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If you like electronic gadgets as I do, you constantly need to charge many batteries, since battery powered devices are still much common. But batteries pollute, disposal has a cost, batteries themselves are not cheap at all if you buy good quality ones.
There is also another issue, many devices work good with alkaline batteries (nominal 1.50V / cell) but have faded display or a weak life with Ni-MH and Ni-Cd batteries (nominal 1.25V / cell) even if just charged.

For this reason is good to adapt your device to be connected to a wall power supply. Wall wart chargers for cellphones can be bought at every electronic fair for a pair bucks, and you probably already have some of them unused at home. Any wall wart from 4.5 to about 30V DC output will be good for this project.

So, let's see how to convert your batteries compartment to receive that wall-wart's plug.

Step 1: Components and Tools

You will be surprised to know how cheap this project is.

The main component is a tiny voltage regulator, the L78L33 in my case, but you can buy one with an output voltage from 3.3 to 24V, to match voltage of your device. This regulator can deliver a current of 100mA, this means that it's suitable for devices which usually drain the batteries (let's say 2400 mA) in not less than one day, so I think all the clocks, alarms, wireless devices, etc.
If this won't work, you can use a powerful voltage subbply as the LM78XX serie (up to 1000 mAh).
Refers to the attached data sheets for pins layout.

Then we need two ceramic capacitors. You can buy them or find on some dismantled electronic stuff. The values of 100nF and 330nF are not mandatory and any similar value is good enough.
Furthermore these two capacitors improve the circuit safety and functionality, but they're not essential, so you can decide to avoid adding one of them or both.

Last component is a female socket suitable for my wall wart plug. Mine is a 2.1mm x 2.5mm power plug, but if you want to use an old Nokia cellphone's charger (for example) it's smaller, 1.3mm x 3.5mm if I'm not wrong. You can find any type of these sockets on eBay.

Step 2: Set Up for Connections

My first version of this wall wart adapter was soldered, but for my other digital clock I tried to avoid soldering to make this project to everyone's reach. If connections are not soldered are probably weaker and we need to avoid forces on them, especially when we move the single-core wires closing the compartment cover. To achieve this you have to transform your linear wire to a more flexible spring.
I prefer to use single-core wires since they're handy to insert behind batteries metal connections.

Step 3: Begin Twisting

As first step twist capacitor's legs as in picture. My 100nF cap is salvaged so it has very short pins, better to find some with longer ones, as the 330nF which is brand-new.

In second step add the TO-92 packaged regulator, so that the 100nF is on the Vout side (pin1). Notice that the schematic in picture (on the datasheet) shows bottom view.
Then twist this central pin to capacitors' twisted legs.

Last step is to twist remaining capacitors' pins to regulator's external legs.

Step 4: Add Wires

Now you have to connect (you bet... twisting it!) the wires ends on the regulator legs. Red one goes on the 100nF side (+) and black one goeas on the 330nF side (-).
Now you can inspect the DC-IN plug to see which is the central contact pin, which coincide with positive connection of the wall wart. Although this is true for 98% of wall warts, it's better you check on the psu label the connection schematic.
The regulator's leg on the 330nF side goes twisted on this pin (input +) and the central leg goes to one of the remain pins (ground).

Step 5: Drill Hole

The benefit to place the circuit in the battery compartment is that you can't insert batteries and plug together. This avoid you to burn the device with a double voltage.
So let's drill an hole with the proper diameter in center of the cover. My female socket has 8mm external diameter.
Insert the socket and screw it up.

Step 6: Close It

Arrange wires to they go toward the batteries positive and negative connectors. You recognize them because they're not connected together and the negative is usually a spring.
I noticed that you can easily push a wire end behind the metal connectors, so that they keep it firm. In this way you have no need to solder anything and you can replace the batteries whenever you wish.
Close the cover and insert the power plug.

Step 7: Partially Cloudy on a Bright Display!

The clock works very good and the display is much more visible than with rechargeable batteries.
If you want to check the voltage out of the circuit you can use a multimeter. As you see the voltage is pretty precise very near to 3.3V).

Step 8: Do You Want to Solder?

If you prefer to solder the pins together you can see my other adapter which works as well very good.
Following this tutorial you should be able to remove batteries from any your electrical device which has no power plug socket yet, so let me know if it worked!
<p>I've purchased a multi-voltage adapter , set the voltage to 3 volts ( 2 batteries ) , connect the terminals directly to the battery terminals for a gas heater, burners worked, but did not open the gas valve.why?</p>
<p>Well done, and very good job on the pictures - very good quality.</p>
<p>Thanks man!</p>
The DC power jack shown has an extra terminal because it includes a switch. Connected properly to the existing battery terminals it will cut off the DC power supplied by the batteries and run the device from the supplied DC power transformer. That would allow you to choose to run the device with either batteries or an external adapter.
are you sure that the switch is not Normally Open? And it closes when I insert the jack? I have to give a try... if you're right I can drill the clock body and leave the battery compartment available.
Female DC connectors use common to switch between power supply sources. You can check the connector pins with an ohm meter or a continuity tester. When the power jack is removed from the DC connector, the two common(-) pins will be connected together. When the jack is inserted, one pin will not be connected to the circuit. Connect your battery pack ground to that open pin so that the battery ground only connects when the jack is removed. <br> <br>So instead of just connecting your power supply circuit directly to the battery wires, cut the battery common wire or add a jumper so it can go to the extra female connector lug. Connect the common wire going to the device on the other common connector pin.Then you can use either or both. <br> <br>Another way to calculate the voltage is to count the batteries and add 1.5 volts for each one. More voltage will require a different voltage regulator. I happen to have a thermometer that requires just one AAA battery. You could make your own regulator for them by using 3 1N4001(to 4007) diodes in series as each one drops the voltage one half volt. Plus would be the anode end from the power supply positive and common would be the cathode end(cathode has the stripe) of the 3 diodes. <br> <br>
<p>Right, if you need 1.5V you can use one diode after the voltage regulator, so to keep the possibility to use any power supply.</p>
<p>Only downside to this is you leave batteries in a device and end up forgetting about them, leading to their corroding over time, leaking and ruining it. If you wanted constant power AND you were mindful to keep track of the batteries, sure, using that sort of power socket is a fine idea.</p>
<p>The female power supply sockets all normally are wired that way. Wire the device to the wrong common terminal and nothing may work. You could also wire a dongle outside of the device using DC connectors with 2 screw on terminals, but that would not look as nice. </p><p>The wall wart replaces the batteries so it would be the user's choice to use both. The cord would have to be unplugged for the batteries to work at all with the female connector. The connector used can accommodate up to 12 volt power supplies too so it might be a good idea to mark the required voltage on the back of the device. Smaller connectors for lower voltage wall adapters are harder to find.</p>
<p>With this voltage regulator you can input a voltage up to 30 V, so any 5V, 9V, 12V, wall wart (common models) is good.</p>
And if the switch is normally closed (NC) so I made a mistake saying to connect to any of the remaining pins... since one disconnect when jack is inserted.
I just LOVE this tutorial, I'm a mechanical engineer, but designing and building electrical solutions fascinates me, is there any way I can learn how to make stuff like this? any pointers?
It seems you have already found the best place where to learn that ;-)
<p>Agree! :D </p>
:)
<p>I would suggest taking classes at your local community college, they usually teach &quot;real stuff&quot; and skip all that &quot;theoretical stuff&quot; that works on paper and not in real life. Or subscribe to an electronics magazine. GL</p>
<p>I was actually considering the college thing, I didn't know if I can subscribe to few classes without being part of a program, have to check... it would be wonderful if I can do that :) </p>
<p>Thats just reallyyyyyyy great just what i wanted but A Quick question.. What if my device uses 3 or 4 AA batteries(most of my devices are).... this curcuit wont work thn right ?? it'd b super grt if u guide me through with same thing with 3 bat. n 4 bat. devices seperately.. thanx</p>
<p>that component exists with output voltages of 3.3; 5; 6; 8; 9; 10; 12; 15; 18 and 24 V.</p><p>I think that L78L05 will work for 3xAA or 3xAAA batteries and L78L06 for 4xAA or 4xAAA, but for powerful devices, as a speaker or a lamp you should use LM7805 or LM7806.</p>
<p>Reread step 1 where he talks about how to select the voltage regulator. In the case of 3 - 4 AA, that would mean 1.5V X 3 = 4.5 V and 1.5v X 4 = 6v. 5v should work fine for most cases with 3 or 4 AA batteries. These means you need a LM7805 with enough current to supply the device.</p>
<p>so those .5v-1v </p><p>less or extra wont damage my device thn... e.g. my wireless game controller</p>
<p>No it should be fine if it was made for alkaline AA or AAA as they will come from factory with a voltage of about 1.7v. So 3 X 0.2v = 0.6 or 4 X 0.2 = 0.8v.</p>
<p>thank you thank you so much... for clearing that out.. that also solvd many questions in my head which were bugging me for a long time</p>
<p>Most batteries put out between 1.3 and 1.5v through their working life, so the 0.5v reduction in the regulator keeps it within working parameters. Great tutorial.</p>
<p>but in case of 3xAA(4.5v) there'd be extra .5v ..thts wat concerns me... n sry for stupid questions bt m kind of newbie in these things</p>
<p>You're not likely to &quot;burn the device with a double voltage&quot;, as you're making parallel connections. Your device will see whichever voltage is higher, the batteries or the wall-wart, which will likely be the wall-wart.</p><p>You may cause the batteries to &quot;leak or explode&quot;, however, if the wall-wart voltage is significantly higher than that of the batteries (i.e. dead batteries), because current will flow into the batteries &quot;charging&quot; them.</p><p>I like to do this with clocks, the wall-wart tops off the batteries, but a power-outage hits, and I have battery backup.</p>
<p>oh, you're right...</p>
<p>So that's how you get 3.3v! I've been looking for a 3.3v regulator chip, apparently the 7800 series doesn't carry voltages below 5v. Thanks for this awesome guide! It's really useful. </p>
<p>your ibles are impressive too!</p>
I needed this for the twins' electric swings some 6 years ago! Excellent instructable!
Good Project!! I may do this one day to my digital clock! Thanks!
plain and simple, great tutorial.
Hey, been doing elecronics for a number of years, and I think this is fairly nifty. <br> <br>But I've noted that when wiring up the regulator, compared to the datasheet you have behind, you've wired up the DC input to the Vout pin. <br>Might be a prob if people follow this without checking.
if you look closely, the to92 is upside down in the picture. pin 1 is on the right as it lies in the picture.
Well that's good then :)
I couldn't find why you made such a big coil of wires. Why?
The coil does not need to be that long but you do need enough to insure that no matter how to remove the door the wires do not become disengaged from the battery connector since they are not soldered on. The coil is used because it is nicer than stuffing extra wire down in the battery compartment. If you do solder it on I'd do it inside the case not the battery compartment, then you could keep batteries in the compartment to take over when the AC power failed. Been there, done that.
Oh haha I gotcha
I think: <br> <br>&quot;If connections are not soldered are probably weaker and we need to avoid forces on them, especially when we move the single-core wires closing the compartment cover.&quot; <br> <br>Nice work Andrea!
right, and... they are photogenic!!
Hi again andrea - Very nice, I covered two other things to do when I published my articles in WorldRadio and QST magazines. In WorldRadio I published an article on how to select an old wallwart for a new device correctly. This was before switching regulator power supply wallwarts. With a transformer one had to match up voltage and current. There was some give and take and my article showed how to select a wallwart that would work for the new device and not kill it with too much voltage. The other article in QST was how to run a 9V radio on a 12-14 Volt battery this was very similar to what you're doing here. You're using a higher voltage power supply, switching wallwart, to supply a linear regulator that drops the voltage to that needed by the device. I used a higher voltage battery 12V which is pretty standard for amateur radio use, to a linear LM7809 1A regulator to supply 9V at between 100 - 200mA (no heat sink needed in this case). I tested my circuit and it worked even when using the original 9V power supply. I use this in my bedroom with one of my trickle charger pluses that I designed to keep the 12V battery charged so even if we lose power my clock radio will still wake me up on time. It has done this at least twice since I put it all together. I'll keep the URL to this &amp; pass it on to my son &amp; others, again well done &amp; thanks!
Good idea Phil :-)
Would you not be able to avoid all the extra parts and wires by just buying a AC/DC adapter whose voltage output matches the needs of the device? <br> <br>I've converted a few things from battery to AC. All I did was buy the appropriate DC adapter and solder the wires to the battery terminals.
Bremus, my project is made in eco-save philosophy, and using a wall-mart you already have saves money and garbage. If you don't have one ask to a friend, or search in used stuff at some electronic fair.
Fantastic solution <br>
thanks Matthew!

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Bio: I'm an Italian freelance structural engineer, graphic designer and photographer. I'm also investigating electronics, robotics and science in general. I enjoy hacking and ... More »
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