Resistor or Fuse? Powering 12v Device from 12v Battery w/ protection

Thread Starter


Joined Oct 13, 2011
Hi Guys,

First off, apologies for the typo in the title.

I'm new to the forum and electronic circuitry in general, so please be gentle with the flaming ;)

I would like to power a 12v DC device with a 12V DC rechargeable battery, pretty straight forward.

The battery provides 8 amp hours - but from what I understand, what matters with DC is the voltage, the amperage is how much power the battery CAN provide, so battery can be of any amperage greater that the requirement, and it won't damage the circuit, provided the voltage is correct, right? So, I can connect a 80 amp hour battery to power a 12v 1 amp device?

The device requires maximum 1 amp. From what I've read on the forum, and other sites, 12v batteries fluctuate between about 10 to 14 volts, depending on the charge.

The device is sensitive to fluctuations, and I would like to protect the device from under/over-voltage/current as much as possible. I.E as the battery gets flat, or when I connect a charge to the battery, while the device is drawing power from the battery, etc

I would like to know, do I need a resistor, a fuse, or a new buzzword I learnt, a linear regulator?

I've been browsing the RS-Online site, and all it's doing is making me more confused, 12 volt at 1 amp is 12 watts, but there isn't a 12watt resistor, does that mean I can connect three 4watt in parallel to do the job? What about a fuse?

Thanks in advance for the help guys, much appreciated! :)

P.S In case you guys were wondering what the "device" is, it's a Router, which I want make portable - either with a dedicated battery, or drawing power from the car battery.


Joined Nov 30, 2010
2 errors. The idea about 12 watt resistor is wrong. Do not try to do that. It is not necessary and your logic is wrong.

A 12V battery will run about 12.0 when it's tired and 14.5 when charging, but the router probably won't care. You are over-worrying. Besides, a linear regulator will eat up 2 volts for it's "overhead" and ruin your regulating idea. You'd need a buck-boost switching regulator to tighten this up, and I think it's over-kill.


Joined Dec 26, 2010
Yes, provided the battery voltage is correct, your load device will draw the current it requires, not more. A bigger capacity battery just allows a longer running time before it becomes run down. There is always the chance of an accident though, i.e. a short-circuit!

A fuse in line with your battery is therefore advisable. A lead/acid battery, even of fairly small amp hour capacity can deliver a high current (or amperage) into a short-circuit, causing a significant fire hazard. This does get worse for bigger capacity batteries, because more amp-hours does generally equate to more amperes into a short-circuit.

You probably won't want to have the fuse of any lower rating than to stop the battery boiling up or the wiring catching alight though, as fuses have a bit of resistance and can drop some voltage. This is quite a consideration for low-voltage systems.

Note this distinction: the amp hour capacity tells you (approximately) the amount of current multiplied by time that can be drawn to run down a fully-charged battery. Current or amperage is either the current actually being drawn, or perhaps for a battery the maximum continuous rating, or the the maximum current that can flow for a short time, e.g. the cranking rating for a vehicle battery. These are related, in that a larger amp-hour capacity battery of a given type will usually be capable of more current (it will have lower internal resistance).

The amp hour capacity is normally measured at a current expected to discharge the battery in a given time, often 10 or 20 hours. Some batteries for things like mobile phones are rated for a 1 hour discharge, but this high drain may not even be safe for certain other types. Your 8 Ah battery might therefore provide about 0.8A for 10 hours, or 1A for 8 hours. If you tried to take much more current you would find that the amp hour capacity was less, (don't expect 4A for quite as long as as 2 hours, even if that's a safe current for your battery) or if you drew only a little current the capacity might be a little more. This website will tell you a lot about batteries - the link goes to a page about capacity, but there is much, much more. Enjoy!


Joined Aug 12, 2011
As already stated, you can probably just hook up the battery with an inline fuse and be done with it. The router itself likely has an internal regulator (the electronics inside actually run on much less than 12V) so it probably can operate within the range of a 12V LA (lead-acid) battery. Though, a 12V LA battery is considered fully discharged at 10.5V. You'd have to verify the router will operate with a supply somewhat below 12V. Voltage right off the charger for a 12V LA batt is right around 13V.

Providing a constant exact voltage is most easily done with a battery of higher voltage range regulated down to the required voltage. You do that with batteries in series or with a single higher voltage battery. A switching regulator is the best type to use since efficiency is much higher, but if efficiency is not a concern, a basic linear regulator can also do the job. There are POL (point of load) buck switching modules made for that purpose such as this one.

It's possible to use a battery of lower voltage and boost the voltage or use a battery of approximate voltage and use a buck-boost type of regulator, but these are much more involved methods usually reserved for applications where running a specific battery or battery configuration is imperative.