Lets suppose, i have a 12 v battery and i want to give 9v to my variable load.can i use a 3 v battery in reverse to get that output ?. If yes, what are pros and cons of this method? Why can't we see such method implemented anywhere?

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    \$\begingroup\$ This will charge the battery. \$\endgroup\$ – user253751 Sep 26 '16 at 20:42


A "perfect" battery might work for this, but any real life battery is going to have big problems. It will work OK to begin with but the current is flowing through the battery in the wrong direction, i.e. backwards compared to when the battery is being used normally. This is the same as if the battery is being charged, so the battery will quickly become overcharged. The consequences vary depending on the amount of current and the battery chemistry but may include:

  • Battery voltages much more than 3V
  • Producing dangerous/explosive gas
  • Fire, thermal runaway or explosion
  • Other damage to the battery

To drop 12V to 9V for a variable load, use a regulator. To get the most stable voltage, use a linear regulator. If efficiency is more important, look for a switching regulator. Regulators are very cheap - my usual supplier has a bag of 10 suitable 100mA regulators for sixty pence (less than a dollar US).

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    \$\begingroup\$ @DavidSchwartz If you are adding two voltages by putting the batteries in series, then both are being discharged. In this case, trying to subtract one from the other, one is being charged. You aren't just gradually reducing the voltage, you're changing the sign too. \$\endgroup\$ – Jack B Sep 26 '16 at 18:12

As others have already said, this is a bad idea since it would be charging the 3 V battery with no mechanism to ensure it isn't overcharged. Note that trying to charge a non-rechargeable battery is essentially "overcharging" it.

I want to point out yet another problem with this method, which is additional series resistance. Even if you carefully monitor the 3 V battery and ensure it is not overcharged, the series resistance of this 3 V battery adds to that of the 12 V battery for the purpose of the combined 9 V battery. For example, if the 12 V battery has 1 Ω series reistance and the 3 V battery 5 Ω, then the effective 9 V battery will appear to have 6 Ω series resistance.

As others have said, use a regulator. That's what they are for.

The common and cheap 7809 will work in this case with up to about 300 mA output in free air, and to it's full 1 A capability with proper heat sinking.

At this relatively low voltage range, there are a number of switcher chips out there with integrated switches. You add the inductor and a few other parts, and you get 9 V from 12 V at around 90% efficiency relatively easily.

You can also buy off the shelf DC-DC converter modules. These are usually quite pricy compared to a switcher chip and the few parts around it, but are really simple to use. This can be appropriate for one-off projects, or the less-skilled in electrical engineering. One side advantage of these is that the output is usually isolated from the input. That means that the 9 V supply is floating relative to the 12 V battery. Sometimes that can be useful. When you don't want that, just tie the grounds together.


If you connect your battery this way it will work. And you start to charge the 3 V battery at the same time. Most probably that is nota good idea since overcharging a battery connected in this manner, damages the battery and can be dangerous in case of a Li Ion battery. Therefore dont connect a battery to lower your supply voltage.

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    \$\begingroup\$ No, this is totally incorrect. The battery is being discharged. If you don't see why, first imagine two 12 volt batteries connected positive to negative. Clearly they're both being discharged. Now imagine we gradually reduce the voltage of one of the batteries. Clearly, it still continues to be discharged. Think about it. \$\endgroup\$ – David Schwartz Sep 26 '16 at 18:06
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    \$\begingroup\$ @David Schwartz but in this case the batteries are connected positive to positive (not negative). So you are wrong. \$\endgroup\$ – Bruce Abbott Sep 26 '16 at 18:13
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    \$\begingroup\$ @DavidSchwartz Decapod is correct here, both in the charging and in the conclusion not to do it. \$\endgroup\$ – Spehro Pefhany Sep 26 '16 at 19:51


In reality for low current 9V output you can do it either way step up or step. they standardize cells sizes and cascade in series to get higher voltages when more watt-hours are needed, which is why portable tool batteries with higher voltage imply more storage power.

Putting batteries in parallel leads to self discharge and potential high current failures when initially connected.

However if your design is low current at 9V, then your choice is based on cost, weight and total V*Ah or watt-hours of storage capacity.

Here are two high efficiency solutions for step up and step down.

enter image description here

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    \$\begingroup\$ Where does battery charging come into the question? Have you answered something different here? \$\endgroup\$ – Sean Houlihane Sep 26 '16 at 18:21
  • \$\begingroup\$ Presumably @SeanHoulihane means controlled, intentional battery charging. \$\endgroup\$ – user253751 Sep 27 '16 at 3:12
  • \$\begingroup\$ I do think he answered something different here, presumably by mistake. Happened to me once. \$\endgroup\$ – PNDA Sep 27 '16 at 13:55
  • \$\begingroup\$ Deleted 1st answer and proposed solutions either way. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 27 '16 at 14:07

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