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Current limiting resistor in SLA battery charge circuit?

rmd

Active Member
Most of the posts I've seen that describe using a diode (Schottky or otherwise) when installing a small Sealed Lead Acid (SLA) avionics backup-battery never mention using a resistor to control the SLA charge current. Is it because the diode is going to limit the current (or fry) and not harm the battery or is there some other reason?

By way of example with some simplified assumptions and a 7.5 Ah battery...

Assume charging voltage or 14V (ignoring voltage drop across diode )
Battery voltage down to about 12V
Desired charging current 1.5A (that's 20% of 7.5)

E=IR => 2v / 1.5A = 1.3 Ohms
P=EI => 2v * 1.5A = 3 Watts

So, right around a 1 ohms / 3 watt resistor would be called for.

Of course, the math changes somewhat if the SLA is more discharged.

What am I missing / misunderstanding?

Thanks in advance,
Rob
RV-7A
N706DR
 
Small SLA charging

My small battery is charged thru a relay. The relay control switch can be selected one of three ways: 1 OFF 2 Auto 3 ON
In the Auto position, there is simple circuit that looks at the Main charging voltage. If that is above 14 volts, the relay closes.

(If there is some interest in this, I might be able to dig out the relevant data.)
 
Desired charging current 1.5A (that's 20% of 7.5)

Of course, the math changes somewhat if the SLA is more discharged.

You yourself pointed out the problem. Lead acid batteries have an optimum charging current. More discharged than 12 volts, the simple resistor allows too high a current. Approaching 13.5 volts, too low a current. Works (assuming the current doesn’t get so large it overheats the battery) but not optimum.
 
Definitely use a current limiting resistor at a minimum. Would be nice if there was a commercially available DC to DC charger made for this purpose.

I would overkill the wattage rating to ensure the resistor does not get too hot.
 
Most of the posts I've seen that describe using a diode (Schottky or otherwise) when installing a small Sealed Lead Acid (SLA) avionics backup-battery never mention using a resistor to control the SLA charge current. Is it because the diode is going to limit the current (or fry) and not harm the battery or is there some other reason?

By way of example with some simplified assumptions and a 7.5 Ah battery...

Assume charging voltage or 14V (ignoring voltage drop across diode )
Battery voltage down to about 12V
Desired charging current 1.5A (that's 20% of 7.5)

E=IR => 2v / 1.5A = 1.3 Ohms
P=EI => 2v * 1.5A = 3 Watts

So, right around a 1 ohms / 3 watt resistor would be called for.

Of course, the math changes somewhat if the SLA is more discharged.

What am I missing / misunderstanding?

Thanks in advance,
Rob
RV-7A
N706DR

Why do you feel you need to limit the charging current? Unless you are applying more voltage than the battery's rated maximum charge voltage, it will be fine with whatever it "decides" to draw for current.

The purpose of the diode is to prevent back flow of power from the backup battery to the main bus, should it fail and drop/short to zero or some other low voltage. Putting a resistor instead of a diode defeats this main purpose. I don't know about your specific configuration, but an equally good reason to use a diode is so that when cranking the engine, you don't pull down the avionics backup battery enough to cause problems or reboots. The voltage drop across the diode is simply an artifact of how they work - in this case, not desirable nor harmful.
 
Why do you feel you need to limit the charging current? Unless you are applying more voltage than the battery's rated maximum charge voltage, it will be fine with whatever it "decides" to draw for current.

.

+1

Most SLA batteries can take a lot of amperage without serious damage and are also somewhat limited on the charge current by their internal resistance. While it is a good idea to use an external charger if the battery gets real low (below 12V), but above that, the battery is unlikely to get damaged by taking as much amperage as it wants. Auto batteries (similar Lead acid composition) see this kind of activity all the time. That said, SLA's can vent some gas when charged a too high of a rate and therefore always recommend chargers vs open current when heavily discharged.
 
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You yourself pointed out the problem. Lead acid batteries have an optimum charging current. More discharged than 12 volts, the simple resistor allows too high a current. Approaching 13.5 volts, too low a current. Works (assuming the current doesn’t get so large it overheats the battery) but not optimum.

Are you sure this is the case here? No matter what the batteries internal voltage is, the airframes electrical bus voltage will always be 14 (regardless of the batteries resting voltage) when the alternator is running and the resistor could be sized for that voltage. A 7ah battery shouldn't be able to consume enough amperage to pull the alternator below its target voltage, assuming that alternator is sized for at least 200% of typical loads. And the resistor would prevent excessive draw anyways. Most of our planes have alternators running at around 20-40% of capacity and should easilly throw out another 10-15 amps without a significant voltage drop. If both the main and BU batteries are heavily discharged, that could pull the voltage down for a bit though.
 
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Battery installation

If the purpose of having two batteries is one starting and one avionic backup,
then the diode works as a one way valve. (volt drop typically 0,3V).
If something goes bad in the electric system i.e alternator, starting battery,
the fault can´t draw current from the backup battery.
As others have said, there is no reason to use a resistor to limit charging
current to any of the batteries.
One thing I would recommend though, is a circuit-breaker or fuse for the (smaller) avionics backup battery. This is to protect the cables from catastrophic failure if the backup battery fails short.

Good luck
 
Originally Posted by AlexPeterson View Post
Why do you feel you need to limit the charging current? Unless you are applying more voltage than the battery's rated maximum charge voltage, it will be fine with whatever it "decides" to draw for current.
+2

Below is a quote from the ODYSSEY Battery Technical Manual, last paragraph on page 17.
https://www.odysseybatteries.com/odyssey/docs/us-ody-tm.pdf
Standard internal combustion engine alternators with an output voltage of 14.2V can also charge these batteries.
The inrush current does not need to be limited under constant voltage charge.


Lithium batteries, on the other hand, do need to have the charging current limited due to their low internal resistance.
This can be accomplished by inserting a silicon diode in series with the charging current if the alternator
is capable of exceeding the recommended charging current. No resistor is needed or recommended.
A diode will drop the voltage, thus limiting the charging current.
 
Rob,

First off the diode only controls the direction of current flow. It allows the backup battery to receive power via a path without the downside of supplying power to that path.

As Bob pointed out ideally you would want a constant current charger. I actually designed one for this application, but I have been distracted for I don't know how long on building my plane. That's another story. Your basic assumptions are correct. The problem is as the battery charges the voltage across the resistor will drop and therefore the current will go down. This means that initially the battery starts out charging quickly and eventually reduces to a very small charge. This is not a good approach for a battery that is used all the time. For a backup battery it can be acceptable since eventually it will get charged and if it is only infrequently discharges then you will be fine.

One last comment on the power rating of your selected resistor. It is always better to use a larger wattage resistor if at all possible. The power rating of power resistors is often when the get very warm.
 
Thanks for all of the thoughtful responses.

To clarify my initial post - I should have indicated the question revolved around using the current limiting resistor in addition to the diode, not in place of it. Sorry about the confusion that caused.

Alex: Regarding why would I want to limit the current... Some technical resources I've come across in the past indicated that when charging SLA batteries (all lead-acid batteries??) the current should be limited to around 20% of the Ah rating of the battery. In some recent searches, I found the desire is 10-15%. To my untrained eye, that seemed a little low so I looked at some battery manufacturer specs next.

The manufacturer of an 8Ah battery I came across spec'd a maximum of 2.4 Amps (30%) at 14.5 to 14.9 volts and a different supplier's 7Ah battery listed 1.75A (25%) at 14.5 to 15 volts. So, even though I don't recall exactly where I'd come across the 20% I used in the initial example, it didn't seem like an unreasonable value.

Avanza: Regarding the use of a fuse - Yep, I agree. Also, yes the purpose is to just power a limited set of avionics (AHRS and EFISs) prior to engine start.

Rolf: I like the idea of the voltage monitor and relay approach.

lr172: I did wonder how car batteries and even our main battery does this all day long without getting harmed. I hadn't even considered the potential gas generation due to too high a charge rate. Thanks for the education.

jwyatt: I misread your name at first as WATT and thought that's an appropriate person to reply! Thanks for the mouser link.

Bob: I guess initial post was pondering why most of the posts I've come across don't even mention the potential for a resistor to go along with the diode. I agree that it's not optimum, but sorta figured that not even having one was less optimum and would lead to the possibility of even higher charging currents than the sub-optimal approach. That's probably as clear as mud, but it seemed logical in my head.

Kind Regards,
Rob
N706DR
 
Another option which I used for many years before switching to a TCW unit, is the Sterling Aux battery charger/maintainer, it charges the BU battery at a constant current rate of 3A until charged, works like a charm. It's better than the diode/resistor arrangement.

2023011715232129-5730468657096437777-L.jpg


For sale if you're interested.
 
Here's the specs for a small 12V VRLA battery from Digikey (as used for 10 years in my Rocket):

Charge Voltage
(25°C)
Cycle Use = 14.4-14.7V (-30mV/°C)
Max Current = 3.6A
Float Use = 13.5-13.8V (-20mV/°C)

The key here is "Float Use". If you assume a charging voltage of 14.3 volts (typical alternator output), a typical low leakage Shottky diode will drop around 0.5V. That means 13.8V to the battery, right on the float voltage requirement.

No resistor required.

Battery.PNG
 
Ray, Thanks. Yep, I understand the "directionality" of the diode and the need to use a higher-then-minimum calculated wattage for the resistor. I should have accounted for that in my initial post too.

Walt - Thanks. I'll stick with the diode approach.

Joe - Thanks for the Odyssey link. It's not the main ship battery that caused me to create this post though.

Regards,
Rob
N706DR
 
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