source stringlengths 620 29.3k | target stringlengths 12 1.24k |
|---|---|
signal eye voltage level too high USB 2.0 Upstream near end I am using the ‘ Allion USB HS Device Electrical Test Fixture ’ with a Lecroy oscilloscope (4 GHz, 40 GS/s) in order to make a USB upstream Near-End Signal Eye measurement (for High Speed Measurement). The fixture connects via two 50 ohm identical coaxial cables, to the oscilloscope terminals. The problem I am having is the following: Even when testing with USB-Compliant devices, the voltage level of the device in the signal eye is too high or too distorted. This is not normal, since it should show without any points touching the red area. An example of the signal eye I am getting is below: Any help on the matter? <Q> That eye looks like its levels are designed for HS operation. <S> LS and FS modes use a higher differential voltage. <S> This waveform looks like it's clamped by diodes between the lines. <A> The quoted Allion set of test fixtures is not designed to test HS eye with naked 50-Ohm cables. <S> The test fixture for device measurements is and high-speed differential probes (or two active singe-ended probes) must be used with this fixture. <S> The fixture with two SMA connectors is used for INPUT from signal generator to test the receiver sensitivity. <A> The solution: I contacted the support from the oscilloscope and they mentioned the firmware version was too old and had to upgrade it since USB IF changed the impedance on the new fixtures to 100 ohms, the oscilloscope was expecting 90 ohms.
| If you are using the signal sensitivity fixture as direct 50-ohm connection to a scope, you will have improper termination, and signals will be severely distorted.
|
Why do French power sockets have a protruding ground? French power sockets have a protruding ground. Why did they make this decision? I can't see any major benefits from it, it saves some pennies on each power cord, but it exposes the ground connection to people nearby, which can be dangerous as in real life grounds are not perfect, they can have a live voltage and are a risk of electric shock if there's lightning nearby. Once someone told me they are safer because the ground connects first, but this is also nonsense as the order of connection depends on the size of the plugs alone and not where they are. In case you don't know what I am talking about: <Q> I'm not 100% sure, but I believe that the reason for the unusual earthing arrangements on french and german plugs is backwards compatibility with existing installations. <S> Appliances can be categorised into 3 categories. <S> Class 0: no earth connection, not "single fault safe" (now banned in most first-world countries) <S> Class 1: "single fault safe" only if an earth connection is present. <S> Class 2: <S> no earth connection, "single fault safe" If your existing installed base of installations has no earth provision and your existing base of appliances is class 0 <S> then it can be argued that it is sensible to let people plug class 1 appliances into their existing unearthed sockets. <S> They aren't any less safe than the old class 0 appliances were. <S> There is also an argument to be made that having some appliances earthed while there are class 0 appliances in use makes things less safe than having no earthing at all. <S> Adding a conventional earth pin disallows plugging new appliances into existing installations, and does nothing to prevent old class 0 appliances from being brought into new earthed installations. <S> The design of the German and French plugs on the other hand allows new appliances to be used in existing installations in a way that is no less safe than before, while keeping old class 0 appliances out of new installations. <A> Once someone told me they are safer because the ground connects first, but this is also nonsense as the order of connection depends on the size of the plugs alone and not where they are <S> You are right. <S> The ground pin could be on the cord itself. <S> But it is how it is. <S> But this is a generic argument, looking at may available type sockets and plugs used worldwide <S> (Type A to Type N). <S> Until there is one universal plug and socket adaption <S> say by the year 2200 (Like phone charger pin types reduced to two now, majorly from many. ), the variants are going to stay. <S> https://en.wikipedia.org/wiki/AC_power_plugs_and_sockets#CEE_7/5_socket_and_CEE_7/6_plug_(French;_Type_E) <S> Plug Type E <S> Used in: France, Belgium, Slovakia and Tunisia among others (see complete list of countries on the right) <S> The Type E electrical plug has two 4.8 mm round pins spaced 19 mm apart and a hole for the socket's male earthing pin. <S> The Type E plug has a rounded shape and the Type E socket has a round recess. <S> Type E plugs are rated 16 amps. <S> Note: <S> The CEE 7/7 plug was developed to work with Type E and Type F sockets with a female contact (to accept the earthing pin of the Type E socket) and has earthing clips on both sides (to work with Type F sockets). <S> https://www.iec.ch/worldplugs/typeE.htm <A> Without the original argumentation how the design was presented to be the one and only right version we can only guess. <S> Grounding at the edges of the plug need only a little material removal with a saw or knife or by melting. <S> This type of grounding needs a hole which requires more complex tools and there can be metal inside the plug in front of the drill. <A> This design forces the user to only plug compatible equipment in that socket. <S> That is, if you get a German-only plug (where ground contacts are elsewhere), you simply won't be able to insert it in the French socket, as there will be no hole to receive the ground pin. <S> If the plug fits, you can be quite certain that the appliance is correctly grounded. <S> Indeed, this doesn't apply to flat plugs which have no ground pin, but appliances having such plugs don't require grounding in the first place. <S> In the opposite case (French-only plug, German socket) the plug will mechanically fit, but the ground will not be connected, giving a misleading impression of safety. <S> When the ground contact goes live, no plug design will make it safe: no matter how concealed the ground pin is, you will get a shock the moment you try to operate the equipment you have plugged in. <S> BTW, I don't see how this design "saves some pennies on each power cord", because you still need to make a female receptacle for the ground pin, which I suspect cost more than a third pin would have.
| My guess: It's difficult to modify existing non-grounded non-double-insulated device plugs to fit.
|
How to determine if given OpAmp model supports single supply (Ex. OPA167x) Reading OpAmp datasheets I often struggle with the task of determining whether a given model of OpAmp can be used with single supply or not. Sometimes, there is a plain statement in the features. But most of the time, it is not explicitly written. For example, in the OPA167x datasheet, the manufacturer only says Wide supply range:– ±2.25V to ±18 V, or 4.5 V to 36 V I guess thats enough information here, but are there any other cues (or words) that I can look for? Due to english not being my native language I might miss something. <Q> All OPAmps can be used with a single supply, but some are easier to use than others. <S> A key criterion for single supply use is that the input common mode <S> voltage should include the negative supply, usually connected to ground. <S> A very useful criterion is that the output should drive to or nearly to the negative supply. <A> If output voltage max 12V with 15V supply (for example with a classic 741 op amp), then it needs 3V of "headroom". <S> This is not a formal specification, but something that is inferred based on the specified output voltage limits at specified supply voltage test condition (VCC - VOHmin) and the (VSS - VOLmax). <S> Running a 741 from 5V single supply doesn't work because its output voltage is too constrained, max output voltage would be 2V but <S> min output voltage would be 3V. <S> This forces the output to sit near the center without any ability to drive signal. <S> Sometimes the output voltage swing is not symmetrical, it may be weaker driving high output and stronger on low output voltage; or there may be "crossover distortion" when it changes from being a current-source to a current-sink. <S> Unfortunately datasheets don't usually call attention to this kind of distortion, except to specify it is guaranteed below some amount. <A> If you use an op-amp, it really just needs large enough supply voltage between the supply pins. <S> So the op-amp does not care if you have 12V single supply or <S> +/- <S> 6V dual supply, from the op-amp point of view it just has supplies, it does not make a difference. <S> However, something else in your circuitry might require a certain configuration for the power supply, like ability to input and output negative DC voltages in respect to common ground. <A> Ground is a concept invented by humans so we can specify other voltages relative to it. <S> Circuits don't care which node humans label as "ground." <S> An op amp that can take a supply up to +12V can be run from +/-6V, and one specified for +/-15V can as easily be run from a single-ended 30V supply. <S> Whether the spec sheet indicates single-ended or differential supplies is primarily a matter of marketing.
| Two parameters affect whether an op amp can support single-supply, low-voltage operation -- minimum total supply voltage, and output voltage swing.
|
How to connect wires coming from bottom side of a single-layer PCB? I intend to use an IMS PCB, that is a PCB with aluminium substrate. It will be a single layer PCB (because of the cost, and heat), so only the top layer. I need to connect approx. 20 wires to the top of this PCB, which are from another PCB, but I need them to come from the bottom face of the IMS PCB. One solution is drilling holes and soldering them, like this: But I would need a better looking solution... Do you have an idea? :) Thanks. <Q> There is a product from Molex that might be designed specifically four your use case. <S> It is called: Bottom-Entry Lite-Trap SMT Wire-to-Board Connector <S> And it looks like that: <S> There will still be machining to do on the PCB, in order to allow the connector housing through, but the result will be solder-free and super clean looking, I think. <S> But for 20 connections it will set you back 14$ per PCB just in connectors, wether it is worth it or not I will let you be the judge of that. <S> https://www.japanese.molex.com/molex/products/part-detail/pcb_receptacles/1042660110 <S> https://www.digikey.com/product-detail/en/molex/1042660110/WM12961CT-ND/5980647 <A> You could use almost any kind of through-hole soldered connector mounted on the bottom with pins upwards, soldered on the top. <S> Such as any one of the hundreds of kinds: Image from Duckduckgo <S> Personally I like things which unplug and would consider a right-angle pin header and ribbon cable. <A> Mmmpf. <S> Somehow I missed that you want to make a single layer board. <S> The following runs into a double sided board. <S> Sorry. <S> I'll leave it here in case <S> somebody else finds it useful. <S> Design your single layer board and use a single pin connector at each point in the schematic where you need to connect a wire. <S> When you place it in the layout, use a through hole single pin connector <S> That'll place a footprint that has a solder pad you can connect to, and the PCB fabricator will drill a hole in the pad for you. <S> On an IMS board, the fabricator will drill the hole oversized in the aluminum before it is laminated together with the fiberglass and copper layers. <S> After lamination and etching, the holes for the through hole parts will be drilled and plated as on any other PCB. <S> Here's an example of what I mean from Leiton.de: <S> The orange is copper, the gray is aluminum. <S> The yellow is insulation. <S> With that stackup, you can insert your wires from the bottom and solder them on top. <S> I found several fabricators that will do an IMS board that way. <S> The insulated through hole thing seems to be standard on Multi cb aluminum core two layer PCBs. <S> Unfortunately, they don't provide online estimates. <S> I didn't spend a lot of time trying to find quotes - I have no idea if that stackup is insanely expensive or what. <S> You might search for fabricators that do double layer aluminum core PCBs with plated through holes (PTH.) <S> That's what Multi CB offers.
| If you really want to solder it, consider an IDC ribbon cable header, which crimps onto the cable and presents pins to the PCB (from underneath) which you solder (on top surface). The holes in the aluminum are filled with a non-conductive material (epoxy or other filler.)
|
Thermal Resistance of PSU's heat sinks I have a non functional PC power supply unit, and I would like to keep its heat sinks, but they have no model or part number in them. Is there a way to estimate, calculate or measure their thermal resitance? The PSU is a Sentey BCP 450-OC, here is a picture of the heat sinks <Q> You work in a controlled temperature environment (i.e. indoors at 20°C). <S> You put a measurable amount of power through the resistors (e.g. by using a lab voltage supply and measuring the voltage over and current through the resistors). <S> You measure the temperature and wait until you've reached a stable state. <S> But: honestly, I don't think this is worth it. <S> Both the shape and the fact <S> they're fixed right in front of a fan means that they are to be used with forced cooling, i.e. with a fan in a specially shaped enclosure. <S> You'd have to measure in the enclosure and with the fans you're planning to use. <A> Is there a way to estimate, calculate or measure their thermal resistance? <S> Yes. <S> I am assuming that you want to know the heatsink to ambient thermal resistance. <S> The calculations are complicated, somewhat easier for forced air (fan), than natural convection. <S> I wouldn't trust any calculation that I would perform without verifying it with a test. <S> So, ... It is easiest to test. <S> This heatsink is designed for forced air, it won't be very good without a fan. <S> Attach something to the heatsink that you can be made to dissipate a known amount of power. <S> A chassis mount resistor will be a good choice. <S> Measure the temperature rise of the heatsink, subtract the ambient temperature, divide by the power. <A> The fluid aerodynamics of forced air velocity overwing surface determines the thermal resistance coefficient in °C/Watt for case temp. <S> rise. <S> The air velocity over the surface, not the fan exit, significantly lowers the resistance by some constant, k (< 1) times the Rca, Resistance case-ambient. <S> This is more of an aerodynamic design requirement. <S> One could come up with a more accurate answer, but based on my exposure to heatsink design and testing I would estimate 8 °C/W and with that airflow 1~2 °C/W.
| Measuring the effective heat resistance should be pretty straightforward: you'd affix load resistors to where you'd later put the components that should stay cool. The difference of resistor temperature to ambient divided by the power is your thermal resistance.
|
what does pull-up resistor and pull-down resistor mean? How do I implement them in my circuit? I am trying to understand the basics of electronics and I often come across the terms pull-up and pull-down resistors. From what I understand these resistors help the switching logic devices like transistors. Could you help understand the basics of these components in simple terms. Edit 1: I am trying to build a small feed back system that uses a transistor as switch. The system uses a buzzer that makes a sound when the yellow and gray components meet. The problem that I am facing is that the buzzer keeps making a low volume sound even when the components are not touching. I was reading about pull up and pull down resistors and would like to know how to implement it here Figure 1. Buzzer schematic. simulate this circuit – Schematic created using CircuitLab Figure 2. The OP's schematic redrawn by @Transistor using the CircuitLab schematic editor. (There's a button on the toolbar.) <Q> A pull-down resistor pulls the voltage down to the "low" logical level (0V or close to it) when the is no signal driving the input. <S> This is achieved because the input impedance of the device being pulled up/down is usually very high (and much higher than the resistor itself, so they won't form a voltage divider). <S> Due to this there is almost no flow of current through the resistor, and the voltage drop is almost zero volts. <S> This means, in turn, that the voltage (either "high" or "low") <S> you are connecting the resistor to then gets passed "as is" to the input of the device being pulled up/down, which helps set a "default" input value when there is no signal driving it. <S> However, when an input signal appears, it will usually have a source impedance much lower that the pull up/down resistor, and will dominate he combination and set the state regardless of the resistor. <S> This diagram (source here ) illustrates the behaviour described: EDIT: Regarding your specific use case, you'll need to pull down the input of the BJT. <S> Use a resistor 20x R1 and it'll probably be fine. <A> A pull-up or a pull-down resistor, when connected to a point in a circuit, provides that point with a weak default voltage which dominates unless some other stronger component forces it to something else. <S> A pull-up means it pulls the point towards some positive voltage (typically the supply voltage), for example +5V, while a pull-down pulls the point towards ground or 0V. <S> You could for example connect a pull-up resistor to the collector of a NPN transistor. <S> When the transistor is off, the voltage at the collector is decided by the pull-up resistor, but when the transistor is on, it is stronger and forces the voltage down. <S> You can also use it with e.g. a switch where you want to set the default voltage output when the switch is in "off" position. <S> You might read the Wikipedia article for some more info: <S> https://en.wikipedia.org/wiki/Pull-up_resistor <S> One important aspect is that since this is indeed a weak voltage source, you cannot use it to drive circuits that needs lots of power. <S> In that case you will need to amplify the signal somehow, for example with an op-amp, or some other voltage follower circuit. <A> I am assuming that you are using a NPN BJT in the diagram. <S> Please connect a resistor of about 10k(higher) between base and the ground(battery negative) of the circuit. <S> Connecting the pulldown resistor provides a known state instead of leaving it floating. <S> Currently, When the yellow and Gray parts are not connected, the base of the transistor is floating.
| It's very simple: A pull-up resistor pulls the voltage up to the "high" logical level (5V, 3.3V or whatever is used as a "high" level) when the is no signal driving the input.
|
Puzzling behaviour from two inverters connected together I'm an electronics newbie and I am puzzled by the behaviour of a circuit I have created. I have connected the output of one inverter to the input of another but the behaviour of the circuit is not what I expected. I've included a schematic and two photos of the circuit. The first photo shows the two inverters unconnected and the circuit behaves as I expect it to, that is, LED1 turns off when I press the tactile button. However, in photo 2 you can see that I have connected the output of the first inverter to the input of the second and LED1 is off, but it should be on. Can you explain what is happening? <Q> The problem here is that the anode (top) of LED1 is connected directly to the base of Q2. <S> When Q1 is off current will flow from R3 through Q2 base to ground. <S> As Q2 is a silicon transistor the voltage on its base will be clamped to about 0.7 V. <S> A red LED needs about 1.8 V to light up <S> so LED1 won't see enough voltage to light up. <S> (Q2 will switch on and off correctly so LED2 will behave as expected.) <S> To correct this, link LED1 to Q2 with a resistor (e.g. 1 kΩ, depending on the value of Vcc) rather than a wire link. <S> This will allow the voltages on LED1 and the base of Q2 to be different. <S> Edit: I've removed text about an earlier version of your schematic which no longer applies to the updated version. <A> However the second inverter will function coupled to the first. <S> So the output LED2 should function as expected. <S> I didn’t look at your breadboard implementation. <A> Schematic seems to be showing only the circuit of second bread board photo. <S> As per second bread board picture, ...LED1 is getting clamped by the VBE drop of Q2. <S> So, LED1 would not glow when its anode is connected to the base of Q2 regardless of the switch position. <S> As per first bread board picture,... <S> The switch is off,The green wire connecting LED1 anode and Q2 base is removed. <S> So, LED1 is no longer clamped by the VBE of Q2. <S> That's why LED1 is glowing. <S> The LED2 is also glowing because Q2 base is left open thus Q2 is OFF.
| The base of Q2 in your schematic will shunt the LED and prevent it from ever turning on.
|
Why can a MCU input floating pin easily change state? I would like a technical ("engineering") explanation of why a MCU input floating pin can easily change its state depending on the outside electromagnetic interference.Is it related to the fact that the pin is in a high impedance state? If so, how exactly? <Q> As an ultra simplified, 1st year EE model, you can consider a disconnected input to a CMOS chip to be an RC circuit. <S> The tiny gate leakage currents represent the R, and the tiny gate capacitances plus the stray capacitance from the pad or pin to the outside world being the C's. <S> Change the external EM field to the outer plate of the capacitor enough <S> (ground noise, other nearby PCB traces switching, bonding pad cross-talk, cosmic rays, etc.) <S> and you can change the voltage on the transistor gate. <A> It is not limited to MCU input pins. <S> Also other CMOS chip inputs work like this. <S> Basically the CMOS input stage is just gate terminals of two MOSFETs. <S> FET gate is basically insulated, but gate needs to have voltage in respect to FET source for the FET to turn on. <S> The FET gate is mainly few picofarads of capacitance load, and draws only very little leakage current, so even a high impedance source will quickly charge few picofarads with small current to have enough voltage to turn on. <A> CMOS gates have very low input capacitance (x pF) which may be dominated by parallel traces with 10 mil track and gaps, but even far less if the input is floating near Vdd/2 unless there is a resistor driver. <S> Thus it depends on the crosstalk capacitance and self-bias leakage voltage. <S> Generally, 1M pullup is adequate in less noisy ground plane environments and 10k is used in noisier logic layouts to avoid ripple effects during transition or avoid false inputs from crosstalk. <S> While CMOS outputs range from 20 to 75 Ohms for 74ALV to 74HC logic. <A> Shortly: Noise (outer electromagnetic signals) can create current over conductors. <S> This current induces voltage across a resistor according to Ohm's Law. <S> If the induced voltage is high enough then the input buffer will see this is logic-high. <S> Likewise, if the induced voltage is negative enough then the input buffer will see it as logic-low. <A> High impedance means in fact, very high resistance. <S> And EMI can cause dislocation of little charges. <S> Think of a clogged valve at a bike tyre. <S> If you put your pump with an manometer on and give pressure, the pressure easily rises to very high values without performing a lot of work. <S> When you release the piston of the pump the pressure will drop as fast as it has risen before. <S> There was only a little amount of air able to move and had nowhere to go. <S> If the pressure is what you are watching for, (which is somehow similar to voltage) you will detect sharp changes while little has been done. <S> If the valve isn't clogged (i.e. not floating) you have to move significant more charge into the input to change the pressure. <S> To keep up this analogy: <S> A non floating input (i.e. with pull-up or pull-down resistors) is a bicycle tube with well defined holes. <S> After having moved some charges in it, a certain pressure builds up but will only prevail, as long as you keep charges flowing by constant pumping. <S> An EMI is e.g. a crow landing on the piston of your pump. <S> If your valve is clogged, pressure will rise and stay elevated until the crow sees a mouse or a cat and decides to move away. <A> If the input pin is floating and “short”, it is electric field interference that causes a voltage fluctuation on the pin. <S> This may arise from electro magnetic fields but the problem is dominated by the electric field perturbations of that EM field and capacitive coupling to the floating MCU pin. <A> When the pin is in high impedance, it presents as a very small capacitance (a few pF) and practically infinite DC resistance. <S> The pin is vulnerable to noise coupling from outside sources because it takes so little charge to change its state. <S> Noise coupling can be magnetic (inductive) or capacitive. <S> The case of bringing one's hand near the floating pin is an example of capacitive coupling: the air gap between the pin and the hand forms a capacitor, in series with the pin's capacitance. <S> The closer the hand, the smaller the dielectric, the larger the coupling capacitor, and the more coupling occurs. <S> Inductive coupling occurs when a current flows in a nearby circuit, creating a magnetic field. <S> If this is close enough to the floating pin, it too can couple onto it and cause problems.
| The input impedance may be easily biased by leakage to some DC voltage or by trace capacitance to a mutual trace with a pulsed voltage.
|
Is a floating pin equivalent to a very weak pull down? I know that if a MCU pin is configured as "input" and it is not connected to any pull up/ pull down resistor, it is said to be floating.In this situation, we could say that the pin is in series with the (very) high input impedance of the MCU, which is eventually connected to ground.So we could say that the pin is in a (very) weak pull-down condition, with the MCU input impedance acting as the (very high) pull-down resistor.So basing on this assumptions, if we suppose that no significant interference acts on the pin, I would say that (if we wait long enough) the pin voltage will be eventually "pulled to ground". Where am I going wrong? <Q> Have recently been characterizing an input pin on a PIC microcontroller. <S> The pin can be alternately set to CMOS in/out, Schmitt input, or analog input. <S> When set for Schmitt input: <S> Data sheet specification of leakage current is +/- <S> 5 <S> nA. <S> Measured leakage current at room temperature is less than 3 pA. <S> Similar results were measured years ago on a different PIC microcontroller. <S> Be aware that leakage current is very temperature-dependent, and can be in either direction . <S> So you cannot assume that it floats to a logic low state. <S> It is possible that electric field of a hand-waving over a floating pin can change its logic state. <A> A CMOS chip's input circuit generally includes both P channel and N channel transistors. <S> If the chip is powered, then the (tiny) gate leakage currents can be in either (or both) directions: to ground, or to some internal voltage <S> rail(s). <S> The process variation may not allow one to pre-determine which direction of leakage is greater (unless the chip was designed with specific direction and process+operating margin). <A> tl; dr version: <S> The input impedance of a CMOS input is high, practically an open circuit. <S> It needs to be terminated, either by tying to a supply or by driving it with a signal. <S> Why? <S> A CMOS input is a capacitance : the gate-source capacitance on the input pair. <S> Your intuition tells you that even though it is a capacitance, the charge on the input should bleed off somehow over time. <S> In fact CMOS gates do that: they have a small Vgs leakage. <S> There is also leakage in the input protection circuit - a pair of reverse-biased diodes that catch voltage above VDD and below Vss. <S> However, there's two confounding things going on. <S> First, the two FETs (P and N) have their sources tied to the Vdd and VSS rails, respectively (as do the protection diodes.) <S> So their leak paths are to opposite rails: P FET to Vdd, <S> N FET to Vss. <S> If the leakages were equal, then in theory the input would find a bias point in between. <S> In reality this is very hard to predict because gate leakage varies wildly with process and temperature. <S> It cannot be counted on as a bias. <S> Second, this Vgs gate leakage is very, very small, and so is the gate capacitance (a few pF typically.) <S> On the other hand, the influence of outside electric fields can easily couple onto the floating pin and cause it to change state. <S> This is in fact how a basic capacitive touch sensor works: hum pickup from the body. <S> Bottom line: never leave a CMOS input floating. <S> Always tie it off. <A> In this situation, we could say that the pin is in series with the (very) high input impedance of the MCU, which is eventually connected to ground. <S> Where am I going wrong? <S> You are going wrong in forgetting <S> it is also eventually connected to Vdd, not just GND. <S> So why would it eventually pull to GND over Vdd? <S> simulate this circuit – <S> Schematic created using CircuitLab <A> Where am I going wrong? <S> ... <S> to the fact that a pin can be connected to two CMOS in push-pull configuration, and they could leak current in an unpredictable way... <S> Thinking better, even if the pin is a pure CMOS input, you can not know in advance what will happen. <S> May be that for a short time our CMOS input behaves like a load. <S> A (non-CMOS) <S> transistor can be very different, but if you restrict the question to MCUs, I believe it is difficult to find normal transistors. <A> No, a floating pin cannot be considered to be equivalent to having a defined pull-down. <S> It could also tend to float high or at midpoint or any level.
| It cannot be left floating.
|
Thoughts on storing electricity in water vapor for an engineering class I had to create an electrical energy storage device and I began thinking about thunderstorms (I think it was a cloudy day) and how they create lighting. I know it's not feasible, but would it be possible to store a charge in water vapor (maybe like a capacitor). Could that water vapor then be stored in a container and be discharged whenever you connect the container (I am thinking like a metal lead coming out of a soda bottle) to a ground? The thought of a thunderstorm in a bottle seems like a very dangerous but cool idea (not that I would use it for my project, I was just wondering if this would be possible). <Q> A "bottle capacitor" is called a Leyden Jar. <S> The original one was full of wine, and held in a damp human hand. <S> The wine is one capacitor-plate, the hand is the other. <S> Thunderstorms themselves are storing charge on their raindrops. <S> Not on vapor. <S> That way, if all the raindrops get charged the same, they can't zoom rapidly outwards from repulsion (because they're large heavy objects, not tiny molecules.) <S> And, if they did start repelling outwards, well, the region of falling rain might be a half-mile wide. <S> The drops won't be repelled apart significantly, compared to a half mile. <S> Inside a tiny bottle, any charged droplets (or vapors) will repel outwards within about a second or two. <S> They end up stuck against the glass. <S> You'll have a charged bottle surface, with neutral gas inside. <S> What solves this problem? <S> CAPACITORS! <S> If you put any region of negative charge right next to a region of positive charge, then insert a barrier so they cannot attract together and cancel ... <S> then you can store some electrical energy (stored in the barrier, stored in the strong e-fields between the two.) <S> Thunderstorms do this by having positive charge on small droplets, and negative charge on larger drops, then letting the positive droplets lag behind as they're falling. <S> Gravity and air friction acts as the "virtual barrier" between the two regions of falling, opposite-charged drops. <A> Thoughts on storing electricity in water vapor <S> If that read storing electrical energy in water vapor , it would have been a viable concept. <S> Because vapor of a few hundred degrees Celsius can store tremendous amounts of energy. <S> It's also cheap and reasonably safe, and easy to turn back into electricity. <S> The technology already exists, too. <A>
| No, you can't capture lightning in a bottle.
|
Does this design smell bad? Can I pull a pin simultaneously down and up? I am building a simple circuit using PT7C4511 PLL Clock Multiplier . This chip has an OE pin that stops the output when LOW. By default (with no external signal driving it) it stays HIGH because of a built-in pull-up (270K). What I want to do is to turn on and off the chip's output using a 2-pin header. However, I want it to be working when the header is shorted, and stop the output otherwise. To achieve this, I have connected 40K pull-down to OE pin, strong enough to overcome the internal pull-up, but not too strong, so that when OE is shorted to VCC, the pin can easily go HIGH again. I showed this to a friend with more experience in EE and while he agreed that it should work, he wasn't too excited about the design. He could not point to a specific issue, but the whole thing "smelled" to him. Is he right? Why? <Q> Yes, it smells. <S> Use a transistor. <S> This is the normal way to invert a signal. <S> The OE input is a sourcing input. <S> These are designed to be used with a switch which is either open or switched to common. <S> It's not meant to have Vcc connected directly to the input! <S> You have a switch (your header), which would be perfect, but you want the behaviour inverted - so invert it with a transistor. <S> That's why it's called TTL. <S> The resistor to ground also increases the noise sensitivity of the circuit needlessly. <S> Here when the header is open the base of the transistor is high and OE is pulled low. <S> When the header shorts the transistor turns off and OE goes high via the internal pullup. <S> I've shown 10k to the transistor base here, which is a bit greedy for power - a lot of values would work here, though. <S> The higher you go the less power it draws, but the more sensitive to noise you become. <S> Absent a compelling reason not to, giving the input a lower pullup resistor here is probably a good idea. <A> Well, you can't really remove that 270K pull-up, so that just means you have to use a significantly smaller (stronger) pull-down. <S> Also, as an on-chip resistor, the precise value of that pull-up is not going to be very well controlled and could vary by quite a bit. <S> I would recommend going even smaller on the pull-down, perhaps 10k or even 4.7k or 1k. <A> Your 40.2k pulldown is probably fine. <S> The datasheet, 'DC Electrical Characteristics' table on page 2 gives you all the info you need for this. <S> The V IL line tells you that 0.8V is the maximum value which the OE pin will recognize as being 'low'. <S> The R line tells you that the OE pin has a 270k Pullup. <S> You know you have a 3.3V supply <S> , so with this info it's possible to calculate the maximum value pulldown resistor you can use and still have the input recognized as pulled low - and that value is 86.4k. <S> So since your 40.2k is less than half that you're well into the 'safe' range <S> (you should expect about 0.43V). <S> The only other thing you might want to consider is putting a cap on that pin (since you're connecting it to a header and that might pick up some noise). <S> I'd probably put a 100n there. <A> I agree that this is not the best design. <S> However I would not think that it smells like rotten cheese. <S> A much better approach is to lower the impedance all around and simply use a 10K or 12K ohm pullup on the pin and use the two pin jumper to GND to disable the output. <A> 40K is not a very strong pull-down, it will be prone to noise and with a wire attached will have a poor fall time when the switch opens. <S> 5K or less would be more appropriate. <A> Actually an elegant solution compared to the one requiring an additional transistor. <A> If you don't mind having it the other way around (short to disable it), then an obvious solution would be to have just the jumper between OE and R3. <S> When the jumper is not connected, the internal pull up will pull it high. <S> When the jumper is connected, OE will be pulled to ground via R3. <S> No other components required.
| If you can spare the current and want a more noise resistant circuit you can also tie OE to Vcc with a parallel pullup (Rp) to the internal 270k. I don't see any problem with this providing the 40k resistor is sufficient to provide the low voltage for the input.
|
Magnetic flux induced in toroidal core and conductor position I was just reading Electric Power Engineering by Elgerd when I came across these statements about flux I couldn't fully understand: "By experiment, we can confirm a most interesting characteristic of the flux. If we keep the iron core position fixed but move the cable to any position inside the core window, the flux will not change. (The reader should contemplate what happens to the flux in the air core under the same conditions.)" Why is this so? Shouldn't changing position increase flux in the regions of the core close to it and decrease in the far away areas? Is this the case for air core? "The flux will not change even if we bend the cable into any conceivable shape as long as the total current Ni remains unchanged and as long as it passes through the core window. In fact, the magnetic flux will remain unchanged if we replace the N-strand cable with an N-turn coil carrying a current i as long as the product Ni stays constant." I have trouble imagining what the N-turn coil he's referring to would be placed like. Wouldn't it cause NI going in and coming out both? Also, what would happen if I change the shape of the iron core itself keeping the cable straight? How would flux change then? <Q> The magnetic core has low reluctance. <S> Low reluctance means that it is a very attractive path for magnetic fields to flow through compared to air hence, most of the circulating flux in the toroid will be constant largely irrespective of the conductor position. <S> This is because the magnetic field strength is dictated by current flow and it is assumed, for this explanation, that current flow is constant. <S> There will be some flux in the air but, for a high permeability core (low reluctance) it will be a small percentage. <S> If you made the toroid extremely large, the path the magnetism travels along gets extremely long and reluctance increases linearly with the circumference. <S> If you continued this train of thought, you will get to a point where air carries the most flux because it is seen by the flux as being the preferred path due to it having lower reluctance than the massively oversized and distant toroid core. <S> See my answer here for a mathematical explanation of how reluctance works in a similar type of problem. <S> If N parallel strands carried N amps, this is the same as N looped strands carrying 1 amp. <S> Parallel strands are counted as one loop. <A> If you have a finite permeability core with an arbitrary shape, then things are very complicated, you'll need to do 3D integration in mixed materials, not fun. <S> In general, the flux varies somewhat with the cable position. <S> We can make some simplifications under certain circumstances. <S> 1) Core has infinite permeability. <S> All the flux travels in the core, the cable position does not matter at all. <S> The flux is constant due to the topology, that it loops the conductor, and takes all the flux. <S> 2) <S> Core has very high but finite permeability. <S> As the relative permeability of the core gets closer to one, the flux becomes more sensitive to cable position. <S> As a rule of thumb, if the core is iron or ferrite with a relative permeability of (say) 2000, then the change in flux as the cable is moved from middle to edge will be less than 1/2000th, which for most purposes is negligible. <S> 3) <S> Core is air, and the winding is a uniformly wound toroid. <S> This is called a Rogowski Coil . <S> Although the flux in any part of the torus varies as the central cable moves, the total flux linkage, and therefore the voltage induced in the winding when the cable current changes, is constant due to the symmetry properties of the torus. <S> Even with a circular core, a finite permeability means there will be some small sensitivity of flux to cable position. <A> Regarding the position of the wire not affecting the result: This can be proven mathematically, but the math is very complicated. <S> A good college professor can most likely do it (I recall seeing it done once). <S> Sometimes, it is best to just trust the conclusions of the experts. <S> Regarding N-turns: It is the current that passes through the hole that matters. <S> If the wire loops around and passes through the hole again and again (you have multiple (N) turns of wire going through the hole), <S> the equivalent current that the torroid sees is the product of the number of turns and current. <S> If the shape of the core is not circular, then the symmetry is lost, and the position does matter somewhat. <S> Regarding an air core: The rule still applies for the area defined in the figure. <S> However, the amount of flux in this area is much, much less. <S> The flux is not contained in this area. <S> An iron core is useful because the flux will mostly be contained within the iron core (path of least resistance). <S> Then, secondary windings can do something useful.
| Almost all the flux travels in the core, and the core flux barely changes as the cable position changes.
|
Are heatshrink tubings toxic for human health (e.g, using them in a "prolongedly-touched" place)? Are heatshrink tubings toxic for human health (e.g, using them in a "prolongedly-touched" place)? E.g. if I use them for supporting the tip of my phone charger or the tip of a 3.5 mm Audio jack (for strengthening purposes), will there be any long term health hazards (toxicity etc.)? I am worried because I touch those places many times over the course of daily activities. <Q> Shouldn't be. <S> Heatshrink tubing is just plastic. <S> Certain plastics just shrink when heated just like the thin clear plastic wrap you see in that wraps tightly around boxes and other packaging. <S> The material used for most heatshrink tubing <S> is polyolefin, in case you are curious. <S> There is even medical grade polyolefin heat shrink tubing. <A> Unlikely. <S> They certainly contain things that aren't per se healthy for consumption (plastic softerning agents etc), but you're not eating them... <S> Anyway, whether or not an industrial product has health hazards is answered in their manufacturer's safety datasheets, if there's any danger at all. <A> If you're really concerned, you can get FDA CFR 177.1330 food grade shrink tubing, which is polyolefin similar to other common types of shrink tubing. <S> Personally, I have zero concern for touching. <S> There are many other materials available such as PTFE (Teflon), PEEK, PVC. <S> Some of the highest shrink ratio material is PTFE, however it is very expensive. <S> In general, plastics themselves are seldom harmful, but there have been some harmful additives identified in recent years, in particular some flame-retardant agents. <S> Additives, particularly plasticizers, as well as fillers, can constitute a substantial percentage of certain plastics. <A> Probably fine. <S> If you're concerned and don't want to get the food-safe stuff, get your tubing from a source reputable enough to also have an MSDS for the specific part number of tubing you're buying. <S> That should cover prolonged contact.
| You might like to know there is also food-grade polyolefin heat shrink tubing as well. They don't really have to do anything weird to it to make it shrink.
|
How to wire a Rpi, a relay, a DPDT and a PWM module to a DC motor, allowing to control its direction and speed? I am currently using a Raspberry Pi to control the speed of a 12V DC motor and it works perfectly fine right now like this: This works: I can start, stop and ramp up the DC motor speed. But obviously, it only goes in 1 direction and it cannot spin in the other direction unless I physically reverse the red/black wires at the output of the PWM module. I tried it and it works, the motor can spin in the other direction with -12V instead of +12V. Then, I ordered a DPDT 8 pins relay (LY2NJ) because I saw the following YT video and that is exactly what I want to do: https://www.youtube.com/watch?v=xTGzcN8JrUk This is how I connected everything, but I think something is off in the connections I made, because there was smoke when I connected the battery, so I immediately unhooked everything and I'm now asking for your expertise to help me find what's wrong. Note that the bottom relay enables the RPi 3.3V GPIO pin to trigger the 12V DPDT relay. What's wrong with the connections I made? (DO NOT REPRODUCE THE ABOVE) EDIT: WELP, after frying my older Raspberry Pi (It's now RIP instead of RPI, get it? Yeah it's not funny) I have followed @brhans advice to power the DPDT relay AFTER the PWM and also I'm freaking out about grounding to the RPi ever again, so I grounded all the relays to the -12V instead of to the RPi: Do you think this will fry my second RPi or should it be good like that?? <Q> simulate this circuit – Schematic created using CircuitLab <S> It was a leap of faith to add PWM to the video you saw which was incorrect for unipolar PWM. <S> Shielded twisted pairs will help reduce the likely EMI from the motor to RPi. <S> The motor must be stopped before changing directions and PWM set to 0. <S> This is just a conceptual solution and motor DCR and power ratings must be considered with load inertia, speed, & acceleration needs. <A> OK I was drunk yesterday. <S> Thank god my 25Ah <S> (yes, you read that correctly) lithium battery did not explode. <S> I wired the relay wrong! <S> That is why the RPi fried. <S> Today, I unsoldered the mess I did yesterday and rewired the relay as it should have been to make it a reverse polarity switch. <S> For any future noob (hopefully not as much as I am) coming here to look up the way to wire a LY2N-J DPDT relay for reversing polarity, THIS is the correct way <S> AND NO SMOKE <S> THIS TIME :) <S> I also tuned some other things as well, while I was at it, such as replace the relay for a true 3.3v one, which is less prone to sticking. <S> I also connected every single 12V relay, PWM and DC motor to the battery's -12V "ground" instead of the RPi's ground, because I'm freaking out due to what happened yesterday. <S> I still would feel better protecting the PWM signal pin with a diode, which I don't have right now. <S> BONUS: <S> Anybody knows what extremely fast inverse polarity recovery diode <S> I could order to protect the GPIO port "just in case" the PWM decides to short to the signal pin? <S> Thanks! <A> Question <S> How come my advanced motor circuit keeps frying Rpis? <S> Notes <S> (1) @tlfong01 is casually refining and drawing the circuit without prior approval by @that-ben. <S> * (2) <S> The Rpi's PWM signal with duty cycle <S> 0% might actually cut of the 12V DC motor power. <S> In other works, the first stage 5V relay switching on/off 12V motor power supply might not be necessary. <S> Answer <S> Well, there are many causes frying the Rpi's. <S> Perhaps we can redraw the circuit diagram showing the power, signal, and ground lines. <S> One ounce of prevention is better than one pound of cure . <S> You might like to try the following prevention tricks: (1) Tidy up <S> the 12V motor power ground and Rpi signal ground lines so that they don't stick or share common segments. <S> Add ground stubs freely. <S> (2) Use optical isolation and better <S> still "Total Optical Isolation" to "completely" separate signal and power grounds. <S> /to continue, ... <S> Discussion Recommendation <S> (1) Use H-bridges instead of DPDT relays. <S> Apopular choice is L298N, heavy current H-bridge include BTS7960B. References <S> Part 1 / to continue, ... <S> Part 2 (1) BTS 7960B <S> High Current PN Half Bridge Product Brief - Infineon <S> (2) BTS 7960B Half Bridge Datasheet - Infinion (3) AliExpress H-Bridge Drivers (4) AliExpress <S> Dual BTS7960 43A H-bridge <S> Driver Module - US$4 (5) TaoBao Dual BTS7960B H-bridge 43A 25kHz PWM DC/BLDC/Stepper Motor Driver Module - ¥22 <S> (6) BTS7960 43A Motor Driver Arduino Instructable - Mohannad Rawashdeh, 157,992 Views / to continue, ... <S> Appendices <S> Appendix A - Old design notes
| A full-bridge with PWM is a better way with a CNC bridge for Arduino or RPi.
|
Current leaks at 500 VDC Background: I have a small soldered joint on a 500 VDC (0.1 A) lead that's exposed to air. The edges of the joint are fairly ridged and have a few significant edges. In the higher voltage range (40 kV) these edges can potentially cause a loss in current due to the ionization of air. Unfortunately, the joint is at a point that I cannot cover with liquid electrical tape or transformer oil. Question: Will the current drop due to this exposed joint (the exposed lead and joint measure 2 cm) be significant? How much of a current drop should I expect due to the ionization of air, if any? EDIT: Here are the picture of the exposed sections. Please note that only the white cable has the 500 VDC, others are just ground or 12 VDC inputs. The two sections that are exposed to air are the connection to the board and the end section (the part covered in solder). I used a gold US dollar for reference. <Q> I wouldn't worry about air ionisation unless your 500 V circuit was exceptionally high impedance (many MΩ). <S> The wire to resistor joint would ideally have some heat-shrink insulation over it, more for strain relief and protection against electric shock than concern over ionisation. <S> The connection to the screw terminal has a large margin of uninsulated wire which looks tinned (though it might just be out of focus). <S> If stranded wire is soldered together to go under a screw terminal the solder will spread under the pressure resulting in a loose connection developing over time. <S> It's better to either leave the wire unsoldered (but with the strands twisted together) or crimp on an insulated ferrule. <A> You will not get corona discharge at 500VDC using any kind of sensible setup (like you have). <S> For higher voltages it's good to avoid sharp points which increase the electric field gradient. <S> That terminal strip (5mm pitch?) is probably not rated for 500VDC, but it's probably okay enough. <A> Current leaks at 500VDC Question: Will the current drop due to this exposed joint (the exposed lead and joint measure 2 cm) be significant? <S> How much of a current drop should I expect due to the ionization of air, if any? <S> Like other answers show, you don't need to worry about corona discharge. <S> If you worry about leakage currents, you'd better worry about leakage due to creepage , especially between the solder pads of the blue terminal strip. <S> High voltage power supplies typically are (non-ideal) voltage sources which try to maintain a steady voltage and provide as much current as required. <S> Leakage currents will therefore cause an increase in current, not a drop in current.
| The wire insulation should continue up to the edge of the screw terminal (but not inside it) or finish inside the insulated boot of a ferrule. Unless this 500VDC is a current source, a leakage current will not cause a drop in current.
|
Speed up or slow down motor of cassette recorder using a potentiometer Background : There is a cassette recorder ( RadioShack CTR-80A ) and its owner would like to have the ability to speed up and slow down the motor to record and play the sound in a distorted way. We've done a similar thing already to a Sanyo TRC-3460 voice recorder, but its diagram was easier to understand. Question : Where in the following diagram one would put a potentiometer to achieve a distorted effect (slow down or speed up a bit) when recording or playing audio from a cassette? What value should that potentiometer have? Also , could you please point me to the resource describing the symbols which are used for switches in this diagram (the ones with white and black triangles). I've search high and low and do not understand what they mean. Well, not well enough, anyway. <Q> I built this: it worked well enough to be a fun toy: <S> simulate this circuit – <S> Schematic created using CircuitLab <S> it just plugged into the 'Cue' socket ( J4 REM on your schematic) placing it in series with the motor. <S> It did not give stable or repeatable speed control <S> Any reasonably big power transistor should work eg: <S> TIP31 <S> the 2N3055 <S> was just what I had on hand. <A> This motor in the lower right corner is connected directly to the 6V supply on top right, so there is no speed control. <A> The old school recorder is designed to run on four 1.5 Volt dry cells or a simple unregulated mains transformer .The <S> nominal 6VDC motor is designed to give small speed variations for large voltage variations .This means that reducing motor volts using a 100 ohm wirewound pot or any other means will not give good results <S> .Consider changing the motor pully or getting inside the motor and modding the internal speed governor .
| In many of these old-school devices the motor speed was controlled by a centrifugal governor on the rotor of the motor, so faster was not an option only slower.
|
Is it possilbe to make up some sort of circuit that would allow you to vary the Ah provided by a battery? I suspect the answer to this question - "Is it possible to make up some sort of circuit that would allow you to vary the Ah provided by a battery?" - is "NO, Not Possible" but here goes. I want to test a device with varying Amp Hours from a rechargeable battery. For example, I have a fully charged 4.5Ah battery but I want to make it look like it only has 2Ah or 1.5Ah or some other value under 4.5Ah - is that even possible? I'd recharge the battery before each test. In effect, I want to limit the time the device would operate before the battery was exhausted by altering the Ah of the battery. This is sort of like using a timer to control the time but that does not meet my requirements. I want the device to operate until the battery is exhausted and I'd like to determine times vs Ah so that I can substitute a battery with an appropriate Ah rating for the approximate time I want the device to operate. If this is possible, can you give me at least an idea of the circuit needed? Or - Can a power supply be constructed on which you could not only set the voltage but also the Ah it would provide before shutting off power output? That would be another way I could do what I want to do. It would not actually emulate a battery in that it would produce a constant voltage where as a battery's voltage would decrease with use but it would be a way to at least approximate what I want to do - if my first idea of varying the Ah of a battery is not possible. Perhaps a device to measure the amperage draw and time and calculate the Ah and shutoff power to the device at a set Ah value?? This is all probably quite impossible. I may have to get batteries of the correct voltage with different Ah and just time how long the device operates with each battery. <Q> Take one 4.5Ah battery. <S> Fully discharge in accordance with the manufacturers instructions. <S> Charge it to 2Ah. <S> You now have a 2Ah battery. <S> It's discharge characteristics will not be identical to a real 2Ah battery, but for your purposes they are probably close enough. <S> The easiest way to charge the battery to a known capacity is with a constant current charger. <S> 1A for 2h will charge a battery to nearly 2Ah (due to losses). <A> If this is possible, can you give me at least an idea of the circuit needed? <S> Yes it is possible. <S> The circuit needed would be a digital power supply. <S> It will mostly be software doing the complicated part. <S> I suspect is should be possible to add battery emulation to an open source power supply. <S> Like the (soon to be OSHW) <S> uSupply from EEVBlog. <S> If you also want to emulate charging you'd need a 4 quadrant supply, then things get complicated. <S> You can also buy one. <S> Keithley Series 2281S Battery Simulator <A> If this is for testing realistic lifetimes with different batteries, presumably you need more accuracy than simply scaling the results as in "this battery has 1.5Ah instead of 4.5Ah, so divide the actual life by 3". <S> Otherwise you would simply measure once and scale the results for other batteries, and get a simple approximation to the correct life. <S> If you DO need accurate results you need to do much more than simply scale results, OR charge a big battery to some lower capacity. <S> First, study some battery datasheets. <S> Actual capacity varies a lot with discharge rate and discharge history, as well as temperature and other factors. <S> For example, dry cells fail early under a heavy load (like a motor), then recover some capacity while resting, then fail slightly earlier next time under load, and so on. <S> The failure can be modelled as an increasing series resistance, which will be much higher for a small battery than for a large battery at any state of charge. <S> The effect of the series resistance depends on the load ... <S> my "economical" father would run a clock for six months or so from a battery that was "dead" for any other purpose... <S> Then the suggestion of programming a power supply (connected via a programmable resistor aka "dummy load" to your application) or battery simulator to model each battery, is a good one.
| If you have a PC control power supply you can make a DIY a rudimentary one yourself with some scripts. So you first need to learn (either from datasheets or actual load tests) what your batteries do, and learn how that interacts with your load under all conditions (standby, operating, motors running etc).
|
Why aren't antenna signals inside smartphones stopped by the metal case? Our smartphones or other device are surrounded by metal, yet there are plenty of antennas inside it (NFC, 4G, Bluetooth, ...). How can it work ? Isn't the metal supposed to block RF signals ? <Q> Smartphone cases aren't metal. <S> Cheap ones are made of plastic (ABS or polycarbonate). <S> The very high end ones which have glass front and back and metal edging will often use the metal edging itself as the antenna. <S> This caused problems on the iPhone 4 when used by left-handers. <S> In that design, the metal bezel was split into carefully sized pieces that were not connected, so that some could serve as antennas of the correct length. <S> It can be surprisingly difficult to stop a phone working by putting it in a metal box, because the waves are quite short and good at diffracting out of small gaps. <S> What normally happens is it turns up the radio power to its maximum in order to communicate with the base station. <A> Only certain configurations of metal block RF. <S> A enclosing Faraday cage for instance. <S> Other configurations of metal (or other conductors or magnetics) radiate EM when driven. <S> A wire dipole antenna for instance, or the reflectors and directors in a Yagi Uda. <S> They radiate EM when electrical charges (electrons, et.al.) accelerate along the wire. <S> Slots or separations in a solid metal plane can also be used to radiate (or re-radiate) EM. <S> Most cell phones have these openings, usually of very carefully selected dimensions. <S> Slots or cuts radiate EM when charges are made to accelerate around the insulated opening. <S> Lots of other possible geometries. <A> So if you place your cell phone in a cage like that I can assure you that it's not going to be able to communicate with the outside world. <S> You can use an empty paint gallon as a cheap DIY Faraday cage, if you want to test it by yourself. <S> The simple presence of metal structures nearby the cell phone antennas won't block the electromagnetic waves. <S> That's not to say it doesn't affect the generation and propagation of the waves at all. <S> Actually, when designing the antennas the engineer has to take the presence of these metallic parts into account in order to properly tune the antennas. <S> In some cases it's fair to say that certain metal structures are actually part of the radiating structure of the antenna.
| Metal is efficient in blocking electromagnetic waves when surrounding an entire volume forming what is called a Faraday cage.
|
Internal oscillator drift and its effect on UART I am eager to know whether a 150kHz drift of internal oscilator clock (mentioned in a STM32 MCU operating with 8MHz internal clock) could destroy a UART connection? <Q> UART is very robust when it comes to clock accuracy requirements. <S> This is a consequence of the fact that each byte is processed independently and any time difference associated to clock mismatch doesn't propagate for more than 10 bits (start bit + 8 data bits + stop bit) in time. <S> This analysis from Maxim concludes that 2% in clock frequency mismatch is still tolerable. <S> In your particular case you have 150kHz/8MHz, what is equivalent to 1.88%. <S> If the other side of the link has a much more accurate clock <S> (let's say a crystal oscillator with 50ppm, or 0.005%), then the clock mismatch is below 2% and you should be OK. <S> If both sides of the link have a clock that is 1.88% inaccurate, then in the worst case scenario (one side has 8MHz+150kHz, the other has 8MHz-150kHz) <S> you may have 3.8% of clock difference, in which case you run the risk of starting to see problems. <A> No, because the UART stands for "asynchronous". <S> You have somwhere in the datasheet a ratio baudrate vs sampling clock, so it is oversampled, then filtered (decimated). <S> The asynchronous clock is reset once the start bit is detected, so there is plenty room for clock inequalities. <S> For the standard oversampling ratio of 16 , the clocks can be off by +/-5.11 <S> % ( Source ) <S> For your case : $$E=\pm 0.05=\dfrac{X}{8M}$$ <S> $$X=\pm 400kHz$$ Supposing that two similar devices communicate over UART, then each device can deviate maximally for +/-200kHZ, which is within tolerance band of proposed MCU. <A> However that error should include initial oscillator error, drift error (temperature, Vcc, time etc.) <S> and any error in divider digital ratio.
| Especially if you have a crystal-controlled UART at the other end, you are reasonably safe with as much as +/-5% error.
|
MOSFET source measures 12V, still nothing will turn on Please, bear with me, I'm a real rookie at electrical engineering. So please be easy on me. ;-) I've got the following circuit. I have a 5 seconds HIGH / LOW interval on D5 and am seeing the voltage drop to near 0 and rise to 12V when measuring with my multimeter on the terminals of the speaker. Still, my Piezo speaker won't make a sound. When connecting it straight to the 12V source, it works as perfectly. What am I missing? Current at the source of the MOSFET I guess? But, why? Did I connect it wrong? P.s. also tried connecting a LED with a 470 ohm resistor after the MOSFET, but that does not turn on either. And this also simply works directly on the 12V power source. Transistor used: IRF520N ( datasheet ) Voltage Regulator used: L7805CV ( datasheet ) Piezo speaker used: BU2 M28 ( datasheet ) <Q> First, you should start by drawing the schematic in the "standard" way: from left to right. <S> The first item should be the power supply and then the rest of the components. <S> Also try to isolate the circuits in small functional parts. <S> I mean, one can suppose that the Wemos and the 7805 will work as expected. <S> So I would draw the schematic for your questions like this: <S> Here you will see that the N-MOSFET is not in the lower side <S> so First problem <S> : You are using an N-MOSFET as a high-side switch Changing the components, and placing the IRF520 as a low-side switch gives us this circuit <S> However, the IRF520 if not guaranteed to work with 3.3V of the Wemos so it may or may not turn on. <S> The Vgs of the IRF520 is between 2 and 4 volts, sometimes 2, sometimes 4. <S> To sum up: Problem 1 <S> Your IRF520 should be used as low-side switch Problem 2 <S> You will have to use a buffer for the IRF520 or use another transistor. <A> I see 2 problems: You seem to be trying to use an N-channel MOSFET as a high-side switch. <S> To avoid a complicated driver, N-channel MOSFETs are more easily used as low-side switches. <S> (As commented by DKNguyen , the existing connections are wrong, even for a high-side switch.) <S> That MOSFET has a Vth of up to 4V. <S> The Wemos D1 mini Lite documentation states that it has 3.3V logic level outputs. <A> You could use a pull-up resistor divider on the MOSFET gate to say 10V, and use another transistor like a 2N7000 (yes, high Vth, but it doesn't have to sink much current), or even BJT 2N3904 or whatever, to pull the gate low (inverted). <S> An opto-coupler also works. <S> Most logic-level MOSFETs for 3.3V with a 1V or less <S> Vth are going to be surface mount.
| So you need a "logic-level" MOSFET, with a much lower Vth.
|
How to damage the components with ESD? I have a task to present the ESD safety practices in lab and at work in general. I want to make this interesting and for the freshers out of college, I want to let the importance of ESD precautions sink for long time. I am planning to spoil the devices as a part of demonstration. Question : How can I do this? The image tells me to start with FETs.. may be a simple timer and a LED connections and a FET. I am open to use other components as well. <Q> ESD damage can be very subtle and not easily demonstrable. <S> The eventual failure may take years to surface. <S> Or is only measurable by some change in parameters. <S> Exmaple, only visible by electron microscope: https://nepp.nasa.gov/index.cfm/6095 <S> Or here: <S> http://www.ti.com/lit/an/ssya010a/ssya010a.pdf <S> I do not think a demonstration will be very effective. <A> As I recall, the old 4000 (not 4000B) series ICs were quite sensitive to latchup and ESD damage compared to more modern parts. <S> They're probably fairly hard to get ahold of in 2020. <S> You could try a standard CD4001B for example, cheap and easily socketed for replacement. <S> You can start with the usual human body model and make it less friendly (lower resistance, more capacitance, higher voltage) as necessary. <S> For demonstrating damage, I suggest a SPDT switch to a gate input with a high value resistor in series (maybe 22M or more), and the output driving an LED, so that any damage that causes somewhat increased leakage to either supply rail will be evident. <S> Add a series resistor to the Vdd <S> so latchup doesn't kill the chip if you want to zap it with the power applied. <S> Note that the ESD discharge that is guaranteed safe may be considerably less than what is required to typically cause measurable damage. <A> In our electronics lab at university somebody thought it was a good idea to buy cheap plastic chairs. <S> Every time you stood up you would get zapped by sparks up to about 1cm in length. <S> Painfully. <S> So, find an old chair like that, and get someone wearing nylon or some insulating polymer clothing and insulated trainers to sit on it. <S> Then get them to stand up and hand a circuit board to someone who is already grounded. <S> The spark to circuit board will definitely not be good for it. <S> [Although I have seen boards survive that] Then tell them it is good practice to touch the hand of the person you are handing the board to to discharge any static first.
| Charging a small capacitor to high voltage and discharging it through the part with a low value or no capacitor may be able to demonstrate the damage.
|
Do capacitors waste power as heat or whatever on AC? Capacitors store energy then give it back once required. A perfect capacitor is nearly lossless on DC power because you only fill it once then it keeps energy in it until you discharge it so no power loss to mention, but on AC the capacitors will be charged then discharged all the time which in my theory seems to waste power, is that true? If it is true then how can I measure that loss for a 250V 8000µF capacitor connected to 170V AC power line? <Q> There would be power loss because with real life capacitors, there are parasitic losses, this means that the capacitor can not act as a pure capacitive load in real life, this is mainly because of the building process/materials/sizes. <S> In real life capacitors have an ESL (Equivalent Series Inductance), an ESR (Equivalent Series Resistance), and a Leakage Resistance in parallel with the capacitor which is commonly notated as Rleak . <S> You would need to know the ESR to calculate power loss. <S> But keep in mind that this is a parameter that degrades with component use. <S> As per the industry standards EIA-463-A, MIL-C-62F, under nominal operating conditions, capacitors are considered completely degraded and not usable in the circuit when it's ESR value reaches 2.8 times of the initial ESR value and capacitance decrease in excess of 20% of the initial value. <S> Link to original text. <A> There are three loss mechanisms within the capacitor, all of which are fairly minor, and one that it causes to the power supply, which depending on how you're billed for your electricity, may or may not worry you. <S> Within the capacitor, the electrodes have resistance, which causes \$I_{terminal}^2R_{electrode}\$ losses. <S> The dielectric has a conductivity which is usually very very small indeed in plastic dielectrics, more significant in electrolytics, also causing \$I_{leakage}^2R_{dielectric}\$ losses. <S> The dielectric also has hysteresis loss, also negligible in most plastics, more significant in electrolytics. <S> The charging/discharging current flows in your supply lines, but there's no nett power transferred to you by this, so most domestic meters will not charge you. <S> This is usually called reactive power, or VAr. <S> Commercial customers may be charged for VAr, as it causes the electricity infrastructure supplier to have invest capital in thicker supply cables than would otherwise be needed. <S> However, as most commercial users tend to have inductive loads like motors, capacitors across the line will tend to reduce their VAr. <S> This is why there is a market in large power factor correction capacitors to be used in shunt with large factories. <A> An ideal capacitor is lossless, the energy released during discharging is equal to that stored during charging. <S> However: Real capacitors are not ideal, the losses in a real capacitor on AC will likely be much higher than on DC, because on DC you only have leakage losses, while on AC you have losses caused by currents flowing in and out of the capacitor. <S> The plates of a capacitor are usually very thin and so can have noticeable resistance. <S> Even if the capacitor itself was lossless, the current flow caused by the capacitor can change the losses elsewhere in the system. <S> In the simple case consider a capacitor connected to the grid by a long cable, current flow will cause resistive losses in the cable. <S> On the other hand currents from capacitors can cancel with those from inductors, so it's possible for an appropriately sited and sized capacitor to actually reduce losses in the cable feeding an inductive load. <A> No. <S> Pure capacitance does not waste power as heat. <S> Which is why inductive loads such as motors or fluorescent light ballasts are often compensated with capacitors for the loads to look like pure resistance to achieve better power factor.
| However, non-idealities such as series resistance and dielectric losses do consume energy, so practical capacitors you can buy do waste energy some amount.
|
What's an easy way to connect multiple pins between multiple breakout boards? I'm building a circuit by screwing multiple breakout boards into a wooden panel and using jumper wires to connect header pins between different boards. Most of the time this works fine, but sometimes I need to connect a pin to more than one other pin. This can easily happen with i2c, power, or ground. On a breadboard this would be easy, but I don't have room for one, or at least not full-sized. I could also use solder to bridge some header pins together and make a sort of bus connector, but I don't see a nice way to mount it to keep it moving around. Or maybe I could solder multiple wires together, but that seems likely to result in a tangle. It seems like this is something people would commonly want to do and there should be a more elegant way, but I don't know the search terms to look it up. What am I missing? <Q> Add in even more breakout boards to serve as connective splitting junctions. <S> The strip patterns are probably longer than you need so use a knife to cut them up into smaller segments. <S> Busboard <S> If using male headers or solid-stranded hookup wire, then you can just buy small miniature solderless breadboards. <S> Solarbotics <A> What I've done in the past is to take multiple jumpers, cut the ends off, strip and solder them together. <S> Add some heatshrink. <S> Basically a jumper-wire version of a standard electrician's connection (they would use wire-nuts). <S> (e.g., https://www.do-it-yourself-help.com/how-to-splice-electrical-wires.html ) <A> You could use a piece of solid wire as a bus bar and use electrical tape to insulate it after you soldered the taps on it. <S> Just slit the tape at the tap locations and wrap it along the length of the bus bar. <S> I would use a 16 or 14 ga wire if you have any house wire laying around.
| If you are using female crimp housings at the ends of your wire then solder rows of headers onto little protoboards with a strip pattern.
|
Type A USB 3.0 with Type C USB 3.0 orientation: What if I use a USB A - to - USB C adapter cable, with the A side plugged into a USB 3.0 Type A receptacle, and the C side able to accept a slave device, say perhaps a USB C flash drive. Will the C side of the cable still be rotationally symmetrical, as would be the case for using type C USB 3.0 throughout? My guess is no, because there are no CC pins on the Type A 3.0 connector, and only one set of Tx and Rx pins. <Q> The Type-A receptacle doesn't have CC pins, it is true. <S> But the CC pins are present in the other end (Type-C) cable. <S> The A-to-C adapter you described is called "legacy cable assembly", where the Type-C end represents host functionality (because Type-A receptacle is normally a USB host). <S> To make the Type-C cable end to look like a host, the C-end includes a 56k resistor pulled to VBUS. <S> The resistor is mounted inside the Type-C overmold. <S> This signals to a Type-C device that a standard 500/900mA host is behind the cable, and everything works, in both orientations. <A> Yes, it works. <S> There is a CC pin on the USB-C side of the cable, and the cable connects it to VBus through a termination resistor (as the host device would do for a C-to-C connection). <S> This lets the downstream device know its intended role, as well as the orientation of the cable. <A> Adapters with C sockets are not allowed by the standard, if someone was to make such an adapter anyway, then you are likely to find that superspeed only works in one orientation (or doesn't work at all).
| If a USB 3 cable with an A plug and a C plug is used, then it is up to the device with a C socket to determine the orientation of the cable and perform usb 3 signaling on the appropriate pins.
|
Will a capacitor draw charge/current if it's not feeding a load? I have a number of capacitors connected in parallel, as in the following circuit: simulate this circuit – Schematic created using CircuitLab If the switch is open, will the caps still draw charge/current? Will this only happen when the load draws current from/through them? (Is the load required for the caps to charge?) I have a basic understanding of how caps work and don't have an LCR meter or scope with which to test this. <Q> By them selves the capacitors will have slight self discharge and with the battery connected there will also be a small leakage current. <A> If the switch is open, will the caps still draw charge/current? <S> Yes, it will be very small current. <S> See below for an example. <S> Will this only happen when the load draws current from/through them? <S> (Is the load required for the caps to charge?) <S> When you close the switch the load gets connected, depending on the amount of current drawn by the load, the capacitor might contribute to the load current too. <S> If the 9V supply is very far (inductance of the cable or with a higher internal resistance). <S> Soon after load current is reduced, the capacitors will charge back to the supply voltage. <S> And then, only leakage current seen above will be there. <A> Theoretically, zero current will flow when the switch is open. <S> This is because once the capacitors fully charges up, they will not conduct any further. <S> Or you can say, Capacitor blocks DC. <S> Capacitor don't need a Resistor to charge. <S> They charge theoretically instantly as soon as you apply a voltage to them. <S> Resistors are mostly included, and that too in series, to delay the charging time of capacitor. <A> Yes, since real capacitors do have a finite resistor parallel to the capacitance. <S> The value of this resistor depends on the type and quality of the capacitor. <S> Electrolytic capacitors as shown in the circuit usually do have a low resistance compared to film capacitors thus drawing more leakage current. <A> simulate this circuit – Schematic created using CircuitLab Figure 1. <S> OP's circuit redrawn. <S> If we redraw the circuit to lay it out in a conventional fashion (reading left to right with positive supply at top <S> it becomes a little more obvious what is going on. <S> If the switch is open, will the caps still draw charge/current? <S> It should be clear now that there will be an initial current to charge up the capacitors. <S> As they charge the current will fall reaching zero when fully charged. <S> Closing the switch will not affect the capacitor current. <S> 2, Will this only happen when the load draws current from/through them? <S> (Is the load required for the caps to charge?) <S> With the redrawn schematic you can see that the capacitors will charge whether the load is connected or not. <S> Note 500 kΩ on a 9 V supply will limit the LED current to \$ <S> \frac {9}{500k} = 0.018 \ \text <S> {mA} \$ <S> which isn't enough to light an LED. <S> 1 kΩ would give about 7 mA to split between the two LEDs but usually we would use one resistor per LED to balance the currents or connect the two LEDs in series to reduce the current required by the whole circuit.
| Because you resistor is not in series but in parallel to the capacitor, it will charge instantly. Practically, a negligible amount of leakage current flows through capacitors, which may be insignificantly low for most applications.
|
Simple CMOS switch with weird voltage in LTSpice I am new to all this, so please forgive me, as I feel the answer to this must be so simple. I tried to build a basic CMOS switch. It has a supply of 5 V. There are periodic pulses of 5 V supplied to the input. I expected the output to alternate between 0 V and 5 V. After running the simulation, I was surprised to find out that it alternates between 0 V and 620 mV. Either I am not understanding things correctly, or there is something wrong here. Why am I not seeing 5 V on the output? <Q> You should rotate the NMOS 180 degrees as shown in left picture. <S> But to draw a schematic correctly, you should actually put Vcc on top, and GND at the bottom (right picture). <S> Note the pictures for the mosfets in LTspice have the both an arrow. <S> If you cut the arrow loose from the gate and reconnect it to the other terminal of the mosfet (drain) the arrow shows how the body diode is connected (shown in red in left picture). <A> Try this way <S> simulate this circuit – <S> Schematic created using CircuitLab <A> The ONLY functional error is the NMOS source connected to the output instead of ground=0V which logically must at the bottom of the schematic and +ve supply at the top. <S> There is slight cross-conduction at Vdd/2 but both body diodes are always OFF. <S> When we have complementary transistors arranged this way, we call it an inverter, as “ All single transistor switches (of all types) are always inverting” and this example has 2 switches not one. <S> So it is a misnomer to call the circuit a switch. <S> In a different arrangement with complementary drivers we can make both conduct or not, then we call its function an Analog switch. <S> So 3 corrections in all; functional, logical presentation and semantics Good question, but learn to verify your assumptions for accuracy. <S> When I started in ‘75, I always had an EE who was not as technical, yet far more careful to check my drawings <S> and he would often find one or sometimes many things in error or missing. <S> A good drawing is understood by any level of skill and obeys the polarity of common sense. <S> “P ‘drain’ supplies from Positive, N ‘source’ Sinks from Negative.”
| The body diode of the NMOS is conducting, clamping the voltage to 620 mV when the PMOS is turned on (also conducting).
|
Why Relays waste power and how to overcome that? Let's say that I am going to: Use a relay to switch between 2 DC +12V power sources (one may interrupt so the other source is a backup). I will use a capacitor on the load with minimum leak possible I can find to supply power during the relay switch operation. The lines thikness is around 8 mm. The current may go as high as 23A and as low as some mA. I have been told that relays waste power, but it is not clear to me by how much as I don't know the best kind of relays to use for my application. I also read somewhere that an IC can be used to control relays so they waste less power, but that IC may also use power too which may result in more power loss probably. Anyway, I want the most efficient solution for my specific applications. <Q> Coil-held relays do indeed "waste" power as you say. <S> However, you're paying for some value. <S> The value is: <S> It can make a connection even when fully de-energized. <S> In fact, at total loss of power, it makes a connection. <S> These are the "NC" contacts. <S> This makes it a good choice for backup power; you connect normal power to both the coil and the NO contacts. <S> You connect emergency power to the NC contacts. <S> Positive interlocking. <S> There is no case where NC and NO contacts are ever connected to each other. <S> That is absolutely mandatory in certain power transfer applications due to the need to prohibit backfeeding. <S> Positive multi-pole throw. <S> There is no case where pole 1's NC contacts can be connected whilst pole 2's NO contacts are connected. <S> That serves a whole bunch of applications where this must be positively so: to name a few, Power transfer as discussed reversing Series-parallel switching <S> Remember, the relay does all that, and, when coil power is entirely lost, the relay actively throws with no power or assistance whatsoever. <S> Now, if you are willing to sacrifice the auto-throw-on-power-loss functionality, you can get all of the above with a latching relay where you throw with a momentary impulse, and then the relay throws. <S> However, the relay will remain in this position indefinitely; it takes active action to throw it back. <S> Further, the 'active action' is a bit hard to manage. <S> You need an impulse of energy for some interval of time. <S> So you need to have electronics that sends the pulse for the required time interval to throw the relay over fully. <S> Worse, you can't use extra contacts for that, because most such relays do not have the special contact needed: which would need to be closed until the relay is almost at end of throw. <A> Look for a relay (Farnell, Digikey, Mouser, RS etc.) <S> that can handle a DC current of circa 25 to 30 amps through the contacts. <S> It's important you choose the "DC rating". <S> Then reduce your candidate list to those requiring a coil voltage of 12 volts. <S> Then reduce your list further to those that require the least amount of coil current to meet your efficiency needs - the coils waste power when activated. <S> Then decide what your switchover circuit will consist of i.e. op-amps, comparators etc. <S> and decide how much power this circuit might waste (it should be a fraction of what the relay coil wastes by the way). <S> If your efficiency still looks poor then maybe you go for MOSFET switching instead of relay switching <S> but you might only get something like up to a 10 times improvement in power losses on light to moderate loads and maybe only a 2:1 improvement on your heaviest load for this extra complication. <A> Relays need lower current to hold the state once turned on. <A> As mentioned already, there is a big danger of sticking/welded contacts if an AC rated relay is used for DC. <S> Above 15A, a contactor is normally better suited. <S> Another solution would be a circuit with 2 or more Power MOSFETs, which are controlled by voltage and not current - <S> the tiny control current is neglegible (nano or micro Ampere). <S> But there is some more lost since the on-resistance is higher compared to a contactor. <S> And special emphasis should be made to the order of switching, i.e. can both sources be parallel during switch over or is a "break before make" needed in the MOSFET control circuit.
| There are latching relays which only need a pulse input to switch the state So a solution would be to search for a latching DC contactor rated for 25A , but those might be hard to find.
|
Why lines A and B are HIGH when idle (rs485) I'm using modbus RTU over RS485 for a project.I use a RaspberryPI as master and Arduino Pro Mini as slaves. My problem is the following :I plug a sensors on the bus and it returs its value.I plug another sensor, this one returns its value but the first sensor doesn't return its value anymore.So it looks evident something makes them incompatible. I tried to diagnose it using a logic analyser and the results are the following :When idle, the lines A and B from the master are respectively HIGH and LOW.When idle, the lines A and B from the slave are both HIGH. I wonder if the line B shouldn't be LOW..Looking why this line was HIGH, I discovered that the line is pulled HIGH because of the library I use on Arduino. This library pulls DE and RE HIGH when transmitting data, LOW otherwise.(On the max485 module, I tied DE and RE together. I'm not sure it's a good practice) Here is the library I use on arduino : https://platformio.org/lib/show/1727/ModbusSerial Here is a schema of a modbus slave module : EDIT : On logic analyser (1 slave connected to the master) : I tried with 2 differents usb/rs485 transceivers Transceiver 1 - 3 seconds record Transceiver 1 - Zoom on request / response Transceiver 2 - 3 seconds record Transceiver 2 - Zoom on request / response So my question is: Is it normal to get HIGH on both lines A and B ?And if not, what can I do to correct it ? Thanks in advance !! <Q> RS485 (TIA485) can be tristated when idle. <S> If it is, the terminations would determine the state of the lines. <S> It may be that the Pi terminates them differently than the Arduino. <A> When idle, the lines A and B from the master are respectively HIGH and LOW. <S> When idle, the lines A and B from the slave are both HIGH. <S> This is because the slave is set to receive data and the RS485 chip has no control on the lines . <S> External pull-ups are the likely cause of this. <S> If master and slave were connected then master (although being in idle) would drive the line high and low and, you would see high and low at the slave. <A> Your connection to /RE and DE is incorrect. <S> If you check the datasheet of MAX485 & friends, you'll see that /RE is active low, but DE is active high. <S> If you connect a pull down to both of these pins, one of them will act up. <S> Simply tie /RE to ground and DE to Vcc.
| That's because the RS485 interface is outputting an idle condition i.e. B is the inverse of A because the transmitter is enabled.
|
How much of the loss of battery capacity in higher currents is caused by its internal heating? For example this page give this data for one AA battery: 0.5A : 1.84Ah1A : 1.411Ah2A : 1.18Ah Is it mainly b/c of the internal heating? <Q> The temperature rise actually helps generate more power because the chemical diffusion and reactions are faster at higher temperatures (between 20 degrees and 30 degrees C <S> the capacity goes up by 10%). <S> The loss of capacity is due to reaction products clogging up the surfaces, side reactions, and limited diffusion of reactants. <A> Not quite internal heating. <S> But pretty close. <S> It is due to the internal resistance bleeding off quadratically more power as heat with a linear increase in current draw. <S> This will cause the battery to heat more, but you will see a decrease in capacity even if you if you magically keep the battery at the same temperature. <A> The rate of discharge is directly proportional to the internal Ohmic losses inside the battery, because of which, the temperature of the battery also rises (decreasuing the output voltage in loaded condition). <S> So, the available capacity reduces. <S> same case with the aging of the battery where for a same contact load, there will be higher drop (loss) across the older battery than the new one. <S> Source: https://batteryuniversity.com/learn/article/rising_internal_resistance <A> What you are showing here is loss in capacity that is related to but not the same as the loss in energy. <S> Fresh chemical need to diffuse into the active region of the cell for the voltage to be maintained. <S> A secondary reason for the capacity loss at higher currents is that the end-point is defined at a specific cell voltage - <S> because of the drop across the cell's internal resistance will increase with the discharge current this will occur sooner at a higher discharge current. <S> For lead-acid batteries the capacity follows Peukert's Law .
| The main reason is just that chemical reagents in the cell have limited speed of diffusion - putting an electrical load on the cell causes a chemical reaction.
|
Why is a horizontal rod a full wavelength oriented radially to a broadcast antenna not the best receiving antenna I have always struggled to get a good physical understanding of Electromagnetic waves. I've done the whole Maxwell's equation approach, wrote boundary element solver tool but cannot rationalize that with a phenomenological understanding of E/M. The waves propagate radially from a broadcast antenna, so you have a time/distance varying electric field in a radial direction. A full-wave horizontal rod with feed taps at 1/4 and 3/4 of its length seems like it should be the best receiver. Help me get a better understanding, if I am wrong. <Q> The waves are not only varying as they propagate radially, but, for a vertical transmit antenna, the EM field will vary vertically as well. <S> That's because a vertical antenna has finite height, thus the top and bottom are not radiating at the same phase (due to the speed of light of propagation from the bottom to the top of the antenna). <S> Thus the radiating EM field will be vertically polarized, and the highest gain will be with an antenna of that same polarization (thus accelerating receive electrons in the same orientation as the accelerating electrons generating the EM waves). <S> There will be some reception with a radially oriented antenna, that's the basic theory behind a long wire Beverage antenna. <S> But the gain will be lower because the pair of antennas will be trying to accelerate transmit and receive electrons orthogonally to one another; and the field propagating from the top and bottom of the transmit antenna will be seen out of phase and thus partially cancel at the receive location. <A> It is true that the variations in field strength propagate radially from the transmitter. <S> However, as your diagram shows, the fields themselves are vector fields, and the orientation of the vectors (both electric and magnetic) is at right angles to the direction of propagation. <S> They have no effect on a conductor oriented radially. <S> In a comment, you mention: <S> Now, I'm thinking about the receiving antenna as a wave guide. <S> Yes, that's a good analogy. <S> In fact, that's the principle behind the Beverage antenna alluded to by hotpaw2 . <S> It works only because it is physically close to the ground, which makes it part of a transmission line — a kind of wave guide. <S> The propagating E field induces a voltage between the wire and the ground. <S> This wave of voltage moves along the wire at roughly the same speed as the radio wave, reinforcing signals coming from the correct direction. <S> Signals arriving from other angles interfere with themselves destructively. <A> Second, because the directivity of a lambda dipole is an —8— with the dipole as its mirror plane.
| First, because the broadcast tower sends out vertically polarized waves and your receiver antenna is horizontally polarized.
|
getting access to more UARTs on micro controllers I will need to interface a micro controller to some breakout boards that only talk through UARTs, so confronted with the eternal 'need more hardware UARTs' problem. I have seen 2 kind of solutions out there: add a barebone ATmega328P or similar, connected by I2C to the main micro controller, and use it as an UART. Sounds like using a hammer to kill a fly. find a dedicated barebone micro controller breakout for doing this, for example: https://www.instructables.com/id/SPI-to-4-x-UART-Bridge-MULTIUART/ This is a bit the same as using a barebone ATmega328P, but at least I get 4 UARTS. I looked for established shops where to buy these already assembled but could not find. Any suggestions? Any other idea? Is there really no hardware bridge I2C <-> UART available as breakouts out there? <Q> There are 7+ mainstream STM32F0 MCUs with 8 USART interfaces. <S> There are several high performance STM32F4/F7 chips with 4 USART + 6 UART (for 10 total) <S> interfaces, and three dozen chips with 4 USART + 4 UART (for 8 total). <S> I am pretty sure you can find similar options from other manufacturers if you stop looking for DIP package. <S> The adapters from LQFP to DIP are freely available and cheap. <A> This does not require dedicated UART hardware. <S> Note however, that transmitting on a bit-banged interface takes more time than a hardware interface. <S> If you use Arduino, this might be of interest to you: https://www.arduino.cc/en/Reference/SoftwareSerial <A> Solved problem. <S> There are bridge chips that can be controlled over I2C or SPI and have a UART. <S> Examples for the phrase "i2c uart ic": <S> SC16IS750/52/60/62 <S> MAX3107 <S> XR20M1172 <S> (double UART) <S> Example breakout: http://www.aliexpress.com/item/32772833676.html <A> The STM32F413 looks like it has 10 UARTS! <S> https://www.st.com/en/microcontrollers-microprocessors/stm32f4-series.html <S> Even better for you, it looks like the STM32F413 is available on a Discovery Board, <S> https://www.st.com/content/st_com/en/products/evaluation-tools/product-evaluation-tools/mcu-mpu-eval-tools/stm32-mcu-mpu-eval-tools/stm32-discovery-kits/32f413hdiscovery.html , meaning you don't have to worry about interfacing. <S> I'll leave it to you to verify that enough UARTS are broken out on the Discovery Board.
| You can also bit-bang a UART interface in software.
|
Controlling cardreader numpad input with microcontroller I am trying to automatically input a pin code into an electronic card reader so I can automate the process. First thought was to build a small robot finger and control that to input the code, complexity vs benefits was too high for this option. Second thought is to directly deliver power to the pressed buttons with a microcontroller. Assumption : when a button is pressed, a switch is closed so current can flow through the circuit. Question : how can I measure the current when a button is pressed? Where do I connect the multimeter on the electronics shown below. Assumption : knowing the amount of current a pressed button delivers to the circuit, I can deliver the same amount directly via a microcontroller. Question : where - given the layout of the board below - do I connect the microcontroller to each button on the circuit? I am a computer science engineer with limited knowledge in electrical engineering, sodirections and resources to help me solve this problem are greatly appreciated. If you think there is a better way to being able to input numbers softwarematically into this card reader, do let me know! The card reader looks as follows: Electronics look like this: With the rear of the buttons looking like this: The rear: <Q> Those inputs are probably matrixed, since I don't see any obvious ground traces. <S> Matrixing means that you drive one side of some switches low, and read the other side to see which one is pressed, then drive some others low and read them with the same input lines. <S> Matrixing allows this particular keypad to be read with 9 I/Os instead of 20, but it means driving high and low in an uncoordinated manner can result in errors. <S> You could figure out the matrixing sequence--which side of each button is driven vs. which side is read, and use AND or OR gates to handle the multiplexing--but it would probably be much easier to use analog switches driven by your controller. <S> Those would simulate the operation of the mechanical switches. <A> Yes, you could possibly measure the Voltage and apply it to the corresponding Button, but: <S> It's easier than you think:The black circles on the inner side of the Buttons are made out of conductive material, wich close the circuit on the PCB. <S> You can exploit this by using some sort of "external Button" to activate a Button. <S> You can see four conducting rectangles for each button, where both on the top and both on the buttom are connected together. <S> My best guess is to solder two thin Wires (one to the upper and one of the lower) to all Buttons each. <S> If you hold them together, the corresponding button should be activated. <S> To achieve this with a microcontroller you can use some sort of a Relay (wich would work), but you should use something like a "Optocoupler", because you would hear the clicking of the Relais and Optocouplers are cheaper. <S> You'll need to connect both wires of a button to one Optocoupler and the other side to a microcontroller like an Arduino and GND. <S> (Just an Example for an Optoucoupler) <S> https://www.ebay.com/itm/30PCS-PC123-DIP-4-Optoisolator-Photocoupler-Optocoupler-Triac-Driver-IC/401774709561 <S> You can use a multiplexer if yout microcontroller has not enough Pins. <A> I'll somehow extend Cristobol's answer. <S> #1. <S> More hardware, less software <S> If you can spend 21 I/ <S> Os (assuming you want to simulate key presses on all the keys) you can use the CD4066 IC. <S> It is an analog bilateral switch and each package includes four of these. <S> You can connect every key's two contacts to one of those bilateral switches and activate the switch using a single I/O. <S> This is a generic solution and you don't need to worry about the keypad being multiplexed or not. <S> It will also cost you a few ICs, microcontroller I/Os and ~100 solder points. <S> The software is as simple as it can be: you simply time some pulses on the correct pins simulating the key presses. <S> #2. <S> More software, less hardware <S> You most probably have some sort of multiplexed array. <S> The traces I've highlighted below suggest common row connections between keys. <S> Multiplexed matrices look like the image below. <S> Just like Cristobol said, scanning it involves actively driving one side (either columns or rows) and reading the other side (rows or columns, respectivelly). <S> You might want to attach an oscilloscope to it and figure out which traces are being driven. <S> After that, with a multimeter, try to understand how these keys are connected between each other. <S> If the traces are being driven slowly enough (say, <S> 1kHz) you can wait for them to be selected (reading with the microcontroller) and then writing on the specific lines being read by the keypad controller. <S> You'll also need to figure out the polarity of this "select" signal, and if the lines being read have pull-up or pull-down resistors. <S> This will require minimal external hardware (only the microcontroller and some wires, assuming the keypad controller and your uC operate at the same voltage) and about 20 solder joint (9+1 for colums/rows + GND on the keypad, then double that on the microcontroller).
| The analog switches can be connected across the top and bottom (redundant) terminals, or if your soldering skills are very precise, connecting across the fingers is also an option.
|
Air circulation and condensation in custom enclosures for outdoor project I am designing 6 custom enclosures in mild steel to be placed outdoors for 5 months in London UK.They will be interconnected, only one will contains 4 power supplies, all with conformal coating and keeping cool (don't overheat after numerous testing). There will also be two Raspberry Pis inside.Up until now I thought I would use Gore vents to avoid condensation. Unfortunately the enclosures won't be perfectly sealed, and the enclosures will be painted with a dark colour.What type of fans (or else) should I install in order to:- provide decent air circulation- avoid condensation (fan heaters?) Should I add vents? Also, as the boxes will be connected with holes through which I will run cables etc, should I install such vents in each one of them or just in one of them? <Q> Vents do not prevent condensation. <S> Condensation is prevented by keeping he temperature inside the enclosure above the dew point. <S> The only way to do that is to keep the temperature inside of the box higher than the temperature outside the box. <S> To some extent that will happen naturally for any box that contains something that generates heat. <S> Everything that uses electrical power generates heat. <S> You need to determine the maximum safe operating temperature for each item in each box. <S> You need to determine how much box surface area is required to dissipate the internal heat at the temperature differential between inside and outside that is tolerable. <S> If the boxes are exposed to the sun, you need to calculate the heat gained by each box due to the sun. <S> You need to determine if there will be a problematic temperature difference between areas in each box. <S> If that is a problem, a fan may be needed to circulate air in the box. <S> You may need to design each box individually according to what is inside. <A> If usage time is limited as you mention, you can just use some desiccant to keep humidity within acceptable limits. <S> Let's say your biggest box (180 l air Volume) will suffer a temperature difference of approx. <S> 20 K during a day. <S> It then will cycle 5% of its air content through openings of the housing. <S> This is approx 10 l. <S> Let's say the incoming air has always 100% rel. <S> humidity at an average temperature of 15 °C and you want to bring it down to 60% to prevent any condensation. <S> So you have to remove 40% of 13 mg/l from the air, resulting in 52 mg water every day. <S> Given 150 days that's only 8 g water. <S> If you use silica gel, you should not expect it to take up more than 10% water under non ideal circumstances, so 80 g silica gel should be sufficient to keep the box dry. <S> Considering the dimensions of that box, you probably can put 10 times that mass inside to cope with eventualities and false assumptions. <A> I do not agree that ventilation does not play a role. <S> It also plays which one. <S> When there is airflow then condensation evaporates faster. <S> The holes will need to be done on the sides, and most importantly from below, so that moisture can drain. <S> In a short time, you can determine how much moisture is formed inside the box and take action. <S> And I would put a small fan to blow air so that the circulation goes faster.
| You can safely install a dehumidifier in each box to protect yourself. You need to determine how much heat is generated inside each box.
|
How to deal with D+ and D- on the end of a device that will be charged only? This is the opposite of the question posted here: What is the ideal way to handle data pins D+ and D- on a USB power adapter to be compatible with fast charging on devices? While that question is focused on how to design a charging port so that devices will be charged from it, I'm curious about designing an end device that I want to be charged with maximum current from all potential ports, if I don't need to use the data ports of the USB for anything else. My device has a Li-Ion IC that self limits and can charge my Li-Ion battery up to 2A. How do I tell chargers to give as much current as they can? If I short D+ and D- on my board, I will appear as a dedicated charging port. Does that mean that wall outlets will give 1.5A or more if they can, and that laptop hubs or other sources may limit out at 500 mA? Is there any chance of a proprietary charger requiring something else other than D+ and D- being shorted in order to maximize charge current? Edit: I'm looking to make my own variant of this PCB from Adafruit . While the part is made to only supply up to 500 mA (despite the schematic labeling), I want to make a version that can charge a 4,000 mAh battery at 0.5C. I also want to remove the USB connection to the AtMega so it can't be reprogrammed. So take the BQ2425 series for example. I know this is also a "power management" chip which also takes care of regulation in addition to battery charging. But there aren't any equivalent MCP73831 chips I could find that do simple battery charging, just with a higher current. The point still stands though. If my USB is meant to sometimes pull 2A if the chip is charging, or much less if it's fully charged and maybe just powering the MCU, what do I so with the USB D+ and D- lines on my PCB? Already the two answers I've gotten don't agree, one says short them and one says leave them open. <Q> In the answer you linked, the DP/DM termination is shown for the ‘host’ (downstream facing) port, be it a regular USB or a power-only port. <S> This termination is how an endpoint can determine the host’s charging viability without having to enumerate as a USB device: <S> the ’device’ (upstream facing) port senses the terminator and limits its current draw accordingly. <S> The exact details of this are a bit of a mess - there’s the USB way, the Apple way, the Sony way, etc. <S> Fortunately, for a simple power connection you don’t have to do anything in the endpoint. <S> Just leave the DP/DM pins open. <S> The host’s current limiting will kick in if your device exceeds what the downstream-facing port can deliver. <S> More here: https://www.maximintegrated.com/en/design/technical-documents/tutorials/5/5801.html <A> The linked question and the answers really cover the topic (my opinion). <S> The shorting of the datalines is not universal but works. <S> Going forward, if your device is still in a stage where you can modify or add some circuits er using a standard charging port IC. <S> One such example is http://www.ti.com/product/TPS2511 To be able to charge from most of the host/charging slots consider using a dedicated controller. <A> The device to be charged (let's call it DUC - Device Under Charging) cannot draw more than 500mA from an ordinary USB 2.0 port of a PC, probably the controller on the MoBo will limit. <S> Wall type chargers <S> /adapters can provide 2.4A or even more. <S> So the DUC should be smart enough to detect the port type that it's plugged in. <S> You may want to take a look into BC 1.2 Specification . <S> Or, simply, read the section 7 of the TPS2511 datasheet that @User323693 have shared in his/her answer. <S> There is an explanation about how a DUC performs USB charging port detection: <S> The hand-shaking process is two steps. <S> During step one , the primary detection, the portable equipment outputs a nominal 0.6-V output on its D+ line and reads the voltage input on its D– line. <S> The portable device concludes it is connected to a SDP if the voltage is less than the nominal data detect voltage of 0.3 V. <S> The portable device concludes that it is connected to a Charging Port if the D– voltage is greater than the nominal data detect voltage of 0.3 V and less than 0.8 V. The second step , the secondary detection, is necessary for portable equipment to determine between a CDP and a DCP. <S> The portable device outputs a nominal 0.6-V output on its D– line and reads the voltage input on its D+ line. <S> The portable device concludes it is connected to a CDP if the data line being remains is less than the nominal data detect voltage of 0.3 V. <S> The portable device concludes it is connected to a DCP if the data line being read is greater than the nominal data detect voltage of 0.3 V and less than 0.8 V. NOTE: <S> Please refer to BC1.2 specification for what SDP, CDP and DCP are.
| So it's obvious that you shouldn't cut or disable the D+/D- pair.
|
Solutions to fix LED light RFI/EMI? Almost every LED light I've tested (hold an AM or shortwave radio antenna nearby) emits RFI or EMI across a vast frequency range. The EM noise comes either from the base of the bulb, or from some box or wall-wart supplying a DC voltage to the LED fixture. Is there any common solution to fixing or reducing this EMI? Other than throwing all these LEDs away and going back to hot incandescents? (which may or may not be legal in some settings.). Putting a opaque Faraday case around a light bulb makes them useless for lighting. Maybe some sort of filter boxes on the AC wiring, or the DC wiring if accessible? A box of ferrite cores? Does there exist any certified power efficient low EMI lighting? (e.g. more than just generic required regulatory labels on the boxes which may or may not indicate anything about what's actually in the box). <Q> You need an EMI choke as close to the socket as possible. <S> This also could include the house wiring in the walls. <S> At shortwave frequencies, the lamp itself is too small in relation to a wavelength to radiate very far, unless your shortwave receiving antenna is close to the lamps. <A> View these LEDs as EMI emitters just like black-brick battery chargers are EMI emitters. <S> If the LED is fed unfiltered non-DC, with fast edges, then only the faraday cage will suffice. <S> And if the black-brick should be poorly shielded (to block outside views of internal 200volt/100nanosecond edges), then that would need the faraday cage. <S> Otherwise, input shielding (coils, caps, ferrites) will improve the EMI. <S> Happy DXing. <A> My guess is STP cable or equiv. <S> In a heavier guage will be EMI quiet with 12V on any earth grounded DC <S> should not interfere with DX from a PC PSU. <A> Differential-mode emissions along the wiring is the most likely source. <S> Slug each line with an RF choke, as close to the source as possible: outputs if a wall wart, inputs if it's in the base. <S> (No good doing one big choke round the whole cable, as this will not address differential-mode RF) <S> If the source's mounting has metal bits, earth them all to each other.
| The wires supplying power to the lamp are acting as transmitting antennas and radiating RFI.
|
Can the length of a wire impact SPI transmission? I'm trying to wire a DotStar LED strip to another one. I have soldered on both ends of the added strip: VCC to VCC GND to GND CO to CI DO to DI I'm using SPI to control the LEDs. For some reason, the added strip does not work. When I test for continuity from the end of the original strip to the beginning of the added strip, I have a closed circuit on VCC and GND, but not on CO to CI not on DO to DI. I have soldered, resoldered, cut the strip shorter in case the copper paddings were damaged, but I still don't get any continuity. Having assembled many of those strips together with success, I don't understand what's going on this time. The only difference I noted is that the wires I used to connect those strips are longer than I usually do (about 8 inches long). So my question is: Can the length of the wires impact the good functioning of SPI? Thank you in advance for your help and suggestions. EDIT: Here's the (unorthodox) wiring schematics, hope this clarifies the question. <Q> Longer wires give more capacitance, which is harder to switch high and low at high data rates. <S> But longer wires will still conduct current, and a continuity test is at low frequency, so something else is going on herre. <S> Try measuring the actual resistance of the connecting wires instead of using a continuity checker. <S> It would also be helpful if you drew a schematic of how you are connecting the strips and exactly where you are measuring continuity. <S> Also, the link you provided is the top-level link to the vendor, not to that specific product. <A> Wiring length can have a negative impact on SPI communication, which is best for communication between devices on the same PCB (short runs of copper). <S> However, I do not think this is the cause of your problem. <S> It sounds like you may have a defective strip, or defective cable, since you aren't getting continuity on your communication lines. <A> Yes, in particular, the clock signal of SPI is a problem at longer lengths. <S> Think about it: The master clocks data sent to the slave and those bits travel alongside each other down the cable so that is fine. <S> But the master also clocks data sent by the slave which means that the slave can only send its bit when it receives the clock pulse. <S> That the wires have to be short enough so that bits sent by the slave are able to get back to the master before the master sends the next clock pulse, or else the master will associate the received bit from the slave with the wrong clock cycle. <S> But this should not be an issue at your lengths.
| Yes, the length of the wires used for SPI can affect the functionality.
|
What is limiting my BLDC Motor power Output? I am using a BLDC 3 phase hub motor and recently installed a motor controller rated for 3000watt (I originally had a 1200Watt controller installed). The motor is rated for 500w nominal 800w peak but is currently probably producing approx 1000w power at the wheel (with the 3000watt controller installed). I was wondering what design aspects of my BLDC motor would be the limiting factor (in terms of max useful output wattage). I think it's one of the following things: Stator magnetic core is saturated (I need a wider or bigger diameter stator?)** winding diameter too small * permanent magnets need to be bigger? (not sure if this would do much) *** motor needs better cooling * *(probably not a big factor since motor performance when motor is cold is not far off from the wattage output (useful) I got with 1200watt controller)) **What design changes (for example more layers, other shape of magnetic core) would positively influence the magnetic saturation point (make it so that this point is reached later/at a higher amount of magnetic flux produced)? *** the permanent magnets (36 pieces if I counted correctly) are approximately 3mm thick, 40mm in height. I'm running the motor at 72v (84v when battery fully charged), originally was intended for 36v/42v . The power output/torque produced by the motor did increase after switching to the 3kW controller but not by a lot. It appears most of the extra watts going to the motor are going into heating up the motor. The motor gets significantly hotter (gets hot in shorter amount of time) than with the previous controller. The motor is running with hall sensors connected to controller (motor controller supports hall sensors) Wires from controller to hub motor (3 phase wires) are 16mm2 each so probably not a limiting factor. The performance appears a tiny bit better when the motor is cold but the output is still quite comparable to the 1200watt controller.. Does the number of windings have any influence on the efficiency of the motor? I'd think longer overall path length of windings within motor would result in more voltage loss thus less efficiency. The winding wire diameter is approx 1.5-2mm. I'm not sure how many layers of windings there are unfortunately (have not rewound this motor yet). Stator's magnetic core outer diameter is approx 25cm, stator's magnetic core inner diameter is approx 15cm. See image: ( Note: currently used wires are much thicker than the ones in the photos, it's an old photo ) What changes could I theoretically make to my motor to get more power output at the wheel? Thank you! <Q> All of the points mentioned will limit getting more power from the motor than it is designed for. <S> The most likely avenue for getting more power from a given stator and rotor would be to operate the motor at a higher voltage and a higher speed, but it appears that you have already done that. <A> There is essentially nothing you can do to get more torque out of it. <S> Torque controls current, which means trying to get more torque increases the current, which has two negative effects. <S> (1) More I^2*R losses = more heat; this you are observing. <S> (2) instead of increasing magnetic flux to produce more torque, you simply saturate the iron, which reduces the inductance allowing even more current and more I^2*R losses (see 1). <S> You may get some small improvement - maybe 10 or 20% torque increase - before you run into these limitations - but not much. <S> The cure for these is more copper (thicker wire, lower R, reduced I^2*R) and more iron to allow more flux without increasing flux density. <S> Both of these mean starting again with a bigger motor. <S> Get a 100 Amp shunt and ammeter so you can monitor the battery current to establish what it does with the original battery, and where the limits are with the new one. <S> However there is no such fundamental limit on speed. <S> (Ultimately there are - insulation breakdown in the windings, or disintegration due to centrifugal force, or bearings limited to 20000rpm, or some such - but these are material or bearing choices not fundamental to the mass of copper and iron) <S> So, run the motor at a higher voltage to get a higher speed. <S> The danger here is that you need a higher torque to reach that higher speed, taking you back to (1),(2). <S> As you noticed ... not much improvement, much more heat. <S> So, run the motor at a higher voltage to get a higher speed - AND gear the motor down so that the torque load doesn't increase. <S> (Test this with the above ammeter) <S> Doubling the voltage but gearing down 2:1 should double the speed while keeping the torque constant : this doubles the power. <S> (Of course, if the motor is built into the hub, gearing down may not be so easy. <S> It would probably be an epicyclic gear system, a bit like a Sturmey-Archer. <A> Your correct observation that the stator core is saturating is indeed the main limitation. <S> In a motor design such as this, there is a trade of between the volume of copper vs iron in the stator tooth space. <S> If you make the tooth thicker, it can support more torque, but then there is less space for copper, so it is less efficient and vice versa. <S> The motor will have been optimised for a specific peak torque, and attempting to get more will saturate the iron. <S> Note, that even when saturated, you can still drive higher flux densities through the iron, and that is why you are able to get more torque with the bigger controller, but as you note, the efficiency in the generation of this extra flux is extremely poor. <S> If you do not want to change the stator design, then the only way to improve the torque output of the existing design is to reduce saturation from the armature reaction by increasing the air gap. <S> You can do this by simply moving the magnet radius outwards, but this will also reduce the stator flux generated by the permanent magnets (which will further reduce efficiency) so it would be better to combine this with an increase in magnetic material thickness. <S> However, you must be careful that you do not saturate the core by adding too much magnet material and to small an air gap. <S> In some cases, it might be beneficial to use a lower performance magnetic material, as if the air gap becomes too large you will have much higher flux leakage effects between the stator poles.
| So to summarise, you need to add more magnet material/airgap without changing the stator flux density, and then you should be able to get more torque from the existing stator. Motors that are well designed are usually close to the design limits in all of the aspects mentioned.
|
What does "For disconnect use only" mean on a connector? I recently bought some Andersen connectors for connecting high-amp devices to my camping setup in my truck. The idea is they hold together through much rougher conditions than a regular cigarette lighter 12v plug would, as well as being superior in virtually every other way as well. However they have this notice on them: FOR DISCONNECT USE ONLY I am unclear what that notice actually means. The French above it NE PAS EMPLOYER POUR LA RUPTURE DU COURANT Google Translate tells me means DO NOT USE FOR CURRENT BREAKDOWN This seems to indicate that the connector should not be unplugged when the system is under load, which makes a lot more sense... What does this warning signify? <Q> The French means 'do not use to break the current'. <S> In other words, this connector is not to be used as a switch. <S> Only pull them apart when the circuit is drawing no current. <S> With a 50A rating, and used with DC, breaking the connector circuit is likely to cause an arc as the contacts separate for long enough to damage them. <A> "For disconnect use only" should read " <S> For disconnected use only" in the sense of "Only 'use' this socket (i.e., 'put a plug in or out of this socket') <S> when the the power is off." <S> A good translation would perhaps be <S> or what about <S> "You must turn off the power, before plugging or unplugging this plug." <S> The basic confusion here: <S> If you say you are "using" a socket, it usually means that, "today, all day, something is plugged in to the socket". <S> The socket is "in use", it is "full", it is "being used by appliance X". <S> But the other way to "use" a socket is literally the act of plugging in and plugging out . <S> (As in "use a screwdriver" or "here, use this".) in this sentence: <S> "Only 'use' this socket (i.e., 'put a plug in or out of this socket') <S> when the the power is off." <S> The 'use' means 'the action of putting a plug in or out of this socket'. <S> That's what's going on here! <S> It's somewhat clearer what is meant in French. <A> I have worked as an electrician for over 20 years in the U.S. <S> I also hold an associates degree in electrical technology. <S> This is a very common terminology. <S> It means that this only acts as a "disconnect", which means you use this to "disconnect" the power, to either work on the equipment, or as a way to wire in portable equipment. <S> As opposed to a breaker, which means that it is both a disconnect and an over current device, which automatically stops all current from flowing when it pulls too many amps. <S> This is there to stop the wires from overheating in the case of a malfunction. <S> Also from my high school french class I learned that when "ne pas" is used it means "not". <S> As in the statement being expressed is not true <A> Assuming from the photo that this is an Anderson SB50 connector , it IS rated to be used to disconnect under load. <S> From their data sheet: <S> UL Rated for Hot Plugging up to 50 Amps Great for battery or other applications where the ability to interrupt circuits is required <S> So why it say "Only" is the puzzling thing, as in "What CAN'T you do with it?" <S> Most of the time I have found that statements like that which are molded into the housings are just some sort of CYA statement for when something doesn't get used in the intended way, they can say "See? <S> We told you..." <S> I wouldn't get hung up on it, the data sheet is the definitive authority. <A> The wording is entirely correct. <S> The english "Disconnect" is a noun in electrical terms. <S> It is not a verb or the act of disconnecting. <S> It is the means by which the device can be removed from the circuit, such as a battery when it needs replacing. <S> It is the device itself and is to be used only for breaking the connection when the ciruit is not in use. <S> The French have the noun for circuit breaker, which is written, but that is for a device that interrupts a live circuit and the warning is not to use this device in that manner. <S> It makes complete sense to someone who is familiar with electrical connections.
| "Do not use this plug as a switch" or "Do not use unplugging as a means to turn off."
|
What is the number of decaps used in each power pin? The preferred decaps for the power pins in the Ic is one smaller value, say 0.1uF and one bulk value like 10uF. But for FPGA there are several power pins in one IC, if I have to provide decap for those power pins:1. weather one smaller and one higher value capacitor is enough for all the power pins? or2. Should i have to give two value(smaller and bulk) capacitor for each power pins?3. weather i can give smaller value to each pin and single higher value decaps as shown in attached image? <Q> It is generally recommended to have decoupling on each individual power pin. <S> If two pins are next to each other or very close then you can also let these pins share the same decoupling capacitor(s). <S> But if there is enough space on the PCB, the preferred option is still to have decoupling on each individual power pin. <S> It is also recommended to use a small value (100 nF for example) capacitor and a large value (1 uF or larger) in parallel to get better decoupling. <S> Why this is so is explained by Dave in this EEVBlog video . <S> It is recommended to place the small value capacitors as close as possible to the IC's power pins. <S> The reason for this is to keep high-frequency loops small to minimize supply ripple and RF emissions (EMI). <S> The small value capacitor works better at high frequencies so its placement is more critical. <S> You can compromise on the large value capacitors and place fewer of them and only place them where it is convenient. <S> Decoupling (or bypass) <S> capactitor selection isn't an exact science, most experienced designers just follow the basic rules as stated above, look in the datasheet of the IC what is recommended, copy what is done in other designs and use common sense. <A> You want a capacitor pair(small and bulk) for each power pin, the value depends on many factors so I cannot tell you what is best with the current info, the values you mention do sound typical to me. <S> https://www.intel.com/content/www/us/en/programmable/support/support-resources/operation-and-testing/power/pow-integrity.html <S> I suggest you read this if you can, there is a physical purpose why you place a pair as close as you can to each power pin. <S> edit: there are other posts discussing similar topics worth a read <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> In general you want 1 small decoupling cap per power pin. <S> You might be able to get away with 1 for every 2 or 3 pins if there is a very large number of pins and they are all close together. <S> The main thing to keep in mind with decoupling caps is you want to get them as close to the pins as physically possible to minimize the inductive loop and path lengths. <S> In some cases, multiple capacitors of different values connected in parallel are recommended. <S> In this case, the smaller one must be placed closest to the pins. <S> Bulk capacitance should be placed near the chip, but doesn't necessarily need to be near the pins. <S> You may also want to provide separate/more careful decoupling for more sensitive supply rails such as analog supplies, PLL supplies, serializer supplies, etc. <S> As usual, check the documentation for recommended decoupling methods.
| If there is a very large number of pins, you might be able to get away with fewer of the larger value cap, as they help more at lower frequencies and as a result can tolerate a bit more inductance from the interconnecting traces.
|
Choice of resistor in half wave rectifier I found following schematic of a half wave rectifier (from here ) and I was wondering how to choose the resistor, and I have following questions: Is it correct that this resistor is basically functioning as a pulldown for when the AC is in the reverse polarity (top negative)? When connecting the output of this rectifier to the input of another device I think we would lose current across the other device when R1 is too small, so my guess is that we have to choose R1 at least as large as the input impedance of the connected device (or even a lot larger), is this correct? And if so, is there a more precise way to compute a value? simulate this circuit – Schematic created using CircuitLab <Q> Is it correct that this resistor is basically functioning as a pulldown for when the AC is in the reverse polarity (top negative)? <S> Correct. <S> And it is only there to discharge any capacitance in your measuring circuit which might otherwise give a false reading. <S> (The input capacitance on your measuring circuit - an oscilloscope, for example - would tend to hold charge between DC pulses.) <S> When connecting the output of this rectifier to the input of another device I think we would lose current across the other device when R1 is too small, so my guess is that we have to choose R1 at least as large as the input impedance of the connected device (or even a lot larger) <S> , is this correct? <S> If you have a load then R1 is not required. <S> The load will ensure that the voltage falls to zero. <S> And if so, is there a more precise way to compute a value? <S> If you're charging a capacitor in your circuit then look at the discharge time constant. <S> Figure 1. <S> Capacitor discharge curve. <S> The capacitor discharges by 63% per RC time period. <S> [Image by @Transistor.] <S> A handy rule of thumb is that the "time constant" of an RC circuit is given by multiplying R and C. τ=RC. <S> After τ the voltage will have decayed by 63%. <S> After 3τ it will have decayed by 95% and 5τ, 99%. <A> Yes, pull-down resistor (level 0, to ground); The value of the resistor should be as large as possible if linearity of the rectifier is intended; <A> I guess you want some nasty fuzz-type distortion in a synth or sound amplifying circuit. <S> You are not worrying about wasted energy or other things which are a problem in high power AC rectifiers which handle watts or kilowatts, you are going to handle micro- or milliwatts. <S> Pull-down action of the resistor = <S> OK <S> but the resistance must be low enough to sink the leakage of the diode and what's stored in the non-ideal capacitances in the circuit. <S> One of the capacitances is in the diode. <S> This makes easily high current 50... <S> 60Hz rectifier diodes useless in high speed pulse circuits. <S> I guess 1N4148 works well in whole audio band with say 3Vpp signals when R=500...500000 Ohms. <S> There's a trap. <S> If the signal source has an output series capacitor, it will get charged and it stops the whole signal. <S> To prevent it you should have also a resistor to GND at the left end of the diode. <S> Learn to use circuit simulators. <S> This site has quite good one for free. <S> Audio signals are often only 1Vpp or less. <S> The voltage drop of the diode (about 0,7V) can cause serious volume loss. <S> You should consider to use a rectifier circuit which has an operational amplifier to compensate the voltage drop in the diode. <S> An example: OpAmp Precision Half Wave Rectifier <A> As a large signal AM detector the voltage drop depends on the diode type and current. <S> As shown with 0.6V silicon signal diode drop means the draw was around 1mA peak. <S> If 10uA is ok for stray current noise immunity, this drop reduces in half with logarithmic proportions.
| You must have so big resistance that the signal source isn't pulled on its knees
|
How Does Kirchoff's Law Make Sense I understand the algebra, and I can also use Kirchoff's Law in calculation. But I struggle to understand how it makes sense logically. The law states, that the sum of the currents entering a junction must also be the same sum for the currents leaving that junction. Now, this would make sense in a circuit without resistors, but if I am to apply resistors, the current would be slowed, and therefore, the amount of coulombs passing through per second (amps) would be lowered, and the current on the other side would surely be lower, than before the resistor? How does this make sense with Kirchhoff's current law? <Q> But that will cause a "traffic backup" which will also slow the rate of current flow into or out of any junction feeding that resistor to that same slow rate. <S> Basically, in steady state, electrons can't "bunch up", because the increased charge concentration will repel them away from the traffic jam location. <S> And electrons don't "fly away" due to heating unless the resistor is an emitter, such as the heated filament in a vacuum tube, with an attractive charged anode nearby. <A> if the sum of electrons entering a node is not longterm zero, then eventually there will be arcing. <S> Does your phone arc, even if you leave it on? <A> Current is the amount of electric charge passing through a cross-sectional area per unit time. <S> A voltage can be thought of (but isn't!) <S> a force. <S> Electric charge is carried by electrons. <S> Voltage provides an energy differential to cause the electrons to move in one direction overall. <S> With that picture, we can now see that what is being converted into work isn't the electrons, it is the energy potential given to it by the voltage that results in the electron flow to occur. <S> Electrons are a conserved physical quantity, thus what enters a junction must be equally conserved by what exists a junction. <S> What is being converted to heat is the potential, the voltage. <S> Hence why electrons are not converted to heat, still exist in equal number, and exit the nodes in the same quantity. <S> However, you do get a voltage drop by the resistance. <S> To give a mechanical interpretation. <S> If you roll a big rock up a hill and then push it down the hill toward a wall. <S> The wall will provide resistance, the rock will push through the wall and lose energy in the process, but the rock is still a rock, same physical quantity, same physical properties...etc. <S> What has changed is the energy potential given by some outside entity (you pushing it up the hill) was converted to heat and kinetic energy. <S> The rock wasn't converted to heat energy, it remains a rock, the energy potential given to it was. <S> Now, if you rolled multiple rocks down a hill one after the other, despite that resistance and energy loss that occurred, the rocks behind it haven't experienced this loss and will push the first rock forward, and the one behind that rock will have itself pushed forward... <S> etc. <S> Since current is a connected chain of such rocks, your flow over the entry and exit will be equal. <S> What comes in must come out. <S> You simply must keep presenting an energy differential to keep propelling them forward (voltage or gravity in the rock example).
| Yes, a resistor will slow down the rate of current flow (for a given potential).
|
I can't understand the function of this diode On a past project, I've used the LM2576 buck converter with success. However, I've never understood how the catch diode works (D1 on the diagram). The datasheet explains the importance of the diode, and why it has to be a Schottky. TI LM2576 Datasheet My doubt is around the workings of this diode on the example circuit. Being a DC output, when the IC is shut down, the inductor will release charge flowing in all directions. IC output will prevent sinking that current, so it must take the load way. Is the voltage drop of the diode what is important here, maintaining the orientation of current? Does this means that current will flow to the load, after shutdown, and return via GND multiple times, using that voltage drop to slowly drop to zero? Does this solution prevent damage to the load, or the voltage source? Thanks in advance for your time. <Q> Figure 1. <S> Buck converter . <S> Inductors don't like you to change the current suddenly. <S> When you try large voltages are induced. <S> In the on-state current flows through the inductor to the load and the filter capacitor. <S> In the off state there is no current supplied by the PSU. <S> The inductor tries to keep the current flowing and since the right side of the inductor is held at the output voltage the left side of the inductor is driven below zero volts until the diode starts to conduct. <S> This will maintain the current for a short period of time. <S> As the inductor current decays the feedback (in your circuit) will trigger another pulse from the converter and the next cycle begins. <S> A few points: Being a DC output, when the IC is shut down, The IC isn't shutting down. <S> It's just turning off its output. <S> ... <S> the inductor will release charge flowing in all directions. <S> Not really. <S> It will flow in the same direction as when the converter output was on. <S> IC output will prevent sinking that current, ... <S> Yes, the output is switched off so no current flows back into the converter. <S> ... <S> so it must take the load way. <S> Correct. <S> Is the voltage drop of the diode what is important here, ... <S> The voltage drop of the diode is important to keep the efficiency high. <S> ... <S> maintaining the orientation of current? <S> Does this means that current will flow to the load, after shutdown, and return via GND multiple times, using that voltage drop to slowly drop to zero? <S> In the on-state the power supply provides energy to the load and to charge the inductor. <S> Current will flow around the red loop. <S> In the off-state energy is released from the inductor to power the load. <S> Current will flow around the red loop. <S> Does this solution prevent damage to the load, or the voltage source? <S> A successful design has to do both. <S> The voltage is usually the critical parameter for most electrical loads. <S> The maximum current may be a critical factor for the power supply feeding the converter. <A> D1 is not best described as a 'catch' diode. <S> That's a better description of the diode across a relay coil, which 'catches' the high voltage transient when its driver turns off. <S> In a buck converter, best to call it a 'freewheel' diode. <S> During the on phase of the IC, the output is taken up to the input voltage. <S> Current flows from left to right in L1, increasing as L1 sees the excess of the input voltage over the load voltage across it. <S> When the IC switches off, the output pin goes high impedance. <S> Current is still flowing from left to right through L1, but it can't be sourced from the output pin. <S> For a very brief time, L1's current is sourced by the stray capacitance of the output pin and D1, and the voltage there falls rapidly. <S> Eventually, the voltage drops to -0.4v, and D1 starts to conduct, sourcing L1's current. <S> This allows the current to continue flowing (it freewheels) through L1 to the load, although the current is now falling as L1 sees the load voltage plus the D1 voltage drop across it. <A> It is a switch mode regulator. <S> It first turns on to allow current to flow into the coil, and when it turns off, current must keep flowing to output to release the energy to output, so the diode allows it. <S> The less voltage drop over diode, the more efficient the buck converter is.
| The orientation of the diode ensures that current flows in the correct direction.
|
Can a USB cable be called a charger? Having a conversation where a colleague asked another to pass the charger, only to find out a cable was handed over. Decided to wonder whether a USB cable really qualifies as a charger, since we are used to plugging it into an adapter (the charger) which connects to the power outlet. However, you have USB ports built into power outlets nowadays and also USB ports in computers which serve the function of a 'power outlet'. So is it technically correct to call a USB cable a charger? <Q> So is it technically correct to call a USB cable a charger? <S> No. <A> So is it technically correct to call a USB cable a charger? <S> No. <S> Technically <S> what is known as a "charger" is a fixed voltage power supply which knows nothing about batteries. <S> The cable knows nothing about batteries either. <S> The battery charging management is built into the phone / tablet. <S> This circuitry monitors the state of charge of the battery and regulates the voltage and current from the power supply and to the battery. <A> Nah. <S> Charger cord , yes. <S> But not a charger. <A> The "USB cable", per se, provides no power itself, it is just a means of connecting one type of adapter to another. <S> Therefore the use of the word "cable". <S> A "charger" is a device used to put energy into another device or rechageable battery. <S> That's the pedantics, what teenagers chose to do with it is out of our power.
| The charger may need a connecting cable from it's output terminal to the device to be charged, this is where the "USB cable" is used.
|
How to clean solder flux paste after marks I want to given a motherboard circuit board to local repairing professional he traced and soldered some chips but its not still working and advised to show motherboard to manufacturer service center. But my motherboard have yellow stains after marks of Soldering or may be soldering flux paste I am attaching photographs of the soldering marks for your reference Before showing to service center I want to clean those yellow marks as some some service center technicians make excuses if is been soldered by local technicians as they may refuse to even touch or find the problem I have tries to clean by isopropyle by still marks are there So kindly if some one can advise or suggest how and.by which chemical can I clean the yellow after marks solder or solder flux paste <Q> What percentage isopropyl alcohol? <S> 70% is not enough. <S> Even 90% is a bit low. <S> I use >99.9% which can sometimes be found in drug stores. <S> Sometimes even that feels low. <S> Ethanol works maybe 50-100% better and smells better too but costs 4x. <S> You don't really need ethanol as long as you can scrub though. <S> It takes rubbing with FRESH areas of a paper towel. <S> Flux saturated yellow/brown areas of paper towel won't rub away flux anymore. <S> Move around to clean areas of the towel. <S> You can rub a thick, sticky flux spot and it will look like nothing has changed, but flip the paper towel over and see its colour. <S> If it is bright yellow or dark brown then you did remove some residue but it was so thick you couldn't tell. <S> So move to a fresh piece of towel. <S> It can take 5 passes or more for really thick residue. <S> Just keep flipping the towel over, looking at it, and moving to a clean spot soaked in alcohol. <S> Brushes help too but mainly as a substitute for your fingers to press the paper towel down into nooks and crannies. <S> I never found a brush (stiff like a toothbrush or even stiffer) alone to be very effective unless submerged since it can scrub off residue but can't pick it up. <S> Helps keep the brush clean too if you use it through a towel, especially if you are using a $30 ESD brush. <A> Use pure (better than 90 percent) isopropyl alcohol (or ethanol) and a brush. <S> When I did that kind of thing often, I had a small (1 cm or 3/8 inch wide) paint brush. <S> I cut the bristles to about 1 cm long as well. <S> Dunk <S> the brush in a container of alcohol, scrub the flux spots with the brush. <S> Repeat until clean. <A> The mildest approach is 99% or better isopropanol. <S> It's possible to use stronger solvents such as lacquer thinner <S> but they can attack plastic and remove markings if they get on the solder side. <S> Use the solvent and one of those free toothbrushes that the dentist gives you on every visit- <S> the bristles are just stiff enough. <S> Obviously you don't use it as a toothbrush after that.
| Drug store rubbing alcohol often contains additives as well as water that leave a nasty white residue.
|
Measure 25A Current In Series? I am working on a DC battery charger project, and there is a dispute that the battery charger does not charge up to the stated 12V / 25A rating (300 watts). We've connected the DC battery charger to an electronic DC load at a constant voltage of 12V, and it draws 20A maximum (which is below the stated limit, and sets a firestorm of regulatory issues since the product need to meet 25A as per the stated ratings). I understand there is variation in battery voltage vs. the nominal voltage (so the current can change), but we do not see this device ever output anything close to the stated power output. If we connect two or three nominal 12V batteries simultaneously then we can coax the charger up to 24A. The designer thinks we have set up the measurement incorrectly and is waffling on why the watt output is so flaky. I am thinking to illustrate the point further, we will get a 12V battery and remove the electronic DC load. I wanted to nail down the measurement connections and terminations with the designer so there is no dispute about "the test setup" and then wire in a ammeter in series to measure the current. I am seeing on high-end Fluke multimeters they are only able to measure 10A of current safely. What can be used in series to safely measure up to 25A of current? I'm seeing clamp style multi-meters, is there anything else like a more traditional meter that can be wired up in series for the measurement? <Q> You should use a 4-wire current sense (shunt) resistor. <S> This is a 0.001 ohm, probably a good choice for what you are doing. <S> https://www.digikey.com/product-detail/en/murata-power-solutions-inc/3020-01096-0/811-1094-ND <S> Be sure to connect the high current to the top terminals and take the measurement on the side terminals. <S> It is only calibrated if you use the side terminals for the measurement. <A> You can buy Hall-effect current measurement modules - they're cheap, fast to deliver, and plug-and-play provided you have an ADC or attach a Voltmeter at the analog output. <S> (ignore the sentence at the top, even if it might be true if the measurement is to be used by a machine) <S> Allegro is a classic for these applications ; the ACS758LCB-050B-PFF-T handles +/-50A <S> with 40mV <S> /A sensitivity. <S> 120kHz bandwidth / 3us is a bonus in your case, but 100µOhm series resistor might actually be more than a nice touch. <S> If you're planning on doing this on a regular basis <S> a "cl-Ampmeter" might be more suited to you though (next day delivery is a bliss there also), just make sure it is not transformer-based, but Hall-effect based, and wind enough of your wire through the clamp to get the sensitivity you want. <A> A Hall-effect DC clamp-on probe can perform a non-intrusive measurement. <S> They’re a bit costly but if you do a lot of this work it will save you time, money, and argument. <S> And... a hacktastical solution. <S> You can 'cheat' and use your feed wire as the sense resistor. <S> Pick two points in your feed wire, one at the source and one at the end <S> Force <S> a known current through it <S> (you already have the programmable load.) <S> Measure resulting the IR drop Calculate R for your known I. Connect to your DUT load and measure the IR drop at the same points Divide by R, carry the one... <S> there's your current. <S> This will be as accurate as any sense resistor solution, provided the wire doesn't get hot. <S> It will be even better as it will develop many times the IR drop of a milliohm sense resistor: <S> lead resistance will be in the tens of milliohms. <S> Cost? <S> Practically nothing. <S> Invasive? <S> Not if you have access to the wire terminals. <S> Even if you don't do this, and opt for a sense resistor inserted in-line, you will still need to calibrate it by forcing a known current. <S> Same will be true for a clamp-on probe. <A> I am seeing on high-end Fluke multimeters they are only able to measure 10 A of current safely. <S> You could use three identical 10 A meters in parallel and sum the readings. <S> Most of the Fluke meters have 20 A capacity for short duration. <S> Check the manual. <A> Frame challenge. <S> It would be a rubbish battery charger if it charged batteries at 25A with any frequency. <S> Proper multi-stage battery charging requires you to deliver an appropriate voltage at each charge stage. <S> The battery will then draw what the battery wants to draw . <S> If your battery charger is just murderously bashing 25A into the battery, then either a) <S> your charger is very, very stupid and will destroy batteries, or b) <S> your charger is very, very smart, has an enhanceed relationship with the battery (e.g. Thermal monitoring), and <S> the battery is quite large for the charger. <S> An eexample is Tesla, which really can Tyson-punch 50kw into a battery that small, because the battery system is designed soup to nuts to do that, with impeccable monitoring and cooling. <S> For instance, I have a (25A, in fact) charger that charges a 1200 pound battery. <S> (That's 550kg, not 1200 quid). <S> It's a starting battery, so it spends most of its regime above 90% SOC. <S> The battery charger operates between 6 and 12 amps at those SOCs, and I consider that normal <S> and I do not consider myself "ripped off". <S> Whoever is expecting to attach it to J random battery at J random state of charge, and immediately see 25A charge rate, is an idiot who should be drummed out of the battery business. <S> -- <S> Now if you are driving into a static DC load, e.g. A resistor bank, not only cant battery chargers do that, but neither can any DC power supply. <S> If you put a 0.6 ohm (1.66666 siemens) load onto a 12V power supply, the right amount of current to expect is 20 amps. <S> If you want 25 amps, you need a lower resistance (higher conductance) load.
| A proper battery charger will charge a deeply discharged battery at a low rate, then increase the charge rate in the midrange, achieving maximum charge rate somewhere in between 20% and 80% SOC, and then feather off charge rate as the battery nears full.
|
Developing a Hardware Test Plan for a Microcontroller - Past Experience I am using this Microcontroller using 3.3V for power. Using GPIO pins to drive some LEDs and send some ports for sending UART, LIN signals. I'm using a 16MHz crystal. Can someone tell me what the main parameters are that I should test in Hardware Testing of the microcontroller? For example, suppose I have an I2C interface between the MCU and a slave. I would check the Vih, Voh levels, rise time, fall time, frequency, and period of the interface. I am asking along similar lines, what are the major test cases that one might overlook while validating the MCU section in testing? Please help with your past experience working with Microcontrollers which you might have overlooked during testing and had some issues with it. <Q> The scope of testing you take on for a pre-bought component focuses on looking for defects at the board level: opens, shorts, possible ESD damage. <S> This would be done on every board. <S> Typically this a combination of functional tests and some electrical parameter tests. <S> How much you do of this depends on the expected defect rate of your components and the product as a whole. <S> You would also validate your system with voltage margin, thermal and shock+vibe testing to ensure that your board meets its intended environment spec. <S> Chip testing is a bit different: you apply test vectors to find faults in the internal logic. <S> This is done when the part is manufactured. <A> For design validation: scope the heck out of it. <S> Clocks can be especially troublesome, but for a self-contained MCU, you probably won't have any really high frequency clocks. <S> Over-test for margin (without stressing anything too much). <S> If your Vcc is +/- <S> 10%, test at -15%. <S> If your temperature range is 0 - 70 degC, test at -10 to 80 degC. <S> If you don't have any margin, it is hard to be confident that the 10th unit won't fail at your normal range. <S> Testing a interface: try to increase the repetitions as much as possible. <S> Automate the data flow in a loop. <S> Example: send and verify a million messages overnight. <S> Try to crash it with crazy input. <S> If you had a keyboard for input, a manager I knew would perform a lunch box test. <S> He would set his lunch box on your keyboard, pressing most of the keys simultaneously, attempting to crash the system. <S> Try to invent something similar for whatever I/O you have. <S> Don't let someone else find a bug like this. <S> This is your reputation on the line, don't let it go to production until you are satisfied. <S> I occasionally had managers that wanted to cut me off after the product appeared to be working, don't let this happen. <A> 3.3V for power. <S> Line and Load regulation Tolerance of 3.3 V Ripple Output current - MCU and other peripherals <S> Using GPIO pins to drive some LEDs <S> Pulse width PWM Frequency <S> \$ V_{OH}\$ , \$ <S> V_{OL}\$ of the GPIOs and send some ports for sending UART, baud rate rise time, fall time <S> ON time, OFF time <S> \$ V_{OH}\$ , \$ <S> V_{OL}\$ , for UART Tx Pin <S> \$ V_{IH}\$ , \$ <S> V_{IL}\$ for Rx LIN signals. <S> Timing and voltage levels as per LIN protocol. <S> similar to UART Tolerance and worst case bus voltages from external device Bus short condition and open condition <S> I'm using 16MHz crystal freq. <S> measurement - Direct or via buffered divided output <S> Voltage levels <S> If 12 V is from a battery (inputs from markus ) Idle system current <S> Sleep mode current Active mode current RF or Full Power ON Mode current <S> Almost all the above tests at lower limit of the battery Stable temperature of the board (after running for 8 to 10 hours), especially the power section. <S> Power on.
| Perform detailed signal characterization on any critical signals. Reset single timing Clock stabilising timing Power supply timing LED currents GPIO rise time and fall time
|
What are the reasons behind marking neutral and line terminals on a power supply AC input? Even though AC has no polarity, almost all the power supply AC inputs or most AC powered device input terminals indicate line and neutral terminals as L and N just like in the following example: Is there a fundamental reason to wire line to L and neutral to N but not the other way around? Could you give an example what might go wrong if one connects neural to L terminal and line to N terminal? <Q> One example where Live/Neutral does make a difference are some lamp fittings. <S> For E27 and E14 fittings the "screw" part is more likely to be touched by the user when changing the lamp. <S> That means that the screw part should connect to the neutral instead of live. <S> The other fittings are symmetrical so there is no difference. <S> Regarding electronic devices I have yet to see an example where the mains connection is not symmetrical. <S> That makes sense because as remarked in Justme's answer, it depends in which country you are if the Neutral and Live are always connected in the same way (UK plug) or that they can be swapped (EU plug).To be on the safe side, the equipment is designed such that the live-neutral connection can be swapped. <A> In some circuits, especially permanently installed circuits, it might make a difference. <S> Even if it does not make a difference, regulations can require clear markings how to connect equipment, so that there is consistent notation. <S> But for normal equipment it does not matter much, as in some countries you can plug in the mains cable in any orientation so it will not be guaranteed which wire is live and neutral. <A> There are some good answers to this question at <S> Why are some AC outlets and plugs polarized? <S> In particular https://electronics.stackexchange.com/a/15232 <S> The currents through both pins are equal and opposite, but the voltages on each pin are not. <S> The neutral is roughly 0 V relative to the Earth at all times. <S> The hot alternates between positive and negative. <S> As Bimpelrekkie points out, depending on the appliance, there's a chance of coming into contact with one of the two sides of the circuit. <S> You want that to be the neutral side not the hot side to lessen the chance of shock/electrocution. <S> (The two comments on your initial question essentially say the same thing and had they been posted as answers instead of comments you should mark one of them as the accepted answer.) <A> In the event of a catastrophic internal failure causing a short to the PSU output the circuit will be dead when the fuse blows. <S> With reverse connection the fuse might not blow and even if it did the internal circuit would be live. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Left: <S> correct. <S> Right: incorrect. <A> It is true normally polarity of ac doesn't affect the performance of ac instrument . <S> Normally ac instrument control by one way switch <S> and it's gives small capacitance. <S> when it is connected in neutral (N) point to control the ac device. <S> Then it is consume small power when it is closed.
| One reason is that typically the internal fuse is on the live wire.
|
Why would a part be available as 4.99k and 5k? I was looking at chip resistors, and came across the MORNTA1001AT5 , which comes in many resistance values. The resistor tolerance is 0.1%, so what is the logic of having the part available in both 4.99k and 5.0k versions? Generally, if you needed accuracy like that, you'd use smaller parts as they are easier to match, so I have no idea why there would be a 4.99k and 5k version. I'm just trying to determine if I'm missing something fundamental. Edit: Could I get an example of where a 0.1% 4.99k would be used over a 5k at 0.1%? <Q> The E96 series of preferred numbers contains the 4.99 values. <S> E96 values (1% tolerance) 1.00, 1.02, 1.05, 1.07, 1.10, 1.13, 1.15, 1.18, 1.21, 1.24, 1.27, 1.30, 1.33, 1.37, 1.40, 1.43, 1.47, 1.50, 1.54, 1.58, 1.62, 1.65, 1.69, 1.74, 1.78, 1.82, 1.87, 1.91, 1.96, 2.00, 2.05, 2.10, 2.15, 2.21, 2.26, 2.32, 2.37, 2.43, 2.49, 2.55, 2.61, 2.67, 2.74, 2.80, 2.87, 2.94, 3.01, 3.09, 3.16, 3.24, 3.32, 3.40, 3.48, 3.57, 3.65, 3.74, 3.83, 3.92, 4.02, 4.12, 4.22, 4.32, 4.42, 4.53, 4.64, 4.75, 4.87, 4.99 , 5.11, 5.23, 5.36, 5.49, 5.62, 5.76, 5.90, 6.04, 6.19, 6.34, 6.49, 6.65, 6.81, 6.98, 7.15, 7.32, 7.50, 7.68, 7.87, 8.06, 8.25, 8.45, 8.66, 8.87, 9.09, 9.31, 9.53, 9.76. <S> So the question is really, who wants a 5.00 value? <S> I've never seen 5 kΩ specifically <S> but I have seen specialty values used for things like ADCs, voltage dividers for multimeter voltage ranges, etc.. <S> Many PLCs use a 250 Ω resistor to convert 4 - 20 mA to 1 - 5 V for their analog inputs. <S> This too is not a standard value. <A> The difference between 4.99kOhm and 5kOhm is of the order of 10 ohm, ie a 0.2% change. <S> The resistor tolerance required, as you mentioned, is 0.1%. <S> So if 0.1% tolerance is allowed, a change of 0.2% would disrupt the accuracy. <S> This means that the separate valued resistors are necessary. <A> I agree with the posts above. <S> All the other values listed in the Datasheet except 500Ω, 5kΩ and 50kΩ are part of the E-series of Resistors. <S> Because the resistor is used for "unity gain operational amplifier circuitry" or "voltage references" <S> I assume there are cases where an integer resistor ratio is needed e.g.: 20kΩ / 5kΩ = 4.00. <S> Which otherwise is not easily achievable without combining multiple resistor values in series/parallel. <S> Therefore they introduced the 0.5 value in addition to the values from the E-series. <A> 5.00 would be part of an E768 series, if anyone made such a range print(Eseries(192 <S> *4))4.96 <S> 4.97 <S> 4.99 <S> 5. <S> 5.02 <S> 5.03 <S> 5.05 <S> 5.06 <S> 5.08 <S> This would represent 0.125% against the Renard scale, where the Eseries numbers are derived from. <S> Now in practice 0.1% resistors appear in the E192 series, even though the E192 series is 0.5% as per Renard scale. <S> This does mean not all possible values can be realised in 0.1% E192 series, but economics comes into play. <S> Why produce every single resistor value when the larger jump between the tolerance extremes can be managed by design engineers. <A> Last year, I designed a Li-on Charger based on LT8490 for the 16S8P battery package. <S> In the input side, 72V is applied and needed to be measured via a voltage divider. <S> Things are getting messy in here. <S> If little valued resistors such as 10 ohms or 100 ohms, then neither IC could handle current applied to pin nor the resistors. <S> I am not a professional but It is obvious that someone uses that rare components you see as unnecessary.
| IC demands high-precision resistors for voltage and current configurations.
|
capacitors burn at discharge due to low current-limiting resistor; how come they don’t burn at charging, when there is no current-limiting resistor? I apologize if the question is superfluous or already answered. This is what I know: the charging of a capacitor from a dc source happens instantaneously. Since there is no series resistor to limit the current, then what actually prevents the current to become infinite and burn the capacitor at charging time? On the opposite is the discharging. If a charged capacitor is shorted, it will burn. Otherwise said, the discharge can not happen as fast as the charging; why the capacitor can charge instantaneously but it can not discharge as fast? What limits the charging current effectively preventing the burning of a capacitor, in the absence of any current-limiting resistor? <Q> The DC source has some internal resistance, aka it can't physically source infinite current (even if you directly short the terminals). <S> Additionally Caps have ESR (Equivalent Series Resistance) that also limit current. <S> Do note that sometime it is prudent to put current limiting resistors at the gate of MOSFETS for example, to reduce current spikes. <A> No, charging of the capacitor does not happen instantaneously...not in real life, and not in a properly constructed circuit analysis. <S> Connecting a discharged capacitor directly across a voltage source violates the definition of "parallel" and KVL. <S> Likewise for discharging. <S> There must be resistance. <S> In essence, you are correct that if there is no resistance then the current, whether charging or discharging, must be infinite. <S> So, if you want to ask a meaningful question you must assume that there is resistance in the circuit. <A> This is what I know: the charging of a capacitor from a dc source happens instantaneously. <S> Since there is no series resistor to limit the current, then what actually prevents the current to become infinite and burn the capacitor at charging time? <S> Alas, if a capacitor will really "burn" at high currents, it will burn regardless of which way the current is going. <S> On the opposite is the discharging. <S> If a charged capacitor is shorted, it will burn. <S> Otherwise said, the discharge can not happen as fast as the charging; why the capacitor can charge instantaneously but it can not discharge as fast? <S> I've never seen this happen. <S> I suppose there are might (might, I say) be caps which self-destruct in the event of a single event, but frankly I've never run across one in 50 years of messing around with electronics. <S> What can happen is that, if a capacitor is repeatedly charged and discharged at high frequency, some types can overheat and blow/burn up. <S> In this case, you use a capacitor with a lower internal resistance (called ESR, short for "Equivalent Series Resistance"). <S> For any capacitor, internal power is produced by the internal resistance of the cap, which produces a power related to the square of the current divided by the resistance, and if you make a thicker internal structure the resistance decreases. <S> (This gets complicated at high frequencies due to phase shifts caused by both capacitive and inductive effects, but you don't need to worry about those just yet.) <A> What limits the charging current effectively preventing the burning of a capacitor That would be my question to you, because I don't know your power supply source or the wiring from the source to the capacitor. <S> This impedance matters. <S> If your wire is 6" of 2/0 AWG wiring from a truck battery, that will have a very different impedance (effective resistance) than 10' of #26 wiring from a 1A wall-wart. <S> The load side of the capacitor will have impedance also. <S> Both impedances factor into the actual current the capacitor will flow.
| A single charge or discharge cycle will not realistically destroy a capacitor.
|
Op Amp with positive feedback (hysteresis) driving N-MOSFET oscillates I have a Supercap voltage control circuit that is based on a On Semiconductor NCS333 Op Amp that's configured with hysteresis (using positive feedback) that drives two N-MOSFETs - one that is actually discharging Supercap when voltage goes over some pre-defined limit (~2.5V in my case) and another one to indicate that discharging is active now. I've used DMN1019U MOSFET to discharge the Supercap as it can tolerate up to 10.7A of current with a very low \$\mathrm{V_{GS(th)}}\$ and \$\mathrm{R_{DS(on)}}\$ , so it won't overheat at high currents. Current through MOSFET is limited by 2512-case resistor, \$\mathrm{R_S}\$ , so most heat would be generated by resistor and not MOSFET. When I'm using any resistor down to ~1.8Ohm - system works fine and correct, but if I want to increase discharge current by placing two 2.2Ohm resistors in parallel resulting in 1.1Ohm total resistance (for example) - output of Op Amp starts to oscillate and MOSFET starts to act as a variable resistor and heats up very quickly as its resistance becomes higher than \$\mathrm{R_S}\$ one. I have tried to use snubber circuit for Op Amp, it helped a little bit, I was able to reduce \$\mathrm{R_S}\$ to ~1.5Ohm but if I go lower - oscillation starts again. Is there any way to stabilize this circuit? I know that DMN1019U has a very high gate capacitance of 2588pF @ 10V, but I need to choose a MOSFET with lowest available resistance and \$\mathrm{V_{GS(th)}}\$ so power dissipation would be occurring in \$\mathrm{R_S}\$ rather than MOSFET. <Q> After a reading your question and having a quick look at the datasheets of the NCS333SN, I am sure that the problem is the heavy capacitive loading of the amplifier by the DMN1019. <S> The two details motivate my belief The "Dynamic performance" parameter table at p. 6 states in each entry (apart from the slew rate SR entry) <S> a value <S> \$C_L=100\mathrm{pF}\$ <S> for the load capacitance. <S> This suggests that it is not advisable to increase too much beyond that limit the load capacitance. <S> When you have a look at the "Typical characteristics" section, looking at figure 1, p. 7, you notice that the nice phase margin at \$f= <S> f_T\$ of the amplifier is a more than respectable <S> \$\phi_M\simeq <S> 60^\circ\$ <S> but again when the load capacitance is \$C_L=100\mathrm{pF}\$ . <S> And if look at figure 13, p. 9, you see that the small signal overshoot is over \$60\%\$ <S> when \$C_L=1000\mathrm{pF}\$ . <S> Since the typical gate capacitance of the DMN1019 is \$C_\mathrm{G}>2500\mathrm{pF}\$ , we'll surely find troubles if we connect the OpAmp output directly to it: and even using a series gate resistor may not get you out of troubles, if this resistor is significantly lower respect to e.g. the standard load stated for the slew rate test, i.e. \$R_L=10\mathrm{k\Omega}\$ , as you have noticed with your tests. <S> What could you do? <S> Since you use the MOSFET as a means for discharging a supercap when the voltage across it starts to rise above the safe \$2.5\mathrm{V}\$ level, you do not need to be particularly fast in turning it on. <S> Therefore you could try to put a \$10\mathrm{k\Omega}\$ gate resistor and see if the amplifier remains stable. <S> Otherwise, if you desire to have nevertheless a quick response, you should try to find an optimal value for the gate resistor \$R_\mathrm{G}\$ , by starting from \$R_\mathrm{G}=1000\Omega\$ and rising it until the circuit is stable for all desired resistor loads. <S> Final note <S> Just to give some examples, devices like <S> TLV3691 or NCS2200 could be a nice choice. <A> The NCS333 cannot operate well as a comparator at 2.5V with an insufficient swing to get the rated RdsOn. <S> Thus at high RdsOn, and low loads the hysteresis reduces to the point where you have almost none. <S> Put a level shifter or inverter to get a full swing on the output and swap inputs or try more hysteresis. <S> You should at least pay more attention to the output swing specs on the Op Amp and Vgs vs RdsOn and use a better comparator instead of a precision Op Amp. <S> edit: <S> This is what concerned me about the gate drive voltage using this Op Amp as a low voltage comparator. <S> p7 https://www.onsemi.com/pub/Collateral/NCS333-D.PDF <S> Adding a large gate resistor will help decouple reactance effects with positive feedback that you have being reduced by the output swing. <S> The table implies a rail to rail CMOS swing output with 50mV rail offset, while the curves imply the output swing is far from Rail to Rail yet <S> Vdd <S> , Vss implied it is still CMOS. <S> After starting at it for a while, I saw the error in my interpretation now. <S> Where the graph says Vs=5.5 meaning supply, they are using bipolar supplies here, but only define it as Vdd-Vss= <S> Vs, so the graphs all start with a 50% swing. <S> Conclusion <S> My corrected analysis indicates you have excess gain at the resonant frequency due to 3rd order effects reducing phase margin and AC hysteresis with the load R load and C of supercap approaching the open-loop pole of the Op Amp. <S> , which may be corrected by suitable compensation. <S> There are 3rd order effects with RC capacitances in 3 parts (IC, FET-Ciss, Supercap) <S> each add a pole and reducing load <S> R pushes the pole up near the IC open-loop breakpoint. <S> Non-linear effects of IC: Hysteresis results in zero gain, yet the IC open loop has 130dB typ gain with a breakpoint well below 10Hz. <S> With 10F and 1.1 Ohms, that's a bit higher. <S> Recommendations <S> The improve your stability, you can add phase-lead RC compensation or switch to a bipolar comparator or decouple the gate Ciss with a large series resistor to lower the gain at the resonant frequency with Ciss using say 10k ohm. <A> One guess: <S> There's a feedback path which should be blocked. <S> Replace R8 with two 2,2 kOhm resistors in series and connect a capacitor, say 10uF between the joint and the GND. <S> Also a cap simply placed over the reference IC can work. <S> Another guess: Your circuit is built on a breadboard <S> and it's full of unwanted parasitic parts.
| Following mkeith's comment above, I think is a very good idea to find a low voltage comparator and use it instead of the NCS333: despite their circuit topology can be (also very) similar, OpAmps and comparators cannot be used interchangeably without any care.
|
Why doesn't the output of the opamp change with the source impedance? The datasheet of this opamp says the input impedance is typically 100Meg. So I expect the voltage gain to be halved with a source resistance of 100Meg. But the following simulation doesn't reveal that and gain almost doesn't change: Above the voltage gain is two. Applied input is 1V. So with a 100Meg source resistance I was expecting 1V output(half of the gain). But the plots show the gain is not affected from the source impedance and is still two. Why 100Meg source impedance is not halving the output gain if the opamp input impedance is 100Meg? <Q> There are other factors to consider such as input bias current (quoted at anything up to 4.5 nA). <S> With 4.5 nA flowing from the input into 100 Mohm, that produces an offset error of anything up to 0.45 volts. <S> That blows the effect of the input resistance out of the water. <S> But, realistically, with a simulation, it all depends on what is set up in the model parameters. <S> Not all models accurately reflect all op-amp artifacts. <S> The datasheet of this opamp says the input impedance is typically 100Meg <S> That figure is the differential input impedance and not the input impedance of one input to ground <S> so, it will have much less of an effect. <S> If you look on page 6 of the data sheet you will see that the common mode input impedance is graphed at about 280 Gohm. <S> But like I said, input bias currents will always blow things apart at extremes. <A> The circuit input resistance is different from the opamp input resistance. <S> With negative feedback you need to multiply the opamp input resistance by the loop gain to calculate the resistance seen by the source at the non-inverting input. <A> If you look at this circuit, the following can be understood: <S> The feed-back ensures that. <S> Any wrong feed-back circuit configuration will not work. <S> 2.The minimum input difference between (+) & (-) inputs required on open loop condition is very very low, more the open loop gain, less the difference voltage required to swing the output. <S> The input current. <S> required is is almost negligible , since all of Op-Amps have differential amplifiers at inputs, and many even have Mos-Fet at input differeintial amo stage. <S> In the above mentioned non-inverting Op-Amp circuit, you assume both (+) & (-) inputs are at at same voltage. <S> If you have V(i) at (+) input through the 100 meg ohm, then the same voltage is maintained at(-) input. <S> In order to ensure that, the voltage divider between output ,(-) input and ground requires V(-)= V(O) <S> × ( R÷(R+R)),i.e V(-) = <S> (1/2)×V(O)Thus means, gain is (V(O) ÷V(I))= 2 , no matter what resistance you use at input. <S> Of course if you try a 100 G ohm input resistor, it may not work. <S> That may not ensure minimum input current required (pico amps) !
| First understand, all Op-Amp cirçuits are designed so that the device operates to ensure almost same voltage at it's (+) &(-) inputs.
|
What is affected by DC/DC converter frequency selection? I planning to use the LMR33640 DC/DC converter which exists in 400kHz and 1MHz switching frequency versions. I know that the 1MHz version allows to use a smaller inductors. But I think that the 1MHz version has higher EMI emissions as a drawback (not sure)? What other PROs and CONs do exist with switching converters concerning switching frequency selection? <Q> The choice of frequency is usually constrained by 4 primary factors: 1 Inductor size (as you mentioned). <S> 2 Higher frequency operation increases switching losses in the switching FET. <S> In synchronous buck converters we usually choose the switching FET for lowest total gate charge and the sync FET for lowest \$R_{DS(ON)}\$ . <S> 3 Compensation. <S> This can be a bit more challenging with a higher loop cutoff but is often worth it as the output error response time will be faster which can make ripple easier to deal with. <S> 4 Switch minimum on time; at higher frequencies this may be more difficult to meet the datasheet parameters. <S> You can read an excellent application note from TI on this subject. <S> Those are the primary issues that require a compromise as they have competing interests. <S> Whether there will be higher EMI or not is very device and layout dependent. <A> You've pretty much got it. <S> The efficiency is usually better with lower switching frequencies as well. <S> Possibly the most important criterion is that, if possible, your switching frequency and its harmonics don't overlap with signals of interest. <A> Possibly an unrecognised “pro” ( <S> Generally) <S> If your primary emitted noise is caused by a high level of inductor peak to peak current, then running at a higher switching frequency allows a smoother level of CCM (or actually running in CCM from DCM) and, as a result, the peak to peak current will reduce for the same output power. <S> This will of course reduce emitted noise but it isn’t necessarily a panacea to noise reduction generally in switching converters.
| A higher frequency requires a smaller inductor with lower copper ( \$I^2 R\$ ) losses as there are fewer windings for a smaller inductance with equivalent saturation current.
|
In an intrinsic semiconductor, why don't electrons go out from both valence and conduction bands? My textbook ( " Electronic Principles " by Malvino and Bates ) seems to suggest the free electrons in conduction band move left and reach the positive terminal of the battery. Does this mean electrons in valence band are not allowed to move left(hole moving right) and enter the positive terminal of the battery? Similarly, it says the electrons from the negative terminal of the battery enter from the right directly into the valence band holes. Does this mean the electrons from the battery are not allowed to enter into the conduction band? In summary: Electrons in an intrinsic semiconductor always leave from the conduction band and always enter from the valence band. This doesn't feel right. What am I missing? EDIT: Saying electrons in the valence band are not free, so the battery cannot pull them doesn't make sense because of the following scenario: If a p-type semiconductor is connected across a battery, the battery's positive terminal has no problem attracting the electrons from valence band. <Q> If it's intrinsic, there's as many electrons in the conduction band as holes in the valence band by definition, because each electron leaves a hole. <S> The electrons in the conduction band are bound to their atoms by (relatively weak) electrostatic forces, and can move fairly freely. <S> Bottom line is, all the electrons are moving to the left--some in long unfettered paths, and some from one atom to the next. <S> Also, in the valence band, it's kind of tag-team... <S> an individual electron moves to an adjacent atom leaving a hole, then another electron from the other side fills that hole. <S> It's just as easy to model it as a hole propagating than a tag-team of electrons. <A> Saying electrons in the valence band are not free, so the battery cannot pull them <S> doesn't make sense <S> The valence band is localized around an individual atom or molecule, so nothing in the valence band can be moved without first being removed from the valence band (which takes energy from the battery, thus contributing to resistance). <A> In an intrinsic semiconductor like pure silicon crystal, the valence band is entirely occupied at 0deg K, thus can't contribute to current flow, just as an entirely full bottle of water can't slosh about inside. <S> At room temp, there are a relative few free electrons in the conduction band and holes in the valence band, though not enough to be a good conductor. <S> I think Cristobol's comment "Electrons in the valence band are bound more tightly, and largely move only from one atom to an adjacent one that contains a hole. <S> " is just wrong. <S> And both are technically electron current, although hole current is more like a line of people (electrons) moving forward in a queue in the direction opposite to the hole current. <S> There just aren't many free electrons and holes in pure silicon. <S> The addition of P and N type dopant atoms greatly increase the population of free electrons in the conduction band and free holes in the valence band to serve as carriers of current. <S> But beccaboo, doesn't a high concentration of holes in the valence band imply electrons being free to move in the opposite direction of hole current (see above).
| Electrons in the valence band are bound more tightly, and largely move only from one atom to an adjacent one that contains a hole. It's my understanding that existing holes move as freely within the valence band as existing electrons do in the conduction band, which just means electrons are moving freely in the opposite direction.
|
Capacitors for 7808 and 7908 regulators I was looking at datasheets of the 7808 and 7908 regulators from STMicroelectronic. The mystery for me is why positive voltage regulators have capacitors without polarity (for example ceramic capacitors) and negative voltage regulators have capacitors with polarity (so electrolytic) and also why sizes of these capacitors are different. Schematic for 7808 is: Schematic for 7908 is: I also found a schematic for a +-15V power supply in one of these datasheets where capacitors sizes and types are also different: Thanks in advance <Q> The negative regulator may show polarized capacitors just to emphasize which way to connect them if polarized capacitors are used. <A> The datasheets generally show the minimum size of capacitor recommended for stability. <S> Have you ever wondered why there seem to be many more NPN transistors available than PNPs? <S> NPNs are 'better' transistors, in silicon at least. <S> The way the physics works, with the dopants available, they have better parameters, so are cheaper to make for any given specification, or not as good at the same price point. <S> So although the 7915 is 'just' the inverted polarity version of the 7815, it's been designed down to a similar cost, and that means it's not quite as fast or stable as the 7815, so needs larger capacitors at the terminals to control it. <A> As Justme says, the 78XX and 79XX series have different internal circuits and therefore have different requirements. <S> The 79XX requires a higher value capacitor on the input for stability than the 78XX series, and appears to need a low equivalent series resistance (ESR) in the capacitor: From the 79XX datasheet. <S> The low ESR is implied by the suggestion to use a tantalum part or an aluminum electrolytic at least ten times larger than the suggested tantalum. <S> ESR is higher in aluminum electrolytics, and is lower in larger parts. <S> The reason behind suggesting tantalum or large aluminum electrolytics is that the 79XX (and the 78XX) series parts are ancient. <S> They've been around since the 1970s, and were probably developed in the late 1960s. <S> Back in those days, a large value capacitor had to be an electrolytic capacitor. <S> There were no practical large value, physically small capacitors around. <S> A non-polarized capacitor with a few microfarads capacitance would have been several centimeters long in every dimension. <S> The 79XX probably doesn't care whether the capacitor is polarized or not. <S> It just "wants" a capacitor with a large enough capacitance and a low enough ESR. <S> You could probably use a modern, large value ceramic capacitor just fine if you remember to allow for the effects of the expected input voltage on the ceramic capacitor. <S> (Large value ceramic capacitors lose capacitance when operated near their rated voltage.)
| They are different regulators so they have different internal circuitry, requiring different capacitance for stability, and perhaps different capacitor ESR requirements for stability as well.
|
Amplifying the output of a thermistor I have a 220V heater which is equipped with a thermistor. Its value at room temperature (20-25 °C) is 10Ω . I'm trying to read its value with an Arduino but the output voltage is as low as 0.007V . Analog values on Arduino go from 0 to 1023 , where 0 is 0V and 1023 is 5V : I need to amplify this value or I won't be able to read it properly. I bought an operational amplifier ( LM358P ) and made the following circuit: simulate this circuit – Schematic created using CircuitLab It's my first ever time working with OpAmps and I'm not an electrical engineer, so the schematic could be completely wrong. If so, I'm sorry for that, it was produced by various internet searches as I'm not an expert. A few question you might have: Why R1 is 10Ω? Because I seemed to understand that it's easier to read values if the resistor between the thermistor and ground has the same value as the thermistor at room temperature. Why R2 is 10kΩ and R3 is 2.2MΩ? Because as far as I understood the amplifier gain is calculated by using the formula 1 + R3 / R2 , so in this scenario I should have a gain of nearly 220 (0.007V should become 1.54V). Why is the circuit like this? I tried to combine the way I usually read voltage value coming from a sensor with a non-inverting OpAmp circuit I found online, and that was the result. I'm not sure this is the right way to do it though, copy-pasting often leads to errors. Nothing is exploding at the moment, but the issue with this is that on the Arduino side I get random values between 250 and 300 . This also happens if I unplug the analog input, so I believe it's not working at all. Did I get it completely wrong? How should I modify the circuit to make it work? <Q> We have a XY problem here. <S> OP says they have a heating element with a sensor, which they believe is a PTC. <S> It is not. <S> After reading the long, long comment thread, I've come to the conclusion that what you actually have is a thermocouple embedded in the sensor. <S> Unlike an NTC or PTC, a thermocouple outputs a voltage proportional to temperature . <S> How to interface it? <S> The thermocouple outputs in the tens of millivolts range. <S> With this low voltage, what's needed is to use a differential amplifier, using both leads from the sensor. <S> But there's more besides just amplifying the voltage. <S> A common type is K (Nickel-Chromium / Nickel-Alumel) but there are many others with different output curves. <S> They are not linear, and need to have that corrected to get the right temperature value. <S> Here's a link to interfacing Arduino to Type K thermocouple: https://www.electronicwings.com/arduino/thermocouple-interfacing-with-arduino-uno <S> Lots of other ideas can be found <S> - thermocouple interfacing is a popular Arduino project. <S> [edited away - use NTC and voltage divider] <A> I doubt that you have a 10 Ω thermistor. <S> But here are a few problems with your circuit to keep you busy. <S> Figure 1. <S> The OP's schematic. <S> The voltage at points (1) and (2) is half-supply, <S> 2.5 V. <S> With negative feedback the op-amp output will adjust to try to get 2.5 V at the inverting input (3). <S> To get 2.5 V at (3) with a gain of 220 the output would have to swing to 550 V DC. <S> That can't happen with any op-amp I know of - even if it wasn't limited by a 5 V supply. <S> Try swapping the thermistor and R1 but change R1 to 490 Ω. <S> This will give you 0.1 V across the thermistor which when amplified by 220 will give you 2.2 V into your ADC. <S> This gets you started but not finished. <A> A thermistor converts temperature to resistance... <S> but you need voltage. <S> So, you have to convert its resistance to voltage keeping the current constant... i.e., you have to supply the thermistor by a constant current source. <S> A non-inverting amplifier with constant input voltage is such a source if you connect the load (thermistor) in the place of R3.
| Thermocouples also have different characteristics depending on the metal-metal junction they use. There are dedicated ICs for this, one of which appears in the Arduino project link below.
|
Confusion about NPN base connection I found an over-current protection schematic, but I am confused about how the base of the Q3 is wired. Is it connected to a potentiometer? If so, to what terminal? <Q> The base of Q3 is connected to the wiper terminal of potentiometer R4. <S> On schematics, potentiometers are normally drawn as a resistor with an arrow representing the wiper drawn at right angles to the resistor. <A> This is a voltage regulator with current limiting. <S> Here are some typical values with current limited to a bit over 100mA and regulating from 24v down to about 12V. R6 represents the load. <S> R2/R3 in my schematic replace the pot. <S> In general you usually model a pot as two resistors with the total equal to the element resistance (ignoring the usually small wiper resistance). <S> So if the pot element is R ohms you have one resistor <S> \$\alpha <S> \cdot R\$ and the other resistor \$(1-\alpha <S> \cdot <S> R)\$ <S> where 0 \$\le <S> \alpha\ \le 1\$ is the pot rotation as a fraction of the full electrical travel. <S> If you don't actually need a pot, the two resistors can just be absorbed into the series resistors R2/R3 in my circuit. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> You have dug up an ancient circuit; something that would have been used sixty years ago. <S> It works tolerably as a regulator ( assuming you have low performance requirements ), but there is a serious error in the current limit section. <S> A resistor should be put in series with the base, about 1K. <S> You can get Darlington transistors that would have Q1 and Q2 in one package. <S> A 6.2V zener is ideal here because it has a positive temperature coefficient that nearly cancels the negative coefficient of Q3. <S> For a stable output a capacitor needs to be connected across the output.
| Q1 should have a low value resistor (100 ohms) connected from base to emitter. Connecting Q4 straight to the current sense resistor will allow unlimited current to flow into Q4 and it will be destroyed before Q1 and Q2 turn off.
|
Lighting up old vacuum tubes and making them turn on/off I know there were older posts about lighting up old vacuum tubes but I would also like to turn them on and off. I am a complete newbie with no real electronic experience here so please bear with me. I bought an old cabinet full of brand new but very old tubes that I would like to use/light up as decorative items. Some gave great information on how to light them up, thanks so much, but I would also like to program the lights to go on/off in an interesting way. I know Amazon sells simple kits that you can create these kinds of things (at least I think they would.) But also what would you all recommend if this was my goal with these tubes? Again complete electrical newbie but very much appreciate any tips, tricks or insights on this project. Here is a possible Amazon product that I might be able to use. <Q> The filaments in electron tubes have a long time constant. <S> This means that when you apply the power so current flows it can take many seconds for the filament to reach its full glowing hot temperature. <S> This means that any "interesting pattern" that you may have anticipated will have a very slow rate. <S> I suspect that your thinking may have been based somewhat on what you have seen with LEDs that can switch on and off quite fast in comparison. <S> Switching the filament on a electron tube on and off in a repetitive manner can greatly shorten its life span due to the thermal cycling that the tube will experience. <S> To deal with the actual switching you need to find the voltage rating and current at that voltage that the filament operates at. <S> A cold filament will draw a lot of current until it heats up over the course of seconds. <S> You will need to ensure that any power source and switching circuitry that you use is fully able to deal with this inrush current. <A> The tube filament would normally be supplied AC, but DC will work fine also. <S> The parts kit that you referenced isn't that useful IMO. <S> Maybe somewhat useful for learning electronics, but not that useful for this project. <S> You need a transistor switch for each tube. <S> A MOSFET is a good choice. <S> It needs to have a low gate threshold, like the one in the schematic. <S> The filaments draw a lot of current, you will need a hefty power supply if you are powering more than a few tubes. <S> You need to research the tubes to determine the filament voltage for each. <S> 6.3V is common for tubes, but not for modern electronics. <S> 5V is common for modern electronics, you might see if 5V will light the tube satisfactorily. <S> Otherwise, you will need a adjustable power supply that can be set to 6.3V or whatever. <S> If you want to control more tubes than the quantity of GPIOs on your MCU, you will need a DIO expander. <S> All of this is fairly simple, but will still be challenging for a beginner. <S> Get some tube sockets, you don't want to solder to the tubes, they will burn out eventually and need to be replaced. <S> Cycling the tube filaments will reduce the life somewhat. <S> Maybe someone else can quantify it. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> There are some commercial products that light vacuum tubes (particularly those that don't have a Bakelite portion at the bottom) from below with LEDs, especially blue ones. <S> A bit jarring for the ancient EEs who remember a blue glow as a symptom of something very wrong in the tube. <S> The actual normal dull red filament glow of a receiving tube such as the once common 12AU7 or 12AX7 is pretty underwhelming, in my opinion. <S> However, if you want to try it: Here is a typical datasheet. <S> For 12V operation you could use a 12V power supply. <S> The filament current of this particular tube is 150mA between pins 4 and 5. <S> DC will work as well as AC for making it glow. <S> Suppose you wanted to control 8 on at once, which is 1.2A you could use a wall-wart switching supply with perhaps 1.5A capacity, and MOSFET such as the NTD4815N , which can be easily driven by the Arduino directly.
| If you have more than one type of tube filament voltage, you will need multiple power supplies. You could use a simple Ardunio in combination with some logic-level power MOSFETs to control a number of filaments.
|
Problems reading a signal with a MCU (TM4C123GXL) input pin I'm tring to read a signal generated by an op. amp. with a microcontroller through an input digital pin. All signals I'm working with are DC. The op. amp. is connected to a light sensor so that it outputs 0V when the sensor is shaded and between 4.8 and 5.2 volts (fluctuates all the time) when the sensor is detecting light. I have measured this signal with a multimeter. The microcontroller's pin is configured as a digital input with a pull-down resistor. The voltage levels for the pin are a logical 1 if the input voltage is 3.3V < voltage < 5V and a logical 0 if voltage < 3.3. The problem I have is that when I connect the op. amp.'s output to the input digital pin, and the sensor is enabled, the voltage drops to something between 0.8V and 1.5V, and it doesn't reach a steady level (always fluctuates, wider range than before). The microcontroller is supposed to change the color of an LED depending on the digital input state. What happens is that the LED blinks all the time. In addition, However, I have connected a steady 3.3V signal coming out of another microcontroller's pin (output) and everything works as expected. In addition, I have measured the voltage at the output pin and it doesn't drop as it happens with the op. amp. I'm not sure I can connect the op. amp. output directly to a digital input pin. My questions are: Is the op. amp. output signal considered analog? If so, if an analog signal has two known 'steady' states, can it be considered a digital signal? Can I connect that signal directly to the input pin, given that it is still DC? Do I need to connect it to an ADC? Note: The microcontroller is a TM4C123GXL . I'm configuring the pin PB4 as a digital input and PB6 as a digital output (3.3V). When connection PB6 to PB4 , everything works as expected. EDIT The schematics is here: Notice that R3 is a photocell. The op. amp. is a OPA2344 Currently my MCU is powered over USB, from my laptop. Notice that I'm connecting the op. amp. output directly to the digital input pin. Not sure if this can be done. As you can see I'm using a voltage divider . This is because, if I power the op. amp. with the 3.3V coming out of the regulator, its output is always below 3.3V, so the MCU will never interpret it as a logic 1 . I would like to do this without the help of a transistor. EDIT 2 I have tried it again powering the op. amp. with the steady 3.3V coming out of the regulator. The output of the op. amp. is steady at 3.3V when the photocell is not shaded, as soon as I connect it to the MCU digital input it drops to 3.29V, so the MCU doesn't detect it as a logic 1 . I have observerd one listake I was doing: I was connecting the MCU GND pin to the common ground. That makes everything 'go mental', I think that is what caused the voltage drop at the op. amp.'s output I saw before. I deduce that it's wrong to connect the MCU GND to the common ground, why? <Q> From your explanation, it sounds like you have configured the Op-Amp to output 0-5 V and the microcontroller's digital pin is trying to read a 0-5 V signal. <S> This appears to be the cause of what you are seeing at the output. <S> Have a look at the microcontroller <S> TM4C123GXL's datasheet here : http://www.ti.com/lit/pdf/spmu296&ved=2ahUKEwjP1-DeqPrnAhVkTxUIHYoLDiEQFjAAegQIBxAC&usg=AOvVaw0XulafQFE2JHx9LJ_xC5p-&cshid=1583103090057 <S> As @Ron and @Steve suggested, please provide a schematic of your circuit. <A> First of all you need a common ground. <S> If you only connect the output from the op-amp to PB4 of the MCU, you have established a common reference voltage. <S> But no current will flow before you make a second connection between the two so that there is a loop between the two circuits. <S> Also you can use one of the ADC channels of the MCU and read the photocell directly from it without the op-amp. <S> Use the photoresistor in a voltage divider. <S> I would also have have used only one power source. <S> So either power everything from the regulator or from USB. <A> The problem was that I was powering the whole circuit with a 9V battery (converted to 3.3V through the regulator) but the MCU through my laptop's USB port. <S> Disconnecting the op. <S> amp. <S> output from the MCU input <S> pin <S> I could measure 3.3V steady. <S> In the moment I connect the MCU <S> GND (ground) <S> pin to the bradboard ground (which is connected to the battery's negative terminal) the op. <S> amp. <S> output goes mental (starts to jump all the time from 1.5 to 2.3 V aprox.) <S> What I have done is disconnect the MCU from the laptop and connect its GND pin to the breadboard common ground, <S> the VBUS pin to the regulator's output (3.3V) and the op. <S> amps. <S> output to PB4 <S> (input pin). <S> By doing so, everything is working as expected. <S> EDIT <S> I have tried a different thing: connect the MCU to my laptop and power the board with its 3.3V pin. <S> At the beginning it didn't work (the op. <S> amp. <S> output dropped to 0V), and I have discovered that that was caused by the multimeter. <S> I have disconencted it and everything works.
| Please note that the microcontroller TM4C123GXL, which you are using, operates with I/O logic levels of 3.3 V and you are using a 5 V logic to drive the microcontroller's digital pin.
|
Best approach for sharing SPI bus across multiple sensors I'm designing a few PCBs for a project: Microcontroller PCB (SPI master) Sensor PCB, where each sensor PCB has several SPI slaves on it. In a full system, there would be 3 sensor PCBs directly plugged into 3 connectors on the microcontroller PCB. In certain scenarios, not all 3 sensor PCBs will be present. My question is about best practices/approach for sharing a single SPI controller across all 3 sensor PCBs. I'm planning to run the SPI bus at 20 MHz and the longest end-to-end trace length between the microcontroller and SPI slave will be about 8 inches. Some ideas I had: Use a 'SPI bus multiplexer' to isolate which Sensor PCB is connected to the SPI bus at any given time, but haven't had much luck turning up results. Was hoping to find something similar to a TCA9544A (I2C multiplexer). Use 4-bit tri-state buffer on the microcontroller board for each Sensor PCB interface. In this case, I'd have 3 of these on the microcontroller PCB. Using the output enable to connect/disconnect a sensor PCB from the shared SPI bus. Update: Here's a diagram of what I'm trying to do: I suppose I'm more concerned with signal integrity and EMI/reflections. I believe I have an understanding of the functional connection aspects. How should I handle the inter-board connections, given that a sensor board may or may not be attached? Should I use tri-state buffers at the interface? <Q> question is about best practices/approach for sharing a single SPI controller across all 3 sensor PCBs. <S> The standard SPI configuration in most of the MCUs support multi slave configuration. <S> Post the datasheet of the actual MCU and we can together see it. <S> The slave select lines will define which space device will actively use the rest of the bus. <S> The other three lines can be shorter together. <A> I would definitely avoid running a single SPI bus on 3 different boards without some sort of switching method. <S> I would suggest using the TS3A5018 4-bit analog switch from TI if your SPI's logic level falls between 1.8 and 3.6V. Using a similar circuit to this one: <S> And the following logic: <S> You can effectively isolate the rest of the bus from the active path (eg. <S> to the selected sensor) and avoid SI related issues. <S> You also avoid running high-frequency content all over your system at once which should mitigate the emitted energy, and therefore, EMI concerns. <A> I believe the best approach would be to use SPI-controlled switches like ADGS1408 to select slave device on sensor PCB. <S> This way same bus will be used for both communication and multiplexing. <S> And you only need 1 additional wire per board - when it is low it selects switch itself, when it is high the inverted low signal is used as multiplexed SS for sensors. <S> So, to access certain sensor you first pull board SS low, send sensor address to a switch, then pull SS high and communicate with sensor. <S> In this case you first send board address to local switch, then sensor address to selected board, and then communicate with selected sensor. <S> Note that if bus capacitance becomes a problem you can use switches to multiplex bus wires too. <S> You only need two dual ADGS1409 to switch three bus wires and board SS.
| On master PCB you can select active board directly from MCU (if you can spare pins) or you can add switch(es) to select between as many boards as you like.
|
RC Joystick Question: What is this component? I am a electronics novice and I had a question regarding a small RC project I am working on. I am trying to control my transmitter with a raspberry pi. The RC unit is a twin electric engine boat. The RC transmitter is very simple. There are two joysticks that move in only one direction (UP or DOwn). Center position engine is off, as you move the joystick forward, the engine moves forward. If you pull the joystick back, the engine goes in reverse. When the joystick unit is taken apart, what you have is a double metal spring plate that each is over some sort of coil or resistor. I have taken a picture. I am not sure what the component is in this picture and that is the first question: What is this component? I have added a blown up pictures of the component here as well Analysing the joystick switch and joystick circuitry, there are three wires going to the joystick(blue, black and yellow). The middle (black) is ground. Checking the voltage with a voltmeter after having confirmed which wire is GND, I get -3 Volts on each of the non GND Pair (blue and yellow) when checked against GND. I am a little confused about the -3 volts. How do you get negative voltage? I understand if you get the GND and positive reversed, but I confirmed the GND wire was indeed the GND. If you move the joystick forward one of the metal strips moves closer to the component and the voltage on the blue wire moves closer to zero while the yellow stays at -3V. At max forward, the voltage is zero on blue and -3V on yellow. If you move the joystick back to center, both are at -3V. As the joystick is pulled back, blue stays at -3V and Yellow heads toward zero. Again at Max joystick back, the volt value is zero. Second question is about the -3 volts and how that is achieved? Can someone explain how this circuit is working to me? The -3 Volts is what is throwing me... <Q> The component is a simple jumper (notice the letter J), it's there so those "contacts" can make contact when you turn it. <S> like a switch. <S> Three wires are; two positive and one negative. <S> so when it turns it pulls down probably an MCU pin to ground when you turn it. <S> I think you're measuring wrong with you're multimeter, measuring the polarity reversed <S> so it shows you negative voltage. <S> If you're sure of measuring the voltage correctly, you have to show us the whole circuit/schematic to be able to say how -3 V is generated. <A> (that's why the board designation is 'J2'). <S> It is being used as a contact for the switch. <S> When the the spring is pressed down on it a connection is made between the common wire and one of the other wires. <S> I am a little confused about the -3 volts. <S> How do you get negative voltage? <S> I understand if you get the GND and positive reversed, but I confirmed the GND wire was indeed the GND. <S> Three possibilities. <S> The wire that you confirmed as ground is not actually ground. <S> The circuit generates negative voltage for some reason (unlikely). <S> The circuit has positive ground (even less likely). <S> To understand what is really going on you should trace out the entire circuit , then draw a neat schematic showing all component designations, part numbers and/or values, wire colors etc. <S> If that doesn't tell you why you are getting negative voltage then post the schematic here. <A> The reference designation J <S> * means "jumper"; it is a zero-ohm resistance. <S> When the user pushes the joystick handle to its maximum travel, the metal spring plate makes contact with the jumper's metallic end-cap which generates an "ON" output signal—i.e., <S> ON forward, ON reverse <S> , ON left, ON right. <S> This assembly effectively functions as a limit switch . <S> In this configuration, the joystick control produces an ON_DOWN | OFF | ON_UP output signal rather than a continuously varying output signal. <S> Without knowing how you determined the black wire is ground, there are two possible circuit topologies: <S> First, the black wire is actually +3 VDC, not ground, and the circuitry for each spring clip looks something like Fig. <S> 1 where the right side of switch SW1 represents the spring clip + jumper cap contact point. <S> Each joystick control has two of these circuits: one for ON_DOWN and one for ON_UP. <S> The ON_X signals are digital signals that are connected to two digital inputs on the controller/radio component. <S> When the joystick is in the neutral position, SW1 is open and ON_X is <S> 0 V. <S> When the joystick is at full travel, SW1 is closed and ON_X is +3 V. Resistor R1 could be a circuit board component, or it could be an internal pull-down resistor within the controller/radio component. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 1. <S> Second, the black wire is ground, and the circuitry for each spring clip looks something like Fig. <S> 2 where the right side of switch SW1 represents the spring clip + jumper cap contact point. <S> Each joystick control has two of these circuits: one for ON_DOWN and one for ON_UP. <S> The ON_X signals are digital signals that are connected to two digital inputs on the controller/radio component. <S> When the joystick is in the neutral position, SW1 is open and ON_X <S> is +3 V. <S> When the joystick is at full travel, SW1 is closed and ON_X is 0 V. Resistor R1 could be a circuit board component, or it could be an internal pull-up resistor within the controller/radio component. <S> simulate this circuit Figure 2.
| The component is a 0 Ω resistor, AKA a jumper
|
How do I correctly power the l298n motor driver with 6v? Recently, I have purchased the L298N Motor Driver with a couple of 3V-6V motors. I am planning to make a robot buggy to hold other projects I may do with the Raspberry Pi. Currently, I have been unable to figure out how to properly power the motors and the driver at the same time. For now, I have taped four AA batteries together to test the motors on with the driver. On my version of the L298N, the power inputs include 12V and 5V. After scouring the internet for hours, I have not been able to figure out exactly what to do with my dual-motor setup. For now, both of the jumpers on ENA and ENB have been attached. I tried connecting the 12V input on the driver to the temporary AA batteries, but the red power LED on the driver didn't turn on. The red LED would turn on though when I supplied 5V through the 5V punch block. That had not powered the motors sufficiently. What is the proper way to do this?Here is a schematic of the wiring I have set up.Keep in mind the pins that the wires are connected to on the Raspberry Pi are random (Couldn't "pin" them down correctly) <Q> Why do you have 12V on the battery?Pin 9 of the L298N <S> should be 4.5V to 7V for the logic. <S> Pin 4 is the 6V for it to power the motors. <S> Connect the +6V to pin 4 and pin 9 and the ground of the 6V is 0V for pin 8. <S> Pin 1 and pin 15 should also be connected to 0V. <A> 4 batteries add up to 6V, apparently this board requires 9 to 12 volts power supply. <S> The 5V next to the power supply input is an output which you can use to power your raspberry, but make sure to not mess up the ground as you have in the picture you've shown. <S> Ground is physical pin 6 on the raspberry, not 5. <S> If you really insist on powering it with a 6V supply, then get a boost converter. <A> You need to connect 5V for the logic FIRST, then connect the motor power on the second power block. <S> Here is some proof: The image is from the Arduino forums where someone had the same problem as me here .By <S> the way, you cannot power the L298 motor driver module with 6V. <S> The minimum required for it to work properly is 7.5V.
| For anyone else looking for an answer:You cannot expect the motors to be powered by connecting 5V to the driver board. Try throwing a couple more batteries in series or get a 9V battery instead.
|
Why do charges not lose potential as they travel through the circuit before reaching a resistor? Voltage is defined as: Given the circuit: Why would potential not drop, from the point differentially away from the positive terminal on the battery, to the point I have labeled. I understand that charges lose energy in resistors due to inter lattice collisions, where the acceleration generated due to thermal motion and voltage bias is translated from kinetic energy to heat energy. Before the resistor the resistance is neglible so the drift velocity will be high, and collsions will be minimal. Even so as the charges move from the positve terminal the Electric field will be strong and we will be losing the potential energy as we travel in the direction of the field. With this so why dosent potential drop as we move in the direction of the field? Is the energy associated with the charge essentially translated from potential to kinetic energy, where very little of this kinetic energy is translated to heat energy, and we assume the total potential is associated with the total energy the charge has at this point (KE + PE)? I am really looking for an answer to my question. It would be much appreciated if we address the core question I asking instead of side discussions! After the initial answer, I encourage side discussions if wanted! <Q> The resistance in a real circuit would be far higher than the resistance of the lead wires, so the bulk of the electric field would be seen across the resistor. <S> Now we need to interject some reality. <S> In a schematic, the resistor is a lumped element defined by properties at its terminals. <S> IRL, however, it is a volume of resistive material. <S> Two of the three dimensions of this volume are typically width, and the third is length. <S> If you take some fraction of that length, it will have a smaller electric field across it than the whole device does. <S> If one side is ground, different parts will show different potentials depending on how close they are to the non-ground terminal. <S> This is, in fact, how analog potentiometers work--the center terminal moves from one end of the fixed resistor to another, changing its potential depending on its position. <S> So the electric field is distributed around the circuit. <S> The trick is that most of it is distributed across the resistive element, with very little distributed around the lead wires. <A> Imagine that there was nothing in between the source and the resistor. <S> They are still connected, but there is no "circuit" between. <S> No resistance at all...not just negligible, but nothing. <S> There could be no loss in potential energy, no E-field in the space between the source and the resistor, because there is no space between them. <S> That is the situation depicted in the schematic. <S> We draw the voltage source and resistor a little bit away from each other because it makes the schematic easier to understand, but effectively there is an ideal conductor of zero length between the two elements. <S> There is a strong temptation to take what you know about real circuits and impose that on an ideal circuit, but you have to understand that the schematic is just a representation of a theoretical ideal situation...without making this leap we couldn't apply rigorous mathematical analysis to the circuit. <A> Taken from Matter and Interactions 4th Edition. <S> This is the exact solution I was looking for @all. <S> The battery electric field dimishes with distance but the surface charge density of the circuit rearranges itself due to feedback. <S> Around the resistor charge builds up applying a field counteracting the field of the battery and other surface charges. <S> Credit to @jonk for the tip. <A> Your resoning sounds logical. <S> Consider also the fact that if the E-field in the perfect conductor were large then the current throuh it which is proportional to the product of conductivity and E-field , would be huge which is not the actual case. <A> The E-field in perfect conductors is zero under steady state conditions (such as your circuit). <S> So, the integral in the equation you supply, vanishes. <S> Thus, ΔV=0, i.e. the potential is constant along a perfect conductor.
| Since chagres flow through the resistors slowly the surface charge distribution from the battery to the top surface of the resistor is essentially uniform, creating a very small E field.
|
Power supply capability of the circuit I do not want to include a dedicated power supply for the operation of a low current demanding microcontroller chip in one of my projects. Instead I wish to use something which would just work fine. I couldn't think anything simpler than this circuit. Though it may seem a very silly circuit and may fail at higher current requirement, but I guess it can power a microcontroller and a few LEDs(2 or 3) which will require just few milliAmps of current. Also, I can prevent the resetting of the IC due to minute voltage irregularities by adding some ceramic cap near the it. This is what I can see as of now. Is this feasible to make and use such a circuit for practical purposes. What could go wrong? <Q> The current coming from the bridge rectifier will charge all the capacitors equally (until the total voltage is about 282 volts. <S> But the microcontroller only discharges capacitor number 1. <S> An equilibrium will be reached when capacitor 1 is empty, and capacitors 2-10 have about 282 volts in total. <S> Then your microcontroller won't work any more because it has no power. <A> In theory, it is possible to make this circuit work. <S> And I actually saw a similar design used on an old wall-rechargeable flash light. <S> As for your question "What could go wrong", my answer is: 1 - Personal safety concern, as stated in the comments, is a major one. <S> There is serious risk of electrocution. <S> 2 <S> - In case of any cap failing short, the rest of the capacitors will have to split that cap voltage, bringing them closer to their max rated voltage, thus making the whole circuit even more likely to fail. <S> 3 - Linear regulator (IC7805 in your circuit) will need biasing current which reduces the current budget of the power supply. <S> I refrain from proposing solutions just in order to stay within the limits of your questions. <A> What could go wrong? <S> The biggest problem is that this circuit is not isolated from the AC mains. <S> This means that if you were to touch part of the circuit you could get an electrical shock. <S> If you happened to be in the wrong situation (which is more common than you would think) <S> the electrical shock can cause a few milliamperes of current to flow through your body and stop your heart. <S> Result - you die. <S> The second problem, after your gone, someone investigates your device and suffers a like result. <S> The third problem is that using direct AC attached circuitry in an experimental manner can lead to sparks and fire when something accidental happens. <S> That could be when a test lead comes loose and flops into the wrong place, you connect a meter incorrectly or even getting distracted by something like your child or cell phone. <A> On the DC side you get something like a constant current source that can be stabilized w/ parallel stabilizer. <S> A capacitive divider on the AC side is also an option and will be more like your picture. <S> Then again, looking at how you are asking, you are better off (and way safer!) <S> with something factory-made. <S> p.s. <S> where is this 200v mains? <S> If it is Japan, you are playing with something in the kilowatts range, adding a fire to the electric hazard. <S> Please, improve your competence first. <A> That you are building is called "transformless power supply". <S> They were popular in the past, when switching power supplies were still complex and expensive so the only alternative used to be the large 50 Hz transformers. <S> They were both lighter and cheaper. <S> You can read more about them here . <S> The capacitors should be placed before the bridge rectifier, not after. <S> And you need one more bigger electrolytic capacitor across the output of the rectifier to convert pulsating current into smooth and also some kind of voltage stabilizer. <S> These supplies do have advantages but they are quite dangerous devices that must be properly wired, grounded and isolated, with many things that can go wrong. <S> I would strongly advice against using this design for exposed circuitry in amateur experiments. <A> Look at the notes on the circuit I've added and consider: -
| A lot of power supply circuits for similar purpouses are made with capacitor in series on the AC side.
|
CAN: Difference between an MCU with CAN embedded peripherals and one that does not? For example, an MCU (stm32, esp32, Pic18f .. etc) with CAN embedded, sends logic level signals through pins Tx and Rx. They only need a signal converter. (example: MCP2551). My question: Why can't any MCU with tx and rx pins use just one signal converter? Why does this also need an adapter? (Example MCP2515). <Q> You can avoid a lot of confusion if you get your terminology right. <S> To access CAN bus the MCU needs two things: CAN Controller and CAN Transceiver. <S> Controller has Tx and Rx pins. <S> Transceiver converts these pins into CANL and CANH bus signals. <S> When you say "MCU with CAN embedded" what you really mean is that CAN Controller is already part of MCU, so you only need external Transceiver (e.g. MCP2551) to make it work. <S> When MCU does not have embedded controller it needs both external Controller (e.g. MCP2515) and a Transceiver. <S> Finally, when people say "adapter" they usually mean external device that has both Controller and Transceiver on one board. <S> Note that MCP2515 is not an "adapter", <S> just like MCP2551 is not a "converter". <S> So, to your actual question: Why can't any MCU with tx and rx pins use just one signal converter? <S> Why does this also need an adapter? <S> the answer is: MCU with embedded <S> CAN controller does NOT need an adapter. <S> However "any MCU with tx and rx pins" does not describe such an MCU, because having Tx and Rx pins is too ambiguous, it can refer to UART interface, for example. <A> There are CAN controllers which have built in CAN transceiver as well. <S> Only an example: http://www.ti.com/product/TCAN4550 <S> The driving factor for the dedicated controller to not have a transceiver built in is price, size of the IC. <S> It is expected to use any already available cheaper transceiver and to be able to have wider options. <S> From Renesas note: <S> Definitely not compliant to CAN protocol <A> You seem to be confusing two different things. <S> Microcontrollers also have UARTs that have TX and RX pins. <S> That only talks messages over the asynchronous start stop protocol and is not compatible with CAN protocol, even if they both have pins named TX and RX. <S> If a microcontroller has a CAN peripheral with TX and RX pins, it means it has the necessary hardware to send CAN frames, or message bits and bytes with appropriate headers at some defined baud rate, but as the microcontroller IO pins use for example 3.3V CMOS logic levels, it needs a physical interface (PHY) chip, also known as a tranceiver, to convert the 3.3V CMOS levels to the necessary 12V CAN bus levels. <S> The CAN controller may or may not contain the physical interface, and thus it may also need an external PHY chip. <S> The MCP2515 is also only a CAN interface and it still needs a CAN PHY chip like the MCP2551. <A> The CAN protocol requires a specific collision detection mechanism that is difficult to implement in software, because all stations that have outgoing data start sending at the same time and then one after the other drop out as they determine that a station with a higher priority is also sending. <S> Having a CAN controller inside the MCU means that this protocol is built as a dedicated circuit, so the controller can be connected to a CAN bus (through an external transceiver) with little effort, while an MCU without a CAN controller would require a complex interrupt routine that would take a significant amount of processing time and use a bit more power.
| If microcontroller does not have a CAN peripheral, then it can't directly send CAN frames, and in such a case usually a separate CAN controller chip (e.g the MCP2515) would be connected to the microcontroller via some other bus like SPI.
|
What is the noise figure of a signal generator? I'm wondering what's the noise figure or the overall noise contribution of typical signal generators? For example, if there is a test system with a signal generator that outputs 3 dBm signal (50 ohms), an amplifier with 3 dB noise figure and 10 dB gain and a spectrum analyzer with lets say -140 dBm noise floor with 100 Hz RBW. How much does the signal generator raise the noise level and how much is the noise floor when the generator and amp are connected to the analyzer? <Q> How much does the signal generator raise the noise level <S> What noise level? <S> The Spectrum Analyzer (SA) has a noise level of -140 dBm, but that doesn't change. <S> What this -140 dBc means is that you cannot measure below that level <S> , it is the limit of your SA. <S> Your signal generator (SG) will also have a noise level (look that up in its datasheet). <S> Let's assume it is -90 dBc <S> (90 dB below the signal). <S> Your signal is + 3 dBm <S> so that means that the noise is at -87 dBm. <S> The 10 dB amplifier amplifies <S> both signal and noise by 10 dB. <S> On top of that the amplifier will add 3 dB of noise (NF = 3 dB). <S> So the signal at the output of the amplifier will be +3 <S> dBm <S> + 10 dB = <S> +13 dBm (note how I can just add a signal level in dBm and a gain in dB ). <S> The noise at the output of the amplifier will be -87 <S> dBm <S> + <S> 10 dB <S> + 3 dB = <S> -74 <S> dBm <S> (the 3 dB is from the noise figure). <S> So at the Spectrum Analyzer will just show a + 13 dBm signal with a -74 dBm noise level. <S> That -74 dBm noise level is much higher than the -140 dBm noise floor of the SA so the noise floor of the SA can be ignored. <A> Well specified signal generators should have some data how much the signal contains noise and how it's distributed across different frequencies. <S> You should calculate the total noise power in watts, not in dBm <S> :s. <S> You should add together in watts what's coming from the generator and amplified, what the amp outputs (=theoretical thermal noise amplified by the gain and noise figure) and the equivalent input noise of the analyzer. <S> All must be limited to the bandwidth in use. <S> Without knowing how much your generator creates noise as summed to the payload the calculations must be based on guesses. <S> Noise figure isn't useful quantity for signal generators as others have already stated in their comments. <A> I think that your question is academic. <S> And if what you are really wanting to know is how much noise is contained in your Signal Generator's output I would use a spectrum analyzer with the ability to digitally noise blank the noise floor of the SpecAn then directly input your Signal Generator output into the SpecAn's input and see if it is clean or not. <S> Perhaps this answers what you are wanting to know.
| If the whole output of the signal generator is considered to be the wanted signal, the generator adds no noise.
|
Unity Gain Buffer as a Current Source I keep having issues with creating a unity gain buffer amplifier to isolate a voltage source. My voltage source is very limited in current ~1-2 mA and I need that signal to drive a transformer at 10 VAC. I thought it would be as simple as using an op-amp, but I am finding out that op-amps do not like to source more than a few milliamps. My transformer draws about 20 mA at 10 V. I thought I could use a BJT or two to help source this current, but I'm finding out they do like my 15V rails because all they do is get hot and output nothing. I have them in a push pull configuration. They are darlingtons rated for much more current than what I'm drawing. I am going to be using an NI cRIO module to create a 10 VAC sinewave and all I need is to find a way to buffer the current output, so that I can drive this unloaded transformer. Someone mentioned the LT1210 but I seem to be missing something since I lose half of my input voltage. ignore the .01 Resistors. I have them there to measure current through that trace. <Q> You have to use a power amplifier or maybe a line driver. <S> These type of opamps can supply more current than the standard opamp. <S> (You can find some e.g. at TI website under Amplifiers/Line drivers or Amplifiers/Power Opamps). <S> If you are sure that you don't need more than 20mA, you might also find some standard opamp. <S> Take care, that the higher the output current, the lower the output swing - check that in the datasheet. <S> Extra tip, because I failed to do this a few times: You also need to take care of the self-heating of the amplifiers. <S> It is usually the sum of the intrinsic heating (quiescent current times supply voltage) and the load dependent heating (output current times the voltage drop on the opamp). <S> Multiply this power with the thermal resistance and check whether you are in the operating junction temperature limits. <S> Edit: as pointed out by electrogas also check out the TI AN-31 OpAmp Circuit Collection . <S> You find the current sink on page 21. <A> My power supply is some weird Chinese thing that is switching voltage at 2MHz. <S> This switching voltage is near a full volt +&-. <S> This is the root cause of all my worries. <S> The LM1875 can handle this switching much better than my other op amps I was using. <S> This still causes problems because the sensor I am trying to test is a CT sense which uses the voltage drop across a .1 Ohm resistor. <S> At this switching voltage it would immediately ruin my sensing controller. <S> Luckily I never actually tried to run it, I've only been using test resistors and a multi meter. <S> My original problem with the transformer issue was solved fairly quickly once I found a suitable heat sink and changed my design from a buffer to a simple unity gain inverting amp. <S> I believe it was this switching frequency causing me issues with that first design. <A> So you are driving a transformer from an opamp. <S> The transformer only requires 20mA which the opamp should be able to provide, yet it overheats and fries. <S> First suspect is DC offset. <S> If the transformer requires 20mA AC current but the opamp has a DC offset on the output, and the transformer winding has a low DC resistance, then a lot of current may flow. <S> So you should measure DC output voltage. <S> You can use two electrolytic caps in series back to back to make a non-polar electrolytic cap. <S> Capacitor value should be large enough to pass your signal frequency. <S> You can also add a resistor in series with the opamp output, maybe 50 ohms. <S> At 20mA it won't drop too much voltage and it will protect the opamp. <S> If this is not a DC offset problem, then you need a more powerful opamp. <S> Since you need 3 channels, the cheapest and simplest high current opamp option will be LM1875, but you can also use a 5.1 home theater amplifier. <S> You will probably need to add a few ohms resistor or a cap in series with the transformer if the primary coil resistance is too low. <S> This is again to avoid high DC current due to DC offset, which will make your amplifier hot even without signal and may trip its protection circuit. <S> Now, that said, it's quite weird that your opamp fries with only 20mA current. <S> Maybe it's oscillating (check with a scope). <S> In this case, add supply decoupling caps and a resistor on the output. <S> Add a 50R resistor on the output of each to make sure their different DC offsets don't cause extra DC current.
| You can also parallel several opamps to increase current. The fix is to add a coupling capacitor in series with the transformer.
|
Is there a rule to calculate the correct voltage for a light bulb? Say I have an unknown led light, independently several person know it could be 12v or 24v or 110v even 220v Say It's rated for 9v (and you dont know), let's say you start applying fewer volts until it start turning on and stronger, but how can you know the PROPER and CORRECT voltage? Does it apply for a small motor? Rated say at 6v but you think it's rated at 12v or 9v. What's the rule resistance? watts? amps? Thanks in advance <Q> Rated voltage is chosen or defined based on material and/or component properties and also on endurance tests of multiple samples. <S> Therefore, without knowing what's in the "black box" you cannot know the rated voltage. <S> E.g. different LED's with the same voltage rating (can) have different intensities. <S> If you apply the rated voltage to a ultra bright LED and apply an overvoltage to a 'normal' LED, they may have the same intensity. <S> But the normal LED may get damaged due to the overvoltage. <S> So, checking the intensity of LED's cannot be used to trace back what the rated voltage would have been. <A> Here is a sample link which addresses this thing: https://www.micromo.com/technical-library/dc-motor-tutorials/motor-calculations <S> And to be considered:For a DC motor which run below its nominal voltage has slower no-load speed and a lower stall torque. <S> This not cause any problems with the motor. <S> If anything, the motor will last longer because it won’t be subject to the same level of stress, heat, friction and general wear as it is when running at the full nominal voltage. <S> And if it runs above its nominal voltage, it is going to magnetically saturate the steel in the motor’s core. <S> This increases the current drawn causing the motor to run hotter and shorten the motor’s life. <A> The current/voltage chart of the LED is somewhat J shaped: <S> LED conducts much more (and emits light) above the certain threshold voltage. <S> This voltage is about 1 V for infra red, 1-2 V for the visible light and can be even higher for UV LED. <S> The operating current of the LED should normally be above that threshold while it is not obvious how much can you drive it up. <S> You can take a current vs voltage chart with a suitable milliamperemeter (0 - 50 mA range should work for the most LEDs) and regulated 0 - 4 voltage supply, start from the 0 voltage and do not overdrive. <S> You can do this safer if you have a second voltmeter: put a 100 ohm resistor between power supply and LED and measure the voltage directly on the LED. <S> It is more tricky if the thing has multiple LEDs and additional electronics inside. <S> But it this case it is a device, not a component. <S> It should have some name, or number printed on it that may tell more if googled on the web.
| Without having a number of samples, you cannot know test/deduce what the rated voltage is.
|
Circuit for another circuit proximity detection I just want to ask for some ideas, are there any known low-current circuits to detect direct proximity (max 1-3 cm) of another specially designed device? Must be difficult to "trick" by any other common-world influence, if possible. My goal is to design two PCBs which are only activated by close presence of each other (maybe by some EM/inductive means), both battery-powered. <Q> One idea that immediately jumps to my mind is to use an RFID tag in one of the devices and an RFID reader in the other. <S> The reader should be able to be keyed to a specific RFID tag. <S> You could also support a set of more than one RFID tag or make categories of types of tags. <A> Look for transponders, also known as key fobs from NXP. <S> https://www.nxp.com/products/security-and-authentication/secure-car-access/tiny-single-chip-passive-keyless-entry-go-solution:NCF29A1MHN <S> They do not need any power and will be able to work just like normal RFID tags. <S> Other options are PKE (passive keyless entry) concept used in cars. <S> As soon as the transponder enters into the range, the transponder will be able to ACK and communicate. <S> The range can be controlled with a 3 D antenna for longer distances (upto 100cms) or design without an dedicated antenna to about a few cms. <A> If both devices need to be battery powered and you need more than a few hours/days of battery life, you probably need a two-level system: A first level needs to detect the presence of the other device using as little power as possible. <S> RFID tag systems or equivalent solutions are not a good idea at this level, as the reader will draw quite a bit of power all the time (or very frequently) just to detect if a tag is present (it would need to actively scan for tags). <S> Instead, you could use a reed switch like this one for instance on one device, and a small magnet on the other device. <S> A reed switch is normally open, and closes when a magnetic field is present. <S> This is what is used for door sensors (like this one for instance). <S> This would allow you to have your MCU in sleep mode, drawing very little current, and wake it up when the contact closes.
| Then, to identify you have the "right" other device, you could then use an RFID/NFC reader/tag pair, which would then be activated once presence of the other device is detected.
|
Why would you use a feedback resistor on a logic buffer? Please excuse my poorly drawn photo-interrupter circuit. It is used for external home sensing where a flag will interrupt the light source, turning off the phototransistor. I'm just curious as to why somebody would include the feedback 47.5k resistor here. From my understanding, it is parallel to the 20k resistor when the home signal is high, and is in parallel with the phototransistor when the home signal is low. This would increase the current when the phototransistor is off and have a negligible effect when the phototransistor is on. The 74LVC1G17 already has hysteresis so I am just asking what the benefits of this resistor are. <Q> The resistor adds additional hysteresis beyond what the chip provides. <S> With the nature of what the internet is these days quite possibly you may have found a circuit idea that was designed before the time that the a Schmidt Trigger type IC was used in the circuit. <S> It is quite common for photo detectors to operate with very long rise and fall times. <S> When a non Schmidt Trigger type IC chip buffer is used the slow transition signal at the output of the opto coupler can cause the buffer chip to transition multiple times. <S> In some applications this is not a problem bit in others such as a counting application it can be a big problem. <A> Since the switching point (voltage level) for most digital devices are only roughly defined (as opposed to an analog comparator, for instance), this falls into the category of a hack, IMO. <A> The combination of the two resistors could be used to set the flag positions at which the circuit turns on or off. <S> To use a homing switch like this, you can move the actuator toward it fairly quickly, until the switch closes, and then move it back slowly so you can precisely determine the position at which it opens again. <S> The distance between those two positions needs to be big enough to take up any backlash in the actuator mechanism, and let the actuator accelerate to constant speed before the switch reopens. <S> In addition, you may want to set the position at which the switch reopens to the point at which the circuit is most sensitive (max dV/dx) so that the position measurement is as accurate as possible. <S> If you care about the precise positions where the switch turns on and off, then it's unlikely that the hysteresis build into the Schmitt trigger will be exactly what you want.
| Looks like it's somebody's attempt to "improve" the Schmitt trigger performance of the device, which already has Schmitt trigger inputs (inputs with hysteresis), as you noted.
|
What is a good resistor value for my circuit? I'm trying to tap into my car's brake circuit to switch between receiving a wireless camera signal and powering my device. For this purpose, I connected a relay to the circuit via a fuse tap. It should connect the phone to the receiver (displaying the camera feed) when the brakes are applied and charge the phone otherwise. Unfortunately, I'm having some difficulty figuring out how to get the relay to switch properly. I wired the relay to the circuit in the fuse box that powers the car's instrument lights and the brake lights. The current draw for the instrument lights (which are always on) is 0.7 amps. The brake lights draw 1.15 amps. Without adding additional resistance between the car battery and relay the constant 0.7 amp draw will always keep the relay in the ON position (connecting phone and receiver). Hence, the need for additional resistor between battery and relay. The relay's coil draw is .5 amps. simulate this circuit – Schematic created using CircuitLab Here is my most likely flawed resistor calculation. Please point out any errors. First I calculate the internal circuit resistance V=IxR12 = .7xRR = 17.143 ohms // with the brake lights off12 = 1.15xR R = 10.169 // with brake lights on To keep relay in OFF position, I need to supply a current less than the relay's coil draw (.5 amps). Adding an 8 ohm resistor should do the trick: R = 17.143 + 8 = 25.143 //brake lights offI = V/RI = 12/25.143I = .477 < 0.5 // relay switch in off positionR = 10.169 + 8 = 18.169 // brake lights onI = 12/18.169I = .66 > 0.5 // relay switch in on position It seems like 8 ohms is enough to keep switch OFF with 0.7 amp draw, but my relay always stays in the ON position, even after I increase resistance to 16 ohms. So what gives? <Q> Operate the relay from the signal/voltage that powers the brake lights. <S> This gives 12V when operated and open circuit (= 0V) when off (OR ground when operated and open circuit when off in some less usual situations). <S> The 12V power line from the battery will remain at about 12V at all times under all usual loads except starting. <S> The starter motor MAY load the 12V line down to 10V or less, but nothing else will. <A> In a brake light circuit the fuse is typically before the brake light switch, by tapping the voltage at the fuse box you won't get a useful answer. <S> simulate this circuit – <S> Schematic created using CircuitLab <A> If the car is set up like this: simulate this circuit – Schematic created using CircuitLab <S> And you want to tap into the fuse and make it like this: simulate this circuit <S> Then it would make sense that once you reduced the voltage through the circuit enough, by dissipating the power through R1, then the relay would turn off. <S> The problem is probably that the Relay doesn't require 500mA to switch but uses 500mA at its operating voltage. <S> As the voltage decreases to turn the relay off it probably switches at a voltage threshold much lower than its operating voltage and thus a much lower current draw. <S> In theory, it would work with a high enough value resistor. <S> But the lights will be very dim, you will burn a lot of heat through the resistor, and the relay won't last as long being partially energized all the time. <S> You would be better trying to connect the relay in parallel with the rear brake light or at least in series with only the rear light. <S> If connecting to the fuse is the only option then putting a current sensor in series with the fuse and triggering the relay with an MCU (which reads the current value) would be the best option.
| Easisest is probably to tap the voltage after the brake light switch, on most cars this is somewhere near the brake pedal lever, but some use a pressure switch on the hydraulic brake line.
|
How can I calculate the minimum switch time between main battery and back-up battery for my board? I have UpSquared board for an embedded system. I have a main power supply(battery) to feed it and I also have a back-up battery for some situations. For example, something happened and main battery is not able to feed my board anymore. My question is that how many minimum time is required to switch from main battery to back-up battery because I don't want my board to shut down. Or how can I calculate that time? Edit: Here is schematic of my entrance power of my board:My dV is 5V and I is 4 Amper <Q> When the battery is disconnected, the load will have to only rely on the capacitors on the board. <S> The voltage will drop at a rate depending on the total capacitance available. <S> If allowed drop on voltage is <S> dV <S> , then time will be Time = <S> dV <S> * C/ <S> I Where I is the load current. <S> Share the actual values of load current, allowed dip and capacitance available, so that we can see how much will be the time. <S> Edit 2: <S> Example: Load current - \$800 mA\$ Change in voltage allowed = <S> \$5 <S> V\$ Capacitance available: <S> 4 \$70 mF\$ calculated time will be = <S> \$293.75 ms\$ <A> Assuming that the board in question has a SMPS (Switch Mode Power Supply) at the input (due to efficiency for example), it will act as a constant power load. <S> Meaning, that in case of a supply interruption it will be supplied mainly by the output capacitors of the main power supply, and the time taken for reaching a specific voltage level at the input of the board is given by: $$ t = \frac{C_{MAIN,PS} \cdot (V_{B,NOM}^2 - V_{B,MIN}^2 )} {2\cdot P_{BOARD}} $$ <S> Where: $$V_{B,NOM}$$ is the nominal voltage of the board $$V_{B,MIN}$$ is the minimum allowed voltage, before triggering and the backup supply <A> You don't need to switch Since your system is DC, You can have an electronic version of an "overrunning clutch", i.e. The gadget that lets <S> a helicopter's other engine(s) keep turning the rotor when one quits. <S> This is called a diode , and you put one above each power source. <S> The one with the higher output voltage (allowing for the diode's own voltage drop) powers the device. <S> The switch is instantaneous. <S> Diodes have a relatively constant voltage drop through them, you are paying that voltage drop "fee" at all times. <S> So favor diode types with very low voltage drop, e.g. Schottky. <S> I can't think of a zero-cost way to do this; even a mechanical relay needs to be held actuated by the primary supply. <A> It may fully be there are no big capacitors on the board that is designed to be fed by external power supply. <S> Fortunately if you connect the spare battery via diode, you only need enough time for that diode to open. <S> Conventional rectifier diode may have the switching time of somewhat 20 microseconds only ( source ) <S> so you only need a small capacitor of few μF, or <S> even existing (unknown) capacitors on the board may be sufficient. <S> Some capacitor is probably required as even transient power out may reset the board (read here for more).
| Fast recovery diodes are available that can do this faster.
|
Why is it preferable to use NPN transistors for both high-current-output devices in a push-pull amplifier? With reference to the below circuit why is it preferable to use NPN Silicon transistors for both high-current-output devices in a push-pull amplifier? <Q> Don't have the time to write an extensive answer, but it's desirable that both chains are symmetric; thus, using the same components makes sense. <S> It's easier to find to matched NPNs than if you tried to build a matched NPN+PNP pair – but both exist on the market. <S> I hear that in the 1970's, matched NPN+NPN pairs were significantly cheaper. <S> I don't think that's overly true these days. <S> For the same size, NPNs are easier to make high-current capable. <S> In other words: once you buy a million, NPNs are cheaper than PNP able to carry the same current. <S> Same applies if you buy very large-current transistors <S> : both NPN and N-channel are generally easier to get than PNP / P-channel. <S> Generally, be wary when literature claims things without referring to why something is. <S> Things outdate, and if you don't give reasons, you'll simply have a lot of " <S> I don't even know if this is still true" statements in your literature. <A> To add to Marcus' answer, depending on the frequency of operation, NPN transistors are faster than PNP. <S> This is because electron conduction is faster than hole conduction . <S> In a NPN transistor, current is carried by electrons moving from the emitter to the collector. <S> In a PNP transistor, current is carried by holes moving. <S> Since the holes are just the lack of an electron, for the 'hole' to move toward the collector, really, the surrounding electrons are moving out of the way. <S> The way to visualise this is to think of a bubble of gas in a fizzy drink. <S> The bubble looks to rise, but really the drink (electrons) is moving downwards, forcing the bubble (holes) up. <S> This is clearly visible in part datasheets. <S> For example, the BC547 (NPN) and BC557 (PNP). <S> See the two snippets from the Fairchild datsheets showing the fT (frequency at which the transistor has unity gain). <S> BC547 (NPN) has an fT of 300 MHz. <S> BC557 (PNP) has an fT of 150 MHz. <S> Just another point to consider. <S> In this example, the PNP transistor is half of the speed of the NPN. <S> This is not uncommon. <A> This question makes me feel very old indeed. <S> When this type of audio amplifier was first proposed in the mid-1950s by Lin and then by two engineers Toby and Dinsdale published in Wireless World in 1961 (Google will find them) there were no PNP power transistors to complement the ubiquitous 2N3055. <S> Hence the quasi-complementary arrangement was the only game in town. <S> Later PNP power devices began to be available and later still complementary power darlingtons too which became the preferred arrangement of the day. <S> Today the go-to solution is probably complementary MosFets.
| If your application is high-speed, you'd prefer NPN.
|
PCB Design: Strange 'whiskers' from PCB pads? Regarding the attached PCB image, you'll notice that there are odd 'whiskers' from PCB pads. I'm curious as to their purpose. The board isn't high frequency, so I doubt it's for any fancy RF-magic. My best guess is that because it is a home-etched board, these are intended to aid with component placement in the absence of a silk-screen layer: i.e., pointing out which way resistors go, etc., but there don't seem to be many places where components match. Equally, it doesn't seem to match with another guess that they'd identify pin-1 on an IC or a common pin on switches, etc., so I'm at a bit of a loss. I've never seen anything like it before. Any other suggestions welcome. <Q> Is there copper on the other side? <S> If so, it's possible that these whiskers indicate that "vias" should be inserted in these holes. <S> I'm seeing two nets on the right labelled "A" (no whiskers) and "E" (all whiskers). <S> There used to be push in / snap off pins for this purpose ( <S> made by Harwin) though spare resistor legs, soldered both sides, are a perfectly acceptable substitute. <S> In fact, searching the callsign (plus BREAN) brings up the author's website where you can buy kits of slightly more recent manufacture. <S> So why not email the author , and ask him? <S> Adding BREAN to the search term yields some hints that this may have been a DSB transceiver published in Practical Wireless - possibly December 2006 edition. <S> Re: comment. <S> It's possible some of the supposed "all via" nets are either guard bands, or they might connect different islands of ground plane. <S> Just a hypothesis at the moment. <S> OK there IS copper on the other side ... cross-referencing the tiny holes on the top of the board (without spot face cutter insulation rings) with the whiskered pads in the question, quite a lot of the traces match <S> (I can't see every one) <S> As @DewiMorgan points out in a comment, these holes are ones where the top side copper hasn't been stripped. <S> Then it would be good practice to treat the remaining ones as vias (insert wire, solder both sides). <A> No. <S> This is a typical pcb from this UK supplier of ham radio kits. <S> He has consistently made his boards with 1970s era Bishop Graphics, a sticky tape and donut system that became outdated with the advent of the first PC based design tools. <S> But, being unwilling to adopt any other approach, he still supplies these woefully poorly hand drilled boards. <S> They are generally double sided, the top side being unetched, forming an earth plane. <S> Any ground circuit traces on the solder side have these 'whiskers tags to remind the builder to solder these leads on the top component side too. <S> It's as awful a system as can be imagined, and makes building the kits very difficult, especially in the absence of overlay markings and with hand drawn circuits provided. <A> <A> This is just pure speculation but <S> your hint that it's a home-etched board (it does look manually drilled, and badly so) might point to the thin strips being indicators of over-etching. <A> Looks interesting! <S> It might help with adhesion of the copper to the FR1 PCB. <S> I imagine this pcb is single layered. <S> These pads can break away from the pcb. <S> With these "whiskers" it might be more robust. <A> These whiskers are only present in wider traces. <S> Most probably these traces are ground net. <S> The designer has just used these whiskers as markers for the ground traces.
| These 'whiskers' seem to indicate that pads belong to a common net, probably 'ground'. Given there's an English ham radio callsign lower left, I'm guessing those are "Antenna" and "Earth" (the latter connecting to something on the other side, resembling a ground plane as far as possible on this style of PCB. Some of these holes may be populated by a component leg, which should (if practical) be soldered both sides.
|
Why there is a high inrush current when the microcontroller start-up(CC430f5137 from TI)? simulate this circuit – Schematic created using CircuitLab In my application, I need to measure the current consumption of the mcu (cc430f5137). To measure the current of the mcu. I insert a 10ohm resistor at the Vcc. Then, I connect the power supply to the Vcc and observe the voltage across the resistor so that I =V/10. I found that the current at start-up is around 80mA@1.8V and 200mA @3.3V. It will last for a while and decrease gradually. Below is the picture of the waveform under 1.8V and the diagram of the board. The division of the waveform is 2 ms/div . What may cause such a high current at start-up? Is this normal and how to solve this problem? <Q> 10 Ohms is way too big for a current sense resistor and is slowing the charging of the decoupling caps. <S> Try 10 milliOhms. <A> At a first estimate, you're trying to push 90uC into 50uF worth of caps at 1.8V. <S> Your inrush above, using a triangular approximation, shows about 104uC in the first 4ms or so. <S> Most of it will be the caps. <S> (The total charge delivered is the area under your scope trace.) <S> If you reduce your sense resistor, you will likely see a higher peak and a shorter tail, as the capacitors will charge faster. <A> What may cause such a high current at start-up? <S> Is this normal and how to solve this problem? <S> You are charging up about 40 uF. <S> Initially, capacitors can be considered as a short circuit. <S> Only the ESR of the capacitors and the impedance of the voltage source and the impedance of the wiring limit this initial charging current. <S> The high current will exist longer the more capacitance has to be charged. <S> However, since you are measuring with 10 Ω this resistance limits the current significantly (considered to the other impedances). <S> The real inrush current will be quite higher. <S> how to solve this problem? <S> You could also limit the inrush current with an inductor (because current through it cannot change instantly) but if not damped well enough, it may give undesired oscillations. <S> You could also reduce the capacitance. <S> Is 40 uF really needed? <S> (Assuming the 10 uF are equal, this only reduces the duration, not the height of the inrush)
| You could limit the inrush current by a resistance, but this might give you an undesired voltage drop in normal operation.
|
Is PCB testing of all nets after assembly required? That is my first time asking an assembly house to produce 200 units of a PCB (and not the usual 3 or 5 PCBs). The assembly house came back to me saying that the testability of the board was bad and that they need to have 1.2 mm pads on the bottom side of the PBC for all nets... They require such large pads because beds of nails are a much more economical option than flying probe for 200-400 units since they can do it in-house. The PCB we are talking about : 4-layer PCB (a few power supplies, a battery charger, connectors to screen and other accessories) only surface-mount components all components on top, 60 euros/PCB (bare PCB + assembly) for 200 units. we are targeting 200 units at first, and probably 400 units total. (It is already revision 2 of the same board). consumer product I'm a bit suprised by this request from the assembly house. Using large pads or adding vias on all nets seem to be extreme to me. I'm also a bit surprised not to have noticed such testing pads on other products (if that's even possible to fit them sometimes?). My questions are : Is testing of all nets required for consumer products? Is it common practice? I was more thinking about some higher level tests (if the screen works, if all buttons works, if the battery charger delivers the good amount of current, etc.). Is adding 1.2 mm pad on ALL nets common practice for PCB? is it normal that the assembly house increases the assembly cost if they don't do the bed of nails? The cost of the bed of nails is almost as much as burning (fully) 40 PCBs... I may be a bit naive there but they seem to say that they want to test all nets in order to check if resistors/capacitors/others are good. Is it so common that an assembly house mixed up components on a few boards? <Q> Putting in those pads first time round is called DFM (Design For Manufacture). <S> And it sounds like they're giving you a good price on the bed of nails. <S> Either get this flow working (it's perfectly normal) or ask your next fab house if they have a flying probe tester in house. <S> One way to economise is to use bed of nails ONLY for nets that don't already have connectors attached, and to break out test cables from those connectors to their ATE. <S> I do this for my own projects for small runs, where I don't mind adding the time spent plugging/unplugging connectors to the test time. <S> It can save money on the fixture. <S> but it's a pain and makes the test more expensive. <S> You can ask them if this is an option, but I wouldn't be surprised if they refuse. <S> but you don't want to know the price! <A> One figure of merit we use to evaluate how good or bad <S> we're doing <S> goes by the acronym FTTTY, or First Time Through Test Yield. <S> It's a measure of how many boards/assemblies pass their acceptance testing the first time through. <S> In order to improve FTTTY, we like to do as much testing of things before they reach the final test phase of production. <S> Hence all of our PWBs are 100% tested by the manufacturer, and their ability to do this kind of testing, of which bed-of-nails is only one type, goes into our evaluation of them as a vendor. <S> Now most of our products are expensive, aimed at high-end Mil-Aero-Space applications. <S> It's not unusual to have a single component such as a high capacity, high ball count FPGA cost upwards of $25K. <S> One thing we don't want to do is to scrap such an expensive assembly after having mounted all of the components because of a flaw in the PWB that is not discovered until board test. <S> Therefore we (and our customers) are more than willing to spend money early in the build phase to improve our chances of FTTTY. <S> A lot depends the end/sell cost of your product, what your yield would be with and without testing of bare PWBs, and how that impacts the average cost of your product. <S> If the sell cost of your product is, for instance, $20, and your building 200 of them, it may be more cost effective for you to NOT test the bare PWBs if it means investing several thousand dollars up front to set up PWB testing by the vendor. <A> We call it 100% test points and is a absolute need from us to provide it without fail. <S> Every net which we fail to connect to the net pad needs a clearance from the whole team. <S> Accidents are common in line assembly <S> Components fail sometimes after soldering (due to chemicals or high temperature) <S> Solder itself can get stuck creating short are lower resistance paths The wrong components can also get into the board due to manual error <S> Supplier can supply Wrong items <S> The components can also be counterfeit <S> The ICT even for a board with 200 to 250 components doesn't take more than 10 seconds. <S> So it is not a big hurdle timewise. <S> The exception will be where impedance control is strictly needed such as RF and filtering sections. <S> Further to bed of nails test, camera test can also help check the mounting direction and misalignment or poor alignment of components on the pad. <S> When the number of products grows, and the complexity of the board, investing on such a solution is in best interest of quality and intime Delivery at a little higher cost. <S> Larger vias are good for plant to realise the bed of nails at lower precision and lower cost <S> Designer should consider the ill effects of the via or extra Routing <A> You should thank your PCB supplier! <S> Design for testability is a major issue in electronic systems. <S> Usually the managers want to save three cents by omitting test and control points. <S> Then 5 years later they find that they have a unit that is untestable and must be disassembled before a fault can be diagnosed, costing hundreds of dollars and causing the ATE engineer to curse the manager. <S> (At least that is what I did when I used to do that kind of work.)
| Yes there are other test options : as well as flying probe, there are 2-sided versions of bed of nails which saves the need to bring all nodes out to the test side ...
|
Using laptop charger with slightly higher voltage I am wondering if it is safe to use a MacBook charger on my Lenovo laptop. Both devices charge via USB C PD. When looking at the voltages I can see that the Lenovo adapter is 20V - 3.25A The MacBook charger has 20.2V - 4.3A I am wondering if the .2 increase in voltage with the MacBook charger should be cause for major concern. <Q> The extra 0.2 volts is not a cause for concern. <S> The power brick will not provide power until the laptop asks for it, and it will only supply a voltage that the laptop is capable of accepting. <S> If you look closely at the Apple power supply you will see it also lists 9 volts and 5.2 volts output. <S> This is to maintain compatibility with iPads and iPhones that use lower voltages. <S> If the Lenovo can't handle the 20.2 volts then it will step down to 9 volts at up to 3 amps. <S> That's 27 watts instead of 87 watts but its better than a dead battery, dead from not enough power or too much. <S> The reason they specify the extra 0.2 volts is to account for voltage drop on the cable. <S> At the power brick end of the wire it will be 20.2 volts but at the laptop end it could be 20.0 volts, 19.8 volts, or whatever happens to be lost to heat in the resistance of the wire. <S> First thing is that the laptop is unlikely to be so fragile to be damaged by such a small variation in voltage. <S> Second, the laptop and power brick are smart enough to not burn themselves up. <S> These aren't like the "dumb" power bricks in the past, they talk to each other to find out what the other is capable of before power is put on the wire. <S> Oh, and a third thing, Apple and Lenovo make good stuff and they stand behind their products. <S> The USB group will crack down on them hard if they violate the USB-PD spec to the point something is permanently damaged. <S> There was a lot of sketchy USB-C devices out there when it first came out but big name companies with lots to lose worked hard to clean that up. <S> Big names like Apple, Lenovo, Microsoft, Google, and Amazon. <S> Don't buy stuff with "too good to be true" prices <S> and you should be fine. <A> If everything is USB, then theoretically it will be fine. <S> If there is some incompatibility then it should just not work without damaging anything. <S> Regarding the 0.2V difference, that's tiny and if there is a problem that's not likely to be it. <A> I would be cautious when dealing with non-OEM chargers. <S> Since we don't know the internal circuitry, the Lenovo laptop may only be designed to take 20V and nothing more, since overvoltage can damage electronics. <S> Realistically, it may just not accept the 20.2V if it finds a fault.
| That said, I don't think that a 0.2V increase would damage the laptop. The way USB-PD over USB-C works is that the power supply and device will negotiate a mutually compatible voltage before power is delivered. That being said, in practice compatibility issues aren't unusual so anything is possible. If you want to be safe you should stick to the manufacturer's recommended charger.
|
Supplying power to a 12V motor with 5V and 2A adapter A very elementary power supply question but i couldn’t find the answer with Google, so I would appreciate a response. I’m using a 12V, 340W DC brushed motor for a personal project. While the motor is relatively powerful, I don’t need the full capacity of the motor, but I do not want to buy a new motor as I already have this lying around. For the purpose of my project I only need it to operate it at 5V (required RPM from the data sheet). I have a 12V, max 3A, cont. 2A, 30W driver lying around too. I was wondering if I used a 5V, 2A power adapter (that I already have), will it only supply 2A even if the motor tries to draw more than that and will this damage the adapter by any chance? I really don’t want to buy any additional parts and I have all these components lying around. If I supply 5V to the driver, even if the motor tries to draw more than 2A, will the power adapter limit it, or will it supply more and damage the adapter and driver? Is this okay to do or will I damage anything? I will be using PWM to slow down the motor and average it out to around 3V or less. The motor draws around 2.4A at 3V when I checked with a bench top PSU and DMM. I don’t want to supply more than 2A and I don’t mind it losing a bit of torque. It only spins a threaded rod and the load is pretty much static once it starts. Do I have to take any additional methods to be safe or will the power adapter limit the current supplied? <Q> Motors want lots of current to start, they want lots, given the opportunity they will take lots, and this will cause your 2A powersuply to quit. <S> I was wondering if I used a 5V, 2A power adapter (that I already have), will it only supply 2A even if the motor tries to draw more than that and will this damage the adapter by any chance? <S> Typically they go into over-current shoutdown if over-loaded. <A> Most (if not all) modern 5V phone adapters will limit the current to what they're rated for. <S> If not, the voltage will most likely fall apart and it's <S> probably not going to burn the adapter. <S> The current your motor will draw depends on the mechanical load it has to overcome, so if you're not asking much of it <S> it will probably not exceed the adapter's ratings. <S> Furthermore you can always add a fuse just in case, it costs cents. <A> You've already established the motor draws 2.4 A at 3 V, which is 7.2 watts. <S> If you restrict it to any less than that, you may well find the motor doesn't turn, through insufficient torque. <S> However, your 5 V power supply will deliver 2 A, which is 10 W. <S> If you use a buck DC-DC power converter to get from 5 V to 3 V, or your PWM operates so as to use the motor's inductance to be a buck converter overall, then you will just manage it. <S> In the second case, you may need a large input capacitor, to supply the output pulses of current, to prevent them from overloading the 2 A PSU.
| But motors don't need lots of current to start they only want it, if you use a DC-DC converter that has a current limit you can keep the power supply happy and get the motor spinning.
|
Power spectra density for line code Why do we need to find power spectra of different line codes? why we relate it to to the auto correlation of the signal? how does it help us in transmission of digital signals? <Q> Digital data are stored in the form of bits, i.e: 0 and 1. <S> The binary data is stored in memory cells usually as static charges (DC Voltage). <S> When transmitting a series of binary bits, we need some sort of waveform to be able to send them on a specific channel. <S> Now, to be able to allocate resources such as bandwidth, power, etc, we need to know the spectral behavior of our Transmitted Signal . <S> This signal is produced using a line code acting on a baseband signal. <S> Therefore, we need to know the effect of the line code spectrum on the signal spectrum. <S> Both of these spectra contribute to the final transmission spectrum, which is meant to be transmitted. <S> Over the years, different line codes have aimed for bandwidth/power efficiency, as well as reduction/control of inter-symbol interference (ISI). <A> Others have already said that the spectrum helps to estimate does the signal fit into the line. <S> But there's more. <S> Extracting the right timing in the receiver is essential for optimal decision making. <S> Peak in the spectrum makes it easier. <S> If there's several data lines it's possible that there's crosstalk. <S> Knowing the transfer function of the crosstalk and the spectrum gives a possibility to estimate the effect of the crosstalk. <S> Data communication theory textbooks show how the power spectrum formula can be derived. <S> The final formula can be divided to 3 major sections: the power spectrum of the used pulse the coding rule "how 1 and 0 are presented with the used pulse" data stream statistics <S> The statistics can be shown as discrete autocorrelation function in the spectrum formula. <S> If the data stream contains some periodic patterns the timing in the receiver can get distorted and the decision making isn't optimally timed. <S> Autocorrelation function reveals the perodicity . <A> To recover data, the stored energy (in the communication path) must be low enough that acceptable ISI Inter-Symbol-Interference results. <S> Narrow bandwidth ensures lots of ISI, and thus ensures lots of bit errors. <S> To design the communication system, the required bandwidth (for tolerable ISI and BER)must be known.
| Line codes help us achieve the goal of transmitting the data over specific passband/baseband channels.
|
Sound through the earphone is also sent to the amplifier when using a jack splitter after a audio mixer I implemented the following audio mixer circuit with just 2 input audio sources: http://www.theorycircuit.com/audio-mixer-circuit/ Both sounds got mixed properly. Then, at the end of that audio mixer I connected an audio splitter like the one below: which provides me two female connectors. Then: on one female I connected an amplifier with a speaker on the other female I connected a headphones (without microphone) (for more details, please, check the image below) My problem is: If I speak through the earphones of the headphones, then I can hear that on the speaker and I don't want that to happens. I just want to listen what's comes up from the mixer but nothing else. Any idea on how to avoid this to happen? (which is actually weird, I could not expect that to happen). Thanks in advance! <Q> A dynamic microphone consists of a permanent magnet, and a coil of wire attached to a diaphragm. <S> If you speak into the diaphragm, it generates a small current in the coil. <S> A headphone speaker consists of a permanent magnet, and a coil of wire attached to a diaphragm. <S> If you pass an alternating current through the coil, it vibrates the diaphragm and makes a sound. <S> So a dynamic microphone and a small loudspeaker are actually the same thing. <S> This means that you have effectively connected a pair of microphones to the input of your amplifier. <S> As others have said in their answers, the solution is a small headphone amplifier between the mixer and the headphones. <A> You don't say how you connected the single output of the mixer with a stereo jack. <S> I suppose (and I have good reasons) <S> it is in parallel (L+R shorted). <S> Now, when you speak into the headphone, the loudspeaker (of the phone) modulates the line and the amplifier amplifies exactly that. <S> The impedance of the headphone, which should/could be about 70 ohm (I mean, not 10 times less, not 10 times bigger) is too much comparable with the output impedance of the mixer (about 320 ohm?). <S> I think that the only way to solve your problem is to put an active element in the middle, may be a transistor or another <S> op-amp - <S> but I am not an expert. <S> Another way is to reduce a lot the sensitivity of the amplifier, and/or use a squelch control if it has it. <S> If the amplifier is able to amplify a microphone, then it is too much sensitive to be used after a "line output", which your mixer is (or should be) supposed to have. <A> Which means that voltage will make it move but movement will induce voltage (with little current sourcing capabilities but still). <S> If you present this movement induced voltage (if you speak into your headphones) to the high impedance input of your monitors amplifier it will amplify it, since it does not requires much current from the source (your headphone) to do so. <S> What you should do is to buffer the signal going to your headphones. <S> So that the signal presented to the monitors and the headphone sees an high impedance on both side. <S> Which gets buffered on one side to drive a 25-250 Ohms headphone load, and to a 8-16 Ohms speaker load on the other. <S> I don't know how you could continue using the splitter and eliminate your problem though. <S> It would be simpler to duplicate the output signal, send one to the amplifier (as you have done), pass the second one through a buffer and out another port to the headphone. <S> Some good audio buffer candidate to drive your headphone could be: http://www.ti.com/lit/ds/symlink/lme49600.pdf <S> https://www.analog.com/media/en/technical-documentation/data-sheets/AD8397.pdf
| Your headphone is an electromechanical system with some input impedance (can be quite high in high fidelity headphones).
|
Why do Tesla and other manufacturers use AC motors? I know Tesla and others use AC motors. I read a lot and no one really provides a good answer. I know they are using an AC induction motor, and with Model 3 also an AC synchronous permanent magnet motor also referred to as brushless DC. Basically my question is why does almost everyone opt to use frequency drive to control their motors? I know they could use synchronous permanent magnet motors and run in from DC but they still run it on AC. What advantage are they gaining? Better control, cost, torque? <Q> I know they could use synchronous permanent magnet motor and run in from DC <S> This is the problem in your understanding. <S> You cannot feed DC into a motor with no brushes. <S> It will lock up and not spin. <S> All spinning motors are AC at heart, even brushed DC motors. <S> You need AC (i.e. currents that change direction) in some form at the lowest level to produce a rotating magnetic field to spin. <S> It is just whether you use brushes or electronics to convert the current from your DC power source into the AC the motor needs, or whether you run it off AC directly. <S> Think about what commutation really is, whether with brushes and sliprings or electronics, in any "DC" motor. <S> It's just reversing DC currents to simulate AC. <A> There are two main reasons for using AC rather than DC in an electric vehicle:- A DC motor has a commutator and brushes which switch power to each armature winding at the correct angular position to make the motor rotate (ie. <S> it converts DC voltage to synchronous AC at the windings). <S> This causes extra loss due to friction and brush/contact resistance, and wears down the brushes and commutator which eventually have to be replaced. <S> Wear is higher at high current and/or rpm, which limits the maximum power and speed the motor can produce without reducing its lifespan. <S> A brushless motor can have its timing adjusted 'on the fly' by the controller rather than being fixed mechanically by the brush/commutator position. <S> The BLDC/AC controller can optimize timing to suit what the motor is doing. <S> This allows using a fixed gear ratio rather than having to change gears to keep the motor in its most efficient speed range. <S> The combination of lower maintenance, higher power, (slightly) higher efficiency, and lower cost due to simpler motor and gearbox construction makes a BLDC/AC motor more attractive. <S> This is partially offset by higher controller costs, but since a sophisticated control system is desirable anyway <S> it isn't that much more expensive than a similar DC controller. <A> Before Tesla existed, enthusiasts were doing conversions on a variety of cars to turn them to an all-electric drivetrain. <S> Most of these used traction motors that you'd find in fork trucks on under railway rolling stock. <S> There are a couple of limitations with these motors. <S> As other comments have noted, all motors need to have a current that rotates relative to one of its components, the armature, which in the case of a DC motor is the rotor, and this is achieved by the commutator - the cylindrical copper component on the left end. <S> The brushes (made of a carbon and metallic composite) that run on the commutator are subject to wear, and would require replacement, probably several times during the life of a typical vehicle, so this is the first disadvantage. <S> The construction of the armature, and especially the commutator poses a maximum speed limit on the motor, before you risk the assembly bursting - windings or commutator bars flying out, which is fairly catastrophic, since it'll destroy pretty much all of the motor. <S> Since the maximum torque you can get out of a motor is dictated by the rotor volume, and the flux density you can achieve, the way to get maximum power out of a given size of motor is to run it as fast as it will physically tolerate. <S> BLDC motors, which have no windings on the rotor, can be made to run at much higher speeds. <S> The windings on induction motor rotors are usually a squirrel cage made from bars either cast through the laminations or fabricated from rods, and again can be made to tolerate high speeds. <S> That leads to the inevitable issue with higher power density - the need to cool the motor. <S> are mostly air cooled, though a few low voltage types can run with the rotor submerged in non-conducting fluids - though generally at low speeds. <S> Overall, given the weight incentive to have higher power density, and the need for lowering maintenance, DC motors are not a good choice for EVs.
| Since BLDC and induction motors have the windings on the stator (this is now the armature, since it is fed by currents alternating in a fashion to produce a rotating field), it'a far easier to fluid cool this - commutator motors
|
Naming convention and list of ICs There are some parts that use the same part number across different vendors. e.g.: 23cxx: SPI SRAM (only Microchip?) 24cxx: I2C EEPROM 25cxx: SPI EEPROM 27cxx: parallel EPROM (UV light erease) 28cxx: parallel EEPROM and so on. Is there a list of all the devices in that naming convention like for 74xx series ICs? <Q> There are a lot of examples of this <S> and it was / is done to convey equivalence of functionality. <S> The devil is in the details, though. <S> The xx1117 low dropout regulator is made by at least 3 companies, but their guidance on the output capacitor is different in all 3 cases <S> (the capacitor ESR is critical for proper operation). <S> For logic, there is the xx4000 series (CMOS); <S> CD4xxx was originally from RCA and <S> (what was then) <S> Motorola SPS had their equivalent MC14000 series (the only difference is the leading '1'). <S> The world of op-amps has many examples of this (particularly the OPxxx series) <S> but the details are often different (they are often referred to as 'improved' which to an engineer means 'different'). <S> Serial port interfaces (such as the MAX232 and variants) are widely multiple sourced. <S> The key issue is that the operational voltages and various timings are not necessarily guaranteed to be the same, although devices with a committee specified pinout and operational features will meet that specification (such as in DDRx SDRAM); this is why designing to the specification rather than just the datasheet (which may be far better than the requirement) can be advantageous in case one source of supply becomes unavailable. <S> The venerable (and ancient) <S> 741 op-amp is also available from multiple vendors, as is the 555 timer (in TTL and CMOS variants) as are many transistors (such as the 2N series) and diodes (1N series for example). <S> Bottom line: <S> Yes, there are numerous examples of multiple vendors using the same naming convention across analog and digital parts, but beware and read the datasheet carefully . <A> For there to be a definitive list there would also need to be some body that supplied and enforced the definitions. <S> There is currently no such body. <S> In the early days of integrated circuits, the Department of Defense acted as such a body in the U.S. <S> They published and maintained specifications for certain IC part numbers that were used in military applications, such as the 54LS series of TTL devices. <S> But market forces and commercial applications have overwhelmed any ability or desire to control part numbers in that way. <S> Caveat emptor. <A> If you're after current parts, then do a search at a site like Digikey.com <S> For example "SPI SRAM" leads to 5 pages of part numbers from 3 different vendors. <S> If you're after historical numbering, that's a little more time consuming. <S> Going how far back? <S> as one example. <S> I used UV Erasable PROM back in the 80s, when Intel was the main source.
| So, no, there is no list.
|
Is exceeding Opamp Common Mode Voltage safe? I understand that when you exceed an Opamps Common mode voltage limit that the output will no longer be reliable. And that this limit is often a bit less that the supply rails. But would slightly exceeding this limit to perhaps the supply rail cause the Opamp to blow up or overheat? Or would it just cause the output to be inaccurate? <Q> A few op-amps allow the supply rails to be exceeded without damage. <S> Many require the inputs to remain within the supply rails (or no more than a few hundred mV outside, or current limited). <S> It really depends on the op-amp. <S> Read the abs. <S> max ratings for what might damage the op-amp even if momentarily exceeded. <S> Read the recommended ratings for what to do in normal operation. <S> Note that the ratings apply even when no supply voltage is applied. <S> For example, with the ubiquitous LM358/LM324 op-amp you can exceed the positive rail (up to 32V) but going more than a few hundred mV below the negative rail can destroy the chip. <S> Most CMOS op-amps eg. <S> TLC27L2 do not allow either supply rail to be exceeded by more than a few hundred mV. <S> Sometimes 0mV is specified. <S> As you noted in some cases the output will go to the opposite rail than expected, which can cause some poorly-designed circuits to latch up. <S> Also note that you should use "worst case" common mode range, which means not just using the worst case figures on the datasheet, but also allowing for temperature if that is not accounted for. <S> For example, the LM324 has a common mode range that includes 0V to V+ -1.5V worst case, but only at 25°C. <S> The manufacturer has not seen fit to grace us with even typical characteristics of Vcm with temperature, but a quick glance at the schematic would tend to indicate that low temperatures will be worse, by at least 4mV/°C, so one might allow V+ -2V <S> (and verify at least typical characteristics in a test chamber if low temperature operation is required). <A> "Safe" in my view means: no person gets hurt Do persons get hurt if you exceed the Maximum ratings of an opamp? <S> I have yet to see that happen <S> so my answer would be: <S> Yes, it is safe. <S> But you might actually be asking if the opamp can get damaged , right? <S> Well, that's simple. <S> Look in the datasheet. <S> If you exceed the Maximum ratings <S> then you risk damaging the opamp. <S> If you don't exceed the Maximum ratings then no damage should happen. <S> There is no point in asking what will happen, what the damage (if any) there will be as that depends on so many circumstances and will be different for every model of opamp. <S> So there is no general answer to what the damage will cause as it depends. <S> And the answer is irrelevant as you simply should not be using an opamp outside its maximum ratings. <A> There are two considerations in the data sheet. <S> The first is in the electrical specification section. <S> If you keep to the specs in this section, the chip will perform as predicted. <S> The the next is in the Absolute Maximum section. <S> If you leave the bounds listed in the electrical spec, bu you don't leave the bounds outlined in the Absolute Maximum section, the IC is pretty much guaranteed to keep functioning to spec once you return to the parameter bounds listed in the electrical spec. <S> If you leave the bounds of the Absolute Maximum section, the IC is no longer guaranteed to perform to spec ever again. <A> Some opamps even reverse the polarity of the output if the common-mode voltage range is exceeded causing severe distortion. <S> Image source: <S> Analog Devices - Application Note <S> AN-849 - Using Op Amps as Comparators
| If you exceed the common mode range, you cannot expect the op-amp to behave properly (though it might in some cases). Exceeding the common-mode voltage range causes no damage.
|
Design of bypass capacitors in circuits I have a question when designing a circuit. I want to power two sensors with a 3.3V LDO.My question is the following, in the LDO datasheet they recommend me to place a 10uF capacitor at the output of this one, but in the datasheet of each sensor, they recommend me to use 10uF and 0.1uF capacitors. My question is, do I have to use as many capacitors as each datasheet asks me, or can it be solved in another way? The same happens with the LDO input, it is recommended to use a 10uF bypass capacitor, but that same input that goes to the LDO, goes to another IC that advises to place a new capacitor. Do you know which would be the best solution to these cases? <Q> The datasheets tend to play safe and show a typical circuit that should work in most cases. <S> The datasheets can't know what other components there are, what their requirements are and what is the distance between components. <S> As you don't mention either what components you have and what is the distance between them, there is no way to say what is the best solution. <S> It may be hard even if it was known. <S> So design with all the capacitors, and you can then try it two ways: either mount all capacitors and start removing them if issues happen, or mount only minimal amount of capacitors and start adding them until issues go away. <S> If the chips are really close, perhaps one set of 10uF and 0.1uF will do. <S> But there may be other requirements, such as what type of capacitors the datasheets suggest for each chip. <S> LDOs may not like ceramic caps as they can have too low ESR, while the other two chips might benefit from ceramic caps. <A> Put a 100nF as close as possible to each pin that requires one, if that means multiple parallel capacitors then so be it. <A> The general idea of bypassing is to minimize the loop area of noise, that is, to suppress it as close to the source as possible so as to prevent the noise from affecting other parts of the system. <S> at the device pin, and use a bulk bypass at the regulator. <S> This works fine for most systems. <S> More demanding applications, like fast memories or big SoCs, mix bypass values to obtain broader frequency coverage. <S> This must be done with care, as caps of different values can interact in a way called anti-resonance that will actually make things worse. <S> For an LDO, it is usually not necessary to use a 0.1uF at the output, just bulk to improve transient response (check the LDO datasheet.) <S> There are even LDO regulators that don’t need output bypass at all, bulk or otherwise (TI ‘Capfree’ LDOs for example). <S> Murata has a good app note on bypass design which covers not only the use of capacitors but other kinds of filtering like ferrites and feed-though caps. <S> See here (pdf): https://www.murata.com/~/media/webrenewal/support/library/catalog/products/emc/emifil/c39e.ashx
| For larger capacitors like 10uF where high frequencies are less important, you can usually use a single capacitor for multiple devices and then put them a little further out of the way. The time-honored rule of thumb is to bypass load ICs using 0.1uF
|
Driving RGB Leds from ESP8266 I want to create a schematic for the following behaviour: I want an ESP-12e (ESP8266) microcontroller similar to an arduino, to be able to control 10 RGB leds , changing their colours from blue to green to red and all shades available in between those colours. I have designed the following schematic to achieve this behaviour having researched options: The question I have as per the schematic note is - which 9V regulator would I use , given I need around 600mA from it I am finding it hard to find one within the schematic libraries and additionally, finding an actual regulator to buy appears difficult. I have a question about the design and that is as follows: As I understand it, I can control the amount of brightness of Red, Green and Blue from my LED by controlling the amount of voltage sent to those pins, however, the 2N2222 is on, or off - with no space in between, so does that mean in this case as my GPIO pins fluctuate to control shades of colours, the LEDS in this case will be Red ON or OFF, Green ON or OFF and Blue ON or OFF - or have I misunderstood how this bit works. Many thanks! <Q> To control your LED brightness, you should pulse-width (PWM) modulate your LEDs at a rate faster than the human eye can see. <S> Modern MCUs often have built-in PWM circuits. <S> If not, they will always have timers, and you can use timers and interrupts to do it with minimal software. <S> https://randomnerdtutorials.com/esp8266-pwm-arduino-ide/ <S> Many different sizes of power adapters are available from amazon, add some margin, don't run near their limit. <S> I would use 1A minimum. <A> The question I have as per the schematic note is - which 9V regulator would I use , given I need around 600mA from it ... <S> You don't really need a regulator! <S> LEDs are very sensitive to voltage but with the current limiting resistors you have correctly applied to each LED the variation in current and brightness will be quite acceptable for a reasonable variation in voltage. <S> ... <S> I can control the amount of brightness of Red, Green and Blue from my LED by controlling the amount of voltage sent to those pins, however, the 2N2222 is on, or off - with no space in between, ... <S> Yes. <S> You are describing pulse-width modulation. <S> With your transistor half-way on (to vary the voltage) it will get rather warm as the power dissipated in it will be the product of the voltage across it (the transistor, not the LED) and the current through it: \$ <S> P = <S> V_t <S> I \$ . <S> We generate much less heat if we switch the transistor fully on (V is very low) or fully off (I is zero). <S> We just need to do this quickly enough to fool the eye that the light is dimming. <S> Figure 1. <S> PWM signal transitioning from high pulse width (75%) to low (25%) and back again. <S> Note amplitude remains constant. <S> This will result in a control input of 7.5 V and 2.5 V respectively giving 75% and 25% of rated current. <S> Source: <S> LEDnique.com . <S> Figure 2. <S> Figure 2. <S> By cross-fading the colours an RGB LED can generate a wide color spectrum by mixing light. <S> Source: RGB LED . <S> Each colour can be generated by varying the relative pulse-widths of the RGB LEDs. <S> All on will give white(ish). <A> That’s the hard way, and it will have flicker artifacts if the PWM rate is too low. <S> The easy way is to use WS2812 -interfaced RGB LEDs which come singly or in strips. <S> WS2812 protocol allows daisy-chain connection of multiple LEDs.
| You can use an unregulated power supply (transformer, rectifier and smooting capacitors).
|
Do function generators produce AC signals in the sense we know? When I examine function generators, there are a wide range of products. I see products between 7 USD and 600 USD. If they can do the same job, why is there so much difference in price? I can't make sense of it. I don't think these cheap products can actually produce a sine wave with a negative voltage portion. Maybe they are producing PWM like Arduino (no negative side). For example, in the video below, do it yourself function generator. link 1 , Link 2 I take the function generator to use in diode tests. When I give an AC voltage without disconnecting the diode on the circuit, I will decide whether the diode is solid or not with the oscilloscope. This is the function generator that I think can generate the AC signal that I intend to buy. I think there are a few good brands such as RIGOL, UNI-T <Q> An AC signal is a signal that Alternates , which means it changes direction over time . <S> Even the cheap DIY "function generators" can generate such a signal (if I apply that signal to a capacitor, the capacitor will charge and discharge <S> so the current changes direction) <S> so yes, that's AC. <S> Indeed some might not be able to produce a negative voltage and can only output the AC signal superimposed on a DC signal . <S> You could use a capacitor to block that DC voltage. <S> You could use a DC voltage source as a reference such that it compensates for the DC voltage. <S> For a DIY function generator having that DC voltage present might actually not be an issue. <A> I bought the ultra-cheap function generator last year. <S> It is based on an XR2206 IC clone. <S> It is barely usable. <S> The sine and triangle waves have DC superimposed , you need to add your own cap to remove it (or a bias circuit). <S> the square wave amplitude and duty cycle is fixed. <S> One of the knobs is backwards. <S> The harmonic distortion on the sine is horrible, about 5%. <S> To keep it cheap, they took a lot of shortcuts, they didn't even include the symmetry trim (most users won't be able to set it properly anyway). <S> I thought that I could get by with it, but I couldn't, I just bought a used HP33120A on ebay. <S> What do you get for more money: <S> Better sine waves. <S> Adjustable DC offset. <S> Adjustable square wave amplitude. <S> Lower and higher frequency capability (cheap one is 1 Hz to 1 Mhz). <S> Digital display. <S> Digital parameter control. <S> Proper 50 ohm source impedance (cheap one has about 600 ohms). <S> Sawtooth wave. <S> AM modulation. <S> FM modulation. <S> Frequency sweep capability. <S> Burst capability. <S> Arbitrary waveform capability. <S> Remote programming. <S> I am usually pretty good at work-arounds and making due with what I have, but the cheap version is below my threshold. <S> Depending on your needs, it might be usable for you. <A> Function generators differ in price due to the following factors: <S> - Frequency range - Voltage range - Accuracy - <S> The amount of power it can deliver (this might be the most important for you). <S> Normally a function generator is used through an amplifier which then powers the load. <S> - Whether they can create arbitrary wave forms etc. <S> I can't make sense. <S> I don't think these cheap products can actually produce a sine wave with a negative voltage portion. <S> Why not? <S> Split voltage supplies and amplifiers are trivial.
| Making a "proper" balanced output signal with no DC present requires a slightly more complex circuit.
|
Do linear thermistors exist on the market? I'm working on a project right now that requires me to use a thermistor to sense the tempertaure change with an op amp comparitor with hysteresis. Because of the non-linearity of a normal NTC or PTC thermistor, I wanted to know if there existed any thermistors whos resistance changes linearly to the amount of heat applied. A quick google search shows me that a Silistor is exactly that, but there isn't a product that I found that has that name. So do these linear thermistors exist on the market, and do they exist in a through-hole format? Side note, I'm not making this a product recommendation. I just want to know if they are sold and if they have a specific name. <Q> No, thermistors are always nonlinear. <S> But RTD do exist and they are linear. <S> They are widely used in industry as Pt100 100 ohm resistance at 0 degrees Celsius, since their low resistance are very immune to the environment noise, but difficult (expensive) to measure. <S> On the other hand HVAC use Pt1000 i.e. 1000 ohm at 0 deg. <S> They are easier to measure as Pt100. <A> RTD's (Resistance Temperature Device?) are pretty linear, but they are not anywhere as sensitive a thermistors. <S> There's a whole range of temperature sensing options available if you want to consider an active device. <A> The common "Pt100" sensor has a positive temperature coefficient and is linear within a certain temperature range, see this example datasheet . <S> Note that the Pt100 is a resistance temperature detector or RTD , which although it has a positive temperature coefficient, isn't a PTC. <A> It depends on what range of temperatures you want to measure. <S> If it's restricted to about 10 °C, then form a potential divider with a resistor equal to the resistance of your NTC at the midpoint temperature in series with it, and measure the voltage at the junction. <S> If you are happy to switch resistors, or use several dividers and multiplex their outputs, then you can extend the linear range by stitching several small linear ranges together. <A> Adding to what the other members already wrote. <S> If you are willing to go a bit crazy and need a higher accuracy, you could theoretically "increase" the linearity of a thermistor. <S> This can be accomplished by using a bunch of them and averaging their values (Much like the jar of beans experiment). <S> In practice, you could use a non-inverting amplifier to average the values read out by each thermistor. <S> Consider the following simulation <S> :It consists of three resistors which are modeled as thermistor (In this case the time represents the temperature). <S> Due to the exponential factor and the offset constant, they produce different non-linear voltages. <S> At the right side, you have an average amplifier which also acts as a buffer. <S> As you can seen in the waveform, the output voltage is nothing but an average of all thermistors' voltages. <S> If you are patient enough you can find thermistors which would produce a "differential" behavior, therefore creating a linear temperature behavior if averaged out. <S> Once again, I am not sure whether you really need such a linearity in your project. <S> I hope it helps. <A> You could search for them under "PTC temperature sensors" or "PTC thermistors". <S> Here is what Digi-Key has in that category: https://www.digikey.com/products/en/sensors-transducers/temperature-sensors-ptc-thermistors/550?k=ptc+thermistor&k=&pkeyword=ptc+thermistor&sv=0&pv1989=0&sf=0&FV=-8%7C550&quantity=&ColumnSort=0&page=1&stock=1&pageSize=500 <A> Pt 100, pt 500 and pt 1000 are pretty linear device, there are tables and algoritme for Conversion, but if you need high accurancy at Wide range there are a variety of thermo controlers and 4 to 20 ma transducer. <S> Choice should be made by the purpose.
| All thermistors are non-linear, some more so then others.
|
Connect two computers with USB using two FTDI chips I have a very specific problem. I need to communicate between two computers (Intel NUC and a Jetson TX2, both running linux to be specific). I cannot use wireless, and the ethernet ports are occupied by other devices. I need a very simple low bandwidth communication between those two computers. So my plan is to make a small pcb with two FT232RL chips, and connect them like this:Computer 1 -> USB -> FT232RL <=> FT232RL <- USB <- Computer 2.I assume that this would work like a serial port between the computers. Will this work? Is this a good solution, or is there something better? <Q> SLIP and PPP were designed for this very purpose: IP connections over serial links. <S> Done. <A> Yes, it will work, just like two serial ports connected in a null modem would. <S> But no, it's not a very good solution considering other solutions. <S> These other solutions are more general computing than anything. <S> Use a switch. <S> The devices already have ethernet. <S> If they are one the same network, you are done. <S> This is the entire point of networking. <S> Use a router and port forwarding. <S> Basically complicated version of 1. <S> There are USB to USB bridge cables, but these typically need drivers installed, and are antiquated. <S> There are USB to Serial <S> (RS232 or TTL) that you can just wire to each other, but this is as good as a solution as you home brewing one. <S> You will need to configure low level serial communication at the OS or App level. <S> Usb to Ethernet adapters. <S> Off the shelf these tend to be plug and play, all you do is set up a static IP address for the new network interface, and you're basically at option 1. <S> Finally, the most electrical engineering solution. <S> Use the built in serial TTL ports on both the Jetson and the NUC. <S> Both are IIRC 3.3v <S> so you literally just need 3 wires (TX, RX, Ground) and no extra parts. <S> If they are at different voltages, there are cheap boards to do level translation or you can use a simple transistor based translation. <A> Ten-fifteen years ago there used to be USB-to-USB connect "cables", <S> Type-A on both ends, and some "bridge" in between. <S> Not sure what USB class was employed, but file transferwas available.
| Just get the two adapters and connect them with a ‘null modem’ (crossover) cable, set up pppd or slipd. It's easier to just purchase a solution that's proven.
|
Stepper motor won't turn, just vibrates I'm using 2 x 1.8-Degree Nema 14, 35 BYGHW Stepper Motors with a DRV8825 Stepper Motor. I have tried all 24 combinations of wiring the damn 4 wires but it won't work. I am using a 9v battery as the motor power source, which is fine. It sometimes turns when I have 3 wires connected?? I can guarantee the wiring is correct, and this is the sketch. Here is the code, which I am almost certain is correct too, I just started this project and I can't even get a damn motor to work! /* stepper motor control code for DRV8825 * */// define pin usedconst int stepPin = 9;const int dirPin = 8;void setup() { // set the two pins as outputs pinMode(stepPin,OUTPUT); pinMode(dirPin,OUTPUT);}void loop() { digitalWrite(dirPin,HIGH); //Enables the motor to move in a perticular direction // for one full rotation required 200 pulses for(int x = 0; x < 200; x++){ digitalWrite(stepPin,HIGH); delayMicroseconds(500); digitalWrite(stepPin,LOW); delayMicroseconds(500); } delay(1000); // delay for one second digitalWrite(dirPin,HIGH); //Enables the motor to move in a opposite direction // for three full rotation required 600 pulses for(int x = 0; x < 600; x++){ digitalWrite(stepPin,HIGH); delayMicroseconds(500); digitalWrite(stepPin,LOW); delayMicroseconds(500); } delay(1000); // delay for one second} <Q> There was no need to try 24 wiring combinations. <S> Your schematic shows that you have a two-coil motor. <S> You identify the two coils with your multimeter and connect one coil to A1 - A2 and the other to B1 - B2. <S> The motor should now run if you are using full-step mode. <S> A 9 V battery may not be adequate. <S> Measure the battery voltage while trying to run the motor. <S> You can also increase the step delays by a factor of 100 and see if the motor turns slowly and whether or not you can feel any torque. <S> For further help please edit your question and add a link to the motor datasheet. <A> 9V batteries are very weak. <S> When I was a kid, they were sometimes called transistor batteries, since that is all they were good for, powering low-power transistor radios. <S> Is the enable_n pin low by default? <S> You may need to drive it low. <S> Use full step mode until you get it to work. <S> Try longer delays. <S> If you want to run the motor fast, you need to accelerate gradually. <S> There are drivers to help you for some MCUs. <A> If I had to guess and the motor is really rated at 1.2 Amps, you are fighting a losing battle! <S> There's no way in the world you can suck 1 <S> Amp out of a 9 volter! <S> You basically need to start with a beefier power supply, say 3 Amps <S> or so <S> and you possibly may have to put transistors between the motor and the controller. <A> Two things:Steppers have a lot of mass in the rotor and thus high inertia. <S> Your 1-millisecond initial move may cause it to bounce between one step behind / one step ahead, which would be similar to the buzzing and vibrating you see. <S> You need to allow it time to spin-up or by having longer delays initially. <S> this will allow a more gradual move between poles. <S> As a rule of thumb, to move fast, you want high voltage, for greater torque, you want high amperage. <S> I can recommend a library called "AccelStepper" for controlling if you aren't determined on writing your own. <S> It's always fun and full of important learnings to write it yourself. <A> I'm using a 28BYJ stepper which had a similar problem initially. <S> I corrected it by increasing the delay time from 500s to 2000ms and decreasing the rpm from 100 to 20.
| If using half-step mode (where more than one coil is on at time) you may need to reverse one of the coils. The 9v battery might be too weak. You can try one of the microstepping modes of the 8825 (the Mx inputs)
|
Replace ONE 25v 47uf capacitor with FIVE 60v 10uf capacitors parallel? MY 1000 Watt True Sine Power Inverter blew a 25V 47uf capacitor. The shell was rattling around inside, and the ribbons were exposed. Yay, I found the problem! Sooooo, how can I tell the mounting polarity in order to replace the blown cap? ALSO, would it be a silly idea to replace that cap with FIVE 60V 10uf caps in parallel (I have these available) essentially making a single 60V 50uf cap? ... or should I just buy a direct replacement? Wondering what effect the slight variation in capacitance might have. I'm kind of a novice, but the smoke is already out...what can I lose :P Thanks in advance :) <Q> Polarity? <S> Use an ohm-meter to find what else is connected the cap traces. <S> You may find the polarity on one of those. <S> Ground/Power pours may be recognizable. <S> See if it is connected to any. <S> -ve side of caps, GND pins of IC's, etc. <S> The major difference with 5 caps in parallel to make 50uF may not be the capacitance, but the series resistance. <S> A lower series resistance may do good things (like better ripple), bad things (like instability), or nothing. <S> Depends on the circuit. <S> Like you say, the smoke is out <S> so it doesn't hurt to try. <S> Except that it sent the cap cover flying last time. <S> So'd <S> I'd close the enclosure or wear eye protection before applying power here. <S> The cap may not be the only thing that died. <S> It may just be the most obvious. <S> But try away - with caution. <A> Wondering what effect the slight variation in capacitance might have Almost certainly none. <S> It's 6% <S> Most large-ish capacitors have very wide tolerances in the first place, and circuits which use them are very rarely sensitive. <S> Good background at Wikipedia , which says: Electrolytic capacitors, which are often used for filtering and bypassing, do not have the need for narrow tolerances because they are mostly not used for accurate frequency applications like in oscillators <A> You will find the negative pole where there is a white band with sparrows on the side of the capacitor. <S> I don't think there will be a problem using 5 x 10uF... <S> but one 47uF cap is not that expensive neither and may be easier to mount.
| Check polarity on other devices connected to the pour.
|
Detection of signals buried in noise This is more of a general question to see which methods are the most common to improve the detection of a signal which is buried in noise. Currently, we are building an optical system for medical imaging and the signal that is being detected is 1000x lower than the noise floor. Currently, we are looking at methods such as lock-in amplification/detection and other types of filtering but it was an open question to see which are common methods to improve the detection of a signal in a noisy background. <Q> The only way that a signal that is "buried in the noise" can be detected is if you can run the signal + noise through some filter that attenuates the noise more than it attenuates the signal. <S> At which point the signal is no longer buried in the noise, so "buried in the noise" was just a hasty assumption. <S> In radio carrying digital data, you run it through a matched filter and then a detector. <S> In spread-spectrum radio, you correlate the signal + noise with a pseudo-random sequence, then bandpass filter, then detect. <S> In visual systems, you correlate the noisy image with a 2-D prototype of the anticipated signal, or you run the noisy image through a spatial low-pass filter, then you detect. <S> In all cases, the signal has to be distinct in some way from the noise -- if it is not, then you cannot filter out the noise without filtering out the signal, too. <S> I'll add to this: <S> At the top level, a filter for signals is like a coffee filter or a colander: you have the stuff you want (coffee or fresh-cooked pasta) and the stuff you don't want (coffee grounds, or starchy hot water), but it's all mixed together. <S> So you run the mess through a filter. <S> In the case of coffee, you keep the stuff that gets through the filter. <S> In the case of the colander, you keep the stuff that gets left behind. <S> In either case, you're using the fact that one thing (coffee grounds or pieces of pasta) is bigger than the other (water molecules and all the other stuff you want in coffee, and don't want in pasta). <S> A signal filter does the same thing -- you get rid of what you don't want because it is different from what you do want. <S> If you can't figure out how it's different, and how to build an algorithm to separate it -- you can't filter your signal from your noise. <A> The general concept of detecting a signal in noise is knowledge of something about the signal, and ideally something about the noise. <S> The easiest thing to use is spectral knowledge. <S> If you know the signal occupies some part of the spectrum, then you can safely filter out noise in other parts of the spectrum without losing signal. <S> This is taken to its extreme in the 'lock-in amplifier', which is basically just a method of creating a very narrow bandpass filter at precisely the frequency of the signal. <S> A more general property of the signal is its waveform. <S> We can correlate the signal plus noise with a copy of this waveform, and then average. <S> The noise does not line up with the correlating waveform so adds as power. <S> The signal does correlate, so adds as voltage, leading to a 3dB improvement in SNR each time the number of averages is doubled. <A> Another technique that might be useful is autocorrelation Autocorrelation, also known as serial correlation, is the correlation of a signal with a delayed copy of itself as a function of delay. <S> Informally, it is the similarity between observations as a function of the time lag between them. <S> The analysis of autocorrelation is a mathematical tool for finding repeating patterns, such as the presence of a periodic signal obscured by noise, or identifying the missing fundamental frequency in a signal implied by its harmonic frequencies. <S> It is often used in signal processing for analyzing functions or series of values, such as time domain signals. <S> Different fields of study define autocorrelation differently, and not all of these definitions are equivalent. <S> In some fields, the term is used interchangeably with autocovariance. <A> And then there is the idea of averaging multiple independent observations. <S> In overly simplified terms, the aim here is to increase the signal strength while letting the noise cancel itself out, i.e., the desired signal increases faster than the noise, and the more samples you average the better. <A> If you know exactly what the signal looks like, in the time domain, you can implement Matched Filters to discard noise energy in frequency regions that are not needed to construct the waveform.
| In radio carrying an audio (or Morse code) signal in SSB or AM, you take the signal + noise and you filter it by the approximate bandwidth of the signal, then you run it through a detector.
|
How do laptops seamlessly switch from AC wall power to battery So this is probably a rather simple question but electrical circuits are not my forte. I have been curious about how devices, specifically a laptop, can seamlessly switch between AC wall power (plugged in) and battery power? This has been on my mind after doing some research on Lithium ion batteries and their lifespan in laptops. Some folks say that when plugged in, the laptop is using power through the battery and simply constantly charging it between that 90% and 100% range as it goes. Now this seems rather detrimental for the lifespan of the battery, so I can't imagine that most laptops are actually doing this. In my mind the simplest solution is just running completely on the AC wall power and leaving the battery alone when it is fully charged. So the question lies in how does the laptop switch from using AC wall power to using the battery, without shutting down for that split second when you unplug the machine from the wall power adapter? My expertise is in the software realm, so I am rather ignorant on how the actual electrical engineering works in these cases. My simple guess was that a capacitor(s) supplies the power during that short time when switching between power sources. So, could anyone please explain how this actually works? Feel free to be as technical as you'd like. I have an engineering mindset, so I find the low level details interesting. Thanks! Also, if this is a duplicate question, feel free to close or remove it and link the answered thread. I did some searching and didn't come across a similar enough thread. <Q> So the question lies in how does the laptop switch from using AC wall power to using the battery, without shutting down for that split second when you unplug the machine from the wall power adapter? <S> Short answer: very fast controls. <S> There are many ways of implementing this functionality, but on modern (circa 2020) motherboards this is done by means of a dedicated charger IC that controls when power should come from the AC wall power or from the battery by connecting/disconnecting the battery by means of a MOSFET between the battery and the rest of the motherboard. <S> The example schematic above comes from the ISL9238 datasheet. <S> This IC measures <S> the AC input voltage (which is DC by the time it arrives to this IC) and if present, uses that input to power the motherboard + charge the battery. <S> Otherwise the IC switches to powering the system from the battery only, and the switching occurs within a few microseconds. <S> During that small n-microsecond window the decoupling capacitors on the board hold enough charge to power the motherboard. <S> And should these capacitors, for some reason, be not enough to power the motherboard during the switching period <S> , the body-diode of the series MOSFET will conduct until the IC fully switches on the MOSFET. <S> There are other operating modes available from this IC (such as battery supplemental mode), but that's outside the scope of this question. <S> There's a whole world of charger ICs out there that do this: the ISL9238 is just one of them. <S> Checkout <S> Digikey's Battery Chargers section if you're curious to learn more. <A> A number of laptops work just as you describe, and it is in fact very rough on the batteries. <S> Not only charging to 100%, but just sitting idle at 100%, stresses a lithium cell. <S> Some charging systems divert input power separately to battery and load when plugged in. <S> This is a better system, because it gives more control over the battery current (since it can be measured separately from the load current). <S> However, this doesn't solve the problem of sitting idle at full charge. <S> For this reason, some laptops have a BIOS option to charge the battery to only 80%. <S> Some studies have shown that keeping the battery between 20% and 80% can increase battery life by an order of magnitude as opposed to running down to 0 and charging to 100. <S> Charge levels below 20% are also disproportionately stressful to batteries. <A> The battery and power supply schematics for laptops are very complex. <S> When the laptop is connected to the AC mains power supply, the voltage presented by that unit is provided to both the internal switching power supplies AND the internal battery charger unit. <S> The battery will have FET switches or at minimum be diode isolated on the output so that a laptop running on it's AC power supply IS NOT connected to the battery at all. <S> Diode isolation is rarely ever used because of the Vf/power problem, FET switches or so called Active Diodes get used. <S> To charge (and know the state of) <S> the battery is a complex task. <S> Figuring out the state of charge for the battery involves a complex voltage monitoring or power state IC embedded in the battery pack. <S> If you put some thought into the operational characteristics of a laptop you might think of: The battery voltage is not the same as the external power supply voltage, and typically a down converter would be used. <S> Most typical laptops have a 19V (or thereabout) input level. <S> Most laptop battery (LiIon) packs are 4 cells, so the maximum voltage will be about 16.8V. <S> This allows a reasonable voltage drop (a buck charger) from 19V. <S> Virtually all laptops have a screen indicator for the state of charge, and this would be almost impossible to calculate if the laptop was drawing power from both the external PS and the battery.
| Most laptops will show a percentage charge for the battery while on mains power, and this requires that the battery be isolated from the main internal power supply DC-DC converters.
|
How can I find a correct crystal replacement for this circuit? I need to replace a crystal component (16.257 MHz, one of its legs broke off and cannot solder it back on) from an old video card that is wired up like the schematic below (the resistor / capacitor / frequency values differ, it's just for reference on how it is connected) : Now I read that there is such a thing as a series-resonant oscillator and a parallel-resonant oscillator, but I don't know under what category this falls. However given the schema above, I would like to find a drop-in replacement. When I search for a replacement part on Mouser, by providing the frequency: https://www.mouser.be/Passive-Components/Frequency-Control-Timing-Devices/Crystals/_/N-6zu9f?P=1z0wnq5Z1z0z7l5 I noticed that some of them say "load capacitance = 20 pF" others say "load capacitance = series". I also noticed different form factors. The ones listed here all have a very low form factor, while my broken component had a larger profile: What kind of replacement could I choose? Does the housing (form factor matter) matter, and what about the load capacitance difference or other properties I should be aware of? <Q> You should use the larger profile case (HC-49) because it is more tolerant of overdriving. <S> If you can't get it, you can use the smaller HC-49/U or HC-49/US, but pick the one with the lowest ESR. <S> The allowable drive level is about half. <S> For a video card, probably any of them will work. <S> I would pick something with a 15-20pF load capacitance. <A> Electrically, a crystal looks like this:- simulate this circuit – <S> Schematic created using CircuitLab C1 and L1 produce a series resonance, while C2 and L1+C1 produce a parallel resonance. <S> The two resonant frequencies are very close, but not exactly the same. <S> A crystal specified for operation in one mode will produce a slightly different frequency if operated in the other mode. <S> For many applications the difference (typically less than 0.1%) is not significant. <S> Here's an LTspice plot of impedance in ohms (left scale, solid line) and phase (right scale, dotted line) for the circuit above:- <S> Lowest impedance (equal to R1) is achieved at the series resonant frequency (10.002 MHz). <S> This suits an oscillator circuit with a non-inverting amplifier that needs a feedback element with low impedance and small phase shift. <S> External parallel capacitance has little effect because it doesn't change the series resonant frequency (which is determined by the physical properties of the crystal) and phase varies steeply in this area. <S> Parallel operation is possible in the higher impedance area where phase shift is ~+90°, which suits an inverting amplifier with high input impedance and large phase shift. <S> In this mode the exact frequency can be 'tuned' over a small range with external capacitance. <S> Crystals designed for this usually have the external 'load' capacitance specified. <S> Your circuit is designed to operate the crystal in series resonance, so a 'series' crystal should oscillate closer to its marked frequency. <S> However a 'parallel' crystal should also work provided its resistance is low enough. <S> The larger profile case has a bigger crystal in it which can handle higher power without being overloaded (which might cause misoperation and/or reduce its lifespan). <S> You should replace it with one of the same size. <A> Using incorrect crystal will still work but at slightly wrong frequency, since those crystals run at their rated frequency in a parallel circuit with the rated load capacitance.
| That is a series resonant oscillator circuit, so you need a series type crystal which runs at the given frequency in series operation.
|
What's the use of the average voltage in rectifiers? I've noticed that there is much talking about \$V_{dc}\$ in rectifiers (specifically, the half-wave rectifier) and less focus is given to \$V_{rms}\$ . Why is that? I thought the RMS value was the much more important one (because it's the value of the DC signal that delivers the same power within the same time interval), but what use is the average voltage? Why do we even bother to calculate and emphasize that \$V_{dc} = \frac{V_{MAX}}{\pi}\$ ? <Q> Vrms is a quantity which only makes sense in the heating equivalent in a resistive load. <S> Very few rectifier applications that I have seen have a rectifier which drives a resistive load for any heating application. <S> There used to be true RMS voltmeters which actually would apply the measured voltage to a resistor and would apply a DC voltage (of course using buffer amplifiers in each case) to an identical resistor which was controlled servo fashion to cause a match in the temperature rise in both resistors, and then the DC Voltage measured would be equivalent to the AC voltage (I have one of these). <S> In this way the measured AC waveform could literally be arbitrary in shape. <S> There are true RMS converters that do this without heaters, and add cost to a multimeter. <S> So I think your question is in relation to a average reading, RMS scaled readout of an AC voltage multimeter. <S> This is a very cheap way to accomplish the RMS reading, but is only good for sinusoidal measured input, since the area under a half-sine between zero crossings is 2/π × peak and must be scaled to the .707 × peak. <S> Any other wave shape than half wave sine output will not correspond to the 2/π. <S> BTW the average of a voltage is the DC voltage at the output of a lowpass filter and this is how the rectified voltage pulses are converted into a DC voltage and scaled, in the case of a sinusoid, by the ratio of 1/2 2 <S> : 2/π for a cheap way to read an AC voltage as long as it is sinusoidal. <A> I've noticed that there is much talking about Vdc in rectifiers (specifically, the half-wave rectifier) and <S> less focus is given to Vrms. <S> Why is that? <S> When designing a DC supply the peak and rms voltages are usually much more important. <S> However a DC meter reads average voltage, so when measuring the rectifier output you need to know the relationship between average, rms, and peak voltage. <S> As to the 'much talking' about it, I suspect it is mainly confined to basic electronics courses and multimeter usage tutorials. <S> Few modern devices use power supplies with unfiltered half-wave rectifiers, so the need to explain why your meter is reading much lower than expected is less than it used to be. <A> Power is not the only consideration when designing rectifier circuits. <S> The goal is often to design a circuit that supplies a specific voltage to a load, and so knowing the average voltage is an important part of that design process. <S> If you don't believe me, try charging your beloved mobile device with a 0.1W supply....at 170V.
| Average voltage is also important for calibrating an average reading AC meter (which consists of a DC meter with half wave rectifier) in the equivalent rms voltage.
|
High Voltage (0-600V) DC Voltage Sensing, Shared Ground with Isolated Converter and HV- I'm designing a circuit to detect when the voltage across two DC input lines (HV+ and HV-) is greater than 60V. The voltage will vary from 0-600V. A seperate supply of 12v and 5v is available for use. The output side of the sensing circuit (and if used the 12v and 5v supply) must be galvanically isolated from the high voltage lines. Initial Idea My initial idea was to use a 12v zener diode voltage divider with an opto-isolator as shown in Fig 1. R2 and D2 are modelling the opto-isolator. The value of R1 is set so that once the input voltage V1 reaches 60V, enough current (5mA) gets through to meet the forward current requirement of the opto-isolator diode. The main issue with this circuit is the large amount of power (~43W) burnt across R1 (be it a single large resistor or several smaller ones) once the input voltage reaches 600V. Possible Solution? I was hoping that an isolated DC/DC converter (such as https://uk.rs-online.com/web/p/isolated-dc-dc-converters/1896931/ ) could be used to supply an IC (such as a comparator) to reduce the current through any voltage divider used between the HV lines, as shown in Fig 2. My question is whether the converter can share a ground with the HV line (ie HV- connected to converter ground)? And should you have time and/or are willing, suggest alternative solutions to this issue? Thank you <Q> The isolated DC-DC you picked has an isolation rating of only 1kV. <S> That's not enough to meet safety standards nor is it in my opinion enough to use on a 600V bus. <S> You can read a good explanation of the various types of isolation HERE. <A> I'd use a LM431 as comparator in the HV lines. <S> Because my phone doesn't support the editor, below a modified schematic: <S> The LED represents the LED from the optocoupler. <S> Because the HV can become 10x as big (60V to 600V) <S> I'd choose an optocoupler with min CTR > 100% and a resistor (R lim ) that limits forward current from 1 mA @ <S> 60V <S> (that's what the LM431 needs as well!) <S> and so 10 mA @ <S> 600V. Drawback is this resistor still dissipates 6W at 600V. UPDATE <S> As Jasen points out, the cathode voltage should be limited. <S> This can be solved by adding a resistor parellel resistor (R p ) to the LM431 and optocoupler's LED, so it makes a voltage divider with the current limiting resistor. <S> Choosing the value of this parallel resistor equal to the current limiting resistor, limits the voltage to max 30V (because then the LM431 starts working). <S> The LM431 has a threshold of 2.5V, so, keeping in mind Iref, you can calculate R1B and R2B to get 2.5V when the HV becomes 60V. <S> If I recall correctly, LTspice also has this (or an equivalent) shunt reference in its library. <S> If 6W dissipation is too much, <S> you'll need a DC/DC converter: <S> disconnect the optocoupler's current limiting resistor from the HV lines and feed it with the DC/DC converter. <S> Because this DC/DC converter is locally (between the HV lines) <S> the converter itself does not need to be isolated. <S> But I think this converter will be too costly to be interesting. <A> This is just an idea but may be useful. <S> Figure 1. <S> NE-2 <S> type neon lamp powered by alternating current (AC). <S> Source: Wikipedia [ https://en.wikipedia.org/wiki/Neon_lamp] . <S> A small electric current (for a 5 mm bulb diameter NE-2 lamp, the quiescent current is about 400 µA), which may be AC or DC, is allowed through the tube, causing it to glow orange-red. ... <S> The lamp glow discharge lights at its striking voltage. <S> A small electric current (for a 5 mm bulb diameter NE-2 lamp, the quiescent current is about 400 µA), which may be AC or DC, is allowed through the tube, causing it to glow orange-red. <S> The gas is typically a Penning mixture, 99.5% neon and 0.5% argon, which has lower striking voltage than pure neon, at a pressure of 1–20 torrs (0.13–2.67 kPa). <S> The lamp glow discharge lights at its striking voltage. <S> From memory these breakdown at about 60 V which would suit you. <S> CircuitLab won't load for me right now but the idea is to make a series RC circuit with R connected to V+ and C connected to ground. <S> Connect the neon indicator across the capacitor and make your own opto-isolator by attaching the light sensitive receiver of your choice to the neon. <S> It's going to flicker so you would need to address that in your software. <S> Will it work? <S> I don't know. <S> Figure 2. <S> Only the cathode lights up. <S> +DC (left), -DC (center), AC (right) supplied to NE-2 type neon lamps. <S> CircuitLab is working again. <S> simulate this circuit – <S> Schematic created using CircuitLab Figure 3. <S> The idea. <S> Component values not calculated. <A> if you need precision your "possible solution" is on the right track, but because isolated DC-DC converters typically have poor regulation you probably want to use a voltage reference instead of a divider for the inverting input, or use a LM431 which has an internal voltage reference. <S> simulate this circuit – <S> Schematic created using CircuitLab
| The idea can work, and you can tie the DC-DC isolated ground to the HV-, but you probably want a converter with "reinforced" isolation of at least 5kV.
|
Higher voltage battery able to drive load longer at stable voltage? If I've got two batteries of the same capacity, one is 12V and the other is, say, 18V with a step down buck converter, will I essentially be able to power my load at 12V for longer? I know that a battery's voltage will decrease as it discharges, so it seems to me that a better way to power a load with a more stable voltage (and maybe for a longer period of time at this voltage) would be to power it with a higher voltage battery and step that voltage down. Is this true? <Q> What matters in not voltage , it's energy . <S> The energy in a battery is quoted in Wh (watt-hours). <S> Many still don't specify the watt-hour capacity but do quote the Ah (amp-hour) capacity. <S> With this and the battery voltage you can calculate the Wh capacity from: $$ Wh = <S> V <S> \times <S> Ah $$ . <S> Compare the two. <S> If you use a step-down converter you need to factor the efficiency of that into the calculations. <A> The other guys are right in theory or abstract. <S> And certainly, whichever battery has higher energy storage will likely last longer. <S> So I think you are very much on the right track in your thinking and approach. <S> A simple buck will be easier to find and cheaper and more efficient than a buck/boost. <S> Let us consider 12V. <S> If you use a 3S lithium ion battery pack, your span will be something like 12.6-10V. <S> If you need 12V, then you will have to buck/boost. <S> But if you go 4S, then your span will be 16.8-13.2V. <S> So you can use a buck. <S> That is going to be easier and more efficient. <S> In low-current, low cost applications, a linear regulator may also be considered instead of a buck. <S> Typically linear regulators can be designed with lower quiescent currents than a buck. <S> So the overall system battery life may not be any worse when the low quiescent current is factored in. <S> I doubt that is the case for you, though. <S> Usually 12V applications are not low-power. <A> Assuming we are comparing apple with apples, ie. <S> both batteries have the same energy capacity and power, there is no fundamental difference between them. <S> So you are looking at secondary effects such as converter efficiency and the load's tolerance for voltage variation. <S> The primary advantage of a higher voltage stepped down is that the regulator can maintain constant output voltage throughout the discharge. <S> If your load requires at least 12V for proper operation then this could extend run time significantly (if the '12V' battery drops below 12V toward the end of its discharge). <S> On the other hand the step-down regulator itself wastes some power, so the higher voltage setup may actually perform worse . <S> Achieving high converter efficiency at both high and low currents is difficult. <S> A load that needs high power for short periods (eg. <S> DC motor) <S> and/or spends long periods in standby (eg. <S> telemetry transmitter) might be better powered directly from the battery. <S> Finally you should consider the extra bulk and weight of the regulator. <S> To be truly equivalent the two systems should be the same size and weight, which means the higher voltage battery may have to be smaller to accommodate the regulator. <S> A good example is a drone, where any weight increase results in lower run time. <S> A higher voltage stepped down would suffer from extra loss in the buck regulator, plus the extra weight of the 'oversized' regulator required to handle peak motor currents. <A> You are forgetting that there are converters that can both step up or step down the voltage as appropriate. <S> So you don't actually need a higher battery voltage than your load voltage. <S> But you do need the battery to have enough energy left in it, whatever its actual voltage is relative to the desired load voltage. <S> A dead battery is a dead battery, no matter how high or low its voltage is. <S> But just because a battery has a lower voltage than you need for you load doesn't mean it has no energy left inside it that can't be used to step up the voltage. <A> Only something to add: <S> Higher voltage battery has more cells in series. <S> The total voltage with a regulator probably stands usable when the weakest of the cells is exhausted. <S> That weak cell starts to get charged reversely when the rest of the cells still output current. <S> The weakest cell can get serious punishment during that process. <S> Advanced battery packs have individual cell monitoring circuits to prevent self-destructive usage like this.
| But from a product design perspective it is a good idea to use a battery pack that is a bit higher in voltage than what you need and buck down to your desired voltage.
|
How many times of a frequency of a PWM can properly generate a sine wave to drive a motor? I know PWM with dense and loose duty cycle is used to generate sine wave for applications such as 3-phase motor driving. But, How fast does it require to properly generate a sine wave to drive a motor? I mean, if a sine wave is 1 Hz, how many times of the PWM base frequency is required to properly generate a sine wave PWM? I guess 100 times would be fast enough, 1000 times would be perfect, 50 is barely enough, and 10 is not enough based on my imagined drawing of PWM in my mind. Is there a standard or commonly used number of times to generate a sine wave? Thank you! <Q> At that frequency, motor losses were increased a little, but careful selection of motor ratings to suit the duty compensated for that. <S> The acoustical noise made by the motors was ok for factory environments. <S> PWM at 30 or 40 times motor frequency is sufficient to reduce motor losses to an almost insignificant increase above pure sine wave losses. <S> Offering a customer adjustment in the range of 30 X to 300 X provides the opportunity to find an operating point at which acoustical noise is not a problem. <S> Operating in the upper end of that range may come at the cost of reduced VFD efficiency and the need to upsize the VFD for a given motor. <S> As devices with reduced switching losses become more competitive, higher PWM multiples may become more common. <A> In my experience as a post-doctoral researcher in motor drives, 10x fundamental frequency is generally accepted as a minimum PWM frequency. <S> I searched the literature to find precedents for this when writing papers/my PhD thesis, but found surprisingly little formal research on this. <S> Source: experience ( https://scholar.google.com/citations?user=N7P5-hUAAAAJ ) <A> For practical purposes, the most important limit is to go above the acoustic threshold, i.e. the frequency should be more than 20 kHz. <S> Above that it is a matter of mechanical vibrations caused by torque jitter. <S> Electrically the torque jitter is determined by the motor coil characteristics, most importantly the coil inductance, operating voltage and operating current. <S> For example, for a small brushless motor the coil inductance could be 100 µH, operating voltage 10 volts and operating current 1 ampere. <S> If the PWM frequency is 20 kHz, the low and high duration is around 25 µs. <S> During this time, the coil current will change by 25 µs * 10 <S> V / 100 µH = <S> 2.5 A. <S> The magnetic field follows the current, which causes the torque to jitter by +- 125%. <S> If the PWM frequency could be raised to 100 kHz, current would change only by 0.25 A and the torque jitter would be +- 25%. <S> Larger motors usually have more inductance, permitting lower PWM frequency. <S> The velocity jitter is not as large, because the inertia of the motor will smooth it out. <S> But the torque jitter directly causes vibrations in the surrounding structure, and the strength of the vibrations depend on mechanical properties of mass and elasticity.
| In the early days of VFDs, PWM was successfully used to control motors at less than 10 times the frequency of the synthesized sine wave.
|
Broad VF range of LED... 5-9V? I'm not quite sure what to do with this LED, it's got a forward voltage rating of 5-9V, so I've got a lot of wiggle room and I'm not sure of the ideal (or even if there is an ideal) voltage at which to drive it. Here is the datasheet info. Perhaps someone can tell me how they think I should drive it. I know I'll be using two lithium cells in series, so because of the broad range in this LED's VF I can either drive it directly until the 6V cutoff, or regulate it with a buck to a a certain voltage. Thanks a heap ] 1 ] 2 [ ] 3 <Q> You shouldn't drive an LED with a voltage source. <S> You need to use a current source at up to 150mA <S> A resistor from a voltage significantly higher than the LED voltage is usually acceptable for low power applications although the current won't be as well defined as a proper current source. <S> The data sheet is telling you that when you drive the LED at the specified current (150mA in this case) the voltage across the LED could be anywhere between 5V and 9V. <S> If you use a voltage source such as 9V there could be enough variation between devices <S> such that one operates correctly whole another <S> has much higher current that exceeds the safe limit for the device thus destroying it. <S> Similarly, if you used 5V some LEDs would give out the specified amount of light but another might not give very much at all. <S> The behaviour will also change over temperature or as the LED warms up during operation. <A> You don't drive it with a particular voltage. <S> You use a constant current source set to whatever current level you need for the LED to be bright enough for your use. <S> The datasheet says the LED is rated for a maximum of 150mA. <S> That is a UVC LED. <S> Please make sure that you won't be exposed to it. <S> UVC can damage your skin and eyes - and that seems to be a very powerful LED. <A> You can be safe with 6V, directly. <S> If there's only one LED per power supply you can connect your led directly without any current regulation. <S> But don't try the same with 8 or 9V. <S> At 6V you will have less light than at 9V but you will be within the safe area. <S> If you have several LED in parallel, you need to add 1/4W resistors (exact resistance can be calculated but it should be around 30ohm or so). <S> You can also power two leds in serie with 12V (most common power supply) or 4 leds with 24V. Also without any current control. <S> If you want maximum light for your LEDs then you will need a constant current power supply of ecxactely 150mA, able to give 3W. <S> This is less common but not impossible to find.
| You'll want a constant current driver that can supply at least 9V, but that limits the current to no more than 150mA (less if you need a lower intensity light.)
|
How do I calculate resistance when I have the voltage drop? This is my circuit: 3 volts in with a 1k ohm resistor in series with the LDR. I have the voltage drop across the LDR but I need to find the resistance. <Q> Let the resistance of the LDR be R. R and the 1k resistor form a voltage divider, and we know the total voltage across both is 3V. <S> So we get V = 3 R / (R + 1000) <S> Solve this to find R. <A> You need the current. <S> You could either measure it directly with your meter, or you could use the voltage across the 1K resistor and the fact <S> you know its a 1K resistor to calculate the current through the resistor. <S> Since the resistor is in series with the LDR, the current through both should be the same. <A> It's a voltage divider, basic rule: <S> R1/V1 = R2/V2 with V1 = <S> V - V2 <S> (V = 3V; V2 = Drop) <S> You can derive it using <S> I = <S> V / R = V / (R1 + R2) and V2 <S> = <S> I <S> * R2 = <S> > <S> I = <S> V2 / R2 <S> Just some basic math left and you get the above, and finally: R2 = <S> (V2 * R1) <S> / V1 = <S> (V2 * R1) / (V - V2) <A> Keep the ball flat.. <S> Current in both elements is the same. <S> V = voltage measured RLDR = resistance of LDR <S> I = <S> V / RLDR = (3 - V) / <S> 1kOhm <S> RLDR = <S> (V * 1kOhm) / (3 - V) <A> The voltage from the battery is shared between the LDR and resistor in proportion to their individual resistances. <S> For instance, if the LDR in this circuit has a resistance of 2kΩ, it would have two thirds of the total resistance and so it would drop two-thirds of the battery voltage. <S> Writing it like an equation: $$ \text{Fraction of total circuit voltage drop} = <S> \text{Fraction of total circuit resistance} <S> $$ <S> The fraction of total voltage drop is just $$ \frac{V_\text{LDR}} {V_\text{battery}} $$ and the fraction of total resistance is just $$ \frac{R_\text{LDR}}{R_\text{LDR} + R_\text{resistor} <S> } $$ <S> so if you accept they are equal you have: $$ \frac{V_\text{LDR}} {V_\text{battery}} = <S> \frac{R_\text{LDR}}{R_\text{LDR <S> } + R_\text{resistor}} $$ which is just the potential divider equation for resistances. <S> You can now solve for the resistance of the LDR. <A> You know the voltage drop \$V\$ at the LDR.Substract that voltage from 3 V.The difference is the drop at the 1 kΩ resistor. <S> The total current calculates as \$\frac{V}{1000 Ω}\$ . <S> Let's say LDR is 1 V. <S> Then <S> \$3V - 1V = 2V\$ . <S> 2 volts drop on the 1kΩ . <S> \$\frac{2V}{1000Ω} = <S> .002A\$ <S> LDR is \$\frac{1V}{.002A} <S> = 500Ω\$ .
| Then use the LDR voltage drop with the total current using Ohms Law to get the LDR resistance.
|
How do I solve using nodal analysis when there is a battery before the resistor? I need to use node analysis on the following circuit. I am confused as to how to solve node 1 as there is a battery of 3V between the node and 4 ohm resistor. Should I solve for it? <Q> You will need to use the superposition theorem. <S> That is replace the voltage sources with a short subsequently and then add the resulting values. <S> This explains it in detail: https://www.allaboutcircuits.com/textbook/direct-current/chpt-10/superposition-theorem/ <S> Also: <S> https://en.wikipedia.org/wiki/Superposition_theorem <A> First, if you are going to use nodal analysis you must select and clearly indicate which node is the reference node. <S> You can write one equation for the currents in and out of the supernode, and another equation (by inspection) for the voltage between the two nodes of the supernode. <S> In your specific circuit, you need to name a new node at the right end of the 3V source. <S> Create a supernode comprising that node and V1. <S> Start writing equations, and off you go. <A> Well, we have the following circuit: simulate this circuit – <S> Schematic created using CircuitLab <S> When we use and apply KCL , we can write the following set of equations: $$\begin{cases}\text{I}_2=\text{I}_1+\text{I}_3\\\\\text{I}_1=\text{I}_2+\text{I}_6\\\\\text{I}_5=\text{I}_4+\text{I}_3\\\\\text{I}_4=\text{I}_5+\text{I}_6\end{cases}\tag1$$ <S> When we use and apply KVL , we can write the following set of equations: $$\begin{cases}\text{I}_1=\frac{10-\text{V}_1}{3}\\\\\text{I}_2=\frac{\text{V}_1}{8}\\\\\text{I}_2=\frac{\text{V}_3-\text{V}_2}{4}\\\\\text{I}_4=\frac{\text{V}_3}{12}\\\\\text{I}_5=\frac{6-\text{V}_3}{14}\\\\\text{V}_2-\text{V}_1=3\end{cases}\tag2$$
| Again assuming that you are required to use nodal analysis: If you have a voltage source in your circuit that is not connected to the reference node (either directly or through other voltage sources) then you must create a supernode that encloses both of the nodes of the voltage source itself.
|
I am designing a 7 segment display but I am finding it a little bit hard to associate them I am new in this domain and I want to design a 7-segment display from 0 to 9 and then from A to F. I made the truth table composed of 4 inputs and 8 outputs each output for the corresponding pin of the segment, then designed the k-maps and lastly I wrote the Boolean equation of each output. The problem is that I don't know how to link each equation with each other. So I asked this question like 2 days ago and arrived to a conclusion which ended pretty much what I want to do but there is still one small problem which is I don't know what to put as inputs like in the image below. I tried to put inputs like the ones in the image, some of the logical circuits like "c" and "d" lighted up when the input is 1 and turned off when it is 0. So if someone could tell me what is my mistake knowing that each logical circuit from a to g are all correct. <Q> The main link for the logic for the individual segments is that they must share the same inputs. <S> Optionally, they can share computation of a given expression: for example, inverted values of all the inputs occur more than once, so you can share the output of the invertors ( /A , /B , /C , /D ). <S> Also, A./B occurs twice and could be done with a single AND. <S> If you're implementing in gates, you can pretty much directly convert to logic as shown in the circuit below. <S> (NB: following is for common cathode seven-segment display, common anode would be similar but with some logic reversal, as you generate /outa etc, not outa .) <S> I don't know what gates you consider acceptable, I'm using maximum of 3-input ANDs and ORs just because the schematic capture of stackexchange has those. <S> You might well want 7- and 6-input OR gates, depending on what you're implementing in. <S> If you actually build it, you can really see the value of programmable logic arrays, printed circuit boards, and MSI. <S> Or of course microcontrollers, where the whole thing is just something like portb = segmentmap[x & 0xf]; . simulate this circuit – <S> Schematic created using CircuitLab <S> You might be interested that the datasheets for 74LS47 seven-segment decoders give the following (though note they don't give hex output for 10-15): <S> From Texas Instruments <S> 7447 <S> Datasheet Second Half of Question Since you updated with your separate logic blocks, you need to join all the a, b, c, d inputs to each block together, so that each block calculates the its segment value for the same inputs. <S> I'm not sure what simulator package you're using, but you'll want something like the following, which should display a 9 if the logic is correct. <A> This is the truth table from the DM9368 data sheet, which should confirm the correct decoding. <A> Your inputs are as follows: <S> A B D C are <S> A3 A2 A1 A00000 will show 00001 will show 11001 will show 9 See the second half of the answer by @jonatanjo
| You may well find that some terms can be optimised because they are shared between segments: for example, I've shared /A , /B etc, but you can find shared terms after the AND gates.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.