I am trying to build a switch: I have to batteries both 12V, I want to minimize voltage spikes when switching from one battery to another. the switch is essentially being made as an emergency. One battery will operate, if it goes out then the other will begin to operate.

  • \$\begingroup\$ So you mean a UPS? \$\endgroup\$
    – Passerby
    Commented Mar 12, 2016 at 0:59
  • \$\begingroup\$ So, assuming that I understood your intention correctly: You want to discharge the batteries in turns, so that you can swap the empty pack with a full one without interrupting power delivery. The system should thus only switch between source batteries when the previous battery becomes fully discharged. \$\endgroup\$
    – jms
    Commented Mar 12, 2016 at 1:01
  • \$\begingroup\$ many types of 12V batteries can be damaged by a complete discharge \$\endgroup\$ Commented Mar 12, 2016 at 4:27

4 Answers 4


Simply use two diodes, this does the trick automatically.

In this example V1 and Bat are power sources, and +VE will be the output.


Note that if you have a load which has a serious current draw (e.g. over 1A) then the diodes must be sized accordingly, and you have to prepare for some power loss on the diode. (0.7W @ 1A).

This can be solved by using a so-called "ideal-diode" which in fact a FET with very low resistance, plus a proper driver chip called "ideal diode controller".

This solution scales well, the example below works up to 10A, and basically you can easily get FETs at 50A too, nowadays.

(source: linear.com)

  • 1
    \$\begingroup\$ This falls into the 'why didn't i think of that' category. I like it :) \$\endgroup\$ Commented Mar 12, 2016 at 3:18

The question has nothing about current draw or switch speed, but given that the current draw isn't high and the switching event is reasonably fast, you could simply place a large-value capacitor between the load and the switch.

The capacitor would supply voltage for a small amount of time, that time being proportional to the current that you were drawing. If you know the current draw, you could even do a 5 minute SPICE simulation to find the 'ideal' capacitor value.


You can use an LTC4353:

The LTC ®4353 controls external N-channel MOSFETs to implement an ideal diode function. It replaces two high power Schottky diodes and their associated heat sinks, saving power and board area. The ideal diode function permits low loss power supply ORing and supply holdup applications.

The LTC4353 regulates the forward-voltage drop across the MOSFET to ensure smooth current transfer in diode-OR applications. A fast turn-on reduces the load voltage droop during supply switchover. If the input supply fails or is shorted, a fast turn-off minimizes reverse-current transients.

The controller operates with supplies from 2.9V to 18V. If both supplies are below 2.9V, an external supply is needed at the VCC pin. Enable inputs can be used to turn off the MOSFET and put the controller in a low current state. Status outputs indicate whether the MOSFETs are on or off.

This is called a dual low voltage ideal diode controller, which is part of a category called PMIC (Power management integrated circuit).


First, you aren't building a switch, you will place a switch on each battery negative lead, like a knife edge switch. The spike is natural and you can place a voltage regulator between the batteries and the load which will compensate a little for the brownout.

Voltage regulator

Place the two batteries on the left side of this circuit. The capacitor 4700 micro farads (THIS IS A BIG CAP MAKE SURE YOU GET POLARITY RIGHT ) will store charge and donate it to the circuit as battery drops.

The Diode will adjust for low spikes in reverse voltage situations.

Second, what you are doing switching a circuit from one battery to another is a dangerous impulse condition on circuits. I assume you don't have shunted condition when one battery is opening and closing opposite polarity that would be dangerous as well. You could fry both batteries (batteries are current sources in parallel with resistors), burn the circuits, burn your house etc. I don't know what the rest of the circuit is so I can't prove or disprove this. Solid state electronics can't handle spikes without a regulator.

This switching is making discontinuous conditions that might lead to current spikes that overload it. All it takes is one transistor to blow while capacitors are loaded and they do blow. That's worse than stopping to disconnect and reconnect a new battery.

One way to avoid dangerous conditions like this and I hope I am exaggerating, is to use heavy duty variable resistors - ones that can take many times the wattage of the circuits - that are tunable from 0 to 10 mega ohms with a mechanical rotary switch. A suitable potentiometer can be used. With mega ohm resistors that will decrease current to pica amps before you turn the other battery up. You can ramp one up to 10M ohm and ramp the other down to zero as the switch.

If you don't understand what I am describing here then you shouldn't be playing with batteries and switches, the raw power of even milli amp-hour DC batteries is enough to kill you if you happen to connect them to your heart through your arms. I've known mechanics melting their wedding rings to terminals through their wrenches. I implore you to wear insulated gloves and insulated work boots if these are automotive circuits. Even sweaty hands can make enough connection.

AC is a skin effect electromagentic current, it oscillates along the rim. DC is a bulk current that goes through the centre of conductor. In this case you.

  • \$\begingroup\$ "the raw power of even milli amp-hour DC batteries is enough to kill you if you happen to connect them to your heart through your arms" really? What voltage is this battery? \$\endgroup\$
    – gbulmer
    Commented Mar 12, 2016 at 8:02
  • \$\begingroup\$ He didn't elaborate that's why I advised a little overly cautious. I think it's car batteries. \$\endgroup\$
    – daemondave
    Commented Mar 12, 2016 at 14:35
  • \$\begingroup\$ I do not believe any human (or any mammal?) ever has been killed by 12V through skin contact. So I don't think your answer over cautious. I think it is totally misleading. I would be interested in scientific evidence to the contrary. Further, the fact that a wedding ring might be welded to a wrench by a (large) electrical current is utterly irrelevant to your warning about milli-amp hour batteries killing someone. Finally the comment "If you don't understand what I am describing here then you shouldn't be playing with batteries and switches," is misleading, as it is not based on fact. \$\endgroup\$
    – gbulmer
    Commented Mar 13, 2016 at 17:39
  • \$\begingroup\$ Funny. You don't provide facts but an opinion and yet you are certain of your opinion. Here's a fact. I'm an electrical engineer. Here's another fact: hypertextbook.com/facts/2000/JackHsu.shtml \$\endgroup\$
    – daemondave
    Commented Mar 14, 2016 at 3:05
  • \$\begingroup\$ The reference lists 0.2, 0.1-0.3 A and so on as DC that can kill through your heart So your opinion is faulty and not fact based. And 200 ma = 0.2 A. This is the problem of opinion based systems that people who don't know still have an opinion. \$\endgroup\$
    – daemondave
    Commented Mar 14, 2016 at 3:08

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