Finding Transistor Load Line Endpoints in CE Configuration

The question asks us to find the coordinates of the load line for a transistor in Common Emitter (CE) configuration with a given power supply voltage (\(V_{CC}\)) and collector resistor (\(R_C\)).

In a transistor circuit, the DC load line is a graphical representation on the output characteristics curve (\(I_C\) vs. \(V_{CE}\)) that shows all possible operating points (\(V_{CE}\), \(I_C\)) for a given value of \(R_C\) and \(V_{CC}\). The equation for the DC load line is derived from Kirchhoff's Voltage Law applied to the collector-emitter loop:

\[V_{CC} = I_C R_C + V_{CE}\]

Rearranging this equation to express \(I_C\) in terms of \(V_{CE}\) (or vice versa) gives the load line equation:

\[V_{CE} = V_{CC} - I_C R_C\]

To draw the load line, we typically find two extreme points on this line:

• The Cut-off Point: This occurs when the transistor is not conducting current, meaning the collector current \(I_C\) is approximately zero. At cut-off, the entire supply voltage drops across the collector-emitter terminals. To find this point, set \(I_C = 0\) in the load line equation: \[V_{CE} = V_{CC} - (0) R_C\] \[V_{CE} = V_{CC}\] So, the coordinates for the cut-off point are \((V_{CE}, I_C) = (V_{CC}, 0)\).
• The Saturation Point: This occurs when the transistor is conducting maximum current, ideally when the voltage across the collector-emitter terminals (\(V_{CE}\)) is approximately zero (for ideal transistors). To find this point, set \(V_{CE} = 0\) in the load line equation: \[0 = V_{CC} - I_C R_C\] Solving for \(I_C\): \[I_C R_C = V_{CC}\] \[I_C = \frac{V_{CC}}{R_C}\] So, the coordinates for the saturation point are \((V_{CE}, I_C) = (0, \frac{V_{CC}}{R_C})\).

Now, let's use the given values from the question:

• \(V_{CC} = +12 V\)
• \(R_C = 1 k\Omega = 1000 \Omega\)

Calculate the coordinates:

• Cut-off Point (\(I_C = 0\)): \(V_{CE} = V_{CC} = +12 V\) Coordinates: \((+12 V, 0 mA)\)
• Saturation Point (\(V_{CE} = 0\)): \(I_C = \frac{V_{CC}}{R_C} = \frac{12 V}{1000 \Omega} = 0.012 A\) Convert current to milliamperes (mA): \(I_C = 0.012 A \times 1000 \frac{mA}{A} = 12 mA\) Coordinates: \((0 V, 12 mA)\)

Thus, the two endpoints of the load line are \((+12 V, 0 mA)\) and \((0 V, 12 mA)\). We can represent these coordinates as \((V_{CE}, I_C)\).

Let's compare these calculated coordinates with the given options.

The calculated endpoints \((+12 V, 0 mA)\) and \((0 V, 12 mA)\) match the coordinates provided in Option 1. These points define the DC load line on the transistor's output characteristics. Any valid operating point (Q-point) for this circuit must lie on this line.

Revision Table: Transistor Load Line Calculations

Additional Information: Transistor Operating Regions

Understanding the load line helps visualize the transistor's operating regions:

• Cut-off Region: Lies below the load line, typically near the \(V_{CE}\)-axis (\(I_C \approx 0\)). The transistor acts like an open switch.
• Saturation Region: Lies above the load line, typically near the \(I_C\)-axis (\(V_{CE} \approx 0\)). The transistor acts like a closed switch.
• Active Region: Lies between the cut-off and saturation regions, along the load line. The transistor operates as an amplifier in this region, where \(I_C = \beta I_B\) (for CE configuration). The Q-point is usually set in the active region for linear amplification.

The load line shows the possible DC operating points, and the base bias circuit determines where on this line the actual Q-point is located. Changes in the input signal (base current \(I_B\)) cause the operating point to move along the load line, resulting in a corresponding change in \(I_C\) and \(V_{CE}\).