BJT bias point (Q-point) intuition: A transistor is biased at the middle of the load line. If the base resistance RB is increased (with other conditions unchanged), what happens to the Q-point on the load line?

Introduction / Context:
Biasing a BJT at the middle of the load line aims for symmetric signal swing. The base network (often RB or a divider) sets base current and thus collector current for a given transistor. Understanding how RB affects the Q-point is critical for stability and distortion control.

Given Data / Assumptions:

• Transistor initially biased at mid-load-line (roughly half of VCE across transistor).
• We increase RB while other elements remain unchanged.
• Assume a fixed supply and temperature; small-signal beta variations ignored.

Concept / Approach:
Base current IB is approximately (VBIAS − VBE) / RB (for a simple bias). Increasing RB reduces IB. Since IC ≈ beta * IB in the active region, a smaller IB reduces IC. On the load line IC decreases, moving the Q-point downward (toward cutoff), and VCE rises accordingly.

Step-by-Step Solution:
Original IB sets IC(mid) on the load line.Increase RB → IB decreases.IC decreases → Q-point moves “down” (smaller IC) along the load line.VCE increases as the collector drop RC*IC reduces.

Verification / Alternative check:
Graphically, with the same load line (fixed RC and supply), decreasing IC shifts the operating point downward (lower IC axis value). SPICE simulations of RB sweep reproduce this movement.

Why Other Options Are Wrong:
(b) “Up” would require increased IC (opposite effect). (c) A parameter change must move the Q-point somewhere. (d) It stays on the same load line unless parameters like RC or supply change drastically.

Common Pitfalls:
Confusing the vertical (IC) direction with horizontal (VCE) on the plot; forgetting that less base drive reduces collector current.

Final Answer:
down.