Question

In the first quadrant of V-I characteristics of Uni-Junction Transistor, the slopes of characteristic are in the following sequence

• A. \( \text{Negative, zero, positive, zero and negative} \)
• B. \( \text{Positive, zero, negative, zero and positive} \)
• C. \( \text{Positive, zero, negative, zero and negative} \)
• D. \( \text{Negative, zero, positive, zero and positive} \)

Hint:

UJT Characteristic Landmarks: - Peak Point: Slope goes from positive to negative (passes through zero). - Valley Point: Slope goes from negative to positive (passes through zero). Remembering these two turnaround points makes it easy to spot the full sequence on a test.

Answer 1

The Correct Option is B

Concept: A Uni-Junction Transistor (UJT) is a three-terminal semiconductor device that exhibits a distinct negative resistance region. Its characteristic curve is divided into three distinct operating regions:
• Cut-off Region: The emitter voltage is below the peak point triggering threshold. Only a small leakage current flows.
• Negative Resistance Region: Once the emitter voltage reaches the peak voltage ($V_p$), the device triggers open. Holes are injected into the base channel, increasing conductivity and causing the emitter voltage to drop even as the current increases.
• Saturation Region: The voltage drops to its valley point minimum ($V_v$) and then begins rising again linearly with current due to ohmic resistance limits.

Step 1: Analyzing the sequence of slopes along the V-I curve.
Let's follow the first-quadrant emitter characteristic curve from left to right as current increases:
• Initial Cut-off Phase: Emitter voltage increases with current. The slope ($\frac{dV}{dI}$) is positive.
• Peak Point Area: The curve reaches its maximum peak voltage ($V_p$). At this localized peak, the derivative slope briefly becomes zero.
• Triggered Drop Phase: Emitter voltage drops while current increases (the negative resistance effect). The slope ($\frac{dV}{dI}$) is negative.
• Valley Point Area: The curve reaches its lowest voltage point ($V_v$). At this localized minimum, the derivative slope briefly becomes zero again.
• Final Saturation Phase: The device acts like a standard resistor, where voltage rises along with increasing current. The slope ($\frac{dV}{dI}$) returns to positive.

Step 2: Matching the complete sequence.
Putting the steps together in order, the sequence of slopes is: \[ \text{Positive} \rightarrow \text{Zero} \rightarrow \text{Negative} \rightarrow \text{Zero} \rightarrow \text{Positive} \] This precisely matches the sequence in option (2). Hence, the correct choice is option (2).