Bipolar Junction Transistor Current Gain | Generated by AI
Basics of Bipolar Junction Transistors (BJTs)
To understand the current gain \( \beta \) in this transistor problem, let’s start from the fundamentals. A Bipolar Junction Transistor (BJT) is a three-terminal semiconductor device used for amplification and switching in electronic circuits. It comes in two main types: NPN (most common) and PNP, but the principles are similar. The terminals are:
- Emitter (E): Releases charge carriers into the circuit.
- Base (B): A thin layer that controls the flow of carriers.
- Collector (C): Collects the majority of the carriers.
In an NPN BJT, current flows from collector to emitter when the base-emitter junction is forward-biased (positive voltage on base relative to emitter) and the base-collector junction is reverse-biased (negative voltage on base relative to collector). This setup defines the active region of operation, where the transistor acts as a current amplifier.
Key Operating Regions
BJTs have three main regions of operation:
- Cutoff Region: Both junctions are reverse-biased. No significant current flows (\( I_B \approx 0 \), \( I_C \approx 0 \)). Transistor is “off.”
- Active Region (or Forward-Active): Base-emitter forward-biased, base-collector reverse-biased. Here, a small base current \( I_B \) controls a much larger collector current \( I_C \). This is the amplification mode.
- Saturation Region: Both junctions forward-biased. Maximum current flows; transistor is “on” like a closed switch, but no amplification.
The problem specifies the transistor is in the active region, so we’re dealing with amplification behavior.
Currents in a BJT
In the active region:
- Base Current (\( I_B \)): Small current injected into the base, mostly to provide minority carriers.
- Collector Current (\( I_C \)): Much larger current flowing from collector to emitter, proportional to \( I_B \).
- Emitter Current (\( I_E \)): Total current out of the emitter, where \( I_E = I_B + I_C \) (by Kirchhoff’s Current Law).
The relationship is approximately linear: \( I_C \approx \beta I_B \), where \( \beta \) (beta) is the DC current gain or current amplification factor. It’s a dimensionless ratio, typically 50–300 for discrete transistors, depending on the device.
- \( \beta \) isn’t perfectly constant—it varies slightly with temperature, voltage, and current levels—but in basic analysis, we assume it’s constant in the active region.
- The collector current also has a small leakage component (\( I_{CBO} \)), but it’s negligible: \( I_C = \beta I_B + (1 + \beta) I_{CBO} \approx \beta I_B \).
DC Current Gain vs. Small-Signal (Incremental) Gain
- DC \( \beta \): Calculated at a specific operating point using instantaneous currents: \( \beta = \frac{I_C}{I_B} \).
- Small-Signal \( \beta \) (or \( h_{fe} \)): For dynamic changes (e.g., AC signals), it’s the ratio of small changes: \( \beta \approx \frac{\Delta I_C}{\Delta I_B} \). This is useful when the transistor is biased at one point and we apply a small variation, as \( \beta \) is assumed constant over that small range.
In problems like this, where currents “change from” one value to another, the incremental approach often gives the “approximate” \( \beta \) if the change is small relative to the operating point.
Applying This to the Problem
The scenario: Transistor in active region. Base current increases from \( I_{B1} = 12 \, \mu\text{A} \) to \( I_{B2} = 22 \, \mu\text{A} \). Collector current changes from \( I_{C1} = 1 \, \text{mA} \) to \( I_{C2} = 2 \, \text{mA} \).
First, convert units for consistency (1 mA = 1000 μA):
- \( I_{B1} = 0.012 \, \text{mA} \), \( I_{B2} = 0.022 \, \text{mA} \).
- \( \Delta I_B = I_{B2} - I_{B1} = 0.022 - 0.012 = 0.010 \, \text{mA} \) (or 10 μA).
- \( \Delta I_C = I_{C2} - I_{C1} = 2 - 1 = 1 \, \text{mA} \).
DC \( \beta \) at Each Point
- At initial point: \( \beta_1 = \frac{I_{C1}}{I_{B1}} = \frac{1}{0.012} \approx 83.33 \).
- At final point: \( \beta_2 = \frac{I_{C2}}{I_{B2}} = \frac{2}{0.022} \approx 90.91 \).
These match options A (83) and B (91) closely, but \( \beta \) isn’t constant here—it increased slightly, which can happen in real transistors due to factors like Early effect (base width modulation). However, the problem asks for “its current gain \( \beta \) is approximately,” implying a single value, and the changes are significant (83% increase in \( I_B \), 100% in \( I_C \)), so DC points alone don’t capture a “constant” \( \beta \).
Incremental (Small-Signal) \( \beta \)
Assuming \( \beta \) is roughly constant over the operating range, the best estimate is the slope of the \( I_C \) vs. \( I_B \) line: \[ \beta \approx \frac{\Delta I_C}{\Delta I_B} = \frac{1 \, \text{mA}}{0.010 \, \text{mA}} = 100. \]
This matches option C exactly. Why use this?
- In the active region model, \( I_C = \beta I_B \) is linear, so the ratio of changes reflects the gain directly.
- The average of DC points is \( \frac{83.33 + 90.91}{2} \approx 87 \), but that’s not an option and ignores the linear assumption.
- Option D (183) might come from a miscalculation like \( \frac{2}{0.011} \) or doubling something, but it’s off.
Graphically, if you plot \( I_C \) vs. \( I_B \):
- Points: (0.012, 1) and (0.022, 2).
- Slope = rise/run = \( \frac{1}{0.01} = 100 \).
This incremental \( \beta \) is what circuit designers often use for bias point analysis or small-signal models.
Why This Matters in Practice
In amplifiers, you bias the transistor in the active region (e.g., at mid-point currents) so small signal variations stay linear around \( \beta \). Here, the change is large, but the problem tests understanding of the proportional relationship.
If \( \beta \) varied more wildly, we’d need a full SPICE simulation, but for MCQs, 100 is the intended answer.
For deeper reading:
All About Circuits: BJT Fundamentals
Electronics Tutorials: Transistor as an Amplifier