# transformer less power supply?

I have uploaded a picture of my design.

I needed around 500mA current to run my mini project, so i decided to go through the transformer less power supply instead of normal transformer used supply.Since the other power supplies are quite bulky. But the problem occurred is, the supply doesn't supplies minimal current to run the project. Please, help me with this problem and suggest some alternatives or amendments to be made.

I have uploaded a picture of my design.

I have uploaded a picture of my design.

active| newest | oldestThis would be dangerous, so it is not recommended as a beginner project.That said, I do believe that learning about electronics requires real-world experimentation.The circuit you have posted is essentially a shunt regulator. You are using the capacitive reactance of a capacitor to deliver current. If you do plan to make this, you assume all the risks of dealing with 220V mains, so read up a lot on HV safety. I made a list of things you should add if you really want to build this to handle 500mA:

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* Add a small fuse on the live or hot wire in series with the mains, that will blow at a relatively low current. Adding a varistor or MOV in parallel with the mains (after the series fuse) is also a good idea, or better yet, a crowbar circuit. A crowbar circuit is designed to purposely short out the mains if the voltage never exceeds a specific amount, purposely blowing the fuse.

* Insulate

everythingwith heatshrink or hot glue or tape or that has over 80V on it in simulation. Make sure you make it in an enclosure such that it is very hard to accidentally contact the HV side of stuff. Do not poke around the circuit while it is plugged in, and wait at least 30 seconds after unplugging it to continue working on it, discharge all the capacitors safely before working on it.* Make sure you use a capacitor that is rated for

at leasttwice the voltage of the mains. I highly recommend using a decent X or Y rated MKP capacitor, as these are designed to fail open-circuit rather than short-circuit. If the dielectric ever breaks down due to a power surge, then the metal film will melt away and create an open circuit. DO NOT USE A POLARIZED ELECTROLYTIC CAPACITOR.* To get that much current handling capability, you will need to make some modifications to this circuit that probably make it more practical to use a wall wart. These transformerless power supplies really should

notbe used in applications that need more than 100mA, or in devices where you will come close or in contact with the voltage output. Regardless I will tell you how to figure out and make to be able to deliver 500mA:----------------------------

I'm sure I will get corrected my Icing orsteveastrouk, and hopefully so, I hate giving wrong answers and love learning. This is how Ithinkit can be figured out, everything below is just a my best guess and somewhat verified in LTspice:** First of all, you will need a

lotmore than a single zener diode. because it can only dissipate up to 1 watt of heat before getting too hot to function and burning out (also extra redundancy will ensure that the supply will be reliable.)Remember, you could have over 500mA of current flow through them when no load is attached to the output,and the 1N4735A has a Vdrop of 6.2V, so that means that if just 0.5A flows through a single diode, assuming the voltage across it will be 6.2V (it will actually be higher, you need to look at a IV curve to see what it will probably be or test it) 6.2V times the current, 0.5A, is3.1Wof power dissipation. You should have at least 5 of those all in parallel to spread that power over all the diodes, that way each diode has less current to deal with and only 0.62W of power to dissipate at maximum.----

** Second, we need to calculate the value for the main pass-capacitor. I guess around 5-6uF based on simulations and trial & error. Regardless, I would prefer some calculations to understand why:

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First, knowing that I want the supply to deliver as much as 6.2V at 500mA, I figure out

whatload will draw exactly that by using ohm's law. It turns out that that 12.4 ohms is right. With 6.2 volts across a resistor of that value, exactly 500mA will flow. Likewise with 500mA going through it, it will drop 6.2V. Thats just 2 ways of thinking of the same thing.----

Pretend now we are dealing with DC, 220V*sqrt(2) = 311.1V peek on the supply, a unknown value pass "resistor" (which is actually a unknown capacitor) and connects to the 12.6 ohm apparent load resistor to make a resistor-divider circuit. Knowing the formulas for it will be helpful, though not necessary. Only ohm's law is needed. So the "pass resistor" needs to drop 331.1V - 6.2V = 304.9V, and with that voltage drop, we need 500mA to flow. Again, ohm's law tells us 609.8 ohms.

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You could literally use that for the transformerless power supply by using a 150W 220V lamp as the ~609.8 ohm resistor, but it would be incredibly inefficient. You want to use a series pass capacitor and rely on capacitive reactance instead. Let's replace the 609.8 ohm resistor with a capacitor that has a similar level of capacitive reactance.

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The capacitive reactance formula is Xc = 1 / (2*PI*freq.*capacitance). Xc can be sorta be treated like a resistance (it is impedance technically). Solve for capacitance instead of Xc because we know we want Xc to be close to 609.8 ohms. 1 / (609.8 ohms * 2pi * 50) = 5.22*10^-6 or 5.22uF.

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So you will need to find capacitor that has at least that much capacitance, you may need to parallel up a few capacitors to get that value.

Actually, in simulation, I noticed 5.22uF is not quite enough, I think this is because I did my calculations with the peak voltages on 220V instead of the RMS value, 220V. So redoing some of my calculations, 220V - 6.2V = 213.8V, divided by 0.5A = 427.6 ohms Xc, so then 1 / (427.6 ohms * 2pi * 50) = 7.444uF. This value yield more accurate results.

Also note that the output of a transformerless power supply will NOT be isolated from mains, so there will be potential between the output of this supply and any other grounded object. Also if something goes wrong you may end up with 220V directly on the 6.2V output.

I recommend watching ElectroBOOMs youtube channel as a fun way to learn what mains high voltage and high current is capable of. He basically teaches people what NOT to do when building things. Here some of my favorite video of his:

https://www.youtube.com/channel/UCJ0-OtVpF0wOKEqT2...

Your back with a vengeance.

Your back with a vengeance.

Have you had any electrical accidents that left a scar on your skin ?

Regardless all that ^^^^ was just me messing around with some math formulas, trying to figure out the math for myself, learning about it myself, then documenting as an answer! Thats how I learn and teach at the same time!

You wish :P No, I am not back with vengeance, just got bored with working on some projects and youtube, so came here to annoy you! jk ;-)

500mA is not trivial current, and remember an online powersupply like this is lethal. There is no way I can recommend someone of your age and experience using a circuit like that. It can KILL.

I totally agree with you about the risk of operation of such circuit but if i reduce my minimal current to 150mA from 500mA, then is their any possibility of receiving this value.

And, please suggest some corrections too.

+1 poor idea.

Notice the earth ?

+1

Correct he is half-wave, only using D2 which is1/4 of the DC bridge.

This should help you sit-up !

A 6 volt 500 ma wall wart is $5.95 and can run from 120 to 240 VAC input.

http://www.goldmine-elec-products.com/prodinfo.asp...

BTW your LED is backward !