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

M

#### Mark Brehob

Jan 1, 1970
0
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

E

#### ehsjr

Jan 1, 1970
0
Mark said:
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

J

#### JeffM

Jan 1, 1970
0
ehsjr said:
-----
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

Oct 21, 2006
1,325
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.

Last edited:
E

#### ehsjr

Jan 1, 1970
0
JeffM said:
-----
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

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