USBDeviceTask() more often than every 1ms. <S> On the side, I have tried calling USBDeviceTask() <S> every 10ms, and it seems to work fine!! <A> I have my own powerdown/powerup functions when going in and out of sleep and didn't realize that the default USB suspend callback function put the device to sleep: void USBCBSuspend(void){ <S> Sleep(); // Go to sleep, wake up when a USB activity event occurs //If using the WDT, should go back to sleep if awoke by WDT instead of USBIF while((0 == <S> USBIF_FLAG) <S> && (0 == RCONbits. <S> TO)) <S> //If using the WDT, should go back to sleep if awoke by WDT instead of USBIF { Sleep(); <S> //Entry into sleep clears WDT count, much like executing ClrWdt() instruction } //After <S> the USB suspend event ends, you should re-configure <S> your I/ <S> O pins <S> //for normal operation mode (which is allowed to consume more current). <S> //However, it is recommended to put this code in the USBCBWakeFromSuspend() //function <S> instead of here (so that this function will work with either //sleeping or clock switching to a lower frequency).} <S> This is certainly something I didn't intend to do, so have this commented out for now. <S> While a bit of an aside, <S> a do-while loop really provides a more-readable implementation here better anyway: <S> // <S> USB_onSuspend(void)// Called when the USB host sends USB suspend <S> signalingvoid USBCBSuspend(void) <S> { do { Sleep(); } <S> while ((USBIF_FLAG == 0) <S> && (RCONbits. <S> TO == 0));}
| Without specific details of your application, I would suggest running USBDeviceTasks() as the only high priority interrupt task and set the timer and its tasks as lower priority.
|
How do filament LED bulbs work, looking very similar to incandescent bulbs? Filament LED bulbs use very narrow strips containing the LED,which look similar to filaments of an old incandescent bulb: They look like a normal bulb with large "old-style" coal filaments, and are of about the same size and shape as the common standart E27 socket bulb. There are two basic variants of these bulbs: One "plain" as described, looking impressively similar to a classic bulb The other has a white coating on the glass, about 15mm wide next to the metal socket. How are these filaments build?I could imagine the it's "pretty normal" LEDs, but lots of them,directly bonded to something? (There are references to COB) How does the heat management work?Ii have seen speculations about filling the bulb with either heliumor "some patented organic gas" -could that be enough for cooling? How does the driver - which seems to be very small - work? In the variant with white coating - hiding some components, I assume - is already not that much space, and in the plain variant, there seems to be almost no space for something nontrivial. Below you see some bulbs of the white coating variant, switched on: (Images with the permission of this supplier at Alibaba ) Note how the LED filament bulbs are not so much similar to the tungsten filament bulbs recently being phased out - they are more resembling the long outdated carbon filament bulbs: bd® Antike Edison 220V-240V 40W Retro Vintage Industry Style Deko Glühbirne (kl) <Q> The whole thing is then coated in phosphor. <S> The light isn't completely uniform but it's good enough. <S> There's a spec sheet here: http://www.runlite.cn/en/product-detail-145.html <A> I purchased one of these lamps in Italy: 3W/300lumens/9,00euros <S> I took some pictures with very low exposition to higlights multiple "dot leds": I can count 28 dot-leds per each strip and 4 strips. <S> This gives 112 dot-leds. <S> Other numbers:100 lumen/watt2.68 lumen/dotled75 lumen/strip0,03 euro/lumen <S> Assuming half the cost is for the electronics, we could assume, for easier calculations, <S> 4,00 euros per 4 filaments, hence 1 euro per 28-dots/75-lumen strip. <A> To understand how these devices function, it helps to understand how traditional LEDs function. <S> An LED is a Light Emitting Diode, so basically you can think of it as a simple PN junction that would be used in a diode (although in actuality the structure of LEDs is more complicated, often a double heterojunction structure ). <S> (source: gsu.edu ) <S> Based on images of a similar LED "filament" bulb on Amazon image1 , image2 <S> It seems that each "filament" is a single LED with a radial structure, that is using radial layers rather than the traditional planar layers. <S> So the core layer (cathode for example) would be composed of a conductor/metal, then the next outer layer would be n-type material, then the next outer layer would be p-type material, and finally the outermost layer would be a transparent conductor (to let light pass through it) <S> such as ITO (Indium-Tin-Oxide). <S> It looks as though the LEDs in the bulb are arranged as two lines in parallel, each with two LEDs in series. <S> The driver would depend on the specs of the particular LED "filaments", but a driver could be as simple as diode bridge if the LEDs have a high enough forward voltage. <S> Because these LED "filaments" are so long and thin, they have a large amount of surface-area per volume, so that the heat is more dissipated than in a traditional LED.
| These LEDs are not a single radial die, they are made with a transparent substrate with many LED dies in series (probably 25) placed on it.
|
Quick and dirty LED power from AC I have some extra 1n4007 diodes, and was wondering whether, instead of using a power supply to drop the voltage, could I could power LEDs by just making a full wave rectifier, Using a capacitor to smooth out the power and powering n LEDs in series? Here is the spec sheet: http://www.jameco.com/Jameco/Products/ProdDS/2095171.pdf These LEDs are rated to go as high as 2.6V (red) or down to 1.9V. Could I create a series of 120/2.1 = 58 of them? With voltage going up as high as 128V, that's still 2.2V, well within spec. This would mean that each LED would be less bright than maximum but would run cooler. Would that result in less efficiency? From the spec sheet, it looks like within a fairly large range, light output is pretty linear. The current for this would be somewhere in the 20-30mA range.Aside from the potential for failure of one LED in series to kill the light, is this a reasonable hack? It seems to me that it's at least very efficient. The only losses I can see in this circuit would be in the diodes. I should see perhaps a 1V drop in the 1n4007? Is there anything more efficient I can do? <Q> What you have just described is most if not all cheap led christmas lights. <S> Diode bridge rectifier, capacitor, and a few parallel strings of multiple leds in series. <S> Maybe a fuse. <S> That's literally it. <S> See http://www.ciphersbyritter.com/RADELECT/LITES/LEDLITES.HTM for a full primer. <A> Your hack might work. <S> Then again, you may kill some LEDs. <S> The problem is called "thermal runaway". <S> Look at the data sheet. <S> As the LED gets hotter, its forward voltage drops. <S> For a fixed voltage, this means the current goes up (and it goes up fast). <S> This means the LED gets hotter, since it's dissipating more power. <S> This means ..... <S> That's why you need a resistor or a controlled current source. <S> As described, with no resistor I suspect you'd kill some LEDs. <S> Of course, one advantage to putting LEDs in series is that, with a little luck, you'd only kill one at a time. <A> Basically, you need a full-wave rectifier (unless you want a lot of flicker), the LEDs, and then a dropping resistor (or resistors) to handle the current. <S> You do not want a diode; you need something to set the current. <S> Here's how to do it: <S> Take 110 volts and divide it by the forward voltage of the LEDs, and get rid of any fraction. <S> Multiply the number of LEDs by the forward voltage to get the total voltage dropped by the LEDs. <S> Subtract this number from 120 volts. <S> This is the voltage that you need to drop in the resistor. <S> Figure out the current you want from the datasheet. <S> Figure out the resistance you need, using V=IR == <S> > <S> R = V / I. <S> If you put this resistance in line with the series LEDs, you will get the proper current. <S> Figure out the wattage dissipated by the LED, using W = VI. <S> You will likely have to use multiple resistors so that you don't exceed the resistor wattage. <S> The reason for using 10 volts for the resistor instead of making it as small as possible is to make the current more constant. <S> If the resistor is small, the current gets much more sensitive to manufacturing differences in the forward voltage of the LEDs and the actual line voltage. <S> Oh, and obviously, you need to do a very good job of insulating these things, because you're dealing with possibly fatal voltage. <A> I bought some LED icicle type strings powered by a controller. <S> I reverse engineered the controller and basically it was a full wave bridge rectifier turning 120 Volt AC into DC. <S> There was some gating IC's that controlled thyristors that actually pulsed 2 separate 75 LED series strings from the unfiltered 120 Volt DC. <S> I saw no fuse or current limiting resistor, so probably not the safest design should something fail by shorting out. <S> I did not like <S> the LED's to be pulsed in any manner <S> so I chose to remove the thyristors and replaced them with 1200 Ohm 1 watt resistors. <S> This limited the current in the strings to about 10 mA and produced decent light at night without the annoying pulsating. <S> It has been operating for several days now with no issues. <S> Thinking of adding 1/8th watt 1 Ohm resistors to each side of the AC feed that would likely act as a fuses.
| I've built some ornaments that run directly off of AC.
|
Does every PIC need a bootloader? Does every PIC chip need a Boot-loader like the Arduino chips do? <Q> One reason to have one however is to allow firmware to be upgraded without needing to connect up a programmer like the ICD3 or a REALICE. <S> (Some products are designed with headers that allow a programmer to be connected after the product is put into a case, but most are not.) <S> I have used two different ways of upgrading firmware without a hardware programmer. <S> Both require bootloaders to be flashed into program memory, usually at the very beginning or very end of program memory area. <S> Ideally, once the bootloader is flashed, then this portion of memory can be write-protected <S> , so an errant application program cannot erase the bootloader. <S> One way to upgrade firmware is to provide the application on an SD card, and have a slot in the product that allows this to be accessed. <S> Each time the microcontroller starts up, it checks to see if the application on the SD card is different from the one in flash, and if so update it. <S> (You don't want to blindly download and flash the program each time since flash has a limited number of erase cycles.) <S> This requires the bootloader to have the code necessary to access the SD card (which will probably have a FAT16 or FAT 32 file system on it.) <S> Another way to upgrade firmware is FOTA -- firmware over the air. <S> In this case, the firmware is downloaded over some wireless connection, such as a cell modem, bluetooth, or Wi-Fi. <S> It is then flashed into program memory. <S> A variation is to combine the two, whereas the new program is downloaded and written to an SD card, since it may not be practical to flash the program as it is being downloaded. <S> Note in this case the bootloader doesn't need to have any of the code to access the wireless network, as the application program can download the code and write it to the SD card before it switches to the bootloader. <S> The PIC32 is particularly suited to having a bootloader, as it has a 12 KB segment of memory specifically set aside for this purpose. <A> First, there are no Arduino chips, only Atmel AVR chips. <S> Second, the Arduino bootloader exists to enable the novice user to develop without experience, or professional tools. <S> Atmel chips by themselves do not require any kind of bootloader, nor do PIC chips. <S> Bootloaders exist to facilitate updating firmware in the field, without the use of a programmer. <S> This is by no means a design requirement. <S> If the product in development has no PC connectivity, then adding a bootloader would incur pointless cost. <S> Think about all of the electronic devices in your home that do not connect to a computer, such as microwave ovens and alarm clocks. <S> I will bet you anything these devices use some kind of microcontroller, and do not have bootloaders. <A> If you buy the programming adapter, (PICkit or ICD), you can program the chip directly. <S> After all, that's how you put in a boot loader initially. <S> It uses 5 wires for the connection. <S> A boot loader gives you the convenience of programming it remotely, and removes the need for having the ICD present. <S> But having the ICD for debugging purposes is very handy.
| As others have stated, you don't need a bootloader for PIC microcontrollers (and most others from other companies, in fact I know of no microcontrollers that require one).
|
Using single resistor to receive 5v signal on the 3.3v rated pin I have been investigating ways to safely receive 5v signal on the input pin that can only support 3.3v input signal. I came across an article that states that it's possible to only use a single 10k Ohm resistor to achieve that. I can understand how this can prevent too much current from flowing through the schottky diode, but I am still skeptical this resistor would produce the necessary voltage drop for the signal that goes further into the MCU since the resistance of the resistor that is placed after diodes is much higher than 10k. Will the 10k resistor work for this purpose and if so, what voltage drop would it produce? Thanks a lot. <Q> If you are serious about building reliable designs with todays high density chips you should never place a voltage on a pin that exceeds the range specified in the maximum ratings section of the manufacturers data sheet. <S> There are a multitude of different chips designed with either 5V tolerant inputs or designed to properly level translate from 5V down to 3.3V. Use one of those, or if you want something cheaper that comes with a lower bandwidth, then you can utilize two resistors at each input as a voltage divider to step the 5V down to 3.3V. <A> In fact, Microchip put out an app note a few years ago that demonstrated sensing the phase of your 120V mains by connecting the 120VAC to the PIC pin via a resistor! <S> It was an X10 communications application... <S> However, using the internal protection diodes in this way is kind of sketchy. <S> It also limits any future design changes, for example, if you want to swap out the microcontroller. <S> If you are only receiving data on this pin, and not transmitting, it is a great application for a voltage divider. <S> You can choose fairly large resistors to minimize wasted power... <S> Good luck! <A> Yes... and no... <S> (5V - 3.3V)/10k = 170µA <S> The current will flow through the protection diodes. <S> If the micro is in deep sleep, or possibly even in reset, it will consume less than that. <S> If there are no other loads to use this current, the 3V3 supply voltage will rise to a value close to 5V. <S> At this point if the micro is a 3.3V part, it will die, other 3V3 parts on the same rail might also die... <S> It can also latch-up, or bust an ESD diode, etc, as explained in the other answers. <S> If you're not saving µA <S> then a voltage divider will work fine, but be aware that the extra resistance will slow things down when interacting with the pin's capacitance, so don't do this on a 10MHz signal! <S> It's okay for slow stuff though. <S> If this is battery-powered and you want to save µAmps, or you need the speed, a 74LVC logic gate is a good choice, when powered from 3V3 its inputs are 5V tolerant and it will consume very low static power while being very fast. <S> Also you might be able to hack the source of the signal. <S> If it's an open drain/open collector output, simply put the pullup from 3V3 instead of 5V, problem fixed. <A> I will give you a very easy solution. <S> Add one more resistor connected to the ground, which will draw MOST of the current that you are worrying about to the ground. <S> Of course, the value of the resistor should be a lot less than the input impedance of your port pin. <S> In other words, use the two resistors as resistive voltage divider which steps down 5V to 3.3V. then connect the point where the two resistor meets to MCU's GPIO. <A> Use a 10K resistor from the output of the 5v chip to the input of the 3.3 volt chip. <S> Anode to ground and cathode to the 3.3 volt input. <S> Effectively you made a voltage regulator. <S> The input of the 3.3 volt chip will never exceed 3.3 volts so it will be completely safe to run in that mode. <S> You should be fine going the other way, a 5 volt input should read a high state at 3.3 volts.
| People who promote low cost shortcuts, such as the unitary 10K resistor resistor to isolate a 3.3V input from a 5V source, are not serious about building reliable electronics that can last for a long time unless they are are using an device that has documented 5V tolerance on its I/O pins. Then connect 3.3volt zener diode for across the input of the 3.3 volt chip. This can work if the microcontroller has the same protection diodes that are shown in the article.
|
How can I use/convert the touchpad, keyboard and screen of an old laptop to use them with a desktop PC? Is there a common way for all the laptops' hardware, like screen, touchpad, and keyboard, that can be followed in order to use them with a desktop PC? What I need to do, to use screen, touchpad, and keyboard of my old laptop with my desktop PC?Would be fine if I disassemble my laptop, then I remove the screen connector from the motherboard, and I try solder pin-by-pin of the laptop screen connector with a VGA cable? Is there a standard connector for laptops' screens?Concerning the keyboard and touchpad I was thinking to do the same thing but with an USB cable, would it be ok?The purpose of this question is also to become a sort of guide for all the people that are trying to reuse the hardware of their old laptops. <Q> A laptop screen is usually an LCD screen with an LVDS connection (or two depending on resolution.) <S> This is a differential parallel data bus with data pairs and a clock pair. <S> The protocol is not too complicated, but you won't be soldering directly to a desktop compatible video connector without an interface board. <S> LVDS is pretty standard, but an interface board for your specific LCD panel will provide the proper drive voltages for the LCD and backlight. <S> If you search for your LCD panel, you may be able to find a controller board that will run the LCD as a desktop monitor. <S> That is probably your best hope, if you don't have the capability of making this yourself. <S> Most laptops typically have USB or PS/2 based keyboard and touchpads, although some are raw devices. <S> For USB style or PS/2, you just need to locate the place to connect all wires and you should be set. <A> Laptop keyboards and touchpads typically come in three flavors. <S> Direct interfaces (Raw keyboard matrix and raw resistive or capacitive touchpad matrixes), PS/2 (Older), and USB (Newer). <S> Both the PS/2 and USB types require you to find the pinout, and then can be wired to a USB cable. <S> Some might run on 3.3v or lower instead of 5v, so you need to measure that and provide proper regulation first. <S> Laptop LCDs do not use VGA, or DVI, or any of the external interface types. <S> They use a LVDS (Low Voltage Digital Signal) interface, and varies widely by panel and laptop manufacturer. <S> But in the last few years, universal LVDS interfaces have become cheap. <S> They are basically the same video boards you see in regular external monitors, taking VGA, DVI, HDMI to LVDS. <S> You also need to provide the proper power, which varies by screen type and backlight type (High Voltage Inverter, or LED or whatever). <S> Laptop Cameras are the same thing. <S> Typically all are USB at this point, though some take 3.3v or lower instead of 5v. <A> Laptop screen will be most complicated part, because LCD controller is built into graphics cards in most cases, and you can't extract parts responsible for driving LCD from laptop . <S> That makes connecting Laptop LCD to VGA complicated. <S> This information may be outdated, it applies to laptops that were manufactured few years ago (before HD and HDMI came out). <S> I have not tried this with modern laptops. <S> Much easier way is disassembling desktop LCD monitor , which have VGA to LCD module. <S> See other answers for more information.
| Keyboards and touchpad are USB based in most cases.
|
Voltage divider for battery monitor ADC - loading effect? I'm building a voltage divider to use for feeding into an ADC to determine the battery voltage of a device I'm designing. Is it important to consider the input impedance in the ADC and to account for the voltage drop caused by that extra load or should I not even worry about it? <Q> Yes, you definitely need to consider the input resistance and/or bias current requirements of the ADC input. <S> The usual technique is to keep the Thévenin resistance of your divider low enough that the offset caused by the ADC input is "low enough" (e.g., less than 1% or 1 LSB), depending on your system requirements. <S> For example, suppose that your ADC input resistance is 1 MΩ <S> and you want to keep the error caused by this to less than 1%. <S> Therefore, the Thévenin resistance of your divider should be less than 10 KΩ. <S> So, now you have two equations, one for the divide ratio: $$Ratio = <S> \frac{R1}{R1 <S> + R2}$$ and one for the resistance: $$R_{TH} = <S> \frac{1}{1/R1 + 1 <S> /R2}$$ and two unknowns, R1 and R2. <S> It's straighforward math to plug in your known values for Ratio and R <S> TH and solve for R1 and R2. <A> taken so, put a 100nF capacitor to ground on the ADC input. <A> It would really help to know what ADC you are using. <S> In any case, you need to consider it, but the good news is that most microcontroller ADCs have high impedance inputs so <S> a divider made with a few tens of kilo-ohms won't be affected by it very much. <S> How much accuracy do you need? <S> Beyond a certain point you will need to calibrate anyway, so the impedance of the ADC can be compensated for at that point anyway. <A> ADC input impedance is generally much higher than ESR of the battery, but <S> if too high then imbalanced stray capacitance can inject CM noise. <S> CMRR must be considered over entire BW of the circuit. <S> Calibration is key to verifying Vfref noise and linearity glitches from Digital noise getting into analog measurement, so guarding, grounding , nyquist filtering and shielding improves the results. <S> Then ratio of resistor divider voltage should be linear and accurate. <S> Otherwise, circuit will inject aliasing noise , CM noise and offset input result in inaccurate results. <S> Use a good signal or DAC sawtooth voltage to calibrate linearity , gain and offset.
| Yes, you do need to consider the input impedance and probably more importantly you'll need to put a capacitor across the raw input to the ADC - if you are planning on using high impedances, a lot of ADCs that are built into MCUs will take a little glitch of current that will affect the accuracy of the reading
|
Advantages of DDS over arbitray function generator DDS and arbitrary function generator are digital function generator, and both of them are generate sample from memory, so how to differentiate DDS and arbitrary function generator in term of the operation? and waht is the advantages of DDS over arbitrary as DDS quite popular right now. <Q> They are the same thing. <S> A DDS generator without arbitrary waveform capability will have a fixed waveform lookup table stored internally. <S> A DDS generator with arbitrary waveform capability will have a programmable lookup table. <S> The difference in pricing of test equipment is mostly historical, i.e. they can charge you extra for the arbitrary waveform function. <A> The DDS and AWG are almost 2 different thing. <S> AWG works on the principle of sampling while main concept of DDS is fetching waveform from memory. <S> AWG can generate complex waveform (LTE, xxxQAM...etc). <S> DDS almost always deal with simpler waveform. <S> Complex signal which are more common includes chirp signal. <S> (It is technically possible to generate complex signal with DDS for special purpose application, but this is not too common for some reasons) <S> One big advantage of DDS is inherently, there are phase continuity and repeatedbility after frequency hopping. <S> This is impossible for analog PLL synthesizer. <S> While it is possible with AWG, it is much more painful to implement. <S> There is memory within DDS consisting of sine wave (in fact it is just a quarter of sine wave). <S> By varying how fast you move along the sine wave, you get different frequency. <S> (Try to explain as simple as possible) <S> Above is a block diagram in Keysight website describing DDS of Keysight UXG Signal Generator . <A> Consider a part like the AD9910, in its most common use it runs as a DDS outputting a sine, but it does have on chip ram and there is are ways to configure it for arbitary waveform output. <S> I would note that AFG usually has a much bigger ram to allow lower frequency synthesis as a DDS <S> will rely on the phase wrapping to generate low frequency outputs which does not work if you are for example trying to generate something complex at low frequency because the temporal aliasing only works for a sine wave. <S> If you have a 1GHz clock and say a 10 bit (1K word) LUT you can generate arbitary waveforms down to 1MHz, but for anything lower then that you will be limited because the clock is actually too fast (And the reconstruction filter wrong if you just slow the clock). <S> Such a small LUT is fine for sine generation (Especially if you also do some taylor series correction) but is a serious limit in an AWG.
| A DDS generator without arbitrary capability is a little cheaper, but these days the difference is quite minimal or in some cases non-existent.
|
heat air with nichrome wire 108 rows of grapevines, each row 650 feet long. Suspend a nichrome enameled wire along each row in order to raise the temperature of the air within 18 inches of the wire 6 degrees Fahrenheit. What kind of transformers and controllers? 220 V AC is all that is available. No three phase. <Q> Let's assume you can surround your 4-hectare field with <S> R-1 (SI units: <S> 1 K-m 2 /W) <S> insulation <S> (e.g., approximately 1 inch of foam board insulation), top, bottom and sides. <S> That's a total of 81600 m 2 of heat-transfer area. <S> You want to raise the temperature by 6°F, which is 3.33 K, so the heat flux is over 270 kW. <S> You'll need almost 1200A @ <S> 230V, or about 11A per row. <S> That would be a resistance of 230/11 = <S> about 21Ω per row. <S> You could get this as a single strand of 3.6mm Nichrome (roughly AWG 7) or as, say, 8 strands of 1.27 mm copper (AWG 16). <A> What you are asking for makes no sense just from a power point of view alone. <S> Do the math. <S> Just to pick something, let's start with very conservative wind of 1 mile/hour <S> = 450 <S> mm/s. <S> If the wire is supposed to heat a volume of air within 18 inches (460 mm), then each second for each meter of wire, it must bring a volume of 450mm x 460mm x 2 <S> x 1m = 0.41 <S> m 3 up to temperature (the factor of two is because the 18 inch range extends on both sides of the wire). <S> The density of air at 0°C is about 29 g/mole = <S> 29 <S> g / <S> 22.4l = 1.30 g/l <S> = 1.3 kg <S> /m 3 . <S> That times <S> the 0.41 m 3 from above means 531 grams of air need to be heated per second. <S> The heat capacity of air at 0 <S> °C is 1 kJ/kg°C. <S> (1 kJ/kg°C) <S> x (0.531 kg) <S> x 3.3°C = <S> 1.75 <S> kJ. <S> That's how much energy 1 meter of wire needs to put out every second, or 3.19 kW, just to keep up with a 1 mile/hour wind. <S> You have over 21.4 km of wire that has to put out this much power, so that would take about 37.5 MW. <S> 37.5 MW is clearly absurd, and that is only to deal with 1 mile/hour air movement. <S> Note that this has nothing to do with how exactly the air is heated, whether from a wire or something else. <S> That's the power the heater has to put out regardless of the mechanism. <S> As Dave Tweed pointed out in a comment, this calculation is assuming the wind sweeps away the warm air to where it is of no use anymore. <S> Some of the warm air will be pushed to neighboring grape vines, so the power isn't totally lost. <S> However, due to the fact that warm air is less dense, this warm air will generally rise. <S> This will cause turbulences and up and down drafts in unpredictable locations as the warm air rises and cold air from above falls to replace it. <S> The exact behaviour is impossible to predict, but that hardly matters here. <S> Remember that this calculation was based on a very slight wind. <S> Even if somehow 3/4 of the heated air stays within the grape height per second, that still requires 9.4 MW heating power, which again, was only for a 1 mile/hour wind. <A> Lets pick a Nichrome wire and work through this. <S> The problem is going to be that most wire has data related to heat starting at around 400°C. <S> So data down in the lower temperature regions is hard to predict. <S> 26 AWG Nichrome <S> 2.67 \$\Omega/ft\$ 400°C at 1.58 <S> A <S> 650 ft x 2.67 \$\Omega/ft\$ = <S> 1735.5 <S> \$\Omega\$ \$^1\$ <S> Max current is 220V / 1735.5 \$\Omega\$ = 0.127 <S> A. <S> 108 rows is 13.7 A of power. <S> Not completely out of the realm of possibility. <S> It could get warm. <S> It would make your power meter spin, but maybe. <S> I would hate to have to purchase 650 ft x 108 rows = <S> 13.3 MILES of nichrome wire. <S> Probably not the least expensive thing to buy. <S> Now the practical issues. <S> If you were to tent over all your grapes, there is a possibility of the heat staying within 18" of the wire. <S> Otherwise, the air will heat up and lift into the sky as cold air replaces it. <S> This is general convection. <S> I would guess that tenting with some other type of heated air blown solution makes a whole lot more sense than just stringing a wire. <S> It may be possible, but it just doesn't seem like a good solution. <S> \$^1\$ I excluded the resistance of the return wire. <S> It would make most sense to use nichrome to return and actually use a 1300ft run, so the resistance in the return wire is at least useful. <A> Canopy solutions for over 12 miles of vines is not trivial but effective with gas fired hot air blowers. <S> Machine pruning lowers risk of frost damage if no buds yet. <S> Micromist sprayer Sprinkler solutions offer best economical solutions but must ensure heat absorption by freezing water exceeds heat loss by evaporation. <S> Wind blowers may include helicopters. <S> Low dew point can further exacerbate this problem. <S> To avoid damage under these conditions, sprinklers should be started at 1.1°C (34°F) if the dew point is -4.4°C (24°F) or above; 1.7°C (35°F) if the dew point is -6.7 to -5.0 <S> °C (20-23°F); and <S> 2.2°C (36°F) if the dew point is -9.4 to -7.2 <S> °C (15-19°F). <S> This recommendation should only be followed when a frost is predicted. <S> Sprinklers may be turned off when the air temperature has risen to 1.1°C (34°F).
| Electric heaters require significant power but may be beneficial to heating water conduit lines with controls..
|
What are Guard Rings? I have heard about Guard Rings many times, and I know they are supposed to avoid currents in places where there shouldn't be no currents, but I never found a good text to read more about it. Can someone describe them properly or please recommend some material to further reading? <Q> From my answer to an earlier question : A guard ring is traditionally used to protect high impedance nodes in a circuit from surface leakage currents. <S> The guard ring is a ring of copper driven by a low-impedance source to the same voltage as the high impedance node. <S> This would typically be the input pin of an op-amp. <S> Here's an example of a classic guard ring layout for a metal can op-amp from National Semi's AN-241 <S> : <S> The way this works is, say there's a low-impedance node nearby, like V- in the picture. <S> Current can't flow from V- to the susceptible input pins, because it will reach the guard ring first, and be consumed by the source that's driving the guard ring. <S> At the same time, the guard ring won't drive any leakage current of it's own onto the susceptible node, because it's kept at a very similar potential. <A> When using triax cable the outer braid is connected to ground and acts as a conventional electrostatic shield. <S> The inner braid is the guard and will be driven by the measurement equipment to approximately the potential of the input signal on the inner conductor. <S> That minimizes leakage currents involving the input signal since the surrounding material is very close to the same potential. <S> It also minimizes the effects of coupling noise via the cable capacitance, including some microphonics, by maintaining both sides of the capacitance that ties to the input signal at the same potential. <S> An excellent reference for small signal effects is " Low Level Measurements Handbook " published by Keithley Instruments . <S> It's available from their web site or you might be able to talk a friendly rep out of a paper copy for frequent reference. <A> Rather than shunting common mode noise to ground where coupling capacitance can draw current on AC signals, Guard Rings use the output signal to shunt E fields nearby by eliminating the current flow with no voltage difference to the input, thereby reducing the effective capacitance of stray noise and also reduce loading effects of voltage drop from a Capacitive load. <S> So think of guard as a method to shield and reduce load capacitance effects which induce noise voltage from grounds. <S> Ground noise coupling to signal is significantly reduced with guarding. <A> Guarding is also used on ion chambers for radiation measurement to prevent surface current leakage at the connector. <S> There is an innermost signal connection, middle guard and outermost connection. <S> The inner connection is to a wire inside the chamber and the outer is to the body of the chamber. <S> When a few hundred volts of bias is applied between those 2 conductors then surface and cable leakage can easily swap the nanoamp signals of interest. <S> The solution is to drive the guard conductor to the same potential as the inner conductor. <S> Leakage current only flows from outer to guard and the signal on the inner is protected. <S> It's pretty easy to get less than a picoamp of leakage with a hundred volt bias (it is, however, pretty tricky to measure that leakage accurately). <S> The Keithley handbook is a great reference on the subject and a lot of other things that can cause errors when measuring small signals.
| Guard rings are analogous to the guard conductor for triax cables.
|
Significance of GND in bridge rectifier simulate this circuit – Schematic created using CircuitLab In my book, this diagram is used to explain how a bridge rectifier works, in positive and negative half cycles. But it does not explain the significance and necessity of the ground in the circuit. What is the significance of the ground here? If we consider ground's voltage as zero voltage (generally what I do in circuit analysis) and for any half cycle (just say positive half cycle), there will be 0.7V voltage drop across D1. Doesn't it make positive half cycle start from -0.7 volt? It seems odd to me. <Q> The ground is not necessary and has no significance in reality. <S> The wave form for each half cycle will be offset by the voltage across the diodes, as current will only start to flow then <S> the voltage at the low-voltage side of the transformer exceeds two diode drops. <S> There will be a small gap of zero potential ( V=IR across the resistor ) across the resistor when the AC crosses between positive and negative. <S> At this point, the voltage of the more positive wire of the low-voltage side of the transformer will be increasing from zero to say 1.4V wrt the other wire of that side of the transformer. <S> The voltage dropped across the diodes increases until they start to conduct then remains constant ( to a first approximation ), they don't drop 0.7V if you only apply 0.1V to them. <S> So the half cycle starts at 0V <S> then stays there until 1.4V is applied to the bridge, then tracks the sine-wave less the 1.4V diode drops until it hits zero again, then stays at zero until the next AC crossing. <A> Without a ground (or some path such as a 1G resistor to ground) <S> a SPICE simulation will fail, and it also provides a reference voltage for discussion. <S> The node that is shown as grounded is the one that would normally be grounded in a circuit with a positive supply voltage. <S> Say a \$\mu\$A7805 is used as a regulator, that's where the 'common' terminal is connected. <S> The 'ground', outside of simulation world, is just a shorthand for a common connection, but of all the possibilities, this one is the most sensible in terms of something that is likely to be common with a lot of other points. <A> I think in that specyfic case ground is present on that drawing to show <S> where is point considered at 0V on other related drawings (like voltage vs time graph). <A> Here, I built the circuit which carries resistive load. <S> If we put GND to the bridge side, we get full-wave rectified outputsignal. <S> If we put GND to the source voltage side, we will see dividedalternances. <S> Positives are on the top, negatives are below located. <S> To me, I generally use GND to bridge on my univ. <S> projects. <S> :)
| If the ground is not present, the lower voltage end of the resistor will float with respect to the AC supply, which is fine for most cases.
|
sensors for detecting breathing movements I am looking to find out what sensors can i use in my wearable, to detect if someone is breathing.. we would also like to find out their breathing pattern if it is long deep breath or shallow breath. <Q> There are two main methods you can use -- impedance plethysmography ( http://www.bem.fi/book/25/25.htm ) or a simple belt transducer ( http://www.thoughttechnology.com/sciencedivision/pages/products/respiration.html ) -- but search on "respiration belt sensor" for a variety of methods to do this. <S> Some use strain gages, some use pressure in something like a cuff. <S> The first time I've ever seen this described was simply a rubber tube filled with mercury, but you don't want to do that!! <A> Heart speeds up during inhale and slows down during exhale. <S> To get more information - you can search the web for "respiratory sinus arrhythmia biofeedback". <A> Borrowing @tcrosley's comment (+1) suggesting measurement of thoracic expansion, here <S> is a thesis in which the candidate came up with a home-made optical strain sensor for just this purpose. <S> This was the voltage output: <S> I would think a small load cell, an LVDT, or a capacitive sensor would be better, but he got this one to work well enough to finish his thesis.
| Breath rate can be extracted from heart rate from ECG signal, so wearable ECG sensor may be used.
|
Battery chargers that can recharge disposable batteries. How do they work and are they safe? A few years ago a product called SecondWind appeared, which claimed that it could recharge disposable batteries. To a layman like me that sounded like a great idea, but does such a product really work, and is it actually safe to use? The labels on the batteries promise nothing but certain death if you try to recharge them, and a charger for disposable batteries seemed like something that might be potentially hazardous. <Q> What happens when you recharge any kind of battery? <S> current flows into the + terminal <S> a chemical reaction occurs inside the battery, more-or-less the opposite of the chemical reaction that occurred when the battery was discharged the chemical reaction recharges the battery and generates heat <S> What can go wrong? <S> the battery can go into thermal <S> runaway and overheat <S> , sometimes explosively the battery can develop an internal short and draw excessive current, which can lead to thermal runaway or maybe just wreck your charger <S> the battery can develop an internal open and become a paperweight <S> How to prevent the wrongness? <S> monitor the battery temperature (internal if you can, otherwise external) and reduce current if the temp gets too high charge really slowly ( <S> trickle charge) do not overcharge <S> What else? <S> This overcomes the voltage drop in the internal resistance of the battery. <S> What do commercial chargers do? <S> Commercial chargers typically do some combination of constant-voltage and constant-current charging, the slightly fancier ones monitor temperature, and the really fancy ones may also monitor other things. <S> Why are they not recommended for alkaline batteries? <S> Each type of battery recharges slightly differently. <S> Old ni-cad chargers may not work well for ni-mh rechargeables, and vice versa. <S> You have to buy a special kind of charger for li-po batteries, lead-acid types, etc. <S> The real reason that alkaline rechargers are a niche item is that you don't gain much (in terms of recharge capacity or number of recharge cycles) from recharging alkaline batteries compared to other types. <S> The moral of the story is you can recharge most any kind of battery, but some are better at it than others. <A> I had a low-stress application for reusing alkaline cells (low current, didn't necessarily need a full recharge), so I tried it myself [original research] although [this ain't Wikipedia]. <S> The first thing I discovered was that the available information is so dead-set against trying this, that it is very difficult to determine what the proper terminating voltage would be. <S> I used a bench supply <S> so I could limit both the current and the voltage. <S> On the first try, the results were good. <S> On the second try, the cell leaked. <S> After that, I intentionally kept the charge current to C/10 or less, using cells that had never been recharged, and that were not completely exhausted. <S> After some period of time, the cells began expelling gas and electrolyte. <S> The gas leaking was audible (as in, "What's that sound?", and "Oh, it's the cell I'm recharging") <S> As an aside, cells that are intended to be recharged often have mechanisms targeted at absorbing the gas, or at least safely venting it. <S> Not in this case. <S> Conclusion: <S> The chemical reaction isn't necessarily "reversible" in an exact sense. <S> When run backwards you get other things like heat and gases. <S> Electrolytic reactions involve things like dissolving metal, and the resulting corrosion products don't automatically know where to go back to. <S> In the end, I quit trying to recharge these. <S> No deaths were reported. <S> I believe these results will vary with brand (manufacturer), so if it worked for you, you were just more lucky. <S> But this is an example of why it may not work. <A> Single-use alkalines aren't guaranteed to be rechargable. <S> Why not buy actual rechargable batteries which keep working even over a large number of recharge cycles? <S> It's much cheaper over the long term. <S> I've recharged lots of 9V alkalines for use in DVMs. <S> Trickle-charge 20mA with a cc power supply set to 9.5V max. <S> It doesn't work well with deeply-discharged batteries below about 7.5V. <S> But you can get some extra life from 9V alkalines if they've only been run down to 8V or 8.5V or so. <S> So, probably a commercial charger will accomplish this without tying up your bench supply. <S> Yes, sometimes a cell starts venting gas, little popping sounds. <S> I've only seen this on deep discharged ruined batteries where the voltage remains way low despite all "recharge. <S> " <S> There's one good application for all of this. <S> If you score a huge box of "dead" 9v alkalines, you can raise the voltage most of them. <S> Then hook them all in series and use them as woodburners, electrostatic kilovolt supply, carbon arclamp demos, etc. <S> Or just competing on YT/reddit for highest unwise 9V-bank voltage evar. <S> Get yours from indie theaters FM microphone discards, or office buildings replacing smoke detector batteries. <S> 244 batteries in series <S> https://www.youtube.com/watch?v=8hwLHdBTQ7s 490 batteries in series http://www.break.com/video/ugc/490-9v-batteries-351064 <S> If you have some copper sulfate or copper chloride solution, you can demonstrate some issues behind "recharging." <S> Recharging is basically a form of electroplating, of converting the corrosion products back into solid battery plates. <S> Stick a couple of small copper electrodes into your bluegreen solution, then crank up the power supply to high current. <S> Fast-growing clots of black goo! <S> That's dendritic copper growing on the negative electrode; a forest of nanofilaments. <S> Perform the plating at much lower current <S> and you instead get bulk copper metal. <S> Battery design has similar issues: the physical structures produced by the "corrosion" and the "plating" must not degrade over many cycles, and deep discharge (excessive 'corrosion') must not ruin the battery or reduce the number of charge cycles. <A> There is a blanket safety warning , but chargers with a limited capacity of C/10 would not likely result in overheating losses and take more than 10 hrs to trickle charge. <S> You might be lucky to get out 50% of charge put in up to 90% SOC on the 1st re-use , but this declines rapidly with temperature rise and cycle count on primary cells. <S> Using chargers >= <S> 1C charge rate could result in rapid temp rise should an internal short occur. <S> Monitoring battery temp. <S> is safe if done in real-time.
| In order to "fully" recharge a battery, the charger must actually exceed the rated voltage of the battery (briefly, and at a low current).
|
PCB design consideration for a SMA antenna I am designing a PCB with the transceiver SX1231 (868MHz) with a SMA connector for the antenna. I don't know much about antennas. I would like to know if there is some parameters I need to be careful to keep the circuit adapted (to reduce reflections of the signal's power). In another PCB that a friend developed, I see a lot of Vias (a bunch of them, one after the other) around the pads of the SMA connector, what is the reason of this? Do I need to consider the width of the traces that connect the capacitors and inductors between the transceiver and the SMA connector? How can I calculate this? Do I need to take in consideration the material of the PCB? Is there something I need to take in consideration about the ground planes? Thanks for your help. <Q> You need to learn about fixed impedance traces. <S> A good resource for calculating the trace width for impedance is Saturn PCB Toolkit <S> Most SMA connectors have pins for surrounding the signal with ground plane, or SMT versions have recommended via surrounds. <S> The idea is to surround the signal in ground, just as the outer braid does in a coaxial cable. <S> FR4 PCB material should be fine with the frequencies you are talking about. <S> Designs are working on FR4 with no issues quite a bit above 1 GHz. <S> The important thing about not generating reflected signals is reducing impedance mismatches as much as possible. <A> Here is a useful online tool <S> but there are a few others: <S> - The results show an impedance of 50 ohm with a track width of 2.92mm over a ground plane 1.5mm away. <S> PCB material has a permittivity of 4. <S> These are all values I've entered and you can do the same. <S> Keeping impedances matched reduces reflected power. <S> Vias have an inductance that can be significant at UHF (and above) <S> and so when connecting a top track to a ground plane, several vias are used to "parallel" the inductors and reduce the effect. <A> Another thing to consider, beside impedance, is thermal relief for through hole connectors. <S> Make sure the ground vias have such relief for good solder.
| You should consider making your track impedances match your antenna.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.