Instructables

From battery to wall wart in 10 minutes (with no soldering)

Featured
Picture of from battery to wall wart in 10 minutes (with no soldering)
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.
 
Remove these adsRemove these ads by Signing Up

Step 1: Components and tools

Picture of components and tools
wallplugpsu03.jpg
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.
1-40 of 46Next »
walidlol8 months ago

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?

Tankless-Gas-Water-Heater-Diagram-1.png
bjost10 months ago

Well done, and very good job on the pictures - very good quality.

andrea biffi (author)  bjost9 months ago

Thanks man!

gdaily110 months ago
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.
andrea biffi (author)  gdaily110 months ago
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.

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.

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.

andrea biffi (author)  BurgersBytes10 months ago

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.

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.

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.

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.

andrea biffi (author)  BurgersBytes10 months ago

With this voltage regulator you can input a voltage up to 30 V, so any 5V, 9V, 12V, wall wart (common models) is good.

andrea biffi (author)  gdaily110 months ago
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.
akassab10 months ago
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?
andrea biffi (author)  akassab10 months ago
It seems you have already found the best place where to learn that ;-)
ASCAS andrea biffi10 months ago

Agree! :D

:)
3366carlos akassab10 months ago

I would suggest taking classes at your local community college, they usually teach "real stuff" and skip all that "theoretical stuff" that works on paper and not in real life. Or subscribe to an electronics magazine. GL

akassab 3366carlos10 months ago

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 :)

D3zire10 months ago

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

andrea biffi (author)  D3zire10 months ago

that component exists with output voltages of 3.3; 5; 6; 8; 9; 10; 12; 15; 18 and 24 V.

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.

kbaker4 D3zire10 months ago

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.

D3zire kbaker410 months ago

so those .5v-1v

less or extra wont damage my device thn... e.g. my wireless game controller

kbaker4 D3zire10 months ago

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.

D3zire kbaker410 months ago

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

ianmoore33 D3zire10 months ago

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.

D3zire ianmoore3310 months ago

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

tiger1250610 months ago

You're not likely to "burn the device with a double voltage", 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.

You may cause the batteries to "leak or explode", 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 "charging" them.

I like to do this with clocks, the wall-wart tops off the batteries, but a power-outage hits, and I have battery backup.

andrea biffi (author)  tiger1250610 months ago

oh, you're right...

ASCAS10 months ago

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.

andrea biffi (author)  ASCAS10 months ago

your ibles are impressive too!

dlindstrom10 months ago
I needed this for the twins' electric swings some 6 years ago! Excellent instructable!
bhermance10 months ago
Good Project!! I may do this one day to my digital clock! Thanks!
rayleb10 months ago
plain and simple, great tutorial.
booga00710 months ago
Hey, been doing elecronics for a number of years, and I think this is fairly nifty.

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.
Might be a prob if people follow this without checking.
rayleb booga00710 months ago
if you look closely, the to92 is upside down in the picture. pin 1 is on the right as it lies in the picture.
booga007 rayleb10 months ago
Well that's good then :)
FoamboardRC10 months ago
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
jps1 FoamboardRC10 months ago
I think:

"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."

Nice work Andrea!
andrea biffi (author)  jps110 months ago
right, and... they are photogenic!!
1-40 of 46Next »