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Chapter 2 • Theory & Derivations

Transistors: BJT, JFET & MOSFET Biasing Circuits

Comprehensive analysis of three-terminal semiconductor amplifiers: BJT structure and current relations, Common Emitter/Base/Collector characteristics, DC and AC load line analysis, Q-point stabilization and bias techniques (fixed, collector-feedback, and voltage divider with stability factor S derivation), JFET operation, Shockley drain current equation, and Depletion/Enhancement MOSFETs.

§2.1 Bipolar Junction Transistors (BJT): Structure & Transistor Action

1. Physical Structure of the Bipolar Junction Transistor

A bipolar junction transistor (BJT) consists of a single crystal of semiconductor with three alternating doped regions forming two back-to-back PN junctions:
  • Emitter ($E$): Heavily doped ($N_E \gg 10^{19}\text{ cm}^{-3}$) to inject a massive stream of majority carriers into the base. Moderate geometric physical width.
  • Base ($B$): Extremely narrow ($W_B \ll L_n, L_p$, typically $< 1\;\mu\text{m}$) and lightly doped ($N_B \sim 10^{16}\text{ cm}^{-3}$) to minimize electron-hole recombination so that nearly all injected carriers traverse it safely.
  • Collector ($C$): Moderately doped and physically largest in volume to dissipate heat generated by carrier collection under high reverse bias voltages.
Transistors are fabricated in two complementary polarity types: NPN (majority carriers electrons) and PNP (majority carriers holes).

2. The Transistor Action Mechanism (NPN in Active Mode)

To operate a BJT as a linear amplifier in the forward-active mode:
  1. The Base-Emitter (B-E) junction is forward-biased ($V_{BE} \approx 0.7\text{ V}$ for Silicon).
  2. The Collector-Base (C-B) junction is reverse-biased ($V_{CB} > 0$, typically $5\text{ V} - 20\text{ V}$).
Because the B-E junction is forward-biased, the heavily doped emitter injects a large flux of electrons into the base. Because the base is extremely thin ($W_B \ll L_n$) and lightly doped, only a tiny fraction (typically $< 1\%$) of electrons recombine with base holes, contributing to the tiny base current $I_B$. The remaining $> 99\%$ of electrons diffuse across the base to the edge of the reverse-biased collector-base space-charge layer, where the strong electric field sweeps them across into the collector, constituting the collector current $I_C$.

3. Terminal Current Relations

By Kirchhoff's Current Law:
$$I_E = I_B + I_C$$
The fraction of emitter current collected is the common-base current gain $\alpha$:
$$\alpha = \frac{I_C - I_{CBO}}{I_E} \approx \frac{I_C}{I_E}$$
Typically, $\alpha$ ranges between $0.98$ and $0.998$. Here $I_{CBO}$ is the tiny collector-to-base reverse leakage current with emitter open.

§2.2 Transistor Configurations: CB, CE, CC & Characteristics

1. Common Base (CB) Configuration

The base terminal is common to both input (emitter-base) and output (collector-base) circuits.
  • Input Characteristics: Curve of $I_E$ vs. $V_{EB}$ at constant $V_{CB}$. Resembles a forward-biased diode. Dynamic input resistance is very low ($R_{\text{in}} \approx 20\,\Omega - 50\,\Omega$).
  • Output Characteristics: Curve of $I_C$ vs. $V_{CB}$ at constant $I_E$. In the active region, $I_C$ is nearly horizontal and independent of $V_{CB}$, yielding high output resistance ($R_{\text{out}} \approx 1\text{ M}\Omega$).
  • Current Gain: $\alpha = \Delta I_C / \Delta I_E < 1$ (no current amplification; provides voltage and power gain).

2. Common Emitter (CE) Configuration (Most Widely Used)

The emitter terminal is grounded/common to both input (base-emitter) and output (collector-emitter) circuits.
  • Current Gain $\beta$ (or $h_{fe}$):
    $$I_C = \alpha I_E + I_{CBO} = \alpha (I_B + I_C) + I_{CBO} \implies I_C(1 - \alpha) = \alpha I_B + I_{CBO}$$
    $$I_C = \left( \frac{\alpha}{1 - \alpha} \right) I_B + \left( \frac{1}{1 - \alpha} \right) I_{CBO} = \beta I_B + I_{CEO}$$
    where the common-emitter current gain $\beta$ is:
    $$\beta = \frac{\alpha}{1 - \alpha}, \quad \alpha = \frac{\beta}{1 + \beta}$$
    Typically, $\beta$ ranges from $50$ to $300$. The collector-to-emitter leakage current is $I_{CEO} = (1 + \beta) I_{CBO}$.
  • Input Characteristics: $I_B$ vs. $V_{BE}$ for constant $V_{CE}$. Input resistance $R_{\text{in}} \approx 1\text{ k}\Omega - 2\text{ k}\Omega$.
  • Output Characteristics: $I_C$ vs. $V_{CE}$ for constant $I_B$. Shows three operating regions:
    1. Active Region: B-E forward-biased, C-B reverse-biased ($V_{CE} > 0.3\text{ V}$). Curves have a slight upward slope due to base-width modulation (Early effect). Ideal for linear amplification.
    2. Saturation Region: Both B-E and C-B forward-biased ($V_{CE} < V_{CE(\text{sat})} \approx 0.2\text{ V}$). Transistor acts as an ON switch with negligible voltage drop.
    3. Cutoff Region: Both B-E and C-B reverse-biased ($I_B = 0, I_C = I_{CEO} \approx 0$). Transistor acts as an OFF switch.

3. Common Collector (CC) Configuration (Emitter Follower)

Collector is AC ground. Input applied to base, output extracted from emitter.
$$\gamma = \frac{I_E}{I_B} = 1 + \beta$$
Features very high input impedance ($R_{\text{in}} > 100\text{ k}\Omega$), very low output impedance ($R_{\text{out}} < 50\,\Omega$), and near-unity voltage gain ($A_v \approx 1$). Used ubiquitously for impedance matching and output buffer stages.

§2.3 DC and AC Load Line Analysis & Q-Point Selection

1. The DC Load Line

Consider a Common Emitter circuit with collector supply voltage $V_{CC}$ and collector load resistor $R_C$. Applying Kirchhoff's Voltage Law to the output loop:
$$V_{CC} = V_{CE} + I_C R_C \implies I_C = -\frac{1}{R_C} V_{CE} + \frac{V_{CC}}{R_C}$$
This is the equation of a straight line on the $I_C - V_{CE}$ output characteristic plane, termed the DC Load Line:
  • Cutoff Point (X-axis intercept): Set $I_C = 0 \implies V_{CE} = V_{CC}$.
  • Saturation Point (Y-axis intercept): Set $V_{CE} = 0 \implies I_C = \frac{V_{CC}}{R_C}$.
  • Slope: $-\frac{1}{R_C}$.

2. The Quiescent Operating Point (Q-Point)

The intersection of the DC load line with the transistor output characteristic corresponding to the base bias current $I_B$ establishes the quiescent operating point (Q-point) $(V_{CEQ}, I_{CQ})$. For maximum symmetrical undistorted output voltage swing (Class A operation), the Q-point should be positioned precisely at the midpoint of the DC load line:
$$V_{CEQ} = \frac{V_{CC}}{2}, \quad I_{CQ} = \frac{V_{CC}}{2 R_C}$$

3. The AC Load Line

When an external load $R_L$ is capacitively coupled to the collector, the effective AC load resistance seen by the collector signal is $r_c = R_C \parallel R_L$. Because $r_c < R_C$, the AC load line passes through the same DC Q-point $(V_{CEQ}, I_{CQ})$ but has a steeper slope $-\frac{1}{r_c}$:
$$i_c = -\frac{1}{r_c} v_{ce} \implies I_C - I_{CQ} = -\frac{1}{R_C \parallel R_L} (V_{CE} - V_{CEQ})$$
The maximum peak-to-peak undistorted output voltage swing is limited by the distance to cutoff or saturation along the AC load line:
$$V_{o(\text{p-p, max})} = 2 \min\left( V_{CEQ}, I_{CQ} r_c \right)$$

§2.4 Transistor Biasing Methods & Thermal Stability Factor (S)

1. Need for Biasing & Thermal Runaway

The Q-point coordinates $(I_C, V_{CE})$ are temperature-sensitive due to three factors:
  1. Reverse saturation current $I_{CBO}$ doubles approximately every $10^\circ\text{C}$ rise in temperature.
  2. Base-emitter voltage $V_{BE}$ decreases at a rate of approximately $-2.5\text{ mV}/^\circ\text{C}$.
  3. Current gain $\beta$ increases with temperature.
Without proper stabilization, an increase in temperature causes $I_C$ to rise, increasing collector power dissipation $P_C = V_{CE} I_C$. This heats the junction further, causing a catastrophic regenerative positive feedback loop known as thermal runaway.

2. Definition of Stability Factor $S$

The stability factor $S$ is defined as the rate of change of collector current with respect to leakage current $I_{CBO}$:
$$S = \frac{\partial I_C}{\partial I_{CBO}}$$
An ideal bias circuit has $S \to 1$ (complete thermal immunity). Higher values ($S \gg 1$) indicate severe instability.

3. Comparative Analysis of Biasing Topologies

  • Fixed Base Resistor Bias: A single resistor $R_B$ connects from $V_{CC}$ to the base:
    $$I_B = \frac{V_{CC} - V_{BE}}{R_B} \approx \text{const} \implies S = 1 + \beta$$
    For $\beta = 100$, $S = 101$. Extremely unstable; completely unsuitable for linear amplifiers.
  • Collector Feedback Bias: $R_B$ is connected directly from the collector to the base:
    $$S = \frac{1 + \beta}{1 + \beta \frac{R_C}{R_C + R_B}}$$
    Provides negative feedback: if $I_C$ rises, $V_C = V_{CC} - I_C R_C$ drops, reducing $I_B$ and checking the rise. Moderate stability.
  • Voltage Divider Bias (Universal Bias): Two resistors $R_1, R_2$ form a potential divider across $V_{CC}$, combined with an emitter degeneration resistor $R_E$:
    $$V_{\text{th}} = V_{CC} \frac{R_2}{R_1 + R_2}, \quad R_{\text{th}} = R_1 \parallel R_2 = \frac{R_1 R_2}{R_1 + R_2}$$
    The input loop equation is $V_{\text{th}} = I_B R_{\text{th}} + V_{BE} + I_E R_E$. Differentiating delivers the stability factor:
    $$S = \frac{1 + \beta}{1 + \beta \left( \frac{R_E}{R_{\text{th}} + R_E} \right)}$$
    By designing $R_{\text{th}} \ll \beta R_E$ (rule of thumb: $R_{\text{th}} \le 0.1 \beta R_E$ or $R_2 \le 0.1 \beta R_E$), the fraction approaches 1:
    $$S \approx \frac{1 + \beta}{1 + \beta} = 1$$
    The Q-point becomes virtually independent of transistor $\beta$ and temperature variations.

§2.5 Field Effect Transistors: JFET & MOSFET Characteristics

1. Junction Field Effect Transistor (JFET)

The JFET is a unipolar three-terminal device where conduction is governed entirely by majority carriers through a semiconductor channel controlled by an electric field:
  • Construction: An N-channel JFET consists of a lightly doped n-type silicon bar with Ohmic contacts at both ends designated Drain ($D$) and Source ($S$). Two heavily doped $p^+$ regions diffused on opposite sides are connected to form the Gate ($G$).
  • Pinch-Off Voltage $V_P$: Operating with reverse-biased gate-source junction ($V_{GS} \le 0$). Reverse bias widens the depletion regions into the channel. At a critical drain-source voltage $V_{DS} = V_{DS(\text{sat})} = V_{GS} - V_P$, the two depletion boundaries touch at the drain end, a condition termed pinch-off. Beyond pinch-off, the drain current saturates at an almost constant level.
  • Shockley's Drain Current Equation: In the saturation/active region ($V_{DS} \ge V_{GS} - V_P$):
    $$I_D = I_{DSS} \left( 1 - \frac{V_{GS}}{V_P} \right)^2$$
    where $I_{DSS}$ is the maximum saturation drain current at $V_{GS} = 0$.
  • Transconductance $g_m$:
    $$g_m = \frac{\partial I_D}{\partial V_{GS}} = \frac{2 I_{DSS}}{|V_P|} \left( 1 - \frac{V_{GS}}{V_P} \right) = g_{m0} \left( 1 - \frac{V_{GS}}{V_P} \right)$$

2. Metal-Oxide-Semiconductor FETs (MOSFETs)

The gate electrode is physically insulated from the channel by an ultra-thin dielectric layer of Silicon Dioxide ($\text{SiO}_2$), resulting in an astronomical input impedance ($R_{\text{in}} > 10^{12}\,\Omega$):
  • Depletion-Type MOSFET (D-MOSFET): Fabricated with a physical channel. Can operate in either depletion mode (negative $V_{GS}$ repels channel electrons) or enhancement mode (positive $V_{GS}$ attracts additional electrons, boosting $I_D$ above $I_{DSS}$). Follows Shockley's square-law equation for both positive and negative $V_{GS}$.
  • Enhancement-Type MOSFET (E-MOSFET): The cornerstone of modern VLSI and microprocessor technology (CMOS). Has no physical channel at zero bias ($I_D = 0$ at $V_{GS} = 0$). When $V_{GS}$ exceeds the positive threshold voltage $V_{th}$, an inversion layer forms at the $\text{Si}-\text{SiO}_2$ interface, creating an n-channel. In saturation ($V_{DS} \ge V_{GS} - V_{th}$):
    $$I_D = k_n (V_{GS} - V_{th})^2 = \frac{1}{2} \mu_n C_{ox} \left( \frac{W}{L} \right) (V_{GS} - V_{th})^2$$
    where $C_{ox} = \varepsilon_{ox}/t_{ox}$ is the gate oxide capacitance per unit area, and $W/L$ is the transistor aspect ratio.
Solved Problem Example 2.1: Voltage Divider Biasing Circuit Design for Stable Q-Point

Design a voltage divider bias circuit for an NPN silicon transistor operating from a single supply $V_{CC} = 15.0\text{ V}$ to establish the quiescent operating point at $V_{CEQ} = 7.5\text{ V}$ and $I_{CQ} = 2.0\text{ mA}$. The transistor has $\beta = 100$ and $V_{BE} = 0.7\text{ V}$. Allocate $1.5\text{ V}$ across the emitter resistor $R_E$ for thermal stability, and ensure the divider bleeder current is $10$ times the base current ($I_{\text{div}} = 10 I_B$). Find resistors $R_E$, $R_C$, $R_1$, and $R_2$.

Step 1: Calculate Emitter Resistor RE and Collector Resistor RC
$$I_E \approx I_C = 2.0\text{ mA} \implies R_E = \frac{V_E}{I_E} = \frac{1.5\text{ V}}{2.0\text{ mA}} = 750\,\Omega$$

Applying KVL to the collector-emitter loop: $V_{CC} = I_C R_C + V_{CEQ} + V_E$. Substituting values: $15.0 = (2.0\text{ mA}) R_C + 7.5 + 1.5 \implies (2.0\text{ mA}) R_C = 6.0\text{ V} \implies R_C = \frac{6.0\text{ V}}{2.0\text{ mA}} = 3.0\text{ k}\Omega$.

Step 2: Calculate Base Voltage VB and Base Current IB
$$V_B = V_E + V_{BE} = 1.5\text{ V} + 0.7\text{ V} = 2.2\text{ V}, \quad I_B = \frac{I_C}{\beta} = \frac{2.0\text{ mA}}{100} = 20\;\mu\text{A}$$

The required base potential is $2.2\text{ V}$, and the base current is $20\;\mu\text{A}$.

Step 3: Size the Voltage Divider Resistors R1 and R2
$$I_2 = 10 I_B = 10(20\;\mu\text{A}) = 200\;\mu\text{A} \implies R_2 = \frac{V_B}{I_2} = \frac{2.2\text{ V}}{0.20\text{ mA}} = 11.0\text{ k}\Omega$$

Current through $R_1$ is $I_1 = I_2 + I_B = 200\;\mu\text{A} + 20\;\mu\text{A} = 220\;\mu\text{A}$. Thus $R_1 = \frac{V_{CC} - V_B}{I_1} = \frac{15.0\text{ V} - 2.2\text{ V}}{0.22\text{ mA}} = \frac{12.8\text{ V}}{0.22\text{ mA}} \approx 58.18\text{ k}\Omega$ (standard commercial values: $R_1 = 56\text{ k}\Omega$, $R_2 = 11\text{ k}\Omega$).

Final Answer & Physical Insight

RE = 750 Ω; RC = 3.0 kΩ; R1 ≈ 58.2 kΩ (use 56 kΩ); R2 = 11.0 kΩ.

Solved Problem Example 2.2: Thermal Stability Factor S Comparison: Fixed Bias vs. Voltage Divider Bias

A silicon transistor with $\beta = 120$ is used in two different biasing configurations: (a) Fixed base resistor bias with $R_B = 470\text{ k}\Omega$ and $R_C = 2.2\text{ k}\Omega$. (b) Voltage divider bias with $R_1 = 39\text{ k}\Omega$, $R_2 = 8.2\text{ k}\Omega$, $R_C = 2.2\text{ k}\Omega$, and $R_E = 1.0\text{ k}\Omega$. Calculate the stability factor $S = \partial I_C / \partial I_{CBO}$ for both circuits and comment on their thermal stability.

Step 1: Calculate Stability Factor for Fixed Base Bias
$$S_{\text{fixed}} = 1 + \beta = 1 + 120 = 121$$

In fixed bias, any change in leakage current $\Delta I_{CBO}$ is amplified by a factor of 121 directly in the collector current, causing severe temperature vulnerability.

Step 2: Calculate Thevenin Equivalent Resistance for Voltage Divider Bias
$$R_{\text{th}} = R_1 \parallel R_2 = \frac{(39\text{ k}\Omega)(8.2\text{ k}\Omega)}{39\text{ k}\Omega + 8.2\text{ k}\Omega} = \frac{319.8}{47.2} \approx 6.775\text{ k}\Omega$$

The Thevenin base resistance is $6.775\text{ k}\Omega$.

Step 3: Evaluate Stability Factor for Voltage Divider Bias
$$S_{\text{div}} = \frac{1 + \beta}{1 + \beta \left( \frac{R_E}{R_{\text{th}} + R_E} \right)} = \frac{1 + 120}{1 + 120 \left( \frac{1.0}{6.775 + 1.0} \right)} = \frac{121}{1 + 120 \left( \frac{1.0}{7.775} \right)} = \frac{121}{1 + 15.434} = \frac{121}{16.434} \approx 7.36$$

Voltage divider bias reduces the stability factor from $121$ down to $7.36$, providing a $16.4\times$ improvement in thermal stability and completely preventing thermal runaway.

Final Answer & Physical Insight

Fixed bias: S = 121 (severely unstable); Voltage divider bias: S = 7.36 (highly stable, 16.4x better thermal immunity).

Solved Problem Example 2.3: JFET Self-Bias Circuit Analysis via Shockley's Equation

An N-channel JFET with parameters $I_{DSS} = 12\text{ mA}$ and pinch-off voltage $V_P = -4.0\text{ V}$ is biased using a self-bias circuit with supply voltage $V_{DD} = 18.0\text{ V}$, drain resistor $R_D = 1.2\text{ k}\Omega$, and source resistor $R_S = 330\,\Omega$. The gate resistor is $R_G = 1.0\text{ M}\Omega$. (a) Calculate the quiescent operating point $(I_{DQ}, V_{GSQ}, V_{DSQ})$. (b) Determine the transconductance $g_m$ at the Q-point.

Step 1: Set Up the Self-Bias Line Equation
$$V_{GS} = -I_D R_S = -(330\,\Omega) I_D$$

Since gate current is zero ($I_G = 0$), $V_G = 0$, so $V_{GS} = V_G - V_S = -I_D R_S$.

Step 2: Solve Non-Linear Quadratic Equation for Drain Current
$$I_D = I_{DSS} \left( 1 - \frac{V_{GS}}{V_P} \right)^2 = 12 \left( 1 - \frac{-0.330 I_D}{-4.0} \right)^2 = 12 \left( 1 - 0.0825 I_D \right)^2$$

Expanding: $I_D = 12(1 - 0.165 I_D + 0.006806 I_D^2) = 12 - 1.98 I_D + 0.08167 I_D^2$. Rearranging into standard quadratic form: $0.08167 I_D^2 - 2.98 I_D + 12 = 0$. Solving roots: $I_D = \frac{2.98 \pm \sqrt{(2.98)^2 - 4(0.08167)(12)}}{2(0.08167)} = \frac{2.98 \pm 2.227}{0.1633}$. The physical root ($I_D < I_{DSS}$) is $I_{DQ} = \frac{0.753}{0.1633} \approx 4.61\text{ mA}$.

Step 3: Calculate V_GSQ, V_DSQ and Transconductance g_m
$$V_{GSQ} = -(4.61\text{ mA})(330\,\Omega) = -1.52\text{ V}, \quad V_{DSQ} = V_{DD} - I_{DQ}(R_D + R_S) = 18 - 4.61(1.2 + 0.33) = 10.95\text{ V}$$

Transconductance is $g_m = \frac{2 I_{DSS}}{|V_P|} \left( 1 - \frac{V_{GSQ}}{V_P} \right) = \frac{2(12\text{ mA})}{4.0\text{ V}} \left( 1 - \frac{-1.52}{-4.0} \right) = 6.0\text{ mS} (1 - 0.38) = 3.72\text{ mS}$ (or $3.72\text{ mA/V}$).

Final Answer & Physical Insight

Q-point: IDQ = 4.61 mA, VGSQ = -1.52 V, VDSQ = 10.95 V; Transconductance gm = 3.72 mS.

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