PC Trace & Optoisolator

[WCE4]

Hello everyone I have two Questions?<p>1: What size should a pc trace be to carry 5 amps at 120v AC?<p>2: What is the best way with the least amount of heat to reduce a 120v ac supply to turn on a AC input optoisolator Vf (led) = 1.5 max If = 10 mA Part No. H11AA1 <p> Thanks for the help
WCE4

[bodgy]

This varies, depending on the copper thickness, and on a particular line of thought, I have two charts for 1oz copper neither of which agree with each other. <p>The voltage part is the separation of the tracks, track width the current carrying capacity.<p>Chart1
Track Width
mm inch Amps
0.5 0.02 3
0.8 0.031 3.5
1.0 0.040 4.0
1.25 0.050 5.0
1.6 0.062 6.0 <p>Track Separation
mm inch Volts
0.4 0.0015 0-50
0.6 0.025 50-100
1.25 0.05 150-300
2.54 0.1 300-500<p>Chart 2<p>Width
mm A
0.6 0->0.5
1.6 0.5->1.5
3.0 1.5->3
6.0 3->6<p>Separation<p>mm V
0.6 0->150
1.6 150->300
2.5 300->600
4.0 600->900<p>Colin<p>PS I can only quote European standards here, but there should be 6mm separation between the AC and Isolated sides, however you are of course stuck with the physical package size of your Opto<p>[ October 04, 2003: Message edited by: bodgy ]</p>

[cato]

you can use a capacitor as your impedance to limit the current. Figure out how much current you need to turn on the opto. Figure out what resistance you would need to limit the current to that amount at low line voltage (say 90Vac). Figure out what capacitance would give you a similar impedance at 50 or 60 hz, which ever you are using, and find a capacitor of that value that can handle the voltage and ripple current.

[WCE4]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by bodgy:
This varies, depending on the copper thickness, and on a particular line of thought, I have two charts for 1oz copper neither of which agree with each other. <p>The voltage part is the separation of the tracks, track width the current carrying capacity.<p>Chart1
Track Width
mm inch Amps
0.5 0.02 3
0.8 0.031 3.5
1.0 0.040 4.0
1.25 0.050 5.0
1.6 0.062 6.0 <p>Track Separation
mm inch Volts
0.4 0.0015 0-50
0.6 0.025 50-100
1.25 0.05 150-300
2.54 0.1 300-500<p>Chart 2<p>Width
mm A
0.6 0->0.5
1.6 0.5->1.5
3.0 1.5->3
6.0 3->6<p>Separation<p>mm V
0.6 0->150
1.6 150->300
2.5 300->600
4.0 600->900<p>Colin<p>PS I can only quote European standards here, but there should be 6mm separation between the AC and Isolated sides, however you are of course stuck with the physical package size of your Opto<p>[ October 04, 2003: Message edited by: bodgy ]<hr></blockquote><p>Colin <p> Thanks for the reply and the info. The copper thickness is 1 oz. I am also having the same problem each chart I look at none agrees with each other. I need to meet UL standards I look at their web site but no luck

[WCE4]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by cato:
you can use a capacitor as your impedance to limit the current. Figure out how much current you need to turn on the opto. Figure out what resistance you would need to limit the current to that amount at low line voltage (say 90Vac). Figure out what capacitance would give you a similar impedance at 50 or 60 hz, which ever you are using, and find a capacitor of that value that can handle the voltage and ripple current.<hr></blockquote><p>Thanks for the info I hope my math is right at 60Hz I come up with a capacitance of .33uF to give me about 15ma.

[bodgy]

Maybe or may not be of any help.<p>The link below will tell you which part of the standard you need to find<p>http://www.i-spec.com/Product_Design/pcb.html<p>Complete requirements are given in IEC 60950, Clauses 2.10.5.3 and 2.10.6 (2.9.5 and 2.9.4.3)<p>Above from the website mentioned.<p>Vutrax reckon their software produces UL certified tracks .<p>http://www.vutrax.co.uk/xbook3.htm#current<p>Colin<p>[ October 05, 2003: Message edited by: bodgy ]</p>

[Ron H]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by WCE4:
<p>Thanks for the info I hope my math is right at 60Hz I come up with a capacitance of .33uF to give me about 15ma.<hr></blockquote><p>Be aware that the optoisolator will only operate on one half of the cycle, unless it has anti-parallel LEDs. If it only conducts in one direction, you will need a diode connected anti-parallel (anode of diode to cathode of LED and vice versa). Also be aware that the peak currents will be at the zero crossings of the AC line voltage, because the current leads the voltage by 90 degrees.
It is generally good practice to add a small value resistor (470 ohms or so) in series with the cap to limit high currents caused by transient spikes.

[dacflyer]

as far as trace thicknesses...if foil is in doubt..solder some 16GA or 14GA stripped of insulation along the trace path...if not too difficult for you... i have done this many times to repair fried traces in load switches..
*opto-isolated triac moduals*

[WCE4]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by bodgy:
Maybe or may not be of any help.<p>The link below will tell you which part of the standard you need to find<p>http://www.i-spec.com/Product_Design/pcb.html<p>Complete requirements are given in IEC 60950, Clauses 2.10.5.3 and 2.10.6 (2.9.5 and 2.9.4.3)<p>Above from the website mentioned.<p>Vutrax reckon their software produces UL certified tracks .<p>http://www.vutrax.co.uk/xbook3.htm#current<p>Colin<p>[ October 05, 2003: Message edited by: bodgy ]<hr></blockquote><p>Thanks for the links they were of some help

[WCE4]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by dacflyer:
as far as trace thicknesses...if foil is in doubt..solder some 16GA or 14GA stripped of insulation along the trace path...if not too difficult for you... i have done this many times to repair fried traces in load switches..
*opto-isolated triac moduals*<hr></blockquote><p>Thanks for the info But I need it to meet UL

[WCE4]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by RonH:
<p>Be aware that the optoisolator will only operate on one half of the cycle, unless it has anti-parallel LEDs. If it only conducts in one direction, you will need a diode connected anti-parallel (anode of diode to cathode of LED and vice versa). Also be aware that the peak currents will be at the zero crossings of the AC line voltage, because the current leads the voltage by 90 degrees.
It is generally good practice to add a small value resistor (470 ohms or so) in series with the cap to limit high currents caused by transient spikes.<hr></blockquote><p>
RonH
Thank's for the reply the optoisolator I am going to use is a AC input isolator Part# H11AA1 it has two leds in anti-parallel. The cap(.33uf) will be wired from one AC line to one side of the led leg.The other leg of the led to a 1K ohm resistor in series with the other AC line. Do you think this will work?