If I have a system working at 2.5V-0.3mA powered by two CR123 3.7V-2000mAh.

Is my system going to last the same time if I put my batteries in parallel or in series?

Because if I want it to last 12hrs a day during 2 years, at 3.7V (parallel) I need a 1780mAh battery (I have 2*2000=4000mAh) and at 7.4V (series) I need a 880mAh battery(I have 2000mAh).
In both case in parallel and in series it will work.

But in which case is it going to last longer?

Thank you in advance !

  • 1
    \$\begingroup\$ It looks like your project can perfectly run from two AA alkaline batteries (in-series of course). Good brands of AA have 10 years shelf life, and should be fine for 2 years assuming good quality switching downconverter. \$\endgroup\$ – Ale..chenski Feb 23 '17 at 17:10
  • \$\begingroup\$ @AliChen You have to equalize the discharge range when comparing the two types. Lithium range is 3.6 to 3.0V cutoff. Alkaline is 1.5 to 0.8V cutoff, so it will not power this circuit for its entire life span. The alkaline will drop to 1.25V at about 50% of it rated life. \$\endgroup\$ – Misunderstood Feb 24 '17 at 16:47
  • \$\begingroup\$ @Misunderstood, the OP project needs about 7 Watt-hours to run. An AA alkaline cell has typical energy capacity of ~4Wh at slow/pulse discharge, so two of them should do the job. However, you are correct that the voltage drop will cross the 2.5V level of the project requirement. So, instead of crafting a complicated buck-boost switching converter, three batteries will do, with a good buck DC-DC down-converter. How does it sound now? \$\endgroup\$ – Ale..chenski Feb 24 '17 at 19:06
  • \$\begingroup\$ @AliChen I am under the impression that Alkaline AA low drain capacity is 2870mA Hour which will effectively (1400mA Hour) be less than the CR123's 2000mA Hour. \$\endgroup\$ – Misunderstood Feb 24 '17 at 19:17
  • \$\begingroup\$ @Misunderstood, you are confusing current capacity with energy capacity. One AA is 1.25V * 2.87 Ah = 3.6 W*h. The Li-on has more 3.3*2 = 6.6Wh. So you need either 2xAA, or 1xLi-ion for this project, with proper and efficient DC-DC conversion. \$\endgroup\$ – Ale..chenski Feb 24 '17 at 19:32

You actually answered your own question in there..... When you have them in parallel, you get 4000mAh, and in series you only get the 2000mAh.

So with that you can see quite clearly that you get more mAh when they are in parallel.

If you wanted to know which one will last longer then you have all the required information right there in your question.....

Just take the mAh you have and divide it by what is required. Whichever comes out with the highest number is the one that will last longer.

4000/1780 = 2.247 2000/880 = 2.27

So there isn't actually that much of a difference in this case.

Pretty sure this is the answer you were after unless I have missed something from the question?

A good place to calculate battery life is HERE

  • \$\begingroup\$ Thank you, it is the answer I was looking for. I was thinking of this but you confirmed my thoughts. So now I just have to find a DC DC converter with the best efficiency to convert 3.7 to 2.5 or 7.4 to 2.5. \$\endgroup\$ – Tagadac Feb 23 '17 at 10:14
  • 2
    \$\begingroup\$ Just a small addition: Connecting batteries in series will double the voltage. Connecting batteries in parallel (assuming identical batteries) will double the amp-hours. The total power available stays the same due to P=VI. \$\endgroup\$ – CHendrix Feb 23 '17 at 13:09
  • \$\begingroup\$ @Tagadac, for your project (~7Wh) you don't need two cells. If this is a serious project, one non-rechargeable Li-based cell will do the job. Look into Tadrian industrial offering, tadiranbatteries.de/eng and pick up a cell with proper energy capacity, impulse discharge rate, and small self-discharge. If it is DIY-type project, three AA cells will do the same job. \$\endgroup\$ – Ale..chenski Feb 24 '17 at 19:16

You description of the battery is confusing.

A CR123A is a primary (non-rechargable) lithium magnesium dioxide which has a nominal voltage of 3.0V.

A voltage of 3.7 would be a Lithium Manganese Oxide commonly used in power tools.

The CR123A is commonly used in camera flash units as is secondary (rechargeable) 3.6 V Lithium Cobalt Oxide.

Another 3.6V is the Lithium Nickel Cobalt Aluminum Oxide used in medical devices which is similar to the Lithium Cobalt Oxide with higher capacity.

I could not find any CR123A with 2000 mAh. I did find a lithium magnesium dioxide 2000mA hrs in an AA package.

The only 3.7V battery @ 2000 mAh was a battery pack.

So I do not know what you have.

The next two images show the stable internal resistance.

This is a lithium magnesium dioxide CR-2/3AZ. It's a 17 gram lithium the same as a CR123A.

enter image description here

This is a lithium magnesium dioxide CR-AG 22 gram, 2400 mAh.

enter image description here

You asked: The system has to be shutdown when the voltage reaches 3V to prevent battery damage, fire etc. Is that true?

This is the lithium magnesium dioxide 2000mA hrs in an AA package. Probably similar to what you are using as it has a 0.3 mA curve.

Looking at the the above images, the internal resistance begins changing as it nears its end of life.

Below the battery is basically dead at 3.0V.

A 3.7v or 3.6V lithium should considered dead at 3.0V.

enter image description here


As far as parallel or serial. Serial will NOT increase capacity.

Given that most Lithium battery packs use parallel batteries, the consensus in the industry is it's okay to connect batteries of the same chemistry in parallel, and the internal resistance is very stable, parallel would be the way to go. There is the risk of a short (1 in 200,000) which can present a safety issue. Most shorts do no start fires or explode. Severe shorts may.

As far as safety, that is not my area of expertise. I would recommend to read what Isidor Buchmann has to say at Battery University, Why are Protection Circuits Needed? .


I have concerns over buying this battery from Leisure Chow on Amazon.
6 X 2000mah 3.7v Cr123a 123a 16340 Rechargeable Battery

That is not a CR123A. The description says "100% brand new 16340 3.7V 2000mAh rechargeable batteries ". It is the same physical size as a CR123A. A CR12A is not rechargeable. For a 2 year use, I would go with a primary non-rechargeable battery. Secondary rechargeable batteries have self loss of up to 5%. Will never last 2 years.

I have serious doubts this battery is 3.7V or 2000mAh.

I do not know how many you are buying if this is a one shot deal or you are going into production and resale. Either way you want quality. Quality comes from a vendor like Panasonic or another well known name brand. They have reputations and a lot on the line.

I but buy a lot on Amazon and I have criteria for picking a vendor on Amazon. First want vendor to have hundreds of reviews. I then look at the ratio between 5 star and 1 star rankings. I want to see greater than 70% 5 star and less than 7% 1 star. These Leisure Chow batteries 38% 5 star and 30% 1 star. I would never buy a product with even 10% 1 star. I expect to see infant mortality in products so a certain percentage will die early and generate 1 star reviews and many people are so stupid they do not know how to use the product. But that accounts for less than 10%.

I do not know what kind of battery Leisure Chow is selling. At 3.7V it should be a Lithium Manganese Oxide which have high energy and lower capacity. Used in power tools. There are no 2000mAh Li-manganese batteries due to it low specific energy. You do not want Li-manganese.

If you have to have a rechargeable you need a Lithium Nickel Cobalt Aluminum Oxide.

What I think you need is a Panasonic CR-V3
3.0V, 3000 mAh, Manganese Dioxide. Two cells in parallel. When you buy a battery pack the manufacturer has matched the two batteries.
Or any of the 3V on this page:
Panasonic CR Series Primary Cylindrical Manganese Dioxide Lithium Batteries
Datasheet: Panasonic CR-Series Datasheet

  • \$\begingroup\$ Thank you for your response and link. He said "The lowest permitted “low-voltage” power cut-off is 2.5V/cell.", does that mean I can detect 2.8V instead of 3V before shutting off my system ? \$\endgroup\$ – Tagadac Feb 27 '17 at 10:25

In general connecting batteries in parallel is a bad idea. Such a parallel connection violates a basic rule of electricity. Parallel connection works for certain types of batteries, but is a disaster for others.

Here's the problem: A battery is a voltage source. As such it will attempt to deliver its generated voltage at whatever load is connected to it. When two voltage sources are connected in parallel, each will try to deliver its voltage to the other battery. If the batteries both have the same exact voltage, the net current flow between the batteries is zero. If the voltages are not exactly equal, a current will flow from the higher voltage battery into the lower voltage battery. The magnitude of this voltage follows Ohm's Law. The resistance/load in the simple connection of two batteries with no external load is the interconnecting conductor (e.g. metallic battery holder) and the internal impedance of the batteries. Since both of these impedances are quite low (generally fractions of an Ohm for the holder and a few ohms for the batteries, the resulting "cross current" can be quite large compared to the magnitude of current the batteries are designed to deliver to a legitimate load.

The other effect is that the lower voltage battery will have a reverse current flowing thru it (again by virtue of Ohm's Law). Not all batteries can tolerate such a situation. Rechargables generally can because they are designed to be charged, However, one-time use batteries are not designed to withstand reverse current flow.

Typically, the end result is a relatively rapid discharge of both batteries with a variety of side effect depending on the type of battery chemistry. Heat generation is one common side effect. In certain situations the heat can be great enough to thermally destroy the batteries.

The only salvation is that the two batteries might be perfectly matched both in initial voltage (e.g. as manufactured) and remain matched as they discharge over time. Or, the battery chemistry is such that it can accommodate the voltage difference and both batteries will work in unison to develop a shared operating voltage. Lead acid batteries can do this. Common alkaline batteries are not so accommodating. I'm not sure about your CR123 units. However, their current rating is high enough that I would perform an experimental parallel hook-up without a load, and in a safe, contained situation where I could terminate the connection if things get dangerously out of control.

  • \$\begingroup\$ Mmh... This is quit a good answer... I think I will choose to put them in series haha ! Thank you :) \$\endgroup\$ – Tagadac Feb 23 '17 at 11:22
  • 1
    \$\begingroup\$ Do you have any thing to back up these claims? I do not believe this is true. Almost all battery chemistries allow parallel configurations. If one develops a high resistance or opens is less critical in a parallel circuit than in series configuration, but a failing cell will reduce the total load capability. It’s like an engine only firing on one cylinder instead of on two. An electrical short is more serious as the faulty cell drains energy from the other cells. Most shorts are mild and manifest themselves as elevated self-discharge. \$\endgroup\$ – Misunderstood Feb 23 '17 at 15:28
  • \$\begingroup\$ @Misunderstood, the "elevated self-discharge" is the keyword here. If you want a battery to last 2 years, never connect them in parallel. If you run a quadcopter for 15 minutes, do whatever you believe. \$\endgroup\$ – Ale..chenski Feb 23 '17 at 16:45
  • 1
    \$\begingroup\$ I took some extra time to fully understand what you are saying. Nothing you say is true. If the voltages are different the higher will discharge in to the lower at no consequence. From then on they will remain match voltages. If batteries could not take reverse current that would be a major problem, but it is NOT. The only danger (in another application, not this one) is if in the unlikely event one shorted and the batteries had high capacity and high discharge rates they will get hot and possible start a fire. Not going to happen with CR123 batteries. \$\endgroup\$ – Misunderstood Feb 23 '17 at 17:05
  • \$\begingroup\$ @Misunderstood, you say, "If the voltages are different the higher will discharge in to the lower at no consequence." Try to understand the consequence of "discharge". Then you said "they will remain match voltage". This is a highly dubious statement, especially under pulsed discharge mode as in this application. \$\endgroup\$ – Ale..chenski Feb 23 '17 at 17:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.