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I would like to drop a 240V DC voltage to a 5V DC voltage. The 5V stage needs less than 5mA in current.

The solutions I looked at are:

  1. Resistor Divider (Not really interesting because of the power to dissipate)
  2. Voltage Regulator (240V to 5V is hard to find, if it actually exists)
  3. Buck Converter (I don't know well enough those parts to use them in a design)
  4. DC-DC Switching (a bit expensive)

Ideally, I would like something not too expensive. It could be something to build by myself.

Personally, I thought about mixing solutions 1 and 2:

  • dividing the voltage from 240V to 30V with a resistor divider
  • then regulating the 30V to 5V with a voltage regulator

What are you guys thinking about it? Do you have others solutions? Do you know parts that could help me?

Is the solution really different if the 5V stage draws 50mA instead of 5mA?

DC voltage as high as 240V is dangerous, I use it carefully.

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    \$\begingroup\$ wait wait, you want to change 240V all the way down to 5V for a device that will only pull 5mA? You might want to consider using a different input voltage b/c you are going to lose so much power compared to how much you need. \$\endgroup\$ – Funkyguy Jul 10 '14 at 20:46
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    \$\begingroup\$ The first thing a SMPS does is rectify the mains to DC through a bridge rectifier. Putting DC into a bridge rectifier gives you DC out too. What would happen if you just used an off-the-shelf 5V SMPS? Maybe bypass any line filtering first - and maybe the rectifier too? \$\endgroup\$ – Majenko Jul 10 '14 at 20:59
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    \$\begingroup\$ Do you have an old PC power supply lying around? \$\endgroup\$ – Matt Jul 10 '14 at 21:27
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    \$\begingroup\$ If you want cheap, pop down your local shop and buy a USB charger. They offer a very convenient 240 to 5V power supply ready packaged up. \$\endgroup\$ – Simon B Feb 21 '16 at 21:56
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    \$\begingroup\$ Check out DigiKey. They actually have regulators that can do what you need. digikey.com/product-search/en/integrated-circuits-ics/… \$\endgroup\$ – Swarles Barkely Feb 22 '16 at 2:44
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Consider a cascade of converters. Perhaps a zener diode to drop the 240vdc down to the 60-80v range, you can use an LT3010 LDO to drop it down to 5v. The LDO is really good for up to 50mA and requires few components.

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    \$\begingroup\$ Not a good idea: Your zener diode would need to turn (160V*50mA=)8 Watt into heat. The LT3010 would turn ( 75V*50mA) 3.75 Watt into heat. Even with a well designed PCB with large power planes to help with power dissipation, the temperature of the LT3010 would rise by 150 °C and the LT3010 would go into thermal shutdown. \$\endgroup\$ – Jan Lucas Feb 21 '16 at 22:47
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I would actually use a AC/DC wall-wart that I know uses a SMPS inside (most devices nowadays do). It will try to rectify the input voltage (which is already rectified) and after that it's the normal DC-DC conversion it usually does.

A surprising amount of electrical products for AC mains actually work on DC, as this guy demonstrates.

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A modified version of 1 and 2: use a Zener diode in place of one of the resistors to limit the voltage for the regulator and reduce the power consumption under load.

A better solution would be to use a resistor and a low-power Zener diode co create a 5.7 V reference and an emitter follower with a high-voltage BJT, which are not hard to find.

The 50 mA solutions would be the same, except you'd be wasting (and dissipating) about 12 Watts.

Another solution would be to find or build a switch-mode power supply, of course.

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    \$\begingroup\$ The emitter follower is the simplest solution imo. If the OP can't easily obtain high voltage devices, they could cascode ones of lower voltage. \$\endgroup\$ – Bitrex Jul 11 '14 at 6:44
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You coule use an 0A2 and an 0G3 glow tube in series wih an appropriate current limiting resistor. They're readily available from various suppliers online, and with regulated voltages of 150 and 85 volts respectively, would give you an approximately 5 volt output immediately.

Edit: You may need to ad striking resistors to get them to tuen on:http://ultra-fi.blogspot.com/2011/09/get-string-of-vr-tubes-to-go-pop.html

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You can make this very simple with a simple series transistor and a zener diode.

transistor series regulator

The major drawback is that it wastes energy. Dropping 235 volts at 5 mA is around 1 Watts.
What you can do to reduce this waste is drop to a voltage until you can get cheap buck converters.

Works great if you can do something with the heat.

Also described here.

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You should able to find a "wall wart" for under 10 USD that will do this. Consider that 9/10 of the world uses 230VAC and also wants to charge USB devices that take 5V DC. A basic wall transformer that converts 240V AC to 5V DC is more common than ( you fill in the blank ).

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  • \$\begingroup\$ oops. I assumed you meant 240AC instead of DC. \$\endgroup\$ – Craig K Feb 22 '16 at 1:16
  • \$\begingroup\$ What happens if you plug a 240AC charger in a 240DC outlet? \$\endgroup\$ – workoverflow Nov 6 '17 at 13:34
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    \$\begingroup\$ @workoverflow Actually, that would work pretty nicely. Virtually all wall-warts rectify AC to DC as the first step. Arguably, $10 is not that cheap though. \$\endgroup\$ – Dmitry Grigoryev Nov 6 '17 at 14:35
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I'd use a cheep USB wall-wort. They are designed for AC but the vast majority of these just rectify the incoming voltage and have universal input voltage to cover mains from anywhere in the world so they will cope with 240VDC.

Actually anywhere between 127V to 360V DC (90V to 254V AC).

The actual converter is usually a flyback type switch-mode power supply.

This also provides isolation if required though may be less important as your input is DC. Note 240V DC can still kill you so isolation is not a bad thing.

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There is an efficient way of doing this cheaply for AC, using:

  • a high-voltage non-polar capacitor
  • a 400V 1A bridge rectifier
  • a 5V 1W Zener
  • a 220μF 16V smoothing capacitor
  • a 1K 5W surge limiting resistor (to stop it blowing up)

Since you need only 5 mA at your output, calculate the needed capacitive reactance:

240V / 5mA = 48000 Ω (Ohms)

The capacitor needed for this reactance at say 50 Hz:

48000 Ω = 1 / (2 * pi * 50 Hz * C)

C = 0.07 μF

For proper regulation, let's double the capacitance, giving about 0.15 μF.

Connect your bridge rectifier through this capacitor to the 240V mains.

Connect your 5V Ζener in reverse bias mode across the DC output of the rectifier.

Smooth out the 5V across the Zener using the 220μF capacitor.

You may also use a 0.5 A fuse at the input side for protection, in case the high voltage capacitor gives way.

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    \$\begingroup\$ And no voltage isolation. \$\endgroup\$ – Harry Svensson Nov 6 '17 at 14:38
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    \$\begingroup\$ Question is about DC though, where this solution is not applicable. \$\endgroup\$ – pipe Nov 6 '17 at 15:52

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