Final answer:
The base resistor RB for the TIP31C BJT should be 110 Ohms, and the collector resistor RC should be 4.7 Ohms to drive the required 1A through the OP295A IR LED, powered by a +5V source voltage.
Step-by-step explanation:
The student's question pertains to the design of a transistor driver circuit for the TIP31C BJT and an OPTEK OP295A IR LED, with the output of a 555 timer. The target is to drive 1A through the LED with a +5V source voltage.
To calculate the base resistor (RB), we start by identifying the base current (IB) required to drive the IR LED at 1A. The TIP31C has a current gain (beta), which is typically around 25. Therefore, IB = IC/beta = 1A/25 = 40mA. Given that VBE(on) (the voltage between the base and emitter when the transistor is on) is typically 0.7V for silicon BJTs, the voltage across RB will be Vcc - VBE(on) = 5V - 0.7V = 4.3V. Using Ohm's Law (V = IR), RB = V/IB = 4.3V / 40mA = 107.5 Ohms. A standard nearest value of 110 Ohms can be used for RB.
For the collector resistor RC, since the collector-emitter voltage (VCE(sat)) when the transistor is fully on is typically around 0.3V, the voltage across RC will be Vcc - VCE(sat) = 5V - 0.3V = 4.7V. Again, using Ohm's Law, RC = V/IC = 4.7V / 1A = 4.7 Ohms. A standard nearest value for RC would be 4.7 Ohms.
Therefore, the appropriate values for the resistors in this circuit would be RB = 110 Ohms and RC = 4.7 Ohms.