12V 10Ah SLA Charging problem

lbenson

Senior Member
I'm looking for a simple charger for a 12V backup SLA, say 10Ah, from mains (e.g. 18V laptop charger)--LM317 with appropriate components seems simple enough, as in this 6V charger thread: http://www.picaxeforum.co.uk/showthread.php?27999-6-VOLT-1Ah-SLA-Charging-problem

The conditions of use are as follows:

1 criticality of operation: not critical, merely desirable.
2 frequency of use probably much less than .5% of the time, maybe 2 days per year, usually in smaller chunks.
3 duration of use: minutes to days, shutoff when < 11.9V (by picaxe, not shown in circuit below).
4 safety of charger to prevent damage to battery: important
5 efficiency of charger: doesn't really matter (within reason)
6 need to be fully topped off at end of charge: not very important

The devices powered are picaxe 14M2, cable modem (internet), wireless router (the latter two have 12V 2 amp and 1 amp wall warts respectively). The most important thing the picaxe will do once mains go off is send an email that it has, but it will also monitor temperatures and door openings in the vacant house and send email updates.

I have used a UPS in the past, but think that an inefficient way to go (as far as runtime for a given 12V SLA goes). The UPS battery died, so I figured this would be a good time to start this project.

Here is a thread which has the following LM317 charging circuit: http://www.circuitstoday.com/battery-charger-circuit-using-lm317

Battery Charger LM317.jpg

In the comments, the author says this circuit is really for 6V charging, but changing the value of R4 to 1K2 would make it work for 12V (Note in image to that effect added by me). Also, I don't have for R1 a .56 Ohm resistor, 1 Watt or 5 Watt, for current limiting, but think I have a 1 Ohm 5 Watt resistor, which would allow less current.

Would this circuit with the change to R4 and R1 be suitable for my purposes? (The circuit by which the picaxe switches from mains to battery is a different matter--I also want to be able to power-cycle modem and router since they occasionally lock up).
 
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techElder

Well-known member
Some things come to mind after reading your post.

My experience for "long life" of SLA batteries used in "float" projects is to have the most minimal "trickle" charge current you can maintain. I've added 1M resistors in series with chargers before to reduce the "trickle" current in "standby" mode. (That's probably why your UPS battery "died".)

What is the advantage of 12V batteries over 6V batteries in your project?

How will you "shutoff" the project when the battery crosses the 11.9V threshold?
 

lbenson

Senior Member
Thanks for the advice re 1M series resistor in trickle mode.

The advantage of the 12V battery is that it will be cutting in for 12V wall warts for the modem & router when mains drops out. And I have a bunch.

>How will you "shutoff" the project when the battery crosses the 11.9V threshold?

I actually won't shut it off, so the picaxe will still run looking for mains to return, but will turn off the modem and router by switching them back to (the absent) mains power with latching relays. Considering the at-present nearly 8-year life of a picaxe on 3 AAs, I would not expect the picaxe to further drain the SLA to a significant degree. If that seems a danger, I could shut off the devices at a higher voltage.

(I realize that if the modem and router really are drawing 2A & 1A respectively, the runtime I'll get from a 10Ah SLA will be limited--but long enough so that I will get notification of mains dropout. Dropouts in this area are rare, and are usually a matter of hours--but I want to know if it happens in the winter when I may not be around.)
 
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hippy

Technical Support
Staff member
I think the only way to really tell is to build it and test it, or run it through some spice or similar electrical simulation.

I favour constant voltage, current limited charging for SLA and that has worked well for me. This seems to tilt towards being constant current by changing the voltage to control the current. I must admit that I am not sure how it limits voltage and current nor entirely convinced it does.
 

neiltechspec

Senior Member
I have built 4 of these type chargers, Constant Voltage of 13.8v (using cheapo buck SMPSU modules) and current limited to 500mA (using LM317).

2 of them are used on Car Batteries (classic cars), one for a Leisure Battery and one for SLA's.

Something else to consider may be a Solar Charge Controller, generally they float charge at 13.6v and would run off a laptop psu of 18v as the solar panel.

Neil
 

Flenser

Senior Member
lbenson,

What follows is my understanding of how this LM317 two-stage constant-voltage, constant-current SLA charger circuit works plus the steps I would use to design one.

R2, R3 and R5 are used to set the max output voltage. This output voltage is measured across R2, R3 and R5 which is also the voltage across the battery.

The feedback components R1, R4 and Q1 are added to the curcuit for sole the purpose of limiting the maximum current:
  • These components have NO effect on the max ouput voltage of the circuit.
  • When there is current going though Q1 these components operate to REDUCE the ouput voltage.
  • When there is no current going through Q1 these components have NO effect on the ouput voltage of the circuit.
The LM317 operates to keep a nominal 1.2V between the output and adjust pins. Thats it! This is all it does.
(NOTE: This value is specified as 1.25V in some datasheets. These two values are close enough that the difference (4%) is not important for explaining the design process. If you have a chip specified as 1.25V than substitute that value into your calculations.)

So the design process for one of these chargers would be:

1) Choose how much current we want to pass through R3.

The datasheet specifies a value for the Minimum Load Current of the LM317 as 3.5mA (typ) and 5mA (Max).

The LM317 operates to keep 1.2V between the output and adjust pins so if R3 is 120 Ohm then the minimum output current will be 1.2/120 = 10mA. This is greater than the Max value from the datasheet so it looks OK.

I have seen some example circuits in datasheets that have a 240 Ohm resistor for R3 which would be a 5mA current through R3, also OK.

There is actually some extra current coming out of the adjust pin. I have seen this specified as 50uA and as 100uA. Either way this is so low compared with our design current (1% - 2%) that we will ignore it in our calculations as well.
2) Choose R2, R3 and R5 to set the max output voltage for our charger.

In the thread "6 VOLT 1Ah SLA Charging problem" BeanieBots makes the comment "I float my AGM lead acid batteries at 13.8v and they last > 5 years when kept at normal domestic temperatures". A 12V SLA battery has 6 x 2V cells and a 6V SLA battery has 3 x 2V cells so we'll use 6.9v as the max output voltage for this 6V charger example.

Resistor Voltage@10mA
R3 (120 Ohm) 1.2V
R2 (470 Ohm) 4.7V
====
Total 5.9V

Max Voltage 6.9V
R5 1.0V

So our design value for R5 would be 100 Ohm to give a voltage of 1V@10mA. A 220 Ohm pot for R5 seems a good choice as the fine adjustment around the 100 Ohm point should not be too fiddly.

For a 12V charger with a Max Voltage of 13.8V we need to drop 7.9V across R5 at 10 mA so R5 would need to be 790 Ohm. This leads me to guess that the comment at the bottom of this circuit has a typo and may have been intended to read "change R5 to 1K2 for 12V" (i.e. change R5, NOT R4).​

3) Choose R1, R4 and Q1 to set a max current limit for our charger.

The max current we want the circuit to limit the charging to is a design choice that we need to make.

In the "6 VOLT 1Ah SLA Charging problem" thread BeanieBots makes the comment "NEVER charge at > C/10 " so we'll use that for this design example. For a 1Ah battery our max charging current at C/10 is 100 mA.

We will use a nominal 0.6V for the threshold voltage that will turn on Q1 so by using V=I*R we can calculate R1 = 0.6V/100 mA = 6 Ohms.

The 0.56 Ohm value for R1 in the circuit shows that it was designed for a max charging current of I = 0.6V/0.56 Ohm = 1.07 Amps.

Resistors are not just specified with their resistance value. They also come with a power rating so we need to check how much power is being generated in R1. Power = V*I = 0.6V * 100mA = 60 mW so any 0.125W or 0.25W resistor should work OK.

R4 is used to limit the current into the base of Q1 and I don't know how the value for this resistor is calculated. As we always choose R1 so that the base of Q1 is 0.6 V above the emitter I expect that we can probably always use a value of 100 Ohms for R4.

The max current through Q1 is 100mA. The max base-emitter voltage is about 0.6V. The max collector-emitter voltage is the total voltage across R2, R5 and R1, which I calculate as 6.3V for our 6V charger and 13.2V for a 12V charger. Any common small-signal NPN transistor will probably be OK.​

4) Check if we need a heatsink on our LM317 chip.

The LM317 chip is a linear device so the power generated in this chip will be (Vin -Vout)*100mA. Using a 9V supply for this example the power works out to be (9V - 6.9V) * 100mA = 0.21W.

The National Semiconductor Datasheet for the LM317 specifies a rated maximum junction temperature (TJ) of 125C for the LM317 and specifies the calculation for the junction-to-ambient thermal resistance as:
ThetaJA = TJ/Power

So our calculated thermal resistance = 125C/0.21W = 595C/W.

From the National Semiconductor datasheet "If the calculated maximum allowable thermal resistance is higher than the actual package rating, then no additional work is needed." The actual package rating for the TO220 package with no heat sink in the datasheet is 50C/W so no heatsink is needed on the LM317 chip for this 100 mA charger when we use a 9V supply.​
 

techElder

Well-known member
Ibenson, you are making this too difficult for a "float" charging system that uses the battery only sparingly at "rare" dropouts.

I would be tempted to try a two-resistor charging system. Both resistors are in the high side charging line.

One resistor is a high value for a "trickle" charge; very low current charge.

The other resistor is valued for a "maximum" charge that gets the battery to a usable voltage quickly without high-current damage.

The "maximum" charge resistor is switched in/out with a PNP or P-CHAN transistor using the PICAXE monitoring the battery voltage.

Switching on the high-side allows you to use your wall-wart maintaining the current common ground.

The charger is in "float" mode by default.

Again, you described a typical "float" charging system.
 

AllyCat

Senior Member
Hi,

The circuit shown in #1 is definitely "dodgy" and generally I endorse hippy's and flenser's comments.

The reference to changing R4 for 12 volts operation is nonsense, they probably mean R2. But changing R2 to 1k and R5 to 470 ohms would be more consistent with the 6 volt design.

Furthermore, as mentioned by flenser, the current limit is just over 1 Amp, but with 6 volts output and 18 volts input, that puts more than 12 watts dissipation in the LM317. So a very significant heatsink would be required (which is at least mentioned in the supporting text).

I favour constant voltage, current limited charging for SLA and that has worked well for me. This seems to tilt towards being constant current by changing the voltage to control the current.

Indeed, but I do disagree slightly with the second sentence. The circuit is fundamentally a "precision" voltage regulator (the LM317), but if the load current gets too high then the transistor turns on (at only approximately Vbe = 600 mV) to make the LM317 "think" that the output voltage is too high (excess voltage drop across R3) so it reduces the output level.

As for the value of R4: in principle it's not needed at all (i.e. could be zero ohms). However, taking into account Murphy's Law, there's always the possibility that R1 will be incorrectly connected, or fail in use. That potentially allows the full charging current (1A+) to flow into the base which would almost certainly destroy the NPN. But 100 ohms might still allow quite a high current to flow, so I'd choose a rather higher value.

Cheers, Alan.
 

lbenson

Senior Member
Thanks very much to all for the responders, and especially to Flenser for the detailed explanation of the M317, and to Tex for confirming my suspicion that there was an easier way to go considering my specific circumstances.

I have a 15V DC 1.2 Amp wall wart. Would I be ok using that with 2 diodes inline before R2 (or 3, depending on real-life measurement)? And then what value for R2 and what wattage, considering a 10Ah 12V SLA? I have on hand 3, 5, and 10 Ohm 5 Watt resistors.

Note that for this project, the fastest bulk charging is not a requirement.
 
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techElder

Well-known member
Ibenson, your battery is going to determine those parameters. Just pick a reasonable value and start calculating backwards to wattage on resistors, etc.

In a standby application, you need to read the specs for your battery as far as what voltage is best for it to float at. I'm guessing somewhere around 13.8 volts.

And, there are different kinds of SLA batteries. Some are designed to be used in float situations. However, it seems you have a supply of batteries at hand.
 

AllyCat

Senior Member
Hi,

Before you progress too far, I don't think that FET design is going to work unless the FET has a Threshold Voltage just a few volts less than the input voltage (unlikely). IMHO it needs a NPN driver (with pullup resistor) to interface the PICaxe with the FET. You might also want to consider whether R2 should be on the Source or Drain side.

If you have a regulated 15 volt supply, then indeed all you really need is two diodes and a resistor. The resistor value will depend on how deeply the battery may be discharged and how much power can be conveniently dissipated. Even as low as 3 ohms, the peak charging current will be less than an amp (assuming 11 volts), so less than 3 watts dissipation (plus another watt in the diodes). The current will fall to zero at around 13.8 volts, so full charging may take well over a day. That's why an active (LM317) solution is generally preferred.

Then, if you want to trickle charge the battery, a resistor of around 120 ohms (not via the diodes) would give about 10 mA. Even with the battery at 11 volts, it only needs to be rated a little over 100 mW.

Cheers, Alan.
 

techElder

Well-known member
A float current of 10mA will ruin the battery over time.

I only suggested a circuit layout. Charging a battery in a days time is considered fast in some applications.
 

lbenson

Senior Member
I think the latest responses have revealed the source of my hesitation about the resistance and wattage rating of the current limiting resistor: confusion on my part about the voltage to be considered in the Watts equation.

With a 15V DC wall wart and 2 .6V diode drops from 1N4001 diodes, I was thinking I had ~ 13.8 volts relative to 0V with a 5 ohm resistor: 13.8V / 5 ohm = 2.76 amps * 13.8V = 38 Watts = Whoa!

But (as I now understand it) the real voltage in the equation is the charging voltage minus the voltage of the battery, say 12V at a minimum. Then 13.8 - 12 = 1.8 / 5 ohm = .36 Amps * 1.8V = .65 Watts. Now my 5 Ohm, 5 Watt resistor looks good. Or for 3 Ohm, 5 watts: 13.8 - 12 = 1.8 / 3 ohm = .6 Amps * 1.8V = 1.08 Watts

Is this correct?

If so, then as the battery voltage rises, the current drops, e.g.: 13.8 - 12.8 = 1 / 5 ohm = .2 Amps * 1V = .2 Watts.

This also implies that in the LM317 circuit, it isn't magic in the LM317 which causes the current to drop as the battery charges, but the current-limiting resistor.

So is it safe to say that in an LM317 "constant current", "constant voltage" charging circuit, the current is only constant at a given voltage for a particular charge state of the battery? So perhaps, "current limited" might be a better description of this type of charging circuit?

Now to look at Tex's circuit with the picaxe controlling the P-CH power mosfet through a 2n7000 mosfet with "negative" logic (high = bulk charge off, low = bulk charge on).
 
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techElder

Well-known member
Basically, you want to figure wattage of the charge resistor at the maximum voltage in with the minimum voltage out.
 

lbenson

Senior Member
Basically, you want to figure wattage of the charge resistor at the maximum voltage in with the minimum voltage out.
Does this mean, as I have supposed, for the equation, Watts = Volts squared / Ohms, that the "Volts" in question is (maximum voltage in minus minimum voltage out (voltage of the battery to be charged)), and thus for my case of Volts-in = 13.8V (after two diode drops), minimum Volts-out = 12V, Watts = 1.8*1.8 / 5, or .65 Watts?
 

neiltechspec

Senior Member
Interesting document, the key statement for float charging is this :-

The recommended constant float voltage is 2.25-2.30 volts per cell. Maintaining this float voltage will
allow the battery to define its own current level and remain fully charged without having to disconnect
the charger from the battery.
The trickle current for a fully charged battery floating at the recommended charge voltage will typically
hover around the O.OO1C rate (1OmA for a 10AH battery, for example.)


Neil
 

techElder

Well-known member
I realize our dependence on datasheets, but my experience over many years with portable SLA and standby float systems is different from the factory datasheet.

I always adjusted the FLOAT current to some very minimum value. 10mA was always too much.

I used various different AH batteries, so there isn't any one value that I used, but I remember many values in the range of 1 - 2 mA.

allow the battery to define its own current level
Decide on the battery voltage; then play with the FLOAT charge value. There is some current value that will maintain your selected voltage value.

Perhaps it is 10mA, but not in my experience.
 

neiltechspec

Senior Member
How can 10mA be too much ???????????

If you float charge with a constant voltage i.e 13.8v, then the battery WILL define it's own current level.
 

hippy

Technical Support
Staff member
If you float charge with a constant voltage i.e 13.8v, then the battery WILL define it's own current level.
That seems correct, but only for the case where the charge voltage matches the desired battery voltage.

In my experience simple chargers are fixed voltage using a voltage for charging which is higher than the ideal voltage for float charging, and the current limit is being imposed to deal with that. That is; 10mA isn't detrimental even if the voltage is higher than ideal.

If the voltage drops depending on charging or float charging it probably isn't a requirement.
 

lbenson

Senior Member
This is an old thread, but I'd like to return to it with a new wrinkle.

The purpose of the new design is to turn on a 12V LED when mains power goes out. Here is the LED: https://www.ebay.com/itm/48-SMD-COB-LED-T10-4W-12V-White-Light-Car-Interior-Panel-Lights-Dome-Lamp-Bulb/382189873739

The determination that mains has dropped out is based on using CALIBADC to differentiate between voltage supplied by a power plug at 5V and voltage supplied by 3 AA batteries. The potential supplies have in-line diodes so the battery will not be drawn on when mains is available.

Assume that the charging voltage is supplied by a DC-DC boost module, like this here: https://www.ebay.com/itm/DC-DC-2A-Adjustable-Step-Up-Boost-Booster-Power-Conversion-Module-Micro-USB/152892199860

The DC-DC is fed from the 5V mains power plug, and the output is the 13.8V which is the "battery charged" level for the 12V SLA I am using.

When mains dropout is detected, C0 goes high activating the low-side mosfet switch, Q1, turning on the LED.

Here is my proposed circuit.
SLA Battery Charger.jpg
My first question is, is R3 needed if the output of the DC-DC module is the voltage the battery wants to achieve (and assuming that "bulk charging" is not required). If needed, what value would be requred (assume a 7Ah battery)?

My second question is about the monitoring of the battery voltage level so that the LED is turned off if the voltage drops to 11.8 volts. If there is an R3, doesn't the 22K/10K voltage divider for monitoring the voltage level itself become a voltage divider with respect to the DC-DC module if it connects on the right of R3, and a voltage divider with respect to the battery if connected on the left of R3?

Is an R3 needed if mains comes back on after the battery voltage has dropped to 11.8V, and the DC-DC module then is supplying 13.8V at whatever current it is capable of?

I suppose my third question is, does this make any sense at all?

And my fourth question--will the DC-DC boost module work for trickle charging (e.g., supply a very nominal current).

The LED would very rarely be on.

Any outside-the-box suggestions would also be welcome.
 

BESQUEUT

Senior Member
The purpose of the new design is to turn on a 12V LED when mains power goes out.
The determination that mains has dropped out is based on using CALIBADC to differentiate between voltage supplied by a power plug at 5V and voltage supplied by 3 AA batteries. .
Overcomplicated : you can detect mains presence just before D1...
 

hippy

Technical Support
Staff member
Maybe I am missing some context but it seems the original thread was about charging a battery while this seems to be power fail detection and turning on what I presume is emergency lighting. I can't see what purpose the PICAXE serves and it could all be achieved far simpler with a relay.
 

Attachments

lbenson

Senior Member
Overcomplicated : you can detect mains presence just before D1...
I'm not sure why using CALIBADC with a half-dozen lines of code to detect mains presence (5V) versus battery (4.5V--tested to work) is overcomplicated relative to using an additional (scarce) pin. But I'd be happy to simplify.

The 3-AA battery box could be replaced by a 3V3 LDO regulator run off the 12V battery (with adjustment of the voltage divider).

I can't see what purpose the PICAXE serves and it could all be achieved far simpler with a relay.
I like the no-MCU design with N/C relay, but what it lacks is the ability to prevent the battery from discharging below 11.8V. What is the value of R3 (assuming one is content with Tex's suggestion of 1mA trickle charging)?

A bit more complex than just a relay, but a PICAXE and relay together could do this easily
This is what I'm trying to figure out now. Since with the 13.8V DC-DC module I only need to check the battery voltage when mains is out, perhaps the relay would cut in the voltage divider on C4 and be activated by the same C0 pin as the mosfet and LED.

Is it reasonable to suppose that the DC-DC boost module will supply a trickle charge at around 1mA? (That is, does it need a greater current to provide reliable regulation?)

(To further complicate, I also want an LDR on pinC.3--I know I won't be able to use ADC.)
 
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lbenson

Senior Member
Request for comments on another 12V battery circuit.

This is intended to be used with an 08M2, 14M2, or 20M2 or X2. The purpose is to run off of a DC-DC Step Down Power Supply Adjustable Push-button Module with LCD Display like this: https://www.ebay.com/itm/DC-DC-Step-Down-Power-Supply-Adjustable-Push-button-Module-with-LCD-Display/112165057965
DC-DC Step Down Power Supply Adjustable Push-button Module with LCD Display.png
The Step Down module would be fed with a 15V wall power plug and would be set at an output of 13.8 volts. The Schottky diodes would drop about 2V. This would mean that the charging voltage on the battery backup would be 13.6 volts, which is fine with me. (My original load would be a Raspberry Pi Zero-W with camera, which tests indicate will run for about 4 days on my 15Ahr 12V battery.)
12V charger with disconnect.png
The picaxe would monitor the battery and would disconnect the 0V from the load with a low-side mosfet if the voltage drops below 12 volts. It will switch the mosfet back on if the voltage rises to 12.4 volts.

This may not take the battery to its safe extremes either on the high side or the low side, but it would seem to give me some days of usage if the mains power dropped out. If the battery were discharged to 12V, it would take several days to fully recharge, which is not a problem for my expected use.

Is this setup reasonable?

One question I have--my DVM measures 140R across the 1N5817 diode. Does this mean that the 10R R3 is superfluous, since the current through the diode is already limited?

I haven't measured the current drawn by the Pi+camera, but it runs successfully on a 1A 5V wall power plug, so well within the 2A recommended current (at unspecified voltage) of the DC-DC module.
 

SteveDee

Senior Member
...One question I have--my DVM measures 140R across the 1N5817 diode. Does this mean that the 10R R3 is superfluous, since the current through the diode is already limited?

I haven't measured the current drawn by the Pi+camera, but it runs successfully on a 1A 5V wall power plug, so well within the 2A recommended current (at unspecified voltage) of the DC-DC module.
The diode (D6) will not limit current. The voltage drop will only vary slightly as current rises, so I guess you need to keep the 10R resistor.

If you can get hold of the battery manufacturers data, you may find you can run it down to about 30%. The state of charge graph will give you the equivalent voltage. (see one of my posts: https://captainbodgit.blogspot.co.uk/2017/02/birdbox2017-double-zero-deployed.html#more

Pi+RaspiCam may be around 120mA (what model Pi are you using?). With a 15Ahr battery you might expect: 0.7 * 15/0.12 = 87hrs approx
 

techElder

Well-known member
Lance, you might want to review your schematic with polarities of connectors and battery in mind. ;)
 

premelec

Senior Member
PIR? Motion sense?

@lbenson - I'm not entirely clear on your use but suggest you can gain more run time if looking for some moving activity by incorporating a PIR motion sensor and/or simply a variable light detector [during daylight] to indicate some something worth photographing is happening - this could cut your power draw a lot... Good luck with it...
 

lbenson

Senior Member
Nice blog post there, Steve. As it happens, I had never heard of European blue tits until maybe 2 days ago when I read an article on their high divorce rates (show up late, your mate has another date). I can't find it now.

I have a R-Pi3 running off of one of those DC-DC buck modules at 5.09 volts with a camera--the module reports about .5 amps with the occasional bounce up to .85. The R-PiZW is my target. Without a camera, I've measured one which peaked at less than 150mA--and ran at about what you said--90 to 120mA.
 

lbenson

Senior Member
Lance, you might want to review your schematic with polarities of connectors and battery in mind. ;)
Tex--it may be that the "+" on the charger header is misleading--that's not an indication of polarity--it's Eagle's mark of where you need to click to select the part in the drawing program. On that header, pin1 is 0V and pin2 is 12V.

A slightly different perfboard wireup is working.

This is the same regarding wiring, but less confusingly labelled.
12V charger with disconnect 2.png
 
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SteveDee

Senior Member
...I had never heard of European blue tits until maybe 2 days ago...

I have a R-Pi3 running off of one of those DC-DC buck modules at 5.09 volts with a camera--the module reports about .5 amps...
Yes, north Americans find it hilarious that we Brits have birds called blue tits and great tits...(we find it pretty funny too)

Re: current consumption, the Pi 3 will draw more current than a model "A" or a Zero.

But I think you are reading the wrong current. If you are running a Pi via a 5V switch-mode regulator from a 12V battery, it is the current draw from the 12V battery that is important, as this affects run-time. As power = Volts x Amps, and the switch-mode regulator is chopping the supply, the current from 12V supply should be much lower than the reading you get from 5V supply.

If you set your system up to stream video, the current draw will be higher when your remote system (browser or video player) connects to the stream, but lower when you disconnect.

Do you have any lights?

If your system is headless, you can disable the HDMI interface which will save maybe a few 10's of mA.

From my notes for a model "A" Pi:-
with multimeter on 200mA range connected into 12V supply;
- Running the "motion" program: 120mA (P=13.1V x 120mA = 1.57W)
- RaspiVid streaming @30fps: 140mA (P= 1.83W)

I hope this helps.
 
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