BJT Configurations: Characteristics of CB, CE, and CC Modes
Common Base (CB) Transistor Configuration
Circuit Description
In the Common Base (CB) configuration, the input is applied between the Emitter (E) and Base (B), and the output is taken between the Collector (C) and Base (B). The Base terminal is common to both the input and output. For active region operation, the emitter-base junction is forward-biased, and the collector-base junction is reverse-biased.
Input Characteristics
- Definition: A graph of Emitter Current (IE) versus Emitter-Base Voltage (VEB) for a constant Collector-Base Voltage (VCB).
- Shape: Similar to the forward characteristic of a diode because the E-B junction is forward-biased.
- Input Resistance: Rin = ΔVEB / ΔIE, typically very low (20 Ω – 150 Ω).
Output Characteristics
- Definition: A graph of Collector Current (IC) versus Collector-Base Voltage (VCB) for a constant Emitter Current (IE).
- Shape: Nearly flat in the active region, showing that IC is almost independent of VCB and approximately IC ≈ αIE.
- Output Resistance: Rout = ΔVCB / ΔIC, typically very high (≈ 1 MΩ).
Operating Regions
| Operating Region | Emitter-Base Junction | Collector-Base Junction | Application |
|---|---|---|---|
| Active | Forward-biased | Reverse-biased | Amplifier |
| Cut-off | Reverse-biased | Reverse-biased | OFF Switch |
| Saturation | Forward-biased | Forward-biased | ON Switch |
PN Junction Diode V-I Characteristics
Definition
A PN junction diode is formed by joining P-type and N-type semiconductor materials. It allows current to flow primarily in one direction—from the P-type to the N-type material—when it is forward-biased.
V-I Characteristics
The V-I (Voltage-Current) characteristic curve shows the relationship between the diode current (ID) and the voltage across the diode (VD).
Forward Bias (P positive, N negative)
- The depletion width decreases.
- Current is negligible until the voltage reaches the knee (or threshold) value.
- Knee Voltage (Vγ): Approximately 0.7 V for silicon and 0.3 V for germanium.
- Beyond the knee voltage, the current rises exponentially with a small increase in voltage.
Reverse Bias (P negative, N positive)
- The depletion width increases, and the junction resists current flow.
- Only a small reverse saturation current (IS) flows, which is due to minority carriers.
- At a critical reverse voltage, a breakdown (either Zener or avalanche) occurs, causing a sharp increase in current.
Common Emitter (CE) Transistor Configuration
Circuit Description
In the Common Emitter (CE) configuration, the input is applied between the Base (B) and Emitter (E), and the output is taken between the Collector (C) and Emitter (E). The Emitter terminal is common. The CE mode is the most widely used because it provides high current and voltage gain.
Input Characteristics (IB vs VBE)
- The curve resembles that of a forward-biased diode.
- The base current (IB) rises rapidly after the cut-in voltage (approximately 0.7 V for Si).
- The curve shifts slightly with changes in VCE due to the Early effect (base-width modulation).
- Input resistance is moderate (approximately 1 kΩ).
Output Characteristics (IC vs VCE)
- Active Region: IC ≈ βIB. This region is used for amplification. Junctions: E-B (Forward), C-B (Reverse).
- Cut-off Region: IB = 0, and both junctions are reverse-biased. The collector current (IC) is very small (leakage current ICEO). The transistor acts as an open switch.
- Saturation Region: VCE is low (< 1 V), and both junctions are forward-biased. IC is limited by the collector resistor (RC). The transistor acts as a closed switch.
Common Collector (CC) Transistor Configuration
Circuit Description
In the Common Collector (CC) configuration, the input is applied between the Base (B) and Collector (C), and the output is taken between the Emitter (E) and Collector (C). The Collector terminal is common.
Key Features
- Current Gain (γ = IE / IB): Very high, where γ = 1 + β.
- Voltage Gain (Av): Approximately 1 (unity). The output voltage closely follows the input voltage, hence the name “Emitter Follower.”
- Input Impedance: High.
- Output Impedance: Low.
- Application: Primarily used for impedance matching, connecting a high-impedance source to a low-impedance load.
Comparison of Transistor Configurations (CB, CE, CC)
| Parameter | Common Base (CB) | Common Emitter (CE) | Common Collector (CC) |
|---|---|---|---|
| Current Gain | α = IC / IE (≈ 1) | β = IC / IB (20–500) | γ = IE / IB (1 + β) |
| Voltage Gain | High | High | Low (≈ 1) |
| Input Resistance | Very Low (20 Ω – 150 Ω) | Moderate (~1 kΩ) | Very High (~500 kΩ) |
| Output Resistance | Very High (~1 MΩ) | Moderate (10 kΩ – 50 kΩ) | Very Low (< 100 Ω) |
| Phase Shift | 0° | 180° (inverted) | 0° (non-inverted) |
| Main Use | High-frequency applications | General amplification | Impedance matching |
Transistor Current Gain: Alpha (α) and Beta (β)
Alpha (αdc)
αdc = IC / IE. This is the ratio of the collector current to the emitter current in the CB configuration. It represents the fraction of emitter current that reaches the collector, so α is always less than 1.
Beta (βdc)
βdc = IC / IB. This is the ratio of the collector current to the base current in the CE configuration. It represents the current amplification factor, and β is usually large (typically 20 – 500).
Relationship Between Alpha and Beta
Starting from the basic transistor current equation: IE = IB + IC
- Expressing β in terms of α:
β = α / (1 − α) - Expressing α in terms of β:
α = β / (1 + β)
Transistor Circuit Problem Solutions
Problem 1
Given: IC = 2.5 mA, IE = 2.6 mA
Find: αdc, IB, βdc
Solution:
- IB = IE − IC = 2.6 mA − 2.5 mA = 0.1 mA
- αdc = IC / IE = 2.5 / 2.6 = 0.9615
- βdc = IC / IB = 2.5 / 0.1 = 25
Problem 2
Given: Initial IC = 3 mA, Initial IE = 3.03 mA, New βdc = 70
Find: Initial αdc and new IC
Solution:
- Initial IB = IE − IC = 3.03 mA − 3.00 mA = 0.03 mA
- Initial αdc = IC / IE = 3 / 3.03 = 0.99
- New IC = β × IB = 70 × 0.03 mA = 2.1 mA
Problem 3
Given: αdc = 0.98, IB = 100 µA
Find: IC, IE, βdc
Solution:
- β = α / (1 − α) = 0.98 / (1 − 0.98) = 0.98 / 0.02 = 49
- IC = β × IB = 49 × 100 µA = 4900 µA = 4.9 mA
- IE = IC + IB = 4.9 mA + 0.1 mA = 5.0 mA
Problem 4
Given: IC = 1 mA, IB = 25 µA, Target IC = 5 mA
Find: αdc, βdc, and the new IB required
Solution:
- β = IC / IB = 1 mA / 25 µA = 1000 µA / 25 µA = 40
- α = β / (1 + β) = 40 / (1 + 40) = 40 / 41 = 0.9756
- New IB = IC / β = 5 mA / 40 = 0.125 mA or 125 µA
