1
\$\begingroup\$

I'm trying to make a fairly small AC(220V) to DC convertor to power microcontrollers and some actuators. It needs to output a current of 1 amp to the load. I tried looking at using transformers, but all the ones I could find that are rated for 1 Amp output current are too big for my design. So I changed my design to be a capacitor power supply. I'm thinking of using the circuit provided in this website: http://www.circuitsgallery.com/2012/07/transformer-less-ac-to-dc-capacitor-power-supply-circuit2.html

But this one only supports around 150mA, It seems that if I change the X rated capacitor to 11uF from 2.2, I can get an output current of 1 Amp. Does this seem feasible. I am fairly new to electrical design, any help would be very appreciated!

\$\endgroup\$
3
  • 6
    \$\begingroup\$ Welcome to the site. Since you are new to design, I'll be gentle. Don't do it. Using a capacitor instead of a transformer is a shock hazard, and you can kill someone. I mean, really and truly. Don't do it. \$\endgroup\$ Commented May 29, 2017 at 3:01
  • \$\begingroup\$ 1A will make a capacitive dropper supply too bulky. Use a switching topology, and try to buy an integrated solution. Here's a simple app note on transformerless supplies: ww1.microchip.com/downloads/en/AppNotes/00954A.pdf Run through the analysis and calculate what you would need for 1A. It's a false economy here. \$\endgroup\$
    – user133493
    Commented May 29, 2017 at 3:03
  • \$\begingroup\$ The problem with passive series c offline regulator is the ratio of Vin/Vout amplifies the power loss in shared current and hiV plastic caps are low k density compared to electrolytically so beyond 1W get impractical compared to low cost SMPS like those used inside LED bulbs. \$\endgroup\$
    – D.A.S.
    Commented May 29, 2017 at 3:40

5 Answers 5

3
\$\begingroup\$

A capacitor input power supply is only appropriate for specific, low current, non-isolated applications, like tickling a SMPS into life, or driving an isolated low power LED bulb, where safety is not an issue, and a constant output current is OK.

As you want to drive microcontrollers and actuators, safety and isolation is paramount, and this is the wrong design to use. In addition, 1A is quite high, and the load will vary, so this type of supply is triply inappropriate.

If an iron transformer based supply is too large for your application, see if you can find a complete SMPS solution, like a USB charger. If they are too big, then you need to rethink your project dimensions.

When you do your own designs, it's often quite easy to get the behaviour you intend, say 1A output. It's more difficult to avoid the behaviour you don't intend, like electrocuting yourself. That's the reason that people buy isolated power supplies rather than build their own, anticipating all the bad things that can happen, then avoiding them, takes a lot of experience. You're unlikely to get it right and safe first time.

\$\endgroup\$
4
  • \$\begingroup\$ Could I possibly use this? digikey.com/product-detail/en/rohm-semiconductor/BP5728/… \$\endgroup\$ Commented May 29, 2017 at 3:35
  • 1
    \$\begingroup\$ I wouldn't. Have you read the datasheet? It needs a lot of extra stuff, about 2 dozen items. The mains isolation comes from an external transformer and optocoupler. There are a dozen items at mains potential. If you were a manfacturer and wanted to base a 10000+ design run of power supplies on this, then yes, go ahead. For one-off, for a hobbyist, whose lack of experience is indicated by the fact that you're asking on a public forum like this, no way. Buy something complete, mains in, DC output, isolated, and spend your time and energy on the micro and actuator side of the project. \$\endgroup\$
    – Neil_UK
    Commented May 29, 2017 at 3:46
  • \$\begingroup\$ @AbhishekRavi - Here is what I'd do if I were you - Go to local online stores (amazon/ebay). Search for decent brand mobile phone charger (eg - sony, samsung etc) which claim to give 5V 2A output or better. 2A because you might have to attach some other peripherals later on. Cut the USB cable to take out GND and VCC wires and then power up my project using the supply. If you don't want to show the mobile charger, you can open it up, take out the PCB and use it directly. Make sure you take proper precautions while handling high voltages. \$\endgroup\$ Commented May 29, 2017 at 4:14
  • \$\begingroup\$ Thank you so much for your responses! I hope i didn't waste your time. I'll look into fully implemented SMPS circuits. \$\endgroup\$ Commented May 29, 2017 at 19:04
1
\$\begingroup\$

Without knowing what output voltage you actually need, or the exact dimensional constraints of your system, many of these off the shelf power supplies could suit your purpose -

Digi-Key power supplies

I've found it more and more rare that rolling my own power supply is necessary or cost- or time-effective.

\$\endgroup\$
0
\$\begingroup\$

If you felt adventurous, you could use something such as TNYxxx series IC's. They do not need much, but winding a transformer may be hard.

\$\endgroup\$
0
\$\begingroup\$

Use of capacitor's to replace a transformer is not new. One can obtain any output of desirability just by changing capacitor values. You are really dealing with Time-Constants of the capacitor/load resistor. I recall one television maker using a capacitor input to replace a transformer, (about a 200 watt load) and it worked well. Effiency is very high, higher than a switching power supply, so long as the dielectric of the capacitor does not heat up! (due to wrong dielectric and/or input frequency involved.) There is no regulation here. This system is also used today in small multi LED table lamps that are very bright by overloading the LED into a short life! It too uses a 2.2 ufd capacitor. The value is too high! Reducing this cap. from 2.2 ufd to 0.47 ufd reduces the table lamp brightness to a more realistic value without pushing the LED to destruction! I also got rid of the smoothing 20 ufd capacitor on the bridge rectifier outputs as this only increases the output voltage to a maximum of 1.41rms volts, which is not needed. This is driven by the full ripple output of the rectifier but is not seen as a human eye cannot see this, again due to the time constant of the eye-ball! This is a good system to drive LED's of any array, usually 36 LED's as one series string. But can be adapted to parallel strings, but not really necessary! No regulation is needed on an LED, its only a resistor! Small voltage variations cannot be seen. And this power supply uses no electrolytics or regulation which spells extremely long life for the lighting system. Cap's do fail due to very high current demands which drives up internal temperatures. (as in switching power supplies) If this is a problem, use multi cap's in parallel to reduce current load on each cap.

\$\endgroup\$
1
  • 1
    \$\begingroup\$ Please format this answer into more than one large block of text. As it stands, it is incredibly difficult to read. \$\endgroup\$
    – Daniel
    Commented Apr 5, 2018 at 23:21
-1
\$\begingroup\$

Non-isolating Zener supply 5% efficiency or less

enter image description here

Plug-in timers that are microcontroller-based usually use non-isolating power supplies, like this:

Non-isolated DC supply schematic

R1 essentially drops the difference between the Zener diode and the AC mains potential, so it's not going to be efficient for anything except light loads. Also, your load can't change dramatically, as the resistor has to be sized to provide enough current to the zener to cause it to reverse avalanche, without providing too much current. If your load starts pulling too much current, its voltage will drop. If your load doesn't pull enough current, the zener diode can be damaged.

Pros

Very small

Very cheap

Excellent for extremely light loads (MCU + switch device)

Cons

No isolation

Load current isn't flexible; must be fixed within small window

Mains-frequency regulated transformer supply 20-75% efficiency

enter image description here

You can always use a transformer (60:1 or so), a bridge rectifier, and a linear regulator, like this: Regulated DC supply schematic

This introduces a bulky, costly transformer into the design, but it's more efficient than the previous design, and your load can vary quite a bit.

Pros

Easiest to implement

Designed for medium current loads -- a clock radio, for example.

Full isolation

Relatively inexpensive

Cons

Bulky

Fairly inefficient

Fully-isolated Switch-mode AC/DC Converter 75-95% efficiency

enter image description here

Most efficient (and most complex) is a AC/DC switching converter. These work on the principle of first converting AC to DC, then switching the DC at very high frequencies to make optimal use of the transformer's characteristics, as well as minimize the size (and loss) of the filter network on the secondary. Power Integrations makes an IC that does all the control/feedback/driving -- all you need is to add a transformer and optoisolators. Here's an example design: Power Integrations LinkSwitch converter schematic example

As you can see, AC mains voltage is immediately rectified and filtered to produce high voltage DC. The Power Integrations device switches this voltage rapidly across the transformer's primary side. High-frequency AC is seen on the secondary, and rectified and filtered. You'll notice that the component values are quite small, even considering the current use. This is because high-frequency AC requires much smaller components to filter than line-frequency AC. Most of these devices have special ultra-low-power modes that work quite well.

These converters, in general, provide a great amount of efficiency and can also source high-power loads. These are the sorts of supplies you see in everything from tiny cell phone chargers to laptop and desktop computer power supplies.

Pros

Extremely Efficient

Full isolation

High output current: can source 50+ amps of low voltage DC fairly easily.

Small size

Cons

Large BOM (Bill of Materials)

Difficult to design

Requires thoughtful PCB layout

Usually requires custom transformer design

Expensive

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.