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  #11  
Old 09-13-2012, 12:31 PM
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Mark Dickens Mark Dickens is offline
 
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Quote:
Originally Posted by BobTurner View Post
Be careful here. Unless two diodes are perfectly matched putting them in parallel doesn't mean they will equally share the load.
Perihelion advertises them for that purpose and provides a metal clip so you can do that.
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  #12  
Old 09-13-2012, 04:33 PM
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Most TO-220 diodes have the cathode tied to the metal tab, so you will need to use an insulating thermal pad and shoulder washer to provide isolation.

I like the resistor type that Gil showed. Used them many times.

Diodes in parallel are ok if each alone can carry the current. The good thing about paralleled diodes is that the voltage drop will be lower, as each device should be seeing less current. Not always true if the diodes are not matched, but there are diodes that are designed to be paralleled and are reasonably matched. Again, as long as either diode alone can handle the current provided the package can safely dissipate the heat, you will be ok.

For the diodes, I suggest the MBR6045. This is a dual Schottkey diode where the cathodes are tied together at the output. (You would have to tie the anodes together externally.) Each diode will take up to 30 A, and will drop about .5 volts at 10 A. It comes in the TO-247 package, which is larger than the TO-220 but gives better thermal performance and you won't need a shoulder washer, though you will have to insulate the metal base. The BER108 silicon insulating pad (Digikey) would work fine.
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  #13  
Old 09-13-2012, 07:19 PM
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I'll go with Bob on this one. I wouldnt parallel diodes. The lower forward voltage drop diode will be conducting more current. This will result in the forward voltage going up so it does balance somewhat but I would never say that 2 30A diodes in parallel are equivalent to a 60A diode.

I do think this is not required for Axel's original application so the thread has gotten a little off topic.
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  #14  
Old 09-14-2012, 09:06 PM
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I checked the battery info and it says that it should not be charged at over 2.5A and to use a 2A charger. So I did the numbers

V=14.4 (charging system) ? 11 (low aux bat) - 0.5 (diode drop) = 2.9 volts

R=V/I = 2.9 volts / 2A = 1.45 ohms

P=I^2*R = 2^2*1.45 ohms = 5.8 watts

So I am buying this for the aux charging side
Resistor http://www.mouser.com/ProductDetail/...%2fRPVtnecI%3d

Diode http://www.mouser.com/ProductDetail/...R9wwwmLiRmM%3d

Thanks for the help.
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  #15  
Old 09-15-2012, 05:18 AM
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Quote:
Originally Posted by AX-O View Post
If the battery ever gets low that diode will be emiting smoke. Components should never operate at "maximum", derate by 50% minimum normally.
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  #16  
Old 09-15-2012, 09:19 AM
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Quote:
Originally Posted by Walt View Post
If the battery ever gets low that diode will be emiting smoke. Components should never operate at "maximum", derate by 50% minimum normally.
Walt, What spec on the diode needs to operate at 50%? Are you saying that i should use a 1A diode?
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  #17  
Old 09-15-2012, 10:12 AM
krw5927 krw5927 is offline
 
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Believe Walt is talking about the "Forward Continuous Current:2A" spec. If your battery goes below say 10V, then with your 1.5 ohm resistor you would be pulling almost 3A through the resistor and diode to charge the battery. Do that for any length of time, or if the diode is heat soaked from a long day on a hot ramp, the diode rated for 2A continuous might let out its magic smoke.

By advising you to derate by 50%, he means you should normally operate at only 50% of a device's capacity. So in this case if you expect to pull a max of 2A through the diode under normal conditions, buy one that's rated to at least 4A max continuous current. Like this perhaps.

http://www.mouser.com/ProductDetail/...kFNCepudM48%3d
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Last edited by krw5927 : 09-15-2012 at 10:14 AM.
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  #18  
Old 09-15-2012, 02:18 PM
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I would go on and size the resistor for 2.5 A, as the current won't stay at that level for long. As the battery is charged, the voltage will rise, causing the current to drop off.

You picked a good resistor package. Kurt's diode recommendation is good, but you need a way to mount it, and you will need some air space around it to dissipate the initial heat. I still like my suggestion for a TO-247 package as you can bolt it right to the frame and the heat sinking will keep it cool.

Paralleling diodes is fine, as long as each individual diode can withstand the maximum current. The advantage is slightly lower power dissipation.
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  #19  
Old 09-15-2012, 05:25 PM
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Quote:
Originally Posted by krw5927 View Post
Believe Walt is talking about the "Forward Continuous Current:2A" spec. If your battery goes below say 10V, then with your 1.5 ohm resistor you would be pulling almost 3A through the resistor and diode to charge the battery. Do that for any length of time, or if the diode is heat soaked from a long day on a hot ramp, the diode rated for 2A continuous might let out its magic smoke.

By advising you to derate by 50%, he means you should normally operate at only 50% of a device's capacity. So in this case if you expect to pull a max of 2A through the diode under normal conditions, buy one that's rated to at least 4A max continuous current. Like this perhaps.

http://www.mouser.com/ProductDetail/...kFNCepudM48%3d
Thanks Kurt, you nailed it.
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  #20  
Old 10-09-2012, 05:13 PM
Mich48041 Mich48041 is offline
 
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A battery should be charged following the manufactures instructions. According to documention from Odyssey, if run down, the PC310 8AH battery should be charged at 14.4V to 15.0V, then float charged at 13.5V to 13.8V. In either case, there is no current limit. See page 4 of Odyssey Publication No: US-ODY-AM-001 :
http://www.odysseybatteries.com/files/techbook.pdf
Then on page 13 it says to use a MINIMUM three-step charger current 6 amps.
Based on the above, there is no need to limit charging current. Only voltage needs to be limited. Electrical system voltage (14.2V nominal) will be reduced by the STPS160H100TV diode. Therefore in the circuit posted by AX-O, the 5amp breaker, the diode, the resistor, and second diode are all not needed; that is unless the battery manufacture's requirements differ from Odyssey.
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