Zanderist
Member level 3
- Joined
- Sep 16, 2010
- Messages
- 56
- Helped
- 9
- Reputation
- 18
- Reaction score
- 8
- Trophy points
- 1,288
- Location
- New York City
- Activity points
- 1,670
An opamp has a voltage gain of about 200,000 times. If the charging current drops to 3% of 430mA then it is 13mA. A sense resistor value of 1 ohm develops a voltage of 0.013V across it then it can drive the output of an opamp to saturation that will turn off the charging. The battery will not notice the 0.013V tiny voltage lost.Its the Op Amp part I'm a little stuck on because as I've seen in Google searches it involves a sense resistor which I'm afraid will effect the constant voltage and current parameters I'm trying to work to.
DO NOT use this 12V Ni-Cad battery charger to charge a 3.7V Lithium battery cell.Take a look at this: .....
The circuit is this way in order to gain an understanding as to what goes into the making of the chip. In short reverse engineering the ic and putting it on display in an exploded display.I do not know why your circuit has so many parts when simple Li-Po battery charger ICs are available.
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?
We use cookies and similar technologies for the following purposes:
Do you accept cookies and these technologies?