Charging While Wall-Warting

[Robert Reed]

Thanks MrAl
Will download and play around with it. If I hit any stumbling blocks, I will take you up on your kind offer.

[smariotti]

You still have to mention your output currents so we can make sure
your regulators dont overheat.

Thanks for your continued help and analysis. The thought process itself is as useful for me to understand as the information that it carries. There are many independent axes of constraint that I hadn't been considering---not having had any hard engineering background. (My background is in computer science, which has very different requirements specification approaches.)

Here's the board I put together last night. I'm trying to save space as all of this (including batteries) needs to fit inside a portable enclosure.



Mild first degree burns on the thumb and forefinger notwithstanding, it was a fun thing to build. The output wires are on the right hand side of the board going to the right. From top to bottom they are: black (ground), green (3.3v), red (5.5v), blue (8.4v (straight off the battery pack) as I needed to add an additional voltage requirement... 7.5v-ish.)

Right off the bat I can see that I need a better resistor than that 1/4W one that leads from the wall supply's + lead. From what you described, I'll want a 5W one.

Some details about the load:

I'm driving three separate devices: a LCD flat panel display (300mA @ 7.5v (currently 8.4v straight from the battery pack)), an electronic game device designed for battery operation (250mA @ 3.3v) and a PS/2 keyboard (< 100mA @ 5.0v.)

That's ~650mA not counting the regulators themselves, which I'm sure draw something. At any rate, I'm well under 1A for the total load.

The Wall Wart I'm using outputs 12v at 2A I believe (I need to check that when I get home.)

[MrAl]

Hi again,

Just a thought...if you can connect your 3.3v regulator to the main
supply instead of the output of the 5v regulator you might get away
without needing any heatsinks for the 5v regulators. With the
3.3v reg connected as you have drawn, it pulls an extra 250ma
through the regulators so that means they have to pass 350ma
instead of 100ma, which is quite a difference.
I didnt look up your part for the 3.3v reg though so check that out
to find out if it can take a higher input voltage.

[Sambuchi]

Mr Al I would like to say thanks for your input on this topic.. I've been following this thread for the past week. I too am doing the same thing with a small project that is near completion... hope to post it here shortly.

smariotti- your board looks great!

Thanks again.

Tony

[MrAl]

Hi again Tony,

Ok good luck with it, and yes i'd like to hear how the project goes
as you get it completed.

[smariotti]

Just a thought...if you can connect your 3.3v regulator to the main supply instead of the output of the 5v regulator you might get away
without needing any heatsinks for the 5v regulators. With the 3.3v reg connected as you have drawn, it pulls an extra 250ma through the regulators so that means they have to pass 350ma instead of 100ma, which is quite a difference. I didnt look up your part for the 3.3v reg though so check that out
to find out if it can take a higher input voltage.

Sadly my 3.3v ldo regulator only takes 7.5v at the high end. My battery level exceeds this when fully charged (a total of seven 1.2v cells for 8.4v.)

The charger, however, works great. I've been testing it over the last few days by charging it and discharging it and taking voltage measurements at various points. I'm going to integrate it with my device this week and have confidence that I have a nice solution for driving my portable commodore 64 clone (made from Jeri Ellsworth's DTV device revision 3 board.)

My next project is going to be to integrate the new Microsoft Xbox 360 "Chatpad" keyboard with my device. The Chatpad is a plug-in thumb keyboard for the Xbox 360 controller that's driven internally by a PIC 18F883 (?) and I believe uses USB protocol. That'll be a crash course in brain surgery as I attempt to figure out PIC chips and the PS/2 keyboard interface.

Thanks for all of your help, MrAl. And I'm glad that this thread is useful to others. After learning a lot from other people's forum posts in various places, I'm happy to have helped someone else.

[philba]

when using batteries, I can't help but think that a switcher is a better approach.

If I were doing this, I'd look closely at the LTC3555. Integrated switcher, battery manager. runs off bat, usb or ext power. has one LDO output and 3 SWmode outputs. Looks like a very clean solution.

[Sambuchi]

philba, that chip is incredible!
http://www.linear.com/pc/downloadDocume ... 856,D25066

It has tons of smarts built into it. I like the I2C feature and the built-in LED function... Linear took all the fun out of it!

Definitely a smart battery charger.

Little over kill for what I want to do thou...

[smariotti]

All is well in charger and battery pack land, except for one little thing... my 8.4v connection doesn't have the power to drive my device.

I hooked up a (approximately) 500mA load and suddenly the voltage drops nearly in half and the device powers down.

Here's a picture of the schematic, modified for my new 12v wall wart, the 220 ohm resistor between the diode and the battery, and my new 8.4v output.



When I put a load on the 8.4v output, the voltage begins to drop like mad and finally ends up around 5v or so by the time I've disconnected the load and measured.

Does this mean my cells are bad? These are salvaged cells, so I suspect either the cells or bad, or I need some kind of protection to tap the power supply for 8.4v at the cells themselves.

[philba]

well, the first thing I would try to answer is "what's the current draw of your 3.3 regulator?" Secondly, what's the capacity of your batteries? that should tell you a lot.

Now, I hate to say it but your approach is seriously inefficient. Basically, you are using 8.4V to derive 3.3V for an efficiency of around 40%. I would look into using a switcher instead of the 3.3 linear. I'd bag the 7805s and design a 3.3 switcher that could take 14-6V as input. There are also lots of power management chips that have switchers and battery chargers built in, like the one I mentioned above. You should be able to at least double your battery life.

[dyarker]

How long did you have to heat a battery to get solder to melt?

If you didn't file, or emory cloth, the battery ends first; and used a BIG soldering iron you probably "cooked" at least one of the cells.

-OR-

Recharging through 220 Ohms takes about 6 days. When you do your test the batteries may be 1.2V, but at the almost discharged end of the curve.

[smariotti]

well, the first thing I would try to answer is "what's the current draw of your 3.3 regulator?" Secondly, what's the capacity of your batteries? that should tell you a lot.

Now, I hate to say it but your approach is seriously inefficient. Basically, you are using 8.4V to derive 3.3V for an efficiency of around 40%. I would look into using a switcher instead of the 3.3 linear. I'd bag the 7805s and design a 3.3 switcher that could take 14-6V as input. There are also lots of power management chips that have switchers and battery chargers built in, like the one I mentioned above. You should be able to at least double your battery life.

Point well taken. In fact, I think I WILL move over to a much more efficient system once I get this working. It's been a learning process and if I can understand what's wrong and fix it, then I can look at a second design. I have the PDF for that recommended switching power supply chip, though after scanning through it I can't say I quite get how to apply it yet.

My cells are 700mAh each, with 7 in series for 4900mAh?

If I'm reading the data sheet right, when input current is 300mA, "Ground pin current in normal operation mode" is typically 5mA. Is this the current draw you're referring to?

[dyarker]

"My cells are 700mAh each, with 7 in series for 4900mAh?"

No, with cells in series it is still 700mAH at a higher voltage.

[MrAl]

Hi there,

I tend to agree that there is something wrong with one or more of
the 'salvaged' cells. Cells that sit around for a long time often
dont work once used again, however cycling them sometimes brings
them back to life. To cycle, you charge the cell up then discharge it,
then charge it back up again. If this doesnt help after about 4 or 5
cycles then the cell or cells should be discarded. Of course NiCds
are to be recycled.
You should test each cell individually by measuring its voltage.
Some cells die before others.

You probably want to go with new cells anyway, and a larger size too
like sub C's at least. The smaller AA's dont hold as much charge.

You also have to realize that when you increase the nominal voltage
of the battery pack that you also have to increase the input voltage,
or at least decrease the resistor selected for charging, after calculating
the low line and high line charge currents and dissipations in this
resistor to make sure it wont burn up and there is also enough current
flow to charge the cells in a decent amount of time.

It's too bad you cant use NiMH cells here, as they have a bit more
capacity. I did a test recently to see what would happen if a NiMH
cell was charged at a relatively low current for 24 hours 7 days a
week and i found out the cells die after about 6 months. The
cell size was AA and the current was 90ma pretty much constant.
I'll post a graph of its performance (actually two of them) a bit later
today i hope.

In the following graph, the cells are being charged and as soon as the
charge begins (1 amp constant current) the cell voltage jumps up to
a very abnormal level. The terminal voltages follow a very abnormal
path, and the -delta V starts way too early for the rated capacity
of these cells. The temperature starts to rise at a constant rate
as soon as the charge begins too, which also shows that these cells
are not really being charged even though there is current applied
because there is no endothermic reaction.
The discharge profile is even worse, and is not even worthy of a graph
of its performance as it is so bad and can be described easily...
The cells start out at about 1.4 volts each, which isnt too far from
typical for a good cell, and then as light load is applied the cell voltages
immediately dip down to around 0.85 volts, which is clearly not working
as the time between load application and cell dip below 0.9 volts is so
short that you couldnt use these cells to run a fleas motorbike for one
second.





Here's a graph of the discharge to show how bad these cells really are...

[smariotti]

"My cells are 700mAh each, with 7 in series for 4900mAh?"

No, with cells in series it is still 700mAH at a higher voltage.

Ahh, Ok. Well, I'd think I'd probably only get an hour's operation if my total charge is never more than 700mAh.