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Modern Physics
Junction Transistor
A junction transistor is a three-terminal semiconductor device formed by sandwiching either a thin layer of p-type semiconductor between two n-type layers (n-p-n) or an n-type layer between two p-type layers (p-n-p). It functions primarily as a current-controlled amplifier or a solid-state switch.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Working principle of n-p-n and p-n-p transistors, biasing conditions for active, cut-off, and saturation regions, transistor current components (I_E = I_B + I_C), and common-base (CB) and common-emitter (CE) current gains (α and β).
PGT extension
Detailed mathematical analysis of common-collector configuration, hybrid equivalent parameters (h-parameters), thermal runaway, stability factor, DC load line analysis, and transistor biasing circuits (voltage divider bias).
Prerequisites
Energy bands in solids and Fermi level
p-n junction diode forward and reverse biasing characteristics
Minority and majority charge carrier dynamics in semiconductors
Structural Regions of a Transistor
The Emitter is heavily doped to supply a large number of majority charge carriers into the base region.
The Base is extremely thin and lightly doped to minimize the recombination of injected charge carriers before they reach the collector.
The Collector is moderately doped and physically larger than the emitter and base to effectively dissipate heat generated by collecting charge carriers.
Transistor Biasing and Operating Regions
Active Region: Emitter-base junction is forward-biased and collector-base junction is reverse-biased; used for linear amplification.
Saturation Region: Both emitter-base and collector-base junctions are forward-biased; the transistor acts as a closed switch.
Cut-off Region: Both emitter-base and collector-base junctions are reverse-biased; the transistor acts as an open switch.
Current Components in a Transistor
The total emitter current (I_E) is the sum of hole and electron currents crossing the emitter-base junction.
A small fraction of carriers recombine in the base, constituting the base current (I_B).
The remaining major portion of injected carriers crosses into the collector, constituting the collector current (I_C).
Transistor Configurations
Common-Base (CB) Configuration: Base is common to input (emitter) and output (collector); characterized by low input impedance, high output impedance, and voltage gain with current gain (α) less than unity.
Common-Emitter (CE) Configuration: Emitter is common to input (base) and output (collector); characterized by moderate input impedance, moderate output impedance, and high current gain (β) greater than unity.
Common-Collector (CC) Configuration: Collector is common to input (base) and output (emitter); characterized by very high input impedance, very low output impedance, and voltage gain close to unity (voltage follower).
Derivations
Relation between Common-Base Current Gain (α) and Common-Emitter Current Gain (β)
Conditions: Active region operation with negligible reverse saturation leakage current (I_CBO)
Write the fundamental current relation for any transistor: $I_E = I_B + I_C$.
Define the common-base current gain as $\alpha = \Delta I_C / \Delta I_E \approx I_C / I_E$.
Define the common-emitter current gain as $\beta = \Delta I_C / \Delta I_B \approx I_C / I_B$.
Substitute $I_B = I_E - I_C into the definition of \beta: \beta = \frac{I_C}{I_E - I_C}$.
Divide numerator and denominator by I_E: $\beta = \frac{I_C / I_E}{1 - (I_C / I_E)}$.
Substitute I_C / $I_E = \alpha into the expression$ to yield the final relation.
$V_T = Thermal voltage (\approx 26 mV at room temp)$ Siemens (S) or \Omega^{-1}
Active region, room temperature
Graphs and experiments
CB Input Characteristics
Draw: Curves of input current (I_E) versus input voltage (V_EB) for various constant values of output voltage (V_CB).
Interpret: Resemble forward-biased p-n junction diode characteristics. A change in V_CB has a negligible effect on the curves.
Exam use: Evaluates understanding of input resistance determination ($r_e = \Delta V_EB / \Delta I_E at constant V_CB)$.
CE Output Characteristics
Draw: Curves of collector current (I_C) versus collector-emitter voltage (V_CE) for different constant base currents (I_B).
Interpret: Shows three regions: active (nearly horizontal flat curves), saturation (steep rise at low V_CE), and cut-off (I_C \approx 0 when $I_B = 0)$.
Exam use: Frequently tested for identifying operating regions, load line plotting, and determining output resistance.
DC Load Line and Q-Point
Draw: A straight load line equation $I_C = -\frac{1}{R_L} V_CE + \frac{V_CC}{R_L} superimposed on the CE output characteristics intersecting the active region at the Quiescent point (Q)$.
Interpret: The Q-point defines the DC operating voltages and currents without any signal input. Proper placement ensures distortion-free amplification.
Exam use: Numerical and conceptual questions on maximum signal handling capacity and thermal stability.
Solved numericals
TGT core
In a transistor, the collector current is 2.4 mA and the base current is 0.1 mA. Find the emitter current and the current amplification factor α.
Use the fundamental current relation: $I_E = I_B + I_C$.
Substitute the given values: $I_E = 0.1 \text{ mA} + 2.4 \text{ mA} = 2.5 \text{ mA}$.
Calculate the current gain $\beta = I_C / I_B = 2.4 / 0.1 = 24$.
Calculate α using the relation $\alpha = \beta / (1 + \beta) or \alpha = I_C / I_E = 2.4 / 2.5$.
Answer: Emitter current = 2.5 mA, $\alpha = 0.96$
TGT core
The common-base current gain α of a transistor is 0.98. What is the change in collector current if the emitter current changes by 2 mA?
Recall the definition of common-base current gain: $\alpha = \Delta I_C / \Delta I_E$.
Rearrange the formula to solve for the change in collector current: \Delta $I_C = \alpha \times \Delta I_E$.
Substitute the given values: \Delta $I_C = 0.98 \times 2 \text{ mA}$.
Calculate the final product.
Answer: \Delta $I_C = 1.96 \text{ mA}$
PGT extension
A transistor connected in common-emitter mode has a current gain $\beta = 50$. The load resistance R_L is 4 kΩ, and the internal input resistance is 500 \Omega. Calculate the voltage gain and power gain of the amplifier.
Use the formula for voltage gain in CE mode: $A_v = \beta \times (R_L / r_{in})$.
Substitute the given values: $A_v = 50 \times (4000 \Omega / 500 \Omega)$.
Simplify the fraction: 4000 / 500 = 8.
Calculate voltage gain: $A_v = 50 \times 8 = 400$.
Calculate power gain using $A_p = \beta \times A_v: A_p = 50 \times 400 = 20000$.
Answer: Voltage gain = 400, Power gain = 20,000
Assumptions and limitations
Junction transistors operate under low-frequency small-signal approximations unless high-frequency hybrid-π models are employed.
Carrier recombination in the base is assumed constant, though it varies slightly with temperature and operating voltage.
Transistor parameters are treated as linear over small excursions around the operating point.
Common misconceptions
Common error: The collector current is greater than the emitter current because the collector is larger.
Correct: The emitter current is always the largest current in a transistor ($I_E = I_B + I_C)$.
The emitter supplies all carriers; some are lost in the base, so collector current must be strictly less than emitter current.
Common error: A transistor can step up power passively without an external DC power supply.
Correct: Amplification is a process of transferring a signal from a low-resistance circuit to a high-resistance circuit using DC source energy.
Conservation of energy holds; power gain is achieved by drawing DC power from the bias batteries.
Common error: The base region of a transistor should be heavily doped to facilitate current flow.
Correct: The base region is extremely lightly doped and very thin.
Light doping and small thickness minimize electron-hole recombination in the base, ensuring high collector current.
Collector current with leakage in CE configuration: $I_C = \beta I_B + (1 + \beta) I_CBO$.
Transistor transconductance in the active region is given by $g_m \approx \frac{I_C}{V_T}$.
Glossary
Emitter ($I_E$)The heavily doped region of a transistor that supplies charge carriers.
Base ($I_B$)The extremely thin and lightly doped middle region of a transistor.
Collector ($I_C$)The moderately doped and larger region that collects charge carriers.
Alpha ($\alpha$)Current gain in Common-Base configuration, defined as $\frac{\Delta I_C}{\Delta I_E}$.
Beta ($\beta$)Current gain in Common-Emitter configuration, defined as $\frac{\Delta I_C}{\Delta I_B}$.
Thermal Voltage ($V_T$)Voltage equivalent of temperature, approximately $26\text{ mV}$ at room temperature.
Transconductance ($g_m$)Ratio of change in collector current to change in base-emitter voltage.
Transistor Configurations Comparison
Parameter
Common-Base (CB)
Common-Emitter (CE)
Common-Collector (CC)
Input Impedance
Low
Moderate
Very High
Output Impedance
High
Moderate
Very Low
Current Gain
Less than unity ($\alpha$)
High ($\beta$)
High ($1+\beta$)
Voltage Gain
High
High
Close to unity
Phase Shift
$0^\circ$
$180^\circ$
$0^\circ$
Practice Quiz
Choose an option to reveal the explanation.
1. Which region of a junction transistor is heavily doped to supply a large number of majority charge carriers?
Explanation: The emitter is heavily doped specifically to inject a large number of majority charge carriers into the base region.
2. Why is the base region of a transistor made extremely thin and lightly doped?
Explanation: The base is thin and lightly doped so that minimum charge carriers recombine within it before successfully reaching the collector.
3. In the active region of a transistor, what are the biasing conditions for the emitter-base and collector-base junctions?
Explanation: For linear amplification (active region), the emitter-base junction is forward-biased and the collector-base junction is reverse-biased.
4. When a transistor acts as a closed switch, which operating region is it in?
Explanation: In the saturation region, both the emitter-base and collector-base junctions are forward-biased, allowing maximum current flow (closed switch).
5. Which fundamental equation correctly relates the emitter current ($I_E$), base current ($I_B$), and collector current ($I_C$)?
Explanation: By Kirchhoff's current law at transistor nodes, the total emitter current equals the sum of base current and collector current: $I_E = I_B + I_C$.
6. What is the physical significance of the Common-Base current gain $\alpha$?
Explanation: Common-base current gain $\alpha$ is defined as $\frac{\Delta I_C}{\Delta I_E}$ and is strictly less than $1$ because $I_C < I_E$.
7. If the Common-Base current gain $\alpha$ of a transistor is $0.98$, what is its Common-Emitter current gain $\beta$?
Explanation: Using the relation $\beta = \frac{\alpha}{1 - \alpha}$, we get $\beta = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49$.
8. Which configuration provides very high input impedance, very low output impedance, and a voltage gain close to unity?
Explanation: The Common-Collector configuration acts as a voltage follower, characterized by very high input impedance, low output impedance, and $A_v \approx 1$.
9. What is the corrected expression for collector current in the common-emitter configuration including leakage current $I_{CBO}$?
Explanation: The complete collector current equation in CE configuration incorporating leakage is $I_C = \beta I_B + (1 + \beta) I_{CBO}$, where $I_{CEO} = (1+\beta)I_{CBO}$.
10. What is the approximate transconductance ($g_m$) of a transistor operating at a collector current of $1\text{ mA}$ at room temperature?
11. Which statement about the collector region is correct?
Explanation: The collector is moderately doped and physically larger to effectively dissipate the heat generated by collecting charge carriers.
12. When both emitter-base and collector-base junctions are reverse-biased, the transistor operates in the:
Explanation: When both junctions are reverse-biased, almost no current flows, and the transistor acts as an open switch in the cut-off region.
13. In a Common-Emitter amplifier, what is the relationship between the AC voltage gain ($A_v$) and current gain ($\beta$)?
Explanation: The AC voltage gain in a CE amplifier is given by $A_v = \beta \left(\frac{R_L}{r_{in}}\right)$, where $R_L$ is load resistance and $r_{in}$ is input resistance.
14. Which misconception about transistor currents is false?
Explanation: The statement that collector current is greater than emitter current is false; $I_E$ is always the largest current since $I_E = I_B + I_C$.
15. What is the AC power gain ($A_p$) expression in a Common-Emitter configuration?
Explanation: Power gain is the product of current gain and voltage gain, hence $A_p = \beta \times A_v$ for the CE configuration.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. In a transistor, the collector current is $15\text{ mA}$ and the base current is $1.5\text{ mA}$. What is the emitter current?
Explanation: Using $I_E = I_B + I_C$, we get $I_E = 1.5\text{ mA} + 15\text{ mA} = 16.5\text{ mA}$.
2. If the current gain $\alpha$ of a transistor is $0.96$, the collector current changes by $12\text{ mA}$ when the emitter current changes by:
Explanation: Since $\alpha = \frac{\Delta I_C}{\Delta I_E}$, we have $\Delta I_E = \frac{\Delta I_C}{\alpha} = \frac{12\text{ mA}}{0.96} = 12.5\text{ mA}$.
3. For a transistor circuit, if $\beta = 100$ and base current $I_B = 20\text{ \mu A}$, calculate the approximate collector current ignoring leakage.
4. Which of the following configurations is most suitable for impedance matching when connecting a low impedance source to a high impedance load?
Explanation: Common-Collector (emitter follower) has very high input impedance and very low output impedance, making it ideal for impedance matching.
5. In a common-base amplifier circuit, the change in emitter current is $2\text{ mA}$ and the resulting change in collector current is $1.96\text{ mA}$. Calculate the current amplification factor $\alpha$.
6. What happens to the base current if the emitter current increases while collector current remains constant?
Explanation: From $I_B = I_E - I_C$, if $I_E$ increases while $I_C$ is constant, $I_B$ must increase.
Timed Topic Test
Attempt these questions under a ten-minute limit using a separate timer.
1. Which region in a transistor is doped moderately compared to the heavily doped emitter?
Explanation: The collector region is moderately doped, whereas the emitter is heavily doped and the base is lightly doped.
2. In a transistor operating in the active region, the reverse bias on the collector-base junction causes:
Explanation: The reverse bias on the collector-base junction creates a high resistance barrier that sweeps across the injected minority carriers, forming the collector current.
3. If $\alpha = 0.95$, what is the value of the CE current gain $\beta$?
4. The base current $I_B$ in a transistor is generally what percentage of the total emitter current $I_E$?
Explanation: Because the base is thin and lightly doped, only a small fraction ($1\%$ to $5\%$) of carriers recombine, making $I_B$ very small compared to $I_E$.
5. Which transistor configuration exhibits a phase reversal of $180^\circ$ between input signal voltage and output signal voltage?
Explanation: The Common-Emitter configuration inverts the output signal relative to the input, producing a $180^\circ$ phase shift.
6. Why is amplification possible in a transistor despite collector current being slightly less than emitter current?
Explanation: Amplification is achieved by transferring a signal from a low-resistance input circuit to a high-resistance output circuit, multiplying the voltage/power using DC bias energy.
7. If $I_C = 2\text{ mA}$ and $I_B = 40\text{ \mu A}$, find the Common-Emitter current gain $\beta$.
8. What is the primary role of the DC bias batteries in a transistor amplifier circuit?
Explanation: DC biasing establishes the necessary operating point (quiescent point) by properly forward-biasing the emitter-base junction and reverse-biasing the collector-base junction.
9. In the saturation region, what happens to the collector-emitter voltage ($V_{CE}$)?
Explanation: In saturation, both junctions are forward-biased, causing the collector-emitter voltage $V_{CE}$ to drop to a minimum saturation value (typically around $0.2\text{ V}$).i
10. Calculate the transconductance $g_m$ of a transistor operating with a collector current of $2.6\text{ mA}$ at room temperature ($V_T \approx 26\text{ mV}$)?