# Tag Info

1

I asked some colleagues too and got a couple of interesting answers. Cut a USB power cable and solder on wires that can connect onto header pins in the breadboard. (Not off-the-shelf, but doesn't sound tricky either.) Use a Gadgeteer power module. I have several .Net Gadgeteer kits so this one's easy for me. Take the Gadgeteer power module (for example ...

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I have found these breadboard power supplies ($2.12 or so on ebay.com, shipped) very useful for the purpose described: Jumper-selected 3.3 and 5 Volt simultaneous outputs from wall-wart, USB or 9 Volt battery. They have a convenient 2.1 mm barrel jack port on board, besides a power button and an indicator LED. You can choose to have 3.3 Volts on one of ... 0 Really any basic power supply would do. For example, I use this: simulate this circuit – Schematic created using CircuitLab Transformer, rectifier, large capacitor and a variable voltage regulator. You can also split R1 in two (5K and 500R) for extra fine tuning. Gauge the potmeter so you can see the output voltage without measuring. 4 Wall wart power supples - cheap, plentiful, just be careful of their output voltage. Many are odd voltages or unregulated (the voltage changes with load) although many are quite usefully designed. Alternatively, old PC power supplies - Dangerous prototypes do an ATX breakout board, either use one of those or read about how it works & use the information ... 2 The starting point I would take with this circuit is to break the diagram down into functional blocks (or sub systems). Reading the circuit from INPUT -->--OUTPUT (Left to Right as drawn) Block 1. On the left hand side we have 4 DIODES arranged in a conventional (full wave) bridge rectifier configuration converting the AC input into (bumpy) DC. C1 is ... 0 What you have is a full bridge converter in power supply parlance. T1 and T4 are driven ON together, followed by T2 and T3, exciting the primary winding with opposing polarity. The secondary is a simple center-tap rectifier which along with the LC filter gives you DC output. In continuous mode, the output is a function of the input voltage, the transformer ... 3 I agree with others that switchers are a better choice in terms of efficiency, but they can be somewhat complicated to deal with if you're inexperienced, and there can be lots of weird effects that aren't immediately obvious (precharge sinking, beat frequencies, etc.) that can make life difficult. Assuming you've figured out your power dissipation and know ... 4 You haven't mentioned current output requirements but, the biggest problem you'll have is power dissipation and I'd urge you to consider using switching regulators for 12V, 9V, 5V and 3V3 and if necessary use them to generate 13V, 10V, 6V etc. and have Low-Drop-Out (LDO) linear regulators to take the drop to the final voltage you need. The benefits are a ... 2 In my opinion, for the last thirty years, nearly all magnetic induction energy transfer systems place a capacitor to put the secondary coil in resonance. This achieves better power transfer efficiency and everybody knows this. We have been doing this in the medical device industry for [at least] the last two decades. This is just obvious. Witricity and ... -1 With respect to @Kurt E Clothier's answer above, there are some clarifications I feel that need to be made. You can't "overcurrent" a device. It just won't use the excess. If your AC adapter has higher current rating it won't destroy your device. You can certainly exceed the device voltage which is why his AC adapter should be 5V rated. USB is 500mA max ... 0 USB is 5v. if your ac adapter is 5v you can do it but be careful with the wiring 2 Used a double-pole toggle switch (DPDT or DPST) - one pole to switch the USB power, and the second to switch the 9 volts. 1 Hmm, let's start w/ power first: 3072 * 20mA * 2.1 = 130 W! Your power supply can only give you about 5W. You're very much short there. If you can, greatly reduce the size of your matrix. One solution would be to use a desktop ATX power supply. Those have 5V lines through their hard drive connector and can give you lots of power. If you take the 5V you'll ... 2 The LT3083 is a good device and your limitations on drop-out can be improved by considering the following: - As can be seen, the drop-out at 3A load is just a bit over 300mV but at 1.5A it is about 170mV. Given that devices can be paralleled (diagram on right), maybe you should reconsider this device. Also note the larger words at the bottom - to truly ... 1 While many LV circuits are isolated as Anindo says (and should then be "double insulated" and marked as such) there certainly are cases where your circuitry should connect to wall socket ground. If the LV circuitry connects to exposed metalwork, for example, then it ought to be connected to the wall socket ground - not via a high resistance but by a "fault ... 4 To establish the ELV you will need isolation. This will probably be a transformer (Linear or Switchmode does not matter) Transformers can be protected in a number of ways from faults... Self Limiting (Internal resistance limits the available power) The use of Triple Insulated wire (TIW) in the transformer construction. Fuses in the primary or secondary ... 2 It's usually easier to think of the input side safety "ground" as PE, or protective earth - it's more representative of what's actually going on. This is a typical TN earthing system. There is a hard connection between the neutral and the protective earth at the generator location. In real life: In a non-isolated power supply, anything on the primary ... 5 The voltage converter circuit shown, like most (safe) DC adapters in domestic or retail use, is an isolated power supply - in other words, the voltages are isolated from any reference to the power supply lines, including the building earth line. From the circuit's point of view, the only reference voltage is the 0 Volt on the output side. Voltages are ... 2 The ground connection of wall outlets is a safety ground, not a reference ground. I believe you can connect safety ground to the metal enclosure or omit it for double insulated appliances. 0 Your power supply ground is not connected to the scope ground. You can fix that, or use an isolation transformer 220v/220V to connect your oscilloscope to the main supply. 1 In my experience the 24 volts is for AC not DC. Check the markings carefully. I am thinking of using a 24 vac valve ( solenoid ) on dc to make it easier to power with a battery. I expect a greatly reduced voltage. I do worry that I may polarize the magnetic material, but will have to try the experiment. 2 Have you tried triggering the Solenoid with 12v only? I think you will find they are quite liberal with their input voltage range, and with a standard ATX supply, the +12v rail can handle a couple amps no sweat. As to your direct question, most if not all ATX or similar power supplies have their voltage and current ratings listed on a sticker on them. Since ... 2 Here is a random 350W power supply (http://intrl.startech.com/Computer-Parts/PSUs/Dell/350-Watt-ATX12V-201-Replacement-Power-Supply-for-Dell-PC~ATXPW350DELL). It can provide 0.8A on the -12V rail. The answer for your specific power supply will depend on the power rating - try to find the datasheet or look for stickers on the back. My guess is it will ... 3 The maximum load for -12V on a typical ATX power supply (regardless of output rail) is only 300mA. That won't help you much. You'd be better off boosting the the +12V rail to 24VDC with a switching converter and using it to power the solenoids. 3 The ripple when switching on the LCD backlight leads to one suspicion: The backlight is an electro luminescent (EL) panel, driven by a high frequency inverter. A typical EL inverter would consist of a switching circuit, producing a square wave load on the supply. The common way of dealing with this is to decouple that load from the supply rail, at the ... 1 More microfarads in the capacitors bwtween the rectifiers and your voltage regulator. That would be your first strategy. It is also possible that the voltage you have on these capacitors is too low and the voltage regulator is falling out of regulation when the current draw due to the backlight is introduced as part of the load. Addressing the voltage level ... 5 I don't think you need any special kind of UPS, unless you have particularly power sensitive equipment. Nearly any modern switching power supply will handle the transition from mains power to battery of a typical line interactive UPS. This type of UPS will protect you from both power surges and power sags, switching to battery when the mains voltage is out ... 0 There's an article by Morgan Jones in Linear Audio 5, where he shows that 1nF in series with 1k across the secondary is an amazingly generic solution to transformer ringing. 3 Problem: You power the LM317 with 12V and it generates an output of 5V @160 mA. So 12V - 5V = 7V @ 160 mA that the lm317 needs to dissipate (heat). P(watt) = I * U = 0.160 A * 7V = 1.12 W of heat The datasheet of the LM317 says: The Thermal resistance, junction to case is 5 °C/W The Thermal resistance, case to ambient is 80 °C/W (no ... 5 A linear regulator dissipates heat proportional to the amount of voltage it must drop, and the amount of current flowing through it. Input supply = 12 Volts Option A: Output Vout = 3.3 Volts Vfred = 2.1 Volts Current = (Vout - Vf) / R = 7.5 mA per LED = 60 mA total Power dissipated by regulator, P = (Vinput - Vout) x I = 522 mW Over half a watt is ... 2 The quoted maximum current the LM317 can supply requires that you provide a way for the chip to get rid of the heat it must produce to 'boil away' the voltage difference. You don't say what the output voltage of the LM317 is, I assume 5V. That means that each part LED that you turn on caused the LM317 to dissipate 0.02 * (12 - 5 ) = 0.14 Watt. A TO220 ... 2 Yes, the LM317 can handle up to 1.5A, but when dropping 12V down to 5V, it needs to get rid of 10.5 watts worth of heat in the process! There's no way it can do this without a substantial heatsink attached. Even at the 160 mA your red LEDs are drawing, the regulator is dissipating well over 1 watt. Its temperature will rise until the amount of heat being ... 5 You need a voltage regulator that regulates to 3.3V and as the input voltage drops to 3.3V or below, the output remains close to the input voltage despite it not being able to regulate any more - in other words it acts like a <0.25ohm resistor when unable to regulate. The LP3964 has a drop-out voltage of 24mV at 80mA and its output will follow the input ... 2 What is your load current? This influences the best answer. A minority of regulators shut down gracefully and have little Vin to Vout voltage drop. However, if it doesn't say so in the spec sheet it is quite likely not to do it in practice. Yould could achieve close to what you want by using a bypass MOSFET across the regulator and switching it on with a ... -1 You want a Low-dropout regulator. They're a commonly available IC. 3 One way to "bullet proof" your power supply is to add transient voltage suppresors (TVS). See this application note for instance: http://www.vishay.com/docs/88490/tvs.pdf‎ Also check the fuse configuration of the ATmega32. Some of them may help to make your system more robust. In particular, check the Brown-out detection module and the start-up time (SUT0 ... 0 Or you could buy a properly implemented ready-made boost converter with > 90% efficiency for under$5. If you make many of these, vendors like Recom or Murata may actually have pre-built modules. If you only need one or a few, Pololu has a nice 5V 200 mA boost converter with 90% efficiency for \$4.95. Link, while it lasts: ...

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Over a very limited range - yes. You will need to have a return path that is supported by electric fields. The best way to look at this would be like a AC coupled circuit - coupled through a capacitor of which the capacitor is formed by some plate that the circuit is coupled to and another plate that is providing a return path. We know that electric ...

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A device like the Power Integrations LinkSwitch-TN allows you to implement a simple buck converter that converts rectified mains voltage to a reasonably-well-regulated non-isolated low voltage with reasonably good efficiency. If you need low ripple after the fact, you can put a linear regulator on the stepped-down output of the buck and minimize your ...

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Its not so much that they "can't" as much as it is they are just not designed to. Computers use switching regulators that are designed to operate around a certain current and voltage but do not perform well if the current and voltage deviates too much from this point.

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From the article: As it appears, Haswell's C6/C7 states require a minimum load of 0.05A on the 12V2 rail, and many desktop power supply units (PSUs) just cannot provide that low current, reports The Tech Report web-site. Meanwhile, numerous older PSUs, which comply with ATX12V v2.3 design guidelines only called for a minimum load of 0.5A on the CPU power ...

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It is easier in some ways to design a high-efficiency switching regulator if you can assume that it has both a minimum load as well as a maximum load, reducing the "dynamic range" it must handle. Many PC power supplies are designed this way, both the main supply for the box, as well as on-board regulators for the CPU and memory. The new chips violate the ...

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Sounds like the back-EMF from the solenoid is disrupting the micro. There are a few potential issues / solutions here; 1st, read up on snubber diodes/protection used with relays/solenoids, it's been covered many times here on all the hundreds of variations on "how do I switch a relay from my *duino?" questions. 2nd, look at the smoothing and grounding of ...

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I read somewhere he stopped it because of fears of the physical effects the system might have on us. In the end, I think if he said it would work it will work.... gotta go with the guy who invented electricity as we use it to this day....and the radio...and x-ray...too bad he isn't still around, the advancements he would make today!

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You should definitely not be connecting an AC line to terminals labeled "VDC" those are the outputs, not the inputs. Beyond that, there's not enough information to really answer your question. Anything involving the AC mains is not territory in which you want to be experimenting without a very good idea of what is proper.

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EDIT - because of the reshaping of the question. my answer at the bottom isn't that relevant. To the edited question: - "Can you send data usefully over one wire, literally one wire?" My answer is "Not with any level of success". Reasoning - if you are sending signals you need a forward and a return path for the current. If the return path is tenuously ...

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If you want torque you don't need to worry too much about drive voltage : it's the drive current you need to worry about. It's rated at 18 amps; make sure your drive circuitry can handle at least that much current without strain. Basically, voltage controls speed (at light loads), current controls torque. If the motor is heavily loaded, (close to stall) ...

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A bunch of issues: Are you sure you have a brushed DC motor? This kind has only two connections. You simply apply voltage and it spins, which is what you are assuming in the rest of your question. All you said is that the motor has permanent magnets and is "DC", which could still leave other possibilities, like a brushless DC motor. Those are a lot more ...

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Your motor will need more than 30VDC to work (I guess). You can control it with PWM at 95V. PWM is actually a rectangular signal, with variable duty cycle.

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Basis for my unfounded assumptions: I think the magneto is flooding the engine assembly with noise. It isn't dissipative, and there is no connection to a larger ground. Given the insulating nature of design, not even earthed when landed. It's just one side of a circuit with big spikes on it. So: Don't earth your battery powered modules to chassis. You know ...

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