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09-12-2012, 06:18 PM
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Join Date: Jun 2005
Location: SoCal
Posts: 2,452
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Question about resistors and diodes
All,
I am having troubles figuring out how to size some diodes and resistors (look at top left hand of diagram).
This is what I need to do. I have an 8 amp/hour AUX battery and I would like to power my E bus for approximately 4-5 on the ground while I strap in the aircraft and set up the avionics. After that is done, I will turn on the MAIN battery and start the engine. Once the ALT FLD is turned ON I should be able to charge my MAIN battery and my AUX battery (via the Main bus). Diagram below.
Problem 1
The E bus can pull up to 18 Amps (if everything is on), but I won?t get that high on the ground before start. The research I have done says to create a margin of safety from 20-50%. So a 20% safety margin for 14V is approx 17V (total) and a 20% margin for 18A is approx 22A (total). This would be for powering the E bus via the AUX battery only. Would this diode work? http://www.alliedelec.com/images/pro...C/70055670.pdf
Problem 2
Power to charge my AUX battery would also be at 14V. There will be a resistor in-line to slowdown the charge from the Main bus into the smaller AUX battery and a diode to limit the direction of flow. Per someone?s guidance, I need to charge the battery at 3A.
I (Amps) = V (Volts)/R (Ohms)
3A = 14V/R
R=4.67 Ohms
So do I purchase a resistor sized to 4.67 ohms http://www.alliedelec.com/search/pro...x?SKU=70065330 in order to step down the charge into the AUX battery to 3A? Do I size the diode based on the 8 amp/hour rating of the AUX battery? Do I size the diode based on the 3A resistor? What about watts? Can I just use a 3A diode and not include the resistor?
Thanks for any help or guidance you can provide. I am getting close on this electrical stuff.
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Axel
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The information that I post is just that; information and my own personal experiences. You need to weight out the pros and cons and make up your own mind/decisions. The pictures posted may not show the final stage or configuration. Build at your own risk.
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09-12-2012, 08:40 PM
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Join Date: Mar 2008
Location: Dublin, CA
Posts: 1,261
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Axel,
I'll try to help but I'm on my phone so I'll be brief. First you only need a single diode not a bridge with four diodes so I would pick another part. Second a 50V diode should be fine. I would probably pick a 25A diode. the power for a diode is P = Forward Voltage x Current. The wattage required for the diode is based upon the power rating of the diode and the size of the heat sink. A lower wattage diode may require more heat sinking. A larger diode has more heatsinking built into the package.
__________________
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09-12-2012, 09:13 PM
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Join Date: Jun 2007
Location: Merritt Island, FL
Posts: 602
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Your resistor calculation is wrong. When sizing a resistor to limit current, the voltage you use in the calculation is the voltage across the resistor, not the voltage applied to one end.
Also, you need to take the voltage drop introduced by the series diode into account in the calculation. For a regular silicon diode, the voltage drop it causes will be around .7 to 8 volts at 3 A. Thus, the 14 volt source will act like a 13.2 to 13.3 volt source. If the small battery is at, say, 11 volts when you start charging it up, the voltage across the resistor will be 14 - .7 -11 volts, or about 2.3 volts. That is the value to use in your calculation, and would yield a resistor with a value of 2.3V/3A = .76 Ohms. You could use a standard value resistor of .68 or .75 Ohms here. Note, of course, as the small battery charges up, it voltage will increases and the charging current will decrease as the voltage across the resistor approaches 0.
Peak power for produced by the resistor (as heat) would be be around: P = I^2*R = 3^2*.75 = 6.6 Watts. A 7.5 W resistor would probably be ok here, as the power will decrease towards near 0 after a bit.
I would not eliminate the resistor. You might be able to use a smaller value to reduce the power dissipation since, again, as the small battery charges it voltage will rise and the current will do down, hence the power dissipation will go down.
A 5 Amp minimum diode here would be ok. One in a heat-sinkable case would be best. One in a TO-220 case, bolted to some alumimum for heat sinking, would be ok. A Schottky diode would be better, as the power dissipation would be about half.
For the other diode, I would highly suggest a Schottky diode. For a silicon diode at 22 A, the voltage drop will be .9 volts or so, and it will dissipate 20 Watts of power. That is a lot and you will need some significant heat sinking if you don't want to burn the device up. For a Schottky diode, the power dissipation would be about half this. Still enough to get pretty warm, but easier to heat sink. Again, one in a TO-220 or other power package that can be bolted to aluminum would be good.
Battery chargers that use resistors are not very fast or efficient, but they are cheap.
__________________
Jeff Rosson
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09-12-2012, 09:27 PM
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Join Date: Dec 2008
Location: Los Angeles, CA
Posts: 301
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resistor sizing
For the resistor used to drop the charge current, I would use the largest power rating that would fit. This is because if the that battery is near death, or depleted and then is recharged, the resistor will drop more voltage and dissipate more power. In addition, if you fly to Death Valley, and try to charge with the ambient temp at 100 degrees F, the resistor will get pretty warm. Power resistors are cheap, bigger is better. I am not talking about the resistor values, just the power rating. Same thing with the diodes, when they are hot, thier ability to dissipate heat goes down and they could burn, so you might consider higher power rated diodes then what the calculations say. JMO
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Last edited by johnny stick : 09-12-2012 at 11:37 PM.
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09-12-2012, 10:37 PM
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Join Date: Jan 2005
Location: 57AZ - NW Tucson area
Posts: 10,011
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This would be a good resistor to use. 50 watts and about 2 inches long by 5/8 on the body size. Easy to mount on a metal bulkhead for a bit of heat dissipation.
http://www.mouser.com/ProductDetail/...S66idteltCc%3d

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09-12-2012, 11:23 PM
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Join Date: Mar 2008
Location: Dublin, CA
Posts: 1,261
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Axel,
I'm not sure how you determined the desired charging current. Most batteries have a charging current that is usually a factor like C/n. Where C is the capacity of the battery (A/H) divided by some factor n. I would look at the specs for the battery you plan on using and find the recommended charging current. As others have said you didn't have the resistor calculation right. The voltage across the resistor is Vresistor = Vcharging - Vdiode - Vbatt. Then the resistance is = Vresistor/I charging. You need to determine the minimum Vbatt for a functional discharged battery this might be 11.5V. I'm not sure about this number, but again I would look at the battery spec. The charging current will go down as the battery charges and its voltage rises. That means with this scheme as the battery gets more and more charged it will charge at a slower and slower rate.
__________________
Ray Tonks
2020 Donation Paid
Titan IOX-370, Dual PMAGs, 9.6:1 Pistons, FM-150
RV-7 Fuselage in progress
* Cabin Interior - In progress
RV-7 SB Wings
* Both Wings fully skinned
* Fuel Tanks Complete - No leaks finally
* Ailerons Complete
* Flaps Complete
RV-7 Empennage - Complete (a little fiberglass work left)
Vans Training Kit # 2 - Complete
RV-7 Preview Plans
Vans Training Kit #1 - Complete
EAA Sheet Metal Class - Complete
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09-13-2012, 06:45 AM
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Join Date: Apr 2008
Location: California
Posts: 652
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For your consideration
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09-13-2012, 08:44 AM
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Join Date: Apr 2012
Location: Collierville, TN (KFYE)
Posts: 1,433
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Quote:
Originally Posted by cfiidon
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Bought one just the other day for tying the e-buss to the main buss. Tie both diodes together and it can handle 60 amps. It's a hoss.
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09-13-2012, 11:54 AM
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Join Date: Dec 2011
Location: Livermore, CA
Posts: 6,797
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Be careful here. Unless two diodes are perfectly matched putting them in parallel doesn't mean they will equally share the load.
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09-13-2012, 12:22 PM
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Join Date: Jan 2005
Location: Mtns of N.E. Georgia
Posts: 1,322
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Schottky
Quote:
Originally Posted by cfiidon
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Digikey has them for a lot less money. The part # is, DSS2X61-0045A. Made by Ixys. No heat sink, but it's TO-220 form and the base is isolated, so if you bolt it to aluminum, that is more than enough for heat sinking.
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Last edited by mannanj : 09-13-2012 at 08:56 PM.
Reason: More info
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