Please may I ask what the arrangement is when you see a single LED powered straight off of a mains power supply without any transformers or switch mode circuitry? In other words, totally uninsulated or regulated.
I have an old computer with an external hard disk which needs about 30 seconds to spin-up before the computer. With a small timing circuit, 555, or using a PIC even (have dozens) after a set time a relay would be set, thus powering on the computer. A 7805 could be introduced to take into account the voltage swing. Half wave rectification could result in 120v too.
This doesn't need to be insulated from the outside world, safety is not a concern, just that roughly 5v should be available for the small circuit and the 3amp relay.
Any ideas?? Just interested to hear of how this is done. Or would it be easier to just buy a small tordial TX and make the box a bit bigger?
I think you'll find that that is a neon lamp and not an LED. LED's connected directly across the mains will give light in the form of fire.
Why not buy a 5V supply to run the electonics off rather than mess with mains yourself?
SAFETY IS THE PRIMARY CONCERN! Mains is not something to be fooled with or it WILL kill you. Insulate everything or when you (or somebody else) is not at full concentration you will touch something and die. There should be fuses and other protection in the circuit as well. This is why you should seriously think about buying a supply and letting someone else handle the high voltage design - and the legal concerns that go with it.
If it were me then I would use 5V from the Hard drives supply to trigger a PIC (but a 555 or an RC circuit would be just as good) to control the PC's power switch. If the PC has an ATX supply, then great because you can stay low voltage and use a simple relay to replace the PC power switch (remembering that only a pulse is required to simulate a buton press). If the PC supply is AT then the power switch is mains and you could use a mains relay switched by 5V but you need to be sure that the terminals are properly insulated on the mains side.
Seriously, your tone doesn't sound like you are giving mains electricity the respect it needs. I know enough people who have died from electric shocks and each time it was because they thought safety wasn't a concern. Be careful.
Half wave rectification would NOT result in 120V and a 7805 can't drop anywhere near that much voltage. If you don't know this then it would be a very bad idea for you to make any live circuit. I recommend that you use a wall wart for your 5V.
Having said that, a mains rated capacitor in series can be used as a sort of voltage dropper followed by a rectifier to get low voltage. I actually found a circuit like this in a Russian battery charger from the 60s. Yikes! You get small size at the expense of safety and I don't think that its worthwhile.
Typically a series capacitor is used with a capacitive reactance sufficiently large at the mains frequency to limit the current to a desired value. Due to harmonics and other high frequency noise usually present on the mains, the capacitor reactance at these high frequencies are quite low, thus quite large peak currents could flow, thus, it is a good idea to also include a series resistor.
These kind of circuits only makes sense with very low current demand, since at higher currents, the capacitors become quite large.
In order to minimise the load current and hence the required capacitor size, it would be a good idea to use a relay with a large coil voltage (and hence low coil current) driven by a high voltage open collector output or a separate high voltage transistor.
The relay will draw a substantial amount of current and it is challenging to make a circuit that can supply this much current from the mains without wasting (and having to get rid of, as heat) more than a watt (and that's if you use a 48V relay too, low voltage relays will be worse.) A small transformer would be a nicer solution. By the way, a "solid-state relay" will draw less current than the coil of a mechanical relay though it may not like charging the capacitors of a switched-mode power supply.
If you must have something cheap, would it suffice to get rid of the relay (which I guess was to apply power to the computer) and then open up the computer and insert a circuit with a PIC or whatever to hold the computer in reset for 30 seconds after the computer's power comes on? That way there is no new mains wiring, just a bit of logic. If you can't open the computer for whatever reason, is there a connector on the computer which provides +5V from the computer power supply, and also a reset signal which can be pulled low to reset the computer? I am guessing that this is a
there's 5V on the disk drive power connector plug (inside the external case) there's 12V there too which is probably better suited to driving a relay with mains contacts. 555s will operate off 12V, and can drive many relays directly.
Piece of cake if you know the current consumption of your circuit. I run LED's and small circuits directly from the 120 VAC 60 HZ mains all the time.
Use a capacitor to drop the voltage. - no heat and no energy wasted.
You take your circuit - measure the current and voltage it needs. Subtract the voltage from the mains supply voltage and calculate a "dropping capacitor" for the voltage you want to drop with the current you need.. Use the formula for capacitive reactance.
Put the cap and a 100 ohm 1/2 W resistor in series with one side of the mains - the resistor is to limit inrush current as the cap charges and works as a fuse if the cap shorts. That goes to your four diode full wave bridge rectifier.
Here I use ~.47 ufd/250 Volt caps for lighting a LED or two or three in series from the 120/60 mains
If you are just interested in using an LED on the mains supply - you can just bypass the LED with a small diode to keep the reverse voltage from getting over .6 volts instead of using a rectifier or use back to back LED's and light two with no rectifier.
The cap should be a non-polar type with an AC rating for the mains voltage or 3X the DC rating.
Don't use the circuit with no load - the output voltage will be high.
When powering circuits you have to take into consideration your load - if you're driving a relay, for instance, the current will be much higher when the relay is energized. So in addition to solving the capacitive reactance for the relay current + circuit, you also have to give it some means to prevent voltage overshoot when the relay is not energized - a simple shunt regulator with a single 1 watt zener may be all you need -
I posted the data from the Siemens 1990 Optoelectronics data book "Operating LEDs on AC Power, Appnote 6 " on alt.binaries.schematics.electronic a couple of months ago
Another way to steal a little bit of power that may come in handy for your disc spin up timing - a current transformer. Just add a few turns of wire to a transformer with an open core (like a toroid or side by side bobbin transformer) and put the load on the low voltage windings - A trick we frequently use to turn on solid state relays when an X-ray machine or other dangerous piece of equipment is running
- to sound horns or work warning lights. A little trial and error involved - but that ties the output to a load - whenever the load is drawing current the output is there.
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Absolutely. BTW you can connect an LED directly to the mains under a couple of conditions:
1) You use sufficient current limiting.
2) You make sure to wire your LED with another diode (which can be another LED) in antiparallel configuration. This limits the reverse voltage for each of the diodes.
I rigged up a cheap 120VAC motion sensor by tying the LED of an optoisolator directly to the 120V light fixture of motion sensing lights. I used a 10K
2W resistor and a reverse LED as a local indicator. The optoisolator converted the dangerous 120VAC into a safe optoisolated 12VDC.
BTW the current limiting applies to neon bulbs also.
If memory serves, I have seen this done once with an LED using a pair of resistors as a voltage divider - it might have even just been the one resistor. Personally I think it was cheap workmanship though. This will work with an LED, although it will flicker as it's only illuminated for less than half the time, I think it can work with only one resistor but only when you know the resistance of the device it is powering. For hard disks this is a non option - as thier resistance varies wildly as it spins up, settles, and does all its internals stuff, and the resistors would have to be big mothers as hard disks draw a hefty current. No no no no no
forgive me if I seem a little dense here... but if it's an external hard disk, does it not have its own regulated power supply? could this not be utilised in some fashion? but what sort of interface is it? ide, scsi, parallel? that might give us some clues as to what would be a good solution.
safety is always a concern.
I need a bit more information on your set up really, but I can envisage two possible scenarios:
scenario 1) you have a true external hard disk, in an external hard disk caddy with all the wiring and gubbins and some sort of parallel interface whotnot. The drive is powered independantly of the computer. If this is the case I would utilise the regulated 5v line powering the hard disk
however, I think scenario 2 is more likely:
scenario 2) You have a hard disk that is external of the computer - connected via an IDE lead and the computers power supply. You need the hard disk to power up before the computer - although I dont know why at this stage, but i'll pretend for now there is good reason for it. If this is the case, you NEED a good regulated power supply for the drive, and not just the 5v line if its a 3.5" drive - 12v will be required as well or it wont spin up. 2.5" disks generally only require the 5v line. Your best bet is a regulated 5v/12v power supply - which should be fairly simple to construct. then simply run the timer circuit off that supply.
You could use a 12K (12000 ohm) limiting resistor in series the the LED to limit the average current of the LED to around 20ma. The 12K resistor will need to be at least a 5 watt unit and will get pretty warm during operation. You will also need a bypass diode hooked across the LED with the anode of this diode to cathode of the LED and vise-versa to limit the reverse voltage of the LED.
The avoid the loss of power and subsequent heat generated by the resistor you can use a 0.22 uf capacitor in place of the resistor. Make sure the voltage rating of the capacitor is at least around twice the peak value of the AC or around 600 volts. It must be a non-polar capacitor (most electrolytics are polar).
C is calculate from the following formula: C = 1/2*Pi*f*Xc this formula is an algebraic variation of the capacitive reactance formula
We use to use this technique to limit current for indicator lights in 440 volt power distribution switchboards.
NOTE: You can still get shocked by this arrangement if either or both of the diodes open. 20ma of current across your heart is enough to kill you. Don't even try this if your not familiar with the proper electrical safety measures.
The safest technique, by far, is to use a small, step-down transformer because of the increased level of isolation it offers.
that fuse from neutral to earth may look harmless but it'll mess with any ELCB/RCD etc that the circuit is connected to, much better to leave the connection to earth off, at and treat the setup as what it is: a "live chassis", fusing the Live side probably makes more sense too.