Figure 4. The Relay Circuit for Charging and Load Diversion
The relay coil needs several mA of current to activate. And the current gain of the transistor is about 100. It is assumed that the relay
coil requires 40 mA to activate, the value of the base current limiting resistor can be calculated as follows:
Relay coil current = Collector current = IC = 40 mA
The base current I
B
of the transistor TR1 can be calculated using the equation:
IC
= h
FE I
B
I
B
= I
C
/ hFE
I
B = 40 mA / 100 = 0.4 mA
The voltage drop across R2 can be calculated as follows:
V(R2) = 5 V - VBE
V(R2) = 5 V – 0.7 V = 4.3 V
The resistance of R2 can be calculated as follows:
R2 = V(R2) / IB
= 4.3 V / 4 mA ≈ 10kΩ
The current I
C
is enough to drive the relay coil. Therefore the transistor is ON and the current is passing to the collector-emitter
junction of the transistor through the relay coil. Therefore the relay is activated and relay switch is connected to the norm ally open
connection. The load (lamp) is connected between the normally open terminal and ground. Therefore the current from the diode
rectifier will pass through the load to the ground. The current path from the rectifier to the SCR charge controller is cut o ut. The
battery will not be charged any more for this condition. This arrangement o f the circuit prevents the battery from the over charging.
The audible alarm circuit is controlled by the port B pin (RB1). It is shown in Figure.5. The RB1 pin is connected with the b ase of
transistor TR2 through the current limiting resistor. While the b attery is charging the RB1 pin output is logic ’0’ state and transistor
TR2 is OFF. Therefore the buzzer is not activated in this condition. The RB1 output is logic ’1’ state if the battery voltage reaches its
full terminal voltage (full charge). The transistor TR2 is turn ON. The buzzer produces audible alarm.