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.
What changes do I need in this charger to make it work with the 5 series Ni-MH?
Like this?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.
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.
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.If you have a light load, the regulator will drop a large voltage and may get hot (need heat sinking).
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.
Don't worry, I have 'goofy' days - about 7 of them every week. :lol:
Brian.
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