audioguru2
- Apr 6, 2004
- 12,026
- Joined
- Apr 6, 2004
- Messages
- 12,026
Hi Guys,
I was wondering, how can this project charge four NI-MH batteries from only 5V in series with a diode, and how can a resistor in series with a voltage source supply "constant current"?
The AA size NI-MH batteries that I have charging right now at 150mA, which is less than their recommended current of 210mA, measure 1.4V each, and they are not fully charged yet.
So 4 X 1.4V = 5,6V plus 0.7V for the backwards-protection-diode = 6.3V or more to reach a full charge. Additional voltage must also be added for a voltage drop across the current limiting resistors. From a 7805 5V regulator?
The voltage across the batteries increases while being charged, therefore the current will also vary when charged through a simple resistor.
These problems are solved if the 7805 and its current-limiting resistors are replaced with another LM317 set as a current regulator:
1) Connect a current-setting resistor between its output and its reference pins, and take the output current from the reference pin.
2) Calculate the value of the current-setting-resistor: 1.25 divided by the desired current in amps (10 ohms gives 125mA). 1/2W is fine.
Caution: A fully-charged battery that is still charging gets hot and may be damaged! Frequently check them, snce if you start charging batteries that are not completely discharged then you won't know how long to charge them for.
Project link: http://www.electronics-lab.com/projects/power/035/index.html
I was wondering, how can this project charge four NI-MH batteries from only 5V in series with a diode, and how can a resistor in series with a voltage source supply "constant current"?
The AA size NI-MH batteries that I have charging right now at 150mA, which is less than their recommended current of 210mA, measure 1.4V each, and they are not fully charged yet.
So 4 X 1.4V = 5,6V plus 0.7V for the backwards-protection-diode = 6.3V or more to reach a full charge. Additional voltage must also be added for a voltage drop across the current limiting resistors. From a 7805 5V regulator?
The voltage across the batteries increases while being charged, therefore the current will also vary when charged through a simple resistor.
These problems are solved if the 7805 and its current-limiting resistors are replaced with another LM317 set as a current regulator:
1) Connect a current-setting resistor between its output and its reference pins, and take the output current from the reference pin.
2) Calculate the value of the current-setting-resistor: 1.25 divided by the desired current in amps (10 ohms gives 125mA). 1/2W is fine.
Caution: A fully-charged battery that is still charging gets hot and may be damaged! Frequently check them, snce if you start charging batteries that are not completely discharged then you won't know how long to charge them for.
Project link: http://www.electronics-lab.com/projects/power/035/index.html
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