I found this circuit in a cheep rechargeable LED touch. However, as far as I could understand, I saw an issue of this circuit (other than the safety issue). So genius! Please tell me what I think on this, is correct!!
In order to drop the voltage, a Mylar Capacitor is used. Further, the Ni-Cad battery (3.6/700mA) will be charged with 4V/70mA source. However as we know, when the battery-charge go up (after 1-2 hours), the drain current from the source go down (<70mA). However, at the start, the charging current and the voltage was calculated according to V=IR and then a capacitor was selected in order to fulfill the above charging-values. So according to this argument, after 1-2 hours, charging voltage on the battery must be high than starting 4V (Ex; if I=30mA then voltage on the battery should go up to 134V). can this be happened ?? Can it make any harm to the battery?
Please see the attach diagram
In order to drop the voltage, a Mylar Capacitor is used. Further, the Ni-Cad battery (3.6/700mA) will be charged with 4V/70mA source. However as we know, when the battery-charge go up (after 1-2 hours), the drain current from the source go down (<70mA). However, at the start, the charging current and the voltage was calculated according to V=IR and then a capacitor was selected in order to fulfill the above charging-values. So according to this argument, after 1-2 hours, charging voltage on the battery must be high than starting 4V (Ex; if I=30mA then voltage on the battery should go up to 134V). can this be happened ?? Can it make any harm to the battery?
Please see the attach diagram
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