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We use batteries to power a lot of our electronics. But there are some battery powered devices that don't necessarily need to be portable all the time. One example is my son's battery powered swing. It can be moved around but it usually stays in the same general area. In cases like these it might be nice to power these devices with an AC adapter and save the batteries. So in this project, I am going to show you how you can use an old power adapter to power your electronics in place of batteries. I will share how to modify the adapter and two different ways to connect it to your electronic devices.


Step 1: Use a Voltage Regulator Circuit to Set the Output of the Power Supply to the Appropriate Voltage

It is very rare to find a power supply that will perfectly match an electrical appliance unless they are sold together as a pair. So we are going to have to modify our power adapter to match the circuit that we want to power. The easiest way to do this is to use a variable voltage regulator such as a LM317. The typical configuration for this kind of circuit is shown in the picture above. This regulator uses two resistors to set the output according to the formula: Vout=1.25*(1+R2/R1).

For most applications this circuit can be simplified a little bit. The capacitors are only needed if your load circuit is sensitive to small power fluctuations. So in many cases, these can be eliminated. The variable resistor R2 is useful if you want to be able to power multiple different devices. But if you are going to use the power supply exclusively on one device you can replace it with a fixed value resistor. Wire the circuit as shown with Vin connected to the power supply and Vout connected to the circuit that you want to power. The regulator will bring down the output of the power supply down to the value that you set.

Depending on the power rating of your circuit, you may need to add a heat sink. 

Example:
My son's swing normally runs on four C size batteries. So I found an old power supply with a 9V 1000mA output. I figured that would be enough to replace the battery pack. Then I soldered together the LM317 regulator circuit with a 220 ohm resistor for R1 and a 820 ohm resistor for R2. These resistor values give an output voltage of 5.9V. (It would have been ideal to use a 240 ohm for R1 and a 910 ohm for R2 but I didn't have those values on hand) This output is still well within the operating range for a four cell battery pack. Anything between 1.25V and 1.5V per battery will usually work. Since the electronics on the swing just consists of a motor and a speed controller, I decided that the filtering capacitors weren't necessary and I left them off. See the following steps for the best methods for connecting everything together. 

<p>Please also be gentle with me. I know nothing about regulators/resisters etc. I was wondering if you could help me. I am trying to make a wireless Arlo camera<br>by Netgear designed for exterior conditions wired to an AC adapter. It uses (4) CR123<br>batteries which are 3.2v per battery. How can I make an AC/DC adapter work? I<br>found an 13v adapter. Here is a link to it. <a href="http://www.amazon.com/AC-DC-ADAPTER-13VOLTS-POSITIVE-CENTER/dp/B00GLW4PX4?ie=UTF8&psc=1&redirect=true&ref_=oh_aui_detailpage_o00_s00" rel="nofollow">http://www.amazon.com/AC-DC-ADAPTER-13VOLTS-POSITIVE-CENTER/dp/B00GLW4PX4?ie=UTF8&amp;psc=1&amp;redirect=true&amp;ref_=oh_aui_detailpage_o00_s00</a><br>I don't know if the output specs for the battery vs. adapter are compatible.<br>Can you tell me if your process will not damage the camera or help me with what exactly I need? Thanks for your<br>help.</p>
<p>Be careful here. I've read elsewhere that two of the cells are wired in parallel with the other two cells. This means it's only 6.4V at double the amperage.</p>
<p>I think I see my problem. The 13v adapter is only a 750 mA and the batteries are 1300 mAh. Does this matter? Will the adapter overheat or can I add something to increase the mA rating of the adapter?</p>
<p>The adapter will run warm regardless since it's drawing current, especially if it sheds less heat than it produces over some period of time. The closer it gets to that 750 mA window the warmer it will get. It probably won't overheat if your draw is usually 75-80% of the max (600 mA). What you need to know is what the actual current draw is, since it's probably not 1300 mAh since that would be 1300 mA of constant? current for 1 hr.</p><p>You cannot increase the mA rating of the adapter afaik. They're designed with a a maximum output current.</p>
<p>Ok. A CR123 battery is nominally rated for 3 volts. If the camera uses 4 x 3V batteries, then it needs 12 volts to operate. But this doesn't tell you anything about the current requirement. So you can't just get any 12 volt power supply because you have to match the current output. </p><p>So you have two options. One option is to do a lot of measuring and a lot of math to figure out the exact current requirements on the camera and then try to find a power supply with that exact current output. Or the easier way is to use a 12 volt regulated power supply. This means that it will always output 12 volts no matter how much current is required. </p><p>You can buy 12 volt power supplies that are regulated or you can add a regulated to any power supply that you have lying around. The most common one to use is a 7812 chip. <a href="https://www.radioshack.com/products/radioshack-12v-fixed-voltage-regulator-7812?variant=5717601477">https://www.radioshack.com/products/radioshack-12v...</a></p><p>Connect the positive output of the power supply to the input pin of the regulator. Then connect the output pin of the regulator to the positive terminal of the camera. Then connect all the ground/negative leads together.</p>
if im using a 4.5 volt (using 3 X AA 1.5v) table lamp. i have a ac/dc adaptor which can select up to 12v. but how do i know how much mAh current is ouputing?
If you have a power supply where the current output is not labeled, it is likely a regulated power supply. This means that it will output whatever current is necessary to drive the circuit at that voltage. This means that you just match the voltage to the battery pack, and the power supply will automatically adjust itself to set the proper current.
<p>Hi there, I have a very big wall clock that runs on a single AAA battery (1.5V). Could I use the same method as you did with your baby swing? What size power supply configs would I need?</p>
<p>To explain further, the reason for wanting to hook up to ac power is because the single AAA battery seems to only last a few weeks. I think this may be because the big wall clock is outside under our patio and is exposed to the cold and hot temperature. (not weather such as rain or sun as it's under roof)</p>
<p>That seems unlikely. Unless there is a huge temperature difference between day and night (+/- 30 degrees?) I doubt the battery cares all that much. If the clock is analog (moving hands as opposed to an LCD) it is much more likely that constantly driving a motor and gears is what causes it to run down. You could try modifying the clock to take 2 or 3 batteries in -parallel- to make the time until replacement higher.</p>
That could work. You could used the voltage regulator described in thetutorial.
<p>Hi there, going by the tutorial, I would need to use 240/220 ohm for R1 and about 80 ohm for R2 to give me vout of about 1.6v. Would that work using that configuration? Not sure if R2 has to be higher than R1.</p>
<p>That looks right. </p>
<p>Thanks for the feedback. Alternatively, could I just use a 1.5v power supply with the end plug cut off and then just wire the 2 wires into the battery compartment positive and negative?</p>
<p>If it is a regulated power supply it would work. Otherwise the voltage changes depending on the current output. </p>
<p>Correction! Clock use 1 x AA battery - 1.5V (not 1 x AAA)</p>
Bro. I have a battery operated fan. But the battery got spoilt dur to overcharging. The fan doesnt work even while being plugged in to AC. I then removed the battery and plugged it in directly but it still doesn't work. I want it to run on AC power
If you already followed the instructions in this project, I would probably assume that the fan is broken.
<p>ha!</p>
Hey where do u get a voltage regulator is there anything u can take it out of
<p>You can buy them at Radio Shack if you still have one near you. Or you can order it from online electronics suppliers like Mouser.</p>
I own a hair trimmer set that runs with batteries and I also have a charging cable that comes with the trimmer. <br>Battery is dead now but I can't use my trimmer just with the cord. Why? Any suggestions?
<p>It's probably because you rbattery is charged while plugged into the wall and the device is pulling strictly from the battery and not the power from your wall. This means that if your battery is missing or seriously damaged, it will not work.</p>
Thank you for your answers! I just opened it up and remove the battery system and attached it directly through cable! Works fine!
<p>In most cases there is an internal switch that disconnects the motor when the charger is plugged in. It does this because the charger isn't powerful enough to run the motor and can get damaged you try.</p>
<p>cant you just wire it to a wall plugin cord lets say you have an extension cord and take one end offf of it and wire it to the item you need batteries for directly and just plug it in?</p>
No. If you are using just an extension cord, the you would be feeding 120 Volts AC into your device. This would seriously damage any battery operated electronics and could electrocute you if you touch it.
Sir? I own a hair trimmer set that runs with rechargeable batteries and I also have a charging cable that comes with the trimmer. <br>Battery is dead now but I can't use my trimmer just with the cord. Why? Any suggestions?
why don't use directly a 6v ac power to prevent all the job of a voltage regulator circuit?
Wouldn't it be 6v DC? Aren't all batteries DC? Why couldn't I just take an old 6v cell phone charger, strip the wires, and attach them to the end of the battery compartment? Wouldn't that accomplish the same thing?
<p>I think it was a typo...</p>
I'd like to suggest to eliminate the dummy battery thing, why not rig a power jack. i believe that there's a cavity somewhere near the battery holder where you can set the power jack, it wouldn't just be neat, but give you easy accessibility too(plug in-plug out in almost in an instant.
<p>The Batsub (battery substitute) does the job without any modification to your devices. It's exactly what the doctor ordered. Don't miss out, see batsub.com.</p>
<p>Hi, I find your instructable very cool!</p><p>Even more, because I had a similar idea with a cheap cordless drill, years back. I did it, more or less your way, but without the voltage regulator, since I used a matching AC adapter. Alas, the power was not enough, I think the drill needed more current, than the AC adapter could provide. </p><p>Now, I could use the same procedure to power a cordless jigsaw I have (MAKITAJV100DZ http://www.makitauk.com/products/saws/jigsaws/jv1... I already have the drill (MAKITA DF330DWE http://www.makitauk.com/products/saws/jigsaws/jv1... and I use the same Lithium-ion 10,8V-1.3Ah batteries. This is quite practical. </p><p>The thing is, the saw drains the batteries very fast, which is ok for simple jobs, but if I want to cut something big and thick, I need to change the batteries every couple of minutes.</p><p>Your instructable would make perfectly sence here. But I need the jig saw to have the same power with the AC adapter as it has with the batteries, not like the failed attempt I described in the beginning.</p><p>Now, here comes my question. If I use the AC adapter of an external hard disk, with an output say 12V, 2A, can you tell me what voltage regulator, and more important, what resistors and eventually capasitors should be used?</p><p>I am terribly bad with calculating electronic circuits and I don&acute;t want to fry my jig saw. I you could help me on this one it would be great - thanks!</p>
<p>Sweet! :)</p><p>I like this....A LOT! :)</p><p>TY for sharing.</p><p>Q:</p><p>the idea of doing this, the basic idea I mean,</p><p>will it work for all battery powered appliances?</p><p>Granted, everything is different &amp; each item has to be modified independently </p><p>but is this basically how battery operated things are converted?</p><p>I prolly have 10 or 15 of these power pack things I've saved thru the years.</p>
<p>An electronic device doesn't care where the power comes from as long as it has the correct voltage and can output enough power. So if you match it with a power adapter that has the right power output it will work. </p>
AAAAhhh ok....<br>*smoke fills the room as I scheme &amp; think....*<br><br>TY Sir. :)
<p>This exact swing is the exact item that prompted my generic conversion search, Wonderful. Thank you Google and author.</p>
<p>A couple things: 1. For a device that uses 4 C batteries, couldn't you simply use a 6v wall wart instead of taking a 9 and stepping it down? 2. In most devices, you would only need to connect the leads from your wall wart to the first and last battery terminal. There is really no need to create dummy batteries as the spring connectors are really only tabs of metal connecting one battery to the other. Correct me if I'm wrong on either of these.</p>
The problem is finding a 6V adapter that has the appropriate output current. Most power adapters aren't regulated so their output voltage will change depending on how much current is being used. By using a voltage regulator circuit, you don't need to worry about matching the current.<br><br>Yes, in most cases you won't need dummy batteries but some battery terminals can be difficult to make a temporary connection to.
<p>Do the dummy batteries need to connect to each other?</p>
No. They just fill the space. All you need is for the end terminals on the battery pack to connect to the output of the regulated power supply.
<p>I'm puzzled by Step 2. I don't see this in your setup.<br>It seems to me you can't very well have the batteries in your device and connect the power supply unless you use the modified cell outside the battery compartment. I dont see this in your setup. ???.</p>
<p><img src="http://i.stack.imgur.com/SIZXC.png"></p><p>Here is a different picture. The numbers are different but it does a better job of illustrating. The barrel contact 2 is normally connected to the pin 3. But when the power jack is inserted, it presses on the barrel contact and pushes it away so that it is no longer connected to the pin 3. This can be used to disconnect the battery when the power jack is inserted. </p><p>This is the same type of connector that is often used for head phones. When the head phone plug is inserted into the head phone jack, the speakers are disconnected at the same time that head phones are connected.</p>
I am not too well versed in electronics, so please bear with me. I do have a basic understanding of electricity and do design my own lighting units for photography. But I do have some questions:<br>1.What is the source for the value of 1.25 in your formula?<br>Alkaline AA batteries output 1.5v; rechargeable AA batteries output 1.2v (I no longer use them they are a PITA). How does this affect your formula.<br><br>2. In what way do the resistors and capacitors affect the output?<br><br>I am trying to make an adapter for a 6v battery pack for an electronic flash. I have done this for a Nikon SB800 and a Vivitar 285. Just a regular 6v power adapter does the job. But it does not work on an Aputure Magnum Speedlite MG-68. There is a delay when the on/off button is pressed before starting. 4 AA alkalines work fine, but the 6v external power adapter does not activate the battery.
<p>The 1.25V has nothing to do with batteries it is just a ratio of the resistor values and the corresponding output. Maybe the wikipedia article would help give some details on the design.</p><p>http://en.wikipedia.org/wiki/LM317</p>
<p>Thanks much. Appreciate that.But I sill have some questions:</p><p>BTW, I found the problem with my flash and it works in the same way as the others did; fro the usual 6v power supply. The problem was a clever little trick in the + connector to the aa battery: the connector had to be a smaller diameter; otherwise a large screw head would push the connector out of contact. Silly how simple.</p><p>I don't quite know how to interpret (or understand some things in the schematics:</p><p>1. what determines the resistance values in your schematic: why 240 ohm and 910. I know that gives an out v of ca. 5.98. So, if I wanted 12v output, I would use R1=240 and R2=2064. But what if I used larger or smaller resistors in the same ratio? </p><p>2. In another schematic on wikipedia there was a reference regarding limiting current with the notation IQ (subscript Q). Where could I get some information on that kind of notation. There seems to be a bit of that kind of notation and I would like to understand what these references are and where do they originate?</p><p>And I do really appreciate your help. I won't bother again. :))</p>
<p>The formula that I list is a simplified version of the formula </p><p>VO = 1.25 V (1 + R2 / R1) + Iadj * R2</p><p>Iadj is small (less than 100 microAmps. If you use 240 ohm for R1, then this part of the equation is very small and can be ignored for the most part. But if you use a different value for the resistor, then you may need to include this part. If you want a lot of more detail you can read the data sheet for the part. </p><p>http://www.fairchildsemi.com/ds/LM/LM317.pdf</p>
Heck if that was one of them models that leaked acid all over the child that sat or laid in that would be a great fix

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Bio: My name is Jason Poel Smith I am a Community Manager here at Instructables. In my free time, I am an Inventor, Maker, Hacker, Tinker ... More »
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