Hybrid parameters of transistor — input/output characteristics

Question

Explain the input and output characteristics of an n-p-n transistor in common-emitter (CE) configuration. Define the current amplification factor $\beta$ . If $I_B = 20\,\mu$ A and $\beta = 100$ , find $I_C$ and $I_E$ .

(CBSE 12 + JEE Main pattern)

Solution — Step by Step

With $V_{CE}$ kept constant, we plot base current $I_B$ versus base-emitter voltage $V_{BE}$ .

Key features:

Looks like a forward-biased diode curve (because B-E junction IS forward biased)
$I_B$ is very small (microamperes range)
Threshold voltage: ~0.6 V for Si, ~0.2 V for Ge
Input resistance: $r_i = \dfrac{\Delta V_{BE}}{\Delta I_B}\bigg|_{V_{CE} = \text{const}}$ , typically 1-5 kohm

With $I_B$ kept constant, we plot collector current $I_C$ versus collector-emitter voltage $V_{CE}$ .

Key features:

For small $V_{CE}$ , $I_C$ rises steeply (saturation region)
Beyond ~0.5 V, $I_C$ becomes nearly constant — the active region (used for amplification)
Different $I_B$ values give different horizontal curves (higher $I_B$ = higher $I_C$ )
Output resistance: $r_o = \dfrac{\Delta V_{CE}}{\Delta I_C}\bigg|_{I_B = \text{const}}$ , typically 50-100 kohm

Current amplification factor (CE configuration):

\beta = \frac{I_C}{I_B}

I_C = \beta \times I_B = 100 \times 20\,\mu\text{A} = \mathbf{2\,\text{mA}}

Using Kirchhoff’s current law at the transistor:

I_E = I_B + I_C = 20\,\mu\text{A} + 2\,\text{mA} = \mathbf{2.02\,\text{mA}}

Since $I_B \ll I_C$ , we can approximate $I_E \approx I_C$ .

flowchart TD
    A["CE Configuration"] --> B["Input: VBE vs IB"]
    A --> C["Output: VCE vs IC"]
    B --> D["Looks like diode curve"]
    B --> E["ri = ΔVBE/ΔIB ~ 1-5 kΩ"]
    C --> F["Active region: IC nearly constant"]
    C --> G["ro = ΔVCE/ΔIC ~ 50-100 kΩ"]
    H["Current relations"] --> I["β = IC/IB"]
    H --> J["IE = IB + IC"]
    H --> K["α = IC/IE = β/(1+β)"]

Why This Works

A transistor is a current-controlled device. A small base current controls a much larger collector current — this is amplification. In CE mode, the base-emitter junction is forward biased (allowing $I_B$ ), and the collector-base junction is reverse biased. The thin base region (less than 1 micrometer) ensures that most electrons injected from the emitter pass through to the collector without recombining — giving $I_C \gg I_B$ .

The output characteristics are nearly flat in the active region because the reverse-biased C-B junction efficiently sweeps up all electrons reaching the base. Increasing $V_{CE}$ barely changes this collection efficiency.

Alternative Method — Relating alpha and beta

The common-base current gain $\alpha = I_C/I_E$ and common-emitter current gain $\beta = I_C/I_B$ are related:

\beta = \frac{\alpha}{1-\alpha}, \quad \alpha = \frac{\beta}{1+\beta}

For $\beta = 100$ : $\alpha = 100/101 = 0.99$ . This means 99% of emitter current reaches the collector — only 1% is lost as base current.

For JEE Main MCQs, remember: $\alpha$ is always less than 1 (typically 0.95-0.99), $\beta$ is always greater than 1 (typically 50-300). If a question gives $\alpha = 0.98$ , then $\beta = 0.98/0.02 = 49$ . This conversion appears frequently.

Common Mistake

Students confuse the three transistor configurations (CE, CB, CC) and their current gains. In common-emitter: $\beta = I_C/I_B$ (large, 50-300). In common-base: $\alpha = I_C/I_E$ (less than 1). In common-collector (emitter follower): current gain $\approx \beta + 1$ . The most-tested configuration is CE because it gives both voltage and current amplification. If a problem does not specify the configuration, assume CE.

Question

Solution — Step by Step

Why This Works

Alternative Method — Relating alpha and beta

Common Mistake

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