# Latching relay for battery charging

#### 250ptm

Joined Jul 29, 2010
50

I would like to isolate the charger from the battery by a relay.

If I use a 12v 4 pin automotive relay configured to self latch once voltage is applied (by jumping the 12v input on pin 85 to also feed pin 30). Can the relay coil sustain the 6 amp load to the battery or will that overheat or burn the coil ??

Source of power is a 12v 6A battery charger. Load is the 12v battery being charged.

#### crutschow

Joined Mar 14, 2008
27,243
Can the relay coil sustain the 6 amp load to the battery or will that overheat or burn the coil ??
The relay coil just takes the current it needs (as determined by the coil resistance) which has nothing to do with the 6A load.
The 6A is carried by the relay contacts.

But I see no purpose for the relay, as it's always closed when the battery is connected, no different then if the battery were connected directly to the charger.
How will this relay "isolate" the charger from the battery?

#### 250ptm

Joined Jul 29, 2010
50

To clarify, my charger is connected to mains power through a programmable timer, even when the timer shuts off power to the charger, it is still physically connected to the battery terminals. My charger indicator led's draw approx 30mA whenever it is connected to the battery terminals, even without mains power.

My idea was to isolate the charger LED draw while it is still connected to the battery , so once the timer shuts down mains power, the 12v relay opens and isolates the charger parasitic draw. I have also added a diode between terminals 85 and 30. Since the relay is removeable , it allows me to use the charger normally without a mains timer.

The thing i was not clear about is whether the coil would be affected by the 6A draw from the battery. You have clarified that for me, Thanks again.

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#### kennybobby

Joined Mar 22, 2019
75
What does the diode do for you? What is its current rating?

#### crutschow

Joined Mar 14, 2008
27,243
My idea was to isolate the charger LED draw while it is still connected to the battery , so once the timer shuts down mains power, the 12v relay opens
That will not occur in the circuit you show, as the relay will stay pulled in from the battery voltage.

To have the battery disconnected, you could power the relay from the timer's main's voltage output.
You could use an AC mains powered relay, or use a 12Vdc wallwort connected to the timer output to control the relay.

#### 250ptm

Joined Jul 29, 2010
50
hi crutschow/kennybobby. yeah, my diagram doesn't show a diode, but there is one fitted fitted between pin 85 and 30. It is rated at 50v 10amp. When the timer shuts down power to the charger, it prevents the battery from feeding the charger LEDs and also opens the relay contacts. I have bench tested this and it works just as I described. Do you think that the size of diode i used will be OK? I plan on charging the battery for about 3 or 4 hours once a week.

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#### crutschow

Joined Mar 14, 2008
27,243
The forward voltage drop of that diode will reduce the charge voltage by about 0.65V and likely prevent the battery from fully charging.
What diode are you using?

#### 250ptm

Joined Jul 29, 2010
50
Diode P/N is 10SQ050 listed as a Schottky.

#### crutschow

Joined Mar 14, 2008
27,243
Diode P/N is 10SQ050 listed as a Schottky.
So the drop will be about 0.4-0.5V
You'll need to determine if that will keep the battery from fully charging.

#### 250ptm

Joined Jul 29, 2010
50
So the drop will be about 0.4-0.5V
You'll need to determine if that will keep the battery from fully charging.
According to the charger manual its output is 13.7v to 14.8v at max 6A. It maintains the battery at 95% of its fully charged state in float mode with a 2A current. It looks as though it may work if its only losing 0.5V, i'll do some more testing to see. Thanks for your input.

#### LowQCab

Joined Nov 6, 2012
597
The Charging Current, ( which dictates how long it will take to charge the Battery ),
is determined by the Voltage that reaches the Battery.
If you install a Voltage-Drop, ( the Diode ), you will, at the very least,
extend the time required to charge the Battery, and,
the Battery may never reach a Fully Charged State.

To achieve a Fully Charged State,
the Battery must be Charged with a minimum of 13.7 Volts, ( for Lead/Acid Batteries ).

Unless you have an independent, isolated, power source for the Relay Coil,
which turns-off when AC Power stops,
you are stuck with a Diode Voltage Drop,
unless you want to go to the complications of using an Op-Amp, and/or, a Current-Sensing Chip.
As electronic Circuits go, these are very simple, but I'm guessing that you
won't want to go to that much complexity with this set-up.

Which is worse ??
Under-Charging the Battery because of a Diode-Drop, ( which could shorten the life of a Lead-Acid Battery ),
or,
gradually loosing some Charge during the period when the AC Power is off ??

Why is your Battery Charger connected to a Timer anyway ??????

Lead-Acid Batteries "like" to be kept in a Fully-Charged-State, ( 13.5V to 13.8V ), ( Li-Po's do not ) .

If you have an "Automatic" Battery Charger it will only draw a minuscule amount of
Power after the Battery achieves a Fully Charged State,
and can be left permanently-plugged-in, whether the Battery is connected or not.

What problem are You trying to solve ?
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#### Just Another Sparky

Joined Dec 8, 2019
210
K.I.S.S.

Just use a relay with a line voltage coil. Power it from the same circuit as the battery charger. Charger loses power, relay drops out. Power goes back on, relay pulls in. Furnish a fuse and enclosure. Or just power your existing relay off of a wall wart, lose the diode and call it a day.

#### 250ptm

Joined Jul 29, 2010
50
Hi Lowqcab:

My car is used very infrequently; it has a constant parasitic draw of 30mA which is as per OEM specification. This constant draw takes approx. 12 days to reach 12.4v from a full charge. I dont want it to go below 12.4 that's when sulphidation start to occur.

If use a digital wall timer to begin a charge once a week, the battery is fully charged in 3 hours. It’s basically similar to driving the car for a few hours then leaving it. Why would I leave the charger on any longer, its simply using power for no gain.

As I said in an earlier post, even when the charger is disconnected from the mains supply, it continues to draw power from the battery to show status lights on the charger. That’s another 30ma draw on the one I have already, which simply drains the battery faster.

Hence my quest for a means to stop power being drawn from the battery by the charger after a charging cycle has taken place.

Your idea of an op amp sounds interesting; can you expand on that idea?

To : Just another sparky: Can you give a part number for a line voltage relay?

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#### LowQCab

Joined Nov 6, 2012
597
Do the Math and then decide what's wasteful and how much you are "saving".

If your Electrical Rates are really high, you might be paying ~$0.20 cents per Kilo-Watt-Hour, (Kwh). If your Charger stays plugged-in 24/7, that's 24hrs X 30days = 720hrs per Month. Your Charger, if it is a modern Automatic Charger/Maintainer, will consume less than ~5-Watts of Power when not charging a low battery. One Kilo-Watt divided by 5 = 200, so it would take 200 hours of plugged-in stand-by-time to equal one Kilo-Watt, 720hrs divided by 200 = 3.6 X 20-cents per Kwh = 72-cents per Month to leave it plugged in 24/7. At ~$0.72 cents per month, it's going to take you a full YEAR to pay for that ~$12.oo Relay. The numbers don't support your intentions. If you turn off the Charger after the Battery is charged, then let the Battery get low before you turn on the Charger, then the Charger has to work at full-output for ~3-hours, using 150 Watts of Power, and running in a much less efficient mode than simply trickle-charging. Now you've used almost half a Kilo-Watt-Hour for each weekly charging session, adding up to ~2-Kwh per Month, as opposed to 3.6-Kwh, so you are actually "saving" 1.6-Kwh, or 1.6 X 20 = 32-cents EVERY MONTH !!!!!! What if your Car Battery only lasts 2.5 years, instead of 5 ?? That's half the cost of a new Battery ....... lets say a new Battery costs ~$100.oo,
~$100.oo divided by 2.5-years = ~$40.oo,
That's ~$40.oo per year that you might be loosing if you don't take excellent care of your Battery, meaning, keeping it on a Battery Maintainer, 24/7. And you're worried about 72 cents a Month ??? A single, old-school, 100 Watt Light Bulb costs ~20 times more to run, think about it. If you really want to save some Money, and improve your Life at the same time, throw away your TV Set, that's ~150-Watts for ~3 or more hours every day, and it's just a Propaganda Box that gives you severe Brain-Damage. . . . #### kennybobby Joined Mar 22, 2019 75 Bought a new lead acid battery lately? More like$200, the price has tripled over a few years.

i have tested and reverse-engineered numerous "smart" tender devices. Some overcharged at 15.4V, some undercharged at 12.8V, but the best one i found was a 1 Amp version made by Yuasa, which happens to make a large number of OEM (motorcycle and car) lead acid batteries. It won't charge a dead battery quickly, but will help desulfate and maintain a good battery safely.

i have no financial or business interest in them, just used the product and found it works well. live and learn and pass it on.

#### Just Another Sparky

Joined Dec 8, 2019
210
Can you give a part number for a line voltage relay?
Lol, take your pick. Pretty much any industrial relay off the shelf in North America will have either a 24VDC or 120VAC coil. Omron, Square D, etc. all make great 8-pin and 11-pin 'ice cube' relays. Check surplus stores if you're just tinkering.

The strong argument I can see in favor of an arrangement like this is to prevent overcharging of the battery cells. Modern battery chargers and maintainers often seem to do more harm than good by pushing way too much voltage all the time. Caught my newfangled digital one putting out 15.2 volts. Promptly relegated it to the closet of shattered dreams and picked up a nice old transformer-rectifier number to replace it - which produces a much more reasonable 13.6 volts.

Stupid question for you - why not just let the charger cycle daily without disconnecting it? The losses incurred by the car's ECU and the charger's LED equate to pretty much nothing out of pocket. We're talking a couple of cents annually here. The coil in a relay is going to waste more energy than you'll ever save.

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#### LowQCab

Joined Nov 6, 2012
597
Here is an excellent, and extensive, document on Charging Lead-Acid Batteries,
it's aimed towards smaller "Sealed" Lead-Acid Batteries, but the principles are virtually identical.

And, if you want to spend some time and money on a project,
here's how you can build your own Battery Charger .........
This Circuit is geared towards accuracy and control, rather than maximum efficiency.
Maximum overall efficiency requires a "Switching" Topology,
rather than a "Linear" Topology, as used here.
A Switching Regulator is much more complex.
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#### 250ptm

Joined Jul 29, 2010
50
LowQcab/kennybobby/just another sparky:

Propaganda! Depends on what channel you watch (lol)

My unit is a brand new 3 stage charger. It consumes 0.3A just for the LED’s when its connected to a battery with no mains power applied. It uses 2A to power itself and maintain a float charge to the battery, and up to 6A in full charge mode.

If have this connected to the battery 24/7/365 just for float charging I figure it’s ~ $42.00 p.a. as opposed to a single 3hr 6A charge once per week, for a total of$2.24 p.a

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#### Just Another Sparky

Joined Dec 8, 2019
210
0.3A * 12V = 3.6 watts.
3.6 watts * 168 hours = 604.8 watt-hours.

6A * 120V = 720 volt-amps.
720VA * 3 hours = 2160 VAh.

2160 * 4 = 8640 VAh. Assuming that is at unity power factor (it is most likely not, meaning you will be billed for less on a residential service), that equates to a monthly charge of $1.38 at an upper average utility rate of 16¢/kWh. That's a bottle of soda once a month. Figure 66% power factor based on a 2A no-load current and the charge drops to 91¢ per month. I'd be curious to see just what sort of power factor your charger exhibits in it's floating state. You might be pleasantly surprised to find that only 10-20% of those 2A are real power. If you're really dissipating 240 watts to float charge the battery, either the electrolyte is boiling off or the charger is getting red hot. Either way if you're truly that concerned about saving money and/or protecting the environment, a battery disconnect would be the way to go. Last edited: #### LowQCab Joined Nov 6, 2012 597 LowQcab/kennybobby/just another sparky: Thanks for the schematics and all the links, & your comments are much appreciated! Propaganda! Depends on what channel you watch (lol) My unit is a brand new 3 stage charger. It consumes 0.3A just for the LED’s when its connected to a battery with no mains power applied. It uses 2A to power itself and maintain a float charge to the battery, and up to 6A in full charge mode. If have this connected to the battery 24/7/365 just for float charging I figure it’s ~$42.00 p.a. as opposed to a single 3hr 6A charge once per week, for a total of \$2.24 p.a
Are you measuring "2-Amps" at ~12-Volts ??, I could believe that ........
But 2-Amps @ 120-Volts is 200-Watts, SOMETHING is getting REALLY HOT !!!!
And if you have 240-V Outlet Power, that's 400-Watts !!!!!
You would not be able to hold your hand on the Charger,
you would get your fingers burned immediately.

I think you need to re-evaluate how you are measuring things.
And those LEDs, are consuming 0.030-Amps, (30ma), "maybe", not 0.3-Amps,
that would be a 4-Watt LED, which would blind you if you stared at for too long.

The Circuit that I provided above pulls about ~10ma @ 120-V,
when connected to a fully charged Battery,
that's 120V X 0.01A = ~1.2-Watts,
and that's mostly Transformer-Idling-Current,
and it is a very inefficient, "Sticks and Rocks", crude design, with large Heat-Sinks.
Modern Electronic devices are designed to be efficient,
if for no other reason than,
it costs more money to dissipate more Wasted-Power/Heat,
generated by a less efficient circuit design,
so they actually save on production costs by making the device more power-efficient.
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