Maker Pro
Maker Pro

Replacing Batteries with a USB Cable to power Electric Device?

Durahl

Feb 17, 2014
5
Joined
Feb 17, 2014
Messages
5
Greetings!

Electronics Laymen speaking!

I have two LED equipped Devices currently powered by batteries which I'd like to power using a standard USB Cable ( one for each device ) in order to keep them powered at all times or at least while I'm at the PC.

From the looks of it:
Device 1 is powered by three AAA-Batteries placed in serial configuration - 3x 1.5V = 4.5V?
Device 2 is powered by two LR41-Pill Batteries placed in serial configuration - 2x 1.5V = 3.0V?
If Wikipedia serves my right a USB Cable runs at 5.0V?

How can I reach/lower the necessary required Voltage and is there something else necessary to ensure the safety of both the devices AND my Computer?

Like I already said I'm a laymen so I wouldn't mind, actually I would appreciate, a more elaborated explanation of what I need and how it has to be build together.
Being more of a 2D/3D Guy most Mathematical explanations on how to calculate this stuff left me mind boggled which is why I'm asking for help.

Thanks in Advance :)
 

davenn

Moderator
Sep 5, 2009
14,240
Joined
Sep 5, 2009
Messages
14,240
hi there durahi
welcome to the forums :)

The one important thing you haven't told us is ... what is the current required by the 2 devices. You are limited to 500mA maximum from a USB port

Dave
 
Last edited:

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
The device running from 4.5V *might* run from 5V without modification.

The other one may work with a standard 3.3V (low dropout) regulator or you could try to find a lower voltage reg.

Be careful because battery powered devices don't need to concern themselves about outputs that are tied to a supply rail. If you're powering it externally this becomes VERY important.
 

Durahl

Feb 17, 2014
5
Joined
Feb 17, 2014
Messages
5
@ davenn:
That's probably a Question I cannot answer as none of the two devices ( Figures with LED's ) appears to be labeled with this information.

Wouldn't it be possible to calculate the necessary/used current from the standard configuration while in battery mode and then cap the one of the USB Cable to a similar level?

I'm probably just a very simple minded person thinking like that xD

@ (*steve*):
For the price I paid I'd actually prefer not to take my chances in ruining them especially since I have to make the battery adapters anyway causing enough work so not to going that extra step of just adding a little piece like a resistor seems like kind of foolish to me :°

What changes when using an external power source instead of batteries?
 

donkey

Feb 26, 2011
1,301
Joined
Feb 26, 2011
Messages
1,301
Durahl I know you are a laymen so I will run this step by step for you
the current measuring item needed is a multimeter
put that to current or amps setting
connect one lead to the positive of your battery
connect other to the terminal where the battery was supposed to connect.
DO NOT LET THE POSITIVE OF THE BATTERY TOUCH THE TERMINAL. if you do it will read 0.
now if you have followed these instructions and it still reads 0 then you might have one of them multimeters with 20 different current settings, start on the highest and work down.
the other way of figuring this out is a little bit longer
put in new batteries, take note of the mah rating. see how long they last. then do mah divide by hours run to give you the answer... this method is very rough as not all batteries behave the same and there may still be amps left but the voltage has dropped too low for the device to work properly.
 

Durahl

Feb 17, 2014
5
Joined
Feb 17, 2014
Messages
5
Thanks for the Replies!

If I take your wording correctly you're asking me to see how much current flows by installing the Multimeter inside an open circle of batteries and LED's with the Multimeter closing the circle to check the current running through it?

Besides me not having a Multimeter at hand right now I'm curious as to why I'd need one for measuring the mA for a standard off the shelf AAA-Battery - Shouldn't they come with standardized V and mA values and only varying mAh values for their capacity?

I took one out and they're actually rechargeable ones labelled:
Ni-MH 1.2V 900mA <- I assume this should actually read 900mAh?
Charge 15h 90mA <- can charging be considered the same as discharging?

So combined I'm actually sitting at 3.6V, not 4.5V - Sorry about that.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
No, the circuit determines the current. The battery determines the total charge available and the maximum possible current (the actual current should always be a small fraction of this)

How long do a freshly charged set of nicads last?

A reasonable guess at the average current will be 900mAh/hrs (so if they last 20 hours, the average current is:

900mAh/20h
= 900/20 mAh/h
= 45 mA

Notice how everything cancels nicely :)
 

davenn

Moderator
Sep 5, 2009
14,240
Joined
Sep 5, 2009
Messages
14,240
Thanks for the Replies!

If I take your wording correctly you're asking me to see how much current flows by installing the Multimeter inside an open circle of batteries and LED's with the Multimeter closing the circle to check the current running through it?.

Yes exactly


Besides me not having a Multimeter at hand right now I'm curious as to why I'd need one for measuring the mA for a standard off the shelf AAA-Battery - Shouldn't they come with standardized V and mA values and only varying mAh values for their capacity?

No, capacity is irrelevant, how much current is flowing in the circuit is what is needed to be known

Capacity say 2000 mAh only helps you know how long the battery will last with a given current flow

cheers
Dave
 
Last edited:

Durahl

Feb 17, 2014
5
Joined
Feb 17, 2014
Messages
5
Sorry the late reply!

Just had to figure something out before continuing the conversation which took me a while - Busy man after all xD

I was initially running under the assumption that a Battery is the defining factor of how much current is flowing in a circuit and because of this I was expecting to get some easy numbers on how lot to set the USB's mA values.

Apparently it's the resistance of all the built in components inside the current and the Battery, unlike a USB Cable with a fixed mA value, automatically adjusts to it thus generating different values - Is that correct?

If so, that would explain the change of game, hinted in the first two posts, when going from Battery to USB Powered and the need to get myself a Millimeter.
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
A battery (or a USB port) supplies a voltage. The circuit connected to it determines the current.

Obviously there are limits, you can't draw 100A from a USB port.

As long as you stay within sensible limits, the relation holds. Outside of these limits the voltage source usually supplies a lower voltage (maybe even to zero) to reduce the current drawn, however, even in these circumstances the load still determines the current
 

Durahl

Feb 17, 2014
5
Joined
Feb 17, 2014
Messages
5
So I managed to get some values out by using a Digital Multimeter.

Figure 1 sporting 3-6 blue LED's and a set of fresh charged Ni-MH Batteries had 40-45mA shown on the Display.

Figure 2 sporting 1-2 red LED's and somewhat used LR41 Batteries had a whopping 0.6mA shown on the Display.

I can only estimate the amount of LED's because of how they're clustered inside the Figures - Could have used fiber strands to distribute the light of fewer LED's to multiple locations - Not sure though.

Is this something I can work with?
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,510
Joined
Jan 21, 2010
Messages
25,510
I don't see figure 1 or figure 2.

A possible reason for the low current from the LR41 batteries is that the batteries themselves are incapable of supplying the required current.

Whilst the circuit determines the current, this is only true if the battery can supply that current without its voltage dropping. For currents that are large for the battery in use, the battery voltage will fall and the current will be reduced.
 

Jagtech

Feb 22, 2014
43
Joined
Feb 22, 2014
Messages
43
Use caution when experimenting with USB ports on your PC. On some PCs, the USB ports are fused on the motherboard, and if you blow the internal fuse, you may have an expensive repair ahead.
 
Top