Difficulty: Medium
Correct Answer: cutoff point
Explanation:
Introduction:
The DC load line graphically shows all possible operating points for a given collector load and supply. Choosing the Q point determines headroom for signal swings and sets the amplifier's linear range. This question asks you to interpret a Q point located toward the high V_CE/low I_C end of the load line.
Given Data / Assumptions:
Concept / Approach:
At the high V_CE, low I_C end of the load line, the transistor conducts little to no collector current. This is the cutoff region, where the base–emitter junction is not forward-biased sufficiently and the device is effectively off. Moving along the line toward higher I_C and lower V_CE approaches saturation. Midline choices target symmetrical swing.
Step-by-Step Solution:
Identify the endpoints: (V_CE = V_CC, I_C ≈ 0) is cutoff; (V_CE ≈ 0, I_C = V_CC / R_C) is saturation.Locate the specified Q point: high V_CE with small I_C.Match region: this corresponds to cutoff or near-cutoff operation.Conclude that the upper-right Q point represents the cutoff point on the load line.
Verification / Alternative check:
Kirchhoff's voltage law around the collector loop at cutoff shows essentially the full V_CC across the collector–emitter terminals because I_C ≈ 0, confirming the interpretation.
Why Other Options Are Wrong:
Minimum/intermediate/maximum current gain: current gain is primarily a transistor parameter (beta) and bias dependent, but the endpoint in question specifically indicates the cutoff region.
Common Pitfalls:
Confusing the two ends of the load line. Remember: high V_CE with low I_C is cutoff; low V_CE with high I_C is saturation.
Final Answer:
cutoff point
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