2 Ohm Resister

[Wotname]

Originally Posted by tuberider

On boat trailers with the small battery mounted up front in case the trailer disconnects from the car it engages the electric brake. That small battery is charged by the car's battery. Is there something in the car that limits amps to the trailer battery.

Please remember that the charge rate going into the battery (for any given voltage applied to it) is primarily dependant of the state of charge of the battery and it's chemistry and not the charging source.

Additionally the alternator (along with it's regulator) is constant voltage source and not a constant current source. This means it's output will always be the voltage as set by the regulator (say 14.2 Volts or similar). The actual current produced will be a result of the load applied. More loads, more current, until of course, it reaches it designed maximum designed rated current (say 60 amps).

So a 60 amp alternator will only produce that output when there a high enough load on it.

A small battery being charged is only a small load. The charge current is will be higher when the battery is flat and lower when it is fully charged and has nothing to do with the maximum rated alternator output.

[Andina Marie]

Originally Posted by rtbates

A 2ohm resistor with 6 amps running through it will drop 12V!!

E(voltage)=I(current)xR(resistance)

12=6x2

not gonna work as a charger very well.

what really happens is the voltage across the 2ohm resistor is dependent on the voltage at the small battery and the voltage at the source, on the other side of the 2 ohm resistor. Maybe something like 12v at the small battery and 13.8 at the source, for a voltage across the 2 ohm of 1.8v

Then 1.8=Ix2 or an I of .9 amps, ie a trickle charge..

To properly use the equation E=IxR you must first know what is the known in order to figure out the unknown. In correct knowns like 6A through the 2ohm resistor make the whole process in-correct.

An yes I'm a retired EE..

But don't compare apples and oranges, this is not a starting battery or a house battery that needs charging. It is (nearly) in parallel with the source battery so the voltage follows with a delay. It will never need 6 amps of charge.
The original assumption was that the load on the battery could be supported with a trickle charge of 13.6 - 12.6 = 2 volts when in use with the engine running, divided by 2 ohms = a 1 amp load, perhaps a GPS or depth sounder that was being reset by the drop of engine starting.

[Wotname]

Originally Posted by goboatingnow

Th euse of teh 2 ohm resistor is to electrically decouple the small battery , whose function given its size is to power some instruments or a radio. The reisistor as pointed out by Andina does that isolation task quite admirably. It not only performs discharge isolation , but also regulates charge current ( which a diode would not).

Its an unusual setup , mainly because stable instrument power is an oddity on boats, but theres nothing nonsensical about it

dave

Thanks for your explanation ; I guess the main thing I can take from this is that you and I have different concept of "perfect" .

I really don't think it electrically de-couples the small battery very well let alone isolates it and I can't think of many (any?) good reasons of why the charge current needs to regulated by it but each to their own...

In the equivalent circuit as roughly drawn below, all the resistance values are dynamic excepting the 2 ohm resistor. Ri,s.b. and Ri,b.b. are the internal resistance's of the small battery and big battery and determined in part by the battery chemistry and by state of charge. The load resistance's are also quite varied. Thus the fixed 2 ohm value for the coupling resistor is at best a compromise and doesn't provide any significant isolation for the small battery. There are a variety of conditions that could allow it to have to dissipate 24+ watts. Replacing it with a diode would be (IMO) a vast improvement even though not a perfect one.

I would endorse Bill's view of using an EchoCharge as being close to perfect (electrically).

However one advantage of using the resistor is that the small battery load can be shared by the big battery but the converse remains a problem in that the the big battery load is also applied to the small battery.

[goboatingnow]

I guess the main thing I can take from this is that you and I have different concept of "perfect"

Theres no real concept of perfect in electronics, merely , it will achieve the stated aim adequately. This is because all circuit solutions are compromises, just like boats.

The resistor provides a degree of decoupling. It also ensures thats the instrument battery only has access to limited charge current (* given its small size relevant to the main bank).

A diode would not provide that protection, and in certain situations large currents , compared to the small battery could flow into it.

An echo charge or even a more sophisicated battery to battery charger like Sterlings Powers one, is a better solution , but far more expensive.

The resistor does whats required of it and cheaply too.

ddave

[Wotname]

Originally Posted by goboatingnow

Theres no real concept of perfect in electronics, merely , it will achieve the stated aim adequately. This is because all circuit solutions are compromises, just like boats.

The resistor provides a degree of decoupling. It also ensures thats the instrument battery only has access to limited charge current (* given its small size relevant to the main bank).

A diode would not provide that protection, and in certain situations large currents , compared to the small battery could flow into it.

An echo charge or even a more sophisicated battery to battery charger like Sterlings Powers one, is a better solution , but far more expensive.

The resistor does whats required of it and cheaply too.

ddave

Sorry old boy, I have to disagree with you again (but it's not personal it's just electronics )

The charge acceptance is dictated principally the battery chemistry and it's state of charge and not by any coupling resistor. Said another way, it is the equivalent internal resistance of the battery that limits charge rate. I suppose you could argue that the resistor allows one battery to charge (equalize?) the other when there is no charging source - but this is not necessarily a good thing.

In fact, the resistor limits efficient charging. Even the diode limits charging by introducing a constant volt drop (forward bias voltage) into the charging circuit.

[Andina Marie]

The original question was essentially "Why is there a 2 ohm resistor in the charging line to the instrument battery.". The writer questioned if it was needed for limiting charging current.

In my justification of the use of a 2 ohm resistor I maintain that it is detrimental for charging and that was not its purpose. The purpose was to stop a sudden drop of voltage on the instrument panel during engine starting and/or also as a noise filter for a sound system picking up alternator noise.

There is no doubt a $40 diode isolator would do the job just fine.
There is no doubt that a $125 echo charger would do the job just fine.
But if a $1.25 resistor does the job adequately, I can see that as an attractive choice and much cheaper maintenance.

[Paul Elliott]

For what it's worth, I agree that the 2 Ohm resistor may be an adequate solution.

But this is not a general solution, and it only works if the load being supported is quite small. The problem with the series resistor is the voltage drop across it caused by the load current. If the load were 1A, the resistor would drop 2V, and this would prevent the small battery from being fully charged. Not good! But, if the load were 0.25A, the drop would be 0.5V (assuming the battery has reached charge equilibrium and is drawing little current). This is comparable to the isolation diode that many here are comfortable with. The resistor will provide sufficient isolation in the case of *brief* voltage transients on the main bank.

So, if the goal is to provide a transient-protected supply for a low-current load, then a properly-sized resistor will work. So will a diode.

[goboatingnow]

Originally Posted by Wotname

Sorry old boy, I have to disagree with you again (but it's not personal it's just electronics )

The charge acceptance is dictated principally the battery chemistry and it's state of charge and not by any coupling resistor. Said another way, it is the equivalent internal resistance of the battery that limits charge rate. I suppose you could argue that the resistor allows one battery to charge (equalize?) the other when there is no charging source - but this is not necessarily a good thing.

In fact, the resistor limits efficient charging. Even the diode limits charging by introducing a constant volt drop (forward bias voltage) into the charging circuit.

This is an incomplete explanation. LA batteries when there is sufficient Voltage difference , ie the charging source has a lower output impedance will force too much current into the battery. Try charging a 10 Ah lead acid with a 100 amp charger

While the resistor is inefficient , it does play a role in limiting charge current to ensure a degree of trickle charging

Dave

[Andina Marie]

Originally Posted by goboatingnow

This is an incomplete explanation. LA batteries when there is sufficient Voltage difference , ie the charging source has a lower output impedance will force too much current into the battery. Try charging a 10 Ah lead acid with a 100 amp charger

While the resistor is inefficient , it does play a role in limiting charge current to ensure a degree of trickle charging

Dave

Oooo now you're waving the red flag.
You CAN charge a 10 Ah lead acid battery with a 100 amp charger.
A 100 amp charger does not HAVE to put out 100 amps, that is its maximum rating. The actual current it puts out depends on the impedance of the battery and the voltage on the battery.

The impedance of the charger that you referenced is only half the circuit. It doesn't matter if the charger impedance is 1% or 0.01% of the battery impedance, the difference in current between 1.1 and 1.01 is insignificant.

A 10 AH battery has a much higher impedance so there is NO way you could put 100 amps into it from a 12 volt charger, it would take perhaps over 100 volts to force that kind of current into a small battery but the charger will be more or less shut down at 14.2 volts.

[noelex 77]

Originally Posted by Andina Marie

Oooo now you're waving the red flag.
You CAN charge a 10 Ah lead acid battery with a 100 amp charger.

A lot of what you say is true, but there are practical problems with charging small lead acid batteries with large chargers.

A lot of small LA batteries, particuarly the gel types have limitations on the maximium boost charge current. They will accept more charge current at the boost voltage than is permitted by their specifications.

Here is a typical specification: (note this an AGM battery so better charge current acceptance than the gels)

Cycle Applications: Limit initial current to 2.1A. Charge until battery voltage (under charge) reaches 14.4 to 14.7 volts at 68°F (20°C). Hold at 14.4 to 14.7 volts until current drops to under 70mA.
http://www.power-sonic.com/images/po..._11_Feb_21.pdf

If you hook these batteries up to a large charger they will accept a lot more than 2.1A.


Unlike are boat house batteries charging these small batteries often means limiting the current as well as the voltage. With larger boat house batteries this is unnessary and only voltage control is needed. They will self limit at an acceptable current.

Charging small (as in 10AHr) batteries with a large charge source (battery charger, solar controller, or alternator) often needs some for of additional current control (which is not standard with these charge sources).
A simple resistor is often employed a crude mechanism to achieve this.

Simply hooking up a large charger to a small battery is often not satisfactory.

[Andina Marie]

Well I disagree. The initial stage voltage on a 100 amp charger is the same as the initial stage voltage on a 10 amp charger. You can't charge bigger batteries to a higher voltage than you can the small ones just because they are bigger. They ALL approach gassing voltage around 14.2 volts.


So a 14.8 voltage in the first stage will have the same differential voltage for either charger and the smaller battery will draw the same current from either charger.

[goboatingnow]

Originally Posted by Andina Marie

Well I disagree. The initial stage voltage on a 100 amp charger is the same as the initial stage voltage on a 10 amp charger. You can't charge bigger batteries to a higher voltage than you can the small ones just because they are bigger. They ALL approach gassing voltage around 14.2 volts.

So a 14.8 voltage in the first stage will have the same differential voltage for either charger and the smaller battery will draw the same current from either charger.

The initial voltage is not the same. The big charger will be capable of impressing a higher voltage sooner then the smaller charger up to the CC limit. That's the definition of CC. Hence where there is a large disparity between capacity of the charger and the battery , significant damage can be done to the battery.

In the usual case the charger is a percentage of the AH rating , but where the charger is much bigger great care must be taken to limit the charger current to within the spec of the battery.

Otherwise it is exactly as you specified ( and I have argued it myself )

For an experiment take a 100 A charger and a small 12v 4-5 AH LA battery.

Dave

[Andina Marie]

Although it is a Constant Current source it doesn't mean it will always put out 100 amps. They are not true constant current sources, it is a voltage source with a resistance or solid state current limit. During its attempt to put out 100 amps it will monitor battery voltage and as soon as that voltage reaches the threshold it will go to the next state.
Hence with a small battery, the over-sized charger will go into the next stage within a couple of seconds and continue to limit the voltage to safe levels.

Same thing would happen if you put a 100 amp-hour battery on the charger that is already fully charged, it will detect the fully charged voltage and switch to the next stage within a few seconds.

[Teknav]

Hi Andina! You have bent backwards in explaining the 2-ohm resistor circuits. I find that many folks on here did not attend an EE program, let alone study Circuit Analysis. Some of the concepts might be difficult to understand, hence the back and forth arguments. Suffice to say if someone wants to modify their boat's circuitry, it is their prerogative. Just remember if your boat catches an electrical fire because of your circuits modification, it will only be your fault.

Mauritz
Knowledge is power!

[goboatingnow]

Originally Posted by Andina Marie

Although it is a Constant Current source it doesn't mean it will always put out 100 amps. They are not true constant current sources, it is a voltage source with a resistance or solid state current limit. During its attempt to put out 100 amps it will monitor battery voltage and as soon as that voltage reaches the threshold it will go to the next state.
Hence with a small battery, the over-sized charger will go into the next stage within a couple of seconds and continue to limit the voltage to safe levels.

Same thing would happen if you put a 100 amp-hour battery on the charger that is already fully charged, it will detect the fully charged voltage and switch to the next stage within a few seconds.

The problem Andina is that LAs are not completely self limiting , a discharged battery prevents a low impedance , a large charger will , if correctly designed , not exceed the max absorption voltage , but will charge a LA at well above the manufacturers rate of recommended charge. Heavy duty plates can handle that aggressive charging , but many batteries especially lower capability will not withstand such aggressive charging and material shedding can occur.

So in practice where there is a very large charger and a small battery care must be taken.

Furthermore many large chargers , in an expectation of driving large banks , will have quite a high absorption current limit. That limit is often greater then the manufacturer specs as well.

Dave