I am powering a datalogger with a 12-V battery, and every week or two, I must swap out the battery. I'd like to be able to swap these batteries out without ever losing power at the datalogger. Therefore, I'm thinking about connecting the new battery to a different branch of the cable and then removing the old battery. The two batteries would be connected in parallel briefly, and I know their voltages would equalize almost immediately. Therefore, the new battery voltage would drop immediately, which is problematic.

Is there a way to connect both batteries in parallel without this immediate voltage equalization across the batteries?

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    \$\begingroup\$ How much current does your load draw? \$\endgroup\$ Commented Jun 21 at 17:00
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    \$\begingroup\$ Depending on your load current you could consider an ("ideal") diode-or or maybe even a supercap to hold up the voltage while you do the swap. Need some details to know what the best solution would be. \$\endgroup\$
    – John D
    Commented Jun 21 at 17:09

5 Answers 5


The two batteries would be connected in parallel briefly, and I know their voltages would equalize almost immediately.

When two batteries having different voltage (state of charge) are connected in parallel, a large current will flow from the one with more charge to the other. This current depends on the voltage difference between both batteries divided by their internal resistance in series.

If you plan to use Lithium batteries, internal resistance is quite low, so even a small voltage difference could result in an unsafe amount of current. Suppose one battery is at 4.2V, the other at 3.5V, that's 0.7V. Internal resistance of a 18650 cell is 20-80 mOhm, so with two in series that's 40-160mOhm, thus you're looking at 4 to 17 Amps, which is way over the maximum allowed charge current. Because internal resistance and maximum current scale with battery capacity, this should remain unsafe no matter the size of the battery.

If your batteries have BMS for protection then both will simply shut down and your device will be left unpowered.

It would be a much better option to use a pair of diodes, one per battery, to prevent current from flowing from one to the other. If you're concerned about diode voltage drop causing inefficiency, there are "diode emulation" integrated circuits which use MOSFETs instead, to combine two power sources without one backflowing into the other. See "Power supply OR-ing" or "Ideal diode" in the "Power management ICs" category on the usual shops.

Another idea by MikeB: use diodes from both batteries to avoid backflow, and add a SPDT switch to short circuit the diode corresponding to the active battery and avoid losses.

enter image description here

  • \$\begingroup\$ Re: the last paragraph, put a pair of switches across the two diodes - close them for efficiency when running normally, open them to 'activate' the diodes when needed ie when swapping the batteries. \$\endgroup\$
    – MikeB
    Commented Jun 24 at 7:18
  • \$\begingroup\$ Thanks, I've added it to the answer! \$\endgroup\$
    – bobflux
    Commented Jun 24 at 13:12
  • \$\begingroup\$ The last circuit looks really nice and inexpensive. \$\endgroup\$ Commented Jun 24 at 17:24

The electronic industry has the solution for dual battery hot swap, called "Ideal Diode-OR" controller. An example is LTC4235 from Analog devices, enter image description here You should be able to find more adequate (less fancy) controller for your level of power consumption, with integrated transistors etc. This one (LTC4419) will probably work for you better.


Since the datalogger runs for a week on one battery, I assume the load current is below 100 mA in this circuit.

If you implement a mechanical switch for the takeover, there will be a very short period, where none of the batteries is connected.

To avoid this dropout, I added resistors and diodes to allow both batteries to supply the load while the switch changes the path.

The cross current between the batteries is defined by the resistor value. Here I assume to have a fully charged battery with 13.8 V and an empty one with 10.8 V. These are typical extreme values for lead acid batteries.

With 1 Ω the current between them would be limited to 3 A in this example if both batteries are connected.

The diodes may be useful if the load draws pulsed peak currents. The load voltage would not drop more than the diode forward voltage during such a pulse.

Depending on the used cell chemistry and maximum cross current you will need other resistor values and you probably can omit the diodes.


simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ This causes both batteries to die at the same time. Remove R1, R2 and the diodes. And put a 100uF cap across the load to stabilize the switching transient. \$\endgroup\$
    – MOSFET
    Commented Jun 21 at 19:59
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    \$\begingroup\$ @MOSFET It was not intended in this schematic to have both batteries connected all the time. One is connected, the new one is additionally connected after a week or so, the takeover switch is used and the empty one gets removed. \$\endgroup\$
    – Jens
    Commented Jun 22 at 0:11

You can avoid the possibility of connecting two batteries in parallel altogether, simply by adding a three-component circuit across the load, the schematic is below:


simulate this circuit – Schematic created using CircuitLab

This circuit can be either connected permanently, or temporarily during the battery change-over process. The latter is preferable if the new circuit cannot be left out in the field.

This proposal is based on the following assumptions:

  1. The battery is a sealed lead-acid (SLA) "gel-cell" rated 12V 7AH. A typical battery datasheet is available here:

  2. This battery is discharged down to 1.60V per cell = 9.6V, by which time it has delivered 14.2 Watt-hours.

  3. This battery lasts 1 week, so the load is about 84.5mW (~7mA), calculated from: 14.2 Watt-hours / (24 hrs x 7 days).

  4. The load is a constant power load, and at time of battery replacement represents a resistance of: I load = 84.5mW / 9.6V = 8.8mA, R_load = 9.6V / 8.8mA = 1.09kΩ.

  5. The load will continue to operate without any problems (eg: without losing memory) with an input voltage below the discharge voltage of the battery; let's assume this to be 8.0V.

Capacitance required to hold up the load voltage:
At disconnect, the capacitor is charged to same voltage as the battery, which is 9.6V. When the battery is removed, the load current switches over to the capacitor via the diode, so the load voltage drops immediately by the forward voltage of the diode let's assume this is about 0.6V. So the load voltage is now 9.0V. The capacitor now has to support the load current of ~9mA while losing at most only 1 volt until the new battery is connected. We can derive the formula for the capacitance as follows:

$$ I = C\dfrac{dv}{dt} ⇒ C= I \dfrac{dt}{dV}\ $$
Using the values we have for current and voltage, and assuming time is 1 second, we get the capacitance required per second of hold-up time:

$$ C = 9mA \dfrac{1sec}{1V} = 9mF. $$

So for 10 seconds of hold-up time, you will need:
C = 10 x 9mF = 90mF, round this up to 100mF = 0.1F.

That sounds like a lot of capacitance, but these days it is quite common to get such large value capacitance for low DC voltages. For example, here is a standard electrolytic capacitor rated 0.1F (100mF) 16VDC:

Part no: ESME160LGB104MAC0U
Package: can, 35mm dia x 120mm high.

You can add more of these capacitors in parallel to get longer hold-up time. You can also increase the hold-up time by selecting a diode with a lower forward drop, such as a Schottky.

R1 will limit the current into the capacitor when the fresh battery is connected. Select this resistor to withstand the maxium power dissipation expected. Its resistance value is not that critical; the higher its value, the longer it takes to charge C1 up to the battery voltage, which may be a consideration for when you are connecting this only during battery change-over. For your application this could be increased to 1kΩ or even 10kΩ without any problems; but don't increase too high otherwise the leakage current of the capacitor may prevent it from reaching the battery voltage.

If this added circuit is temporary, the procedure will be:
Step 1: Connect the circuit across the load.
Step 2: Wait for C1 to charge up to the battery voltage (within 0.1V will be OK).
Step 3: Disconnect the old battery. You now have a time limit to complete the next step.
Step 4: Connect the fresh battery.
Step 5: Disconnect the circuit.
Step 6: Discharge C1 to be safe for transport.

You may have heard of "super-capacitors", these have values of capacitance measured in Farads; however, when I looked I could not find many examples rated at voltages over 6V that were of reasonable size and cost. I would not recommend connecting these types of capacitors in series to try to increase the voltage (the voltage-sharing issues may be difficult to manage).

I could not seem to find a suitable unit rated at, say, 16V and 1F. The higher voltage types that I found were quite expensive and physically large, here is one example, but it is 58 Farads - over 500 times more than what you need.

OEM = LICAP Technologies, Inc.
OEM PN: SM0058-016-P-1
Desc = CAPACITOR 58F 0%/20% 16V CHAS

Capacitors smaller than 100mF are readily available and quite low-cost. For example, 10mF (10,000uF) 16V in a leaded can, each one of these will give 1 second of hold-up time:

Part no: B41231A4109M000
Package: can, 22mm dia x 27mm high.


If you're outside of the arctic/antarctic circles and above ground install a solar panel

Eg a 15W solar charger costs less than the 7AH lead acid battery it could easily maintain at your 7 days discharge rate assuming one hour of daylight per day. (if you use bigger batteries the cost will still scale similarly)


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