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Can vias and traces conduct electricity on PCB surface? I'm currently designing my first PCB, though when I place a microSD reader on the board I get errors in eagle that the area the reader is over is restricted. I would have thought this would only apply to components on that side of the board itself, but I get errors when I route traces and vias underneath the reader. So is it possible that traces or vias could conduct electricity through the microSD reader (some bare metal may touch the surface) that might possibly touch the PCB's surface? <Q> Even with solder mask, you should have some additional insulation between conductive material and the board. <S> In most CAD programs, you can define "keep out" areas when you build a footprint, so that the Design Rules Check function of the PC layout program will warn you if you place tracks where you shouldn't (or even prevent you from placing those tracks.) <A> It is not good form to run traces under the slot area even if you have a solder-mask layer. <S> You would be relying on the solder-mask to insulate the board from whatever is plugged into the slot. <S> If it is a high-turnover application (frequently plugging and un-plugging), it could wear down the solder-mask and expose the traces. <S> Not recommended. <A> The other answers are right, a metallic part placed directly on the PCB could short traces, sharp edges could also cut traces. <S> The solder mask is a lousy protection because the part can easily rub through. <S> But since you are a novice to EAGLE: In the list of layers, you'll find keepout <S> and restricted layers which are not visible by default. <S> restricted areas will throw a DRC error when any copper (trace / pad / via) is places in them, as you already experienced. <S> They are usually created as filled polygons and only used when necessary. <S> keepout areas will throw a DRC error when they overlap with an other keepout area. <S> This is used to indicate the outline of a part, and to prevent you from overlapping parts. <S> The area can also be a bit larger than the part itself to prevent parts from being placed too close to each other. <S> Of course, you can draw your own polygons in that layers, e.g. to mark areas with too low headroom to place parts. <S> But unfortunately, most parts come without polygons in keepout <S> and you can easily place them in your keepout area :-(
| If the board does not have solder mask covering the tracks, then any metal or other conductive material placed on the board may contact the tracks and cause a short.
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What is the difference between frequency response and transfer function? I would like to understand the difference between frequency response and transfer function. I know the former can be obtained by substituting \$s = j\omega\$. But what is the difference in information I can get from both representations? What are the respective limitations and where do I apply which method? I'd also be glad for some literature recommendations. Could someone explain the calculations of the second answer (by Chu) a little more extensively? I don't quite get how he determines the values of \$ \phi \$ and X, and how he compares it with setting s equal to \$ j\omega \$ in the transfer function. <Q> A circuit's transfer function is a fully mathematical model that can be used to derive the frequency response and phase response (both together are called the bode plot). <S> However the same isn't true in reverse - you can't always derive the TF from the bode plot. <S> Sometimes you can but not always. <S> So, the frequency response is a subset of the bode-plot and <S> the bode-plot is a subset of the transfer function. <S> Hopefully this picture will help: - Along the top <S> are three bode plot views of a typical frequency response for a 2nd order low pass filter. <S> Bottom left is a 3D view of what lies behind the frequency response - in this example there are two poles (only one shown to make it easier on the eye). <S> Bottom right is the standard pole zero diagram and this 2D diagram alone embodies the transfer function. <S> So, if you look at the 3D picture and imagine viewing from above, you get the pole zero diagram at bottom right. <A> Take, as an example, a sinusoid, \$\small \sin(\omega t)\rightarrow \dfrac{\omega}{s^2+\omega^2}\$, applied to a simple first order lag, \$\small <S> G(s)=\dfrac{1}{1+s}\$. <S> The response is: \$\small R(s)=\dfrac{\omega}{(s^2+\omega^2)(1+s)}\$, and this can be expressed in partial fractions: $$\small \frac{\omega}{(s^2+\omega^2)(1+s)}=\frac{A+Bs}{(s^2+\omega^2)}+\frac{C}{(1+s)}$$ Inverse LT <S> gives:$$\small r(t)=\frac{A}{\omega}\sin(\omega t)+ B\cos(\omega t)+Ce^{-t/\tau}$$ <S> The exponential term decays to zero, leaving the steady-state response as: $$\small \frac{A}{\omega}\sin(\omega t)+B\cos(\omega t)= X\sin(\omega t+\phi)$$ Solving for \$\small X\$ and \$\small\phi\$ gives \$ \frac{1}{\sqrt{1+\omega^2}}\$, and \$\small \arctan{(-\omega)}\$, respectively, as is obtained using \$\small s\rightarrow j\omega\$ in the Laplace TF. <A> They are very similar concepts. <S> The transfer function is a relationship between an output and an input of a linear system. <S> The frequency response is how some characteristic of a linear system varies over frequency. <S> The thing that varies might be the transfer function. <S> But it might be something else, like the input or output impedance. <S> It might be the variation of something on a system that doesn't have a distinct output and input, like a one-port network.
| The frequency response is a special case of the Laplace transfer function where the transients are assumed to be completely dissipated, leaving the steady state sinusoidal response.
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Can we transmit FM signal using AM band and AM signal using FM band? What I am asking is if it is possible to transmit a frequency modulated wave in AM range(550-1650 Khz) and similarly to transmit an Amplitude modulated wave in FM band (88-108 Mhz). I would like to know about the technical aspects of the problem like what problems we may face if try to do so. <Q> AM now supports Quadrature Stereo AM but still fits within 10Khz channel spacing with some guardband. <S> FM cannot fit into the AM band , wwhile AM is even used in the ISM band at 928MHz with 6kHz channel spacing, so there is no reason why AM cannot be used at any frequency. <S> FM has higher SNR and signal bandwidth that prevents application in the lower AM bands. <S> However one can use AM or FM for cable communication at any frequency. <S> Can you see how to fit the above baseband FM signal in the 10KHz channel spacing of the AM band? <S> If you do it's worth a ton of money. <A> You could do AM on the FM band .You will be able to squeeze in many more channels <S> but you wont get the SNR advantages of FM .If <S> you tried to do wideband FM on the standard MW AM band the number of channels would be very low .Also <S> the type of Atmospheric fading on the MW band would make the stereo music sound terrible . <A> The reason that AM mode is used on the lower broadcast band and FM on the higher broadcast band is by reason of government regulation as well as signal propagation issues. <S> The broadcast FM band in the '40's was originally around 40 <S> Mhz the got moved to 88-108 Mhz band. <S> Depending on the application a multitude of modes are used. <A> There is no technical aspect. <S> The aspect is standard equipment and frequency licencing. <S> Waves themselves know nothing about it, so you can construct your own equipment to use any band you want. <A> There is no technical reason you couldn't do what you describe. <S> In fact, there is at least one system for AM stereo that is based on modulating the carrier simultaneously with AM and FM. <S> However, depending on what jurisdiction you live in, there may be regulatory issues. <A> AM and FM waveforms: - <S> In case 1 you move the amplitude up and down a bit. <S> In case 2 you wobble the frequency a bit. <S> This can occur at any point in the radio spectrum. <S> Both can occupy the same amount of spectral bandwidth. <S> Narrow band FM is scattered all over the spectrum and, for low bandwidth speech signals is perfectly OK. <S> Wideband FM used in broadcasts of stereo music occupies a large chunk of bandwidth hence it can't really be used in the traditional AM broadcast region of the spectrum.
| Technically any mode of transmission is feasible on any frequency if bandwidth is available.
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Choosing the right power regulator for Battery powered designs I've come across this problem a number of times in my hobby projects, but I am sick of making an educated guess. I often do microcontroller projects and often want them to be powered by a 3.7V Lipo cell charged by a standard microusb cable. This means input voltage can range from 3.0V to 5.0V and I want an output voltage of 3.3V. With thousands of voltage regulators that can fit those requirements, how do you choose? I can ask about each specific project to get the right regulators, but I would rather have the knowledge I need to find the right regulators on my own. I'll add edits as I find more answers. Edit1 : Switching Regulator - This is the only choice if you need to boost voltage. They are the most efficient and produce less heat than LDO's, but produce noise not usable with RF applications including bluetooth and wifi and are generally more expensive. If you want to use this with RF then you will need proper filtering. LDO - LDO's are cheap and are preferred for RF applications because LDO's produce no EMI interference. They generally have poor efficiency but the efficiency depends on the ratio of the input to output voltage. The closer the two are, the more efficient the output will be. In high current applications they can produce a lot of heat so proper cooling may be required. Charge-pump - A subset of switching regulators that don't require an external inductor. They generally have worse efficiency than inductor-based switching regulators but are better than LDO's. They can also only output relatively small amounts of current. These should be used when you need better efficiency than an LDO but are constrained on board space. Hybrid - There are a few companies who make hybrid Switching/ldo regulators. These provide the efficiency of a switching regulator with the noiseless output of an LDO. The downside is there are very view of them and they don't have high output current in most cases. These still produce heat like an LDO. Quiescent Current - In battery powered applications you need to be concerned about the amount of current the regulator uses to function. The more power it uses, the less battery power is left for your device. LDO's generally have much smaller quiescent current than switching regulators, but since switching regulators have the ability to boost voltage when input voltage drops below desired output voltage, which to use depends on your devices voltage requirements. <Q> Battery powered projects (particularly those with periodic events spaced quite a bit apart) usually benefit from using a linear regulator. <S> Looking at your requirements (LiPo 4.2V to <S> Vo + dropout voltage) <S> a linear regulator will be (on average 3.7V battery, regulated output 3.0V) 81% efficient which is close to the SMPS solution anyway. <S> A switch-mode device uses a lot of power (relatively speaking) just to power itself up; it is overall more efficient where a relatively large load current is being used. <S> What you need to look at carefully is \$I_q\$ for the regulator; this is the amount of current it draws simply to power the internal circuitry. <S> (there simply is insufficient physical packaging to permit it). <S> As most modern microcontrollers and many interfaces will run quite happily at 3.0V or even 2.7V, you should consider using that as the regulated voltage. <S> If you really want Vo = 3.3V for Vin 3.0V <S> <> 4.2V you would need a buck-boost or some such arrangement. <S> I would stick with a simple LDO regulator and drop the regulated voltage. <S> Note that regulators designed for large load currents (not the usual case for battery operated projects) will typically have a relatively high \$I_q\$ <S> The most astounding part I have ever seen for \$I_q\$ : <S> TPS783xx <S> Compare that with a buck SMPS <S> that has a relatively low \$I_q\$ (for an SMPS device). <S> There are other considerations, but the no-load power dissipation of the regulator needs to be kept to a minimum. <S> Incidentally, a Li+ / LiPo can be safely discharged (in my experience) to about 2.7V <A> It is not recommended to drain a lipo below 3.0V. <S> see: <S> https://learn.sparkfun.com/tutorials/battery-technologies/lithium-polymer <S> If you stayed above 3.3V you could just use a linear regulator if you don't mind efficiency too much. <S> Otherwise, if you really want to go as low as 3.0V or be more efficient, you have to use a buck/boost converter. <S> They usually have a very high efficiency of around 90%. <S> The following pdf has a design example on page 10/11:Uout = <S> 3.3VUin_min = 2.6VUin_max = 5.5VImax = <S> 2A http://www.ti.com/lit/an/slva535a/slva535a.pdf <S> If you use proper filtering the switched power supply should not be a problem for your bluetooth module (or do you have any other RF circuits on your board?) <A> Yes it is possible for switching power supplies to generate noise, but at low power levels filtering the output is not that hard. <S> I am using switching power supplies in a low-power radio appliation <S> and I am having very good results. <S> With just an RLC filter on the output of the regulator I don't see any RF interference problems. <S> The regulators maintain the output voltages while only drawing micro amps of quiescent current when the board is put to sleep. <S> The regulators work with inputs from 1.8V to 5.0V and can automatically switch between step-down and step-up mode making 2.2V and 3.3V used on my board. <S> The supplies I am using are from Texas instruments. <S> Go to www.ti.com. <S> They have a power supply design tool called web-bench that will design power supply solutions for you within minutes.
| The regulator with the lowest \$I_q\$ that can handle the input and output voltages and required current is usually the best, but temperature rise is something to be considered - the small devices cannot dissipate much heat
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Common cc/cv charger for same volt and different capacity Can I use Li.ion 7.4v 5400mah in place of built in Li.ion 7.4v 1400mah battery of a portable DVD player and use same old charger? <Q> There is theoretically and practically. <S> Theoretically if the chemistry matches then it should be fine. <S> Practically there are more pitfalls than one could count. <S> Is the battery smart? <S> If it is smart then the internal communication to the battery may not match. <S> If the charger is smart then it may know that it exceeded the capacity of the battery that it was designed for. <S> This would cause the charger to shut off with the battery half charged. <S> If the internal battery fuel gauge is smart it may be confused by the higher battery capacity. <S> The connectors may not match. <S> The new battery is going to be larger than the old causing space constraint problems. <A> I agree with @vini_i as far as the pitfalls go. <S> I would not use it with a smaller battery however, as it could charge the battery faster than intended, causing undue strain on the cells and perhaps "rapid disassembly" of the battery. <S> It all depends on the charge current that the cells can take and what ratio to the capacity it is for the specific chemistry. <S> There are always outliers <S> however I would conclude that in general what you are trying to do should work, aside from the possible smart components preventing it. <A> Thank you very much vini_I ,Winny and Jared Good for sparing time to give valuable answer to my question. <S> Now iam confident, I will connect two new batteries in series duly removing the protection board and taking centre tapping (for balance charging) and connect with the pcb as done with original battery. <S> DVD's charger is 12v, 2A rated and player is having LED indication ON for charging OFF for fully charged. <S> Thank you once again. <S> Any suggestions/remarks thankfully welcomed.
| At the core of the question though, if you are using a cc/cv charger with a larger battery than it was designed for you will be fine, even possibly with different chemistries, but not all, as long as the voltage is the same, it will just charge slower.
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Low-power alternative to Arduino, for simple project I have a really plain Arduino project: ultra-sonic sensor HC-SR04 returns distance which is then used as sleep time during a LED-wave effect. So it's a simple distance <-> LED-wave velocity relation. The closer you walk towards the device, the faster the LEDs. That's all. The current being drawn when not running LED-wave and just checking 5 times per second if the distance is lower than 2 meters: 15 mA. This value is in my opinion huge. I then heard about ESP-8266. It can fully replace Arduino. However the power levels are still not good enough. Browsed more, read about Teensy and others, and they all draw power like a crazy – it appears like this if only a person thinks about using batteries for a moment. I think that professionals know the solution. Hobbysts are stuck in very fun Arduino world. Can someone reveal the truth? There must be an answer as e.g. computer mice can run months on two AA. Or, for example, a shopped motion-detection-triggered lamp nicely worked for 7 days before I got bored with it and have put it into drawer. The lamp was giving actual light , not LED blinks, and didn't drain the three AAA batteries it used. What can I use for my plain project? Need ping functionality for HC-SR04, 8 pins for LEDs and that's all. How to do this correctly and have battery lasting for months? <Q> The Arduino has regulators and USB-Serial chips that are not Sleep friendly. <S> If you look at the Datasheet for the ATMega328 <S> (section 32.2.2) <S> however you will see that Sleep states exist down to around 4-8 uA <S> (wake on WDT) for the MCU. <S> that has only the ATMega328 (or other AVR such as the ATMega8L) on it. <S> You could run from 3 AAA batteries without a regulator at all providing you don't need to support higher voltage LEDs. <S> There are many resources that show how to program an Arduino bootloader environment into a standalone AVR MCU. <S> You could start here and here . <A> The chip used in the Arduino Uno (Atmega328P) can certainly do low-power. <S> See my page about power savings for more details. <S> My base expectation is that if you don't need a timer (eg. to do something periodically) you can get down to 100 nA of current, and still wake on an interrupt, such as a switch press. <S> If you need to wake from time to time (using the watchdog timer) your current usage will be around 6.5 µA. <S> This is way lower than the 15 mA you quote. <S> If you use an Arduino board then the on-board LED, the USB interface chip, and the voltage regulator will be the main power culprits. <S> Plus, not going into sleep mode doesn't help. <S> I made a temperature and humidity logger , see picture below, which has been running now for a couple of years taking a reading every 15 minutes and logging it to an SD card. <S> That runs from 3 x AA batteries. <S> It is designed to sleep most of the time. <S> The "peripherals power mosfet" shown is used to "power off" the on-board devices when not needed. <A> I think I've found the answer: Texas Instruments' MCUs. <S> For example this one: MSP430FR6989 has current – at 1 MHz with 0% cache hit ratio – 0.375 <S> mA. <S> For Low Power Mode 0 <S> (LPM0): 0.120 mA, for LPM1: 0.065 <S> mA. Remaining LPMs are having current 0.001-0.009 mA. <S> If software will query HC-SR04 twice a second – i.e. will sleep in a LPM for 500 ms after each complete action set – then <S> I think I will obtain the desired effect of months on batteries.
| Consider simply building your own Arduino clone (if that's the programming environment you want to use)
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neutralize charge in series capacitors I have a simple circuit as in the picture below. My question is why the negative charge -Q at the bottom plate of C1 moves and neutralizes with the positive charge +Q at the top plate of C2? <Q> The charges in the green circle will neutralize. <S> It is impossible for them not to neutralize. <S> The forces pulling them apart and the forces pulling them together are the same. <S> The forces balance each other out and they will neutralize. <A> The area inside the green circle is electrically isolated from the rest of the circuit so there is always the same amount of charge inside the circle. <S> What charge flows into the top of C1, the same amount flows out of the bottom of C1 and into the top of C2, the same amount flows out of the bottom of C2, this is how capacitors work (or at-least a very good approximation) <S> Why does it move? <S> Electric potential pushes it.. <S> See also: <S> Kirchoff's loop law <A> If you removed the top plate of the top capacitor and the bottom plate of the bottom capacitor, that's exactly what would happen. <S> However, the presence of those charged plates creates an electric field that keeps the charges separated. <S> For the negative charge at the top capacitor to flow down, it would have to get away from the positive charge at the top plate of the top capacitor, which attracts it very strongly. <S> Ditto for the bottom capacitor.
| In fact, the attraction between the opposite charges at the two plates of a capacitor are exactly what makes it work - otherwise, you wouldn't be able to keep a lot of charge together.
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Resistor symbol in a schematic with zero value I can't figure out the purpose of the 0Ohm resistor in bellow schematic. It's crystal oscillator configuration and why the designer have included a resistor there with 0ohms ? Any idea ? <Q> it is either HIGH or LOW. <S> That low output impedance is a shortcircuit across (effectively) <S> the load capacitor connected to thecrystal at that output terminal. <S> That has the effect of damping theoscillation. <S> Mainly, modern processes allow weakly conducting CMOS inverters for thisfunction, and most application sheets do not call for a resistor at thedriven node of the CMOS gate. <S> If an unusual impedance of crystal ora different frequency range is used, either the 33 pF capacitors or thezero-ohm resistor may prove unsuitable, so it may be prudent to allowfor changing those values. <A> The chip has internally an unbuffered inverter therefore, for the circuit to oscillate, the external circuits (including the output impedance of the inverter working with the capacitor connected to that terminal), have to produce another 180 degrees. <S> If this extra 180 degress cannot be produced, the circuit will not oscillate. <S> The crystal (together with the capacitor at XTAL_IN) doesn't quite produce 180 degrees phase shift <S> so, a little bit of help is needed to actually get to 180 degrees and that's where the internal output impedance of the inverter AND the capacitor on XTAL_OUT come in. <S> If that internal output impedance isn't high enough, a little bit extra on the outside (10 ohms or so) usually tips the balance and it oscillates. <S> This resistor also governs the peak current that can pass through the crystal so if you are using a particularly sensitive crystal, damage won't occur. <S> But, the primary role is to provide a few degrees more phase shift. <S> Below is a simulation of a crystal oscillator that doesn't use the capacitor at XTAL_OUT (same effect as not having sufficient output impedance) <S> : - You should be able to see that Vout never quite attains a phase shift of 180 degrees and, unfortunately, this circuit won't oscillate. <S> Picture taken from my answer here . <S> And finally, if at all in doubt, do what the data sheet recommends: - <A> So that a resistor can be added to prevent overdriving the crystal. <S> You will probably see it included in the data sheet as an option. <A> It's much easier to change the value of a component than to add a component that there was no provision for in the design. <S> IF the problem is caught between fabrication and assembly then changing a component value is a trivial matter of changing the BOM. <S> If it's caught after assembly then some rework is needed but it's fairly easy rework. <S> OTOH adding a component where there was no provision before can mean much more difficult rework or even scrapping the boards and starting over. <S> So if you think you don't need a resistor but you are not 100% sure on that you put the 0 ohm resistor in the design.
| CMOS crystal oscillators may require such a resistor, because the amplifierthat drives the crystal (a biased CMOS inverter) has low output impedancewhen It would also allow disconnecting the onboard crystal for diagnosis (temporary)or customization (permanent) purposes.
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Different types of LiPo batteries While picking up LiPo battery for my project, I've seen that there are two different types: small and inexpensive one, like this one here that can be charged with a simple circuit like this one . And there are big ones like this one here that require special expensive charger like this . Why is it like this, if the internal chemistry workings are the same? Why do they need special chargers when the unit cell has the same nominal voltage? <Q> The difference is quite simple. <S> The first battery is a single cell. <S> Single cells are quite easy to charge. <S> Also the battery shown is quite small. <S> The second battery is a series pack. <S> In this battery several cells are connected in series. <S> This is where the problem comes from. <S> Also these packs are much larger making them more expensive. <A> Batteries first: <S> The cheap battery is a single small low current cell. <S> Its rated nominal charge/discharge rate is 0.2C (it's 400mAh so that's 80mA). <S> You'd be looking at about 5 (400mAh/80mA) hours to fully charge at the nominal current. <S> It is designed to be used in applications where it discharges at that sort of rate as well. <S> The expensive battery has a far higher capacity and multiple cells. <S> They also have significantly higher rated currents. <S> They are rated at 20-30C rates, that means applications where the battery is drained in a couple of minutes not the 5 hours of the cheap cell. <S> So very different batteries for very different applications. <S> One is for keeping a low power device running for multiple hours. <S> The other could do that but is designed for something that takes a lot of power very quickly. <S> On to the chargers: The small simple cell/charger will charge a single cell at a fixed (fairly low) rate regardless of what that rate that cell can take. <S> The more complex system can cope with multiple cells (up to 6) at multiple different rates. <S> That is adding a lot of extra complexity which means extra components and cost. <S> It also includes a housing and user controls, the mechanical parts and then assembly and testing probably cost as much if not more than the actual charger electronics inside it. <S> And since it's being sold as a product rather than a bare component has to be tested to verify it meets various regulatory requirements. <S> That adds further development cost that needs to be made up somewhere. <S> For 2.5 times the price you get something with well over twice the functionality. <A> Describing the small charger as simple is a mistake. <S> The chip on there is probably quite complex, these chargers are produced in millions for cell phones driving the price down. <S> The larger charger is much more versatile it can handle different numbers of cells and different and higher currents and has much lower production volumes.
| Cells in series need a special charger to balance them while charging.
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Are they connected in parallel? The question was to find the equivalent resistance between A and B. The correct answer is R. I too arrived at R by considering the ends of the lowest two wires equi-potent because the lowest wire has 0 resistance (and so according to V=IR, V comes out to be 0) and redrawing the circuit by taking the equi-potent points as a single point. But then when i was again inspecting the circuit, it seemed to me that the lowest two wires are connected in parallel which would make the equivalent resistance across their ends 0 since r1=rr'/(r+r') and thus the equivalent resistance between A and B comes out to be R/2. I think this is certainly not the case since the correct answer is R. So are the lowest two wires connected in parallel(which is probably not the case and if not then why? Edit: I would like to know how to proceed if the bottom wire also has a resistor of resistance R. I am unable to simplify the circuit for this. <Q> Both pairs are in series between A and B. simulate this circuit – <S> Schematic created using CircuitLab <A> Two resistances are connected in parallel if their positives are connected together and negatives are connected together. <S> In your problem here the first and third wire are connected in parallel as well as the 2nd and third wire as shown in this sketch. <A> One way to consider this is by inserting a test voltage source and finding the resulting current. <S> From there, Ohm's Law can be used to find the equivalent resistance. <S> This is the fundamental method for finding the equivalent resistance from which series and parallel are derived. <S> If you are confused whether something is in series or parallel, I recommend you use this method and analyze the result. <S> Take a look at the schematic below which inserts a 1V test source into the schematic you gave. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Using Node-Voltage method <S> , the equation at node A will be:$$\frac{1V}{2R} + \frac{1V}{2R <S> } - I_1 = <S> 0$$ <S> 1V is the voltage between node A and B. <S> The first term is the current flowing from one of the loops between node A and B. It is a resistance of 2R because regardless of the path you take, you encounter two resistances to get to node B. <S> The second term the current flowing from the other loop between node A and B. <S> The third term is the current flowing through the test source. <S> It is a negative current because I treat each current as flowing outwards from node A. Solving for I1 gives R as the current. <S> With a voltage source of 1V, this yields a resistance of R. This gives the same result as a parallel configuration, so in essence the resistors are in parallel. <S> One way to think of it is that starting at node A, the current must go through two resistors to get to node B. <S> The shorted wire provides another path to get to node B <S> but the end result is that the resistors are in parallel. <S> Note <S> : If one of the resistors connected to node B were replaced with a larger/smaller value than R, the result will be different from a parallel configuration and it will be like the resistors are not in parallel. <A> Draw an extended line through AB and fold the circuit, using this line as axis.
| If you label the resistors clockwise, you will get R1 and R2 parallel to each other; the R3 and R4 will be parallel to each other.
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what kind of three phase full wave rectifier should we use? We have the following generator: Permanent Magnet Generators. For the project we would like to use this converter . Between this generator and converter we've to convert the three phase voltage to an useable DC voltage, and we want to do that with a three phase full wave rectifier. According to wikipedia there are a few types of that kind of rectifiers. Can some of you that has experience in this field give us the right path to follow. Important is that the rectifier has to be able to handle the current that passes trough it. For the output voltage of the generator, we got the following data from it: <Q> I think this is all you need to know. <S> If you are using a Y output generator (rare) then you'd select individual Diodes. <S> here that the maximum DC current is specified for 120deg conduction angle per diode pair. <S> When it comes to 3 phase rectifiers you will also notice that there is a tendency to use high voltage avalanche diodes, so it's often cheaper to get 1000 - 1200 V devices than it is for 200 -300 V devices at any given current rating. <A> You need a three-phase, full-wave bridge rectifier similar to the one in the DB-25 box pictured in the Swea converter literature. <S> It appears to be just a 25-amp rectifier module with three input terminals and two output terminals. <S> It is bolted to the metal housing of the box for heat dissipation purposes. <S> It is probably rated at least 200 volts, peak inverse voltage. <A> 200 Volt diodes will be fine. <S> The nominal output volts is only 50 V. Normal <S> Si diodes will waste <S> say 1.1 V at say 20 Amp. <S> This is a significant percentage of your paltry 1KW. <S> Remember the bridge has 2 series conducting paths so 2.2 volts are gone. <S> In other words 4.4% power wasted. <S> Shottky diodes could be used which waste about half the volts so your losses could be 2.2%. <S> This is easy but you could do better still with a synchronous rectification scheme using low on-resistance MOSFETs which will cost more.
| If you are using a Delta generator (most common) then you could specify a block type 3 phase rectifier such as this .Notice
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Resistor in series with Solenoid Valve (for voltage reduction) I have a question regards the use of a resistor in series with a latching solenoid valve. Specifically, I have a valve that wants to see 14 Volts, but my power supply is 28 Volts. So if I put a resistance equal to the steady-state valve resistance (which is 9 ohms), then the valve will never see more than 14 Volts. However the resistor will also limit inrush current, which is not necessarily a good thing, if it is needed to get the valve moving. So my question is whether valve performance will be degraded by having the resistor in series, versus what these valves normally see, which is essentially an infinite source powering them (i.e., low series resistance)? Any help would be greatly appreciated. Thanks, Jim <Q> When given 14V into an 9 ohm load would give you 1.55A. <S> If you double both, 28V into an 18 ohm load give you the same 1.55A. <S> The DC resistance of the coil is what is going to be the limiting factor to the inrush. <S> All of that works. <S> The problem is the power. <S> The resistor is going to be dropping 14V. <S> That means that the power dissipated is going to be 14^ <S> 2/9 <S> = 21.7W <S> That is a lot of power to dissipate. <A> Instead of using a resistor you can use a DC to DC Adjustable Regulator LM2596, its very cheap and easy to use, because i think using a series resistor will cause the latching solenoid to misfunction. <A> Providing you are only pulsing the solenoid (it is specified as a latching device) <S> then perhaps you have the opportunity to speed up operation too. <S> I assume here that the 14 V specified is the continuous voltage rating for the solenoid. <S> If you look at the datasheet you may find a duty cycle limited voltage specification too. <S> There is no inrush current by the way. <S> This is a simple LR time constant (no matter how short the time), so the current rises from zero to a maximum defined by the circuit resistance. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> In the RHS schematic above the solenoid is given a 1 mS 'kick' of higher voltage. <S> This provides a higher aiming voltage and shortening the time to reach a given current through the solenoid. <S> You could increase the value of the capacitor to increase the time more than 14 V is applied to it. <S> Steady state is just the same as the LHS when the capacitor is charged to 14 V. <S> Remember that either of these cases absolutely depends on this being a pulse applied to the solenoid (in your comment you said it was 150 mS). <S> Steady state power dissipation would be very high (22 W in each). <S> Providing you know you will always pules the solenoid, the R value could probably be a 9 Ohm 2 - 5 W wire wound. <A> Inductors don't have inrush current. <S> The current thru a ideal inductor changes proportionally to the applied voltage. <S> When you first apply a voltage to the solenoid, its current is zero. <S> It will then ramp up. <S> Due to the DC resistance of the coil, the current doesn't keep increasing infinitely, but exponentially decays to the steady state value. <S> This steady state value is the applied voltage divided by the DC resistance of the coil. <S> Put another way, the current thru the coil exhibits the opposite of inrush since it ramps up over time. <S> To answer your overall question, yes, putting a resistor in series with the solenoid equal to the solenoid's DC resistance allows driving the whole thing with twice the voltage <S> the solenoid is rated for. <S> Of course you're going to use twice the power too. <S> The solenoid dissipates the same power as before, but the resistor also dissipates this power too.
| Most solenoids have a broad voltage specification, and you can overvolt them for a short time to speed up actuation.
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Solutions for holding oscilloscope probes (and freeing up hands) I have a problem, I only have two hands and the newer oscilloscopes come with four probes. This leaves me with zero hands to press the trigger button (even if I hold three probes with one hand I still can't use four). I usually have to ask a colleague to press the trigger button, in a cubicle this can get awkward. What is a specific way I can hold the probes and press the trigger button, Without biological modification? <Q> Without biological modification <S> Haha. <S> Does your oscilloscope come with probe hooks ? <S> If so, the next pcb you make, you can create some wire loops on the PCB . <S> Alternatively, you can build a DIY probe holder such as <S> this <S> (Make it open source so people with 3D printers can replicate your ultra helpful probe holders) <A> In modern electronics and multi-point probing you do not hold probes. <S> There are three options: (1) <S> If the signal is low-bandwidth (<100MHz), you solder a piece of flexible (multi-stranded) wire to points of interest, 0.5 - 1" long, and then connect (passive) probes that usually have hooks/clips. <S> (2) If signals are high bandwidth (500MHz - 8 <S> whatever -12 GHz), you use attachments like this ZIF tip , and solder tiny (AWG 38-40) wires to test points; (3) <S> Alternatively you can use probe holders/positioners , with probe tips of pogo-pin style. <S> But they are really not cheap. <A> After seeing comment from @Passerby about using clothepins, I thought I'd give it a try using common wooden laundry line clothepins. <S> One isn't enough, but two seem to give a stable configuration: <A> There are common several methods of getting scope probes to stay in place. <S> These include: <S> Use the probe grabber clip that comes with the probe. <S> Use bare bus wire wrapped around both the ground <S> sleeve (for ground) <S> an the tip (for the signal). <S> Solder they bus wire to the appropriate places. <S> The probes will stay nicely in place. <A> Other people have suggested various mechanical solutions, which are always a good thing. <S> But if it is too late for you to implement them, I'd like to propose a different method. <S> Most oscilloscopes can be configured to trigger once on a certain level on a specific channel. <S> If you know your signal is 3V sin wave, for example, you can set the trigger level at 2V. <S> You then change your trigger to the one off mode (it is called manual on my scope). <S> You then place all your probes except the trigger channel. <S> When you place the trigger channel probe the scope should display the waveforms. <S> It triggers only once, and you will need to reset it to trigger again. <A> Maybe a little bit late but for the record: <S> We use the following tool from probeHolder.com in our lab and we are very satisfied with it. <S> It works well with oscilloscope probes but also with multi-meter probes. <A> I recently used a probe holder and felt sorry for the efforts to solder to pins (No affiliation) <A> This could be connected to the square-wave calibration output, but I'd sling an AA holder (probably with 10K in series just in case) inside the footswitch and set the trigger voltage to about 1V. <S> Just chop off the 1/4" mono jack plug these swicthes generally come with and replace with a BNC. <A> It depends on if this is for personal or professional use. <S> If its professional use, this is what we use. <S> https://www.tek.com/datasheet/ppm203b-ppm100-articulated-arm-datasheet <S> The PPM203B works really well even in very tiny components. <S> And I mean tiny. <S> Not just tiny like a 2000+ pin BGA package, but tiny like the DIE removed from the carrier of thatoff <S> that 2000+ pin BGA package
| This probe holder is another cool idea for you to try. If you've got a spare (or trigger) channel you can connect it to a footswitch (ebay link for picture).
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Current sensing on positive or negative end On a flying drone (hexacopter) I want to measure the current consumption using ACS758. This sensor uses Hall Effect and the load circuit is isolated. Sensor outputs analog voltage referenced to common ground (Batt-). I use MCU (STM32F405) with ADC, which is powered from a buck converter (22.2V to 3.3V). All electronics use same reference ground (Batt-). Frame is also grounded. <Q> There is no fundamental difference. <S> There are practical differences however. <S> Generally, most people like everything connected to ground, I know I do. <S> That then leaves a choice for which circuit elements are not so connected. <S> We almost always connect the PSU to ground. <S> Often, the load is ground connected. <S> In that case, you cannot use the second configuration, and have to use the first. <S> Any load switches, controllers and current sensors all need to be on the positive side. <S> If we have the luxury of a floating load however, then we have the option of using a ground connected current sensor. <A> This is highly situational in real cases. <S> In your particular example, it may not matter, however I've seldom run into a real application that simple. <S> Usually the input power is grounded, there are multiple supplies or other loads sharing the ground, the battery terminal is not easily accessible etc. <S> etc. <S> Most typically you need a current monitor signal that is <S> ground-referenced (meaning, say with a 12V battery, <S> (-) side grounded (say 10-14.5V in) and a derived 5V supply <S> you would much prefer a signal that goes from 0-2.5V for 0 to full scale current to a signal that goes from whatever the input voltage is down to that minus 2.5V. <S> In theory it doesn't matter- <S> current is the same anywhere in a loop. <S> In practical applications, it usually matters quite a bit, and often the high side (the side not grounded) is the one you'll have to use and the complexity will be relatively high as a result. <A> Since the ACS758 uses a Hall sensor to completely isolate the sense circuit from the output circuit it really doesn't matter where in the current loop you put it. <S> Do whatever is most convenient.
| It all depends which components you can get, feel most comfortable using, or what your existing system configuration forces or allows you to do.
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Do 74HCxx chips require a linear voltage generator? Recently I've read a book that says it's absolutely necessary to use a linear voltage regulator like LM7805 in circuits that use any of 74HCxx chips (CMOS logical gates). I thought that this advice is kind of odd (LM7805 requires 9V instead of 5V I most frequently use, 4V are used just to heat an air) and decided to re-check. I didn't find this requirement in any of 74HCxx specs, at least in specs from Texas Instruments which made chips I use. This requirement is also not mentioned in Wikipedia or any other source. All circuits I tried work just fine without LM7805, with only 100 uF capacitor as a filter. So I'm thinking, maybe this advice is outdated (the book was published in 2010) or maybe author just needed an excuse to give an example of linear voltage generator usage? In your opinion is LM7805 required to use 74HCxx? Do you use it in your circuits? <Q> You should read the family spec to assure yourself. <S> Read section 5 on the voltage ratings. <S> In particular look at Figure 16 which shows the family VCC relationship. <S> If you are using 74HC devices then they are rated for operation from 2 - 6 V .... <S> but if you are using 74HCT then the device is only rated for 5 V +/-10% .... <S> many do confuse this. <S> The 74HCT devices can be mixed with 74LS TTL devices, while the 74HC are not guaranteed to interoperate with the same fanouts when mixed. <S> If you are mixing 74LS and 74HCT devices, you may have difficulties meeting the +/-5% VCC tolerance of the 74LS devices (ripple from switching regulators) but should have no problems for 74HC devices. <A> A regulator of some kind is advisable. <S> It doesn't have to be the ancient 7805. <S> Anything that provides a clean 5V should be OK. <S> Provide good decoupling and make sure the regulator can provide enough current for the circuit. <A> Do 74HCxx chips require a linear voltage generator? <S> No you don't. <S> The operating supply voltage range for the 74HC series is quite large, see Mario's answer: 2 to 6 V <S> The 74HCT series is more critical as it is TTL compatible, I found here that the operating supply voltage range is 4.5 to 5.5 V. <S> You can meet these supply voltage requirements without using an LM7805. <S> There are plenty of other solutions (linear and switching supplies) which will be able to generate the required supply voltage. <S> Using an LM7805 is just one possible solution to get the required supply voltage. <S> In a world without the LM7805 <S> I can still use the 74HC or even 74HCT series logic. <A> All the answers here are correct. <S> I still wanted to add mine because I am pretty sure this is written in the context of a comparison of 74xx families versus CMOS 4000 <S> (and I guess this book must be rather old, too). <S> Indeed, the ancient CD4000 family could work with higher voltages (up to 15V). <S> So you could directly power them with a 9V battery, for example. <S> Whereas the 74xx families, in a similar application, absolutely need a regulator to drop the voltage due to the 5V rating. <S> And at that time, regulators other than linear were less commonly found. <S> So this explains why it could have been formulated this way. <A> Most likely you don't need a voltage regulator if you can ensure that the voltage doesn't get too high. <S> From the datasheet :
| There is no requirement that would require a linear regulator for 74HCx.
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Piece of wire as FM antenna OK, this is really annoying me. I have an old bedside radio/alarm clock that's been fine for a couple of decades, and which still works perfectly well at night. Its crude FM antenna is a simple strand of PVC-insulated wire dangling from the back of it, with a plastic lump on the end about 1cm long and 4mm diameter which might or might not contain any components moulded in. Recently in the mornings (only), the audio has been seriously compromised by crackly interference. It seems to make no difference how I arrange the wire antenna, but I get perfect reception whenever I hold the lump on the end of it, and back to crackles when I let go. Any suggestions as to what's going on, and/or what I might clip on the end of that piece of wire to approximate an adult human being and suppress the interference? <Q> The lump at the end of the wire is probably just a weight intended to keep it hanging straighter than it would on its own. <S> When you grab the end of it, you're coupling your own body capacitance to the end of the antenna, increasing its effective length/area and therefore its sensitivity, producing better "quieting" in the receiver. <S> Try to squeeze the jaws so that they cut through the insulation and make contact with the conductor inside. <S> You can clip the other end of the jumper to something convenient on the wall (e.g., curtains or blinds) in order to raise the antenna up in the air and further improve its sensitivity. <A> Have you recently plugged in a phone charger close to the radio? <S> @andy_aka <S> You were spot on. <S> Moving the USB phone charger to the other side of the room has eliminated the interference. <S> I think the problem as to why it interferes in the morning is due to the completed state of charge of the phone's battery. <S> I can envisage the flyback circuit in the charger generating more EMI when it is supplying a light (or non-existant) load. <S> Fly back circuits are somewhat known for not regulating properly on light loads and they can go into "burst" mode where the whole switching regime is somewhat altered. <S> There is another possibility that on normal load (charging the depleted battery), the interference peaks in the spectrum don't coincide with the radio channel <S> you are listening to and, a lighter load (battery fully charged) causes the switching (still present) to hit the band you want to listen to. <S> The EMI produced by switching converters extends all the way through the FM radio band and can still cause problems at 1 GHz. <S> It's worth experimenting to see if removing the phone <S> (when the radio is being interfered with) makes the problem go away <S> i.e. is it the wire from the charger to the phone that is emitting. <A> Andy aka (comments to the question) was spot on. <S> The interference source seems to have been a USB phone charger plugged in next to the radio. <S> The charger, not the phone. <S> Moving the charger to the other side of the room has cured the problem. <S> Which leaves one small puzzle: why interference mornings, not evenings / night? <S> The only thing I can think is that this is the only time that the phone was connected to the charger with its battery fully charged. <S> Evenings it is not connected, late evenings / <S> early night it would be recharging.
| One way to replicate this effect would be to take one or more alligator clip jumper leads and connect them to the end of the existing antenna wire (just before it enters the lump) in order to lengthen it.
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Electrometer amplifier circuits - surface mount or traditional I have looked at a number of electrometer amplifiers and they all seem to use big through hole passive components and through hole DIL packages. Is there a reason for not using surface mount? <Q> The point of electrometers is usually very high impedance. <S> One possible reason for using large components is the larger creepage distances. <S> Physically large components are thru hole. <S> With larger size, the dimensional differences due to temperature will be greater. <S> Surface mount parts don't work well in that case since they are rigidly held to the board. <S> Thru hole mounting provides some inherent strain relief from dimensional changes. <S> Of course if your devices were made in the 1980s or earlier, then thru hole or even point to point were the only options. <A> Well, I don't think this still applies. <S> All you need for an electrometer is essentially something that can amplify very small currents – and considering <S> the (ancient) OPA128 is still sold by TI labeled as "Electrometer-grade Opamp", albeit it having an "incredibly high" 75 pA bias current, a lot of modern FET-based opamps will do. <S> For example: <S> The LMP7721 has something like an expected 3 pA bias current within "normal" operating conditions (ie. not on fire) and comes in a relatively tiny SOIC-8 SMD package, or smaller. <S> Since for the first stage of an electrometer, a simple unity gain voltage follower is sufficient, there's not much problem building a circuit that has little additional leakage current on the observed input. <S> Hence, the problem really is in the design of the measurement circuitry: How does one avoid that the lines between amplifier and observed charge don't form a capacity in the same order of magnitude as the charged object? <S> That might very well be solved by making sure there's no traces above e.g. a ground plane etc, making a very "wire-y" design maybe preferable. <A> Look into the INA116 by Texas Instruments. <S> It's super low bias, and has an input impedance measured in millions of Gigaohms. <S> It has built in guard band amps, and voltage follower inputs. <S> Gain control resistor setting, and other nicities. <S> It does cost about $20.00 per chip, but they are an amazing input amplifier. <S> For amplifying a microphone, there's nothing better. <S> Hum and noise rejection is amazing. <S> Check it out.
| I personally presume that within the physical possibilities, there's probably harder problems to solve than eliminating trace capacitance.
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Best design choice: linear regulator or switch converter I need to power a Thermoelectric module with 10A maximum load current. I'm designing a card with a minimal design in order to change the current or the voltage across the load, given a standard voltage source. I just need a minimal regulation of the load, which is, just be able to decrease the power of the thermoelectric module from 100% to maybe 80-90%. Thus I was thinking to control the module by voltage, since precise regulation is not needed and voltage control is usually much easier than current control. I am choosing among a linear voltage or a switching regulator solution. The point is that the output voltage should be absolutely flat, with Maximum 5% ripple. A potentiometer should be used to "regulate" the power of the system.That's why to start with I was looking into linear voltage regulators. The point is that, to be on the safe side, I would require at least a 15V max output voltage and 20A output current model. Thus I was thinking in designing several regulators in parallel, but I am aware that this is a bad idea. How could I design a safe parallel regulator circuit? If the linear regulator is a bad idea, is there any off-the-shelf step down switching regulator which does not require any external components (no coils or caps) more than a few resistors? As said I'm not concerned about the accuracy of the system but just about reliability, design robustness and cost. <Q> Your ripple requirement isn't that hard – what's more problematic is that your output voltage * output current = <S> 300 W! <S> That's quite a lot. <S> You would not want to burn a couple of volts at 20A over a linear regulator (which, by the way, makes no sense – that will convert the energy that you don't put inot your thermoelectric module to thermal energy, which is kind of what you wanted to regulate in the first place....). <S> So, if the in-output difference is just 2V, at 20A, your linear regulator would dissipate 2 V * 20 <S> A = 40 W of power. <S> That's a terrible thing to cool. <S> If the linear regulator is a bad idea, is there any off-the-shelf step down switching regulator which does not require any external components (no coils or caps) more than a few resistors? <S> Terminology: When talking about regulators, it's not perfectly sure whether you're only referring to the thing that regulates the currents flowing, or mean the complete system including all the necessary energy-storage components. <S> Usually, we'd use the former meaning. <S> For the other thing (controller + switch + power storage (coil)), we'd say supply , or at least module . <S> You can of course buy readily made power supplies. <S> Every laptop has one, and they even exist for the currents you need. <S> Getting one that is adjustable might be a little harder, but you might want to think about just using PWM on the output to reduce the average power going into your module. <S> Of course, that'll technically absolutely break the "5% ripple" requirement (PWM is actually 100% ripple, if you want to consider it that way), but I'm not sure where that requirement came from in the first place. <S> Maybe you'd want to also specify the acceptable/unacceptable frequencies for deviations from the intended current/voltage point, and explain why you'd need so strict regulatory limits for something as slow as a thermal element. <S> * <S> *Update:* nope, not PWM then, according to your comment :) <S> You can also buy adjustable 300W supplies – but these tend to be a little more costly. <S> Regarding modules: The module we're talking about will most probably be sold as "open|closed frame power supply". <A> Just ask your favorite supplier or search engine for "Adjustable Power Resistors" and use one of those together with your load as a voltage divider. <S> Let's assume that your load is strictly ohmic. <S> Your output specs require 300 W of output power, and you want to regulate (burn away) 10-20% of that. <S> That's 30-60 Watts, which is easily possible for power resistors. <S> Just be careful that your load has a positive temperature coefficient (resistance increases with temperature) <S> or else you may experience thermal runaway and your load will start to glow soon. <S> ;-) <S> By the way, a linear regulator would also burn those 30-60 W as heat, so it would need a big heatsink and probably a fan. <A> If you want design simplicity, why not just toggle the power to the thermoelectric module on and off at very slow PWM rate using an N-channel MOSFET driven by a timer circuit? <S> Remember, your home fridge works in similar way, except even slower running every 10 minutes or so. <S> You could use say a 1 second pulse rate, so a 10% reduction would be 0.1 second off, 0.9 second on. <S> Since cooling is very slow and your module can withstand the full supply voltage, then 1 second rate would be just fine and stress the electronics less, etc. <S> You would need to check the load regulation specs of your power supply to see what kind of ripple you might get with switching a 10 Amp load. <S> For example, a load regulation spec of 1% at full load would mean the output will drop 1% in voltage between no load and full load. <S> An N-channel MOSFET would be very easy to direct drive with a microcontroller, a 555 timer circuit, discrete or opamp oscillator or whatever. <S> And you can make the timer circuit use a pot to change the duty cycle from 80% to 100%. <S> Choose an N-channel MOSFET with very low Rds-on that can handle very high currents. <S> Use the Rds-on * load current to calculate the power dissipation of the MOSFET. <S> Choose a low enough Rds-on and you will not need much (if any) heatsinking for the MOSFET. <S> The gate capacitance will increase with lower Rds-on resulting in slower switching times, but at 1 second rate, it won't matter much. <S> If this approach sounds good, I can help with the slow PWM timer circuit. <S> Hope that helps, -Vince
| Linear regulators simply work by having an "adjusting" internal resistance that simply drops the voltage difference that's between your in- and output and converts that to heat.
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How to get rid of voltage spikes in this supply voltage switcher circuit? In my mainboard there are two main supply.One is 12V and other one is 5V. 5V is always ON but there is a risk of 12V crashing and disappearing. So, I want something like this, when presence of 12V main supply always has to be 12V but when it is gone main supply must be 5V which is always ON. The crudial thing here is main supply must never be below 4.8V. I used to get this done by this simple structure: simulate this circuit – Schematic created using CircuitLab But my client does not want that. Diodes get hot and there is a certain voltage drop. So I thought of this soltion: It is doing its job but it can not prevent this annoying spikes: You see, when both rising and falling of 12V, it leads a spike and it causes to voltage drop below 5V. I want to understand why does it happen? And is there any circuit that can do the job with high efficiency? <Q> Your output voltage is dipping because the power PFET is shorting the output node to the 12V supply that is at a lower voltage. <S> Your threshold for turning on the PFET is much lower than 5V, effectively less than 1V. <S> I went ahead and remade your schematic in LTSpice: <S> The first question you should ask is "When does the NPN turn on?" , which is practically immediately. <S> There's a couple reasons for this: <S> Your NPN, a 2N3055, is a big, power NPN (15 A!!). <S> You need maybe 1-3mA at most to turn on the PFET with the R3-R4 resistor divider. <S> Because Q1 is so large, it really doesn't take much to turn it on a little bit. <S> LTSpice has that transistor turning on at about 0.27V, which is probably less than you assumed. <S> I think you have your R1 and R2 values backwards. <S> Even assuming a target VBE of 0.6V, your NPN would be on when the 12V rail was at 0.73V. <S> The other way around, and it turns on at 3.43V or so (but probably lower). <S> Taking those things into account, we can see the voltages of the 12V rail (red), the NPN VBE (green), and the MOSFET gate voltage (blue). <S> We can see the MOSFET turns on almost instantly. <S> For more fun, here's a look at the DC sweep of V2 (the 12V supply) when V1 (the 5V supply) is held at 5.0V: <S> The current through D1 and M1 spikes to 27A, but really that's just what the diode model is limiting it to - <S> the diode would probably blow up if the 12V supply was really an ideal source. <S> The diode is only rated for about 1A. <S> A comparator would be another solution. <S> Keeping the 5V rail from back-feeding the 12V rail will be a bit tricky. <S> To tweak this topology, swap the 2N3055 for a 2N3904. <S> Adjust the R1/R2 ratio to turn on Q1 when the 12V rail is above 5V. <S> I'm pretty sure you don't need R4. <S> The problem you will run into is that the behavior will be fairly sensitive to device variation (bad for products). <A> Your schematic is hard to see, and so is your graph. <S> If so, use what you had before in addition to your active circuit. <S> However, use Schottky diodes for lower voltage drop. <S> The diodes will guarantee the output is help up to the minimum of the two inputs, minus the diode drop. <S> This is regardless of what your active switch is doing. <A> simulate this circuit – Schematic created using CircuitLab <S> The problem is the threshold for NPN switch to Gate is too low <S> and it turned on a 4V pulling down 5V there are better ways to to do this. <S> but I am just showing a rough example to make it work using your design. <A> I had very similar problem. <S> I needed to keep an MCU and display running from either USB or 12V, whichever was available, to supply about 100 mA. <S> I found a simple and robust solution. <S> Instead of focusing on the narrow problem definition, it helped to look at it from bigger perspective. <S> In other words, instead of thinking "I need to switch between 12V and 5V supplies", the bigger issue was "I need to provide constant 5V to this circuit". <S> My solution is below. <S> I used an LDO from 5V USB and set to 5V output. <S> The LDO can supply 200 mA with only a 100 mV drop between In and Out. <S> On 12V side, I used an old LM317 to put out 5.1 Volts since 0.7 Watt of heat was OK <S> and I wanted very good input overvoltage protection. <S> You can use an SMPS for yours if you need low Pdiss. <S> The 5V LDO gets overridden by the slightly higher 5.1 Volts from 12V and the LDO just shuts off. <S> Lots of advantages to using regulators instead of Schottky diodes or MOSFETs: <S> Overload protection Short circuit protection Thermal protection <S> Very fast switching between supplies with little to no dip during switching <S> I hope this helps. <S> -Vince <S> simulate this circuit – <S> Schematic created using CircuitLab
| From the description, it seems the problem is that the output is glitching off during the switching between 12 V and 5 V input. I'll admit, I'm not very fond of this topology, and I would probably look for a power management IC, or maybe some "ideal" diode ICs to handle the switching in a controlled manner.
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12-bit DAC Output to +/-10v Signal Suppose we want a +/-10v control signal, where 0x0 (to the DAC) = -10v and 0xFFF = +10v. Firstly, how do we alter the signal so that we can have a negative voltage, dependant on the DAC input? Secondly, how can we scale the signal so that it fits within the -10 to +10 range? <Q> There's plenty of good material on the web about it. <S> Here's one: <S> http://www.electronics-tutorials.ws/opamp/opamp_5.html Let's say your DAC outputs 0V to 2.048V for code 0x000 to code 0xFFF (fairly common) and you want a swing of + <S> /-10 Volts. <S> So the gain you want is 20 Volts / 2.048V <S> = 9.77 <S> The circuit is below. <S> So you just need R3 that many times bigger than R1. <S> The trick is to reference once side of the input to the "halfway point" of your DAC output range. <S> So here, I'm referencing it to 1.024V. This way <S> , the amplifier sees a difference (V_R2 - V_R1) on its inputs of -1.024V at code 0x000, and +1.024V at code 0xFFF. <S> Then the opamp applies the gain of 9.77 to that difference and, voila, you get your -10V to +10V <S> output swing. <S> For the voltage reference, just do a search on Mouser or Digikey. <S> You will find a ton out there that put out nice "binary" values like 1.024V and 2.048V. <S> The difference amp configuration is super useful. <S> I hope that helps, -Vince <A> Start with a potential divider that maps the DAC output midpoint to 0V. <S> If the DAC midpoint voltage is +1.25 volts, then a -1.25 reference voltage and two 1kohm resistors will produce 0V for the remapped output at the junction of the resistors. <S> Now, if the DAC produced 0V at it's output, the remapped output would be -0.625 volts. <S> If the DAC produced the full 2.5 volts, the remapped output would be +0.625 volts. <S> So, a 0 to 2.5 volts DAC output is mapped to -0.625 volts to +0.625 volts using two 1 kohm resistors and a negative voltage of 1.25 volts. <S> Next, amplify that remapped voltage using a non-inverting amplifier to convert +/-0.625 volts to + <S> /-10 volts and <S> you have your solution (gain = 16). <S> As with any solution to this, the op-amp will need power supplies of a little greater than +/- <S> 10V but this depends on the op-amp. <S> +/-15 volt supplies would be fairly standard. <S> Use a precision shunt reference to generate -1.25 volts. <A> A perfectly straightforward way to address the problem is to choose a DAC such as the DAC0800, although this is an 8-bit DAC, and you have specified 12-bit output values. <S> Looking at the data sheet , Figure 23 gives you the circuit you want simulate this circuit – Schematic created using <S> CircuitLab The op amp must be driven from at least +/- <S> 12 volts, and +/- <S> 15 is a better planning figure. <S> Since the DAC0800 should be driven from +/- <S> 15, that's the way to go. <S> Note that R1, R3 and R4 should be 0.1% resistors. <S> You can use lower tolerances but your DAC output levels will not be defined to 1 lsb. <S> (1/256 <S> ) = 0.4%) <S> And if you do go to a 12-bit DAC you should simply buy one with voltage output. <S> If you insist on using a DAC0800-like DAC with 12 bits, your resistor tolerances should be 0.025%, since 1/4096 = <S> .000244. <S> Just personally, I'd go with buying a DAC with voltage outputs, but it's your choice. <S> Digital inputs to the DAC0800 are not shown, but are necessary. <A> Applying a difference in midpoint <S> Voltage shifts your offset separate from gain <S> adjust. <S> This is done on inverting input scaled down by inverting gain to Vref. <S> This allows any offset from polar to bipolar. <S> For precision, both must be separate and tuneable or self-calibrated with a precision Vref.
| a straightforward solution would be to use an opamp configured as a difference amplifier.
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Cheapest dummy load? 450w, 15V, 30A, only for a few seconds What is the cheapest and simplest way to dummy load 450w, 15V, 30A, only for a few seconds? I would like to measure the max output capability of an MPPT charger controller, basically as a "all losses included" way of measuring actual solar panel maximum power capability. The theoretical maximum output is 440w. In reality, the solar panels probably won't output more than 350-400w on the sunniest day. The controller has a load-side shunt, allowing me to measure current easily. I'm planning to use a MOSFET to short-circuit the load-side through a dummy load. The resistance needs to be very low to allow 30A to flow at 14.4V. I've considered using a ceramic heater, but they are typically dimensioned to have a resistance that restricts current to their dissipation-capability under continuous load, to prevent overheating.I could buy a heater dimensioned at 500w, but it would probably take up a lot of space.Since it only needs to operate for a few seconds, the concern is different. <Q> You may need to replace your Mosfet with a relay. <S> This has the advantage of being easily adjustable (more or different bulb ratings). <S> I use this for battery capacity testing on boats. <A> Automotive headlamp. <S> That will have no problem with 15V since the electrical system is typically at 14.4V with the motor running. <A> Why not use the appropriate resistor? <S> Some heavy duty wirewound like this: You will certainly need to put several in parallel/series to handle all the power, but note that they can be overloaded for a short time without damage. <S> See datasheet. <S> For example, for Tyco THS series (which is available in a wide range of values, from 10 to 75W): <S> So, if, for example, the pulse lasts for 5 second, you can handle 5 times the power. <S> So you just need a 90W resistor. <S> Take two 1ohm 50W resistor in parallel (THS501R0J), you're done. <S> 3,86 <S> € each at mouser. <S> Can't be cheaper. <S> However, you'll need a hell of a MOSFET to switch that much current. <S> And the MOSFET will itself dissipate a lot and probably need a heatsink. <S> I would rather use a relay, as RoyC suggested. <A> The obvious thing that springs to mind is filament lights. <S> Unfortunately, they have this nasty feature whereby the running resistance is often an order of magnitude more than the cold resistance. <S> If you plan to use a nominal 500W of filament lights for a 500W load, then they will tend to overload the output of whatever you connect them to. <S> This may not matter if the source is a solar panel (which can tolerate a short circuit), or a current-limited output inverter, but it does matter if it's a voltage-output supply with a current trip, try testing a PC power supply by plonking cold 12v halogens onto it! <S> If you plan to use you bulbs at perhaps 30% of their rated current, they won't change resistance by more than a factor of 2, but you'll need a lot more of them. <S> As your load only needs to dissipate for a few seconds, you could use something much dirtier. <S> If you have, or can borrow, a 50m length of 3 core cable of 1mm2 cross section, that has a resistance of about 0.8ohms per core cold, increasing to 1ohm for 'not too hot to touch'. <S> Two or three of those cores in series would give you an appropriate sized resistor, for a few seconds. <S> Don't forget to bypass any fuse that's included in plugs if you borrow an extension lead.
| Use an inverter and a few old filament 240V or 120V light bulbs.
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How do energy harvesting ICs like LTC3106 and LTC3105 safely charge lipos? I'm designing a simple device with a single cell LiPo that I want to be chargeable from a single solar power cell. As I start to look through the data sheets, I'm seeing that the battery charging is typically done just by providing a constant voltage output to the battery. For instance, LTC3106 , page 19: Or LTC3105 , front page: Note position of batteries in line with the load?? These circuits seem to be designed to charge the batteries in question by providing a constant 4.2V power to a single cell battery. Is this really safe? Is this equivalent to trickle charging? Everywhere I've read that a LiPo needs a very specific type of charging discipline, such as MCP73831 (which I use in all of my projects without issue.) For instance, is it actually safe to charge a single cell lipo with a constant-voltage, variable current source and no cut-off? Can someone more familiar with battery chemistries and charging process help enlighten me? (PS: I am not EE-trained, I am a home hobbyist.) <Q> These 4.2V are unloaded, i.e. with battery disconnected. <S> The charging current is relative to the voltage difference between the charger output and the battery, and vice versa. <S> High-powered chargers can provide more current than that, so they have to reduce their output voltage so the charging current stays in range. <S> Trickle chargers cannot provide a lot of current, so their output voltage is the battery voltage plus the difference given by the available current. <A> No, you are correct, you must not trickle charge LiPos. <S> If these converters do not have a way to stop charging, then they must be used with something that can. <S> It doesn't look like the 3106 <S> is intended to charge its spare source (quick read, might be wrong). <S> The 3105 doesn't mention LiPos inside the data sheet, showing supercaps and NimH cells, but that front page is very naughty. <A> The details depend on the strength of the source and the Amp-Hr rating of the battery, which you didn't give, but in general, these are low power converters (and usually low power sources, such as a small solar panel) that can safely charge a "typical" battery due to the low charge current. <S> However, it's best to calculate the maximum output current the converter can provide at your expected Vin with a 4.2V output (or whatever <S> you program Vout for). <S> You can find this information in the datasheet curves. <S> Compare this current to the rated charge current of your battery.
| The important bit with LiPos is that you control the charging current, and the limit is usually relative to the capacity of the battery. These parts do have a way to stop charging - it stops when the desired terminal voltage is reached.
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What is the truth about 1.5 V "lithium" cells? At least one manufacturer out there is marketing "Lithium" cells in familiar AA and AAA sizes, as direct replacements for those standard 1.5 V sizes, boasting the typically better than alkaline longevity you'd expect from lithium. But we all know the range of lithium technology cell voltage is expected to be 3 V for single use cells, up to a max of around 4.2 for li-Ion variations of rechargeable at max charge. All my attempts to research what the truth is (short of buying and cutting one open) have resulted in little more than manufacturers hype. Can anyone shed light on what is going on with these? A stretch to think perhaps they actually have embedded buck converters under the hood? Or has a genuine 1.5 V lithium technology actually been invented? I've included one manufacturer's photo as a reference <Q> Lithium batteries come in many different chemistries, and it is the chemistry that governs the voltage. <S> The most common chemistries are on the order of 3-4V, but there are chemistries which have a 1.5V terminal voltage. <S> The wiki page for Lithium batteries has a list of many different chemistries and their voltages. <S> A Lithium anode with an Iron Disulphide cathode (\$\mathrm{Li-FeS_2}\$) is one such example of a 1.5V terminal voltage, and is the chemistry used in the AA replacement batteries as per the datasheet link on the Wiki page, and in @pjc50's answer. <A> Different chemistry: Lithium Iron Disulphide . <S> Open circuit voltage of 1.8V, drops to about 1.5V under load. <A> If it is rechargeable Lithium 1.5V, one company with Chinese patent uses 3.7V Lipo with smart embedded buck converters into an AA/AAA cell to output 1.5V with a cost of about 5~10 cents per cycle. <S> Note the dual anode (+3.7,+1.5V) requires their special charger for 500~100 cycles with 80% DoD. <S> http://www.kentli.cn/product/show.php?id=1# Bottom line. <S> Only use proven, documented reliable sources for batteries and capacitors. <S> Too many fly-by-night battery vendors should make anyone skeptical. <S> This takes years of proven Quality track record with many initial failures. <S> Making a good battery with high Capacity, low ESR at low cost is hard. <S> It may worth investing in these to verify yourself. <A> And yes they are a direct replacement for standard pencil batteries. <S> And no, they don't have little bucks inside them. <S> The AA type present 1.72V open circuit, and can source >4 amps shorted. <S> They also work very well in extreme cold, although the standard advice is to keep them inside your clothing /sleeping bag. <S> The only type I carry for survival gear. <S> Expensive though. <S> They top the list for total storage capacity but you pay for it. <S> Good quality alkaline batteries are good for daily use and they can be bought in industrial bulk quantities (Duracell Procell). <S> I would suggest that lithium only makes financial sense for survival /tactical situations like GPS or gun sights. <S> Life saving tip <S> : Never use rechargeable anything in survival /tactical situations as they are very unreliable. <S> You'll look really stupid in the middle of the Gobi desert trying to fix the buck converter in your dodgy rechargeable lithium batteries. <S> Update: <S> I've reviewed two relevant (and official) datasheets, one for the above lithiums and one for standard Energizer alkalines . <S> I have to admit to being surprised as to the difference in capacity. <S> I honestly thought that it would be greater. <S> So for lithium, we have:- and for a standard alkaline, we have:- <S> You'll notice that Energizer cunningly have two different styles of graph. <S> So conspiracy theorists might think that this is to obfuscate an easy comparison of capacities. <S> Not me though. <S> The only common discharge profile is at 100mA, giving capacities of ~3500mAh and ~2500mAh respectively. <S> That's only an improvement of 40% over common alkaline chemistry. <S> Other interesting comparisons can be determined like (lithium v alkaline):- <S> Energy density: 233 mAh/g <S> compared to 109 mAh/g ( +114% ) <S> Money density: 8.7 pence/g <S> compared to 1.6 pence/g ( +448% )- <S> based on March 2018 Amazon(UK) prices. <A> This is a lithium-iron disulfide (Li-FeS 2 ) battery . <S> It is a primary (non-rechargeable) chemistry that is sometimes referred to as lithium metal ; do not confuse these with rechargeable lithium-ion batteries . <S> It has a nominal voltage of 1.5V and an open-circuit voltage of 1.8V when new, making it a suitable replacement for alkaline batteries in many applications. <S> The Li-FeS 2 chemistry provides very long shelf life (up to 20 years) and long runtime under a variety of discharge conditions (especially under moderate to high drain). <S> Most notably, it is far less affected by low temperatures than most other battery chemistries, capable of delivering nearly an order of magnitude longer runtime than alkaline batteries at 0 <S> °C (32 °F), hence the "up to 9x longer lasting" claim, and can deliver near-full performance at subzero temperatures (-20 °C or lower) where other batteries would fail. <S> It is also considerably lighter in weight than a comparable alkaline battery. <S> The main drawback is its significantly higher cost compared to alkaline batteries. <S> This is due to the use of expensive lithium metal as well as complex physical construction more similar to that of rechargeable batteries than alkaline batteries. <S> As such, it is best suited for applications where the device must continuously operate in a very cold environment, where the longest possible service life between replacements is required, or where battery replacement is inconvenient, such as in a smoke or carbon monoxide alarm. <S> For more technical information about these batteries, see Energizer's application manual . <A> About non-rechargeable lithium batteries, is a matter of chemistry: the Energizer you posted are Li-FeS2 with a voltage of 1.8 V (1.5 V under load). <S> Reference: Energizer Ultimate Lithium AAA (L92) <S> The rechargeable version of Kentli is the well-known 3.7 V Li-Po version with an internal circuit. <S> Take care that the capacity in the specification is measured in mWh and not in the usual mAh, probably a "commercial" trick - because mWh is greater than mAh. <S> Those batteries needs a specific charger. <S> Reference: Kentli AA 2800mWh <S> (Blue)
| The truth is that the Energizer lithiums are probably the best small batteries you can buy.
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Why is it "Vactrols" are not commonly used? I've been trying to get my hands on some Vactrols (opto-variable resistor) for a long time, unsuccessful. Not even digikey has them. some examples: VTL5C2, VTL5C3, NSL32 Can't see them as "too expensive to manufacture" considering you can improvise one using an LED+LDR+Tape, but it's in no way as "linear" or robust as the actual components. Anyone knows the story behind it? Edit: So there are places where you can get them, but at least where I live it's something almost unheard of..They look usefull in many ways (at least to me), yet I've seen them nowhere except some old audio equipment.. Edit2: Ok, so what could be the alternative to a vactrol if I require exactly that, a significant resistance change controled by voltage, (MOSFET as resistor is not an option for me because of the very low resistance range). Is there such a device/equivalent circuit that allows all this with the simplicity of the vactrol? This is a hypothetical question, based on many compressor units I have seen and seems Hard to achieve such functionality in such a simple way, the closest I've seen are OTA based for closed loop circuit, but they arent nearly as simple as vactrols. <Q> The kind of designs that can be (kind of) achieved with vactrols can also be done with op amps and LM13700s, and will yield much better results. <A> Vactrols contain cadmium, so it is not RoHS compliant. <A> I've never used a Vactrol ( or even heard of it till now) nor would I want to. <S> The same goes for LDR's. <S> Photo Diodes are so accurate often <S> 0.5mA/mW + <S> /-1% unlike LDR's +/-10~50%. <S> When diodes are put in OptoCouplers are very stable unlike phototransistors which have wide ranges in hFE. <S> ( but hey , the chinese will sell them to you for disposal in your country) <A> They are not used because other, easier to use alternatives exist -- namely, translinear VCAs and multipliers. <S> These don't require specialized drive to cancel out nonlinearities (in fact, you can get VCAs with linear-in-dB control inputs nowadays, which makes life quite easy), nor do they run afoul of RoHS regulations as they are quite standard bipolar or CMOS ICs, unlike the Vactrol which requires both a CdS photocell and specialized manufacturing processes.
| They are not stocked because nobody is using them in large quantities. But the real reason are the penalties for using the raw materials used inside like Cadmium, which have been banned in most healthy countries around the world for ground water contamination reasons after disposal.
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Converting -3.7V to -32V (both negative), Is it a boost or a buck? I need to get create a split supply at +/-32V, but all I have to work with are LiPo batteries. I was thinking something like the block diagram below. The positive part is simple, but the negative part is not so straight-forward. simulate this circuit – Schematic created using CircuitLab I tired googling this, but there is not much literature on dc-dc conversion with negative voltage. Main questions I have and need help with are: When you convert -3.7V to -32V, is that considered a boost or buck? The output voltage is lower, but the absolute value is greater. Judging by the lack of information, it seems this is not a typical problem, but is it practical? Can the negative "boost" be done using the same positive boost IC? Ideally, I'd like to use the same chip if possible. Any recommendation on reading material or schematics? Or any other method of getting +/-32V? <Q> It would still be called a boost converter even though the output is lower absolute potential than the the input. <S> I would recommend developing both the +32V and the -32V from a positive +3.7V input. <S> You will have more luck finding devices that will do that. <S> An alternative is that you could use a transformer coupled isolating converter to provide the -32V <S> or both outputs <S> , then it doesn't care where ground is. <S> There are off the shelf devices from manufacturers such as PICO who may have something that meets your needs. <S> Some of those have dual outputs. <A> You need a BIPOLAR BOOST circuit. <S> You can take +7.4V from your two LIPO cells and convert directly to ±32V <S> There are many examples of DC-DC boost converters that will produce bipolar output voltages. <S> The polarity is not a factor in the name/function. <S> It is only a detail of implementation. <A> To make things more less confusing, here's are application notes <S> Designing a negative boost converter from a standard positive buck converter and Positive Buck Regulator Makes Negative Boost DC/DC Converter . <S> (The regulators in these app notes are intended to be positive bucks, but repurposed as negative boosts.) <S> For a simple inductive boost converter, it's advisable not to have a boost factor greater than 6. <S> This applies to both positive and negative boost. <S> p.s. <S> We’re situated roughly in the same neck of the woods, interestingly.
| Many of the datasheets for voltage converter ICs will have an example for generating negative voltages. Going from -3.7V to -32V to is a boost, a negative boost. "Boost" refers to the output voltage being greater than the input voltage.
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How to power up 50 LEDs? (DIY 'fireflies' project) I want to install lots of tiny LEDs lights in my garden (at least 50). I DON'T want high voltage cables in my garden so I am thinking in getting a high power, low voltage transformer to power up these LEDs. For example, I have two 24V/1.25A (30W) power supplies from Canon printers. Let's say this is a small scale test. Later I could use a PC power supply for 50+ LEDs. The LED I will use is 1W but I want to run it at about 3.1V/0.1A because at this current it doesn't require a heatsink (it runs at under 37Celsisus - I hope in summer it won't get much hotter). So, I can have 12.5 branches, each branch will have 8 LEDs in series, so 2.4W per branch. Is this a good design? Any alternative idea is welcome as long as it doesn't involve carrying more than 25V per underground cables. <Q> In each branch you will need a current control resistor so use 7 leds instead of 8 which gives you a 21.7V drop. <S> Your resistor needs to be (24-21.7)/0.1 <S> =33ohms. <S> You can have up to 1.25/0.1 =12 branches on each supply. <A> Sounds like a good plan to me, however actual fireflies are subtle - this will be a lot more in your face- <S> the lights at 300mW will cast shadows easily. <S> You might want to consider adding a DC PWM dimmer (easy to make, much cheaper to buy) and experimenting first before you commit to the current level. <S> You may well be able to run all 50 off of a single 24V supply- for example 100mA per string would allow you to run a dozen strings (1.2A) <S> so 84/supply. <S> If 50mA was enough you could run 168 LEDs. <S> Running the LEDs at reduced current will extend their lifetime. <S> I have used some super bright green LEDs in yard accent lighting and they have dropped to about 1/2 brightness over a few years of timed nighttime use. <S> Whether you use a printer supply or a PC supply, keep in mind that they are not designed for outdoor use and may become dangerous if exposed to moisture (so keep the supply indoors in a dry warm place). <S> Definitely make sure the output is connected to earth ground with a 3-pin plug so that it cannot put mains voltage wrt earth on the output even if it fails. <S> I am more at ease with running the Canon supplies near full rating and depending on their safety than I would be with a random eBay Chinese supply. <A> Most minimal LED installations (for example some commercial freezer lighting or cosmetic shelf lighting) use constant current power supplies. <S> In this way, you are free to install as many LEDs in series as the jobs requires. <S> No other components needed. <S> The voltage will change depending on the load. <S> However, the current should always be the same. <S> Here is an example circuit from a Maxim Integrated Semiconductor Constant Current document : <S> Note, this is from an applications white paper. <S> More parts are show because of the intended audience. <S> If you were to buy a COTS LED Constant Current regulator all you should have to do is add the LEDs. <S> The power supplies you have appear to be constant voltage and can not be used in this way. <S> Outdoors is a bad environment for electronics. <S> Guessing, the most likely mode of failure for long life LEDs will be power distribution. <S> To get around this consider sealing a battery, light sensor, LED and processor (yes processor) in a water tight container. <S> The idea is to only light at night, perhaps only flash as if a lighting bug and stop after about 6 hours past dusk. <S> Minimal embedded processor can easily do this and can be less expensive than the battery powering them. <S> Hence the suggestion. <A> While the number of solutions to providing power to LED's is almost endless, you should also look at the bigger picture as well and decide what is 'Safe' to implement in your garden. <S> If you have small children or animals then this could become a critical requirement (apart from feral animals that might chew on your wiring). <S> The regulations for landscape lighting vary between jurisdictions/states but in general there will be a maximum RMS voltage specification, and typically an isolation specification. <S> Where I live <S> it's 30 V RMS @ <S> 25 A in any cable, an isolation transformer and non-grounded supply <S> (so I don't need Earth leakage detection). <S> IMO, you will have trouble over time as connections corrode. <S> Providing AC will at least slow the corrosion problem over most of the wiring but means you may need to provide a rectification solution at each use point. <S> My preference is to run a 24 V RMS center tapped transformer (3 core direct burial cable). <S> I can then run 12 V incandescent, and 24 V incandescent without diodes or regulators. <S> Then I can half wave (1 diode) or full wave (2 diodes) <S> rectify the 12 V for other lighting types such as LEDs. <S> I can even full wave (4 diode bridge) <S> the 12 or 24 V if that works out better. <S> No complicated constant current drivers or regulators. <A> It is not the best design. <S> I will suggest you to go with a fixed current rather than a fixed voltage regulator. <S> This is the design any big led installation uses (offices, street lights, huge facilities with kw of light power). <S> This is done becouse of several reasons: <S> As a LED heats up, its current consumprion changes (usualy increase) and this reaction makes you difficult to limit (or control) <S> the heat. <S> LED diodes are di per se designed for a current supply and therefore if you control it it with a voltage ( that will let current to freely change according to diode temperature, design and outage ) will make them change color, deteriorate faster and eventually burn. <S> Voltage controlling few LED in series if one will burn, and there shortcut will distribute a higher current on the remaining diodes and make all of them burn faster. <S> If you want a reliable solution your configuration of 8 diodes in series and then branches in parallel is a good solution. <S> But you will need to drive a fixed current in each branch, and not connect all your branches in parallel and controll them with one current driver. <S> Current drivers are cheap, dont worry about their price as I can assure you thah 1W LED diodes will be the greatest part of your expenses.
| For running your LEDs, you could run strings of 4 (4 *3.1 V) with a series resistor from 12 V or a string of 9 from 24 V. Running DC voltages round a garden (where you expect years of service) is not a great idea
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What things to consider when designing a single layer board? I currently have a 2 layer FR4 board that is basically a microcontroller with a motor attached to it. There are some discrete switches and connectors as well. Most of the components are SMT components (0805 being the smallest). There are some through hole LEDs and connectors. For a single layer (one side) board would I have to replace my SMT components to through hole ? For a single layer (one side) board would a decent board house still be able to do 10 mil traces or are there new rules for single layer boards ? ( I understand that I should contact the board house directly to find out their capabilities) Any other useful gotchas (or questions I have failed to ask) would be appreciated from your own experience when doing a conversio, design or increasing manufacturing yield. <Q> I've designed around 8 or so single sided mixed SMD/through hole PCBs (most with microcontrollers, one with some radio modules, and a couple with a low speed gate drive). <S> Expect to put a lot of effort into getting your ground planes to be continuous with a single plane of copper. <S> I've heard it said that single sided designs without jumpers are a kind of holy grail for PCB design. <S> The ratsnest on your pcb design tool is actually a very good indicator whether or not your design will be easy to route on one side. <S> Part layout and grounding will be your most difficult challenge with these designs. <S> You'll want to actually trace out the likely path your ground currents will take (or use something like Hyperlynx PI if you have access to that kind of tool). <S> That said, I've successfully mixed through hole and smd components to give myself a two sided board with one layer of copper (through hole on top, smd on solder side). <S> These board were all milled rather than etched and didn't have any solder mask. <S> In fact, I tried my hardest to only go with single sided boards because the people who operated the milling service I was using were not so great at aligning the two sides (and the through holes weren't plated). <S> I guess that's the price to pay for $0.07 per in^2. <S> If you're going with a standard board house, their two sided rules (trace width, hole size, etc) should apply all the same. <S> From their perspective, a single sided board should be the same as a double sided board unless they're doing something wonky. <A> Assuming low enough speed that this method is acceptable, remember that you can jump over things with through hole components (sometimes leaving the leads longer than stock bend to facilitate that), and that you can (and will often need to) use jumper wires when you hit an otherwise insoluble in one layer solution. <A> You can mix through-hole and SMD. <S> Some prototype PCB manufacturers only support SMD, to keep costs down, so make sure to check. <S> The rules for a two-layer vs. four-layer should be similar enough that you won't need to worry unless doing a tricky or complex design. <S> If the design is relatively simple, not super high-speed, and no RF, the layout should be easy. <S> Be generous with your ground and power lines; a ground plane would be best. <S> Make sure you have a few power decoupling capacitors near the micro. <S> It's handy to try going up-down on one layer, and left-right on the other layer, with your traces. <S> Keep as many traces on one side as you can, which leaves the other free for that big juicy ground plane. <S> And otherwise, read and study everything you can!
| If you're doing anything that requires a really good ground plane you should probably stick with the two layer board at a minimum.
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Request for comments on AC mains frequency conversion method - Supply>PSU>AutoInverter>Load Purpose:To supply 110v 60Hz power to a Lionel o-gauge train set from 240v 50Hz mains in New Zealand. While the train itself will run fine on 110v at 50Hz, the digital electronics for sounds can be ruined by the 50Hz sine. This claim is based on much reading online, not based on my own knowledge. Many people have posted about ruined transformers from running ONLY a step down. Due to the cost of the trains, I wish to have it run on it's designed Hz. Proposed method:Use a 240v New Zealand market computer power supply to feed 12v dc to a high-wattage US-market automotive inverter that supplies 110v power. Relevance to stack exchange posting:If this is indeed a viable solution to this problem it will be very useful for many many people wishing to use US-market electronics outside that market. My idea seems to make sense to me - that once converted to DC there isn't any Hz to measure as there's no wave, and the US-market car adapter will put out AC at the American frequency. The computer power supply puts out plenty of Amps, and the higher-wattage car adapters put out plenty for the Amp draw of the train. Please comment to let me know if this is a sensible approach or not? Please focus on the idea for getting the 110v at 60Hz, not on other aspects of the issue.I appreciate the intent, but this question is specifically about whether my proposed method will work. <Q> Does the train set need 60Hz? <S> If the input transformer has been too agressively designed, then it may need some input voltage derating from nominal. <S> If 60Hz is needed, then PC supply to inverter sounds as good as any other way of doing it. <S> It uses low cost commercial components. <S> Be aware that some PC supplies need a minimum load on 5v for their other rails to stay in regulation. <S> A 50w 12v halogen that draws a couple of amps at 5v is my preferred choice of dummy load. <A> Your idea is good except that the PC power supply probably won't work properly without significant load on the other rails, and even then the voltage may not be stable. <S> I would use a dedicated DC power supply such as the MP3079 from Jaycar . <S> This unit puts out 13.8V to match the typical operating voltage of a '12V' battery in a motor vehicle when the engine is running. <S> Mains transformers designed only for 60Hz often run hotter on 50Hz due to increased saturation at the lower frequency. <S> This may be a problem if the transformer was marginally designed (to reduce size and/or save money) and has inadequate ventilation. <S> Older Lionel trains also have a problem due to their peculiar signalling system that rides 60Hz AC on top of the motor power. <S> This has a DC offset applied (using a half-wave rectifier) to trigger sounds. <S> At 50Hz the sound system may not be able to filter out the AC sufficiently <S> and so makes sounds when it shouldn't. <S> The 2009 Lionel CW-80 controller is reported to be 50Hz compatible. <S> It is identified by a sticker marked 'QC50-OK'. <S> The 2009 PowerMax 40W, <S> Legacy Command Control and all 2008 Legacy equipped locomotives are also confirmed as 50Hz compatible. <A> My proposed method works. <S> I have successfully set up and used this method to provide 1500 watts of power at 110v/60Hz using a mains supply of 240v/50Hz. <S> From mains, I used two different power supplies to try this out: <S> One is a 500 Watt ATX supply that came with very long wires before the ATX and other connectors. <S> I internally cut off all wires not needed, leaving only the PSON directly connected to ground, and all of the factory 12v +/- <S> wires. <S> The 12v wires are simply (and literally) braided together then stripped at the ends, there they are all twisted together. <S> The second is an HP server power supply rated at 1300 Watts. <S> On this one I used the factory connector for output. <S> I installed a switch connecting the PSON, PSKILL and ground, then soldered 8ga wires to each of the eight 12v outputs (4+, 4-). <S> These wires are trimmed to the same length and twisted together at the ends. <S> Both PSUs supply 12.6v, and both allow the US market inverter to power up and supply 110v/60Hz out. <S> The inverter is a Cobra brand 1500 watt unit, with 3300 watt peak rating. <S> (Overkill, yes. <S> But now no need to worry about limits.) <S> The ATX PSU hits it's overload and shuts off with only small loads applied to the 110v side of the inverter. <S> Maybe 300 watts on the 110v makes it trip. <S> The HP server PSU handles everything I loaded onto the 110v side. <S> Did not measure that full load, but it was in excess of 900 watts by rough estimate. <S> More than this system will ever be asked to handle (operating multiple Lionel train transformers). <S> So the answer is: YES. <S> February 2017 costs: Inverter- $160NZD (shipped). <S> HP PSU- <S> $50.00 NZD (shipped). <S> Some 8 ga. <S> wire, a cord for the PSU, and some solder- <S> $25.00 NZD. <S> Total cost of this 1500-Watt pure sine wave frequency converter: $235.00 NZD <S> Cost of equivalent retail unit: <S> N/A. Nothing on the local market. <S> Cost of closest retail equivalent (at only 20 Amps): over $800.00 NZD
| Most electronics will handle the small difference between 50 and 60Hz.
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Voltage Conversion from 220V to 5V I want to convert 220 V ac to 5 V ac, and then fed that scaled ac to A/D converter.I don't want to use transformers.How good it is to use OPAMP as inverting amplifier or voltage divider.?I simulated the circuit in LTSpice by using LT1006 ,apparently there is no problem.I also used Voltage divider and buffer combination,that also worked good in simulation.Kindly suggest me the best and small circuit that has little problem associated with it. <Q> How about simply a resistive voltage divider? <S> It has the benefit of simplicity, a reasonably low output impedance to drive the ADC, and a well specified and calculable input voltage withstanding, if you read the data sheets for the components you use. <S> You could put an LT1006 after the voltage divider if you want to. <S> It depends what your ADC input looks like, whether it needs a lower impedance than is feasible with a resistive divider. <A> What are you going to do with the 5V AC output? <S> If you are just going to compute peak or RMS amplitude, phase angle, etc and do not mind some crossover distortion, you could use two optoisolators as shown below. <S> Optoisolators provide the electrical isolation like a transformer, but are much smaller and a tiny fraction of the weight. <A> If you want to interface opamps with high voltages and don't want to use a transformer, Norton Amplifiers are an alternative to using a voltage divider. <S> They work with currents instead of voltages at the inputs, so you can feed the high voltage directly to the input using a high ohm resistor. <S> The kind of old LM3900 is such an amplifier. <S> You'll find examples in the application note for the chip, chapter 10.12: http://www.ti.com/lit/an/snoa653/snoa653.pdf <S> More modern versions of this amplifier architecture are available, mostly build to be used as video amplifiers, but the same principle as the schematics for the LM3900 apply to them.
| The best circuit to use is a small transformer, as it has isolation between the mains supply and you.
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What effect does Permanent Magnet have on an air-core coil? Given an air-core coil of known inductance and a given frequency of applied voltage, you can predict the coil reactance, the current, and the voltage/current phase relationship. What effect, if any, will a permanent magnet placed at one end of the coil have on the predicted properties of the coil? Additionally, if the reactance for the given frequency is altered by the magnet, can the applied frequency be changed such that a new reactance (in the presence of the magnet) can be matched to the original reactance without the magnet. The former paragraph is at the core of what I'm trying to get at, because I want to know if the magnet will have any kind of transformer effect (parasitic or otherwise) on the oscillating coil. Will the presence of a magnet simply change the inductance of the coil, or will it cause other behaviors (similar to those seen when two coils share mutual inductance)? Or is there something else altogether I am missing? The arrangement of coil and magnet is end-to-end as the field polarities would be aligned along their axes like this: coil <--> magnet_//////_ [N | S] The arrangement would also allow for the magnet to slide into the coil, thereby creating a sort of magnet-core coil. <Q> The reactance of the coil will change and there are two opposing mechanisms at play: - The permeability of the magnet will tend to increase inductance <S> I've seen situations where reactance/inductance doesn't change at all. <S> Another effect is that the eddy currents in the magnetic material will cause the magnet to take power from the AC circuit i.e. losses will increase. <S> At high frequencies the magnet will act like a plate of a capacitor and thus there will be a resonance effect. <S> Again, without detail of the experiment this is hard to quantify. <A> If it is assumed that the magnet is non-coductive and has permeability and permitivity the same as air the only effect is a slight resistance increase due to current crowding caused by the hall effect. <A> The fixed magnetic field will not have much of an effect unless you are getting near the saturation flux of any magnetic materials in the coil such as a core. <S> I think ferrite magnets have a high enough permeability to act as a core but Nd and some other magnets are close to nothing so <S> to get an effect you would need to add the magnet to an inductor with a core. <S> Those are a rarely used but are called permanent magnet inductors (PMIs) and the magnet is added to offset the magnetic field of a large DC bias current in the inductor to prevent the core saturating. <S> It lets you make the core smaller without core losses <S> and I think there are other benefits in efficiency or noise or something too. <S> I think they would be useful in DC-DC converter inductors where there is a very high bias current <S> but I think they might not use them because having a closed magnetic path is important for energy storage and EMI <S> so there is nowhere to put the magnet. <S> If anyone knows more it's something I've been wondering about. <A> If you use steel wire (not forbidden by your question), then the magnetic field will bias the iron's magnetic domains very strongly to one side and minor signal currents will not be able to cause flipping of the domains. <S> Since any domain flipping injects noise into the circuit, we do not want that flipping to occur in our low-noise circuits. <S> With various connectors using STEEL pieces for strength, such as RCA audio plugs and jacks, a permanent magnet has its uses therein.
| The conductivity of the magnet will cause eddy currents that tend to decrease inductance To what extent that either increases or decreases the coil's inductance is guesswork and will vary at different frequencies.
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Is there a standard stating how fast a fast fuse is, how slow a slow fuse is? According to answers to this question there seem to be a standardized notation on fuses. For example F8AL250V means that it's a 8A fast fuse. On the other hand there is answers this question stating the importance of checking the datasheet for specifications on when the fuse should blow. Pulling these answers together appears to be contradicting. While F8AL250V says 8A fast fuse you seem to have to consult the datasheet to find out what that means (which makes the notation doesn't mean anything in particular). This would also be unfortunate since it would require an end user to consult datasheets (he's not likely to understand) instead of just insisting him to use the F8AL250V fuse. Is there a standard of requirements on fast and slow fuses regarding time-current characteristics? What does the standard say about the time before the fuse blows for various over-currents? My specific situation is that I'm considering a power supply (Velleman FPS1320M), the load may briefely draw more than 8A. The question is if a fast 8A fuse will be adequate. For example what amount of time can I expect the fuse to withstand 16A at a minimum (I don't have the number, but I think it's just needed to last for a minute or so) and how fast can I expect it to blow (according to the PS manual it has to blow within two hours). How fast can I expect it to blow at 20 or 22A (needs to blow within 30 respectively 1 minute). <Q> Individual manufacturers may have a consistent naming systems to identify slow, really slow, fast, turbo fast, or whatever, within their product line. <S> Some smaller manufacturers may have copied parts of the designations from a dominant manufacturer, but you shouldn't count on that. <S> Ultimately, only the datasheet tells you how fast or slow fast or slow really are. <S> End users shouldn't have to consult datasheets, and can't be counted onto understand them if they did. <S> This is no different from any other component in a device. <S> The way to handle this is to tell them the replacement part number explicitly. <S> For a common type of fuse, you can write something like "1A 250VAC" near the fuse holder, but that should be for convenience at best. <S> Somewhere in your documentation it should say that the fuse is a Acme ICUB4UCME-1A, or equivalent. <S> If they put in something else, the liability is now on them. <A> Every manufacturer has their own numbering system. <S> It seems fairly easy to interpret that F8AL250V means an 8 amp, 250 volt fast-blowing fuse. <S> It is a little more difficult to tell that it is a 5 X 20 mm glass tube fuse. <S> Littelfuse shows 217008 as a replacement and Bussmann shows S500-8-R. Bussman also offers a version that has wire leads. <S> Both Littelfuse and Bussmann give the interrupting rating as 80 amps. <S> It appears that the F8AL250V manufacturer number is used by ShengBaiDe Electronics of China. <S> It is customary for a product manufacturer to mark the product or provide in the instructions brand and number of the original fuse. <S> If the end user can not find that fuse, their best alternative is to find a cross reference chart provided by another manufacturer and use the replacement recommended there. <S> Littelfuse and Bussmann both have online cross reference search tools. <S> They are left with some risk that the selected fuse will not protect the equipment as well as the original fuse did. <S> If a designer wants to find an alternative fuse, it is best to consult the datasheets. <S> Datasheets provide curves and other information that should be used when selecting a fuse for a new design. <A> Fast is a relative term. <S> I wouldn't expect exact meanings for these two words. <S> More specific meanings of the words might be drawn from the context in which the term is used. <S> When using the science of electrical engineering, it is natural that the specific meaning of slow or fast should be found in a data sheet. <S> Which would also give the conditions where the item would be slow or fast. "require an end user to consult datasheets (he's not likely to understand)" <S> Thankfully there are instructors, classes, on line help (such as EE.SE), and Books to help gain understanding.
| Slow is a relative term. Some numbers tell more about the products than others.
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Is there a shorter or L-shaped version of PowerPole connector? if not, is there an engineering reason why not? I've not worked with this connector in many years. I'm working with a robotics team and these connectors are very long and bulky and take up [more, think] space [than they need to ]. I'm trying to find a shorter or L-shaped version. Have found none, wondering if I missed it or if there is an engineering reason why they can't be made that way. <Q> They're built that way because they are optimized for 1. <S> Power handling and 2. <S> Cost. <S> The pins are sheet metal construction (folded from a flat sheet) and typically tin-plated copper. <S> As such cannot rely on a solid core to improve conduction and better materials (e.g. gold) to reduce resistance. <S> All they have to work with is surface area and since the diameter of the pin must remain small enough to retain strength (it's a rolled up sheet) yet large enough to be manipulated by conventional forming equipment, the only degree of freedom left is the length... <S> ergo the longer pins. <A> According to the text you linked to, the connector is rated for 45 A. <S> This requires a fairly large contact surface area between the two parts of the connector. <S> Apparently the designers of this connector chose to make it long and narrow rather than short and wide. <S> However, it also says the wire is "14 or 16 gauge". <S> According to the first ampacity table I found online, 16 gauge is limited to about 22 A for "chassis wiring" (Different standards might give slightly different ampacities, based on different assumptions about airflow conditions and allowed temperature rise of the wire). <S> If your current requirement is less than 25 A, you can almost certainly find a more compact connector able to handle it. <A> I have never used the specific connectors you have linked to, but here is a general idea. <S> Connector design depends on largely three things: <S> How much current do the contacts carry. <S> This determines the conductor and contact area. <S> Higher the current, larger the conductor contact area you need. <S> If the area is too small, the contact will have high resistance and start heating up. <S> Voltage and insulation. <S> The voltage between adjacent pins determines the distance and and the insulation between between adjacent connectors: for instance a 5V connector can have pins a few mm apart, not so for a 400V rated connector. <S> Reliability and mating cycles:Some connectors are designed for repeated plug-and unplug, some are intended to be connected/disconnected infrequently only during install or service, which also determines the size and strength of the connector as a whole. <S> Based on the criteria you can surely find other connectors, selecting one based on your tradeoffs.
| So either the wire is undersized for the connector, or the connector is over-specified for the wire.
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Determining current draw of arduino I am trying to determine the correct batteries required for my Arduino (Atmega 328) project. I have components which have both voltage and current ratings. I am adding up all the maximum current specifications from the components to determine the ideal mAH. I am currently stuck on the Atmega328 chip. I read the datasheet, but I cant seem to find the specific current draw of the chip. My understanding is that the operating voltage is ~5V, and the current draw of the chip would be ~40mA. But if I am using 6 pins (3 analog, 3 digital), assuming they all output 40mA at the same time (which is unlikely as the 3 analog pins take in x,y,z values from an accelerometer), would it be safe to assume the current draw of the chip would be 240 mA? (this sounds wrong) <Q> The datasheet of a microcontroller tells you the current it uses internally. <S> Only you can determine how much additional current there might be going into the power pin or out the ground pin due to other pins sinking or sourcing current. <S> From what you say, the micro itself can take up to 40 mA. <S> I don't know that particular micro, but that's certainly plausible. <S> The datasheet is also telling you the maximum that output pins are allowed to source. <S> This is of no use in a power calculation. <S> You use this information during the design to make sure this limit is not exceeded. <S> The actual current that the external circuit draws from a output pin is dependent on the external circuit. <S> Consider the limiting case where a pin is left unconnected. <S> Clearly no current is going out that pin. <S> For example, let's say you're lighting a LED directly from a digital output. <S> The processor runs on 5 V, and you have a green LED to ground with 300 Ω resistor in series. <S> The current going out of the pin <S> when high will be about 10 mA. <S> The fact that the output could have sourced up to 25 mA, for example, is irrelevant. <A> Most CMOS microcontrollers, ATMEGA328 included, will consume more current from their DC supply when their clock runs at higher speed. <S> On the ATmega328 data sheet, power consumption is spec'd at a speed of 1 MHz. <S> When its clock is running, and the processor is processing, it will pull 0.2 mA current from a 1.8v DC power supply. <S> This would be minimum current, <S> will all input/output pins not supplying current. <S> When the clock is not running, current draw is reduced. <S> Atmel specs "power-down" mode current draw as 0.1uA. <S> With no clock, it cannot run programs. <S> With its time-of-day clock running (32.768 kHz), current draw is 0.75 uA. <S> This real-time-clock is separate from the processor clock. <S> Some processors brag about their speed-power performance, for example "100uA-per-MHz". <S> This is a figure of merit, suggesting that a 100MHz. processor clock would require the processor to draw ten milliamps of DC supply current. <A> The atmega has a CPU, right? <S> So if you make it work hard, you'll spend more energy. <S> (It will get hot, albeit unnoticable) <S> This is why your laptop CPU/GPU gets hot when you're playing computer games. <S> Your doing a lot of calculations, the transistors are switching. <S> Transistors lose energy when they switch on/off due to crossover. <S> but you'll never know for certain. <S> About the pins, 40 mA is around the max rating a pin can provide. <S> It's not what you are actually drawing. <S> So, you can't use that math to calculate the current consumption. <S> And then there is the hardware problem. <S> There are always current leaks in a hardware. <S> You will never be able to calculate this most of the time. <S> Only experiments will tell. <S> But the thing is, usually, for simple applications, only the pin currents matter. <S> The others are usually less than that. <S> If you are researching this for a battery powered application, please consider sleep/deep sleep options. <S> It works like a charm. <S> Oh and analog input pins don't draw much current. <S> They draw just enough to sample the voltage <S> but it's not that much. <S> Only the GPIO pins in source/sink mode will draw/source that much current. <A> You will go blind trying to work out the average mA/h consumption of an Arduino by adding all the component and feature currents together. <S> Buy yourself a USB voltage/current meter. <S> This is placed in series with the USB line feeding the device and tells you the voltage and (average) current flowing. <S> The USB current meters are available from ridiculously cheap on Ebay ($3) through quite expensive ($30+). <S> My personal favorite is the Portapow which directly tells you the mA/h rating of the attached device. <S> All the USB current monitors I have seen have the same problem when it get's to sleep states.... <S> they all have a limited minimum current resolution. <S> When it comes to measuring sleep state current a series resistor can't be beaten.
| Also, the best way to calculate current consumption ( cheap and fast ) is to get a resistor, calibrate it (find its real resistance), put it in series with atmega VCC and find the voltage drop. The datasheet will give you an idea what the CPU normally draws
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What does this bus signal representation mean? Often, when looking at digital system signal plots over time there are these graphs with two parallel lines that often "swap" place. I am wondering what precisely they represent? I noticed they usually are used for buses, but for the rest I'm clueless. It's probably quite trivial. This is an example of the graphs I mean, in this case from a discussion about how CPU's execute instructions: I've tried googling, but I couldn't find an explanation. <Q> Since most logic and microprocessor IC's use edge clocked timing, it's when the transitions from 0-1 and 1-0 occur that are important to timing. <S> The high and low parallel lines simply tell you that a line may be high or low but not some indeterminate level such as tri-state). <S> In your case shown, greyed out means not valid data/address (multiple signals), not greyed out <S> means they will be 0 or 1 <S> but the data is valid. <S> The crossovers graphically show the setup and hold timing in relation to clocking. <S> You would read accurate setup and hold times from a data sheet of course. <S> For example in the image you show When the clock goes 0-1, the program memory address's are being setup after the greyed (invalid address on the lines) time period. <S> Once the address is stable (non-greyed) <S> the data lines represent (but can be 0 or 1) valid and stable data until the access time passes. <S> Then the data bus will contain either 0 or 1 as appropriate for the memory address. <A> The two parallel lines indicate times when the data on the bus is stable, and the crossovers indicate when the data may change. <S> The shaded sections are times we don't know or care what value is on the bus. <A> This bus could be two bits, or 64 bits, or whatever the design calls for. <S> As shown here, the grey regions indicate that the data value is unknown or indeterminate. <S> In this case, immediately after the positive clock transition, the address and data are unknown. <S> Soon, after that, the address data becomes stable and known. <S> Some time after that the data is known and can be read. <S> edit: As Peter says, the grey regions can also mean "don't care".
| The parallel lines indicate a multi-bit bus, and represent any possible combinations of low/high values.
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What difference does the clock make in synchronous vs asynchronous communication? I am new to embedded programming and have very little knowledge in how digital/electronic system works, but I have been given a task to bring up USART communication between two chips. For that, I have started learning from the basics of serial communication etc etc. What I don't understand in serial protocol is that what is the difference between synchronous and asynchronous communication? Let us say I have two devices d1 and d2. They both are aware of their baud rates. d1 sends data at a baudrate of 9600 and d2 receives data at the same rate. Then my question is, how the clock plays role in synchronous communication? Means with baud-rate I think we have all the information of the communication channel. With this confusion I am not able to clearly understand the exact difference between synchronous and a synchronous communication. <Q> With asynchronous data transmission, when there is no payload data to be sent, the data line becomes idle and therefore the receiver waits for a transition that marks the beginning of new incoming data. <S> In this respect there is no definite phase relationship between data previously received and the new data arriving. <S> This is why it is called "asynchronous" With synchronous data, either a clock is permanently present as a seperate signal or, the clock is embedded into the data (as per Manchester encoding or scrambling) so that the receiver is always aware when real payload data could be present. <S> When no payload data is present the clock is still present. <S> Manchester encoded data: - how the clock plays role in synchronous communication? <S> Whether you extract the clock from the data (or have a dedicated clock line) <S> you need a clock; that is fundamental to any data transmission. <S> Asynchronous transmission re-creates a clock internally based on the agreed baud rate and the first transition of data following an idle period. <S> From this point on until the end of the data block it generates a clock internally for the whole transmission. <S> So the clock plays a vital role in both asynchronous and synchronous data communications. <S> with baud-rate <S> I think we have all the information of the communication channel <S> Not quite - knowing baud rate at both ends is useful but <S> it doesn't tell you when bits actually change state in the data - this "synchronization" to the data is performed by the idle-to-start bit transmission in asynchronous data and is everpresent in synchronous data communication. <A> In Synchronous mode, the clock is shared between the 2 devices. <S> Example SPI -- <S> > <S> At each Clock Cycle, the slave sample the data sent from the master. <S> In Asynchrounous mode; there is no clock shared and the 2 devices need to work at the same frequency. <S> Asynchronous challenge lies in making sure that both devices are at the same frequency (whatever voltage, temperature, drift ...) <A> Based on the Clock reference there are two modes of Serial communications. <S> Synchronous Serial Communication <S> In Synchronous communication, the communicating devices should at least need two lines to send the Data. <S> One line for Clock it will be the reference and another line will be the data to be sent. <S> Basically clock is given by any system or on-chip clock. <S> Its Frequency based on the Baud Rate. <S> Refer the Image: Clock is separate and Data is separate and absence of start and stop bits Asynchronous Serial Communication <S> In Asynchronous Serial communication using single line data can transfer between the devices. <S> Here there is no need of clock line separately, this will handled by some Start and Stop bits with in the data block itself. <S> Refer the Image: Clock is absent but start, stop and other supporting bits are presence <S> Note : <S> Synchronous communication suitable for Bulk data transfer. <S> Itmostly transfer the data frames by frames ASynchronous Serial communication suitable for byte transfer. <S> It mostly transfer the data byte by bytes.
| In General Digital Systems to communicate serially between two or more devices we need some reference that is Clock.
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What is needed to switch a relay via logic output (e.g. from Raspberry Pi)? I'm working on a project with a raspberry pi involving some sensor and a relay.Actually I'm Using a board like this one: So I connect GND,5V, and a GPIO to the IN pin to control the relay state: ON/OFF. Now I'm starting to move from the prototype stage.. I'm working on create the PCB, and I'm wondering if I can drive the Relay, with a "naked" version like this: is it possible? Why all the other components are required on the board, figure 1? resistor, transistor etc.. I'm expecting to connect: D to the source and one between A or B to the destination C to GND E to a GPIO ..correct? Maybe I'm using "rocket to kill an ant".. I'm using the relay for control a low voltage (between 12V and 24V DC)..is there any more appropriate component or integrated circuit that you can propose? <Q> 1) You cannot control the relay directly from an Rpi board, you have to use at least a transistor with base resistor (or a mosfet) and a flyback diode. <S> That is what is on that small board. <S> 2) <S> no that is not how you should connect it. <S> It will not work like that. <S> 3) <S> Yes there are other ways to control a different low voltage <S> but I do advise to stick to a relay solution using that small board as it is cheap and reliable <S> and you do not need to think about shared ground connections etc. <S> You could implement the circuit from that small relay board yourself but believe me, it is not worth the trouble. <A> It probably uses this circuit: - By the looks of it the diode is a 1N400x type and the input resistor looks like 150 ohm (marking 151). <S> Maybe I'm using "rocket to kill an ant".. <S> I'm using the relay for control a low voltage (between 12V and 24V DC).. <S> is there any more appropriate component or integrated circuit that you can propose? <S> Yes, if the load shares a common 0V with the RPi, you can use a MOSFET like this: - <S> There can be some simplifications if the supply is only 12 volt - the zener diode can be removed and the lower resistor shorted in most cases. <A> The transistor is used to drive the coil that activates the relay. <S> The diode is a flyback diode which is used to discharge the coil when switching the relay off, otherwise the transistor will be damaged. <S> Basically, you are right <S> but you need a transistor to drive the relay because the microcontroller can't deliver the required ~90mA at 5V. <S> It looks a bit overdesigned if you only want to switch 24 VDC.
| It depends on many parameters (current, voltage, switching frequency) which relay you can use. You need to design those components onto the board if you want to use this relay "naked".
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Why do electrical systems have neutral, live and ground when electronic systems have only VCC and ground? Wouldn't it be simpler if electrical systems take away the neutral and just have live and ground similar to electronic systems? <Q> Wouldn't it be simpler if electrical systems take away the neutral and <S> just have live and ground similar to electronic systems? <S> Ground is localized to a house or installation. <S> Ground is not provided by the electricity companies because that would be unsafe - how could a safe ground be sourced from an electrical generator hundreds of miles away? <S> How could you detect if current was flowing through some unfortunate person (in the process of being electrocuted) <S> if one of the power feeding wires is actually ground - how could you tell that the current was flowing through that person without a neutral wire? <S> How would you distinguish it from normal load current? <S> That's the whole purpose of residual current devices (UK name) or ground fault current interrupters (USA name): - <A> The reason we seperate Neutral and Earth is safety. <S> Consider what happens when a wire breaks. <S> In a seperate neutral and earth system <S> If the neutral breaks the appliance stops working, but there should be no hazard to the user. <S> If the earth breaks then it can no longer protect against short circuits to the case but assuming there are no other faults it is not immediately dangerous. <S> In a combined neutral and Earth system. <S> If the combined neutral and earth core breaks the case immediately becomes live via the load. <S> Most electronic systems don't have this issue because the voltage is too low to present a significant hazard. <A> Actually your assumption is wrong. <S> In electronics where there are signals on cables that need protection in either direction of radiation, we have 3: power/signal, common and shield. <S> Otherwise double insulation is used for safety on 2 pronged units. <S> In some countries where 3 phase power is offered there is no grounded Neutral. <S> The purpose of a GFCI ( ground fault circuit indicator) is to detect power leakage currents and break the circuit open. <S> Often humidity in bathrooms can trip this due to high sensitivity. <S> It does not work by sensing ground current rather as a differential relay operating from very small current differential between Line and Neutral. <S> The theory being, if they are not equal and opposite, then there must be some leakage.
| Shield is necessary to shunt leakage and stray currents for the same reason Earth ground is provided in ac power service for human safety and reduce radio noise.
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Proper ways to disconnect ICs during low power states to avoid parasitic/backfeed supply I’m working on a low-power battery-based AVR-based project that integrates a few different devices including a neopixel strip and an Adafruit pixie . When the overall device is quiescent, I’d like it to draw less than 0.1mA to maximize the LiPo battery life. I got this all working (measured 0.035mA) but I’m not sure I necessarily did it in the “right” way and I plan to build a product based on this so would like to do it right. (Not shown: a flyback diode for the relay) The core concern I have is the “parasitic” powering of devices when VCC is disconnected via current flowing from data pins. For instance, the Pixie (which communicates via serial), has no power down mode and even when “off” drains about a milliamp. So I placed a small relay to disconnect its VCC, and discovered that the serial pin was actually still powering the pixie. Hints elsewhere suggested that many chips have a diode shunting their digital input pins to VCC as power protection. To solve this, I’ve had to suspend the serial library and actually digitalWrite( PIN, LOW ) during sleep. Same thing with the WS2812b strip — disconnecting VCC still allows the device to be powered from the data pin. And in other designs when I’ve disconnected GND with an N-Channel MOSFET, I’ve seen the reverse - a back flow of current through the data line to ground! (This had to be solved with a diode per a post on PJRC.) The WS2812b’s actually take about a milliamp each even when unlit, So the question: Is there a general, “clean” way to disconnect VCC and GND from parts of a project during system sleep when there are data pins in the mix. What is the best practice? Some ideas: Force VCC to GND (not sure how? Hbridge?). (If I do that, what happens to the data pins that are high?) Place a tri-state buffer between all data pins and these devices, and during sleep put the tri-state buffer in a high impedance state, disconnect VCC or GND only with P or N mosfet Disconnect GND only with N mosfet, and place diodes on all data pins Is there some kind of power latch that disconnects both VCC and GND and puts them into a “high impedance” state (like a tri-state buffer for power?) That way current has no way to flow "out" from the data lines. Can someone enlighten me to the cleanest, most repeatable way of handling this sort of “load disconnect” problem? (Needless to say, I have spent hours googling this problem with little luck, although I did find this tech note on load switching but it doesn't address back-feed and parasitic power) <Q> When I do this, I usually use <S> CMOS analog switches on the affected data lines. <S> Something like the ADG812 has 4 channels of SPST switches that are easily suitable for quite fast logic, and provide a really high impedance between the switch nodes when in the off state. <S> The nice thing about this is that the technique works for both unidirectional and bidirectional data lines. <S> These parts also run on a bright smile: <S> The usual sequence for power down: <S> Turn off data path switches Power down domain. <S> Power up is the opposite, of course. <S> [Update] These are indeed known by other names, such as pass gates and transmission gates . <S> These are significantly different from a true tri-state buffer (as you can see in the diagrame in the link above), but for ordinary logic, the effect is better (this is inherently a bidirectional device) but with lower power. <A> If the data signals are connected to your microcontroller, you can simply make them high impedance by configuring those pins as inputs.(If the other chip uses very little power, you can treat its Vcc like a data signal.) <S> Otherwise, you can use analog switches (74x66 logic chips) to disconnect them. <S> For unidirectional signals, 74x125 would work, too. <A> I don't think there's a one size fits all strategy unfortunately. <S> Switch power to subsystems as you've already done. <S> In software, drive pins low for low power states, unless doing so would cause a high power steady state condition. <S> In that case, drive the pin high. <S> Never let inputs float. <S> Sequence power as necessary to establish safe initial conditions. <A> A nasty problem that can cause microcontrollers to do very strange things. <S> The nice solution is to use pull-down serial <S> I/ <S> O like \$ <S> I^2C \$. <S> This requires pull-up resistors on SCK, SDA lines. <S> The pull-up resistors are tied to the switched Vcc line. <S> Ensure that the switched Vcc line falls nicely to zero volts when it is switched off <S> (don't let it float). <S> You don't have that option - you're forced to use asynchronous serial I/O. <S> Some microcontrollers allow a similar approach as I2C to solve the problem. <S> This solution is not as noise-robust as your present approach, but it should solve the problem of back-powering your I/ <S> O modules from the AVR. <S> It isn't really a "clean" solution, but it is far safer for the microcontroller(s) in your IO modules. <S> simulate this circuit – <S> Schematic created using CircuitLab
| If you can program the serial output pin to be pull-down-only rather than the more common pull-up-for-1, pull-down-for-0 , then you can add a pull-up resistor to switched-Vcc to establish a logic high.
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Is 74LS139 a DeMUX? If yes then how can I give it input and select lines? I am trying to learn DEMUX. So, I have a 4 Channel DeMUX Diagram: From this diagram we can clearly see that a DEMUX needs 2 Select Lines and 1 Input to give 4 output lines. After that I have taken a look at the datasheet of 1:4 DeMUX IC 74LS139, which describes a diagram as shown below. Here is the link to take a look at datasheet. In second image we can clearly see that it has two input lines A 0a and A 1a on left DeMUX and A 0b and A 1b on right DeMUX. There is no select lines. Also we cannot see a logical connection between any input lines of Left and Right DeMUX. Can someone explain me, how to use this IC as DeMUX? <Q> In your first paragraph you say that a demux requires: 2 Select Lines and 1 Input to give 4 output lines <S> In the logic diagram of the 74LS139, you have exactly that. <S> Two address (select) inputs A 0a and A 1a , one input E a , and four output lines <S> O 0a , O 1a , <S> O 2a and O 3a . <S> In fact the image you show from the datasheet explains this in the "Pin Names" section just below the connection diagram. <A> This chip is a dual 2->4 demux, not a 3->8 demux, that's why there are no connections between the two halves of the chips internals. <S> The A lines are the address (=select) lines. <S> The E (enable) line can be used as (active low) input. <S> The outputs are active low, hence the outputs that are not addressed are high. <A> and For enabling this part you can use Ea and similarly for Other part use B0b and B1b as input and Eb to enable it. <S> A0a , A1a , <S> Ea - <S> > <S> O0a, O1a, O2a and O3a. <S> B0a , B0b ,Eb -> <S> O0b, O1b, O2b and O3b.
| The chip is Dual 2->4 demux which means that you have two demux in one circuit so you can have two separate Multiplexed input from A0a and A1a
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P-channel MOSFET for 100A? I'm using an alternator to charge a 300Ah lead-acid battery bank. The cable has a 60A fuse, yet measuring a shunt suggest that the current is about 70-90A for the first 10-20 seconds, and after a few minutes it is down to about 30-35A. The alternator provides 14.0V at the car battery, and cable losses further reduce the voltage to about 13.8V at the battery bank. I'm trying to introduce an automatic switch using a P-channel mosfet. The problem I'm facing is that the voltage drop over the MOSFET should be minimal, since a lower voltage will significantly reduce the charge current.A better alternative would be a boost circuit to 14.4V, but at 30-50A, they get very expensive. The P-channel MOSFETs tend to have higher on-resistance that N-channel.But I've found this: AP6681GMT The package is undesirable, and heatsinking is only possible through the PCB. I built a circuit, but when I tried it, black smoke came out of it, and it failed open. I'm wondering what I did wrong, and what my alternatives are. I understand a relay has a high on-resistance? Would it make sense to buy a starter relay for a car? The specification is typically not provided and perhaps the on-resistance will be too much. simulate this circuit – Schematic created using CircuitLab For testing the circuit, I connected the drain-side to a blower motor, and switched the gate-signal on/off every 1000ms. The circuit worked correctly, even with peak currents of 20A and sustained current of 7A. I then connected the drain-side to the battery bank.When I turned on the alternator, it jumped to 14.0V on the source side.Is it possible that the 1k resistor didn't raise the Vgs fast enough, such that it continued to conduct without pulling-up? e.g. a runaway scenario. UPDATE: Using an N-channel MOSFET instead? N-channel MOSFET: IRFB4410Z (100V, 97A, 9mOhm/10V) or AP99T03GP (30V, 200A, 2.5mOhm/10V 4mOhm/4V) simulate this circuit This would allow me to use a heat sink, to have lower on-resistance, bigger package. But with the inconvenience of a boost-circuit. <Q> I'm wondering what I did wrong <S> What you did wrong was not understand the peak voltages that can come from an alternator and how these will easily exceed the gate-source maximum voltage for that device (+/- <S> 20 volts). <S> You should have used a 15 volt zener protection diode across the gate-source and a small collector resistor to limit current into the zener when the alternator produces spikes. <S> The BC337 is also a risky device to use given that it's peak voltage spec is only 45 volts. <S> Automotive circuits like this are really difficult to get reliable without going for 100 volt rated devices. <A> The basic formula for smoking parts depends on the thermal resistance of temp rise per power lost in ['C/W] for the rating of the heatsink. <S> Pout = 50A*14.4V= 770 <S> Watts <S> However when supplying a sine wave into a rectifier with a battery acting as almost a short circuit the peak currents can be be >>10x times as much, thus increasing the requirement now to >> <S> 500A pk <S> With 2.2mΩ and short term currents of >>100A for a drained battery <S> I^2R = <S> 22Watts <S> with good heatsink to chassis with insulator and paste. <S> To minimize the Vds drop, AND have sufficient Vgs <S> now you need a Pch switch with gate to ground control <S> AND have accurate gate control. <S> Use a Pch and decide what threshold to disable the switch. <S> I roughly chose 13.2V <S> R divider with a poor-man's 1.1V Darlington comparator. <S> ( it works) <S> The threshold may need to be corrected slightly to 13.4V to as a threshold. <S> Since the drain current is low the input threshold to the Darlington is also low at 1.1V @1mA $2.20(1pc) <S> 2.2mΩ <S> Pch type <S> 170A simulate this circuit – <S> Schematic created using CircuitLab <S> The Vbat threshold to Vgs voltage gain is > 1k <S> so the threshold amplifies Vgs to quickly move past turn ON threshold from 1.2 to 2.2V with just a few mV change in V+.. <S> the only drawback is the threshold is not very precise (5%) and has a NTC of -2.1mV/'C which translates to 21mV/'C of battery threshold shift or a drift of 13.2 +/-0.4V for <S> +/-20'C <S> so 13.5V is advised. <A> Deciding against P-channel: Unintended activation and thermal runaway <S> Using high-side NPN instead <S> : The low-side cannot be connected via an N-channel, because a constant chassis connection is desired for better earthing and easier cabling. <S> A high-side N-channel thus seems the best option, however requiring a boost circuit to provide a gate voltage 10V above source. <S> Reducing thermal problems: <S> In addition, I've concluded that using several N-channel MOSFETs will allow me to reduce current effects significantly. <S> For example, using two MOSFETs, each experiences half the current, but 1/4th of the power dissipation. <S> At 33A each and 2.5 mOhm, the heat gain will be 2.7w. <S> Solving overvoltage problems: I'll also try to add a Zener diode at 20V through a 500mA fuse to see if overvoltage occurs despite the main car battery being in parallel. <S> If it turns out to be a problem, I'll add a 16V zener with a resistor and test again, to see if the problem has been resolved.
| In retrospect, I think a P-channel with pull-up, where the source voltage changes, is not a good design approach, because the voltage change can trigger a small voltage across the Vgs, which in turn allows current but at high resistance, thus potentially causing thermal runaway as well unintended activation. And since power dissipation is the biggest problem, I'll probably get three MOSFETs with heat sinks.
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Can I draw 2,2A current at 9V with AA batteries? I am new to electrical engineering aand I need a battery for a project of mine. To power the whole system I need 2.2A of current (that is the max current the motors will need, though they will not be running constantly) on 9V voltage. Can I even draw that much current from AA batteries? And can you also tell me how the batteries should be allinged to be as least space consuming and what will the capacity of all batteries be? <Q> 2.2A is not a problem, and is often done in RC cars. <S> This datasheet shows that a 1300mAh cell is rated for up to 3C discharge rate, which is 3.9A. <S> At 2C (2.6A) you should get nearly 30 minutes of use. <S> Higher capacity cells are available. <S> The only issue is that NiMH cells are only 1.2V per cell, so 8 cells give you 9.6V which may be too much for your motors. <S> Check the spec for your motors to see if they can operate from 9.6V. <A> Do note that at a ~800mAh capacity, draining at that rate will mean it dies in ~20 minutes or so (or rather the voltage drops under .8*6, or 4.8v) <S> as for the physical battery placement to minimize space (defined as volume), that would probably be a long chain of 6 all in a "tube", that way you aren't trying to line up circles creating dead space - how practical that is depends on your application, but I would suggest just getting a holder designed for the batteries <S> For how large the batteries should be, that depends on how long you need them to run and at what average current, which you haven't supplied here, but if you are fixed on a AA size, that doesn't give you much play as far as capacity goes <A> You can draw 2.2A from AA batteries, but you will need to select a battery that was designed for a 2A load and has low internal resistance to prevent the output voltage from drooping too much. <S> I would suggest the ENERGIZER NH15-2300 AA battery. <S> It shows a typical run time of just under an hour at a 2.3A discharge rate. <S> It has an internal resistance of only 30 mΩ, which is 1/10th that of a typical cell (such as the Energizer E91). <S> It is an NiMH type battery so its only 1.2V. <S> If you use 8 of them you can get 9.6V, which is probably close enough. <S> The output voltage for the 8 cell stack at 2.2 A would be 9.6 V - 2.2 <S> A * 8 * 30 <S> mΩ = <S> 9.07 V. <S> http://data.energizer.com/PDFs/nh15-2300.pdf <S> You can buy them for $4.20 each on Digikey. <S> http://www.digikey.com/product-detail/en/energizer-battery-company/NH15VP/N703-ND/1040723 <S> Here is why you can't use a typical AA. <S> Lets use the Energizer E91 battery as a typical example (thanks user2813274). <S> http://data.energizer.com/PDFs/E91.pdf <S> The datasheet for the E91 AA battery states that is has an internal resistance of 0.3 ohms and an open circuit voltage of 1.5V when new. <S> You would need 6 of them to make 9V. <S> So the total series resistance would be 6 <S> * 0.3 Ω = 1.8 Ω for the 6 battery stack. <S> At 2.2A the output voltage of the E91 battery stack would then be 9V - 1.8 Ω <S> * 2.2A = <S> 5.04V. <S> The 5.04V would be for new batteries only. <S> The battery output voltage lowers as the batteries discharge, so over time it will be much less than 5.04V. <S> The problem is made worse by the fact that the total battery capacity is much less at high currents. <S> For example the E91 battery that has a capacity of 3Ah at 1mA but the capacity drops to less than 0.9Ah at 2.2A. <S> A 0.9Ah capacity discharging at 2.2A would lead to "dead" batteries within 24 minutes. <S> So a typical cell like the E91 is likely to be discharged to 0.8V each within 24 minutes. <S> At 0.8V point it may not even be possible to get the full 2.2A out of them even with the output shorted. <S> A typical cell won't last long and due to voltage drop your motor won't receive very much power. <A> Yes you should be able to draw 2.2 amps. <S> Assuming your using akaline, they probably wouldn't last more than 5-10 minutes at that current draw. <S> You might get up to 20 minutes from lithium. <S> Ref: <S> http://www.powerstream.com/AA-tests.htm <S> https://www.duracell.com/en-us/techlibrary/product-technical-data-sheets <S> You probably want to look at 2s (7.4v) or 3s (11.1v) LiPo. <A> It's been my experience that AA batteries produce 1 <S> Amp at 1.5 Volts. <S> They'll produce more Amps than that, but not for very long. <S> If you have 6 AA batteries in a series, you'll get 1 <S> Amp at 9 Volts. <S> Although I would recommend using a 9 Volt battery or at least 6 D batteries to get that much power. <S> You don't want to have to change your batteries every 5 minutes. <S> Although I haven't tested it, one 9 Volt battery should produce 2 Amps, but not for very long. <S> Maybe 10 minutes.
| It depends on the battery, I think you will be able to get away with it, but barely: here is an graph from an energizer AA from the Energizer E91 datasheet Notice that it doesn't go all the way to 2.2A (2200mA), but it looks like you should be able to extrapolate the extra bit as it's almost there For high current drain applications you are best to use NiMH rechargeable cells, which have a lower ESR than most alkaline cells and so can deliver higher currents. Use 12 AA batteries in two groups of 6 to get 2 Amps at 9 Volts.
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Arduino I2C communication between 2 master networks I have 2 Arduino microcontrollers, each with a network of I2C devices connected to them (one has 2 ADCs and the other an LCD display and a RTC). How can I use the I2C connection to transfer the values obtained by the first uC from the ADCs to the second ? Both uC are masters on their I2C busses.I was thinking of making a software I2C on the second uC and connect it as a slave to the first one (so the second controller would have 2 I2C ports: one hardware and one software). Problem is, I can't find any software I2C library that works as a slave. All are masters. Waiting for your ideas. Question is, can I make 2 I2c networks using one Arduino Mega2560 ? One as master on the hardware port to communicate with the LCD and the RTC and one as a slave on a software port on 2 other pair of pins (for SCL and SDA) for receiving data from another master arduino... After analyzing all the data, I reach the conclusion that the 2 I2C busses cannot be linked together. On the external I2C port I have available on the data acquisition uC, I will connect another arduino as slave that will receive the information an pass it on by using a wireless adapter (probably a NRF24N01). That way, I don't need to have wires from my solar controller to the arduino that reports the production to pvoutput website. <Q> IIC is not a good choice for peer to peer communication. <S> The easiest way to connect to microncontrollers is probably with UARTs. <S> Note that provides a separate and asynchronous channel in each direction. <S> You also aren't stuck with standard baud rates. <S> Use something fast that can be derived directly from both clocks. <S> You can easily do a MBaud or more between UARTs on the same board. <S> Especially if you crank up the baud rate, it might be a good idea to implement flow control. <S> Some microncontroller UARTS have RTS/CTS built in, but even if not, this kind of capability is easy to add in firmware. <S> Make sure that the reciever can buffer at least as many characters as there are in the hardware FIFO of the sender. <S> That way the flow control line can be used in the sender to simply not write more data to the hardware. <S> Microcontrollers tend to have small UART output FIFOs (usually just one or two, rarely more than 4), so this is not much of a problem. <A> You can only have one I2C master on the I2C bus. <S> (There maybe some advanced protocols where you can switch who is the master.) <S> Many embedded processors can be configured as either I2C master orslave. <S> (I do not believe the stock Arduino I2C library allows for an easy switch between master and slave. <S> There may be Arduinos which use processors with no support for slave mode.) <S> These types of protocols (SPI or I2C) should be used by parts sharing the same board and power supply. <S> They are generally not robust enough to suffer the complications of (for example) partial power failures, hot-plugging or excessively long cables. <S> All that said, look here for how to connect two Arduino Uno boards (one master and one slave) using an I2C bus. <A> What @olin has suggested is something you should consider. <S> Incase, you are in need of two modes of I2C ( both master and slave ), then read on. <S> The arduino can be configured as either master or slave at any given point in time. <S> So, i would try this. <S> The arduino which has to take both roles (call it Arduino A) <S> can be configured as slave by default. <S> By this way, the Arduino will not miss any commands from the Arduino B (the arduino who will master always). <S> The Arduino A can decide itself when to change its role to Master . <S> This depends on the periodic intervals it has to read data from. <S> Once data is read from the sensors or data is written to LCD, the arduino A can take up role of slave again. <S> One catch is that, while Arduino A has taken up role of the master, the Arduino B cannot communicate with the Arduino A. <S> This can be solved in many ways: Using a spare GPIO, <S> Arduino A informs to B <S> that it is busy <S> The Arduino B can retry after a small period <S> Arduino A can inform Arduino B when it is ready for communication (example: letting know the B when the data to be communicated is available via GPIO). <S> Program the Arduino B to periodically write the data what Arduino A expects. <S> Consider <S> http://www.gammon.com.au/i2c <S> if helpful. <S> There are good information about configurations. <S> Everything in one page. <S> Please don't get lost while navigation. <S> Finally, the factors which influence the decision of logic is your own application demands. <S> The frequency of communication -which might be once in a while or contiguous, criticality of response time, not the least amount of resources one would like to spare,in case one decides to implement I2C software slave also. <A> From the comments you want to be able to connect only to the I2C bus on each of your separate projects and transfer data. <S> Wirning lets you run an Arduino as either an I2C Master or a Slave. <S> You can't run both the master and slave Wiring software on a single micro since they both want to use the USI hardware, but since you only have one USI, you can connect to only one bus anyway. <S> There is a very nice <S> TinyWire <S> library available for ATTiny85 from Adafruit....they also have a very small board called the Trinket too that you could use. <S> There are a bunch of ATTiny85 boards (like Digispark) available that can hook up to the Arduino programming environment so this should be a simple and cheap way to create an I2C slave. <S> While some may say this is overkill, it would be extremely simple to implement and would not require any hardware mods to your project.
| I'd suggest an effective way would be to use an ATTiny85 as a I2C slave interface on each bus and then connect them together via a software UART.
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"Where" does back EMF appear in motor? I've been trying to understand EMF, specifically back EMF in electrical motors, such as for example a setup like this: Say that we apply a voltage to the brushes. I understand that once the rotation gets going, a current will be induced by the changing flux inside the loop, and that this induced current is opposite to the current created by the external voltage. My understanding is that this results in a smaller net current in the loop, and this phenomenon is called back EMF. However, I don't exactly understand "where" the EMF (which is a voltage?) appears. One explanation I've seen is this: where the EMF appears as a voltage source in series with the motor resistance. While this may be a good way to explain how the current will behave, it doesn't seem to explain what actually goes on inside the motor. Surely, the EMF doesn't appear "before" the motor, but rather inside it. Should it then somehow be seen as being superimposed on the resistor? Where, in the first figure, would the EMF appear? Does it change the voltage across the brushes? Where does the "extra" voltage drop happen? <Q> When a wire is swept sideways thru a magnetic field, a voltage is generated along the length of the wire. <S> Spin the motor with just a voltmeter connected, and you will see it make a voltage. <S> So yes, the resistance and the back EMF are actually distributed along the wire in the coil. <S> There are lots (infinity, actually) of little resistances in series that each get a little voltage in series with them when the motor turns. <S> Viewed electrically from the outside, this can't be distinguished from a lumped resistance in series with a lumped voltage source. <S> Since this is simpler to draw, think about, and analyze, that's how motors are usually shown. <A> You can simplify things by removing the battery from that diagram, and turning the motor by hand, as a generator. <S> As the armature turns, the wires cut through the magnetic field, and a voltage is generated in them. <S> This voltage is the same whether the armature is turned as a generator, or as a motor. <S> When it's turned as a motor, this voltage is called the 'back EMF'. <A> Think of it like a generator; you spin the rotor and an output voltage is produced. <S> This voltage is in series with the rotor coil. <S> It makes no difference if instead of manually spinning the motor you apply dc to turn it. <A> The back EMF "appears" across the slip ring brushes. <S> Its polarity is such that it always opposes the "driving" voltage, causing a lower current draw than "normal." <S> The "normal" current is found by locking the rotor and measuring the current drawn, then the rotor is allowed to rotate and the "running" current will be less than the "normal" current, due to the back EMF. <A> Good question: This is actually not so easy! <S> It has taken me days to understand. <S> Thanks to your pedagogic picture I finally got it! <S> You need to view it is as if the battery was disconnected from the circuit. <S> So, you have a circuit without current running through it to start with. <S> You spin the coil clockwise with your hands in the same way it would spin if the current was on. <S> In the position you have made your drawing you have the right side of the coil moving downwards. <S> So, you have the wire of the coil moving downwards, which means that also the positive charges in the wire coil are moving downwards (but not moving along the wire). <S> With the rigth-hand-rule no. <S> 1 <S> you point your thumb downwards in the direction of moving charges, while at the same time you have your fingers pointing to the right along the magnetic field. <S> Then you get a force from the palm of your hand pointing along the coil wire towards yourself in the picture. <S> That is, you have positive charges (current in other words) moving in the wire towards the right-hand side brush. <S> So, you get a surplus of positive charges on the right brush, and surplus of negative charges on the left brush. <S> So, you have gotten yourself an emf that works like a DC battery. <S> The voltage potential of the emf is opposed the original emf created by the DC batteries. <S> This creates a current through the circuit in the below part of the picture which is opposite in direction to the applied current by the DC batteries. <S> As you turn the DC batteries on, you get a current that wants to go from right to left in the coil at the same time as it wants to go from left to right by the induced back emf. <S> The faster you spin the coil with the DC current, the stronger the opposing back emf force will be also at the same time! <S> So, this back emf always works against you.
| The back EMF is generated in the wire that makes up the coils of the motor.
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Resistor in parallel to circuit is causing VREG to output higher voltage I am building a circuit to convert a 300Hz PWM signal to a 30Hz PWM signal with the same duty cycle. It's a pretty simple circuit: I'm having a weird issue with the R7 resistor though. It's in parallel to the GND_PWM_IN to 12V_IN part of the circuit. I have it there because the system that outputs the 300Hz PWM frequency tests the resistance of the circuit, and if it is shorted or over ~150ohms, throws an error code. In short, it wants the resistance to be about 100 ohms. When the resistor is put in the circuit, the output voltage as measured on the 5V line of the VREG is actually around 7VDC. When removed, it outputs a perfect 5VDC. I do not understand what is causing this, but it is causing problems for my microcontroller and the rest of the circuit which doesn't like that high voltage. Any insight into why it would be causing that voltage irregularity, or if there is a better way to fool the impedance sensing circuit without having R7 there, would be greatly appreciated! <Q> A 5 volt linear voltage regulator will pull its output up to 5 volts, but cannot pull the output down if something else trys to pull the output up. <S> You should probably connect R7 to Ground, rather than +12. <S> However, not knowing what sort of circuit is feeding GND_PWM_IN, I can't say for sure it that will work - <S> at least it won't continue to risk damaging the ATTINY, and anything else using that 5 volts. <S> If GND_PWM_IN is a 12 volt signal, you will need a voltage divider or level shifter to reduce the signal to the 5 volt maximum that the ATTINY can tolerate. <A> I'm going to assume that you have nothing connected to 'GND_PWM_IN'. <S> When R7 isn't present, PB2 is only connected to 5V through R2. <S> This is a pull-up resistor, pulling it to 5V. <S> That's fine. <S> You have a problem with R7, in that it is creating a voltage divider between 12V and 5V. PB2 is then seeing approximately 11.75V at it's input. <S> The ADC of your MCU should not be exposed to voltages between 0V-VCC (its operating voltage), which in this case is 5V. Vreg is being pulled above 5V by the higher voltage at PB2, which may irreparably damage the ATtiny. <A> The PWM load is wrong. <S> even 100 Ohms is too low for just 300Hz low current signal to AtTiny. <S> That is for a high current switch to a battery. <S> ( good grief 4K7 ought to be fine. <S> What kind of PWM did you buy? <S> A motor speed control?) <S> The incorrect pullup with 100 ohms will pullup the LDO output to 12V since there are only emitter follower outputs ( source only no sink ) <S> Thus AtTiny is exceeding 6V limit with potential internal thermal damage and junction overstress. .
| R7 is pulling that input line towards 12 volts and, via the ATTINY's input protection diodes, is pulling the 5 volt line up.
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Identify Symbols on Circuit Diagrams for VHF Radio Build I found these circuit diagrams for VHF transceiver online, and am having trouble recognizing a number of the symbols. Would you please help me identify the boxed symbols? Here's what I've got so far: Drawing #1 ( original image link ): The red box with CLK should refer to clock, but I don't know which type or if it even matters. Completely clueless on the blue box with L12. Drawing #2: I think that the red boxes labeled CuAg and CuL are likely wire coils of copper/silver and enameled copper wire respectively. The blue box appears to be a transistor of some kind but I don't know which and googling the label (looks like 25k1904) has gotten me nowhere. The yellow boxes should be inductors with a ferrite core but I don't understand the VK200 labelling. The purple box I'm thinking is a trimming potentiometer? The brown boxes are diodes but not sure if the 12V is referring to a specific type or just that they are connected to the 12V GND. And the green box is a switch ('preklop' translates to 'switch' from Slovenian) but RX izhad and +VRX are throwing me off there. Drawing #3: The pink, red, yellow, and orange boxes look like relays, but I have no idea wha kind. The blue box is a straight up mystery as is the green. Lastly, was going to use Amazon & a local RadioShack to track down most of the items, but if y'all have any sourcing tips for small electrical components, they are all welcome. Thanks in advance for any and all help! <Q> Image #1 <S> The two components in the red box are the two halves of a 74HC74 flip-flop. <S> (The part number is a bit out of place, but is present -- look under the right half.) <S> Image #2 <S> The component in the blue box is a MOSFET. <S> Andrew Morton's interpretation of the part number as "2SK1904" looks plausible. <S> The component in the purple box is a 1MΩ potentiometer. <S> (This diagram is using IEC resistor symbols, which look like a rectangle instead of a jagged line.) <S> The component in the green box, on the right, is a coaxial connector of some sort. <S> The component in the left brown box is using an uncommon symbol for a Zener diode; the "12V" marking is probably supposed to be its breakdown voltage. <S> The brown box on the right is drawn as a normal diode -- this may be an error, given that it's also marked as "12V". <S> The symbols in red and yellow boxes all look like inductors of various sorts. <S> The ones with T's across the middle are adjustable; the ones with a line next to them have a core. <S> I'm not sure what "VK200" would mean either; it may be an abbreviation in another language. <S> Image #3 Blue box is an inductor, connected to ground (the inverted T below it). <S> Red, yellow, orange, and purple boxes all look like adjustable inductors and transformers, similar to ones which appeared in previous images. <S> I'm not sure what the significance of the box is supposed to be; it may mean that they should have metal cases. <S> Green box is two crystals. <S> I can't quite read the text, but it probably indicates the frequency and cut. <A> Drawing #2, Blue box: <S> N Channel MOSFET. <S> Drawing #3 Green box: appears to be resonators or crystals. <S> This image should help with a lot of (what appear to be) variable/fixed inductors and ferrite cores: <S> I'm hesistant to try and identify some of the symbols that I can't read clearly, but hopefully this gives you a starting point. <A> In drawing #3 the green box are not crystal but a ceramic filter. <S> A crystal usually is indicated by one of those symbols and would have the frequency indicated. <S> The boxes around the inductors are used as shielding. <S> Those coils are used for tweaking up the circuit. <S> In Drawing #2 The purple box is 1 Megohm variable resistor. <S> In Drawing #1 The Flip-Flop <S> D= <S> Data <S> , S=Set (?), R= Reset(?), CLK= Clock (synchronization pulses), Q= <S> Non inverting output, Q1=inverting output. <S> For the last one there <S> I hope I remembered it correctly its been close to 35 years since I really messed with this stuff... :)
| The component in the blue box looks like some sort of adjustable inductor or transformer.
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Why don't battery manufacturers make 5 V batteries? Why is it so? Is there something that cannot be crossed while manufacturing 5 V batteries? They can make billions though! Still... <Q> They don't have a choice <S> The voltage of a battery is decided by the reactants in the battery . <S> There are only so many viable battery chemistries out there. <S> You can't just pick two random chemicals off this table ; it also has to be possible/practical -- <S> To actually build it. <S> At a competitive price-point. <S> Out of readily available materials. <S> Which are relatively non-toxic. <S> And don't weigh too much. <S> Endure many recharge cycles (if it's secondary). <S> Never explode. <S> Have enough storage capacity to bother. <S> and not hopelessly lag behind other successful batteries in any category. <S> These restrictions soon winnow the thousands of combinations down to just a few, and none of them have a 5V or 2.5V output. <A> The output voltage of a battery varies while charging and discharging, but most devices that run on 5V need a relatively constant voltage (eg. <S> USB is 5V +- 0.25V). <S> Therefore a voltage regulator is often required anyway , to convert the varying battery voltage to a constant system voltage. <S> Battery powered USB devices often use a single Lipo cell that charges to 4.2V and puts out ~4.2-3.5V. <S> This is ideal because it can be charged from USB with a simple linear regulator, and (if necessary) <S> produce 5V with a boost converter. <A> Actually, four Ni-Cd cells in series will produce approximately 5 volts.
| Regulators are generally designed to either reduce the voltage (linear regulator or 'buck' converter) or increase it ('boost' converter), so ideally the battery voltage should be above or below the required system voltage.
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Soldering problem with heavier components I'm having difficulty soldering components with 1 mm thick or larger wires. The solder melts fine, but it does not 'stick' and just falls off. Components with smaller prongs like regular LED's are no problem. Right now I'm trying to solder some diodes to make a bridge rectifier and the solder is just not sticking. What could be the problem? I'm using a battery-powered RadioShack soldering iron that works fine for the small stuff, and lead-free flux solder. Any tips on how to solder the heavier stuff that seems to reject solder? I've looked all over for this issue and don't seem to be able to find an answer. <Q> Short answer: you need a better soldering iron. <S> Longer answer: <S> All of the heating energy from your tiny iron is being conducted into the mass of metal that you are attempting to solder and being radiated away. <S> I would suggest that you need at least a 40 Watt iron to solder 1mm thick wire of any significant length. <S> Others might suggest that you would need more power than that <S> but it all comes down to what soldering iron you use. <S> A really inexpensive iron may not work well just because the design is such that there is significant thermal resistance between the heating element and the object being soldered. <S> But better irons will do the job nicely. <S> A specific example of a soldering iron that works well at 40 Watts is the Metcal MX-500 series with a sttc-137 tip. <S> This iron will solder the entire perimeter edge of a male DB-25 connector to a piece of copper-clad PCB material using 63/37 solder. <S> It takes a while <S> but it will get the job done. <S> Your typical Radio-Shack 40W iron won't even attempt to work. <A> I would put this as a comment to Dwayne's answer, but I don't have the rep... <S> in addition to what he says about having a better iron, I want to point out the following. <S> When soldering, the leads should melt the solder -- the iron should not do it directly. <S> The iron heats the leads, the leads melt the solder. <S> When you say "the solder melts fine, but does not 'stick'" I'm thinking that you are melting the solder with the iron, not with the diode leads. <S> If the leads are not hot enough to melt the solder themselves, you'll get a dry joint. <S> (If you hadn't mentioned diodes, if it was say a heating element, I'd also suggest it might be the metal - some metals just don't wet to copper. <S> But diode leads won't be made out of such metals.) <A> There are generally three primary reasons for solder not wicking properly: Technique - as mentioned by others, heat the lead not the solder Iron temperature - again, mentioned by others <S> Contaminants - make sure the wire is clean of contamination. <S> Oxidized, dirty, or otherwise contaminated wire or leads will prevent the solder from wicking and will glob instead. <S> A decent flux pen will help a lot.
| Your little battery-powered soldering iron doesn't have enough power to heat a large mass of heat-conducting metal up to the melting point of solder.
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If serial ports are asynchronous, why do they have an SCL line? For instance, this tutorial has these two seemingly contradictory quotes: Because serial ports are asynchronous (no clock data is transmitted), devices using them must agree ahead of time on a data rate. and: Each I2C bus consists of two signals: SCL and SDA. SCL is the clock signal, and SDA is the data signal. Can someone help me to understand how these are compatible statements? How can it have a clock signal without transmitting clock data? <Q> I2C is synchronous but the more familiar UART serial is asynchronous. <S> Both have their own advantages. <S> Synchronous serial ports allow for arbitrary timing, while asynchronous ports require precise timing but use less connections. <A> The article you linked toward is trying to compare similarities and differences between three different serial techniques -- the common "Serial Port" that computers used extensively up to a few years ago, <S> the complex Serial Peripheral interconnect, which "every" DSP supports, and I2C, a "faster than Serial Port, but slower than SPI" synchronous serial interconnect technology. <S> The article was not particularly clear in explaining that "not all serial connections are Asynchronous Serial Ports". <A> The asynchronous serial ports may be better known as "RS232 serial ports" or "COM Ports" on PCs - these are normally used for communication between different pieces of equipment. <S> The I2C and SPI synchronous buses are used between components on the same PC board, or possibly between separate PC boards within a piece of equipment.
| Serial ports exist in both synchronous and asynchronous forms.
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Simple, low power way to buffer a 500mV threshold signal to an open collector output? I have a signal that is 0-500mV when low and 600-900mV when high. I'd like to buffer this signal to an open collector output. The output must withstand a maximum of 3.3V be capable of sinking 60uA. The power for the detection circuit is provided by a (CR2023) battery that ranges form 2V to 3.2V. The goal is to find a simple, cheap, low power solution. (By low power, I am shooting for on the order of 1's-10's of uA.) One solution is to use a low power comparitor like the TS881, but this would also require an external voltage reference to compare to, adding complexity, cost, and power usage. Another solution would be to use a reset voltage detector like the NCP303, but these are expensive and don't seem to come with thresholds lower then 0.9V. I feel like there should be a way to do this simply with discrete parts. Maybe somehow taking advantage of the pinch voltage threshold on a JFET? Or the bandgap on carefully chosen LED? Ideas? <Q> Here is a simple circuit using only discrete transistors and a few resistors.. total BOM cost should be less than 1 cent in quantity. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> It's a bit trickier than it might be otherwise because your margin between on and off is rather small. <S> Threshold voltages on inexpensive JFETs and MOSFETs are not that well controlled. <S> This circuit will pull OUT low (say with 10K load to +3.3) when the input is higher than about 550mV. <S> Below that voltage, Q1 hogs <S> the current flowing through R1 and Q2 is starved, thus Q3 does not get hardly any base current. <S> If you want to reduce current further and your load will be no lower than 50K, consider increasing the resistor values by 5:1 (say 470K 5% for R1/R2 and 100K 1% / 20K 1% for the R3/R4 divider). <S> Or use 499K 1% for R1/R2 but precision is not important there. <S> Total current drain should be less than 35uA not counting the load, which is comparable to (or less than) self-discharge of many batteries. <A> You don't want to use FET threshold voltage, very unrelaible. <S> A low power comparator like LM339 would be ideal, or you can build it the hard way from several transistors. <S> Your threshold has to discriminate 500mV from 600mV. <S> It's easy enough to generate a threshold with that accuracy from a 'normal accuracy' logic rail like 3.3V +/- <S> 5%, with a 6:1 resistive divider, like (for instance) <S> a 15kΩ and a 3kΩ resistor, which would give you 550mV from 3.3V. <S> If the rail had +/- <S> 5% tolerance, then the threshold could vary from 525mV to 575mV. <S> Make it worse by two 1% resistor tolerances <S> and it still doesn't breach the 500mV to 600mV specification in your question. <A> I ended up using an AVR ATTINY microprocessor with a built in analog to digital converter. <S> The internal 1.1V band-gap reference gives more than enough accuracy and precision for discriminating the 0V-1V input signal. <S> The MCU spends the vast majority of its time sleeping, only occasionally waking based on the watchdog timer to sample the analog input and update the digital output. <S> Net power usage is less than 10uA. <S> The cost of the chip is less than $1 and no other parts are needed - the chip can run directly off the CR2032. <S> While there is a bit of software involved, I'd still rate this solution as simple and it meets all the other requirements. <S> Still, I do feel like there is a better (cheaper, lower power, simpler) <S> analog solution out there, so <S> please add you answer if you can come up with one! <A> 17 days after I asked this question, the part magically appeared! <S> UB20M High-Voltage, Low-Threshold, Ultra-Low Power Voltage Detector <S> When the input voltage rises to a typical threshold of 0.65 V, the voltage detector triggers an open-drain output. <S> http://www.bristol.ac.uk/media-library/sites/engineering/research/eem-group/zero-standby/UB20M_Datasheet_Rev.1.1.pdf <S> This is exactly the part I was looking for, and targeted for exactly the use I had planned. <S> So much for the patent I was going to file for a novel method of zero-power wake-up. :) <A> A bit more detail about our chip ( <S> the one that bigjosh is referring to):Find the latest datasheet here, this link won't change. <S> bristol.ac.uk/voltage-detector <S> We don't advertise this, but we have different threshold levels <S> , UB20M is 0.6V, UB20L is 0.46V, the former has a current draw of below 10 pA, the latter around 100pA, so with some very high impedance sources, the former will trigger sooner. <S> We have some videos (a layman's video and some demos) here .
| The voltage detector is an ultra-low power, input-powered device, that does not require connection to a power rail. Or just use an IC comparator of your choice with R3/R4 on one input.
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Why optical disk CD/DVD use spiral tracks, as opposed to cocentric tracks used in hard disks and floppy disks? I'm not sure this is the community in which should I ask this question, if it isn't, I'm willing to transfer it. According to this article : Track positioning also follows two different methods across disk storage devices. Storage devices focused on holding computer data, e.g., HDDs, FDDs, Iomega zip drives, use concentric tracks to store data. During a sequential read or write operation, after the drive accesses all the sectors in a track it repositions the head(s) to the next track. This will cause a momentary delay in the flow of data between the device and the computer. In contrast, optical audio and video discs use a single spiral track that starts at the inner most point on the disc and flows continuously to the outer edge. When reading or writing data there is no need to stop the flow of data to switch tracks. This is similar to vinyl records except vinyl records started at the outer edge and spiraled in toward the center. There should be reasons why choosing one strategy over another in each case, Pros and cons. Wich are they? <Q> As the snippet you posted states, spiral is better for continuous audio/video without the need for buffering whereas concentric rings are better for data that isn't necessarily sequential in nature. <S> If you have data like pictures or files or anything else that needs to be read or written in a random manner, then concentric rings would be better because you can have everything indexed at the beginning so that you can jump straight to the ring that contains your specific data. <S> In summary: Pros of spiral : <S> No need for buffering. <S> Ideal for continuous streams of data <S> Pros of concentric : Indexable <S> Random access capability <S> They both have their advantages and when electronics were more expensive, spiral rings were the better engineering tradeoff. <S> Once you can make a cheap buffer, the advantage of spiral ring media is virtually entirely eliminated because a buffer with a concentric ring technology gives the ability to have random access capability while making streaming data equally ideal to use. <A> One track media like CDs and DVDs were specifically designed for continuous playback of media. <S> Their advantages include: <S> Data is available to hardware as needed. <S> Less memory lower hardware cost. <S> Less synchronization and addressing overhead. <S> (More room for data.) <S> There disadvantages include: <S> No way to go back and re-read error data. <S> Really long seek times. <S> Can not easily go back and write more data. <S> Their advantages include: <S> Fast seek times. <S> Able to randomly read and write data. <S> Their disadvantages include: <S> More overhead needed for synchronization and addressing. <S> (Less roomfor data.) <S> More hardware needed to access data. <S> So, from this we see that a 1 track paradigm is better for media while a concentric track paradigm is better for data storage. <S> And since CDs and DVDs are usually created for media, they are made using 1 track. <S> While HDDs are created to store and retrieve data. <S> So HDDs use a concentric tracks. <A> CD and DVD were designed for media recording - constant data rate, with relatively few random seeks (which cause high seek times, usually hundreds of ms). <S> As a result, a standard CD player have constant linear velocity, and the tracks are arranged in a spiral. <S> Adjusting the read head (Laser) location is relatively easy, just follow the track. <S> You have to remember that the CD first appeared in 1982, the electronics were quite primitive at the time. <S> HDs allow for more random access patterns, spin at a constant speed (5400, 7200 and even 15000rpm) with relatively low seek time (few ms), so <S> adjusting the head location is more complicated when random seeks are thrown into the mix.
| Concentric tracks like hard disk drives use were designed for error free random data access.
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Drive a common anode bi color LED using single microcontroller pin I have a bi color Red-Green LED with 3 legs (Common Anode). How can I drive the LED using a single pin from microcontroller (ATtiny45) without using any external logic gates? I only need red or green at a time mutually exclusive. Combined color/Off state not needed. They are just for status indicators and need to be as bright as possible. Any schematic using discrete components would be helpful. <Q> Given the Forward voltage differences between Red and Green LEDS allows for a very simple circuit and using your Common Anode configuration. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> You can set your LED current to any value you want, but there will always be a variation between Red - Green current that is unavoidable. <S> How well this works for you will depend on how the Red/Grn LEDs are built. <S> For example with <S> this Common Anode LED <S> the difference between the Grn/Red VF is as shown below: <S> With this bi-color at any current below 30 mA through the Red LED, the Green LEED will be essentially off (uA). <S> If you are using discrete LED's this will work for almost all devices. <S> but note that some newer SMD bi-color LEDs use a mixed technology and the VF's are much closer together, for those devices this drive architecture may not work. <S> Update: after doing some tests with various LEDs, here's a way to handle diodes with no difference in Vf. <S> In addition this can be used if you are using diode currents (typically <20 mA) that can be handled directly by a microprocessor pin. <S> simulate this circuit <A> This breaks out of your parameters slightly, but I feel like it is by far the simplest solution. <S> If you use discrete LEDs, or LEDs with independently accessible anodes and cathodes, you can use this circuit: simulate this circuit – <S> Schematic created using CircuitLab HI: D1 OnLO: <S> D2 OnHiZ: <S> OffToggle fast: D1 & D2 on Note: <S> This will only work with a pair of LEDs whose Vf is greater than 3.3V! <S> Even so, the LEDs may still very slightly glow all the time. <S> You just need to evaluate this tradeoff and whether it's worth it for the low part count. <A> Wouldn't this simple circuit below work well for the required 2 color states only? simulate this circuit – Schematic created using CircuitLab <S> ATtiny datasheet shows it can drive low 10 mA to 0.2V, so take advantage of that. <S> Logic high = D2 turns on <S> Logic low = D1 turns on Tristate the output <S> = D1 is teeny tiny bit on, D2 is on <S> So pick the color for D2 which will be on whenever your processor is still in reset. <S> Also, if you want different LED currents then short Rlim and put different resistors on the LED cathodes. <A> This is not the simplest solution, but if the other methods above don't work as expected, try the following, since the above ones (as I can see) seem to only work on certain conditions. <S> You can't use BJTs instead of MOSFETs in this circuit because the LED can go through the bases and they can be both unintentionally turned on at the same time. <S> If you don't have MOSFETs, use this circuit: <S> This circuit works by pulling the NPN down via the input tag to disable the green and at the same time powering the red. <S> The diode is to prevent current going through the base and powering the red LED.
| For Q1, you can use a digital transistor (integrated resistors) or you can use an N-channel MOSFET without a gate resistor.
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Can a lead acid battery charger and the battery be used as a UPS? I want to connect some DC devices(< 1.5 A rating) to the output of a 12V/7Ah battery. Will this system work as a UPS if I connect a battery charger to this battery? i.e the DC devices, Battery and charger will be connected parallely .Will this work as a UPS? The charger is auto-cutoff enabled to prevent over-charging.See : http://www.amazon.in/Battery-charger-12V-1A-Adapter/dp/B01GZRBWLW Will the auto cut-off be affected for above setup? <Q> This sounds like it will work, but there may be some pitfalls. <S> First, ensure the devices are very flexible in their input voltage, lead-acid batteries can vary a bit in their voltage. <S> Put a voltage regulator in between if you're not sure. <S> Second issue from the top of my head could be that the charger puts a load on the battery when turned off, but the auto-cutoff makes this unlikely. <S> This could be fixed with a heavy diode. <A> A battery (ideally deep-cycle) <S> A DC power supply <S> A switch-over circuit that switches the load from the DC supply to the battery when AC power is lost A low voltage cutout to protect the battery from over-discharge (typically at 10.5-10.8V, depending on load). <S> Switch mode voltage regulators for the various voltage outputs you need. <S> I built such a DC UPS for my networking devices using a CCTV battery backup unit to take care of the first 5 points. <S> The final point was handled by a boost converter to 17V, followed by a buck converter to 12V. <S> This arrangement ensures you always get 12V output regardless of where the battery is in the 10-14V range, plus putting the boost to 17V first ensures the input to the buck converter is always high enough for reliable operation. <S> A single buck-boost converter could be used too. <A> In theory it can, <S> But, many chargers charge the battery with 14.2 ~ 14.4V, and when they detect that the battery is nearly charged, (the charge current drops to lets say 200mA), they lower their voltage to 13.8V. <S> The 13.8V is the ideal float charge voltage for many lead based batteries.the <S> 14.2V is to much, if you charge your battery for too long with this voltage, they will dry out. <S> And will be damaged. <S> If you add load to your system, your charger will see always a current greater than 200mA (or whatever your charger trip point is). <S> and always will supply 14.4V. <S> This can be fixed, to add a diode in series. <S> the diode will drop around 0.7V <S> so your battery will never be overcharged. <S> But it takes significantly longer to charge. <S> secondly you need some sort of low voltage cut-off.depending <S> on how long you want your battery to last, somewhere between 10.8V <S> an 11.8V.(the higher the cut-off voltage <S> the longer you battery will last, but the less run-time you have) <S> (also you can use around 40~60% of your rated battery <S> Ah if you want to use your battery more than once).
| For a DC UPS, you need the following components: A battery charger (ideally a smart charger that will float charge the battery when full to prevent over-charge).
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Is it sufficient to disconnect the negative terminal of a voltage source? I have a scenario where I would like to use N-channel MOSFETs, because they have lower on-resistance and are cheaper. simulate this circuit – Schematic created using CircuitLab Would there be any problems with this approach? I've read this answer but I understand the main concern is somehow accidentally forming a closed loop. But if the MOSFET is safely at the negative terminal of each voltage source, I don't see how there would be any risk? Am I correct that the positive terminal of either voltage source cannot conduct until the negative terminal is also connected? In effect, as if it wasn't connected at all? <Q> My original thought was that your N channel MOSFETs are upside down because of the way you have the on/off <S> switch arrangement. <S> However, after @ThePhoton pointed out my blindness, another problem arose in the "turning off" area. <S> To turn-off the MOSFET you do need to have a different arrangement for the "switch" you have drawn. <S> The disconnected pin on SW3 (or SW4) has to be connected to the source terminal on the MOSFET or it won't properly shut down "off" - they'll conduct with 1 or 2 volts across them. <S> Generally most MOSFETs are OK with a 12 volt gate-source signal but some may not be. <S> Presumably your batteries are safe to be used in parallel. <S> Other than that <S> The extra connection due to the wrongly connected switch will of course be another connection so some care is needed here to make sure it works correctly and doesn't leak current and discharge the battery. <S> Is it sufficient to disconnect the negative terminal of a voltage source? <S> It's sufficient in many applications. <A> Disconnecting the negative side is sufficient provided that the MOSFET is the only connection to the negative side of the supply. <S> I have designed electronic circuit breakers for batteries that work that way. <S> They worked fine and were even certified for use in an explosive atmosphere. <A> Am <S> I correct that the positive terminal of either voltage source cannot conduct until the negative terminal is also connected? <S> In effect, as if it wasn't connected at all? <S> A common problem with trying to power down part of a system is "phantom power": a "signal" line which is held high ends up powering the device. <S> Similarly if you disconnect the power supply ground but forget about grounded signals, cable shields tied to ground, etc., then you will have problems. <S> It doesn't look like this applies in your scenario because the MOSFET is right next to the power supply and will disconnect all possible routes to that negative terminal.
| , I'd have no problem doing it this way providing there isn't an electrical path from the negative battery terminal going somewhere else that you have not shown in your picture. Yes, but you have to watch out for all possible routes.
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How to detect very hard impacts with piezo? Normally I'm a software guy, but I've recently decided to try my hand at some electronics projects. What I want to do is make my own impact sensor for target shooting. What I'm picturing is a piece of AR-500 steel for the target, and a sensor that can detect when the round hits the target. I've started by using an Arduino Uno, a piezo and 1MOhm resistor, like in the arduino tutorial. My main problem is lessening the sensitivity of the piezo, so that I don't get false positives. For example, if the round hits in front of the plate and splashes dirt on it, I don't want it to pick that up. Or if the round passes by, I don't want the sonic boom to trigger it either. Currently I have it set that so that a light tap on the piezo sensor housing won't set it off, but a harder tap will (with a threshold of 100 from my arduino adc). My question is, will the piezo output max voltage with a hard tap from my finger, or will it send out a really high voltage if I were to hit the steel hard with a hammer (to simulate a rifle round). If the voltage will keep increasing as the impacts increase (is there a limit?), then I guess it's as simple as reducing the voltage coming out of the piezo. If the voltage hits it's maximum output prematurely, is my only option to isolate the vibration with rubber mounts, or padding or something so that it takes a good wallop to set it off? <Q> Yes, the output of a piezo-electric sensor should keep increasing as the target is whacked harder. <S> There should be a substantial difference between tapping the target with your fingers and hitting it with a hammer or a bullet. <S> One problems with piezo sensors is to keep them from frying the input stage if accidentally whacked hard. <S> Of course, the obvious thing to do is simply try it. <S> Set up something that triggers if a signal gets over a couple of volts or so. <S> Then feed the piezo signal thru a pot so that you can adjust the attenuation, then into the comparator. <S> After some experimentation, you should be able to find a setting that detects hard whacks but not casual bumps. <A> Any Piezo-electric sensor can be used including an Electret mic. <S> a piezo sensor (or even a ceramic capacitor with signal amp). <S> All being high resistance with some capacitance with varying sensitivities, the output is a current proportional to sound vibration and the voltage is proportional to resistance. <S> Thus the gain depends on the load pullup R value . <S> I suggest 1K is small enough to attenuate signal. <S> Then add tape over orifice. <S> The signal is AC coupled and pulled up to Vcc for "half-wave negative peak detection" pulses down with an envelope burst of noise. <S> The cap voltage drops with this current pulse when the diode conducts, dropping from Vcc to some voltage below Vcc/2 which is enough to drive the next stage to a logic "1" out. <S> The 10M resistor pullup R3, slowly "rearms" the circuit back at Vcc. <S> This produces a +ve voltage on the noise impulse with a duration of C2 * R3 or 100 ms in this case. <S> This may be extended to a 5 seconds by increasing C to 1uF. ( with a good quality cap > 20M leakage resistance ) simulate this circuit – <S> Schematic created using CircuitLab Attenuation of gunshot or any noise is easily muted with tape to also vary sensitivity over the sensor and proximity of sensor to target. <A> May be put a resistor with the correct value at the out of the piezo, to find the correct value you could solder a potentiometer (in the correct Ohm range of the piezo) and do some test until the arduino will detect the signal. <S> Then you mesure the potentiometer with your ohm/voltmeter and replace it with the correct resistor. <S> , I think may be it can damage your piezzo. <S> Also may be you can replace the piezzo by another device less sensitive like a small headphone speaker (in the correct ohm range).
| The other simple solution is to isolated the piezo from external sound, for that you can use a cylinder of fixing paste and/or some layer of cork wood, I think ity will be the best one because the impact of a bullet make a really strong noise
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Why are copper cables round? Why are cables only made of round shapes? What is the advantage of it? Why aren't there triangular or quadrilateral or pentagon shapes of cables? <Q> Like Jessica Rabbit- because they are drawn that way. <S> Wire production involves pulling (drawing) the wire through successively smaller dies (often with annealing in between). <S> The dies are most easily made with round holes (they are typically made from very hard materials such as diamond). <S> Not all wire is round - rounded rectangular wire is sometimes used in inductors and transformers. <A> When you move away from circular cross sections you run into two significant problems: First, corners would exert more pressure against adjacent things, and are more likely to result in insulation damage. <S> Second, for high frequency use the skin current effect would result in higher impedance in a cable with the same cross sectional area but with corners. <S> Further, the greater amount of insulation required to cover that conductor would also change the impedance. <S> While this could be dealt with, we already have a great understanding of the skin effect, impedance, and insulation effect at high frequencies for round conductors that it would be limiting for designers to use alternate configurations without a good reason (such as fitting more copper in a smaller area for efficiency in inductors). <S> There are many other considerations - wire drawing is already difficult and strenuous, having a circular cross section reduces die contact for a given area of copper. <S> Insulation would increase due to the additional circumference of a triangle or square compared to a circle (again, for the same cross sectional area of copper). <S> When bending or flexing the corners would see more stress, would harden more quickly, and would crack more readily than circular cable. <S> When discussing cables it becomes even harder to make other cross sections. <S> Imagine taking 10 conductors to make a cable in a triangle. <S> When bent, twisted, or squeezed you'd find the individual wires in the cable moving out of their place and deforming to a more circular shape. <S> How would you keep those wires in place to maintain the desired cable cross section? <S> You could go to a single wire, but then it will be much harder to manage, much less flexible, and more prone to breaking and damage. <A> Flexibility A round cross section allows the wire to flex in any direction equally. <A> Not strictly copper but for larger cables you do get triangular cores. <S> They tend to use aluminium for larger cores like these. <A> In industry, solid (retangular) metal (usually copper) buss bars are used, generally for high current applications: ... <S> so, not always round. <A> From a production point it's easily the most simple shape one can go for, which I would presume is the dominant factor in all of this. <S> It's also mechanically favourable if you consider the strain when you start bending things. <S> And if you think about it, its also is the most efficient way to pack as much cross-sectional area (hence conductive material) in a limited space. <A> They are not only round. <S> Anyway dies holes are round cause they are easier to make and the most economic. <S> Also round cables are more versatile in some general situations, for example to insert them into corrugated (or not corrugated) pipes, to bend or curve them in little space, for ex. <S> when is needed to connect them to electrical home-office standard outlets etc. <S> So the round section is in general the most used for practical and economic advantages. <S> Anyway the round section is not the only standard. <S> Other shapes are used. <S> For example lastly is spreading the use of adesive flat cables to home or industrial electrical systems: you just attach them to the wall with their adesive and you can putty and paint directly on them without the use of corrugated pipes, without broaching the wall, without plastering. <S> Also flat wires more effectively dissipate the heat produced during their operational phase causes they have a bigger surface exposed to air compared to the section needed for a specific electric load. <A> Due to the skin effect, alternating current tends to travel on the surface of wire. <S> DC through the whole core. <S> A round wire maximizes the cross-section per unit of material used, so it is the most cost effective at providing a given gauge or capacity. <S> It also has the side-effect of providing the most surface area on the surface of the conductor for a given cross-sectional area of the overall wire. <A> For low voltage cables (240 V, 440 V) the shape of the conductor isn't critical. <S> For high voltage cables (3.3kV and above), the voltage stress needs to be controlled so the insulation doesn't break down. <S> Voltage stress is strongest at sharp corners. <S> So The easiest way to control this is to make everything round.
| While the wire making die process itself is largely the reason for it's shape, dies could make it in a variety of shapes, but the circular cross section is also the most cost effective.
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Reducing the number of pins needed to read a 12-key keypad where the buttons are grounded? Unfortunately, I'm not sure what the right question is; I'have a 12-key keypad. I wanted to utilise a matrix based approach, except that they all seem to be connected in such a way, that they're floating when not pressed; when you press one, it connects to ground. This seems to me as if it's prepared for a set-up with a pull-up resistor on the wire over to the MCU, and not a matrix-style reading, where I (if I understand correctly) sequentially pulse on the rows, and sequentially read the voltage on columns. However, I don't have 12 spare pins on my MCU, so I'm wondering how to minimize the number of pins needed to read this 12-key keypad. Are there any ICs that I might use? Any approaches? If I recall correctly, the board it used to be connected to had several (de)multiplexers connected in cascade, but I couldn't figure out the way it worked. However, I'm pretty sure the keypad itself is wired like this (it has its own connector as it's on a separate board which I'd like to keep), and that makes me confused. <Q> If the common connection is floating, an approach using an output pin and an ADC would be to tie one switch input terminal the output pin and the common input to the ADC input. <S> Then connect another switch input to ground and connect all the other switches to points on a string of eleven 1K resistors. <S> If one needs low-power wakeup and the processor's ADC pin doesn't have weak pull-up that can be pulled low through about 3K of resistance, add a 100K pull-up to the ADC pin. <S> When idle, ground the isolated I/ <S> O pin and set the other for weak pull-up. <S> That pin will be pulled down if any button is pushed. <S> To find out which button is pushed, set the output to VDD and read the ADC (preferably with the pull-up disabled). <S> The voltage will then indicate which pin is pressed. <S> If the switch makes even halfway decent contact (e.g. 3K or less of resistance) <S> the voltage should be largely independent of switch resistance. <A> There are many solutions. <S> Just how many IO pins do you have available? <S> Plenty out there. <S> If your MCU has no I2C, you can bit-bang the interface. <S> Downside is this would require 12 pull resistors. <S> If you have 4 outputs and 1 input available, use a demux IC (e.g. 74HC154) to "scan" the keypad. <S> Each demux output drives a key and the common output of the keypad goes to your MCU input. <S> When your scan hits a key that is pressed, the input will be driven high. <S> This takes 5 IO lines but advantage is it needs only 1 pull resistor on your MCU input (or none if your MCU has built-in pull). <S> It might need 12 diodes or a demux that has open-collector outputs if keys can be pressed simultaneously. <S> A totally cheesy design is to have a different value resistor for each key. <S> Then use either a divider or a current source/sink read by an ADC so that each key presents a different voltage to the ADC. <S> I've seen this used in actual commercial products but use only if you are desperate and can only spare 1 IO pin; this takes a lot of good SW to debounce and get working reliably. <S> It also very difficult to detect simultaneous key presses. <S> I can draw you a schematic if Option 2 circuit is not obvious to you. <S> Also, you need to consider: 1) debounce algorithm, 2) tracking the press and release of multiple key presses. <S> For example, when someone is typing very fast their keystrokes will overlap. <S> You need to consider if you need to allow for this or not. <A> Here ya go. <S> Schematic below. <S> I've only shown 8 buttons, but 74HC4067 has 16 analog channels for up to 16 buttons. <S> Select which button you want to check by driving the 4 bit line to the desired channel, then read the MCU input. <S> High = button not pressed. <S> Low = button pressed. <S> Simple debounce algorithm is to only scan say 30x per second max. <S> simulate this circuit – <S> Schematic created using CircuitLab
| Use a 16-bit I2C expander IC.
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segment on 4 digit 7 segment not displaying Segment E on digit 3 has stopped working for some reason. It was working fine when I first got but just recently stopped working. This segment displays fine on the other digits except digit 2, that one has started to fade. Here is a picture of the display. Any ideas on what could be causing it to be not working and/or how to fix it. <Q> You have most probably burned the segment in question. <S> The use of only one resistor per digit is the main reason. <S> You need use a resistor for each segment. <S> Furthermore to prevent losing a nex display again use a larger resistor to begin with. <S> With 330 or 390 ohm per segment it will work fine. <S> With the use of one resistor per digit you can not control the actual segment current and since no diode is equal there will always be one that takes a larger current than the remaining segments. <S> Therefore change over to one resistor per segment. <A> From the pictures you seem to be using the one-resistor-per-digit approach. <S> When one of the digits or one of the output pins change somewhat this will probably not work well. <S> Try to use one resistor per segment . <A> Any ideas on what could be causing it to be not working and/or how to fix it. <S> However, I have done similar things (using 1 resistor per digit) or worse (no resistors at all) and they work just fine. <S> the simplest and surest way is to take out the display and put the segments to a power source (with a resistor) to test. <S> if that particular segment doesn't light up, you have found the reason. <S> after that, I would look into the code first and to make sure that there is nothing funky going on there. <S> try to light up just that segment for example to see if indeed there is a problem. <S> then, I would go after the wiring. <S> those boards aren't that greatest in securing contacts. <S> maybe the pins are bent, maybe the wires aren't where they are supposed to be, ... <A> Copy and paste the code shown below, this will help you to: Configure board pin Display pin Error in board pin, display pin or jumper wire by replacing all of components one by one, you can add more pins for other configurations. <S> Code: <S> int ledDelay = 1000 <S> ; int pin2 = 2;int pin3 = <S> 3;int <S> pin4 = <S> 4;int <S> pin5 = <S> 5;int pin6 = <S> 6;int <S> pin7 = <S> 7;int <S> pin8 = <S> 8;void <S> setup(){pinMode(pin2, OUTPUT);pinMode(pin3, OUTPUT);pinMode(pin4, OUTPUT);pinMode(pin5, OUTPUT);pinMode(pin6, OUTPUT);pinMode(pin7, <S> OUTPUT);pinMode(pin8, OUTPUT);}void loop(){digitalWrite(pin2, HIGH) <S> ; // turn the red light ondelay(ledDelay); // wait 50 msdigitalWrite(pin3, HIGH); <S> // turn the blue light ondelay(ledDelay); // wait 50 msdigitalWrite(pin4, HIGH); <S> // turn the blue light ondelay(ledDelay); // wait 50 msdigitalWrite(pin5, HIGH); <S> // turn the blue light ondelay(ledDelay); // wait 50 msdigitalWrite(pin6, HIGH); <S> // turn the blue light ondelay(ledDelay); // wait 50 msdigitalWrite(pin7, HIGH); <S> // turn the blue light ondelay(ledDelay); // wait 50 msdigitalWrite(pin8, HIGH); <S> // turn the blue light ondelay(ledDelay); // wait 50 msreturn;}
| a device / segment failure is certainly a possibility.
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How to reduce current? Let's say I have a power source of 4.2V/1A. I know how to reduce the voltage using resistors, but if I would like to be able to 'divide' current in half for example, how can I do that using standard components? My component is a battery, which should not be able to draw more than 150mA from the power source. I know I can buy already built components for this, but I want to understand how to do this using only the basic components. I know that charging battery is dangerous, but I do this daily by hand (from bench power supply with current limiter enabled), but battery is just an example. All I'm trying to do is to reduce the 'max current potential' from the power supply. I'm familiar with 'voltage divider', I want to do exactly that, but only with current. Can I do this purely with basic components , such as diodes? Thanks! <Q> My component is a battery, which should not be able to draw more than 150mA from the power source. <S> So, trying to make one from passive components or diodes isn't going to be effective as a protection against over charging with too high a current. <A> Well, as others in this thread have already pointed out, a (decent) current limiter requires some active circuitry. <S> A very simple current limiter using only one basic component, a JFET, is the following: simulate this circuit – Schematic created using CircuitLab <S> That circuit limits the current into the load to the value of \$I_{Dss}\$ of the JFET. <S> Just select a JFET having an \$I_{Dss}\$ of 150mA and you'll be fine. <S> There are a couple of drawbacks, though: <S> Current limiting is not very stable: due to its simplicity there will be small variations of the current with varying supply voltage. <S> Idss is not well specified and has wild tolerance: usually a datasheet will give you a (broad) minimum and maximum value, or sometimes just only a minimum. <S> For example, the J107 has a specified minimum \$I_{Dss}\$ value of 100mA, and no maximum (see image below). <S> So, in the end, you will end up selecting a specific specimen by hand using trial and error. <S> Power dissipation: JFETs are usually low power devices (<1W); if the voltage drop across the JFET multiplied by the current is higher than that (assuming a reasonable safety margin), you are out of luck. <A> There are two ways to provide a current-limited supply to charge a battery. <S> a) <S> The current limiter way. <S> Use an active current limiter. <S> The simplest of these, if you have the voltage headroom, is an LM317, which maintains 1.2v between its output and adjust terminals. <S> If you connect (for instance) 12ohms between them, it will limit at 100mA. <S> Choose different value resistors to get between 5mA and 1A. b) <S> The poor-man's limiter, designed for battery charging. <S> As your battery voltage doesn't change quickly, and as power supplies are often adjustable, a resistor of an appropriate value will limit current from a supply to the battery. <S> As the battery voltage rises, the drop across the resistor will fall. <S> This will reduce the current, unless you are monitoring it every few minutes and adjusting the supply up to compensate. <S> Quick to do as a one-off, very tedious if you have to do it more than twice!
| You aren't looking for a current divider but rather a current limiter and this is an active device because, in principle, it doesn't restrict current into (or out of) a device until the "limit value" is reached.
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Can an electric heat mat cause fake touches on a capacitive screen? After installing one of these heated desk mats on my cold work desk, I've noticed some erratic behavior on touchscreen devices I develop software for when they're laid down on the mat. Most noticeable on iphones, when the heat is on, part of the touch grid becomes unresponsive and sometimes receives ghost touches. It happens regardless of device temperature. I'm posting here in EE, because I have a suspicion the cause might be some sort of electromagnetic interference. I'm happy to move it elsewhere. There seems to be a cause-effect relationship, because it's happening on multiple devices from different manufacturers. What is going on here? Here's a video demonstrating the effect: https://youtu.be/0jz8HaytlEs <Q> The problems you are describing are typical of EMI disturbance. <S> Your tablet/phone with touch screen is a typical class B device which must accept any radio disturbance (up to a level) but may misbehave or stop functioning properly when a disturbance is applied. <S> EDIT: <S> With your updated information that the problem still exists even when switched off but plugged in tells us that the power supply is the root cause of your problem. <S> It will increase with load. <S> You can try using clamp-on ferrites or of the wiring is detachable, get a toroid core and wind several turns around it to form a common mode filter. <S> This will also increase you chances if suing the manufacturer who most likely bought and bundled a power supply with the mat and you would not believe how many no name Chinese manufacturers of non-EMI compliant manufacturers they have to choose between. <S> Your phone <S> /tablet will be subjected to higher and higher field strength the closer to get to the mat and at some point something will misbehave. <S> Exactly what will "give" first it difficult to say but something physically large, like the screen falls into the high probability category. <A> Capacitive touchscreens are sensitive to radiated electrical noise. <S> They need to sense pico-Colulombs of charge, so any electrical interference can cause problems. <S> Interference from your heat mat could easily cause touchscreen malfunctions (bad performance, phantom touches, erratic behavior). <S> I'm most familiar with touchscreen problems from my charger investigations. <S> If your phone's touchscreen doesn't work while charging, you probably have a dangerous, cheap charger. <S> If the wall wart for your heat mat contains a switching power supply, that could be what is generating the interference. <S> And then the wire in the mat will be acting as a huge antenna. <S> If you have an oscilloscope, try waving the probe near the mat and see what you get. <S> For a detailed discussion of capacitive touchscreens and their noise problems, see Noise Wars: <S> Projected capacitance strikes back by John Carey of Cypress Semiconductor. <S> It explains in detail how cheap power supplies leave out components that would reduce the noise. <A> You have described the problem as being particular to the touch-grid. <S> While the EMI theory proposed is certainly valid, it could also be due to a capacitive effect as mentioned in your comments. <S> A simple experiment to try would be attaching the mat surface to earth ground. <S> This is commonly done for ESD work mats to allow static charge to slowly dissipate. <S> The earth ground terminal is accessed through any mains wall plate screw, using a ring terminal connection at one end of an insulated wire. <S> The other end is usually riveted to the ESD mat, but in this case do not puncture your heat pad! <S> An alligator clip to the surface should work fine. <S> It's a bit of a shot in the dark but easy to implement, assuming that you're comfortable removing the wall plate, connecting wires, etc..
| Your heat mat is most likely PWMed to regulate the temperature and if the implementation isn't good in this regard, the dV/dt and dI/dt flanks are high with the load acting like a large antenna, you have an EMI problem. Cheap chargers generate a lot of noise - both radiated and in their output voltage, and this can cause touchscreen malfunctions. If the power supply is something simple, like a 12 V "dumb" supply and all control is done is the mat itself, you can try a different (approved) power supply.
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How can I tell how many power supply's it is safe to put in series? So at university, to provide both negative and positive supply rails for op amps, we were told to set two independent supplies to a required voltage, lets say 15V, then plug the positive terminal of one of them to ground, and the negative terminal of the other to ground and thus achieve -15V and +15V. This is a diagram of the sort of thing we did: My question is this, if you wanted to achieve a higher voltages by adding even more power supplies, what would be the limiting factor at which point you could no longer add supplies safely (Without blowing them up)? What specs on the data sheets of my power supplies determine this and I should look out for? Also does the relative voltage to ground have any effect, or just total potential difference. For example does it matter if I do this: or this: <Q> The insulation transformer breakdown rating might be one limit if approaching 1kV and this is degraded if the DC output is earthed due to leakage capacitance stress across insulation. <S> The other is DC load regulation of the worst power supply will limit the amount of load current in order to prevent overloading the weakest link. <S> This is similar to series batteries. <S> Finally short circuit protection might in fact exceed device OCP capability due to the possible reverse voltage on the weakest series device, from a short circuit. <S> The same is true for back EMF say from a large motor or negative resistance arc. <S> Otherwise you can string as many as you want in series if you observe the above potential failure modes. <A> It depends on the isolation potential each supply is capable of. <S> That's the maximum allowed voltage difference from any output to any input. <S> For most commercial power supplies, that's often in the 1.5 to 2.5 kV range. <S> Let's say you have a bunch of 15 V supplies running from 120 VAC, and they are rated for 2 kV isolation. <S> Since the input is a sine, it will range from ±170 V from ground, so let's use 200 V. <S> If one of the outputs is grounded, then we have up to 215 V difference from any output to any input. <S> From the above, there is (2 kV)-(215 V) = <S> 1785 V of isolation potential left. <S> For each supply from the one that is grounded, another 15 V is used up. <S> That comes out to 119 supplies pn either side of one that is grounded. <S> Since two supplies in the middle can share a ground connection, the absolute maximum in this example just from isolation voltage is 119 + 2 + 119 = 240. <S> So basically, the answer is "a lot". <S> Note that in the maximum string of supplies above, the total end to end voltage would be 3.6 kV. <S> You'd of course have to be very careful with that, and make sure any insulation you are using can handle that. <S> Ordinary off the shelf wire doesn't usually have 3.6 kV insulation rating. <S> I also want to make it clear <S> I'm not actually recommending or endorsing this. <S> This is a bad idea. <S> One problem is that the isolation from output to case is probably less than output to input. <S> This means that effectively the cases of the string of power supplies must be considered "live". <S> You can't, for example, have the string of 240 supplies sitting on a metal rack. <S> All but a few of the supplies in the middle would be unsafe to touch. <S> The math above also doesn't leave any margin for voltage spikes on the power line and the like. <S> Again, more than "a few" supplies in series is a bad idea. <A> The data sheet ought to give this information. <S> I've had a quick look through a well known online supplier at PSUs from £40 to £800, and none of them do. <S> In order to be able to be connected to mains, most instruments will survive the mains input going to +/- <S> 1.5kV with respect to ground. <S> Unfortunately, this tolerance is rarely matched on the output side. <S> As the output is designed to handle only low voltages, the clearances between tracks and grounded elements on the PCBs may only be appropriate for hundreds of volts, not thousands. <S> If you do contemplate connecting several PSUs in series, then buy one, strip it down and examine the clearances. <S> Power it up taking the output to 50% more than your target voltage above ground/chassis. <S> Check that nothing breaks down to chassis, to programming connectors, to any accessible metalwork on the case. <S> You may be pleasantly surprised. <S> You may save yourself the cost of buying many supplies to find they weren't suitable. <A> Since most floating output power supplies (the only type you could connect in series) have a limited breakdown potential to mains Earth, this would appear the only safety issue. <S> It's unlikely that you'd ever hit this since it's typically in the 1500 <S> + V range. <S> The other issue that you more likely would hit is that short circuit and fold-back protection is now a cascading series of partial events, and the overload voltage may not be non-zero. <S> For example, if you had just two 30 V supplies that fold-back from a set 1 A maximum output value. <S> The current trigger levels will be different by some small amount, so one supply will fold-back first taking you from 60 V to (30 + fold-back voltage). <S> So your output voltage no longer follows the expected fold-back or constant current voltage profile. <S> This could mean you burn rather than protect your external circuits you are powering. <S> As you connect more supplies in series, the problem gets worse.
| Generally, if you want to stress a component, and the datasheet stays silent about how much you can stress it, then prudence says you shouldn't, until you've conducted your own tests.
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Can a resistor of known value be used to induce a predictable current, in accordance with Ohm's law? CONTEXT I have a 100A shunt and an opamp, which I'm going to use for measuring current through a battery bank. But in order to calibrate it, I need to know the actual current flowing through the circuit. I don't have a 100A (or even 50A) capable meter. I could use my 10A-capable multimeter, but I wouldn't be able to measure at the higher current levels. METHOD Can I place a known resistor over a known voltage, and then use Ohm's law to reliably calculate the current? simulate this circuit – Schematic created using CircuitLab E.g. if I measure the resistance of a dummy load with a multimeter, and then measure the voltage drop over the resistor, can I then reliably calculate the current? Also, since my cheap voltmeter can measure up to 10A, I'm thinking to test it with an adjustable power supply, by gradually increasing the voltage until the current reaches maybe 5.00A, and then measure the voltage with a separate voltmeter (so that the circuit remains constant), and then use this to calculate the resistance more accurately. Is this a stupid way to approach it? <Q> If you have a 100A shunt it is already calibrated (probably within 0.25% or 0.5%). <S> Just disconnect the shunt, apply 0mV to your readout circuit and make it read zero, then apply 75.00mV and make it read 100.0A (repeat if necessary). <S> There is no need for a high current source unless you doubt the provenance of the shunt. <A> Either of those methods is a reasonable way to approach it. <S> A clamp on current meter may also be a good option. <S> If you decide to use your multimeters current function, you'll want to get close to 10 amps to get the best resolution. <S> 9 amps will give you better results than 5. <S> Keep in mind that at 100 amps, resistive heating of your current shunt may throw off the calibration unless you compensate for it, or actively cool the shunt. <S> One issue you may have with the shunt is that at 100 amps, even a 0.1v signal will result in 10W of power disipation from the shunt. <S> A 1/4W resistor will quickly burn up, and won't be accurate anyway due to the heat. <S> You'll need to go for an even smaller signal, use a very burly resistor, or run the system for only a short burst (like 1 second). <S> Calibrating at a lower voltage using the current function is probably an easier option. <A> Op-amp and additional resistors aside for now... <S> Since current in any series circuit is the same, simply measure the voltage drop across a resistor, and then divide that by the resistor's value. <S> (I=E/R) <S> Ohm's law
| Measuring the current in that circuit can be done with ONLY a voltmeter, provided that you know the value of at least one of those resistors.
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The concept behind a shunt resistor Why do we use "shunt resistors" and is there a difference between a standard resistor and a shunt? Or is this context based, e.g., as in pull-up or pull-down resistor? Addition: What is meant by shunt? I'm sort of missing the fundamental concept as well. <Q> The term "shunt" is strictly context-based. <S> However, shunt resistors most often have a very low resistance, because they are most often used to make current measurements. <S> Such low resistances may have four terminals: two to make external circuit connections, and two for measurement: <S> This shunt has two large terminals on top for external circuit connections, and two smaller screw terminals on the side for measurement. <S> It may be connected to a specified amp meter (ammeter), or voltage can be measured from screw-terminal-to-screw-terminal. <S> But its basic electrical property is the resistance of the flat metal bridge between the large terminal blocks. <A> I know that shunt resistors can be used in parallel with an ammeter to allow larger currents to be measured(larger <S> than what is the ammeter rated for) <S> and that's the proof for "low resistance and high power rating" simulate this circuit – <S> Schematic created using CircuitLab <S> Say that the ammeter depicted in the schematic can handle at <S> most 10 A.Now Rshunt must have a smaller value than the ammeter's resistance. <S> That's: <S> \$R_{shunt}=\frac{R_{ammeter}}{n-1}\$ <S> Where n is the number of times I1 is bigger than the maximum AM1 current. <A> A 10Ω shunt resistor can be used as 10Ω pull-up resistor if you need one. <S> It would just probably be bigger than 10Ω resistor you'd use for a pull-up. <S> I personally like the medical definition of shunt ( an alternative path for the passage of the blood or other body fluid ) for explaining shunt resistor's purpose <S> - it provides a path for high current to pass through without damaging other elements (often expensive equipment that does not like high currents flowing through them).
| To shunt means to redirect part of the current elsewhere which would otherwise flow through another path. Yes, the name depends only on resistor's placement in circuit.
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Why change the quiescent current in an op-amp? (LM4250) I am trying to understand the LM4250 op-amp. Datasheet: http://www.ti.com/lit/ds/symlink/lm4250.pdf Main description: Pin-outs for DIP: Example circuit: It appears that pin 8 (quiescent current set) can be used to change the quiescent current, according to this table: My question is: why would you want to do this (change the current)? My understanding of quiescent current is that this is the minimum amount of current consumed by the device (not under load). Why would you want this to be any higher than the minimum the device is capable of? Also, what is this symbol? <Q> You would change the current in order to modify the tradeoff between performance (higher currents mean higher slew rates on internal nodes) and power consumption. <S> That allows this single design to address a wider range of applications. <S> The odd symbol just seems to be another name for the negative supply (pin 4), which is where the other end of the resistor needs to go. <A> why would you want to do this (change the current)? <S> Also the lowest quiescent current amps have severely limited bandwidth. <S> [1] The resistor(used to vary the current) is used to bias the LM4520. <S> This is used to vary the characteristics of the amplifier over a limited range. <S> A single external programming resistor determines the quiescent power dissipation, input offset and bias currents, slew rate, gain-bandwidth product, and input noise characteristics of the amplifier . <S> See section 3 in page #5 of the applicaiton note AN-71 <S> Micropower Circuits <S> Using theLM4250 Programmable Op Amp. <S> here is the link . <S> The symbol is V- , the negative supply input (Pin 4). <S> The purpose is below: In applications where the regulation of the V+ supply with respect to the V-supply (as in the case of tracking regulators) is better than the V+ supply with respect to ground the set resistor should be connected from Pin 8 to V-. <A> Tradeoffs for bipolar Op Amp Q current are; - Slew Rate GBW Product, Open Loop gain, Input Noise current.
| For general purpose op amps, some performance parameters (noise anddistortion in particular) tend to improve with higher current.
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Measuring mOhm resistance using LT006 precision amplifier and 4-Wire Method I am trying to use the 4-wire method to measure the resistance of small strip of carbon steel, which will have a resistance of about 0.3 mOhms. I am currently trying to test my circuit with a known 0.3 mOhms resistor. I am using the Arduino to make the measurements. I have a 5V supply and the circuit has a resistance of 150 Ohms, giving 33 mA current. The potential drop across the 0.3 mOhms resistor is expected to be 9.9 uV. This is amplified by the LT1006 by a gain of 1000. I am using oversampling ( https://gumroad.com/l/eRCaGuy_NewAnalogRead ) to achieve a resolution of 16-bits, so I a gain of 1000 should be sufficient. A schematic of my circuit is shown below I expect the resistor of the sensor to be given by V_sensor/V_100 * 100 Ohms. I have used the circuit to measure 100 Ohms and 220 Ohms resistances successfully, using a gain of unity on the LT1006. However, the issue is that the V_sensor I am measuring with the 0.3 mOhms is equal to my input offset voltage (0.002), i.e. its the same value as when the inputs of the LT1006 are shorted together. I have used a 10k pot to null the offset. But it only reduces it to 0.002 V. Also, is it possible to achieve a higher gain on the voltage across the sensor, by replacing the 10 kOhms resistors on the LT1006 inputs by 100 Ohms. I read here ( Kelvin "4 Wire" Resistance PCB Design Questions ) that the input resistance for the 4-wire method should be 10 k. <Q> Use an AC current source, and AC preamplifiers on the small developed voltage. <S> There's no 'offset' if you use an AC signal instead of direct-coupled DC. <S> AN-98 <S> See Fig. <S> 27 <S> The use of FFT oscilloscope, or a phase-locked amplifier, might accomplishthe desired result without having to wire up a dedicated measurement circuit. <S> A low-voltage measurement with DC sensitivity will always show artifacts, if not from offset voltage, then from thermocouple effects. <S> AC measurements are theonly good solution. <A> Don't increase the gain of the first stage beyond what you need to escape the input noise. <S> Better put in a cheaper difference amp in the second stage. <A> If you want to measure microohms, you need much more measuring current. <S> A commercialy available microohmmeter with the range of 320 µΩ and <S> a resolution of 1 nΩ uses 10 <S> A current <S> , that is 300 times more current than you tried. <S> 10 mA is used for the range of 3.2 and 0.32 Ω, 100 mA for 320 and 32 mΩ and 1 A for 32 and 3.2 <S> mΩ. <S> Several measurements are done for one single resistor value: <S> zero current and current in both directions and both polarities of the voltage at the measured resistor. <S> All these measurements are used to compensate offset error. <S> But a lot of know how is necessary to minimize errors due to thermoelectric voltages caused by different temperatures in the measurement circuit.
| I see the following ways (ordered by increasing impact on your circuit) Integrated auto-zero/zero-drift amplifiers, which can give you an offset of below a microvolt out of the box: http://www.analog.com/en/products/amplifiers/operational-amplifiers/zero-drift-amplifiers.html , http://www.linear.com/products/Zero-Drift_Amplifiers AC measurement (as mentioned by @With3rd) Self-designed chopper-stabilized amp (if you need more control over your circuits noise)
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Which TWO wires to disconnect to disable USB mouse As there is more choice in DPDT switches than there is for 4PDT switches I am wondering which TWO connections are the best to disconnect to disable a USB mouse? I need an easy to use switch that allows the mouse to be 'disconnected' without pulling the USB connector out of a laptop so I want to use a DPDT rocker or paddle switch. From what I have found so far the pin-out is: VCC (+5V) Data -ve Data +ve GND If my switch only allows two of these to be disconnected is it better/safer to disconnect both data lines, both power lines or one of each? (Note: I've attempted to do the disabling programmatically in Windows but unfortunately Windows APIs identify the multiple input devices on the machine all as 'Generic HID' (Human Input Device) so it's not easy to determine the correct one to disable.) (Note 2 - added in response to some comments and answers: The reason for this question is that I need to supply a solution to a user in a user-friendly manner. To this end registry hacks, Device Manager etc are not user-friendly enough. A simple toggle switch is easy to understand and, so long as it doesn't fry the MoBo or the USB hub, then is safer than sending a user into system settings.) <Q> However, disconnecting <S> VCC alone may be enough , and even if not, it should at least be safe. <S> Take a look at this figure from USB specification: <S> The host will only communicate with the device when it sees D- line pulled up. <S> However, once VCC is disconnected, Rpu is effectively eliminated from the circuit, leaving both D+ and D- at ground potential. <S> Once you reconnect the VCC line, the host will again see D- line pulled high, and will react as if a new USB device was just plugged in. <S> In the best case, your mouse will just work. <S> In the unlikely worst case scenario, the mouse won't have enough time to initialize (since normally VCC is connected before D- and the device has more time to start) and will fail to reply to the host within the allotted time. <S> In that case, you'll have to keep VCC and GND connected and put a DPST switch on D+ and D- . <A> The safest way is to disconnect the Vbus line. <S> This powers off the device but keeps the Ground connected for RFI shielding and ESD protection. <S> If the ground is disconnected then the mouse becomes an antenna connected to the D+ and D- lines radiating as the host initiates a USB training sequence. <S> There is no problem if the data lines are connected as long as the ground is not disconnected <S> all USB PHYs are designed for this because some devices are not bus powered. <S> But there is a problem if the Vbus line is connected while the ground is disconnected because the data lines D+ and D- have ESD devices connected to the substrate in the die and this may originate latch-up or stress the ESD devices above the specifications. <A> As Tony Stewart commented you can use the registry directly .... <S> you can use Device manager to find the VID/PID for the particular mouse: <S> You can also just manually disable the mouse directly in Device Manager by right clicking and 'Disable' the device or run a Regedit macro, WMI or Powershell script to toggle the registry values.
| Leaving VCC and GND connected is obviously safe for both the mouse and the host: VCC and GND leads in USB connectors are even made longer to insure this safe state is reached before D+ and D- are connected. My first reflex was to say that you should disconnect D+ and D- lines. As a result, no further communication will take place.
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name for relay that holds state and only consumes power when toggling states I am wondering if there is a certain label or naming convention for a relay that can mechanically hold its contact state after the coil is de-energized. I thought this is what a "latching" relay does, but after some googling, it looks like latching relays hold their state by opening/closing a current path for a secondary coil. I would like for there to be no power consumed by the relay once the switch has changed state. Does such a device exist? <Q> Latching Relay coils tend to be high power and demand short periods of power. <S> Whereas Non-latching relays can have a high effective current gain of I_contact(max)/I_coil = 1000 to 5000. <S> Faster power <S> Relays tend to be lower ratios and Latching Relays even lower ratio. <S> They come in 2 versions; single coil with bipolar voltage +Set , <S> -Reset <S> , 0V idle <S> these are used all types of "relays" aka ... solenoid, relay , Contactor or transfer switch body. <S> dual coil with +Set and +Reset being exclusive inputs that are permitted to operate one at a time. <S> Here the common coil end is often metal case grounded. <S> There are other ways to make a latching relay too. <A> As others say, a latching relay will stay in its last set poition without power. <S> One application for this sort of latching system is as a relay controlling a motor - when you push the "start" button, the relay pulls in, and one contact closes to maintain power on the coil when you release the button. <S> If power fails, the relay releases, and the motor will not start unexpectedly when power is restored. <A> I thought this is what a "latching" relay does <S> Wiki <S> says this: - A latching relay (also called "impulse", "keep", or "stay" relays) maintains either contact position indefinitely without power applied to the coil. <A> Such relays are known as "latching" or "bi-stable". <S> They mechanically stay in their current state when power is removed. <S> You seem to have gotten the wrong idea what "latching" means for a relay. <S> These do exactly what you are asking for. <S> Such relays either have two coils, one to drive the relay to each state, or a single coil and the direction of the current selects the state that the relay is driven in. <A> There are also relays called "magnetic latching relay" that use a coil to pull the relay contact arm into one position where a magnet then can hold the arm in position even though power is removed from the coil. <S> This type of relay has three typical methods that are used to release the magnetically held contact arm. <S> A second relay coil is energized and can pull the contact arm back away from the magnetically held position. <S> This second coil exerts a stronger pull on the arm to overcome the magnet. <S> The primary pull in coil is used with a reversed current flow direction. <S> This produces a repulsive field to cancel out the magnet field and allows the arm to return to its non-latched position (usually by a spring). <S> The relay may have a mechanical button or lever which can be physically activated to release the magnetically held contact arm position. <S> Again a spring is used to return the arm back to the unlatched position.
| The latching relays you mention, that need power to hold in one state, are better described as relays with a self-latching contact or circuit. It is called a latching relay and all the ones that I have seen retain their state of latchedness through any power cycle.
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What PCB components typically require hand assembly or are associated with difficult assembly? With regard to PCB design for manufacturing (DFM), I've come across a guideline of 'Avoid selecting components that need to be hand placed'. However, this guideline didn't go on to state what components generallyrequire hand placement. So my question: What component types typically require hand placement? And maybe a bonus question, what components can almost always be automatically placed? <Q> Some features I've seen on parts that require hand assembly: <S> Unbalanced part. <S> If the center of gravity of the part isn't over the parts of it that contact the board, it will fall over before it can be soldered down, unless specially handled. <S> Low-temperature part. <S> If the body of the part is made of (some types of) plastic, or contains some low-melting point metal, it may not be able to withstand the heat of an automated soldering process. <S> And it will often also have a spec about the maximum time to keep the iron in contact with it during hand soldering. <S> Mixed mode assembly. <S> If through-hole and SMT parts are mixed, it may be more cost effective, in a small production run, to hand-solder the through-hole parts. <A> Most surface-mount components are designed to be machine-placed. <S> A Pick-and-Place machine uses a system of cameras to precisely place components, via suction. <S> Some components are difficult or impossible to pick up with suction. <S> In those cases, it is common to see something like this: <S> The flat plastic cap provides a surface area for the machine to pick up the part. <S> After assembly, the cap can be slid off. <S> Generally, through-hole parts require hand placement, or specialized equipment. <S> Sticking with surface-mount parts will make it a lot easier for manufacturing. <S> Also, as was mentioned in the comments; if the part comes in a reel, that's a pretty safe bet that it can be machine-placed. <S> When in doubt, ask your board assembler for details about their capabilities. <S> They may be able to help recommend alternatives to hand-placed parts. <A> In my opinion it highly depends on your assembler. <S> As Daniel Giesbrecht mentioned - most of the SMD components nowadays come in a "pick and place friendly" packaging. <S> That is - they will be supplied in tape, with plastic caps or masking kapton stickers so the machine can pick it up. <S> But unfortunately it comes with a price - literally. <S> Most of the time parts in tape will be more expensive (sometimes very significantly) than parts in bulk (in trays, bags etc.). <S> Second problem is the assembler - or, to be exact, it's capabilities. <S> Yes, you can buy, for example, a 80 pin SMT connector in tape. <S> But this tape will be very wide (72mm or even more). <S> To feed this tape into the pick and place machine you need a feeder that is 72mm wide. <S> And not many assemblers will have that kind of feeder because it is very expensive and rarely needed. <S> So when you go to a smaller assembler, they will either order the same part in tubes/tape - and place it manually, or cut the tape to strips and stick it (sometimes literally) inside the machine and use it as a tray of components. <S> There are also problems with the height of the tape, weight of the parts and nozzles used to pick up the part. <S> All of them in tapes ranging from 8mm to around 32-44mm (to be on a safe side). <S> If you have any "non standard" parts - like very large IC's, large connectors, mechanical parts (studs, spacers, terminals), switches, heavy parts, large electrolytic capacitors etc. <S> - contact your assembler and ask if they will be able to place them automatically.
| When it comes to components that will be placed (probably) without a problem - all of the standard "chip" components, IC's, tantalums, electrolitics, inductors, basic connectors. TL;DRIn summary - what you consider a "component placed by hand" is highly dependent on your contract assembler.
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Can you solder a jumper cable to an LED then to a resistor directly without breadboard? The project involves lighting a simple LED from a Raspberry Pi, and I am planning on ditching the breadboard and soldering parts directly onto to each other. For instance: 3.3v pin - LED positive leg resistor - LED negative leg - ground header pin I am considering the health of the wires if I directly solder them together without a breadboard. Will they be damaged easily and shall I invest in heat shrink to cover the soldered parts? <Q> Soldering resistors and LEDs using wires (aka flying leads) is very commonplace. <S> Heatshrink is best for insulating them and providing a strain relief, the latter is especially important where vibration is an issue. <S> Otherwise the joints will flex and work-harden, become brittle and eventually break. <A> Breadboards are a tool to allow for rapid prototyping without semi-permanent applications like crimping or soldering. <S> They are not required. <S> Jumper wire, which is typically 20-24 AWG wire, is ideal for small current applications like a single led. <S> Depending on how you mount that led, you should consider heat shrink or any other type of non-conductive covering. <S> Tape works. <S> Hot Glue works fine. <S> If you mount the led in a panel or hole of some type, so it doesn't flop around, then you really don't need to insulate the leads, but it still helps. <S> Heat shrink or other insulators do help improve the mechanical strength of bare wire. <S> Solid wire is stronger, but more brittle than stranded wire. <S> Even with heat shrink on, solid wire will break at the bend point if you bend it too much (metal fatigue), while stranded wire will last longer (but will break eventually too). <S> Heat Shrink makes both last longer. <A> If the wire is stranded, provide some kind of strain relief or ensure both ends are mounted rigidly. <S> Soldered stranded wire will be likely to break at the end of the soldered region if it is flexed repeatedly. <S> Heat shrink is a fine way to insulate the soldered joints, and can help with the strain relief. <S> (This is not harmful in itself; it just means the wire is less insulated than you intended.) <S> Use flux, and tin the wires before attempting to make the joint. <S> Soldering wires can be trickier than soldering <S> through-hole PCB mounted parts because there is nothing firmly holding the two wires in the position you want. <S> (At least, I find it so; some make it look easy.) <A> The answer is yes. <S> Circuits can be soldered together with only components or on copper breadboards (solid plane with cuts or the kind with through holes). <S> One is only limited by ones creativity. <S> In fact this is one of the best prototyping methods because by making a direct connection you are reducing parasitic effects. <S> (jumper breadboards have lots of stray capacitance and connection resistance) <S> By soldering directly these can be eliminated Jim Williams who was the lead engineer of one of the leading manufacturers of IC's uses this method to prototype circuits extensively. <S> Shown here and here <S> So put that soldering iron to good use, make good clean prototypes and use copper plated breadboards instead of lossy breadboards with jumpers. <A> Yes. <S> As for the health of the wires, I wouldn't worry. <S> As long as you use decent quality materials and solder correctly you'll be fine. <S> But of course you'll still want to insulate them if you can. <S> Bare wires are an accident waiting to happen
| Also, make sure your soldering tools and technique are efficient, so that you do not apply heat so long that the wire insulation melts away from the hot wire. Breadboards are for prototyping and testing, so you don't have to solder and resolder Everytime you want to change the circuitry.
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Create Circuit to come ON when momentary contact is RELEASED, not pushed I am trying to create an electronic circuit, that is sort of a "not?" When the switch is pushed, nothing happens. When the switch is released, after a momentary press, then the Led would light. This will drive a simple light, and finally the entire process would be reset with a second button. The reset can happen at the momentary press of switch #2. A friend of mine at work seemed to think this would be a good application for a 556 timer chip, which I gather has two 555 timer circuits in it, which are somehow linked, and the output of one of them controls the start of the second timer. I'm not exactly sure how you would apply that to my project, but I suppose it makes sense, the first button would trigger the timer circuit, for as long as button #1 is pressed, and then AFTER that button press, trigger the second timer, which would go on forever until a signal from Switch #2. I assembled and created this circuit. I don't currently have a switch that does what #2 does, but with a jumper wire, I can simulate it. It doesn't work. :( My friend from work seems to think it was in my description - But I re-read the description and it's pretty straight forward. LOAD TO BECOME ACTIVE WHEN SWITCH #1 IS RELEASED. SWITCH #2 TURNS OFF THE POWER TO THE LOAD. Doesn't SOUND that complicated to me, although I wasn't able to figure out a way to make it work. In my mind, logically, sure, you have a latch that's activated by the release of the first button, and is de-latched by the second. Easy peasy! Tons of Thanks in advance guys! (and/or gals) <Q> Bradman175 is on the right track, but perhaps a diagram that shows what he's talking about would help: simulate this circuit – Schematic created using CircuitLab <A> Here is one of many possible solutions. <S> This one is very straight forward and uses 1/2 of two edge-triggered D-flip-flops inside a 74HC74 IC. <S> Basically, whenever there is a rising edge on its CLK (clock) input the flip-flop transfers whatever is at the D (data) input to its Q output. <S> Here, D is tied to VDD <S> so it is always logic 1. <S> When START button is pressed, nothing will happen because CLK only responds to a positive (rising) edge. <S> When START is released, R1 pulls CLK high and the flip-flop transfers the 1 at the D input to its Q output. <S> Additional presses on START do nothing since Q is already high. <S> When STOP is pressed, /CLR is pulled low and clears the flip-flop <S> so Q goes low. <S> /CLR <S> input is negative logic (denoted by the bubble/circle on the input). <S> Note that the LED is wired to /Q. <S> /Q is the complement of Q. <S> But the LED is powered from VDD <S> so it still lights up when Q goes high. <S> The reason this is a bit better configuration is that most CMOS ICs can drive low better than high. <S> I think this is what you would want if it was controlling some kindof machinery. <S> You always have to design your circuits to take unexpected inputs from users. <S> Hope that helps, -Vince <A> You basically want to use an arrangement of transistors along with a circuit called the silicon controlled switch . <S> You want a SPDT button that charges a capacitor. <S> Then you can use the off button to turn the SCS off. <S> The motor in the diagram is the load.
| When released, the capacitor would go where the on button is to latch the SCS on. Also, if the user is pressing the START and STOP at same time, the STOP always overrides the START button and holding STOP down will cause it to ignore START signals.
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What is the least number of 8 Ω resistors you can arrange to have a total resistance of 5 Ω and how do you do it? 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> I never faced this particular question before. <S> But faced with it now, I'd start out (after mentally verifying that a bridge arrangement would be pointless -- and it is) by asking myself what, in parallel with \$8\:\Omega\$ would yield \$5\:\Omega\$? <S> It turns out that the answer is <S> \$13\frac{1}{3}\:\Omega\$. <S> Then I'd wonder about what might make that. <S> Well, if I had \$8\:\Omega\$ already, then I'd need another \$5\frac{1}{3}\:\Omega\$ to get that total. <S> Well, luckily \$16\:\Omega\vert\vert 8\:\Omega\$ makes that. <S> So that's my answer. <S> It's not a general purpose algorithm to get from \$X\$ to \$Y\$, exactly. <S> But it's a thought process to find an answer here. <S> And it suggests an algorithm (discussed below.) <S> The answer I'd give is: $$\left(\left(\left(8\:\Omega+8\:\Omega\right)~\vert\vert~8\:\Omega\right)+8\:\Omega\right)\vert\vert ~8\:\Omega$$ <S> or, simulate this circuit – <S> Schematic created using CircuitLab <S> The algorithm might be to realize that 8 and 5 are relatively prime and that it is likely that you'll need to reach their product, \$8\cdot 5=40\$, in order to find an answer. <S> Intuitively, this makes some sense thinking that you are probably looking for a least common multiple of the two values. <S> So it does strongly suggest that 5 resistors will be the minimum you can use here. <A> It's possible <S> but I cannot tell you with the fewest resistors possible. <S> 16 ohm parallel to 8 ohm would get you close at 5,3 ohm. <S> \$5,3 = <S> R||2R\$. <S> Depending on the tolerance it can already be a good answer. <S> Series: \$Rt <S> = R1 + R2\$ <S> Parallel: <S> \$Rt = \frac{R1 <S> \cdot <S> R2}{R1 <S> + R2}\$ also written as \$Rt = <S> R1||R2\$ <S> The easiest and exactly 5 is <S> \$Rt = <S> R||R <S> + R||R||R||R||R||R||R||R\$ where R = 8 ohm. <S> 2R parallel gives R/2. <S> 8R parallel gives R/8. <S> In this case a total of 5 ohm. <S> I would write a program to check the first 100k combinations of parallel and series resistors. <S> My best with trial and error is. <S> \$5,05 = R||3R||4R\$ <A> You can find it mathematically. <S> It only works for rational numbers though. <S> Note that this does not mean it's the solution with the least amount of resistors required. <S> So far for 8 and 5, it's only 1. <S> For 8 to reach 1. <S> You need 8/1 which is 8. <S> Thus you put 8 resistors in parallel. <S> Now you have this. <S> Then you need to put these jumble of resistors in series to add to the amount of resistance you want. <S> Since each jumble of resistors is 1, 5/1 is 5 <S> so we need 5 jumbles of them. <S> Now you have this abomination. <S> Congrats <S> you got your desired resistance... now count them up yourself.
| You first need to find the highest number that when multiplied by any positive integer, it can equal both numbers.
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How do I connect a PIC microcontroller to a computer? I am still a beginner, middle school student, and I recently started taking interest in Arduino, Raspberry Pie, etc. Now I want to play around with the PIC microcontroller, but I can't find a way to start since I can't find an online source for learning. I want to know how to make it execute programs, because I want to play around with LEDs. Can anyone help me? Also, can you give me a link to a place where I can learn online, please? <Q> Progressing from the Arduino to a real system like a PIC requires more up-front investmant. <S> Primarily you need to invest in a programmer. <S> Many development boards come with a programmer built in, but in general it is very handy to have one to hand regardless. <S> You can buy cheap Chinese cloned PICkit2 programmers on eBay, but I have found that the USB connectors on the ones I have used have been dodgy and needed replacing. <S> The "normal" budget programmer these days is the PICkit 3 which deals with all the modern PIC chips, like the PIC32MZ series. <S> For high end work and debugging you might end up wanting to invest in a better programmer, like a RealICE or something similar, though they are expensive. <S> They give you much better debugging though, such as external trigger breakpoints which you don't get from a budget programmer. <S> A good transition from the Arduino's AVR chip to the PIC chips though is to maybe look at the chipKIT system - especially if you want to jump straight to the PIC32 chips. <S> This is an Arduino compatible system based around the PIC32MX and PIC32MZ chips. <S> Programming is the same as Arduino initially, but you can branch out and write raw PIC32 code if you so wish and program through the bootloader instead of needing a hardware programmer. <S> It's still handy to have a hardware programmer to hand though... <S> For 8 bit PIC there are other boards available, such as the CHIPINO, but again that requires an external programmer to work with it. <S> A friend of mine also makes the Firewing which comes in a variety of forms, including a PIC18 version which has a bootloader and USB interface. <S> It's designed to work with his own version of Basic, but there is nothing to stop you using MPLAB-X and XC8 to program it. <S> One caveat with PIC chips is that it can be tricky to work with a bootloader on them. <S> You have to know how to work with linker scripts in order to fit in with the bootloader. <S> Again, it is handy to have a hardware programmer to hand to reinstall the bootloader in case you manage to break it. <A> Try the The MPLAB Xpress Evaluation board ($12). <S> It is integrates seamlessly with MPLAB Xpress cloud-based IDE and MPLAB Code Configurator for the quickest development. <S> Additionally you don't need to buy the PICKIT to program the board. <S> With MPLAB Xpress Evaluation board you can just drag-and-drop with USB Interface. <S> For the learning purpose, It will be good start for you in less investment! :) <A> I think what you're asking about is how to get code on to the PIC? <S> If so you need a programmer, a pickit3 is the cheapest of the Microchip ones.
| So in short: buy a hardware programmer, such as the PICkit3.
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Pros and cons of low-power integrated series voltage regulators vs. shunt regulators TL431 (and equivalent) shunt regulators are jellybean parts nowadays and quite cheap. The same can be said of TL317 (low-power version of LM317) adjustable series regulator. When applied to supplying power to light loads (<100mA) they are both quite effective and simple to use. Despite their different internal topology, it seems that they can be used in the same role quite interchangeably. I was wondering what are the pros and cons of using one or the other, or, more generally, what are the pros and cons of using a low-power shunt vs. series linear regulator. What are the design decision that can make a designer choose one over the other? I'm aware that, for example, series regulators may have considerable drop-out voltage (2.5V for TL317), something that shunt regulators don't have (but may have higher minimum bias/quiescent current, instead). But taking into account LDOs regulators, this doesn't seem to be so compelling nowadays. In the end, are there conspicuous advantages of one topology over the other (and in which case), especially when powering low-power loads? Note: direct comparison of the two jellybean parts I've mentioned is appreciated as well as a more general comparison between topologies. You choose. The important thing is I'd like practical considerations based on actual design experience or knowledge on reasonably current products or systems. In other words "why and when a design engineer would use one over the other". <Q> To me the critical difference is how they handle loads with varying power requirements. <S> In some scenarios this could be a positive feature. <S> A series regulator will take less power from the supply if the load current drops. <S> In some scenarios this could be a positive feature. <S> This also makes the series regulator more tolerant of sloppy design --- you don't have to know exactly the maximum and minimum power that might be needed by the load in order to design with the series regulator, and your design is not penalized (with extra heat generation) if you overestimate the load's current requirement. <A> A shunt-regulator, fed from a constant-current source, can produce some of the quietest voltages on earth, with excellent rejection of ripple at all harmonics of 120Hertz FullWaveRectification. <S> Example is the diyAudio forum/thread "Simplistic NJFET RIAA" phonograph PreAmp. <S> Their goal, with the ShuntReg, is not precise VDD but very quiet VDD. <S> I've read many of the early posts; there are 16,000+ posts over the last 11 years, and I've concluded the noise floor of their ShuntReg is approximately 1 nanoVolt for 60/120Hz hum. <S> The very first post of that thread has a PDF with schematics for the RIAA Preamp and for the ShuntReg. <S> Lots of design effort goes into raw_DC design, transformer selection, and Grounding. <S> Some of the DIY folk write about the absolute silence coming from their speakers, with the Gain at maximum and the phone cartridge lifted from the vinyl. <A> Here is one of the ShuntRegs, used to provide power to the 2-stage NJFET RIAA; the 24v DC output would be for a MOVING MAGNET preamp needed lowgain. <S> Here is an example of the RIAA preamp, for 250uV MOVING COIL, needing highgain thus high VDD (large Rdrain values) <S> In the ShuntReg, planning the Ground is big-deal, so the rectifier-diode current surges do not cause <S> I*R drops in copper foil shared by the "clean" output voltage. <S> The "diyAudio" experimenters have converged on remote-boxes for the power-cord, transformer, rectifiers, and first raw-DC filtering; often they'll spend $$ on inductors for raw-DC filtering, to further slow down the current surges. <S> Then a 2 meter DC_power cable brings the unregulated DC into the PreAmp box, where ShuntRegulators (one for Left channel, one for Right channel) reduce the DC down to extremely quite (1nanovolt) noise floors with excellent stability so the PreAmp NJFET transconductances remain constant and <S> the LeftRight gain balance <S> remains constant (within 0.1 dB, to maintain sound stage).
| A shunt regulator will always take the same power from the supply, even if the current drawn by the load is reduced.
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Do electronic devices consume more power when the ambient temperature is cold? Just a casual observation: when it is winter, my personal devices such as cellphone and Ipad consume power more quickly compared when it is in the summer. But this could simply because of change is usage habit: staying at home more leads to using your gadgets more often. At the same time I have noticed that the back of my Ipad heats up a lot more during the winter, which could be caused by greater power usage. Does anyone have any insights into how ambient temperature dictates power consumption in electronics (maybe even consumer electronics)? <Q> It's the opposite, actually. <S> Your devices are made up of individual electrical components, while datasheet for the devices are harder to come by, we can look at component datasheets easily. <S> The above is a power consumption over temperature plot for a processor, one of the major power consumers in your devices. <S> Most components have a positive coefficient for power consumption and temperature. <S> Some have less linear curves where consumption is minimum at some middle value. <S> But for the most part, hotter devices use more power to operate normally. <S> As others have commented, this doesn't take into account the battery capacity when it gets cold. <S> But, the answer of "do they consume more power when cold?" is no. <A> Actually, they have less battery range. <S> The major factor in cold is on the batteries. <S> They lose a lot of capacity. <S> It's quite likely that your iPad "heating up a lot" is relative to ambient <S> : it's colder outside, so the iPad feels like it's hotter. <S> Ever go out in the cold for awhile and run your hands under "cold" 45 degree (7 deg C) water to warm up? <S> It feels like the water is boiling hot , doesn't it. <A> CMOS transistors conduct better when cold. <S> That means they'll demand more current during switchover from Logic_low to Logic_high; that current is called "crowbar" current or "shootthru" current.
| The lower capacity of the battery due to temperature might be large enough to make the devices seem to last less time in the cold.
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Why do RCDs / GFCIs sense differentially rather than directly? As far as I know, all RCDs / GFCIs sense leakage current by differential measurement between active and neutral with a current transformer. Both the active and neutral wires will pass through the transformer. I have tried without success to find references on why we cannot simply pass the earth wire through the current transformer instead. Is this because some other kind of leakage or noise can occur on this wire that does not represent a fault condition? <Q> I feel the GFCI (Ground Fault Circuit Interrupter) is slightly mis-named, as it is intended to trip if any current leaks in or out of the circuit, regardless of the path it takes. <S> If there is a fault, the "missing" current need not flow through the ground wire that goes by the GFCI - it might flow from a kettle, through you, to the water pipe, and then to ground, with no current flowing through the ground pin in the outlet. <A> why we cannot simply pass the earth wire through the current transformer instead <S> [?] There are 3 wires: hot, neutral, earth. <S> Hot and neutral are intended to power the load. <S> Earth is untended as a safety ground. <S> So far, so good. <S> Let's imagine that you are measuring current only though the earth lead, and you are not measuring anything else. <S> A victim accidentally touches the hot and the cold water pipe. <S> The residual current will go through the victim, but will not happen to go through the earth lead on which you are sensing current. <S> By the way, there are plenty of appliances that don't have the 3rd prong for earthing at all. <A> Sensing current on the earth/ground wire does not detect leakage current to ground not to the ground pin of the outlet. <S> For example, a hairdryer used in a bathroom by someone sitting in a bathtub and it gets dropped into the water. <S> The current leakage path is through the water and/or body to the plumbing. <S> Also, the used arrangement allows a GFCI to safely replace a two-prong outlet where there is no ground connected. <A> It may be passing through you, or something else. <S> The current through the line and neutral should always be equal, and when they are not, there is some kind of fault, which may or may not involve the ground line.
| It's because if there is a fault, the current may not be passing through the ground line.
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What is the shape of an antennas output? I've been told that EM beams are "gaussian" beams that spread nonlinearly, so the spot size grows faster and faster as you travel further away. However, when you talk about an antenna, it is defined by a beamwidth expressed in degrees, which makes it seem like there is a constant growth of the spotsize. So do antennas not output gaussian beams, or does the beamwidth express something different than I am thinking? <Q> You can express the solution to the electromagnetic wave equation using various different basis functions. <S> Usually we choose a basis set that is convenient for matching up some boundary conditions that apply to our problem. <S> The gaussian beams are a convenient basis set when working with a beam emitted from an aperture. <S> But you might prefer the spherical harmonics if your antenna radiates more isotropically, for example. <S> TLDR: Gaussian beams are just one choice of how to express the emission from an antenna. <S> The gaussian beams do diverge with an angle, which for the 0th order beam is given by $$\Theta = <S> \frac{\lambda}{\pi <S> w_0}$$where \$\Theta\$ is the full angle of divergence (not the half-angle), and \$w_0\$ is the beam's waist diameter. <S> So there's no contradiction between saying that a radiation pattern is a gaussian beam and saying that it diverges with a known angle. <A> (in addition to ThePhotons succinct answer) Antenna can have many beams depending on shape, Lambertian response from Pad Antenna like SMD LEDs, or Torroidal response from Dipoles. <S> Gaussian is not a common shape for EM Antenna but very common for light sources. <S> A few exceptions perhaps similar are very high gain Helix or high precision parabolic dishes but not perfect Gaussian shape. <S> Gaussian shapes are more likely to define light beams such as some LED's or Lasers with a Gaussian lens.e.g. <S> Full Width Half Maximum, FWHM , is equivalent to -3dB power or 50% <S> Iv peak. <S> where full angle \$\Theta=2\theta\$ of half angle <S> True Gaussian response FWHM can be related to Standard deviation by \$\Theta_{50\%} = <S> 2ln(2)\sigma ≈ <S> 2.455\sigma\$ and <S> \$ \Theta_{10\%} <S> ≈ 4.22 <S> \sigma \$ <S> Above are 3 different hypothetical Gaussian profiles. <S> The offset one is not as bad as above but common for IR LED's which is why Half Beamwidth is used for these device specs and Full Beamwidth is used for visible LEDs. <S> This is because IR LED parabolic lenses are very small and chip angle may be offset. <S> Visible LED's use transparent substrate with parabolic reflectors and only a few are Gaussian in profile. <S> ( although some are very smooth Gaussian-like) <S> Perhaps some IR LED OEM's are using Full angle now to avoid this confusion. <S> Here is a smooth Gaussian laser response. <S> 5mw Green laser For advanced principles on Gaussian Light http://marketplace.idexop.com/store/SupportDocuments/Gaussian_Beam_PropagationWEB . <A> The shape of antenna output also depends on the type of antenna. <S> While in the case of Directional antenna ( like antenna arrays), electromagnetic radiation or energy is more in a particular direction compare to other direction so the shape is like a pencil beam. <S> There are many other specific shapes generated by shaped beam pattern antennas.
| In the case of an omnidirectional antenna, electromagnetic radiation is uniform in all the direction and shape is spherical in nature.
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Verification of SCR Driving Circuit Several years ago I attempted to build a coil gun, but gave up in frustration after burning out several SCR's when designing the trigger circuit. The SCR is used as the switch that dumps the energy stored in the capacitor into the coil. I've since brushed up on my fundamentals and tackled the problem once more. Below is the circuit I've come up with. V2 applies a 3.3V, 10ms pulse to the gate of M1. This pulls the gate of M2 low and causes current to flow into the gate of the SCR. Since V2 is a logic-level signal, I used this configuration so I could drive M2 without needing a level-shifter. R1 R3 is a current limiting resistor used to limit the gate drive current to 40mA, which is the rated maximum gate current required to turn on the SCR. R4 is used as a pull-down resistor to prevent spurious signals from triggering the gate of the SCR. Since C1 is charged to 200V, I've included D1 to prevent the drain of M2 from seeing large voltages when the SCR is triggered. L1 is the coil and D2 is used as a flyback diode to prevent reverse charging of C1. I know I've left out some details, but is the overall design of this circuit good practices? <Q> The major problem appears to be that your SCR Cathode will start to rise as the device turns on and critically reduce the gate drive. <S> You certainly won't meet the minimum t(gt) of 2 uS and may find the SCR turns on very slowly. <S> Perhaps more like this: simulate this circuit – <S> Schematic created using CircuitLab <A> I haven't scrutinized your triggering circuit and won't dwell on it as it may be somewhat secondary to your SCR <S> burn out problem (the trigger circuit does seem unintuitive). <S> You have not indicated the nature of and how the capacitor charging source V_CAP is connected. <S> If V_CAP is a DC source with no zero crossings and remains connected to the capacitor when the SCR is fired, the SCR will remain in conduction once triggered, ultimately allowing a steady DC current to flow through the SCR and coil and likely destroying the SCR (essentially a short of the SCR). <S> If V_CAP is isolated from the capacitor prior to triggering the SCR, then you should not suffer from SCR burnout (assuming that withstand voltages are respected). <S> As pointed out, the cathode of the SCR should be connected to ground potential. <S> You can do this as suggested or you could arrange the circuit similar to a flash lamp circuit with the capacitor in series with the coil and V_CAP. <S> The capacitor charging current path would be through the coil. <S> At the point the capacitor is charged, V_CAP would be isolated from the capacitor and the + terminal of the capacitor would be connected to ground through a switch or SCR dumping the charged current into the coil. <S> If V_CAP happens to be the unfiltered rectified output of something like an inverter circuit, the rectified waveform may eliminate the need to isolate the SCR to allow it to shut off and the capacitor to resume charging. <S> I also don't think you need the recirculating diode D1. <S> If the SCR was instead a power MOSFET and you were using the MOSFET to switch off the coil current, then you might need the recirculating diode. <A> A typo on my part. <S> I meant D2, not D1. <S> I had a slight brain lapse. <S> I was thinking triggered circuit where the capacitor dumps its energy into a circuit where the energy is dissipated (flash lamp or capacitor discharge ignition). <S> Your circuit has no apparent resistive load <S> so is theoretically a resonant undamped circuit. <S> The resonance means that the current will cross through 0 so the SCR will attempt to cease conduction. <S> Have a look at this discussion of a resonant LC circuit with an initial voltage on the capacitor: <S> https://www.khanacademy.org/science/electrical-engineering/ee-circuit-analysis-topic/ee-natural-and-forced-response/v/ee-lc-natural-response-example <S> Of course, there is no such thing as zero resistance. <S> However. <S> ignoring resistance, your peak sinusoidal current is 7.61xV_CAP (based upon 50 uH and 2900uF). <S> If your V_CAP is 20 volts, your 'theoretical' rms current would be 107 amps. <S> That might exceed the rms value of the SCR; but, probably not the non repetitive (single half cycle) <S> surge current of the SCR since the current should extinguish at the first current 0 crossing. <S> However, if the gate circuit remains high during this, it is not immediately obvious how this effective 1/2 wave circuit is going to operate. <S> Modeling with SPICE might prove some insight: http://www.ecircuitcenter.com/About.htm <A> In the case with no D2 and assuming no resistance in the inductor - capacitor circuit. <S> The initial voltage on the cap is V_cap. <S> At the moment the SCR is triggered into conduction, the voltage across the inductor matches the capacitor (assume an ideal diode) and the current starts to increase in the inductor following a sinusoidal form with the characteristic frequency W = <S> (1/LC)^0.5. <S> The current lags the voltage by 90 degrees and reaches a peak magnitude just as the voltage across the inductor and capacitor go to 0 volts. <S> Current continues to flow but starts to reduce as the voltage across the inductor and capacitor increases in reverse polarity. <S> The current passes through 0 at the point that the voltage on the capacitor and inductor has reached its maximum negative value of -V_cap. <S> Once the current passes through 0 the SCR ceases to conduct because it will be reverse biased. <S> When the capacitor voltage hits -V_cap, current in the SCR is zero, the SCR ceases conduction and the SCR is reverse biased and will not go into conduction if the gate is triggered again. <S> If you were to then reconnect the capacitor to the V_cap source the capacitor would effectively have a voltage of 2 x V_cap applied to it <S> and it would indeed go Kaboom. <S> With the addition of the diode D2, as soon as voltage across the capacitor / inductor starts to go negative, D2 should go into conduction shorting out the capacitor / inductor and preventing charge from building up on the capacitor in the reverse direction. <S> This should prevent the capacitor kaboom problem. <S> However, you are requiring the diode to dissipate all of the stored energy in the circuit which, depending on the capacitors initial voltage, may be significant. <S> I might be inclined to use a power resistor as a dampening circuit; but, that will depend on the operating frequency of the device and how much energy you have stored in the capacitor. <S> The other thing to consider is that if this nail coil is actually 'doing anything', there will be an effective load resistance in parallel (or series depending on how you model it) with the inductor. <S> If the load is large (parallel resistance is low), the circuit may become over damped in which case the current in the inductor will reach a peak and then may decay exhibiting no oscillation.
| A resistor in series between V_CAP and the capacitor may be desirable to limit charging current to something reasonable. You should move the SCR to the ground side of the inductor.
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Why does physically touching a specific pin eliminate circuit issues? This is a strange question and I don't know if I will be able to describe it correctly. There have been several instances working with a system that basically consists of a board camera, an FPGA, and TFT display where I can eliminate image noise by physically touching a specific wire or pin. For example, the current system I'm working on has pretty strong red noise in the live camera image. If I specifically touch the CAM_DATA[7] pin (the MSB for red), the noise disappears. It also disappears when the oscilloscope probe is connected to that pin, but not when the ground clip is attached. What effect does a person touching a specific wire/pin have on a circuit and how can I reproduce this "solution" with a basic components? ( I have verified that the issue is not just a loose wire or loose connection that is coincidentally going away because I'm pushing on it.) <Q> Touching a pin does not eliminate the circuit issue, it just hides it. <S> Your circuit still have an issue, and a big one - this particular CAM_DATA[7] <S> bus signal has insufficient timing relative to bus clock, likely an insufficient hold time. <S> A finger, or tweezers, or scope probe all do have some capacitance, 3pF, 5pF, or 10pF. <S> Actually, the fact that you have discovered this kind of sensitivity is a good sign, since it likely means that the entire bus has a serious problem that must be fixed. <S> Under some voltage or temperature or component variation the this effect can show up on some other data line. <S> You need to check the entire bus for violation of hold time and make sure the data-clock relationship meets specifications with some margins. <S> If the timing is tight, it could be that the bus does not have trace length matching, or something else. <S> Since it looks like the bus is an input to FPGA, advancing internal clock will fix the problem - it is likely that the clock insertion delay was uncounted for relative to the data I/O cell delay. <S> ADDITION: After the bus timing is fixed, I would recommend a quick and dirty test: when your system is running a relevant test, firmly press a wet finger over the bus traces, and see what happens. <S> A good high-speed bus, with typical 50-Ohm traces and solid timing will continue to function, while a marginal bus will break down and show some garbage artifacts. <S> Just applying a finger over solder mask increases the signal propagation delay along the bus transmission line (by 50-100ps), which might cause data integrity problem on a poorly implemented bus. <S> Of course, the "wet finger test" does not replace a thorough validation of bus margins by clock timing margining and data skew checking. <S> There is an excellent elaboration on the "wet finger test". <A> The effect is to add some resistance and capacitance to ground (or to +V if you happen to touch that as well). <S> Mostly you are shifting (delaying) the edge of a signal. <S> Sometimes it is due to lengthening the rise time of the signal. <S> Sometimes it due to attenuating the signal, so that it takes longer to "get up the rising slope" to the level where the edge is sensed. <S> If you touch +V instead of ground, you can sometimes decrease the rise time and shift the edge the other way. <S> In any case, for signals that are switching very close in time to their reference clock, this sort of disturbance shows up in the result. <S> If you could put the clock and the signal in question on a scope and watch what happens when you touch the pin, you would readily see what I've described. <A> I don't have an answer for what's wrong <S> but I have a quick and dirty work around if it has to go out the door. <S> Take 2 pieces of small gauge solid insulated hook up wire (14-20 ga.) and tightly twist it together. <S> Solder 1 end to the pin and the other to ground. <S> Take your wire cutters and start trimming off the twisted wire until the image is stable. <S> If the image never stabilizes start with a longer piece of twisted wire. <S> That's always worked for me. <S> What you did is called a gimmick capacitor with about 1 picofarad per inch depending on wire size. <S> They were SOP for tuning the finals amplifier on RF rigs to get the feed back from the plate to the grid just right. <S> It is still useful anywhere you need a very small capacitance, particularly a variable one. <S> If it still won't stabilize I would try solding one wire to the pin and the other to V+ and start trimming again. <S> If that does't fix it you do have a really weird problem. <S> When you touch a pin you also inject 300 to 600 microvolts of ~60 cycle AC on the pin as well.
| Attaching a probe (or touching a pin) increases signal propagation delay (or elongates signal edge) on that particular signal, so it gets properly latched, and the overall data don't have intermittent corruption anymore.
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50 Volts on the Neutral (180 Volts on the Live) bad or safe? We have an off-grid UK home on occasion powered by a generator (MOSA GE 6000 SX/GS). Due to the design of the generator, it has two modes: centre-tapped-earth 110 V, and 230 V. When running in 230 V mode, the two windings in the generator are connected in series; in 110 V mode, the windings are operated in parallel. The result is that when the house is powered (generator operating in 230 V mode) there is 50 volts between the neutral and earth, and 180 volts between the live and earth. I am told by an electrician that the neutral should be 0V (tied to earth) while the manufacturers of the generator unsurprisingly say that 50 volts on the neutral is fine. Which is correct? System Diagram Note: Generator is connected to earth by an earth stake connected to the case. Portion of internal circuit diagram of generator: Note 1: BL=Neutral (230V), BR=Live (230V).Note 2: The manufacturer added that the two windings are: connected in series when the generator is in 230 V mode, and connected in parallel when in 110 V mode. <Q> The portable generator is designed primerally to provide a 110V center-tapped earth supply as is normally used on UK construction sites. <S> The 230V output is something of an afterthought which is why you end up with <S> it referenced to earth in a weird way. <S> In the standards for modern appliances in Europe there is no expectation that the "neutral" pin is at earth potential. <S> So powering modern appliances off the generator should be safe. <S> OTOH in UK house wiring <S> it is normal to assume that neutral is at earth potential. <S> We don't normally put any overcurrent protection in the neutral and we frequently work on circuits with only the live isolated. <S> As I see it you have a few options, each with it's pros and cons. <S> Make sure all circuits are RCD protected with a double pole RCD (note that most RCBOs are only single pole isolating) to mitigate the lack of overcurrent protection in the "neutral" and place warning notices <S> so people don't try to work on circuits that are only single pole isolated. <S> Set up the consumer unit with double-pole breakers. <S> This is a good option electrically but can get kind of costly as double pole breakers are not widely used. <S> Modify the generator, remove the existing earth reference and add a new one at the neutral end of the winding. <S> Downside here is you will almost certainly be voiding the warranty and the 110V output will no longer be the center-tapped earth supply expected on UK construction sites. <A> It is safe as long as you don't use any 115V in the 230V series position. <S> But what it tells me is your connection is offset by the generator Y to ground connection , which makes no sense in the 230Vac position as it would force a Neutral offset of 115V/2 out of 230V. ?? <S> In North America we have 3 wire plus ground service with 240Vac and Neutral grounded at outside transformer giving 120 and antiphase 120 for small items and 240V for large items (Stove + dryer). <S> But you wanted to anyways... <S> I would Remove the generator 110V coil centre tap to ground and move gnd to the common connection between the series windings. <S> Then in fact it would be just like North America(3wire+Gnd, 240V single split phase) <S> you could then also use this as the Neutral connection for 120V uses if any. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Fix the wiring so the correct end of the output winding is Earthed in your configuration. <S> The US National Electrical Code 210.19A.1 states in Informational Note 4: ...limits the voltage drop in a branch circuit to 3% (5% total for feeder and branch circuit) for a reasonable efficiency of operation. <S> Based on this the N-G voltage limit on a single phase 120 VAC circuit is 3.6 VAC and for a single phase 220 VAC circuit 6.6 VAC (equal to the allowable voltage drop on the hot wire). <S> This informational not in the NEC makes the following assumptions: Earth/Ground and Neutral are bonded at a single point in your system <S> The voltage differential between Neutral and Earth is due to current flow in the Neutral wire (voltage drop).
| If you have no need for 110V, there no necessity to change. There is definitely an expectation in most standards that there is little to no voltage between Neutral and Earth.
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Arduino best way to control external potentiometer of device I have a consumer equipment which has a knob and turning potentiometer to control the heating element. I have not taken apart the device yet to see the exact value of the pot but looking for some advices what is the best way to go with making this both arduino controlled and keeping the manual functionality intacted. I come up with 2 ideas: Stepper motor control (this would obviously work but there is not much space inside the case and mechanical elements can break over time) Using some sort of FETs to emulate the potentiometer's resistance If I want the arduino set values and the external pot values to be in synchron then the only way I see to do this is to completely take out the original pot from the circuit, hook up the arduino control there and then use (preferably the same) pot to do the same through the arduino so I can override the values from software when I like. Also I would add an additional on/off switch to decide which value should be the relevant, the arduino's or the external pots. Obviously at reboots it would be always overwritten by what's the external pot giving to the Ardu at the beginning. <Q> How about using a continuous rotation encoder for manual control and using LEDs for indication purpose. <S> Digikey has some of those rotary encoders: http://www.ttelectronics.com/sites/default/files/download-files/Datasheet_RotaryEncoder_EN12Series.pdf <S> http://www.digikey.com/product-detail/en/tt-electronics-bi/EN12-HN22AF18/987-1193-ND/2408771 <S> These can be rotated continuously and will give you certain number of pulses per revolution. <S> Once full level has been achieved, rotation in that direction will have no effect. <S> However once user starts rotating in opposite direction, you go ahead and turn off LEDs in sequence to give a feedback for level reduction. <S> Once minimum level has been achieved, rotation in that direction will have no effect. <S> When automatic control is desired, you can simply set the level and light up that many LEDs to give level indication. <A> Be aware that most consumer gear is designed in a way to achieve the absolute minimum build cost. <S> In the case of a heater product with a variable temperature control there is an extremely high possibility that the control is not a potentiometer at all. <S> More likely it is a rotary thermostat control with a mechanical set of contacts that open and close relative to the set temperature. <S> These contacts directly switch the AC power to the heater element. <S> Such controls are rugged, known reliable from decades of usage and suitable to minimum component count in the heater. <S> Replacement would require that you provide for switching the AC power to the heater on and off. <S> Also needed will be a temperature sensor to provide feedback to the MCU of your Arduino so that its software can implement closed loop temperature control and know when to turn the SSR on and off. <S> Also be aware the circuitry in the heater, <S> whether the control be a rotary thermostat or an actual analogue potentiometer, is highly likely to be directly connected and referenced to the AC line power source with no isolation. <S> Do use much care when playing around with such circuits to not get maimed or electrocuted. <S> Anything you do should provide isolation for safety sake. <S> In the case of using an SSR for example select one that has optical isolation between its control terminals and the AC switching terminals. <A> Under the assumption that you open the device and you find a simple potentiometer inside:The output from the SIG-pin will be a continuous voltage between GND and VCC. <S> The device reading this output is most likely a high-impedance analog input. <S> A simple way to override the signal from the SIG-pin would be to hook up an analog output from the Arduino to the SIG-pin. <S> When you want to use the knob normally, simply set the pin as an input-port, and when you want to override the knob, set the pin as an output-port and output a signal between GND and VCC from the analog pin. <S> There are a few gotchas you must be aware of: You should have a resistor between the SIG-pin and the analog output-pin on the Arduino to avoid pulling too much current from the analog output-pin when the knob is at either extreme. <S> You should make sure that the maximal output from the Arduino does not exceed VCC as measured on the VCC pin on the knob. <S> I would recommend that you probe VCC and GND on the knob inside your device and make sure that VCC is below 5 volt.
| When manual control us desired, you can capture the rotation using arduino and light up certain number of LEDs to indicate the current level. If your heater has this type control it will not be immediately straightforward to just replace it with a simple analogue potentiometer. A solid state relay (SSR) may be applicable to this usage.
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Understanding difference between Interrupt Address and interrupt vector address In my undergrads, I was taught 8051 and in interrupts my professor had said that on interrupt SP points to the address that is hardcoded and Program goes to excute those instructions. But in between two address space is only where I can write interrupt handling code and I wrote more than than byte available it will overwrite the next interrupt instructions. Now when I look at Atmega or LPC2148 Datasheet it has interrupt vector address and I am assuming that it points to a address where instructions for interrupt handling instructions are written( not exactly on those addresses). Is my assumptions correct? P.S Due to being on low bandwidth net, I am not able to provide links to Datasheet. Would do that when I get back. <Q> The Interrupt Service Routine (ISR) is the program that's to be executed when an interrupt occurs. <S> Some CPU architectures have fixed addresses which the CPU will execute a subroutine call to. <S> This is true of the MCS-51 (8051). <S> The ISR must start at this address. <S> It is not uncommon to just put a Jump instruction at this address that takes the CPU to the rest of the ISR elsewhere in memory. <S> Other CPU architectures use interrupt vectors. <S> The vector is a memory location at which the address of the ISR can be found. <S> The location of the vector is known to the CPU, either by being fixed or in conjunction with a special CPU/hardware register. <S> When the CPU services the interrupt, it reads a vector value from memory and executes a subroutine call to the vector value. <S> This is true of the ARM, 6502 and 68000 family. <S> External hardware may have a hand in specifying the particular vector to use within a table of vectors but the principle still stands. <S> So the handling of interrupts in the 8051 CPU and in the ARM CPU seem different to you because they use fundamentally different schemes for finding the address of the ISR. <S> But these two methods (hard-coded address versus vector in memory) are pretty-much the only schemes you'll come across in all the CPUs you'll see. <S> (There's the occassional oddity, like the Z80 in Interrupt Mode 0 where it expects to read an instruction from external hardware that'll take it to the ISR, but I wouldn't muddy your water with that stuff while you're getting the hang of it all.) <A> The program address for an interrupt vector is the program address the CPU jumps to when an interrupt is triggered. <S> It's just like other program addresses. <S> In the case of AVR, adjacent ones only differ by 1 instruction, which means you can only put 1 instruction for each interrupt vector if you want to use all of them. <S> Usually, it will be a rjmp instruction to the rest of the program that handles the interrupt. <S> If you use a high level language like C, you can tell the compiler to put the program in charge of handling the interrupt in the right place. <S> Here's how: http://www.nongnu.org/avr-libc/user-manual/group__avr__interrupts.html <A> Interrupt vectors are the addresses on the MCU from where the interrupt service routine's address would be loaded to Program counter on occurrence of a specified/intended event. <S> For example, the timer counter's overflow can cause an interrupt if its configured for that. <S> If there is an interrupt service routine (ISR) associated with that timer module, the same would be called just like any normal task switching (except few exemptions).
| Your understanding is partially correct,When interrupt occurs, the current execution status is backed up into the stack pointer and the program counter is loaded with the ISR's address which is basically stored in the interrupt vector address.
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What are the extra lines on this farfield plot? This is a plot from CST microwave studios of the farfield beam pattern of an antenna. I'm having trouble understanding why the plot isn't single-valued (2nd red circle) and also why there are blue lines. Any help would be much appreciated <Q> Here's how I see it: - I've extended the blue lines and they form an angle that is 75.8 degrees and coincide with the -3 dB points from a peak magnitude of 7.28 dBi. <S> I'm having trouble understanding why the plot isn't single-valued (2nd red circle) <S> I have no idea <S> which 2nd red circle you mean and why you expect something that is single valued. <A> What type of antenna was this? <S> My guess would be that the red line is the antenna pattern on one axis. <S> The green line is probably the pattern on a different axis, it looks a little too uniform for real world data but it depends on the antenna type. <S> I can't think of any other reason to have a circle at around -2dB. <S> The blue lines seem to be indicating the 3dB points of the antenna beam. <A> The angle between the light blue lines matches what is written at the right: "angluar width (3dB): 75.8 deg" I'm not sure what you mean by not beeing " <S> single valued" because actually for each angle there is exactly one intensity value. <S> Do you mean the fact that there is a second lobe? <S> That means that the antenna radiates/receives also backwards (i.e. opposite to its main direction) a little bit. <S> Its sensitivity is least at direction 100°, not 180°.
| Just from looking at the plot: the light blue lines could be the angle at which the level intersects the 3dB circle (which is not shown in the diagram but you can imagine).
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DC Buck Converter for more than 3A I want to use an Arduino Nano to controll an LED strip using an WS2812B IC. The strip takes 5V at 3.6A. I want to power the arduino and the strip using two 18650 Li-ion batteries (LGDBHE41865), which can deliver up to 20A per cell. These batteries are unprotected and I think I need a protection board, that allows enough current to flow. However, I was unable to find one. It would be nice if someone can point me in the right direction. In addition I need a buck converter to step down the voltage from the batteries (2*3.7=7.4V) to 5V. I found some, but all of them have 3A limit which isn't enough for me, so I need some suggestions there too. I'm new to this stuff, so maybe there's a simple solution I can't think of. It would be nice if someone can tell me how to do this right. Thanks. <Q> I found some, but all of them have 3A limit which isn't enough for me, so I need some suggestions there too <A> Use the TI Web Bench Designer. <S> It will give you like 50 circuit diagrams that meet your criteria. <S> http://www.ti.com/ Just enter your required input voltage range, output voltage and output current and hit go. <S> Its as easy as that. <S> The LM25085 comes up as a good choice. <S> You can select... 1) Input voltage ragnge. <S> 2) Output voltage. <S> 3) <S> Output current. <S> 4) Ambient operating temperature. <S> 5) <S> Prioritize between smaller design area, lower cost, or higher efficiency (or abalance between those). <S> 6) <S> Required efficiency 7) <S> Required maximum footprint 8) <S> Maximum allowed BOM cost 9) <S> Various other power supply features <S> The TI Web Bench designer will give you... 1) <S> A list of multiple auto-generated designs that meet your criteria. <S> 2) A complete schematic for each design. <S> 3) <S> A bill of materials. <S> 4) <S> 5) <S> An analysis of the whole circiut. <A> The simplest solution would be to buy something off the shelf and don't mess with making a custom PCB. <S> If you can tolerate the heat, try a linear regulator module. <S> You would generate 2.4V*3.6A = 7.2W worth of heat through it, a little more if you are also powering the Arduino. <S> If you don't like wasting power, I would suggest a switching regulator module. <S> Something like this step down module from Polulu would work. <S> Additionally would look in the drone / RC hobbyist sites like Hobbyking because drone motors require similar if not high current draws compared to your application.
| A couple of buck regulators that appears suitable: - Don't start quibbling on cost, if you want something cheap and possibly unreliable and noisy go ebay. For some designs you can actually have a PCB fabricated for the design and ordered through Digikey.
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Why are relays still used in electric ovens? I bought a new electric fan oven recently. It has a digital thermostat and control system. Yet much to my surprise, I can hear a relay clicking on and off inside it to control the power to its heating element. The oven is rated at 4kW (230V). I would have expected it to be using a triac to turn the power to the element on and off. So why not? I don't think that the answers here duplicate the question about using relays in automobiles. The design criteria for switching 230V AC are very different for 12V DC. To start with, LVDC would use a MOSFET whereas mains AC would use a Triac. Considerations concerning voltage drop across the semiconductor device and dissipating the waste heat are different. Safety regimes are different. The operational environment is different. And so on. <Q> Advantages of relays over triacs: <S> Very little voltage drop when on. <S> This means they don't dissipate much power. <S> For high power devices, the cost of dealing with the heat often outweighs the cost of the component that dissipates the heat. <S> Good isolation. <S> Making that isolation withstand normal power line voltages is pretty easy and cheap. <S> Able to withstand high temperatures better than semiconductors. <S> Silicon stops being a semiconductor at around 150 °C. <S> It's not too hard to make relays that can withstand substantially more. <S> Better input noise immunity. <S> Stray capacitive coupling even from nearby power spikes, RF pickup, and the like aren't going to trip a relay. <A> Adding to the points of Olin's answer: If you don't need the fast switching times of semiconductor devices, relays are pretty robust and cheap, compared with the circuitry needed to implement a solid state switch capable of switching the same amount of power. <A> Additionally, when a triac fails, it is often "stuck" in the conducting state. <S> It won't turn off anymore. <S> Might not be a good idea to have a semiconductor which, when damaged by (for example) a voltage or current spike, turns on your oven at full power while you are on vacation. <A> To be clear regarding the important point I think that Chue X might be making: a relay has excellent isolation between the line and load terminals while a triac does not. <S> For example, the BT136-600 datatsheet shows that this 4A triac a max leakage of 0.5 mA. <S> That's a triac that would be suitable for a regular wall dimmer. <S> Unless the dimmer includes a mechanical switch, then you would measure 120 VAC on the load side when the triac is off if there is no connected load. <S> If there is a load connected you will measure a much lower voltage which would be equal to the leakage current times the load resistance. <S> As a rule of thumb, you would expect that a higher power triac capable of 4 KW would have higher leakage current due to its much larger active area. <S> That would create a substantial shock hazard in the oven when the element is burned out or needs to be removed for service. <S> There would be 230 VAC with significant current capability exposed on the heating element connections. <S> Using a relay insures that the element is safely isolated from the line when the oven is off. <S> Regarding opto-isolated triacs: that is referring to isolation between the line/load connections and the control connections. <S> A good overview of optocouplers, including opto-isolated triacs is available here Optocoupler tutorial . <S> Opto-isolated triacs still have substantial leakage current and are frequently not suited for controlling certain loads. <S> This is also the sort of leakage that a relay provides between its coil and load, as mentioned by Olin's answer. <A> Adding to the point of Olin's answer, there is insulation between switched and the control sides. <S> Whereas a triac requires a small amount of current between the two circuits. <S> Ref: <S> Triac versus Relay
| That is needed to prevent dangerous voltages and currents flowing back through the control input signal to electronics that are controlling it. That can be quite useful when in a device that is intended to get hot. The relay coil is inherently electrically separated from the relay switch.
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How to protect valve heaters from inrush during cold starts? Some valves on the market today are rare and expensive. GEC Marconi KT66's retail for about £300 /pair today for example. Some others clearly show a bright flash as their heaters initially power up which is worrying when you see it for the first time. One concern I have is that the traditional constant current approach may not be sufficient to initially raise the temperature of the heater element sufficiently. I understand that some over current is necessary to begin the thermal resistance increase. Without over current, the heater will just sit there and may not get sufficiently hot. I'm not sure about this statement though. One modern aspect of this question is that I'm not trying to be totally authentic with all of the circuitry. Only the valves. I'm prepared to consider any modern form of protection in what you might call a hybrid design. An anachronistic design featuring relays or micro controllers would be fine as well as DC drive. What modern or traditional means can be used to reduce the initial current /thermal surge on a valve heater to extend it's life? NOTE. A single small valve might require 300mA @ 6.3V RMS. <Q> In order to limit the rushing current in the cold heaters + the current in the empty capacitors, use a 33 ohms inrush current limiter like a negative thermistor (NTC) just made for that purpose in the primary circuit of your power supply transformer. <S> Here is an interesting article about that. <S> When it’s cold (at start) <S> the NTC presents maximum impedance to the rushing current. <S> After few seconds, the NTC heats up and its impedance lowers down to a ohm or so. <A> Sure an NTC could provide some benefit if it was sized correctly. <S> Greg has stated this. <S> An old school approach is to use a separate filament transformer which has just one secondary winding of usually 6.3 VAC. <S> This transformer is rated for the job. <S> It is not much bigger than it has to be. <S> A grossley oversize transformer makes this cold inrush current worse. <S> On most valve systems the anodes and screens take much more power than the filaments. <S> This means that the traditional multi secondary transformer makes inrush worse. <S> What was sometimes done was making the leakage inductance greater to limit the cold current even more. <S> When you encounter such a high leakage filament transformer the Volts does depend more on the loading so you should use them within their spec. <S> Of course you would not overload it but if you wanted to run it at say 30% capacity you may find that the filament volts are too high. <S> The traditional approach here is to use a series resister to bring the filament volts into spec. <S> Some filament transformers had primary tappings to achieve this. <S> Even if you need a resistor you wont be wasting as much power as the LM317. <S> If this high leakage filament transformer is <S> unobtainium and the NTC can’t do enough then bypassing a primary current limiting component with a relay will work .Another <S> possible modern solution is to employ 2 back to back low on resistance mosfets and slowly ramp up the gate volts at power up. <A> Even AC heaters always coupled a little of the AC to the signal. <S> With DC, it's easy to apply PWM at startup. <S> It doesn't need to be anything fancy. <S> A open loop ramp is fine. <S> Just ramping the duty cycle linearly from 0 to 100% over a couple of seconds or so is good enough. <S> If the 6.3 V (or whatever your heaters use) DC supply is a switcher, then it is already essentially run from PWM. <S> That causes the supply to ramp up slowly. <S> If you are making your own supply with a spare microcontroller PWM output, then it's even easier. <S> Just ramp the maximum allowed duty cycle up slowly, regardless of what the control algorithm says it wants. <S> All these methods have the advantage of basically not being in the way during normal operation. <A> There is no requirement to run the heaters on 6.3 V AC, so firstly I'd suggest you convert to run them on DC.You <S> don't need to run them constant current either, though I have seen that done. <S> A simple voltage output DC supply with a long risetime (say 5 seconds or so) allows the heaters to come up to temperature smoothly. <S> You could use an LM317 with a circuit like this (from the datasheet ) which will ramp up from 1.25 V to the final voltage set by the R1, R2 divider. <S> R2 set to 910 + 56 Ohms (966 Ohms) <S> would give you 6.3 VDC instead of the 15 V shown.
| First, in a modern circuit it's better to run the heaters from a DC supply. Some integrated chips have a soft-start feature, or you can use heavy capacitive coupling from the output to the feedback input of a integrated IC.
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Which is give better result for find angular position of device using 9-axis IMU snesor.? I'm Using LSM9DSO 9 axis Sensor Which gives output of Accelerometer , gyroscope , Magnetometer Output data. i want to find the angular position of the device using this sensor. currently i'm trying to find an angle of device of x-axis but data is fluctuate so , i have refer some websites and documents which show that it is find using Accelerometer data also but i have little confuse about to which is better solution to find angular position of device. <Q> Generally you determine angle by using the accelerometer readings. <S> This only works is the unit is not accelerating at the time <S> but you can do a quick sanity check that sqrt(x*x <S> + y <S> *y <S> + z <S> *z <S> ) = 1 <S> g <S> +/- <S> some error margin. <S> If that is true then you probably aren't accelerating. <S> For any given axis the amount it is away from pointing straight up will be given by angle = <S> cos -1 ( <S> reading in g). <S> If you know how far each axis is from vertical then you can work out your orientation relative to the ground. <S> However anything calculated this way will drift away from the correct answer so you will need to correct it once you stop accelerating. <S> You can use the magnetometer as a sanity check and a method of tracking rotation <S> but it's not very reliable for calculating absolute heading without first being calibrated for your current environment. <S> All of the readings from an IMU will be very noisy unless you are using a very expensive one, expect to have to average and filter a lot to get smooth results out. <S> In this situation expensive means the price has at least 4 digits before the decimal point. <A> You need some kind of algorithm to fuse the acceleration, rotational rate and magnetic field readings. <S> There are several different ways to do this, from simple complementary HP/LP-filters to more advanced quaternion based non-linear Kalman-filters. <S> I have, with great success <S> , used Sebastian Madgwicks open source AHRS-algorithm with a LSM9DS1 and STM32F405 processor. <S> You can download a paper describing it and some source code from here: http://x-io.co.uk/open-source-imu-and-ahrs-algorithms/ <A> Thanks a lot to all for your greater Support. <S> I have found some interesting Reference website. <S> which give the best method to implement both to find meaningful data like, pitch & Roll. <S> Guide to using accelerometer Guide to using gyroscope <S> Hope <S> it's helpful to others <A> Consider an I + Q magnetic sensing set of coils. <S> Digitize the output of each coil; knowing the amplitude and the phase, against a reference phase, is an alternative sensor. <S> Use a single wire on the wall, to provide a spot source of Hfield.
| If you are accelerating then assuming you knew your orientation before hand you can calculate your current orientation by using the gyro readings to measure the change in orientation.
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What is the function of AVDD,AVSS and Vref of ADC in Microcontoller? I want to know the function of AVDD,AVSS and Vref in Analog to Digital converter in a Microcontroller. If AVDD and AVSS represents the range of input voltage ,what is the function of Vref? <Q> I'm going to assume based on the names you are talking about Atmel AVRs. <S> "AVDD" and "AVSS" are basically power supply and ground for the analogue portions of the circuitry. <S> These must be connected to "VDD" and "VSS", but they are brought out separately so that you can add a filter on to "AVDD" for noise reduction, and so that digital noise doesn't get coupled onto the analogue ground. <S> "AREF" is an analogue reference voltage which allows you to configure the upper range of the ADC. <S> When specified by register bits, AREF is used instead of AVDD to provide the ADC reference. <S> The ADC range will be between AGND and AREF allowing you to use the full dynamic range of the ADC even if the peak amplitude of your input signal is smaller than the supply voltage. <A> AVDD and AVSS are the supply voltages. <S> Most ICs cannot operate with voltages exceeding the supply rail voltages so that is how AVDD and AVSS determine the input voltage range. <S> Exceed this voltage range and the ESD protection diodes will start to conduct causing all kind of weird (if you are unfamiliar with this) behavior. <S> The value of the supply voltage is assumed to be inaccurate and is thus not always suited as a reference voltage for an ADC. <S> An ADC needs some kind of reference value because it outputs a number and that number is related to the value of the input. <S> If the input voltage is 1.00 V but the ADC has no idea "how much" 1.00 V is then it cannot output the correct number. <S> For that a reference voltage is needed. <S> The supply could be used but sometimes the supply voltage varies. <S> So then a different (more accurate and stable) reference voltage can be supplied via the Vref pin. <A> Usually the formula is something to the effect of: LSB (V) = Vref/(2^(ADC bits)-1) <A> VREF has the burden of supplying the fast surges of current, as the ADC plows through its binary-search approximation. <S> An ADC operating at 1Million conversions per second, with Cref of 10pF and with VREF of +5volts, needs F <S> * C * Vref average current; that is 1e6 <S> * 1e-11 <S> * 5, or 5e-5 amps or 50 microAmps. <S> Thus the source resistance of VREF is an error, appearing in FullScale accuracy. <S> If VREF is provided by an OpAmp, that OpAmp needs to recover from the sudden surge (100pS turnon) and charging demands of the 10pF cap. <S> To ease this burden, place a substantial capacitor right by the VREF pin, and ground the bottom of the cap to AGND pin. <S> Pay attention to currents flowing in the GND traces, because 100mA and 10 squares of trace (100mils by 10mils = 10 squares) is modeled as 100mA <S> * (10 <S> * 0.0005 ohms) = <S> 100mA * 5 milliOhms = 500 microVolts. <S> If your ADC is 16 bits, you likely have LSBs of 60 microvolts or smaller, and some error --- 500uV/60uV <S> == 9 LSBs --- will appear in FullScale or in ZeroScale performance of the ADC.
| In standalone ADC' Vref is used as a reference point to determine the relative (in voltage) size of an LSB.
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AC not rectified to DC properly So I am trying to make a temperature controlled DC fan using IC as voltage comparator. The IC I am using is LM741. This is the layout of my circuit: Now when I use a 9V battery as power supply, the circuit works fine. But I want to run this circuit using wall socket. So I used a step down transformer to get 12V AC and used this to rectify it: But the circuit does not work properly. The fan is always on and is not switched off by adjusting the trimpot. Can anyone tell me why this is happening?I checked the output of the rectifier circuit using DMM, I get around 15.6V DC. Edit: I am sorry I messed up while drawing the circuit... I did not actually connect the ground of the circuit the way I drew it.. This is how I connected the circuit: <Q> Your grounding is messed up. <S> You only show one of the transformer outputs, labeled "12 V 50 Hz AC", so the other is implied to be connected to ground. <S> However, the "+15V" node is then not 15 V with respect to ground. <S> You should consider the negative side of the full wave bridge to be ground, then NOT ground either transformer output. <S> This would have been more obvious if you had drawn the schematic properly. <S> Try to put high voltages at top, low voltages at bottom, and logical flow left to right. <S> Here is the circuit you really want, properly shown to be as little confusing as possible: <S> Your overall circuit could be better too: <S> Since the 741 opamp can't get all the way to the negative supply, use a voltage divider to drive the transistor base. <S> At 15 V, there is plenty of voltage available. <S> Arrange the resistors so that the transistor doesn't come on until the opamp output is at least 5 V or so above the negative supply. <S> The 741 opamp is not a good choice for comparator, but can be made to work in this situation. <S> As WhatRoughBeast mentioned in a comment, a little hysteresis would be good too. <S> You want to make sure the fan is driven either fully on or fully off, not in-between due to noise when the opamp inputs are nearly equal. <S> A high-value resistor from the opamp output to its positive input will provide some hysteresis. <S> This will have the effect of "snap action". <S> That keeps both the fan and the transistor from being run partly on, which could cause a lot of dissipation and damage them. <A> The only problem is your Vpp swing specs on 741 is <S> +-10Vmin for a +-15V supply rated at 1k load. <S> You need a swing that is <= 0.5V from V+ regardless of ripple with a 741 load of about 5% of fan current of est. <S> 150 mA (due to hFE reduced when saturated) or ~8mA. <S> Given you have a single supply then swing at high current on 15V <S> is reduced so that with NO Rbe pullup plus Vout cannot reach Vcc-0.5 so PNP CANNOT turn off. <S> Adding 1 to 2 diodes is optional to improve noise margin and thermal drift of Vbe. <S> Since this is an emitter follower Fan voltage also drops from 15 to 12. <S> adding caps to O.A. inputs, to reduce stray noise is desirable, even essential with long wire hookups. <S> choosing Diode bridge rectifier ought to be based on load <S> Req*C>=8/f for 10% ripple. <S> with f=120 or 100Hz and R depending on fan <S> \$Req=\frac{V^2_{cc}}{Watt}\ <S> \$ <S> e.g. 2.8W fan @12V simulate this circuit – <S> Schematic created using CircuitLab \$12^ <S> 2/2.8= <S> 51\Omega\$ <S> thus \$C4>8/{100Hz*51\Omega}=1500\mu F\$ <S> (nearest) <S> I drew battery for laziness but imagine diode bridge 12V transformer <S> A far simpler more efficient driver uses an NPN open collector with similar diode series or a comparator that drives to ground with Pullup R 10x load R. Suitable caps to reduce noise. <S> Hysteresis not necessary , in fact some negative feedback with 1M gives a smooth speed control over a narrow range. <A> There are many good reasons not to use the LM741: - Minimum recommended power supply rails are +/- <S> 10 volts (a definite problem) <S> Input voltage range is typically from -Vs + 2 volt to +Vs - 2 volt <S> Input offset voltage is typically 1 mV ( <S> 5 mV maximum) <S> Input offset current <S> is typically 20 nA (200 nA maximum) <S> Input bias current is typically 80 nA (500 nA maximum) Input resistance is typically 2 Mohm (300 kohm minimum) <S> Typical output voltage swing is -Vs + <S> 1 volt to +Vs - 1 volt Guaranteed output voltage swing is -Vs + 3 volt to +Vs - 3 volt Supply current is typically 1.7 mA (2.8 mA maximum) <S> Noise is 60 nV/sqrt(Hz) for LM348 (quad version of 741) <S> GBWP is 1 MHz with a slew rate of 0.5 V/us <S> The LM741A is slightly better but still a dinosaur in most areas. <S> So point 1 and 2 are being contravened due to low voltage supply of only 9 volts and pot wiper going down to 0 V and up to positive supply. <S> Regards your bridge <S> rectifier you really need to set this up properly - at the moment with the capacitor shown as it is, it is reverse polarized. <S> And of course, I hear you say " <S> but it works at 9 volts". <S> Well that's not the way to design things - you design to the data sheet and work within the restrictions they impose. <A> The only problem is your Vpp swing specs on 741 is 10Vmin for a 15V supply rated at 1k load. <S> You need a swing that is <= 0.5V from V+ regardless of ripple with a 741 load of about 5% of fan current of est. <S> 150 mA (due to hFE reduced when saturated) or ~8mA. simulate this circuit – Schematic created using CircuitLab
| Instead of a emitter follower, I'd use a low side NPN in common emitter switch configuration. Because the output cannot swing down to 0V (guaranteed to swing to 0 volts plus 3 volts) your fan's top speed is severely limited (may not be a problem of course).
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Can an ADC be damaged by applying a higher voltage than VREF but that is still within specs? Take the following example:A SAR ADC that is rated for 5v and 3v3 operation is operated at 3v3 (for both VDD and VREF). Someone is a bit careless and applies 5v to one of the inputs. Can this permanently damage the ADC in any way? A slightly detailed explanation for why or why not would be appreciated. <Q> That would be written in the datasheet. <S> If the absolute max rating that applies for the input pins is given as "5.5V", you're fine. <S> If it is given as something like "VDD + 0.5V", you're not fine. <S> Every bit of information and the way it is written makes sense in a datasheet. <S> Always. <S> 1 <S> 1. <S> Unless there is a mistake in it. <S> Which happens, too... <S> And when it does, that sucks. <A> Logic ICs, even mixed-signal parts like an ADC, typically have undershoot and overshoot protection diodes that clamp input pins voltages to the positive and ground/negative supply rails. <S> Inclusion of these diodes is usually stated in the data sheet, sometimes obliquely by stating 'Absolute Maximum' input pin voltages as 'Vdd+0.6V' and 'GND-0.6V' or such like. <S> The current capability of these diodes is often not stated but, as a guide, clamping diodes in 74HC/74HCT families could handle 20 mA continuous. <S> It is advisable to include a series input resistor before your ADC input pin that will limit the overshoot/undershoot input current to 1 mA or less. <A> Very related to this are overshoots and undershoots during operation of the ADC; the injected charge, as the ESD diodes turn on, has to find a RETURN path. <S> If that moving charge upsets the charge on the sample-hold capacitors, or upsets the comparator decisions, your 10 bit ADC will be flakey. <S> There is a famous IC design house in Plano, I believe, that placed 16-bit ADC on chip with a hard-working MCU; the ADC was only good for 8-bits unless the MCU was put to sleep; this is what I was told. <S> You CAN integrate noisy digital with quiet analog, on the same silicon die, but you must provide "quiet time", often 100 nanoseconds long, or accept poor analog decision-making. <S> There is a famous chip company in Dallas that acquired a famous DAC/ADC company; the Dallas company integrated some of the newly owned IP --- a 24bit ADC --- with a 32MHz MCU; I did not believe the 24-bit ADC performance would remain...24 bits; careful reading of the datasheet reveals the truth: the MCU must be throttled down, to 8MHz or 125nanoSecond period, to achieve the ADC's 24 bits. <S> If you examine the datasheets for the LLTC 24 bit (standalone) ADCs, you may notice the SPI pins (3 of them) are given special isolation, not directly connected to any computer bus. <S> Are these caveats -------permanently damaging? <S> Do we call it "damage" when the 16-bit ADC only provides 8-bit performance?
| In a 16 or 20 or 24 bit ADC, the mere presence of a few hundred milliVoltPP on SPI pins, even with logic levels being "0" or "1", will inject charge and upset the performance.
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Right kind of resistance wire for near-skin heater? I'll be up-front and say I know jack squat about electrical engineering aside from "conductors heat up when you run current through them." I'm looking in to building a low-voltage (~12-24 volts) personal heater that I can attach to the inside of a jacket; basically a small battery-powered electric blanket with no control system other than just unplugging the battery when it gets too hot and plugging it back in when it gets too cold. I'm not sure how possible this is to do safely though. My main question is, what should I use for the heating coil? I was hoping to have a long wire that zig-zags over most of the back of the inside of the jacket. Kanthal wire is really cheap right now 'cos of all the people wanting to build vaporizers; is that an appropriate material? If so, what specific mixture and gauge should I be going for? Looking at the specifics it seems like it's meant for extremely high temperatures, whereas I want something that stays around 30-40 degrees Celsius on low voltage. I can't seem to find anything that says it's designed for that kind of application, though. Am I barking up entirely the wrong tree? Is this a practical idea at all ? <Q> Your desired temperature is low, but the wire is small <S> so will get significantly hotter than the overall temperature increase. <S> However, the real problem is the total energy required and the size battery that will take. <S> Maintaining let's say 10°C additional temperature rise over the surface area of a jacket will take some power. <S> You'll have to measure that. <S> My guess is it would take 10s of W. <S> You'll probably wrap the wire in a zig-zag with the spacing being much larger than the wire's diameter. <S> This means the wire itself will be significantly hotter than the average temperature rise. <S> That could be a problem if it directly touched the skin, or it could melt, otherwise damage, or even catch the insulation on fire. <S> Yet another issue is that you don't want the inside of the jacket to be 40°C. <S> That's above body temperature. <S> That will get uncomfortable fast and cause sweating, which is really bad in a winter situation. <S> Go get a better jacket instead. <A> Totally boring but https://www.makita.co.nz/products/model/DCJ200 would solve the problem. <S> It will heat for a couple of hours off an 18v rechargeable pack. <A> It probably is feasible. <S> But: you need to realize that even for electricity, conservation of energy applies! <S> So if you want to produce 10,000 Ws (wattseconds) = 10 kJ of heat, you need to have that much energy in your battery. <S> So, in your place, I'd start by calculating what amount of heat you want to produce. <S> That's really easy. <S> Calculate the amount of material (ie. <S> flesh) <S> you want to heat up, multiply that by the specific heat of the material (use the specific heat of water, flesh is mostly that stuff), and then you know how many Joule you need. <S> One Joule is 1 W · 1 s. <S> Then divide that by the time (in seconds) you want to allow the heating to heat up that much material, and you've calculated the power (in Watt) necessary to heat up that amount of material that fast. <S> Now, I don't believe you know nothing. <S> You've probably heard of 1 W = 1 <S> A · 1 V. <S> So if you have a 12 V battery, in order to produce 1 W, needs to make \$\frac1{12}\text{ A}\$ current flow. <S> If you need 120 W, you'll need 10 A. <S> It's that simple. <S> Now that you know how much current you need to spend, you can calculate two things: <S> how long your warmer should last with a single charge. <S> Most batteries are rated in "Ampere hours", Ah, so take one battery that you think would be of acceptable weight and size, and figure out how long it will be able to source that current. <S> how much resistance your wire needs to have, because, Ohm's law, U = R · I -> <S> R = U / <S> I (the resistance is the ratio between voltage U=12V and the current I). <S> Heater wire has a fixed resistance per length. <S> You can thus calculate the length of wire necessary. <S> This doesn't address things like maximum power impedance matching etc, but it does in theory give you a good idea of what your system needs to look like.
| Nichrome is probably a good choice. Another problem will be the temperature of the wire.
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Is an external oscillator for MAX2769 GPS receiver needed? I am designing an RF frontend for GPS using MAX2769 chip. However, due to lack of theoretical and practical background in signal processing, I am unsure how to interpret the documentation when it comes to the usage of an external oscillator. The docs say in the Crystal Oscillator section: The MAX2769 includes an on-chip crystal oscillator. A parallel mode crystal is required when the crystal oscillator is being used. If an internal ostcillator is included, why would I need an external one? Can I not use the internal crystal oscillator (to avoid needing an external one) but still achieve the desired functionality or receiving I and Q data? Do I need an external oscillator to keep the internal one in check because of its lower quality (PLL)? Does XTAL (pin name) stand for external? <Q> Maxims 2769 does not include a crystal. <S> Load capacitance for an external crystal can be tuned with configuration registers to slightly pull it to counteract aging. <S> An active external oscillator can also be used, but pulling is not supported in this setup. <S> The wording in the datasheet is somewhat sloppy about this. <S> If the unit would include a crystal, you can be sure MAXIM would advertise the fact more clearly and also give performance characteristics for it. <S> Choose a (low-end) <S> TCXO for GPS <S> , a cheap uC crystal will boost phase noise (might work, but will degrade link budget). <A> Can I not use the internal crystal oscillator (to avoid needing an external one) but still achieve the desired functionality or receiving I and Q data? <S> You will need some thing that oscillates to generate a baseband clock. <S> So, no, you won't get around using an external oscillator; the fact that you can use a relatively cheap crystal oscillator instead of an active oscillator is a feature. <S> If an internal ostcillator is included, why would I need an external one? <S> Because that's how the device was designed. <S> The internal oscillator is used to excite the external one, which in generates a stable clock. <S> Do I need an external oscillator to keep the internal one in check because of its lower quality (PLL)? <S> well, the PLL links the different frequencies, but there's more than one PLL in the chip, so I can't answer this question. <S> Does XTAL <S> (pin name) stand for external? <S> it stands for crystal. <S> See the pin table on page 10 of the datasheet. <S> Seriously, this is a relatively complex chip, and making a GPS receiver might really not be the project of choice for a signal processing and RF beginner. <S> Maybe you'd want to start with understanding how the synthesizer inside the device works if looked at in isolation. <A> There is no internal XTAL or Ref oscillator only external. <S> this is necessary to locate the accurate GPS carrier frequency. <S> The internal VCO has high f error and phase noise and only reduced when using the external XTAL ref. <S> It is used to synthesize other frequencies as part of the PLL <S> and it's <S> output is as stable as the XTAL and then very stable when locked to the Rx signal. <S> The external XTAL must be parallel type with required load caps (1 big and 1 small one twice the load in pF from Xtal datasheet ) <S> then the 5 bit XTALCAP register can tune the XTAL as a VCXO with internal capacitance and presumably a coupled reversed diode acting as a series Varicap from a DC bias using a 5 bit DAC which then is used to fine tune the external ref. <S> freq. <S> This Xtal goes to gnd rather than feedback but internal feedback provides the gain to oscillate shown as the sine oscillator symbol in block diagram.
| It does include an oscillator that is able to excite an external crystal.
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What does this BJT-Transistor do? We discussed BJT-transistors in the lecture and this circuit came up, the professor said that this circuit would act as some sort of amplifier for the transimitted signal between the two transistor but I dont understand how it works. Later I discussed it in a training group and was told that the signal would bounce between the two transistors because of the capacitor, but this confused me even more. <Q> Given your instructor's remark, I suspect that you were actually told about "debouncing", which a very useful function. <S> However, the circuit shown is a bad, bad example of how to do it. <S> Let's think about a switch making contact. <S> Inside, you have two pieces of metal making contact, and on a timescale of milliseconds the two pieces can actually bounce, perhaps repeatedly, before settling down. <S> If this happens the switch output will apparently show multiple activations where only one was intended. <S> To combat this, a circuit like this can be used: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> You've noticed that it has the same general outline as your circuit, but a few more resistors. <S> Let's say that the switch is open, and gets briefly closed, then open, then closed for good (a single bounce). <S> Current through R1 turns on Q1 and pulls the collector low. <S> This also discharges C1. <S> As a result Q2 is turned off and Vout goes high. <S> When the switch bounces, Q1 is briefly turned off and its collector rises. <S> However, the resistor charges up much more slowly than it discharges (for appropriate resistor values of R2 and R3), so Q2 never turns back on. <S> As a result, the output Vout has been "debounced", and R2/R3/C1 can be selected for any desired bounce time to be ignored. <S> The original circuit is not very good, since the capacitor voltage swing is quite small, due to the clamping effect of the Q2 base-emitter junction. <A> (comments to answer) <S> The workings of the "circuit" The voltage at the collector of the first BJT will be clamped at 0.7V (approx) by the base-emitter junction of the second BJT. <S> BJT(2) is most likely in saturation (so only a small DC output say 0.2V). <S> BJT(1) has no bias arrangement so it impossible to tell what Ue will actually do to BJT(1). <S> (Is Ue a DC or AC or mixed signal? <S> - no one knows. ). <S> Its a terrible example to give to a class who are just learning to come to grips with the basics. <S> I'm not surprised you were confused. <S> What does the capacitor achieve in this circuit ? <S> Very little <S> and I can't understand why anyone would put it across BJT(1). <S> As the voltage across it is clamped (by BJT(2)) <S> If Ue is taken to +0.7V then BJT(1) would be in saturation (Vc-e <S> ) = 0.2V <S> causing the voltage across the capacitor to fall (discharging rapidly through BJT(1)). <S> In this case Ua2 goes high because BJT(2) is turned off. <S> If Ue is taken low (0V) <S> the voltage across the capacitor rises to 0.7V with the time constant = RL*Cp and the Ua2 falls to Vsat (about 0.2V). <A> This circuit converts a current at its input to a voltage at its output. <S> It is typically used for digital signals, and with a base resistor at the first transistor. <S> (In that case, if the base resistor is larger than R L , the output impedance is lower than the input impedance, i.e., the circuit amplifies the current.) <S> U e and U <S> a,2 are not actually interesting, because both transistors are either off or saturated. <S> The capacitor, if it has been put there deliberately and does not just represent a parasitic capacitance, slows down switching on the second transistor. <S> This might be useful if you want to balance the switching times (a saturated transistor is slower switching off than switching on).For example, such a capacitor is used in several Roland MIDI devices to ensure that the falling and raising edges of the MIDI signal are delayed by approximately the same amount:
| all it can do is act as a 'smoothing' capacitor rather than a coupling capacitor (collector to base).
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Pulling analog DC output from a laptop I am using a laser source that can accept analog input (0-5 V) to set light intensity levels. The input impedance of this driving voltage is 50 Ohms. Typically this is connected to a function generator to create sinusoidal modulations of the light level. I don't care about AC light signals; I just want to set a DC light signal by providing a level somewhere between 0 (off) and 5 Volts , and I want to be able to do this remotely. I could use an Arduino or Labjack to drive this signal, but I was wondering if there was a component-free solution that could just use the laptop, analog circuit elements, and some programming. A couple of ideas I daydreamed about were: Program the speakers to play a tone that I could connect to the laser source via the headphone jack. (Soft tone would be faint light, loud would be bright). This would seem to involve a bit of work as the headphone jack voltage is too low (~0.3 V) and the output impedance is also about 10 Ohms. Program the RS232 port in conjunction with a voltage divider. This would involve a bit of work as I don't know how easy it is to program pins directly. Program the USB output power. After googling around for a bit, this seems tricky to do with Windows. If anyone has a good idea how to generate an analog 0-5V output from a computer, or could comment on the ideas above, please advise. <Q> The 0..5V control voltage needs to drive 50 ohms, meaning that it needs to be capable of supplying <S> 0.1 A. A PC can provide that current from the USB connector but not from the serial port. <S> However, the serial port is easier to programme with than the USB port, or certainly has a shorter learning curve. <S> You could use an USB-to-LVTTL-serial-port adaptor. <S> This plugs into a PC USB port at one end and gives the 5V supply and the RS232C TxD/RxD signals at LVTTL levels (3.3V). <S> You can get these as a module (look at FTDI) or built into a cable, the latter being less than a fiver. <S> You can then attach a microcontroller or other IC to receive your RS232-signalled bytes and convert them into an 8-bit parallel byte. <S> This can go to a DAC (may be internal to your microcontroller) and the DAC analogue voltage can go through a power op-amp buffer for the 0.1 A drive strength. <S> This is one way of doing it and hardly meets your requirement of a few analogue parts. <S> But it does let you make something PC-powered and controlled. <S> You may find microcontroller demo' boards on the Internet that contain some or all of this circuitry for under a ton - have a search around. <S> If you're not microcontroller-minded, this is all somewhat moot. <A> They are CTS (Clear to Send) and DTR (Data terminal Ready) . <S> There was an answer on how to tweak these signals here . <S> You could also use VB, Python serial or even Powershell scripts to tweak them. <S> Depending on the chips used to generate the RS232 signals these will swing between +/-3 to <S> +/-12 <S> V and typically have series resistor of 300 Ohms. <S> If you set both high you may get a substantial enough signal level for your laser. <S> You'd get 25% of the open circuit high voltage <S> so might get as much as 3 V across 50 Ohms. <S> You'd have to be careful and not set both DTR and CTS to the negative voltage unless you are quite sure that your Laser input could withstand this. <S> If you SET DTR high and CTS low you should have about zero volt output with both outputs connected to your Laser, so you only really need to toggle one signal to get a simple Laser on/off signal. <S> One other thought is that you could use TX data, a low pass filter and a serial diode. <S> If you configure the serial comm port to send 8bit, NP then you could send a continuous stream of characters at say <S> 115k Baud and make yourself a very crude DAC. <S> If you set one of DTR or CTS high then the output would be close to zero. <S> Then characters sent via Tx would give a positive input to the Laser. <S> For example here is the datasheet for the Max32xx <S> Rx/Tx. <S> The device output drivers can supply: <S> The output voltage is likely to be somewhat non-linear (some sort of active shutdown) <S> but we know we can supply 35 mA into a short and that the output resistance is 300 Ohms. <S> From that we can deduce that the driving voltage must be approximately 10.5 V. <S> From that I might expect to be able to drive the 50 Ohm load to about 1.5 V with a single driver. <S> So for the CTS/DTR solution you would expect around 3 V into 50 Ohm. <S> For the Tx solution 1.5 V and you could use CTS/DTR to scale to 3 V or higher. <A> You would need a cable of some sort to connect ANY type of solution to the laptop so consider starting with a cable of the type suggested in the answer by @TonyM. <S> Such cable could be the TTL-232R-5V-WE available from the likes of Mouser. <S> This cable is a USB to serial port converter with 0-5V swing at the output signals. <S> It can also source current from the the laptop USB port out a 5V pin. <S> My proposal is a bit simpler that that posed by the other answers. <S> The idea is to use the serial port TxD line to provide a poor mans PWM to the end of the cable. <S> You can send appropriate characters out the serial port such that the average output if mostly low, mid-range or mostly high. <S> For example for mid range send 'U' characters (0x55), for lows send out NULs (0x00) and for highs try DELs (0x7F). <S> The TxD waveform can then be fed through a series diode into a low pass filter. <S> The filter would have its own discharge resistor to establish the low level bias of the filter output. <S> The baud rate at which you stuff data out to the TxD pin would want to be high with respect to the time constants of the low pass filter. <S> This filtered signal would then be used to drive the input of an op-amp driver component that is powered from the 5V line and capable of driving the input impedance of the laser device. <S> A rail to rail output op-amp is the best choice here.
| If your laptop has a serial port then you have two digital output pins that may help you ...
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AM Radio gets static when driving near PV arrays, but not FM My school has PV arrays that act as shades in the parking lot. Whenever I'm within 10ft of them, I get overpowering, humming AM static, but the FM signal is clear as day. Why? <Q> It doesn't matter what you are driving next to. <S> AM is more susceptible to noise. <S> An AM transmitter is basically a filter with gain. <S> If there is noise on that channel you are going to hear it. <S> FM has a carrier frequency, when you tune your radio with FM the radio in it <S> 'locks on' to that frequency <S> an only information around the carrier gets sent to your speaker. <S> This is not as susceptible to noise. <S> If you drive next to power lines you get more noise on AM when you are close to the poles <S> (power lines don't have as much regulations as consumer devices as far as emissions go) <S> but FM comes in clear. <A> EMI - Electromagnetic Interference. <S> The panels are almost certainly feeding into an inverter/battery-charger, which will likely be drawing current in sharp-edged PWM waveforms, because that's how inverters/battery-chargers work - glorified switch-mode power supplies. <S> The frequency & rise/fall-time & mark/space-ratio of the current waveform caused by the inverter/battery-charger will be creating EMI at certain frequencies; actually a series of frequencies in certain integer multiples of the base frequency of the PWM frequency. <S> Whether those frequencies correspond to certain stations on the AM band, or the FM band, or any of the hundreds of MHz between and around them, depends on the specifics of the frequency & rise/fall-time & mark/space ratio of the PWM waveform. <A> There might be some EMI being generated by inverter circuits associated with the solar panels, but the main reason you lose the AM signals is the same reason you lose them under bridges and in parking garages — AM signals simply do not penetrate into/under metallic structures that have apertures significantly smaller than the signal's wavelength. <S> So what you are experiencing is probably a combination of severe signal loss accompanied by a rise in EMI. <S> If the solar panels are grid-tied (possibly using microinverters), it makes sense that the EMI would be amplitude modulated at line frequency. <S> FM signals don't experience the same kind of attenuation in the first place, and the EMI at FM frequencies is probably a lot weaker.
| PV arrays are probably noisy because of DC to DC inverters and poor EMI control and which could also be harmonics of them converting to AC which would be lower frequencies around AM also.
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MOSFET doesn't put out desired voltage I have an IRFB7430 which I want to use to switch 12V. There will be minimal load on the MOSFET, since it is only used to switch a signal. I know it is not its intended use, but it's what I have lying around. The signal on the Gate is only 5V, which should leave the MOSFET with a higher-than-intended R_DSon, but since I only want to switch a voltage, not a current, I thought it might not be a problem. However, the output is only around 3.6V, and even if I switch with a 12V signal (which should get the R_DSon to single digit mOhm), I only get 10.5V on the Drain. I have a 26kOhm pull down resistor on the Gate, and a current limiting resistor of 220 Ohm. Removing either of these did not bring significant changes. +12V is connected to Drain and I am measuring between ground and Source. I guess there is something in the design of the MOSFET which I don't understand and it is its regular behavior. What am I missing? <Q> Are you by any chance trying to use it as a high-side switch? <S> I believe so - in which case you're actually using it as a source-follower instead of a switch. <S> This is confirmed by your measurements, where Vs = Vg-1.5V approx. <S> It's not switching, but operating in linear mode and dissipating a lot of power, supplying a fairly accurate Vg-1.5V to the load. <S> You have 3 options: (1) generate a gate voltage ABOVE the supply (18 to 20V) to switch ON, and 0V to switch off. <S> (2) use it on the low side of the load with 12V gate voltage. <S> This is the simplest - a conventional low side switch - and offers the best performance. <S> (3) replace it with a PMOS FET as a high side switch. <S> Then 12V turns it off and 0V turns it on. <S> But it might be difficult to find a PMOS with such a low ON resistance. <A> +12V is connected to Drain and <S> I am measuring between ground and Source <S> To turn on a regular N channel MOSFET, the gate voltage has to be higher than the source voltage. <S> It will still usually turn on when the source is at 1 or 2 volts but is becoming weaker in that the on-resistance is beginning to rise. <S> With 3 or 4 volts on the source, the MOSFET is hardly turned on at all. <S> This is what I recommend for a 0 volt tied load: - <S> See also this SE.EE Q & A <A> If all you want to do is amplify the voltage for a signal from 5 V to 12 V rather than switch power to something on and off <S> then you can do this: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> This has a couple of big drawbacks over Andy aka's answer. <S> It will invert the sense of the signal, a low input will give a high output. <S> With a high output you have a series resistor in line with the signal, depending on what you need to drive this could be a big issue. <S> The lower R1 the less of an issue this is. <S> When a low output you are dumping power through R1/M1. <S> This is both a waste of power and depending on how 'on' M1 is could cause a lot of power loss and heat in M1. <S> The larger R1 is the less of an issue this is. <S> When switching from a low to high output there will be a ramp up time based on R1 and the output capacitance. <S> This may or may not be an issue depending on the load and value of R1. <S> You can avoid the inverting issue by putting two of these stages in a row. <S> In that situation R1 for the first stage can be nice and large (say 10k) since there is no significant load. <S> The resistor in the final stage and how practical this approach is depends a lot on the load being driven. <S> Calculate a maximum for R1 based on the expected current draw and maximum acceptable voltage drop at the load and then check if that maximum is too low in terms of acceptable efficiency and M1 power dissipation. <S> If you look at Andy aka's answer you will notice that it is using this exact circuit to generate the 12V signal needed to drive the gate of the p-channel part.
| So, if you put 5 volts on the gate, the MOSFET will turn on when the source is at 0 volts for sure.
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What's the purpose of an array of same VALUE capacitors? I'm looking at the schematic of the AM3354 EVM and I don't know why I see so many of the same value capacitors used in parallel and typically around the processor or memory. I understand that different capacitors will have different self resonant frequencies and perhaps that is advantageous for noise filtering but why so many of the same value? Isn't one sufficient? This question is based around the same value of components not simply combining different value capacitors into one larger capacitance part. TI AM3354 EVM Base board schematic <Q> If you count the power pins on the IC, and the supply decoupling caps of the same value, you'll probably find a 1:1 correspondence. <S> This indicates you are expected to place one per power pin, as close as practical to the pin, to minimise the effects of trace inductance. <A> They are usually just multiple decoupling capacitors on the various pins of a IC that are cumbersome to show in their physical location to each pin on a schematic. <A> This is not typical, but... I once had to change a design to use multiple smaller value caps because the mechanical team made the fit so tight that the larger valued caps we could get were simply too tall. <S> We could get shorter caps, but not in time for our deadline, so we punted and used smaller valued, shorter caps and got the prototype ready for its demo. <S> As far as I know, the smaller caps were retained into production. <A> Because large caps also mean large internal resistance and large inductance. <S> By putting small caps in parallel, you can still get the same capacitance, at the same time, smaller total ESR compared to a single big cap. <S> Possible duplicate here: <S> Why too many capacitors in parallel for Vdd supply net? <S> Can't we just add all to replace with one big capacitor? <A> Other answers have covered the one-decoupling-cap-per-power-supply-pin point. <S> You'll also notice that one of the caps is a 10uF. <S> Typically this would be an electrolytic cap. <S> Electrolytics are good at storing a large amount of charge (they have high capacitance for their physical size) but typically also have a high parasitic impedance which slows down their provision of that stored charge to the device. <S> The latter handles larger changes in voltage slowly, and the former handles smaller changes in voltage quickly. <S> It's a fairly intuitive principle, and one which in practise works very well.
| When dealing with devices drawing significant current and/or at the end of long power supply tracks, it's common to use a small high-speed capacitor in parallel with an electrolytic.
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Are silicon microbolometers inherently more expensive than conventional CMOS light sensors? We're finally starting to see practical thermal imaging sensors ( microbolometers ) entering the consumer market. However, they are still vastly more expensive than comparable visible imaging sensors. The most basic 384x288 17µm pixel (i.e., 32mm 2 ) thermal imagers run about $500, whereas $500 will get a 6000x4000 2µm pixel (i.e., 96mm 2 ) CMOS sensor ... plus 5-axis sensor stabilization and more. My question: Assuming the same economies of scale were applied as are already used for conventional CMOS sensors, is the manufacturing process for silicon microbolometers inherently more expensive? Or in the limit is it still just some (similar) number of photolithographic steps? To elaborate: Thermal cameras look for radiation with wavelengths between 7-14µm, whereas visible light is in the range 0.4-0.7µm. Based on the physics alone, at the diffraction limit microbolometer pixels will have an order of magnitude greater surface area. Apparently commercial sensors are at the diffraction limit for both visible light (at 1 micron pixels) and thermal light (at 17 micron pixels). So, to make it fair, we might compare a 1" 24Mpx visible sensor with a 1" 300kpx thermal sensor. Both sensors can be made from silicon using a CMOS process. The structure of microbolometers looks a little trickier than state-of-the-art visible spectrum CMOS sensors, requiring a thermal bridge for each pixel as well as vacuum encapsulation of the sensor. But I know little of large-scale manufacturing processes, so are these variables significant in the limit on a per-unit basis? <Q> In order to suspend the sensor pixels on thermal bridges the sensor layer has to be put on a substrate that is subsequently etched out from underneath the ~17-micron panels. <S> Many procedures that are commonly used on CMOS to do something quick (such as fairly harsh cleaning steps, CMP, etc.) have to be replaced by more complex, slow alternatives, and even those have a significant enough failure rate that more chips have to be scrapped. <S> For example, immersing the wafers into liquids or any process that involves a flow of some medium over them has to be done extremely carefully and slowly. <S> Surface tension is a huge problem when it comes time to get the liquid etching reagent out from under those pixels without breaking them: You can't blow it off with pressure or boil it off with heat. <S> To give an idea of how sensitive they are: Here is an electron micrograph of an array that was hit with compressed air to remove dust: <S> The 3-D fabrication problems with microbolometers are comparable to the ones involved in making DLP chips, which remained relatively expensive even in large-scale production for consumer devices. <S> (Microbolometers are made that use only amorphous silicon, but for performance a vanadium oxide sensor is preferred. <S> Adding VO necessitates a separate and more expensive fabrication line because it is a hazardous substance.) <A> Yes is the answer to the question, as it was asked. <S> Your "problem" (not your question) is: "How can I see LWIR (~ 7-14µm)" utilizing a Technology (in the Future) that would permit lower cost Sensors if "the same economies of scale were applied as are already used for conventional CMOS sensors.". <S> This image was produced using an uncooled ("hot") <S> T2SL MWIR (3-5 μm) Detector. <S> It has better contrast than a high quality LWIR Image. <S> Using SWIR permits seeing through common glass (and using conventional lenses) but only very hot objects (Engines, Fire, etc.) <S> are readily discernable without reflected light which is required to see anything that isn't very hot. <S> Using LWIR is better for exact measurement of temperature but requires expensive Optics and unless you're using Microbolometers you'll need cooling. <S> MWIR cameras are employed when the primary goal is to obtain high-quality images rather than focusing on temperature measurements and mobility. <S> The MWIR band of the spectrum is the region where the thermal contrast is higher due to blackbody physics; while in the LWIR band there is quite more radiation emitted from terrestrial objects compared to the MWIR band, the amount of radiation varies less with temperature (see Planck’s curves): this is why MWIR images generally provide better contrast than LWIR. <S> For example, the emissive peak of hot engines and exhaust gasses occurs in the MWIR band, so these cameras are especially sensitive to vehicles and aircraft. <S> Instead of confining your choice to microbolometers you need to look at QWIP , Type-II Strained Lattice ( T2SL ), or even Cooled LWIR all of which are more 'similar' to CMOS than a microbolometer; and thus have a better future potential for scaling (assuming enough interest in seeing LWIR radiation). <S> More Info about alternatives to Microbolometers: <S> http://www.ircameras.com/articles/infrared-imaging-new-ir-detector-materials-challenge-existing-technologies/ <S> and http://www.laserfocusworld.com/articles/print/volume-51/issue-07/feature/photonics-products-mwir-and-lwir-detectors-qwips-capture-lwir-images-at-low-cost.html . <A> The answer is in economies of scale. <S> CMOS <S> imaging sensors get produced much more than infrared sensors . <S> Therfore CMOS imaging sensors get more R&D funding, <S> more companies producing them and a bigger supply chain ect. <S> Infared sensors are starting to get cheaper and adopted more widely. <S> MEMS are not inherently more difficult to <S> manufacture and sometimes use the same processes, and can even use older cheaper lithography equipment because the size is bigger.
| Microbolometers are inherently more expensive than other silicon ICs specifically because of their required 3-dimensional structure.
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Why are VCC and GND on diagonally opposite pins? I noticed that most of ICs (DIP or SOP) have VCC and GND pins on diagonally opposite pins. Decoupling capacitors must be connected with shortest traces. Placing power pins in the corners makes the traces (inside the chip and outside on PCB) longest possible. It doesn't make sense to me, what's the purpose of doing it this way? <Q> I believe the answer to your question is historical rather than logical. <S> I am old enough to remember when even double sided circuit boards were new. <S> Before this, using a DIP packaged IC (which was what there was then) on a circuit board was made easier by having VCC and GND on opposite ends of the package. <S> This meant that the power to supply the device could come from the top and bottom 'rail' on the PCB and still leave the logic circuits some space to route to each other with the minimum of bridges. <S> Decoupling capacitors were generally placed beside logic chips, with the legs of the capacitor left long to ensure minimal hole-drilling. <S> Remember then that there were no CAD packages to help design your layout and test it either. <S> It was all done on paper, or in my case because I was lazy, usually a masking pen directly on the board with some meths for stuff-ups. <S> You could also use the meths for cleaning the circuit board ;-) <A> I agree with robert. <S> It made routing of power easy in early boards. <S> Such as this one: <S> You can clearly see the power and ground rails above and below the chips. <S> And the decoupling caps can just use the ground of the row above to keep the distance short. <S> Nice and simple to route. <S> It is so much harder these days with all the seemingly random placement of power and ground pins on chips. <A> Decoupling capacitors are meant to reduce variations in voltage on the power line caused by changes in current draw, as well as reduce high frequency noise introduced elsewhere in the system. <S> Just because the GND pins are far away on the IC, does not mean that your decoupling caps need to go between those pins - place them close to the VCC pin, and route a GND line close by (or place vias to a GND plane). <S> The reason VCC and GND pins may be far away, as @PlasmaHH put it, is because it makes sense for certain ICs due to the die layout. <S> Consider a circuit: power -> <S> a bunch of stuff -> GND; it makes sense to have those pins be far away to accommodate for internal circuitry. <S> Additionally, it may reduce the amount of capacitance created (externally and internally) between those two pins. <A> Nearly all modern PCB's have at least one ground plane. <S> In other words, unoccupied space on both sides of the PCB is filled in with a big "ground wire". <S> When the engineer lays out the PCB, he places each decoupling cap as close as possible to its VCC pin. <S> He connects one of the capacitor's pads to the VCC pin with a short trace. <S> He connects the other pad to a via into the GND plane. <S> Because the ground plane is so wide, its impedance is very low. <A> Having VDD and VSS pins separated by a significant distance can only be afforded in not so critical applications. <S> In such cases it could be more important to match established pin layouts instead of using the optimal placement. <S> An integrated circuit circuit is enclosed by the so called pad-ring. <S> I took an arbitrary image from Google to show this: Among the yellow traces that can be seen in the corners <S> are the supply lines. <S> They run around the whole chip because they are needed for the ESD protection circuitry and to supply digital I/O pads. <S> It should be clear from this that the ring could be supplied from any two pads in the ring (assuming there aren't multiple supplies). <S> From the IC designer's point of view critical signal lines and supply pads are placed in a way to minimize the inductance of the bondwires and the leadframe. <S> In particular VDD and VSS would be placed close together and sometimes even with double-bonding to reduce parasitics. <S> So for an optimum placement VDD and VSS would not be placed on opposite corners.
| With a very short trace between the VCC pin and the cap, the inductive parasitic of the wire is minimized.
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Should input power be greater than output power in a buck converter? For example, in order to provide for 10W (5V*2A), the input must be greater than 10W? I asked this because in datasheets, only input voltage is mentioned. There is no mention of the input current. I assumed, as the voltage decreases, current increases. Of course, the power will be the same. 12V*1A = 12W. 6V*2A = 12W. Else, I am wrong again. <Q> Yes, there's no such thing as a converter with 100% or better efficiency. <S> A reasonable buck converter will have an efficiency between maybe 70% and 92-93% <S> So input power will be the output power divided by the efficiency. <A> The manufacturer is assuming that everyone knows and obeys(1) <S> the Laws of Physics, and that the input current at the input voltage must provide enough input power to cover the output power + losses. <S> (1) unlike speeding/parking laws, or theft/murder laws, you don't get to choose whether to obey the Laws of Physics or not! <A> There will be graphs in the datasheet. <S> One will give efficiency at different loads. <S> Find the worst case efficiency for your range of loads. <S> Then divide the maximum output power you desire by that efficiency percentage. <S> Also add a little more on top. <S> A bit of de-rating never hurt anyone.
| That is the minimum input power you need to provide.
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How to invert a sine wave so that the output is 180 degrees out of phase with the input? I'm sure there is fairly simply op-amp circuit that will do this, however I lack the experience to design it myself. <Q> The inverting amplifier is one of the first op-amp circuits you should learn: ( image source ) <S> If Rf = Ri, the gain will be -1. <S> This circuit is probably good up to 100's of MHz, if you choose the right op-amp. <S> You'll want to watch out for both the gain-bandwidth product and the slew rate specs to find the right op-amp. <A> You can use a transformer. <S> If you swap the two output wires it will invert your signal. <A> To invert a sine wave, you can make a simple circuit using just 2 resistors. <S> The resistor version of the inverter has limitations : <S> beware of where you are connecting ground. <S> Also, the circuit has a high output impedance. <S> If the load to be connected is capacitive, the angle (180 degrees) will be affected. <S> simulate this circuit – <S> Schematic created using CircuitLab <S> Also, the output voltages Vo1 and Vo2 will be half of the voltage of V1. <S> Vo1 and Vo2 are 180 degrees out of phase with each other. <A> With an inverting amplifier. <S> See picture for an example. <S> Do you need any gain in the circuit or just invert it? <A> It's a classical op-amp configuration: <S> There are online "Inverting op-amp gain calculator" applications that can simplify component selection for you. <S> There's too much to explain, so you need to do some training...try <S> this as a simple starting point. <A> Several of the canned examples in this tool present INVERTING OP AMP circuits. <S> You will learn about tuned/resonant circuits, about opamps as low-pass-filters, about ClassB power amplifiers, if you download and install (Windows 7/8/10) <S> the free tool Signal Wave Explorer at the website <S> robustcircuitdesign.com <S> Just click on "examples", "scope probe ringing", "run" { <S> 3 clicks}.You can edit the input signal; find the "signal [trapezoid]" and edit that menu,to have a sinusoidal input, or other wave shapes if you wish. <S> I think there is a UserManual; yes, its under HELP. <S> There are 14 canned examples, with editable parameters. <S> For the LRC circuit (the scope probe that rings), you chose inductance, capacitance, resistance. <S> We particularly need feedback on how understandable the examples are, and how clear the menus are. <S> Don't be shy. <S> And enjoy the learning. <S> There is a 2nd tool, focused on OpAmps and interference: Signal Chain Explorer.
| What you want is an "inverting" amplifier.
|
How do hotswapping computer parts work? Computers support hotswapping so that a user can replace a harddrive while the system is running. Is that mostly software just powering off the harddrive or is it some special hardware involved? If there were not special hardware then I suppose that you could hotswap any drive but it seems that the drive must support it in hardware. <Q> It requires several features, some of which put the cost up. <S> These include additional transient suppression on signals, which may put additional capacitance on bus signals thus requiring additional engineering/testing and may reduce performance below what's otherwise possible. <S> Also, connectors which longer and shorter pins that ensure that some pins mate before others and disconnect after others. <S> This is to guarantee that signals are connected and disconnected while power and ground are already connected. <S> In some cases, a control pin can be made to connect last and disconnect first, and this pin will tri-state all other signal pins. <A> You may need software to ensure the device is free'd so that hotswap can occur. <S> The software needs to be able to integrate (size, format, initialize, rebuild) <S> the new device back into the environment. <S> For example if you lose a drive in a redundant array, you need to be able to rebuild the array after changing a drive. <S> You may need hardware support to turn off the power within a chassis to remove and replace drives. <A> Modern hot swappable cards usually have a hotswap controller to limit the inrush current when a new card is plugged in. <S> The issue is simply that when a card is plugged in, any bulk capacitance must be charged up, which can cause system voltages to droop; this issue is then exacerbated by the fact that most cards have switchmode power supplies. <S> These can best be understood as true power converters (to a first approximation, at least). <S> If the system power rail(s) droop, then to deliver the power the card circuitry needs, the input current must increase, adding to the load current on the system power. <S> This can cause brownout if not properly dealt with. <S> Techniques have evolved over the years for these applications. <S> As already noted, there are special connector arrangements for hot swappable cards and in many cases system software needs to manage the event.
| Reliable production level hot swap typically requires both software and hardware support.
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