What is the right width of trace should I choose for 220VAC

Hi Anthony,

I'm going to start by a very strong word of caution. Please be extremely careful when working with mains voltages. If you don't know you can do something safely or are unsure in any way then don't proceed.

So, that said, you say 10V output but you don't state the current requirements. What are they? What is the power supply type? Is it linear? Is it switching? All these things will affect the current required on the mains input. It's the current requirement which drives the required trace width. On the other hand, the voltage drives the required separation which must be maintained between the live and neutral traces, and also to any other traces, not just the continuous mains voltage but any safety standards which require a particular level of isolation. See the following link for some further explanation: http://www.ni.com/white-paper/2827/en/

For the clearances specified in IPC-2221 take a look at the following link: http://www.smpspowersupply.com/ipc2221pcbclearance.html

To calculate the trace widths required take a look at the following link: http://circuitcalculator.com/wordpress/2006/01/31/pcb-trace-width-calculator/

You don't need to worry about the radius setting in the route tool in EAGLE for now, that's for mitred corners. You can use that to smooth out 45 and 90 degree bends if you wish though.

Best Regards,

Rachael

If I have this kind of circuit  :

That what is the trace width should I use from the N and L that is connected to the diode bridge ? Because I cannot measure the current flow. The output for the diode bridge is 2 Ampere. Then what is the trace width should I choose ?

Hi Anthony,

You might need to check out the design of your circuit. As drawn the maximum current that it will draw is 2 mA. R1 100K and R3 10K are in series with the regulator. The data sheet on the LM317 indicates that it needs at least 3.5 mA load to maintain regulation and the circuit as drawn can not deliver more than 2 mA and this would be in a default state where the input of the LM317 is shorted to ground. To accomplish what you want to do you will best use a transformer with a 220 volt primary and a 12 volt secondary, then the bridge rectifier and then directly into the input of the LM317 without R1, R2, or R3. C1 and C2 can be combined and should be a lot bigger like 4700 uF if you want to keep ripple down at 2 amps draw. Back to your original question, if 2 amps at 10 volts is being delivered to the load then the 220 volts will be supplying roughly a tenth of an Amp and the traces to the transformer primary will not need to be over sized.

John

Additional:

The only practical way to go from 220 volts to 10 volts at 2 Amps without a transformer is to use a switching power supply and it is generally smart to buy these rather than try to design your own.

John

Actually Sir, after the diode bridge I use CRC filter as shown in the schematic, and R2 and R3 is the load, but I used voltage divider for the input of the regulator. Also sir, I would like to know why the regulator is shorted ? I know that the schematic isn't that good ,but when I tried in the breadboard it works out, the output of the regulator is connected to the attiny.

And also sir, it is not possible for me to your transformer, beacuse I only have a small space which is not enough for a transformer to go in, so what can I do is only using a refctifier, filter, and regulator for the input supply of attiny, or do you have other way sir ? Thanks your for the feedback sir

And also sir, it is not possible for me to your transformer, beacuse I only have a small space which is not enough for a transformer to go in, so what can I do is only using a refctifier, filter, and regulator for the input supply of attiny, or do you have other way sir ? Thanks your for the feedback sir

Hi Anthony,

I have copied your schematic and left off the regulator for simplification.

If all you want is voltage the voltage divider works very well but if you want current it doesn't work well. Current is determined by putting a resistance across a voltage. We then divide the voltage value by the resistance in ohms and the answer is the current that we expect to flow. I have assumed that you are using 220 Volt mains and this will rectify to 308 Volts DC though a full wave bridge. If we use your voltage divider and put it between the 308 volts and ground we will get about 160 volts at point "A" where you were attaching the regulator. If we want to calculate the maximum current that is available at point A we must put point "A" at ground potential as see how much current will flow when the 110K of R1 and R3 combined is put across the 308 volts. This works out to about 2.8 mA. If this is the max current that is available you can see the problem if you need 1000 times as much out of the regulator. Another problem is that there can not be more that 40 volts between the input of the LM317 and its output according to the data sheet. If you want 10 volts out then there is 150 Volts and it will damage the LM317.

When I told you before that the regulator's input needed to be shorted to ground this was just a thought experiment to determine the maximum current that would be available through R1 and R3.

There are only two ways to easily get from 220 volts AC to 10 volts DC with a 2 Amp capability. One is to use a circuit similar to the one that you drew (without R1, R2, and R3) and use a transformer capable of 25 VA of power before the bridge. This is called a linear power supply. The second method is called a switching power supply and it uses electronics to rectify the 220 volts to DC. It then switches the DC on and off at a higher frequency so the 10 volts can be obtained at the output.

Both of these can be fatal if you are using 220 volts and one does not handle it properly. My recommendation would be to find an existing 15 volt power supply and then attach the LM317 to the 15 volts and regulate the voltage down to the 10 volts that you need or find a 10 volt power supply and use it directly. I have seen laptop computer power supplies that operate in the 10 volt area.

John

Yes sir, all I need is the voltage. theoritically I have considered that calculation though. But sir when I test it on breadboard that the output of the lm317 is connected to attiny85 its alll diferent. I only got around 8 to 12 V DC (I forgot the exact number). And I didn't got till 160 V DC. Do you think something wrong with the addition of attiny because the voltage drop is so big (although it is really helpul).

Also sir, the calculation of the current, can I directly divide the output voltage of diode bridge with 110K Ohms without considering the 120 K Ohms ?

anthonylauly  wrote:

 

...when I test it on breadboard... And I didn't got till 160 V DC...

Anthony, are you building a mains connected device (in this case not isolated) on a breadboard?

On 24/06/2018 7:54 p.m., Anthony Lauly wrote:

Yes sir, all I need is the voltage. theoritically I have considered that calculation though. But sir when I test it on breadboard that the output of the lm317 is connected to attiny85 its alll diferent. I only got around 8 to 12 V DC (I forgot the exact number). And I didn't got till 160 V DC. Do you think something wrong with the addition of attiny because the voltage drop is so big (although it is really helpul).

 

Also sir, the calculation of the current, can I directly divide the output voltage of diode bridge with 110K Ohms without considering the 120 K Ohms ?

 

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A few thoughts.

This forum is for suggestions for the Eagle CAD program so this

discussion is inappropriate here and should be on a basic electronics

forum elsewhere.

Your design is bad, dangerous and should not be continued with.

John showed the input resistors limit the input current to 2.5mA and so

the LM317 minimum input current needs have not been met and so the LM317

cannot be expected to function correctly.

If you read the data sheet for the LM317 you will see it can have a

maximum of 40volts between input and output. You have subjected it to

155volts so it may have been damaged. Any output voltage is likely to be

the result of R5 in parallel with R2. You could confirm this by removing

the LM317 and metering between  the IN and ADJ pins. I suspect you will

find very low ohms. The LM317 is robust and contains a lot of protection

circuitry so you may have been lucky but I would be surprised if it is

still good.

There is a technique where you can create a transformerless power supply

but... it looks simple but requires a great understanding of all the

factors involved. You may care to read articles on the subject.

This one below is something to get you started on understanding the

risks to yourself and the repercussions when components fail.

https://hackaday.com/2017/04/04/the-shocking-truth-about-transformerless-power-supplies/

Hope this helps

Warren

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Hi Warren,

Thanks for your good input. FYI these EAGLE questions also show up on the element 14 forum where we routinely answer nuts and bolts questions. Most of the times they are about CAD problems but sometimes, like this one, they demand a response.

John