Help with a battery pack design

Yeah, that's spell-check for you. It's a typo - it should be 'consumption'. Like I said, use the supply you have, it's fine.

Ken

Sorry, I lost track of what you were asking. Get four Lithium or other hi-capacity rechargeables and use them. You'll probably find that four will do the trick (4.8V versus 5V required) or if needs be get a fifth and use that as well, and a diode (say, 1N4001 or virtually any other power diode) and put that in series.

Sorry about the side-tracks. That sorta day. :-(

Ken

What the......

Not Lithium, NiMH. Sheesh! Or the newer Alkaline rechargeables, for that matter. There's some pretty beefy types out there (1800mAh or more).

Cheers.

Ken

Or even easier yet: get a 12V sealed lead-acid battery, and a pre-made DC-DC converter. That way you don't have to build the LM2592 circuit yourself.

Both items available from

A 12V to 5V 1A converter is about $25. See their part number 216936CH.

--- If your current requirement is 900mA and you want to supply that current for three hours, that means you'll need at least a 2700mAH (2.7AH) battery.

Typically, Fully charged NiMH cells exhibit a terminal voltage of about 1.3V which falls to a fairly flat 1.2V nominal over most of the discharge curve and finally falls to 1.0V, which is considered to be the "cutoff voltage" and they should be discharged no further in order to avoid damaging the cell.

Panasonic makes a 3000mAH cell, with the data sheet available at:

It seems like four of them in series would get you very close to the 3 hour operating time you need, even with your minimum voltage requirement of 4.6V. That is, since your device quits when the input voltage falls to 4.6V, with four 1.2V cells in series the full capacity of the battery (2800mAH minimum) won't be available, but you'll be within about 90% of it and with the average capacity of the battery being 3050mAH, 90% of that is 2745mAH, which is 45mAH better than what you want. Note also that the typical discharge characteristics are given for C, so since you'll be discharging the battery at 0.3C you should get a capacity better than C.

If you want even greater capacity they make a "D" size monster, the HHR650D, with a capacity of 6.5AH.

-- John Fields Professional Circuit Designer

--
Or, you could get yourself a small 12V sealed lead-acid battery
(cheap, easy to charge, universally available, and 5AH would give you
about 5 hours of operation)  and try this:

http://cache.national.com/ds/LM/LM2592HV.pdf

There's a couple more flaws with a zener diode approach, besides wasted power: try to find a 5 watt zener. Unobtanium comes to mind. :-) Also, if you could get one, it would need an additional component - a 1 ohm, 10 watt resistor in series with it, with your device wired in parallel with the zener.

You can start to get a feel for this if you consider the mah (miliampere hour) rating of the battery like a bank account. Take a battery pack rated at 2400 mah. Your device uses 900 ma. If you run it for 1 hour, it is like making a withdrawal of 900 mah, leaving 1500 mah in the account. Run it for another hour, and that is an additional withdrawl of 900 mah. That leaves you with 600 mah, so you can't run for another full hour. You can run for 600/900 of an hour, or 40 minutes. So the total run time with a 2400 mah pack computes to

2 hours and 40 minutes. In actual use, you can have a large variation from the computed number, for a variety of reasons. But the analogy shows you what you are up against: the amount of money in the bank, versus the amount being withdrawn. If you want the money to last longer, you need more of it. (In theory you could re-design your device to use less than 900 ma, but that is not possible in your project.)

There are some other factors that come into play. The specific

2400 mah battery pack discussed is already lower in voltage than the device uses as nominal. The pack is 4.8 volts, versus 5 volts for the device. The 2400 mah rating applies to the discharge of the cells in the pack to 1.0 volts per cell, which equates to 4.0 volts for the pack. So you don't get the full 2400 mah. Or in other words, you don't really have 2400 in the bank.

Solutions:

What this will come down to is a trade off between size and weight of the battery versus run-time. The bigger the battery, the longer the run time.

Most nominal 5 volt devices will work fine on 5.3 - 5.4 volts without a problem. Responders can't say for sure, but it is highly likely your device will not be harmed by

5.4 volts. So a 6 volt gel cell with a series diode is likely a viable choice - and you can select one with a much larger capacity than 2400 mah NiMh cells.

You have proven that Luhan's suggestion works, and will find out how long it lasts when fully charged.

John Fields provided a solution using a 12 volt gel cell and a DC-DC converter that is a bit over 80% efficient - roughly the same efficiency as the 6 v gel cell/series diode solution, but with the added benefit of exactly 5 volts. His solution would work with any 12 volt battery.

Bottom line: if Luhan's solution, with freshly charged batteries, doesn't last long enough, you will have to get a bigger battery.

Ed

Why not a $15.00 6-volt 5 Ah battery, and a silicon diode or two?

Or an ordinary diode in series with a Schottky?

Cheers! Rich