Need a large (20ish) number of battery packs at 5-6V charged every night.

[Mark Brehob]

Hello,
I'm doing some work where we will need about 20 to 40 5V battery packs
and I'm trying to figure out exactly what to do...

Our restrictions include:
Getting 4.8V is probably okay, but 4.5V won't work -- we need to stay
at least at 4.6 V the whole time. We can be high (up to 20V) but
anything above 5V is just wasted.

We need at to be sure to get at least 800 mAh. Again, if it's rated at
1000mAh but only 500 is likely, that's a problem for us. We will
probably hit at least 100 charge/discharge cycles. The discharges
probably won't be full discharges and the number of cycles could be as
high as 500.

We will need around 20 packs and would like to cheaply and safely
recharge ALL the batteries over night. We have plenty of power
supplies (electronics lab) so I'm quite willing to use them (up to
1Amp at 6 V and I have 5 of them I can use if needed).

My current plan is to use 6V NiMH packs (made from AA batteries) but,
I'm not sure how to go about charging them (all) safely at night. I
don't mind spending a bit more for packs (rather than making my own or
something) but I don't want to buy chargers at $40.00/charger if I'm
gonna need 20 of them. And having the power supplies in the lab seems
like it should be useful, but I want to be very very very safe (public
school) with this. I'd like to avoid building my own chargers unless
it's fairly simple (low time) and very safe.

Suggestions?

Thanks,
Mark

 

[BobG]

..> My current plan is to use 6V NiMH packs (made from AA batteries) .
====================================
1800mah cells can be chared at 180ma for 10 hrs. Hang as many packs on
a lab supply as you can (lets use 10...need 1.8A). use a 1 ohm R in
series with ea pack, turn up the volts till you get 1.8 amps, come
back in 10 hrs?

 

[Mark Brehob]

.> My current plan is to use 6V NiMH packs (made from AA batteries) .
====================================
1800mah cells can be chared at 180ma for 10 hrs. Hang as many packs on
a lab supply as you can (lets use 10...need 1.8A). use a 1 ohm R in
series with ea pack, turn up the volts till you get 1.8 amps, come
back in 10 hrs?

yeah,
But I'm more worried about what happens if I don't get back in 10
hours. Do they explode? Stop working? Create a chemical spill?

Mark

 

[Puckdropper]

.> My current plan is to use 6V NiMH packs (made from AA batteries) .
====================================
1800mah cells can be chared at 180ma for 10 hrs. Hang as many packs on
a lab supply as you can (lets use 10...need 1.8A). use a 1 ohm R in
series with ea pack, turn up the volts till you get 1.8 amps, come
back in 10 hrs?

Would adding a diode and an outlet timer be a good idea? That way, the
packs won't discharge and when done the charger would be shut off.
(You'll probably want to measure the current AFTER the diodes, though.)

Puckdropper

 

[Jon Slaughter]

yeah,
But I'm more worried about what happens if I don't get back in 10
hours. Do they explode? Stop working? Create a chemical spill?

How much money can you spend. They have very inexpensive charging chips that
you could use that detect overcharging and all that mess. It was some new
chips from maxim. They monitor several batteries at once but don't remember
the details. You could always build a simple circuit for the charging using
one of these chips for a few bucks then send it off to a pcb manufacture for
replication. (you could use several chips on the same board to reduce
cost... might be able to get several dozen on one more and just end up with
a few boards).

Not necessarily the easiest way but might work for you. The chips of course
were designed for this sort of thing and luckily they have very little
external component requirements(and they have interface for monitoring
through a processor so you could even work on something like that if you
wanted).

Might look into it and see what its about if your looking for something
serious. The nice thing about it is that you just have to design one and
then duplicate it, they are cheap(I would imagine), few extra components,
and have all the saftey issues taken care of. (I think they even monitor the
batteries temperatures)

Jon

 

[neon]

look up LM317 regulator and connect it as a curent source that can be cheap $.5 per charger pluss some resistors need a 8v to 30v dc source and 180 ma per cell got 10 then 2 amp source no bigy 120v to 12 v transformer @2amps and rectify it. don't worry about voltage batteries can care less they charge with current not voltage a souce of 120vdc is fine provided the current source is 180ma

 

[ehsjr]

yeah,
But I'm more worried about what happens if I don't get back in 10
hours. Do they explode? Stop working? Create a chemical spill?

Mark

Not a good idea to do it that way. Some packs could get
a charge at a rate that is too high, others, too low.

If you want to charge at a limited current, here's how
you can avoid the problem. Ten circuits consisting of
an LM317 and resistor in this configuration, each charging
a single pack:

-----
V+ ---Vin|LM317|Vout---+
----- |
Adj [R]
| |
+----------+---> To battery pack +

Gnd -----------------------> To battery pack -

Compute the value for R as follows:
1) Charge rate (C) is 1/10 the mAh rating of the pack. In the
example of an 1800 mAh, that's 180 mA
2) R = 1.25/C
In this example, that's 1.25/.18 which equals 6.944
That's an odd value, so use 7 ohms. 7 ohms will cause
the charge rate to be ~178.5 mA which is close enough.

There is always a loss when charging, so you need to
add 20% - 40% to get the minimum charge time. The typcal
charge of this type is a "14 hour" charger, meaning
you should charge the packs for 14 hours to ensure they are
all 100% charged. It will not harm the packs if they are
left on the charger for several days.

You need to set the power supply voltage above the terminal
voltage of the pack by about 3 volts. Your 5 volt packs
will require 5 cells. Each cell has a terminal voltage
of 1.43 volts at room temperature, or 7.15 volts total,
so set your power supply at 10 volts.

Ed

 

[JeffM]

-----
V+ ---Vin|LM317|Vout---+
----- |
Adj [R]
| |
+----------+---> To battery pack +

Gnd -----------------------> To battery pack -

Compute the value for R as follows:
1) Charge rate (C) is 1/10 the mAh rating of the pack. In the
example of an 1800 mAh, that's 180 mA
2) R = 1.25/C
In this example, that's 1.25/.18 which equals 6.944
That's an odd value, so use 7 ohms. 7 ohms will cause
the charge rate to be ~178.5 mA which is close enough.
Ed

Some minor points:

1) 7 ohms will likely be difficult to find.
6.8 ohms (6R8) is a standard value. (Calulates to 183mA.)

2) 1.25V^2 / 6.8 ohms = 0.23 watts
Rather than running a 1/4W component so near its limit,
you'll want to go to a 1/2 watt part.

3) To assure that you will always be
above the specified Dropout Voltage range,
make sure the output of the power supply feeding these
is at least 2.5V more than the fully-charged voltage of the packs.
(A *lot* more than 2.5V
will make the LM317's dissipation unnecessarily large.)

 

[neon]

the person trying to help you is not informed of battery iditotsyncrosis YOU CANNOT ty any two or more battery across any power source period. that is one shure way to end up with all dead battery in a short time peroid. each battery must be insulated by a diode and/or resistance to do so.try one lm337 with 1 ohm current limit and tie a diode inseries to all battery to achieve isolation. or buy 6 lm337. my choice is one lm337 unless you want to trow the kitchen sink into it.simplicity is the mother of all nature.

 

[ehsjr]

-----
V+ ---Vin|LM317|Vout---+
----- |
Adj [R]
| |
+----------+---> To battery pack +

Gnd -----------------------> To battery pack -

Compute the value for R as follows:
1) Charge rate (C) is 1/10 the mAh rating of the pack. In the
example of an 1800 mAh, that's 180 mA
2) R = 1.25/C
In this example, that's 1.25/.18 which equals 6.944
That's an odd value, so use 7 ohms. 7 ohms will cause
the charge rate to be ~178.5 mA which is close enough.
Ed

Some minor points:

1) 7 ohms will likely be difficult to find.
6.8 ohms (6R8) is a standard value. (Calulates to 183mA.)

---[1R]---[1R]---[5R]--- or ---[2R]---[5R]---
Either = 7 ohms. Standard power resistors. Easy.

2) 1.25V^2 / 6.8 ohms = 0.23 watts
Rather than running a 1/4W component so near its limit,
you'll want to go to a 1/2 watt part.

1/4 watt was not specified. The point of using
at least 1/2 watt is good advice with the 6.8,
and it automatically covered making 7 ohms out
of 1 or 2 power resistors. (Mouser's smallest
5 ohm R is 5 watts.)

3) To assure that you will always be
above the specified Dropout Voltage range,
make sure the output of the power supply feeding these
is at least 2.5V more than the fully-charged voltage of the packs.
(A *lot* more than 2.5V
will make the LM317's dissipation unnecessarily large.)

You must have missed this sentence: "You need to set the power
supply voltage above the terminal voltage of the pack by about
3 volts. " and the rest of that paragraph: "Your 5 volt packs
will require 5 cells. Each cell has a terminal voltage
of 1.43 volts at room temperature, or 7.15 volts total,
so set your power supply at 10 volts. "

Ed