dual batteries

Hi
I'm not sure exactly to what you are referring to as ""Mechanical" solenoids.,could you please explain.?
Also where / how they give voltage drop & why they will give less charge than an "electronic" system.?

Oldtrack,
A mechanical solenoid is "mechanical" ie; stated very simply it is a lump of material inside a cylinder, held in position by a spring, when electrically charged, 12v electricity _Affordable_Storage_Drawers.aspx this lump of material against the preasure of the spring up to form a bridge between two silver posts, therefore forming the electrical circuit between the two batteries.

The following posts were written in only ONE thread confirming the Voltage LOSS of a Redarc "mechanical Solenoid"

Submitted: Tuesday, Mar 24, 2009 at 21:06
greybeard posted:
Decided to change out my battery isolator with a Redarc smartstart.
I'd point out that it is operating as per the specification, apart from the current drain of the actuated solenoid being around the 0.8A mark …..
In that time the smart start has used up ~ 10Ah of battery capacity.

AnswerID: 356014   Submitted: Wednesday, Mar 25, 2009 at 03:57
Member - JohnR (Vic) replied:
Redarc will take 0.6 amp every hour it is open on it's own too. So will any other solenoid while it is open, not just your Redarc. They can take a potential 15-17 amp hours a day of battery capacity.

FollowupID: 626149   Submitted: Saturday, Apr 04, 2009 at 14:49
ctaplin posted:
There is also the 0.8 amp drain of the Redarc's solenoid coil wasting your precious battery power for no good reason.

The above 3 posts were by 3 different guys in the same thread, none were posted by myself.

Maîneÿ . . .

Hi Mainy

I agree with your description of a mechanical relay.
However there seems to be a problem with the differance between voltage loss & amphrs loss. The manual relay cetainly use a higher current [ amps] than a solid state[electronic]
But there is no,or very little voltage loss across contacts of a mechanical relay unless its contacts have become burnt.
In fact a fully electronic relay will more likely have a higher voltage drop between input & output voltages [around 0.25v] due to the loss in the switching components

Mainey,
most people refer to "voltage drop" as being the voltage difference between the two positive terminals when a significant current is travelling through the cable. Voltage drop is proportional to the current flowing through the cable.

The posts you quote refer to the power consumption of Redarc isolators - their coils have a parasitic loss which is a different issue.

Phil,
You say: "The posts you quote refer to the power consumption of Redarc isolators - their coils have a parasitic loss which is a different issue"

Yes, do you believe the "parasitic loss" in the Redarc Isolator does not causes any Voltage drop?

Maîneÿ . . .

"Mainey,
most people refer to "voltage drop" as being the voltage difference between the two positive terminals when a significant current is travelling through the cable. Voltage drop is proportional to the current flowing through the cable. "

Voltage drop can be measured anywhere, in a length of cable or in this case in a Redarc battery Isolator.

The posts I used above confirm there WILL be voltage drop in the Redarc isolator, caused by the Isolator, that is the reason the three guys originally made their posts, they are not my words but theirs.

If less Amps are flowing then it is taking longer to charge if it will charge at all.

Maîneÿ . . .

Mainey - you are using incorrect information.
There is virtually no voltdrop accross the contacts of a mechanical soleniod in good condition (correctly sized for its application).
When the soleniod is open it is using next of nothing.
If you applied your reasoning - once the isolator opened and the batteries were isolated from one another - the auxillary battery would be flattened in a few days - I have a Redarc and this does not happen.
Regards Dennis

Dennis,
I notice you use the terminology *virtually* no voltdrop however are you prepared to actually state there is *NO* (NIL or ZERO) Voltage drop at all, somehow I don't think so.
Because all I've done is copy posts from three other forum users who (as anyone can read) have ALL stated, similar to, they have "current drain of the actuated solenoid being around the 0.8A mark ….. " these are the exact words Greybeard used and also the sentiment of the other two guys who have posted of their *stolen* current by their own mechanical solenoid.


Dennis, you say: "When the solenoid is open it is using next of nothing"
that is the exact point of the three guys I've quoted, yes their mechanical solenoid battery isolator is "using next of nothing" however it IS using, or "stealing" as one poster has said, *something* and that 'something' is actually draining electrical current that should be used to charge the Accessory battery much faster and more fully.

The “mechanical solenoid” either uses 'no' or 'some' current, in this case, as you have said, it does use (steals) 'some' – exactly as the three guys I quoted have posted !!

Maîneÿ . . .

Mainey - You get a volt drop accross any electrical component because there is no such thing as a perfect conductor. Next to nothing or virtually no voltdrop means exactly that - people shouldn't get thier nuts into a nut about a very small effect that makes no practical difference to thier battery performance.
Regards Dennis.

Dennis,
(a) do you believe having the loss's Greybeard and the other two posters have is acceptable ?

(b) do you believe Elontronic Isolators have a much lower loss (as a %) than Mechanical Isolators ?

Maîneÿ . . .

Mainey this could go on for ever - so I will finish with this comment.
If comparing similar rated isolators; the solid state device will have a larger volt drop than the mechanical device.
Regards Dennis

Dennis,
your entitled to your opinion, however the information posted on the thread nominated contradicts your opinion

Maîneÿ . . .

Mainey, I don't think you understand what Dennis and I have said.

#1 "You get a volt drop across any electrical component". The fact that you say you have "zero drop" in your system means you don't know how to measure voltage drop, or don't have a voltmeter with sufficient resolution.

#2 "Voltage drop" is not the same as what you have posted. The posts you quote refer to the loss of power by the coil on a Redarc isolator which continues after you turn off the motor and until the starting battery drops to 12.5V. This is just wasted power. It is not "voltage drop" as we know it.

Phil,
#1 does this picture clarify your sufficient resolution concern?
Image Could Not Be Found
So obviously you believe I "don't know how to measure voltage drop"
somehow I don't believe that statement is correct either, but photographically difficult to show on here.

#2 if you read the thread it was copied from, it raises the Voltage drop scenario, I did not copy the entire thread only relevant information as to the *reason* why the Voltage is less than it should be.

Remember I have stated: "The following posts were written in only ONE thread confirming the Voltage LOSS of a Redarc "mechanical Solenoid" as a heading.
Yes, the guys quoted have nominated their loss as Amps created by their Redarc battery isolators.

I'm sure you will agree any loss of Amps will cause the battery not to be *efficiently* charged, so in fact it may hold less Voltage than a fully charged battery.

Maîneÿ . . .

Mainey - You are getting your electrical terminology all mixed up - transposing amps, voltage and voltdrop, willy nilly, shows a bit of confusion.
The statement - "Yes, the guys quoted have nominated their loss as Amps created by their Redarc battery isolators". Shows that your source of information is just as confusing.
Regards Dennis

Dennis,
I did not change any of their electrical terminology because I believe to change any of their wording would be seen as discrediting their relevant information.
I only quoted it here to show that there are guys out there that do have LOSS'S that they say are clearly caused by the Mechanical Battery Isolator.

I understand & believe Greybeard's post is clear:

"Submitted: Tuesday, Mar 24, 2009 at 21:06
greybeard posted: Decided to change out my battery isolator with a Redarc smartstart.
I'd point out that it is operating as per the specification, apart from the current drain of the actuated solenoid being around the 0.8A mark …..
In that time the smart start has used up ~ 10Ah of battery capacity"

Greybeard is telling us he is loosing 0.8 Amp, he is saying it is caused by his Redarc SmartStart Mechanical Battery Isolator.

You don't need to have an English teachers degree or be a scientist to understand the information Greybeard has posted as his message is simple and clear!

Maîneÿ . . .

Hi Mainy

I hate to sat this BUT you are a very confused man with your electrical terminology.
Voltage loss / voltage drop is the volts lost due to resistance in a cable or contacts WHEN current flows.
I am not sure what you are trying to prove with your pix, yes you do get a voltage ACROSS a battery or between pos & neg in a circuit,THAT is not the same as voltage drop / loss thro the length of cable or across contacts.
Any elecrical device in operation will consume power to operate.
This includes mechanical relays & solid state / electronic relays'
AS I posted before a mech relay will have very little VOLTAGE LOSS /DROP across its CONTACTS unless they are burnt.
IT will draw a small current .
A solid state relay will have a larger but still small VOLTAGE DROP drop across it's switching components which is why many have a heat sink / radiator surface & it too will draw a current but will be lower.
The posts you refer have all used the term amps correctly .
It is you who seem to have bought the term volts loss into the thread.

Hi
If any one is realy concerned about the amphrs lost from the battery when engine not running {IE in standby }, A simple solution is to switch the earth connection of the Redarc via an ignition switch controlled relay , this will over come the problem of loss of AMPHRS.

i Mainy posted "I'm sure you will agree any loss of Amps will cause the battery not to be *efficiently* charged, so in fact it may hold less Voltage than a fully charged battery."

Maîneÿ
Here again you seem to have your terminolgy confused , a battery holds amphrs that is how they are rated. I guess what you are trying to say is that THe more AMPS you can put into a battery the quicker it will charge & that is correct but it is AMPS .
However with engine running 600<800milliamps is virtually nothing .less than 1amp compared to the alternator output of 40<,80 amps depending on size.

PLEASE GET YOUR TERMINOLGY RIGHT you have had this problem in some early threads.

OT, you say,
"Voltage loss / voltage drop is the volts lost due to resistance in a cable or contacts WHEN current flows"
Are you suggesting in a 12v battery, when checking battery Voltage OR checking Amps there is NO current flow ??

It is the flow of current that shows the Voltage or Amps numbers on the meter. If no current flow then no result.

Yes I posted: "I'm sure you will agree any loss of Amps will cause the battery not to be *efficiently* charged, so in fact it may hold less Voltage than a fully charged battery."
Yes, a battery can have less Voltage, ie when not fully charged.
I think a fully charged battery has or "holds" or contains more Voltage than a half charged battery.
Example: 12.1v in a patrially charged battery & 12.6v in a fully charged battery, one battery has or holds or should I say contains more Voltage than the other.

Maîneÿ . . .

Maîneÿ . . . posted:
#1"Are you suggesting in a 12v battery, when checking battery Voltage OR checking Amps there is NO current flow ?? "

#1 answer This is where I find it hard to see your reasoning,The current that digital votmeter uses to read voltage is in microamps & imposes no load on the battery, THe amp readings of course are only available when current flows, & i have not suggested other wise.

#2Mainy said"It is the flow of current that shows the Voltage or Amps numbers on the meter. If no current flow then no result. "

#2 Answer Same as above voltage reading do not require flow except for very small microamps to operate meter.
VOLTAGE DROP DUE TO RESISTANCE LOSSES are a differant matter & do require current to flow thro the LENGTH of cable OR acros the contacts of a device being checked

#3 Mainy said" "I'm sure you will agree any loss of Amps will cause the battery not to be *efficiently* charged, so in fact it may hold less Voltage than a fully charged battery.
Yes, a battery can have less Voltage, ie when not fully charged.
I think a fully charged battery has or "holds" or contains more Voltage than a half charged battery.
Example: 12.1v in a patrially charged battery & 12.6v in a fully charged battery, one battery has or holds or should I say contains more Voltage than the other.
Maîneÿ . . .''

#3 Answer This is where your terminology gets confusing, yes amps are what charges a battery & it needs volts to push it & yes voltage does give an indication of it's state of charge, BUT a battery is not considered to hold volts all calculations are based on amphrs of capacity.
A point to show this is a sick 12v battery can be easily charged up untill it reads 14.5+v & although it may be rated as 100amphr it's actual available amphrs could be down to almost zero.
This is why it is so importent to ensure that CORRECT terms are used .
VOLTS are volts [electrical pressure]
AMPS are amps [rate of current flow]
Amphrs is the storage capacity of a battery[ like gallons of water in a tank.]

Gday Mainey,
I admire your persistence!!
Do a google on "potential difference" or "Ohms Law". Include the inverted commas and you might rediscover what you missed in school physics.
This thread reminds me of the Overlander Forum!

Cheers
Phil

Yes, It's simply turned into a pedantic 'play on words'
it's not that the 12v battery is ½ empty, it's probably ½ full !
When your labelled 'intoxicated' it's said your ?issed or ' full '
Now the terminology of a ' full ' moon may be incorrect too !


Why not put your collective knowledge to good use, as no one has explained why we should even consider using a Mechanical battery Isolator that we are fully aware does waste our 12v battery power !!
Maybe that should be: waste Amps, waste Voltage or maybe even waste Current ??

Particularly when there are more efficient choices available !

Maîneÿ . . .

Mainey,

If you want to be technical, you need to use the right words. If you want to have a chat around the campfire, you can say anything you like.

You clearly are not interested in learning from the many people on this forum who have the patience to answer this thread and try to stop you using the wrong words. We're not having a go at you - just trying to stop you repeatedly using the wrong terminology which can get perpetuated on forums.

To say "waste Amps, waste Voltage or maybe even waste Current ??" is yet another of your classic statements.

You cannot waste Voltage!!!!!!! Volts are units of "potential difference". So how do you waste something that is "potential".

And amps and current are the same thing. Maybe you really wanted to say it was losing Charge or Amphours????

Phil,
I don't want to be seen as a technical person, that's the only reason I posted:
" Maybe that should be: waste Amps, waste Voltage or maybe even waste Current ?? "

See the question marks (??)
It's only a question - it's NOT a statement, it's not meant to be factual, it IS a question starting with the word "Maybe" !!

Political correctness gone to the extreme ( or is that to the Delkor? )

# Wondering why we should consider using a Mechanical battery Isolator ??

( # yes, the above is a question & still waiting for an answer )

Maîneÿ . . .

Mainey,
I'll answer your question:
"# Wondering why we should consider using a Mechanical battery Isolator ??"

Firstly, it is unfair to call the Redarc a "Mechanical Battery Isolator". It is a solenoid (electrically powered switch) that is controlled by an electrical circuit.
A boat switch or cutoff switch are purely mechanical, and is what 99% of forum readers think of when you use that term, so you may want to stop using that term when referring to a Redarc.

I'll assume you mean the Redarc and we'll compare it to a Rotronics MH10 solid state isolator that uses FETs to switch current.

In favour of the Redarc:
#1 Cost. The Redarc costs about $120. A Rotronics MH10) costs about $300.
#2 The Redarc Isolator offers a battery override switch - so in the event of you leaving your lights on and flattening your starting battery, you can simply press a button and start from the aux battery. AFAIK, none of the mainstream electronic isolators can handle the starter current so don't offer this option.
#3 Some people have had bad experiences with failures of the expensive electronic isolators - especially if they do water crossings.

Against the Redarc:
#1 Solenoids can eventually make poor contact
#2 It doesn't isolate the batteries until the starting battery drops to 12.5V. The coil remains energised so consumes quite a lot of power which is unnecessary. This power loss can be overcome by fitting a small relay in the Redarc's earth lead, so its a simple problem to fix.

The advantage of keeping the batteries paralleled until they both drop to 12.5, is that you'll be utilising maybe 10Ah of capacity from the starting battery when camping, so reducing the draw down on the aux battery. The disadvantage is that when not camping, you are wasting power, your starting battery will sit on 12.5V, which will reduce its life because it is more likely to sulphate.

I don't use a Redarc, but I would be happy to use one, but would fit the relay to the Redarc's earth to prevent the power loss problem.


AND, While we're at it............
"Voltage Drop":
In reference to dual battery systems: All conductors (wires, connectors, relays, fuses, earth leads etc etc) have resistance. We minimise resistance by fitting fatter cables and quality connectors. We can never reduce resistance to zero, so your claims of zero voltage drop are impossible.

To measure "Voltage Drop" you need:
#1 A digital multimeter set on the 2V scale (not 12V). One lead goes on main battery pos terminal and the other lead goes on the aux battery pos terminal. (Don't put either lead on earth).
#2 A high current ammeter in line between the two batteries. DC Clamp ammeter is ideal.
#3 The auxillary battery needs to be discharged to say 50%, and nothing powered off it.

Start motor and have it run for for a few minutes so the glow circuit is off. Idle it at 2000+rpm so the alternator has good output. Measure the voltage difference (drop) between the two pos terminals and at the same time, note how many amps are going into the auxillary battery. The aux voltage is always lower than the main voltage when current is flowing into the aux. If you have 30-40 amps going into a aux battery, it indicates the battery is fairly well discharged, is in good condition and the alternator output is good. The "voltage drop" for a well designed and built system in this situation may be 0.5V. If the aux battery is fully charged, the current will be less than one amp and the voltage drop will be small - maybe 0.02 volt. Voltage drop always have to considered with how much current is flowing at the time.

So I use "voltage drop" to tell me how much current is going into my aux battery and hence, how well recharged the aux battery is. My vehicle has a voltage sensing lead at each pos terminal and use a dash mounted digital voltmeter via an on/off/on switch to tell me what the difference is.

Cheers
Phil

Rod,
the Manual systems are not for everyone, the driver has to be very aware of the battery voltage of the Accessory battery system and turn on the Manual switch system when your driving.

The biggest mistake is made by those who may *forget* to turn the Manual switch OFF again when stopped and the fridge will then drain ALL the vehicle batteries, possibly leaving the vehicle stranded and unable to start.

Maîneÿ . . .

Yep I agree Mainey... but they'll only do it once.

I agree with Rod - you'd only do it once. You'd have to be a dill to flatten a pair of N70 batteries - the compressor fridges pull relatively little power now and you need a voltmeter. I wouldn't spend much money on an isolator for a 60series - can't have the 2nd battery system costing more than the vehicle :-))

well Phil it might only be a 60 series but with all the money i saved i could afford a winch and good spotties which i see is sadly lacking from your piece of ex miner bleep e :)

Sorry Frank, I was being too flippant after a couple of reds!
But I have a winch and spotties for sale :-))