Solution for 2012 Test 2, Question 2

Solution for 2012 Test 2, Question 2 Given Data: • VCC = +10 V • Base resistor, RB = 100 kΩ • Collector resistor, RC = 3 kΩ • Current source provides IE = 2 mA • Transistor gain, β = 100 (b) Mode of Operation if RL is a Short Circuit When RL is a short circuit, it means RL = 0 Ω. In this case, the emitter is directly connected to ground. Step 1: Calculate Base Current, IB IB = VCC − VBE RB Assuming VBE ≈ 0.7 V : IB = 10 − 0.7 9.3 = = 0.093 mA 100, 000 100, 000 Step 2: Calculate Collector Current, IC IC = β · IB = 100 × 0.093 mA = 9.3 mA Step 3: Check Saturation Condition For the BJT to be in saturation, the collector-emitter voltage VCE must be low. Let’s find VCE : VCE = VCC − IC RC VCE = 10 − (9.3 × 3) = 10 − 27.9 = −17.9 V Since VCE < 0, the BJT is in saturation mode. (c) Mode of Operation as RL Increases As RL changes from a short circuit (0 ohms) to a high value (open circuit), the mode of operation of the BJT changes. Let’s analyze how: Step 1: When RL is Small (Short Circuit) - The BJT is in saturation because the emitter is effectively grounded, and the current source is forcing a constant current through the transistor. Step 2: As RL Increases - As RL increases, the emitter voltage VE also increases due to the voltage drop across RL . - This increase in VE reduces the base-emitter voltage VBE . - A 1 lower VBE decreases the base current IB , which in turn decreases the collector current IC . - When VCE > VBE , the BJT transitions from saturation to the forward active region (F.A.). Thus, as RL increases from a short circuit to an open circuit, the BJT moves from saturation mode to the forward active mode. 2