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Simple overnight charger for 5 x AA batteries

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neazoi

neazoi

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If the batteries are new, same make and batch, it is perhaps ok to charge them in series. Perhaps if you charge them at a low enough current, you are fine. But forget about fast charge. If you consider the AA cells are 1000mAh capacity, you should charge them at a constant current of 50-70mA and the current should drop to a trickle when the voltage reaches 1.4V*5=7V approx. Many small radios use in circuit charging and they use a simple const current source with a low compliance voltage (i.e., if the batteries are missing, the voltage will not go above 7v). You can make a const current source with either discrete or integrated components. But you will have no warning if one cell goes bad (refuses to get up to full charge and voltage) and more elaborate circuits should monitor the voltages of all the individual cells.
 
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c_mitra said:
If the batteries are new, same make and batch, it is perhaps ok to charge them in series. Perhaps if you charge them at a low enough current, you are fine. But forget about fast charge. If you consider the AA cells are 1000mAh capacity, you should charge them at a constant current of 50-70mA and the current should drop to a trickle when the voltage reaches 1.4V*5=7V approx. Many small radios use in circuit charging and they use a simple const current source with a low compliance voltage (i.e., if the batteries are missing, the voltage will not go above 7v). You can make a const current source with either discrete or integrated components. But you will have no warning if one cell goes bad (refuses to get up to full charge and voltage) and more elaborate circuits should monitor the voltages of all the individual cells.
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I am looking for something simple here. 5 x Eneloops will be used, 1900mAh each. The simple charger suggested in post #1 has been tested. It is an overnight charger (actually more hours than a night). It keeps the batteries a bit under charged though (0.1v or so) but it's ok. I do not need high performance, just a way to avoid having to remove the batteries all the time from the device to charge them.

What changes do I need in this charger to make it work with the 5 series Ni-MH?
 

What changes do I need in this charger to make it work with the 5 series Ni-MH?
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Your broad ideas are sound.

How about using a single FET as a const current source? How about a FET switch to turn off /on charging the battery?
 
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The other thread which you linked has Brian's suggestion to install a shunt (two diodes in series) across the battery. It's an easy safety feature and it can be expanded so your 5 cells do not go above a certain voltage. Some kind of voltage regulation (in addition to current limiting). Between 6.5 and 6.75 V total. This should maintain each cell at 1.3 or 1.35 V (assuming they are new, same brand, same A-Hr capacity (as post #2 recommends).

If you think you'll leave them charge for a couple days, then 100 mA is a good charge rate (C/20). Eventually it tapers to a tiny amount of trickle charge as the batteries settle to a resting voltage.
 
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Expanding on the shunt diode trick - you could fit two diodes across each cell. It isn't the best solution of course but as long as the charging current is kept at a safe level, it will help by steering a little more current to cells with lower voltage, especially ones below 2xVf level. It also protects against the charger going over-voltage if any cell fails open circuit or is accidentally removed.

Brian.
 
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betwixt said:
Expanding on the shunt diode trick - you could fit two diodes across each cell. It isn't the best solution of course but as long as the charging current is kept at a safe level, it will help by steering a little more current to cells with lower voltage, especially ones below 2xVf level. It also protects against the charger going over-voltage if any cell fails open circuit or is accidentally removed.

Brian.
Click to expand...
Like this?

The 78M06 will give 1.2v accross each cell, how can I increase this? Maybe using a 78M08 and somehow decrease the voltage?

Also what is the correct value and wattage of the current set resistor for 100mA as BradTheRad suggested?
 

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You probably want the regulator to have the lower output voltage rather than increase it. The higher it's regulation voltage, the more you need to feed in to it. Think of it's operation like this:
The internal regulation circuit tries to maintain a constant voltage between it's GND and OUT pins, it doesn't actually care whether GND is 0V or not, OUT will always be the regulation voltage above the GND pin voltage. That is used to turn it into a constant current source by providing a fixed resistance load between GND and OUT, a fixed voltage across a fixed resistance implies a fixed current must flow. The output is still from the OUT pin but GND follows the load voltage (battery voltage) so providing a constant current into it.

The resistor is simple to calculate, ignoring the tiny current through the GND pin, it is the value needed to pass the desired current at the regulator output voltage. For example, if you use a 5V regulator and want 100mA the value is (R=V/I) 5/0.1 = 50 Ohms. Similarly, the power loss is V*V/R = 25/50 = 0.5W.

You can see that using a lower voltage regulator works exactly the same as a higher voltage one but obviously with less demand on input voltage. Incidentally, did you realize that variable voltage regulators (LM317 etc.) have an internal reference of only about 1.25V so in the same configuration they work with even less input voltage.

Brian.
 
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betwixt said:
You probably want the regulator to have the lower output voltage rather than increase it. The higher it's regulation voltage, the more you need to feed in to it. Think of it's operation like this:
The internal regulation circuit tries to maintain a constant voltage between it's GND and OUT pins, it doesn't actually care whether GND is 0V or not, OUT will always be the regulation voltage above the GND pin voltage. That is used to turn it into a constant current source by providing a fixed resistance load between GND and OUT, a fixed voltage across a fixed resistance implies a fixed current must flow. The output is still from the OUT pin but GND follows the load voltage (battery voltage) so providing a constant current into it.

The resistor is simple to calculate, ignoring the tiny current through the GND pin, it is the value needed to pass the desired current at the regulator output voltage. For example, if you use a 5V regulator and want 100mA the value is (R=V/I) 5/0.1 = 50 Ohms. Similarly, the power loss is V*V/R = 25/50 = 0.5W.

You can see that using a lower voltage regulator works exactly the same as a higher voltage one but obviously with less demand on input voltage. Incidentally, did you realize that variable voltage regulators (LM317 etc.) have an internal reference of only about 1.25V so in the same configuration they work with even less input voltage.

Brian.
Click to expand...

Very interesting! So a simple 7805 and a 50R can just do it like drawn Brian?
 

Yes!
The regulator and resistor can be thought of as a constant current device that also drops a voltage. That voltage is the one the regulator is designed to produce.

It is wise to add a capacitor (~1uF) across the IN and GND pins to keep the regulator stable. There is a compromise, if the value is too low it won't improve stability, if it is too high, its initial charging current will be carried through the load. For battery charging it makes no difference if there is a brief spike in current but beware of it if you use the design in other applications.

Brian.
 

It would be prudent to add a regulator feeding the const current source. If you use a 5V regulator to produce a const curr source of 100mA, and use a 12V input to the 5V regulator, the compliance voltage will be 12V (minus the regulator drop)- say 10.5V. If you have a light load, the regulator will drop a large voltage and may get hot (need heat sinking). When the battery is fully charged, the diodes will take the full current and get hot (at 100mA may not need heat sinking). It will be wise to feed the 5V (use a low drop regulator) regulator input with 8V that can be got from another regulator (then the compliance voltage will also drop to 6.5V) - in that case you can forget the parallel 10 diodes (saves real estate!) and the regulator will shut off once the output voltage will be 6.5V- you can also use a 9V zener to provide 100mA - to the input of the const current source. A const curr source can be made using a simple JFET and a resistor. Simplify!
 

If you have a light load, the regulator will drop a large voltage and may get hot (need heat sinking).
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Not sure I understand that c_mitra. The resistor dissipation is constant (it is across the regulator output) and the regulator dissipates (Vdrop * I). Given that I is constant, the regulator dissipation is highest when the load is heaviest and the output is pulled low, in other words when the voltage across it is higher. Even under short circuit conditions and say 12V input it will only dissipate about 1.2W. With fully charged cells at say 1.25V, the dissipation with 12V input is only 12-(5*1.25)*I = 0.575W.

I agree you could limit the input voltage to prevent the output going over-voltage but that would only apply to the combined battery cell load. The idea of the diodes is to prevent each cell individually going over voltage while allowing slower charging ones to catch up.

100mA is quite a lot for a JFET current regulator to carry and it still uses 2 components!

Brian.
 

Even under short circuit conditions and say 12V input it will only dissipate about 1.2W. With fully charged cells at say 1.25V, the dissipation with 12V input is only 12-(5*1.25)*I = 0.575W.
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You are correct and I goofed! Taking care of the cells individually is always a better idea (than limiting the overall voltage and current).
 

betwixt said:
Don't worry, I have 'goofy' days - about 7 of them every week. :lol:

Brian.
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I do not think so guys.
You are brilliant and I have learn so much on this forum! A big thumbs up :)
 

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