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Question about resistors and diodes

AX-O

AX-O

Well Known Member
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.

Bus%25201.jpg


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/products/datasheets/bm/MCC/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/productdetail.aspx?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.
 
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.
 
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.
 
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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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.
 
Be careful here. Unless two diodes are perfectly matched putting them in parallel doesn't mean they will equally share the load.
 
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.
 
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.
 
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/Arcol/HS10-1R5-1/?qs=sGAEpiMZZMtlubZbdhIBILIZCpd%252b%252bKVwRK%2fRPVtnecI%3d

Diode http://www.mouser.com/ProductDetail/Vishay-Semiconductors/SB240-E3-54/?qs=sGAEpiMZZMuIUjt4yeP9c1hfdrsm%252b%252b4yR9wwwmLiRmM%3d

Thanks for the help.
 
Walt said:
If the battery ever gets low that diode will be emiting smoke. Components should never operate at "maximum", derate by 50% minimum normally.
Click to expand...

Walt, What spec on the diode needs to operate at 50%? Are you saying that i should use a 1A diode?
 
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...EpiMZZMuIUjt4yeP9c1hfdrsm%2b%2b4ykFNCepudM48=
 
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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.
 
krw5927 said:
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...EpiMZZMuIUjt4yeP9c1hfdrsm%2b%2b4ykFNCepudM48=
Click to expand...

Thanks Kurt, you nailed it.
 
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.
Joe Gores
 
Mich48041 said:
...... 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.
Joe Gores
Click to expand...

Since the AUX battery does not go through the battery contactor and is always hooked up to the Main Power buss, you need the diode to allow charge only and not an unwanted discharge when the master is OFF...
 
Ths may be a stupid question but why couldn't you use a dual batterty relay from a motor home about 12 dollors ?
regards Jerry
 
Really want a current source?

If you are set on DIY current source google 'PNP current source' or simply purchase an LDO current source chip (LT3083). Problem with both of these solutions will be thermal destruction of the current source if the batter gets too low. *The LT3083 may work fine if you drop down to 1a charge and TO-220 with heatsink.


I have never tested it but our famous dollar tool store out here sells float chargers for 10 bucks. These float chargers use a 15v 500ma wall wart for power. May be worth your while just to hack one of these for a test. With a ~14.7V bus voltage, it may have enough head room to float charge your backup and comes in a sealed case!
 
Since the AUX battery does not go through the battery contactor and is always hooked up to the Main Power buss, you need the diode to allow charge only and not an unwanted discharge when the master is OFF...
Click to expand...
Eliminating the parts that I said were not needed, the aux battery is no longer connected to the Main Power Bus. I forgot to say that the wire connecting those parts is not needed either. The aux battery is only connected to the essential bus, through the 20 amp switch. The aux battery will be charged through the STPS160H100TV diode that connects the main bus to the essential bus. Current can not flow backwards though the STPS160H100TV diode to the main bus.
The unlabeled diodes make it difficult to communicate about the circuit. If we could talk in person, I am sure that we would agree.
The aux battery should be protected by a fuse. Even though the aux battery is half size, it can still put out over 400 amps if short circuited.
Bob Nuckolls diagram Z10-8 shows a brownout battery circuit using relays instead of heavy duty switches.
Joe Gores
http://www.aeroelectric.com/PPS/Adobe_Architecture_Pdfs/Z10-8A1.pdf
 
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