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

Question

Hi, I want to design a pcb that convert AC (220VAC) to DC (10 V), so I would use a wider trace for the Neutral and Line. First I would like to know what is the measurement unit use in the radius and width toolbox, second what is the difference of the radius and width of the trace, because when I increase the radius, nothing is happen. Third, what is the right number of radius and width for 220 VAC, thank you.

autodeskguest · Answer

On 6/21/2018 7:15 AM, Anthony Lauly wrote: Hi, I want to design a pcb that convert AC (220VAC) to DC (10 V), so I would use a wider trace for the Neutral and Line. First I would like to know what is the measurement unit use in the radius and width toolbox, second what is the difference of the radius and width of the trace, because when I increase the radius, nothing is happen. Third, what is the right number of radius and width for 220 VAC, thank you. -- To view any images and attachments in this post, visit: https://www.element14.com/community/message/241262 Trace width is determined by how much current is flowing and pcb copper weight as well as how warm you are willing to let the traces get. There are online trace width calculators but I usually find the cross sectional area of copper and use that to determine current carrying capacity. With higher voltage the clearance is important. Conformal coating reduces the spacing requirement. Most important - keep it safe for the user/tech. HTH Doug

rachaelp · Answer

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

jw0752 · Answer

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

autodeskguest · Answer

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 ? -- To view any images and attachments in this post, visit: https://www.element14.com/community/message/241355 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 -- ... use NNTP://news.cadsoft.de and a functional news reader like Thunderbird! --- This email has been checked for viruses by Avast antivirus software. https://www.avast.com/antivirus

jw0752 · Answer

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

anthonylauly · Answer

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 ?

jw0752 · Answer

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

Jan Cumps · Answer

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?

jw0752 · Answer

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

rachaelp · Answer

anthonylauly wrote:

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,

Why do you say the output of the diode bridge is 2A? That will only be 2A if you are drawing 2A elsewhere which you cannot simply because of the circuit you have drawn. The absolute maximum current is limited by the series combination of R1 and R3 and as John quite rightly pointed out, the maximum current which would be flowing would be 2.8mA (actually less if you also consider the regulator).

I'm assuming you probably really want a few tens or hundreds of mA at some lower voltage derived by some other regulator after this circuit as you mention an ATTiny so I am guessing you have some simple low voltage circuit there. Without knowing what you are actually wanting to power with this though it's hard to make accurate estimates.

But, I really think you should do as John suggests and consider using a commercially available plug in power supply to give you your 10V output. What you have drawn above will not do what you appear to require.

Best Regards,

Rachael