Derivation

The Bipolar Junction Transistor: Current Gain, Biasing and the Common-Emitter Amplifier

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Statement

For an NPN bipolar junction transistor (BJT) in the forward-active region, the emitter current is the sum of the base and collector currents, \(I_E = I_B + I_C\). The collector current is proportional to the base current via the current gain \(\beta\), such that \(I_C = \beta I_B\), and the common-base current gain is \(\alpha = I_C/I_E = \beta/(\beta+1)\). The collector current is exponentially related to the base-emitter voltage by \(I_C = I_S e^{V_{BE}/V_T}\), where \(V_T = k_B T/e\) is the thermal voltage. The small-signal voltage gain of a common-emitter amplifier with a bypassed emitter resistor is \(A_v = -g_m R_C\), where \(g_m = I_C/V_T\) is the transconductance.

Why it matters

The BJT is the foundational active device in analog electronics. Understanding its current relationships and biasing is essential for designing amplifiers, switches, and oscillators. The forward-active mode is the operating region where the transistor functions as a current-controlled current source, enabling voltage and power amplification.

The exponential \(I_C\)-\(V_{BE}\) relationship underpins all analog circuit design, from simple biasing networks to complex feedback amplifiers. Mastery of the small-signal model (\(g_m\), \(r_\pi\)) allows engineers to predict gain, input impedance, and bandwidth without resorting to full nonlinear analysis.

Assumptions

The transistor is in forward-active mode.The base-emitter junction is forward biased (\(V_{BE} \approx 0.7\ \mathrm{V}\) for silicon), and the base-collector junction is reverse biased. If the base-collector junction becomes forward biased, the transistor enters saturation and the relation \(I_C = \beta I_B\) fails.

The base current is small compared to the collector current.This implies \(\beta \gg 1\), so that \(I_E \approx I_C\). For low-\(\beta\) transistors (e.g., \(\beta < 10\)), the approximation \(I_E \approx I_C\) introduces significant error in bias calculations.

The Early effect is negligible.The collector current is assumed independent of \(V_{CE}\) (infinite output resistance). In reality, \(I_C\) increases slightly with \(V_{CE}\) due to base-width modulation, introducing a finite output resistance \(r_o = V_A/I_C\) where \(V_A\) is the Early voltage.

The transistor operates at room temperature (300 K).The thermal voltage \(V_T = k_B T/e\) and the saturation current \(I_S\) are temperature-dependent. A change of \(1\ \mathrm{K}\) shifts \(V_{BE}\) by approximately \(-2\ \mathrm{mV}\), altering the bias point and gain.

The signal is small-signal.The small-signal model (\(g_m\), \(r_\pi\)) is a linearization around the DC operating point. For large signals, the exponential \(I_C\)-\(V_{BE}\) relationship causes harmonic distortion, and the linear gain \(A_v = -g_m R_C\) no longer holds.

Derivation

\[ I_E = I_B + I_C \]

Kirchhoff's current law at the emitter terminal: the total current entering the emitter must equal the sum of currents leaving through the base and collector. A

\[ I_C = \beta I_B \]

Definition of the common-emitter current gain \(\beta\): the ratio of collector current to base current in forward-active mode. A

\[ I_E = I_B + \beta I_B = (\beta + 1) I_B \]

Substitute Step 2 into Step 1; the emitter current is \((\beta+1)\) times the base current. A

\[ \alpha \equiv \frac{I_C}{I_E} = \frac{\beta I_B}{(\beta+1) I_B} = \frac{\beta}{\beta+1} \]

Define the common-base current gain \(\alpha\); it is always less than 1 and approaches 1 as \(\beta \to \infty\). B

\[ I_C = I_S e^{V_{BE}/V_T} \]

The diode equation for the base-emitter junction, where \(I_S\) is the saturation current and \(V_T = k_B T/e\) is the thermal voltage. This exponential relationship arises from the Boltzmann distribution of carriers across the junction. C

\[ V_T = \frac{k_B T}{e} = \frac{(1.380649\times10^{-23}\ \mathrm{J/K})(300\ \mathrm{K})}{1.602176634\times10^{-19}\ \mathrm{C}} \approx 0.0259\ \mathrm{V} \]

Compute the thermal voltage at room temperature; \(k_B\) is Boltzmann's constant, \(T\) is absolute temperature, and \(e\) is the elementary charge. A

\[ g_m \equiv \frac{\partial I_C}{\partial V_{BE}} = \frac{I_S e^{V_{BE}/V_T}}{V_T} = \frac{I_C}{V_T} \]

Differentiate the exponential law with respect to \(V_{BE}\); the transconductance \(g_m\) is the slope of the \(I_C\)-\(V_{BE}\) curve and equals \(I_C/V_T\). C

\[ r_\pi \equiv \frac{\partial V_{BE}}{\partial I_B} = \frac{\beta}{g_m} = \frac{\beta V_T}{I_C} \]

The small-signal input resistance looking into the base; it is the ratio of base-emitter voltage change to base current change, equal to \(\beta/g_m\). B

\[ V_{TH} = V_{CC}\frac{R_2}{R_1+R_2}, \qquad R_{TH} = R_1 \parallel R_2 = \frac{R_1 R_2}{R_1+R_2} \]

Thevenin equivalent of the voltage-divider bias network: \(V_{TH}\) is the open-circuit base voltage, and \(R_{TH}\) is the equivalent resistance seen from the base. A

\[ V_{TH} - I_B R_{TH} - V_{BE} - I_E R_E = 0 \]

Apply Kirchhoff's voltage law around the base-emitter loop: the Thevenin voltage drops across \(R_{TH}\), the base-emitter junction, and the emitter resistor \(R_E\). B

\[ I_B = \frac{V_{TH} - V_{BE}}{R_{TH} + (\beta+1) R_E} \]

Substitute \(I_E = (\beta+1) I_B\) into Step 10 and solve for \(I_B\); this gives the base current in terms of the bias network parameters. C

\[ I_C = \beta I_B = \frac{\beta(V_{TH} - V_{BE})}{R_{TH} + (\beta+1) R_E} \]

The collector current is \(\beta\) times the base current; this is the DC operating point (Q-point) collector current. A

\[ V_{CE} = V_{CC} - I_C R_C - I_E R_E \]

Apply Kirchhoff's voltage law around the collector-emitter loop: \(V_{CE}\) is the supply voltage minus the drops across \(R_C\) and \(R_E\). A

\[ A_v = \frac{v_{out}}{v_{in}} = -g_m R_C \]

For a common-emitter amplifier with a bypassed emitter resistor, the small-signal voltage gain is negative (inverting) and equal to the transconductance times the collector resistance. B

\[ v_{be} = \frac{v_{in}}{1 + g_m R_E}\;\Longrightarrow\; A_v = -\frac{g_m R_C}{1+g_m R_E} = -\frac{R_C}{R_E + 1/g_m} \]

If \(R_E\) is not bypassed, the signal current \(g_m v_{be}\) also flows (to within \(1/\beta\)) through \(R_E\), so only part of \(v_{in}\) appears across the junction. The gain falls but becomes nearly independent of \(g_m\), hence of temperature and \(\beta\): for \(g_m R_E \gg 1\), \(A_v \approx -R_C/R_E\). B

Result

\[ I_C = \beta I_B, \quad \alpha = \frac{\beta}{\beta+1}, \quad I_C = I_S e^{V_{BE}/V_T}, \quad g_m = \frac{I_C}{V_T}, \quad r_\pi = \frac{\beta}{g_m} \]

Reading. The BJT in forward-active mode is a current-controlled current source: a small base current \(I_B\) controls a larger collector current \(I_C = \beta I_B\). The exponential \(I_C\)-\(V_{BE}\) relationship enables amplification, and the small-signal parameters \(g_m\) and \(r_\pi\) characterize the linearized response around the Q-point.

Units check. \([I_C] = \mathrm{A}\), \([V_{BE}] = \mathrm{V}\), \([V_T] = \mathrm{V}\), \([g_m] = \mathrm{A/V = S}\) (siemens), \([r_\pi] = \mathrm{V/A = \Omega}\) (ohms). All consistent with SI units.

Limiting cases

Cutoff region. When \(V_{BE} < 0.5\ \mathrm{V}\), the base-emitter junction is not forward biased, \(I_C \approx 0\), and the transistor is off. The collector current is negligible, and the device acts as an open switch.
Saturation region. When both junctions are forward biased (\(V_{BC} > 0\)), the relation \(I_C = \beta I_B\) fails; \(I_C\) is limited by \(R_C\) and \(V_{CE}\) drops to \(V_{CE,sat} \approx 0.2\ \mathrm{V}\). The transistor acts as a closed switch.
Large \(\beta\) limit. As \(\beta \to \infty\), \(\alpha \to 1\), \(I_E \approx I_C\), and the base current becomes negligible. The bias point becomes independent of \(\beta\), improving stability.
Small-signal limit. When the input signal amplitude is much less than \(V_T \approx 26\ \mathrm{mV}\), the exponential \(I_C\)-\(V_{BE}\) curve is well approximated by its tangent, and the linear gain \(A_v = -g_m R_C\) holds with negligible distortion.

Breaks when

Early effect (finite output resistance). The collector current increases slightly with \(V_{CE}\) due to base-width modulation, introducing a finite output resistance \(r_o = V_A/I_C\). The simple current-source model \(I_C = \beta I_B\) is no longer exact, and the voltage gain is reduced to \(A_v = -g_m (R_C \parallel r_o)\).
High-frequency operation. Junction capacitances (base-emitter \(C_\pi\), base-collector \(C_\mu\)) form a low-pass filter, reducing gain at high frequencies. The simple resistive model fails, and the frequency-dependent impedance must be included.
Temperature drift. \(V_{BE}\) decreases by approximately \(-2\ \mathrm{mV/K}\), and \(\beta\) increases with temperature. The Q-point shifts, potentially driving the transistor into saturation or cutoff. Thermal runaway can occur if the bias network is not stabilized.
Large signals. When the input signal amplitude is comparable to or larger than \(V_T\), the exponential \(I_C\)-\(V_{BE}\) relationship causes significant harmonic distortion. The linear small-signal model is invalid, and the gain varies with the signal amplitude.

Failure modes

Confusing \(\alpha\) and \(\beta\). \(\alpha = I_C/I_E\) is the common-base gain (less than 1), while \(\beta = I_C/I_B\) is the common-emitter gain (typically 50–500). Using \(\alpha\) where \(\beta\) is required (or vice versa) leads to errors in bias and gain calculations.
Ignoring the emitter resistor in bias calculations. Failing to include \(R_E\) in the base-emitter loop equation (Step 10) overestimates \(I_B\) and \(I_C\), shifting the Q-point into saturation. The emitter resistor provides negative feedback and stabilizes the bias point against \(\beta\) variations.
Assuming \(V_{BE}\) is exactly 0.7 V. \(V_{BE}\) varies with \(I_C\) and temperature. For \(I_C = 1\ \mathrm{mA}\), \(V_{BE} \approx 0.7\ \mathrm{V}\); a tenfold rise in \(I_C\) raises \(V_{BE}\) by \(V_T\ln 10 \approx 60\ \mathrm{mV}\), to about \(0.76\ \mathrm{V}\) at \(10\ \mathrm{mA}\). Using a fixed 0.7 V in precision designs introduces error.
Forgetting the negative sign in voltage gain. The common-emitter amplifier inverts the signal: \(A_v = -g_m R_C\). Omitting the negative sign leads to incorrect phase analysis in multi-stage amplifiers and feedback networks.

Discussion

The BJT's exponential \(I_C\)-\(V_{BE}\) relationship is the cornerstone of analog design. Unlike a resistor, whose current is linear in voltage, the BJT's current grows exponentially with voltage, enabling a small change in \(V_{BE}\) to produce a large change in \(I_C\). This nonlinearity is harnessed for amplification: a small input signal modulates \(I_C\), which in turn modulates the voltage across \(R_C\), producing a larger output signal.

The DC biasing network sets the Q-point, ensuring the transistor operates in the forward-active region for the full range of the input signal. The voltage-divider bias with an emitter resistor is widely used because it provides negative feedback: if \(I_C\) increases (due to \(\beta\) variation or temperature), \(V_E\) increases, reducing \(V_{BE}\) and pulling \(I_C\) back down. This stabilizes the Q-point against device-to-device variations.

The small-signal model (\(g_m\), \(r_\pi\)) linearizes the exponential curve around the Q-point, allowing the use of linear circuit analysis (superposition, nodal analysis) to predict gain, input impedance, and bandwidth. The transconductance \(g_m = I_C/V_T\) is the key parameter: it relates a small change in \(V_{BE}\) to a change in \(I_C\), and it scales with the DC bias current. Higher \(I_C\) yields higher \(g_m\) and higher gain, but at the cost of increased power consumption and reduced headroom.

At a deeper level, the BJT’s behaviour is governed by carrier transport. The forward-biased base-emitter junction injects electrons into the base (for an NPN) at a rate set by \(V_{BE}\) through the Boltzmann factor \(e^{V_{BE}/V_T}\); the thin, lightly doped base lets most of them diffuse across to the base-collector depletion region, whose field sweeps them into the collector, so \(\alpha\) is close to 1. The base current comes from recombination in the base and hole injection into the emitter, and is a roughly fixed fraction of the electron current — which is why \(\beta\) is approximately constant. Physically the junction voltage controls the current; the relation \(I_C = \beta I_B\) is an equally valid and often more convenient bookkeeping, because \(I_B\) and \(I_C\) are locked together.

Common misconceptions. “A BJT is purely current-controlled.” Both views are useful: \(I_C\) is set exponentially by \(V_{BE}\), and because \(I_B = I_C/\beta\) one may equally treat the base current as the control variable in DC bias design. “The emitter and collector currents are equal.” They differ by \(I_B\), a fraction \(1/(\beta+1)\) of \(I_E\). “The gain is a property of the transistor.” \(A_v = -g_m R_C\) depends on the bias current (through \(g_m\)) and on \(R_C\). And the transistor does not create power: it converts DC power from the supply into signal power, steered by the small input signal.

Worked examples

Example 1. An NPN transistor with \(\beta = 100\) is biased by \(V_{CC} = 12\ \mathrm{V}\), \(R_1 = 100\ \mathrm{k\Omega}\), \(R_2 = 22\ \mathrm{k\Omega}\), \(R_C = 4.7\ \mathrm{k\Omega}\), \(R_E = 1.0\ \mathrm{k\Omega}\) (bypassed for signals). Take \(V_{BE} = 0.7\ \mathrm{V}\) and \(T = 300\ \mathrm{K}\). Find the Q-point and the small-signal voltage gain.

\[ V_{TH} = 12 \times \frac{22}{122} = 2.164\ \mathrm{V},\qquad R_{TH} = \frac{100 \times 22}{122}\ \mathrm{k\Omega} = 18.03\ \mathrm{k\Omega} \]

Thevenin equivalent of the divider. A

\[ I_B = \frac{2.164 - 0.7}{18.03\times10^3 + 101\times 1.0\times10^3}\ \mathrm{A} = \frac{1.464}{119.0\times10^3}\ \mathrm{A} = 12.30\ \mathrm{\mu A} \]

Base-emitter loop equation with \(I_E = (\beta+1)I_B\). B

\[ I_C = 100 \times 12.30\ \mathrm{\mu A} = 1.230\ \mathrm{mA},\qquad I_E = 101\times 12.30\ \mathrm{\mu A} = 1.242\ \mathrm{mA} \]

Collector and emitter currents. A

\[ V_{CE} = 12 - (1.230\times10^{-3})(4.7\times10^{3}) - (1.242\times10^{-3})(1.0\times10^{3}) = 12 - 5.78 - 1.24 = 4.98\ \mathrm{V} \]

Collector-emitter loop. \(V_{CE} = 4.98\ \mathrm{V} \gg 0.2\ \mathrm{V}\), so the transistor is in forward-active mode and the small-signal model applies. A

\[ V_T = \frac{(1.380649\times10^{-23})(300)}{1.602176634\times10^{-19}}\ \mathrm{V} = 25.85\ \mathrm{mV},\qquad g_m = \frac{1.230\ \mathrm{mA}}{25.85\ \mathrm{mV}} = 47.6\ \mathrm{mS} \]

Thermal voltage and transconductance at the Q-point. A

\[ A_v = -g_m R_C = -(47.6\times10^{-3})(4.7\times10^{3}) = -224,\qquad r_\pi = \frac{\beta}{g_m} = \frac{100}{47.6\times10^{-3}}\ \mathrm{\Omega} = 2.10\ \mathrm{k\Omega} \]

Bypassed-emitter gain and input resistance looking into the base. A

\[ I_C = 1.23\ \mathrm{mA},\quad V_{CE} = 4.98\ \mathrm{V},\quad A_v \approx -224 \]

Reading. The Q-point sits near the middle of the 12 V supply, leaving room for the output to swing both ways; the gain is large and inverting, and depends on the bias current through \(g_m\).

Units check. \(g_m R_C\): \(\mathrm{(A/V)\cdot(V/A)}\) is dimensionless, as a voltage gain must be.

Example 2. A transistor with \(\beta = 200\) is biased by \(V_{CC} = 9\ \mathrm{V}\), \(R_1 = 47\ \mathrm{k\Omega}\), \(R_2 = 10\ \mathrm{k\Omega}\), \(R_C = 2.2\ \mathrm{k\Omega}\), \(R_E = 470\ \mathrm{\Omega}\), \(V_{BE} = 0.7\ \mathrm{V}\), \(T = 300\ \mathrm{K}\). Compare the voltage gain with \(R_E\) bypassed and unbypassed.

\[ V_{TH} = 9\times\frac{10}{57} = 1.579\ \mathrm{V},\qquad R_{TH} = \frac{47\times10}{57}\ \mathrm{k\Omega} = 8.25\ \mathrm{k\Omega} \]

Thevenin equivalent of the divider. A

\[ I_B = \frac{1.579 - 0.7}{8.25\times10^{3} + 201\times470}\ \mathrm{A} = \frac{0.879}{102.7\times10^{3}}\ \mathrm{A} = 8.56\ \mathrm{\mu A},\qquad I_C = 1.71\ \mathrm{mA} \]

Base current, then \(I_C = \beta I_B\). B

\[ V_{CE} = 9 - (1.711\times10^{-3})(2200) - (1.720\times10^{-3})(470) = 9 - 3.77 - 0.81 = 4.43\ \mathrm{V} \]

Forward-active, since \(V_{CE} \gg 0.2\ \mathrm{V}\). A

\[ g_m = \frac{1.711\ \mathrm{mA}}{25.85\ \mathrm{mV}} = 66.2\ \mathrm{mS},\qquad A_v^{\text{byp}} = -g_m R_C = -146 \]

Bypassed gain. A

\[ A_v^{\text{unbyp}} = -\frac{R_C}{R_E + 1/g_m} = -\frac{2200}{470 + 15.1} = -4.54 \]

Unbypassed gain: \(1/g_m = 15.1\ \Omega\) is small beside \(R_E\), so \(A_v \approx -R_C/R_E = -4.68\). B

\[ A_v^{\text{byp}} \approx -146,\qquad A_v^{\text{unbyp}} \approx -4.5 \]

Reading. Leaving \(R_E\) unbypassed trades a factor of about 32 in gain for a gain fixed by two resistors rather than by \(g_m\), which drifts with temperature and bias.

Units check. Both gains are ratios of resistances (\(1/g_m\) is in ohms), so they are dimensionless.

Problems

A BJT with \(\beta = 100\) has \(I_B = 20\ \mathrm{\mu A}\). Find \(I_C\) and \(I_E\).
Solution
\(I_C = \beta I_B = 100 \times 20\ \mathrm{\mu A} = 2.0\ \mathrm{mA}\). \(I_E = I_B + I_C = 20\ \mathrm{\mu A} + 2.0\ \mathrm{mA} = 2.02\ \mathrm{mA}\). Alternatively, \(I_E = (\beta+1) I_B = 101 \times 20\ \mathrm{\mu A} = 2.02\ \mathrm{mA}\).
A BJT has \(\alpha = 0.98\). Find \(\beta\).
Solution
\(\alpha = \beta/(\beta+1) \Rightarrow \beta = \alpha/(1-\alpha) = 0.98/(1-0.98) = 0.98/0.02 = 49\).
A BJT is biased with \(V_{CC} = 10\ \mathrm{V}\), \(R_1 = 100\ \mathrm{k\Omega}\), \(R_2 = 50\ \mathrm{k\Omega}\), \(R_C = 3.3\ \mathrm{k\Omega}\), \(R_E = 1.0\ \mathrm{k\Omega}\), \(\beta = 150\), \(V_{BE} = 0.7\ \mathrm{V}\). Find \(I_C\) and \(V_{CE}\).
Solution
\(V_{TH} = 10 \times 50/150 = 3.33\ \mathrm{V}\). \(R_{TH} = 100 \parallel 50 = 33.3\ \mathrm{k\Omega}\). \(I_B = (3.33 - 0.7)/(33.3\times10^3 + 151 \times 1.0\times10^3) = 2.63/(184.3\times10^3) \approx 14.3\ \mathrm{\mu A}\). \(I_C = 150 \times 14.3\ \mathrm{\mu A} = 2.14\ \mathrm{mA}\). \(I_E = 151 \times 14.3\ \mathrm{\mu A} \approx 2.16\ \mathrm{mA}\). \(V_{CE} = 10 - (2.14\times10^{-3})(3.3\times10^3) - (2.16\times10^{-3})(1.0\times10^3) = 10 - 7.07 - 2.16 = 0.77\ \mathrm{V}\). Since \(V_{CE} > 0.2\ \mathrm{V}\), the transistor is in forward-active mode.
A common-emitter amplifier has \(I_C = 1.0\ \mathrm{mA}\), \(R_C = 5.0\ \mathrm{k\Omega}\), and the emitter resistor is bypassed. Find the small-signal voltage gain \(A_v\) and the transconductance \(g_m\).
Solution
\(g_m = I_C/V_T = 1.0\times10^{-3}/0.0259 \approx 38.6\ \mathrm{mS}\). \(A_v = -g_m R_C = -(38.6\times10^{-3})(5.0\times10^3) = -193\). The negative sign indicates inversion.
A transistor with \(\beta = 100\) is biased by \(V_{CC} = 12\ \mathrm{V}\), \(R_1 = 56\ \mathrm{k\Omega}\), \(R_2 = 12\ \mathrm{k\Omega}\), \(R_C = 3.3\ \mathrm{k\Omega}\), \(R_E = 1.2\ \mathrm{k\Omega}\), \(V_{BE} = 0.7\ \mathrm{V}\), \(T = 300\ \mathrm{K}\). Find the Q-point, check the operating region, and find the gain with \(R_E\) bypassed and unbypassed.
Solution
\(V_{TH} = 12\times12/68 = 2.118\ \mathrm{V}\), \(R_{TH} = 56\parallel12 = 9.88\ \mathrm{k\Omega}\). \(I_B = (2.118-0.7)/(9.88\times10^3 + 101\times1.2\times10^3) = 1.418/(131.1\times10^3) = 10.8\ \mathrm{\mu A}\), so \(I_C = 1.08\ \mathrm{mA}\), \(I_E = 1.09\ \mathrm{mA}\). \(V_{CE} = 12 - (1.081\times10^{-3})(3300) - (1.092\times10^{-3})(1200) = 12 - 3.57 - 1.31 = 7.12\ \mathrm{V}\): forward-active. \(g_m = 1.081\ \mathrm{mA}/25.85\ \mathrm{mV} = 41.8\ \mathrm{mS}\); bypassed \(A_v = -g_m R_C = -138\); unbypassed \(A_v = -3300/(1200 + 23.9) = -2.70\).