Charging While Wall-Warting

[smariotti]

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.

I have a flux pen I salvaged from some trash when a neighboring company cleared out of their office space. I can practically make solder stick to teflon with that thing. I don't think I fried them.

[/quote]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.[/quote]

Hmm... I checked my math and I think 220 Ohms should create a 50mA charge current from my 12v source (11v after diode.)

Also, the time to charge is calculated from multiplying
the Ah rating of the cell by 1.4 and then dividing by the charge
current (low charge rates). For any number of 700mAh cells in series,
700 times 1.4 is 980 and divide that by 50 (for 50ma) and you get 19.6
which is almost 20 hours charge time.

If I use that 50mA number, then with a 12v source, minus the diode drop of approx. 1v, that's R = V / I for R = 11v / 0.050A = 220 ohms.

[smariotti]

It's too bad you cant use NiMH cells here, as they have a bit more
capacity.

I'm going to find a place on line that has inexpensive NiMH AA cells (I need to stick with AA because of their size.. I need that thickness for my case) that are rated at 2500mAh-ish. I see AA's for sale on eBay and on some battery web sites that are listed at 2800mAh and sometimes higher... I wonder if the labelling is honest with those. I will probably stick with a reputable brand like Sony or Sanyo.

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.

This mimics the behavior that I'm seeing with the NiCd's. There's no telling how badly these were abused before I found them. They LOOKED OK from the outside, and were inside a device when I pulled them. They were basically picked out of the trash, though, so the original owner may have known something I didn't. New cells sound like a must.

And I think I may end up moving my 3.3v LDO in my charger circuit. That first 7805 is heating up quite hot during charging.

[philba]

"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.

well, perhaps, if your draw was less than 700 mA. You won't get all 700 out of the beasts anyway. my guess is that at 1A they will last maybe a half hour if fresh.

[dyarker]

I get 16mA charging through 220 Ohms (you forgot to subtract the voltage across the batteries from 12V)

Cheers,

[MrAl]

It's too bad you cant use NiMH cells here, as they have a bit more
capacity.

I'm going to find a place on line that has inexpensive NiMH AA cells (I need to stick with AA because of their size.. I need that thickness for my case) that are rated at 2500mAh-ish. I see AA's for sale on eBay and on some battery web sites that are listed at 2800mAh and sometimes higher... I wonder if the labelling is honest with those. I will probably stick with a reputable brand like Sony or Sanyo.

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.

This mimics the behavior that I'm seeing with the NiCd's. There's no telling how badly these were abused before I found them. They LOOKED OK from the outside, and were inside a device when I pulled them. They were basically picked out of the trash, though, so the original owner may have known something I didn't. New cells sound like a must.

And I think I may end up moving my 3.3v LDO in my charger circuit. That first 7805 is heating up quite hot during charging.

Hi again,

I think you are calculating your charge current based on your old
battery pack voltage of 7.2 volts?

Your new pack voltage (nominal) is 8.4 volts, but the charge voltage
should be calculated first:
v=8.4*1.4/1.2
which comes out to:
v=9.8
What this means is that you have to calculate your charge current
like this now:
i=(12-0.7-9.8 )/R
Doing the math, that current i comes out to only about 7ma which
isnt enough, but luckily sometimes the wall wart has high internal
capacitance so your 12v wall wart might actually put out more average
dc voltage even with some load. The max would be 16.8 volts,
and that would lead to a resistor value of:
R=6.3/0.050=126 ohms,
but again that is max. What you will have to do is make a measurement
once you get the circuit hooked up. Using a value of maybe 100
ohms, measure the dc voltage across the resistor and then calculate
the charge current based on Ohms Law. You should have the normal
load applied too, if it's always going to be turned on, or try it with the
load turned on and off if the load sometimes is switched off and see
what the difference is between having the load connected and not
having it connected.

There are a few variables here and it's wise to check with all the
variables at their min and max values otherwise the cells may not
charge or will overcharge.

[smariotti]

I get 16mA charging through 220 Ohms (you forgot to subtract the voltage across the batteries from 12V)

Cheers,

D'oh!

I'm getting 48 ohms now. 12v - 1.0fv - 8.4v = 2.5v. 2.5v / 0.050mA = 48 ohms.

Your new pack voltage (nominal) is 8.4 volts, but the charge voltage
should be calculated first:
v=8.4*1.4/1.2
which comes out to:
v=9.8

I don't understand the 1.2 in the calculation. That's the per-cell voltage? The 1.4 is the rule-of-thumb for calculating charge, right? So you're saying that I want 9.8v at least to charge that battery?

[dyarker]

"D'oh!

I'm getting 48 ohms now. 12v - 1.0fv - 8.4v = 2.5v. 2.5v / 0.050mA = 48 ohms."

No sweat.

So, were you testing with nearly empty batteries, and that is why the voltage out went to about half right away?????????

C U L -

[MrAl]

I get 16mA charging through 220 Ohms (you forgot to subtract the voltage across the batteries from 12V)

Cheers,

D'oh!

I'm getting 48 ohms now. 12v - 1.0fv - 8.4v = 2.5v. 2.5v / 0.050mA = 48 ohms.

Your new pack voltage (nominal) is 8.4 volts, but the charge voltage
should be calculated first:
v=8.4*1.4/1.2
which comes out to:
v=9.8

I don't understand the 1.2 in the calculation. That's the per-cell voltage? The 1.4 is the rule-of-thumb for calculating charge, right? So you're saying that I want 9.8v at least to charge that battery?

Hi again,

1.4 is the rule of thumb for calculating charge yes, but it is *also* the
voltage when the cell is being charged. Since the nominal voltage of
each cell is 1.2 and the charge voltage is 1.4 we multiply by
1.4/1.2 to get the cell pack voltage, which is 9.8, which means you
would use 9.8 in your charge calculation, not 8.4 volts.

[smariotti]

the cell pack voltage, which is 9.8, which means you
would use 9.8 in your charge calculation, not 8.4 volts.

12v - 1.0v for diode - 9.8 for target cell pack charge voltage = 1.2v / 0.050 = 24 ohm resistor, then.

I've got a 47 ohm in there from overnight that's charging today. It looks like the cells are holding more charge based on my load testing this morning. I'll swap the 47 ohm resistor out for a 24-ish next time around.

Then I have to look at getting the 3.3v LDO regulator's load off the 7805's and retool the circuit a bit. The top 7805 is getting too hot for my liking... it's "OW! dammit!" hot.

[MrAl]

Hi again,

Ok, but once you change the resistor you have to recalculate the
effects at high line and low line, and the power rating of the resistor.
Did you do that yet?

Im not sure if you mentioned this yet or not, but what type package
are your LM7805 ic's, TO3 or TO220 ?

[dyarker]

From photo on page 2, it is TO220.

It needs a heatsink.

[MrAl]

Hi again,

Thanks Dale.

smariotti, you will need a heatsink or else reduce power like
you said. You can check the data sheet to find out how much
power the TO220 package can take without a heat sink too.
If you do end up using a heat sink, you also need one for the
other 7805 because that will get hot once it is being used.
Alternately, isolate one of the 7805's with a si pad and proper
mounting hardware and you might be able to mount
the two of them on the same heat sink. This is of course if you
really do need a heat sink (check the data sheet).

[dyarker]

The tab on 7805s is the same as the common reference pin, so one on one side of heatsink, and other on other side, shouldn't be a problem. From the schematic, only one 7805 works at a time.

I suspect (without doing the math) that if the 3.3V regulator input is moved from 5V to 12V/8.2 (via additional diodes), that it would then need a heatsink. (pay sooner, or pay later )

You could do away with one 5V regulator by feeding one LDO 5V regulator by schottkey diodes from 12V/battery. But don't! First investigate a 5V switching module that accepts like 6V to 15V input. That will almost double battery life.

Cheers,

[smariotti]

The tab on 7805s is the same as the common reference pin, so one on one side of heatsink, and other on other side, shouldn't be a problem. From the schematic, only one 7805 works at a time.

I suspect (without doing the math) that if the 3.3V regulator input is moved from 5V to 12V/8.2 (via additional diodes), that it would then need a heatsink. (pay sooner, or pay later )

You could do away with one 5V regulator by feeding one LDO 5V regulator by schottkey diodes from 12V/battery. But don't! First investigate a 5V switching module that accepts like 6V to 15V input. That will almost double battery life.

Cheers,

I'm looking at ways of doing away with one regulator.. and becoming less excited about linear regulators in general and more and more inclined to look at switching regulators like the Linear Tech. one that was referenced earlier in this thread. The 3 output DC/DC stepdown one with SPI interface, salad shredder attachment, and rotary nose clipper. LTC3555?

Also from earlier in the thread was rshayes:

Sometimes you can use diodes to do the switching automatically.

Use a wall wart with enough output voltage to power a battery charger. The output of the battery charger would have a diode between it (anode) and the battery (cathode). When the wall wart is connected, the battery charger will be able to charge the battery through the diode. When the wall wart is disconnected, the battery charger output will be zero, and the diode disconnects it from the battery.

Add a diode from the wall wart output (anode) to the regulator input (cathode). Add another diode from the battery positive (anode) to the regulator input (cathode). Whichever voltage is higher, the wall wart or the battery, will power the output through the regulator. If the wall wart is plugged in, it will be the higher voltage (since you chose it that way), and it will power the battery charger and the regulators. If the battery voltage is higher than the wall wart output, it will power the regulators.

Since the voltage is only a few volts, you may want to use shottkey rectifiers, to minimize the voltage drop in the diodes. This is most critical on the connection between the battery and the regulators. Using diodes rated for currents 5 or 10 times the actual load current will also lower the voltage drop.

I took a stab at mocking up something like what the venerable and distinguished Mr. Hayes suggested, though I'm unclear still on what he meant by "battery charger." Is that a whole battery charging circuit? A charge controlling chip of some sort? Simply a resistor to cut the current from the 12v wart input to the battery itself? In my mock-up I made it a black-box using a schematic symbol for a fuse.



This schematic makes the assumption that the 3.3v regulator can handle a larger voltage input than I already know it actually can. I'll have to replace that part (LP3876 LDO linear regulator) with one that can take 12v max, I think.

Otherwise, does the diode placement and direction make sense?

And yes, I should stick some caps to ground on the outputs of the regulators. I see that now that I'm looking at it more critically since I'm about to post it. :-0

[rshayes]

You have the diodes in the right directions. I would suggest a schottky diode for D2, this will give you a fraction of a volt lower drop, but that can be significant with a battery pack.

The charging circuit could be a simple resistor, but with only 12 volts in the current control may not be very good, as has been discussed in some detail by others in this thread. A better arrangement would be some kind of current source. A PNP transistor with the base voltage set by two diodes and an emitter resistor to set the current would probably work, since you have about 2 volts available once you take the diode drop into consideration. The charging current will drop slightly as temperature increases due to the use of a diode as a voltage reference. This might actually be a desirable characteristic ("If you can't fix it, feature it.").