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The answer is in the circuit description, did you read it?
When the battery voltage or input voltage falls below 10.5V, a cut-off circuit is used to prevent deep discharging of the battery. Resistor R3, zener diode ZD1 (10.5V) and transistor T2 form the cut-off circuit. When the voltage level is above 10.5V, transistor T2 conducts and its base becomes negative (as set by R3, VR2 and ZD1).
But when the voltage reduces below 10.5V, the zener diode stops conduction and the base voltage of transistor T2 becomes positive.
It goes into the ‘cut-off’ mode and prevents the current in the output stage. Preset VR2 (22k) adjusts the voltage below 0.6V to make T2 work if the voltage is above 10.5V.
Yes but i am confused at that point.
It says that when T2 conducts its base is -ive. and when it does not its base is +ive. but i couldnt understand how can that be possible.
Keeping in mind that tip 127 has legs labeled as 1-base, 2-collector and 3-emitter
The case pinout of the transistor has nothing to do with the schematic.
When the output voltage is below the zener voltage (10.5) the base and collector voltages are almost the same.
When the output goes above that voltage the base has 10.5v because the zener/resistor regulate the voltage while the output gets higher.
When the output becomes 0.6+10.5 the transistor conducts.
The base will either have the same voltage as the output (below 10.5) or it will have a negative voltage compared to the emitter when Vout >10.5
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