hi,
When the battery is being charged the current causes a voltage drop across the sense resistor also the current for the LED [as its ON] causes a small voltage drop across the sense resistor, which drives the transistor harder on.
When the charge current falls to a point where the transistor start to switch off, the LED will go off and the current in the sense resistor will fall even further, giving a faster LED switch off.
My default value for the voltage drop for a fully turned on silicon diode at rated current is 1 v.
The Vdrop doesn't change that much with current near the rated value of current and circuits are designed so that this drop, or its variation, don't matter much.
Just my two cents.
My default value for the voltage drop for a fully turned on silicon diode at rated current is 1 v.
The Vdrop doesn't change that much with current near the rated value of current and circuits are designed so that this drop, or its variation, don't matter much.
Just my two cents.
hi Willbe,
The problem is with this charger, is that the voltage drop across the series diode does matter. He starts off with say a 1A charge current, Vdp at 1A for a 1N4001 is 0.9V so the battery voltage is within limits. However when the charge current falls to a few mA the Vdp is only 0.7V, so the battery charge voltage rises by 0.2V, which can exceed the battery rating.
This is always a problem with using series diodes in chargers.
The original circuit with the auto cutoff looks as though it would work OK.