wow, this one is good, i can directly replace the usb port with my 7805... though, usb port supplies 500mA, 7805 can supply for about 1.5A... so my charging time will be not that long...
Not entirely true
what that circuit is is a
constant current source, which is started 'trigged' by the insertion of batteries, and turned off by the temperature of the batteries.
The time taken to chage batteries depends on two things: 1) battery capacity, 2) charge current.
In the above circuit the charge current is determined by the transistor gain and its base resistor. The comparator pulls the base of the transistor low via the resistor, so the current through the resistor is, according to the designer, 5.2mA. He says the gain of the transistor is 90, so..90 * 5.2 = 468mA.
Now a USB port has a limit of 500mA, so he chose the above value to be well under that, but if you were to use a 7805, sure it can supply 1A (max!!). But the circuit would still draw 468mA. Lowering the value of the base resistor would increase the current flowing through the transistor...or...pick a PNP power transistor with higher gain, say 150. (780mA) this would charge standard 2000mAH batteries in about 3 hours.
That said the comparators output has limited current capability, so if you google 'constant current source' you will find an alternative circuit. You could also use a LM317 as a constant current source! Better regulation, more current capability but you would need a small signal transistor to turn it on/off.
can i replace LM393 with any dual op-amp like lm358? or quad op-amp like lm324?
I don't see why not for this circuit, just beware that the inputs/outputs of many opamps cannot go within a volt or two of the supply rails.
i think this one's good for 2battery in series, about 2.4volts... well, how can i transfor this circuit, because i'm gonna use this one with 6volts charger ???
Use a higher voltage power supply
When charging with constant current the voltage across the batteries is determined BY the batteries. However, if they reach the same voltage as your power supply, no current will flow. So, for 6 volts....you'll want a bit of headroom, so try a 9 volt supply, or a 7809.
The threshold of the thermistor is determined by a voltage divider with the power supply, so you should be able to use the same part values for all the resistors/caps.
If I were you, I would measure it and test it completely before letting it lose on batteries. Because the original circuit was designed to cut off charge for a charge current of 468mA. If you charge at a higher current, the temperature threshold may change, but its still a great failsafe providing the thermistor is in contact with the batteries directly.
Blueteeth.