# Tag Info

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Depending on what kind of mains plugs you have, you can or cannot guarantee which way the plug is inserted so you may end up with Live and Neutral swapped at the power supply. As long as you use three pole earthed mains plug and connect mains earth wire to the third terminal with ground symbol it should be fine, as it most likely earths the metal chassis ...

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I guess COM is a common for all three channels. COM is isolated from chassis ground (earth). You may connect the ground of your load to the COM. There is no separate COMs for each channel. The earth (chassis ground) can be connected to COM also, but I suggest you to not connect it, as it is probably already connected on your device (computer). Avoid making ...

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Converting 240Vac to 12Vdc involves a lot of subsystems. You need an EMI filter, rectifier, and then a step down switching converter (buck, or H-bridge, etc.). Electric blankets just produce heat and you can cost-effectively due that by just putting the incoming AC on a resistive load and avoid power conversion completely and greatly reducing the cost.

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I have a 12 V DC input that I am stepping down to 2 V with 610 ohm resistor. When power is removed the light is staying on for a period of time and slowly dims. I suspect that you mean when you switch off the mains power that the LED fades out. simulate this circuit – Schematic created using CircuitLab Figure 1. An old-fashioned power supply driving ...

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First consider an AC coupled PWM signal. Its peak-peak amplitude is constant at (let's say) 5V, regardless of the duty cycle. And its mean value is ... 0 ... regardless of the duty cycle. Now use a DC restorer to define the DC level of its negative peaks at 0V. The positive peaks are now 5V. All you need to do is low pass filter it, and the output value ...

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By using a negative clamp and a LPF of any order to give desired BW yet attenuate PWM ripple, the result is a DC +AC signal proportional toward the PWM duty cycle.

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If there is an industry standard for it, neither DigiKey nor Mouser know about it. You can buy a variety of adaptors to 5.5/2.1 and 5.5/2.5 connectors on Aliexpress. However, you lose whatever signal is carried over the third wire (e.g., in the case of Dell, what kind of power adaptor you have plugged in so the machine knows how much power to draw safely), ...

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It may not be enough, depending on what you are doing on the tablet. It may try to pull more current than just 1.5A. And the tablet may need the battery sensor connection, as its not just power and ground. Some devices are finicky about that (for example, a Mifi 7700 hotspot). You can certainly try it to see if it will run or not.

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Well if you are lucky and what your switch is doing is grounding a pull-up resistor, you can get by with connecting the high V side to ground with your cable. If it's completing the circuit I don't know if you can do much

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The Lo-PO battery is 4.2V when fully charged so simply add a silicon diode (1N4148) in series with it to have 4.3V which will be fine.

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2 silicon diodes in series will drop 1.2 - 1.4 volts or so depending on how much current your watch needs.

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I do not know the internal charge and battery management structure of the watch. In general: There will be short current surge which charger should be able to handle The 5 V can damage the parts. You have to have a LDO regulator with good amount of capacitors. Setting the LDO output of the typical voltage of battery

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Even if you buy enough copper to move the power you are trying to move, you have a bigger headache - safety. DC arcs are i credibly hard to extinguish compared to AC. A switch rated for 15A AC may well be rated for only 2A DC. Look into the physics and safety aspects before you burn your house down. ...and I doubt your insurance company will back you if ...

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I've thought about this for years. I ended up at why not just use Power over Ethernet? Whether or not you use the network. Cat 5e cable is pretty cheap.

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what would it take to do a DC wiring for an house to work along the traditional AC wiring? One way to learn about some of the practical implications of this would be to do some research on areas where this is already commonly done. The power systems that you describe are commonly used in boats and RVs. There is a fair amount of information available for ...

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FWIW, I played with a 12V distribution around a small flat (apartment). I had a big ATX power supply serving up 12V and setup a sort of ring of thick low voltage cable. I then "tapped" off that ring to provide drops to the places that needed it. Pros: You can do what you like - it's entirely unregulated so no "code" to follow, and it's touch-safe so it's ...

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In metals, such as copper, electrons are free to move. They are mobile. They stay attached to the wire, but they can move around within the wire. What causes DC current to flow is that a voltage potential is applied across the wire. Because of how the universe works, the voltage across the ends of the wire creates a uniform electric field (voltage gradient) ...

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I think your confusion is that it is not true that all charges are on the surface of the conductor. If the conductor has a net positive or net negative charge, the net charge will be near the surface, but there will still remain many charged particles in the interior, they are just balanced and cancel each other out.

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Loss at voltage conversion is just a function of how good your converter is, 5% to 20% losses could be expected at each conversion. Losses in distribution are the real killer. With 120/240 volt distribution, we don't really think about distribution losses until we are running many 10s of metres of cable down to an outbuilding. If the cable is thick enough ...

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If the "engine" is a typical automotive internal-combustion engine then you can look for an increase in the supply voltage as the alternator is charging the battery. Measure the battery voltage with the engine running, then with the engine stopped. Use a comparator to look for that change, and have the comparator signal your computational unit to shut down, ...

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A load needs to be tested under known conditions, minimizing as many variables as possible. When trying to determine maximum power requirements, those conditions need to simulate a worse-case situation. In your case, you would set up your experiment to operate with an input of known frequency and amplitude(the "Control"), and a possibly varying gain(the ...

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Using a dedicated ic for boost topologies for instance is way better, basically 100% of the times. Waaay more efficient. Especially if you are planning to use it powering delicate/sensitive circuits(avoiding spikes). But truly you didnt get the point. Finding a way bigger inductor is. Look at what that denominator means. That or switching frequency aspect. ...

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The problem with using the batteries is that they will always be charging. Draw it out and notice how current will enter the positive terminal of the battery. That's opposite to the way batteries are usually used and means the battery is charging, instead of supplying. Always charging means the batteries will eventually pop. Unfortunately, to convert ...

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I would use a finite element analysis. Break the bar into small cubes of material and then model each cube as six resistors, all connected together at the center and one connected to each face. Assume that your two points are at the exact centers of two cubes. Use SPICE, or construct the resulting resistor network and calculate the current through each ...

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The term you are looking for is "bench power supply". Note that the current setting is a limit. Some reading on the subject: Does a hobbyist need a bench power supply? and What is a Bench Power Supply?.

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It is theoretically impossible for a power supply to regulate current and voltage at the same time. The most that is possible if for either current or voltage to be adjustable with a limit set of the other. If the unregulated parameter reaches the limit then the other parameter value is reduced to whatever value is necessary to prevent the limited value to ...

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Your solenoid might not be the best tool for the job (a 5V solenoid may work as-is* and a 12V solenoid would draw less current from your AA batteries), but if you absolutely want to hack something together to make the pair work, here is what you can do: If you don't easily have access to other solenoids but you don't mind hacking this one: the force of a ...

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These are similar to all the ones found on VGA cables. The DC power also conducts broadband spectrum on unbalanced impedance wires from the switched current transients. The purpose of this Ferrite, Folding clamp, split clamshell, ungapped component is to act as a BALUN or to BALance UNbalanced lines at RF frequencies. The other functional name is a Common ...

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Firstly, it’s not a ferrite bead; it is much bigger than what is generally regarded as a ferrite bead and, as such, it will have a noise reducing effect in the low hundreds of kHz to probably over 10 MHz. Generally, ferrite beads are only useful above 10 MHz. I suspect that the ferrite core is placed on the wire to ensure that the wall wart complies with ...

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As you stated, the ferrite is installed to suppress high frequency noise. As you have also pointed out, the location of it is close to the plug end that goes into the device it is powering. This may indicate that its purpose is to suppress noise generated by the end device from getting onto the cable. Cables can act as antennas, and any high frequency ...

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The permeability of the ferrite bead increases the inductance of the wire going through it. This increased inductance increases the impedance of the wire at higher frequencies. The increased impedance reduces the amount of RF common mode power that can flow in either direction: reducing harmonics from a power supply's DC-DC converter switching frequency ...

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It removes noise, specifically high frequencies. See first paragraph from Ferrite Beads Demystified: An effective method for filtering high frequency power supply noise and cleanly sharing similar voltage supply rails (that is, analog and digital rails for mixed-signal ICs) while preserving high frequency isolation between the shared rails is the ...

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10A is 2.5% of 400A, which is more than 'near negligible'. However a shunt with reasonable loss at 400A will have very low output voltage. The answer is to use a low-drift op amp to amplify the voltage enough to drive a switching device. Let's say you can accept a shunt loss of 4 watts. At 400 A that corresponds to 10 mV, with a shunt resistance of 0.01 / ...

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It's a moot point anyway because the voltage going in on the power connector is regulated in all instances unless the regulator itself can't handle the added watt or two of dissipation or there are capacitors directly across the input rated for 25 volts, either scenario of which is highly unlikely. https://batteryuniversity.com/learn/article/...

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First, instead of regular car batteries to get that ~24VDC, use deep discharge batteries. Second, 12V lead acid batteries are fully charged at between 14.4 and 14.8 VDC gassing point. Third, yes, two 12VDC batteries in series can properly supply this instance with plenty of power to get you through the day. You only have to supply the current requirements. ...

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I had the same idea as Spehro Pefhany use a capacitor to store energy and a thyristor to dump it through the relay simulate this circuit – Schematic created using CircuitLab I came up with this, but I'm a bit worried that it might trigger from a supply ripple and short-circuit the supply. but this isn't going to work with the Omron MK12K because ...

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The circuit below does what you want. It Provides an ~= 100 ms latching pulse when ignition is turned on And similar delatching pulse when ignition is turned off. At all other times the relay coils draw no current. This is a "cut down" version of my circuit from this answer to "Power Latching Relay off/on in response to 12V ignition and 1 second ...

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You need a capacitor large enough to hold enough energy to reset the relay when the +12 disappears. Minimum pulse width is stated as 30ms. Assuming a 1V drop would be okay and allowing for say 45ms time, you'd need a capacitor C = $\frac{t \cdot I}{\Delta V}$ or 0.092A * 0.045s/1V = 4,000uF. So maybe 4700uF/16V. Then you need a control circuit to detect ...

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I had the same issue right now, trying to use a Ryobu BPL3640D (36V x 4Ah) for a hackish project. The output of the battery appears to be extremely high impedance, it floats around the actual battery voltage of 40V (pretty full) but already starts dropping when measured with a high-impedance meter. My circuit wouldn't draw any current when powered with less ...

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How important are these last five digits and how risky is it to install a part whose last digits are different? The last 5 digits are important because they tell you the number of turns and diameter of wire on the armature, which determines the motor's power and torque. Too few turns will draw too much current, which will reduce the motor's lifespan and ...

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Some parts of your question are a bit fuzzy. Improve that next time.

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Verifying your claim of $10\:\text{mA}$ for the lower-left quadrant Here's my marked-up version of your schematic: Before I delve too much into a general solution, let's verify your assertion about $I_3=10\:\text{mA}$. It's not hard to do and you are right about that. By inspection, you know that $V_x=R_5\cdot\left(I_4-I_2\right)$. But also by ...

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When the SmartPass conducts to allow current from the alternator to flow directly into the house battery bank, it shorts out the Smart Charger disabling it. This is similar to placing a copper wire right between the input and output of the Smart Charger. This forces the Smart charger to basically turn off. Note this is different from a short-circuit where ...

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It appears that your signal frequency is 20 Hz. This is too low for the sizes of the input and output capacitors. The input capacitor and the two base bias resistors form a high-pass filter of approximately 480 Hz. The transistor's input impedance changes the actual corner frequency, but it still is way above 20 hZ. A similar situation is happening at ...

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If you apply 36V instead of 24V it will draw more current and you will probably destroy the motor.

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