Help with a battery pack design

[Ryan]

ehsjr wrote:

Ryan wrote:

Luhan Monat wrote:

You need to supply 5V at 900ma to power the device. A set of 4
2400mah NiMH will give you 4.8volts for less than 3 hours.


So if I go with something like 6v @ 2900mah with a 5v zener diode
would I be ok?


No. A zener is the wrong choice for you, unless you have
one that is rated at over 5 watts at 5 volts. Don't know
where you'll find a zener like that. You would not get a
huge increase in run time, even if you could get the 2900
mah cells and a proper zener. With a 6 volts supply, and
a zener shunt regulator, you have to waste about a watt
of power.

The solution Luhan proposed - 4.8 volts using NiMh cells
rated at 2400 mAh - would work, and wastes no power. You
might be able to use a 6 volt battery with a 1N540x diode
in series to drop the voltage to about 5.4 (wasting a bit
over 1/2 watt). Or you could use a battery with a high
efficiency (up to 95% efficient) DC-DC converter. But, if
you need more run time than Luhan's solution, your battery
capacity *must* be increased, no matter what configuration
you use.

So I'm understanding the flaw in using a diode now...enless I use
something that gives me a larging increase in mAhs its a waste.

So - how much run time do you need?

Thats a good question. First test using Luhan proposal gave me about
25min of run time (testing again after recharge complete). I would like
a lot more then I'm going to get for the size I want but I think a good
goal would be 2-3hrs.

Ed

 

[Rich Grise]

On Mon, 30 May 2005 11:02:11 +1200, Ken Taylor wrote:

"Ryan" <ryan.news@rnelnet.com> wrote in message
news:529c2$429a47a2$c690d0bc$18733@TSOFT.COM...
Luhan Monat wrote:

You need to supply 5V at 900ma to power the device. A set of 4 2400mah
NiMH will give you 4.8volts for less than 3 hours.


So if I go with something like 6v @ 2900mah with a 5v zener diode would
I be ok?

No, just use the power supply you have for the device. It has plenty of
capacity for the task.

Will you people please read the original post? Ryan wants to make a
battery pack so that he doesn't _have_ to plug the damn thing in.

The existing power supply has the _ability_ to provide _up to_ 2.5 Amps.
The device itself only needs 900 mA, which is still a lot. Luhan Monat
got it right - a set of four 2400 mAH NIMH cells will power the thing
for 2400/900 hours.

Do they make D sized NIMHs?

I'm thinking, a 6 volt golf cart battery and a silicon diode, if Ryan
wants it to transmit for a full day before recharging.

Thanks,
Rich

 

[Rich Grise]

[crossposted to sci.electronics.design AND sci.electronics.basics,
WITH FOLLOWUPS-TO SET to sci.electronics.basics]

On Sun, 29 May 2005 23:52:41 -0700, Ryan wrote:

Pooh Bear wrote:
4 NiMhs or indeed 4 Nicads will give around 4.8 ~ 4.9 V terminal voltage
( close enought to 5 V ) when fully charged - which is clearly enough to
make your device operate. Trouble is - when discharging - the voltage
will gradually drop.

This may or may not - depending on the design of your device - prove to
be problematic or not.

So, I used 4 NiMhs that tested at 5.1v fully charged. The device ran for
about 25min, I added a simple toggle switch so I could kill the use of the
batteries. I turn the device on and off a couple of time to change
settings. I'm recharging now to do a constant run to get a better time.
When I took them out they tested at 4.6v. So indeed you are right when
the voltage drop it wasn't able to run the device.

Good! You're getting it!

with, no I just need to get battery life to last longer. I guess the
question I'm asking what does the 5.1v zener diode do?

A diode usually conducts current in only one direction. In the reverse
direction, an ordinary diode has a "reverse breakdown" rating - this is
the voltage (remember, voltage is pressure, or force - current is flow)
at which the diode will conduct in the reverse direction.

Zener diodes are almost the same as ordinary diodes, except that their
reverse breakdown voltage is calibrated. But the thing is, once breakdown
is reached, there is nothing to limit the current. It goes from infinite
resistance to almost zero resistance. A 5.1V zener across a 6V battery
will conduct as much current as the battery can provide until the battery
gets discharged to a terminal voltage of 5.1V, which is that Zener's
"switch point", if you will. A big, beefy 5 watt Zener could probably
survive this, but this is NOT what you want to do.

An ordinaly silicon diode, like a 1N4004, has a forward voltage drop
of typically, .7V. This _does_ throw away some of your battery capacity
as heat, but unfortunately, some frat boy decided that 5V was a neat
number to design equipment to run off of, and apparently in his ivory
tower, they never noticed that there's no such thing as a 5V battery.
Maybe it was a conspiracy.

But that's neither here nor there. I digress.

The point I'm getting at is, if you use 6V worth of batteries, and
put a silicon diode, like a 1N4004 (actually, any of 1N4001-1N4007
will work, as will almost any silicon diode) in series, in the forward
direction, that will give you 5.3 volts, which will probably run your
equipment. For a while.

If you put a zener in parallel with the batteries - it'll simply exhaust
them faster. It'll take current ( charge ) that that could be usefully
used by the device.

Does the diode only allow the 5.1v to pass through? And if so how much
tolerance would it have, 6v ok 20v bad.

There's no chance of 20V so why are you worrying ?

My question was more geared to learn the function of a diode. So I get
now that it takes (wastes) current but if I passed 20v to a 5.1v diode
would to still function or would it burn?

You really need to read something about the difference between voltage
and current. Voltage is a difference of potential. It's almost exactly
equivalent to pressure, like water pressure in a pipe. Electromotive
force. Current is "how much is flowing (past some point)". You don't
"pass" a voltage. You "apply" a voltage, and current "passes".
Check some of these: http://www.google.com/search?&q=%2Bvoltage+%2Bcurrent

5v @ 10000mah would be ok, right, it would just mean the battery power
would last longer?

The more mAh you have - the longer your battery pack will last. As long
as you don't put any zeners in there !

So this brings up the question what can I do to make the life last
longer now. Different batteries but I don't know of any that run a 5v
that provide more mAh, thus the need to bring the voltage down.

I answered in another branch here - if you want it to last more than
an hour or two, you're going to need a BIG HONKING BATTERY.
I suggested a golf cart battery, and was wondering if there are D sized
NimHs.

Pls report back !
Ditto!

Good Luck!
Rich

 

[Rich Grise]

On Sun, 29 May 2005 23:59:44 -0700, Ryan wrote:

ehsjr wrote:
....
So - how much run time do you need?

Thats a good question. First test using Luhan proposal gave me about
25min of run time (testing again after recharge complete). I would like a
lot more then I'm going to get for the size I want but I think a good goal
would be 2-3hrs.

<commentary>

Damn!

I've just done a quick google on '"D size" NIMH' and somebody's getting
robbed or something. I have a AA NimH in my hand, and it's got "2000
mAh" emblazoned on its side bigger than its logo. But - wait, here's one:
http://www.batteryspace.com/index.asp?PageAction=VIEWPROD&ProdID=709

Most of the other links were in the 2000-2500 mAh range, which is
a lot less than I'd expect from a D cell. Do they just put an AA in
a D case, and sell them? That's a ripoff, in my book.
</commentary>

Does anybody remember "The Secret of NIMH?"

Thanks!
Rich

 

[John Fields]

On Sun, 29 May 2005 23:59:44 -0700, Ryan <ryan.news@rnelnet.com>
wrote:

ehsjr wrote:

So - how much run time do you need?

Thats a good question. First test using Luhan proposal gave me about
25min of run time (testing again after recharge complete). I would like
a lot more then I'm going to get for the size I want but I think a good
goal would be 2-3hrs.

---
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:

http://www.panasonic.com/industrial/battery/oem/images/pdf/Panasonic_NiMH_HHR300SCP.pdf

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

 

[John Fields]

On Mon, 30 May 2005 15:25:06 -0500, John Fields
<jfields@austininstruments.com> wrote:

On Sun, 29 May 2005 23:59:44 -0700, Ryan <ryan.news@rnelnet.com
wrote:

ehsjr wrote:

So - how much run time do you need?

Thats a good question. First test using Luhan proposal gave me about
25min of run time (testing again after recharge complete). I would like
a lot more then I'm going to get for the size I want but I think a good
goal would be 2-3hrs.

---
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:

http://www.panasonic.com/industrial/battery/oem/images/pdf/Panasonic_NiMH_HHR300SCP.pdf

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.

---
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

--
John Fields
Professional Circuit Designer

 

[Walter Harley]

"John Fields" <jfields@austininstruments.com> wrote in message
news:abum911vbcb9du1fhk0ik3g38k8jhdtggj@4ax.com...

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

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 http://www.jameco.com. A 12V to 5V 1A converter
is about $25. See their part number 216936CH.

 

[ehsjr]

Ryan wrote:

ehsjr wrote:

Ryan wrote:

Luhan Monat wrote:

You need to supply 5V at 900ma to power the device. A set of 4
2400mah NiMH will give you 4.8volts for less than 3 hours.


So if I go with something like 6v @ 2900mah with a 5v zener diode
would I be ok?



No. A zener is the wrong choice for you, unless you have
one that is rated at over 5 watts at 5 volts. Don't know
where you'll find a zener like that. You would not get a
huge increase in run time, even if you could get the 2900
mah cells and a proper zener. With a 6 volts supply, and
a zener shunt regulator, you have to waste about a watt
of power.

The solution Luhan proposed - 4.8 volts using NiMh cells
rated at 2400 mAh - would work, and wastes no power. You
might be able to use a 6 volt battery with a 1N540x diode
in series to drop the voltage to about 5.4 (wasting a bit
over 1/2 watt). Or you could use a battery with a high
efficiency (up to 95% efficient) DC-DC converter. But, if
you need more run time than Luhan's solution, your battery
capacity *must* be increased, no matter what configuration
you use.


So I'm understanding the flaw in using a diode now...enless I use
something that gives me a larging increase in mAhs its a waste.

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.

So - how much run time do you need?


Thats a good question. First test using Luhan proposal gave me about
25min of run time (testing again after recharge complete). I would like
a lot more then I'm going to get for the size I want but I think a good
goal would be 2-3hrs.


Ed

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

 

[Rich Grise]

On Mon, 30 May 2005 14:22:17 -0700, Walter Harley wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:abum911vbcb9du1fhk0ik3g38k8jhdtggj@4ax.com...
[...]
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


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 http://www.jameco.com. A 12V to 5V 1A converter
is about $25. See their part number 216936CH.

Why not a $15.00 6-volt 5 Ah battery, and a silicon diode or two?
http://www.google.com/search?q=6+volt+sla+battery

Or an ordinary diode in series with a Schottky?

Cheers!
Rich