I am trying to build a mobile battery setup that can provide continuous uninterrupted power by connecting multiple battery packs in parallel and hot swapping depleted packs with freshly charged ones as needed.

The project goals are as follows:

• Continuous uninterrupted power to the load
• Quick connect and disconnect of individual power packs
• 3 or more power sources connected in parallel
• Some type of indicator (LED or LCD display) to show when a power source is low
• 5V output (also 12V from same sources, if possible)

I have a basic knowledge of electricity and electronics but I have never built anything like this before. I would like to do this with only NiMh AA rechargeable batteries because I already have a large investment in batteries and chargers of this type.

The batteries are 2900mAh AA NiMh. They will all be charged up using standard smart chargers. My questions are all about powering the load (discharging) in parallel, not charging.

From what I have gathered, I will need a fuse at the output of each source for safety. I will also need a diode at each source to isolate it from the others.

I planned to use 3 or more NiMH AAx4 USB packs connected in parallel. At this point, I am unclear about whether I can use a USB splitter cable to make the parallel connections. Would the USB hub function interfere with the circuit?

Alternately, I considered using a USB to Universal adapter at each power source and connecting these together using a 3 to 1 parallel adapter. Is that configuration valid?

One of the load devices that I will be powering is the transmitter of a wireless AV media adapter. According to the company, an input of 4.8V – 5.2V is required. The transmitter uses 500mW at peak level. I was able to obtain this additional information about the included AC adapter that my battery pack needs to replace:

DC output Voltage: 5V
DC current: 2.0A (1.5A Minimum)
Ripple and Noise: 1%
Line Regulation: +/- 1%
Load Regulation: +/- 4%

Finally, is there a way to supply other voltages from the same setup (specifically, 12V) using some type of DC-DC converter like the this one? Even if the 12V is not feasible, should I use the DC-DC converter anyway to ensure that the output stays at a consistent 5V?

Any suggestions/corrections would be greatly appreciated.

Thanks.

 

U will need DC to DC converters if you want 12V, but I think it will put a pretty heavy load on NiMHh. Since they can supply 2900mA, I think you might have thought of that too.
As for 5V, it depends on your load.
Can the load tolerate 4.8V to 6V, if not then u will need a low dropout regulator to get a 5V from a 6V NiMh.

As for redundancy,you can parallel the batteries with low Vf Power diodes together with fuses.

As for monitoring with LCD, you have the option of using PIC's. This way you can monitor which bank is depleting and which bank needs recharging, monitor the discharge rate and so an so forth

By the way..Welcome to AAC

 

Step one draw out a block diagram of the equipment - what goes where?

Break each block in your diagram out into its own subpage and draw out a logic flow diagram of its function, and its inputs and outputs.

You should have a very clear picture of the device by the end of that exercise and we will help you fill in the missing pieces.

Hint: Diodes are one way electric switches and each packs input point should probably have one, or something like it, to facilitate the hot swapping battery pack option you desire. Anywhere you need a steady source voltage for circuit operation, a regulator is usually a very good idea. A 'core', operational power source should be in it, so your device can operate to await an input source connection, when all its primary battery pack connections are empty.

 

Yaa..I forgot...A primary source should always be there...

 

<edit> Sorry, missed that you were a beginner. Don't do a boost-buck variant.

 

Thanks for the quick replies. My first post was mainly to validate the concept and to see if there were any fundamental flaws in the idea before I spent any more time on it (I saw a lot of discouragement at various sites regarding the viability of connecting batteries in parallel).

The main reason for multiple packs is extended battery life. The redundancy part is secondary. Since I can't find anything to buy that will do the job I need, building is the only option. I will get to work on a block drawing as suggested. As my electronics skills are limited, I am really looking for modular plugin type devices if at all possible.

Some questions:

I won't have access to any power sources except 4xAA battery packs. As long as at least one sufficiently charged battery pack is connected at all times, will this satisfy the "primary source" requirement?

I understood that the diodes were needed to keep battery packs with different output voltages from running inefficiently. Regarding hot-swapping, the battery enclosures I will be using have a manual on/off switch. I had planned to have them off when connecting or disconnecting. Is that correct? Or are you saying that with the diodes in place, it will not be necessary to have the packs off when connecting or disconnecting?

I don't really need to monitor extensively. I mainly need to know when a battery pack needs to be swapped out. Is there a type of LED that will change color when the output voltage falls below a certain level? Or some other at-a-glance indicator that will be simple (from my beginner's perspective) to install and use?

The DC-DC converter that I linked to is a regulator that claims 2.6V-14V input and 2.6V-14V output with a max of 500mAh. The AV transmitter uses 500mW at peak. Can I assume that .5W / 5V = .1A required for the load? Is 100mAh correct? Or am I missing something?

Does anyone know if a USB splitter cable that includes a USB hub would interfere with this setup?

Can I buy a cable that has an inline diode and fuse? What size fuse is required?

 

500mW @ 5V is 100mA or 0.1 Amp

A 1A max design would suffice, unless you have more than one purpose for it.

 

I think these last two posts from tyblu and Kermit2 are about a different thread.

No, this is the right one. Doing a block diagram helps us understand what you want done and makes explaining what we're talking about simpler, too.

 

<edit> Sorry, missed that you were a beginner. Don't do a boost-buck variant.

What is a "boost-buck variant"?

 

Er.. I was going to suggest you implement one so the system could compensate for lower voltages (batteries discharged below output voltage + regulator drop), but it's complicated, so I deleted it.

 

Er.. I was going to suggest you implement one so the system could compensate for lower voltages (batteries discharged below output voltage + regulator drop), but it's complicated, so I deleted it.

I understand now (didn't see the post before you edited it).

In any case, could you look at the link to the regulator that I posted? The specs say that it would keep the voltage constant even if the batteries discharged down to a low voltage (as low as 2.4V total).

 

I don't think a battery discharged to2.4V can boost to 5V or 12V for that matter.

What you are trying to do is pretty simple.
You will need some comparator to indicate the voltage levels via LED. but remember this puts an extra load around say about 80mA, for the comparator and indicator LED. An LCD driven with a PIC will consume much less power than that. LCD and PIC function can complicate a design cause you have to do the programing part.

You can buck boost a 6V to 12V at around 90% efficiency if I am not mistaken.
As for 5V, u have to use a low drop out regulator.

Taking into account about the redundancy, you will be using say 2 packs simultaneously. A battery 6V with a power diode will give you at least 5.8V at most. There are diodes that drop 0.1V as Vf. You have to use these. A better approach is to 6 cells and go for 7.2V. This will give sufficient margin and longer discharge.

I am having a hard time getting the USB hub part.

 

I believe he wanted to use the USB hub as an interconnect. Pre-made, purchased solutions. Off the shelf kit bashing.

Either that, or I need more coffee.

 

Either that, or I need more coffee.

U and me both

 

Step one draw out a block diagram of the equipment - what goes where?

Attached is my attempt at a block diagram.

The regulator I am considering is here. Your opinion would be appreciated.

 

What you are trying to do is pretty simple.
You will need some comparator to indicate the voltage levels via LED. but remember this puts an extra load around say about 80mA, for the comparator and indicator LED. An LCD driven with a PIC will consume much less power than that. LCD and PIC function can complicate a design cause you have to do the programing part.

I am not opposed to doing some programming for an LCD to keep the current drain to a minimum. What is involved in doing this?

 

I believe he wanted to use the USB hub as an interconnect. Pre-made, purchased solutions. Off the shelf kit bashing.

The USB Hub was built into the USB splitter cable that I was going to use to make the parallel connections between the battery packs. You are correct that I would prefer a pre-made solution for each "module" of this project. But I am willing to build what I have to in order to make it work.

 

Attached is my attempt at a block diagram.

The regulator I am considering is here. Your opinion would be appreciated.

There must be a catch to that converter.

See it says 2.6V to 14V.. it could be that if the input drops below say 6V, u might not get 5V out. I may be below that.

To get 5V from 2.4V,, u need a buck boost. Seeing that I don' think it will satisfy you once the battery goes below 5V.

U do need to go above 6V and build a low dropout regulator using a LM117 or something like that. It is quite easy and you will be sure of what you have built. They adjustable too.

I suggest that u discard the boost method, since it will help in power loss more than the drop down regulator

U need to go above 6V to have a better working period.

This will compensate for the diode drop too, and if you decide to use a small LCD, it can power it too. far better than 6V.

 

There must be a catch to that converter.

See it says 2.6V to 14V.. it could be that if the input drops below say 6V, u might not get 5V out. I may be below that.

To get 5V from 2.4V,, u need a buck boost. Seeing that I don' think it will satisfy you once the battery goes below 5V.

U do need to go above 6V and build a low dropout regulator using a LM117 or something like that. It is quite easy and you will be sure of what you have built. They adjustable too.

I suggest that u discard the boost method, since it will help in power loss more than the drop down regulator

U need to go above 6V to have a better working period.

This will compensate for the diode drop too, and if you decide to use a small LCD, it can power it too. far better than 6V.

Could you explain what you mean by "a catch"? The datasheet (attached) claims to be able to supply any of the output voltages even if the input is as low as 2.4V. In my specific case, if the voltage was down to 3.3V, the regulator claims to be able to supply 5V@248mA or 12V@103mA.

I read the LM117 specs you attached but I'm unclear as to which version I would need. It says maximum supply voltage 15V but I could not find a minimum voltage. I need the regulator to ensure a 5V output whether the battery pack is at 6V or 4.8V. My power sources are limited to whatever voltage I can get from 4xAA NiMh batteries.

How much should I expect for a diode drop? Can you suggest a specific diode or LCD?

 

I Suggest that you first get that converter and confirm tht you can get the preferred output with an input variation of 4.8V to 6V, which is your input voltages.

If it satisfies you, then we won't have any trouble continuing, if it does not then I can suggest a circuit using LM117

Here is the LCD I was talking about, you can choose any one you like