Semiconductor Electronics • Topic 3 of 3

Transistors & Logic Gates

A bipolar junction transistor (BJT) is a three-layer, two-junction semiconductor device. It comes in two forms: npn (a thin p-layer between two n-layers) and pnp (a thin n-layer between two p-layers). The three regions are the emitter (heavily doped, supplies carriers), the base (very thin and lightly doped), and the collector (moderately doped, larger, collects carriers). In normal active operation the emitter-base junction is forward-biased and the collector-base junction is reverse-biased.

Carriers injected by the emitter mostly cross the thin base and are swept into the collector; only a small fraction recombine in the base. This gives the fundamental current relation $I_E = I_B + I_C$, where the base current $I_B$ is small and the collector current $I_C$ is large. Because $I_B$ is tiny, the transistor can use a small base current to control a large collector current — the basis of amplification.

The transistor can be connected in three configurations: common-base (CB), common-emitter (CE), and common-collector (CC). The common-emitter configuration is the most widely used because it provides high current and voltage gain. Two key gains are defined: the common-base current gain $\alpha = \dfrac{I_C}{I_E}$ (just below 1) and the common-emitter current gain $\beta = \dfrac{I_C}{I_B}$ (typically 50-300). They are related by $\beta = \dfrac{\alpha}{1-\alpha}$ and $\alpha = \dfrac{\beta}{1+\beta}$.

As a switch, the transistor is driven between two states. With no (or insufficient) base current it is in cut-off — no collector current flows and the output is high (OFF). With a large enough base current it goes into saturation — maximum collector current flows and the output is low (ON). Operating only at these two extremes, the transistor acts as an electronic switch, which is exactly what digital circuits need.

As an amplifier, the transistor is biased in the active region (between cut-off and saturation). A small AC signal applied to the base produces large changes in $I_C$, and a load resistor in the collector converts these into a large output voltage. The AC current gain is $\beta_{ac} = \dfrac{\Delta I_C}{\Delta I_B}$ and the voltage gain in CE mode is $A_V = -\beta\,\dfrac{R_L}{R_{in}}$; the negative sign shows the output is $180^\circ$ out of phase with the input.

Logic gates are the building blocks of digital electronics, processing binary inputs (0 = low, 1 = high) into a binary output by a fixed rule, summarised in a truth table. The three basic gates are: OR ($Y = A + B$, output 1 if any input is 1), AND ($Y = A \cdot B$, output 1 only if all inputs are 1), and NOT ($Y = \bar{A}$, the inverter, output is the opposite of the input). Combining a NOT with the others gives the universal gates: NAND ($Y = \overline{A \cdot B}$) and NOR ($Y = \overline{A + B}$). They are called universal because any logic function whatsoever can be built using only NAND gates or only NOR gates.

Solved Examples

Worked Example

In a transistor the emitter current is $I_E = 10\,\text{mA}$ and the base current is $I_B = 100\,\mu\text{A}$. Find the collector current and the common-emitter current gain $\beta$.

Solution

Use $I_E = I_B + I_C$, so $I_C = I_E - I_B$.
$I_C = 10\,\text{mA} - 0.1\,\text{mA} = 9.9\,\text{mA}$.
$\beta = \dfrac{I_C}{I_B} = \dfrac{9.9\,\text{mA}}{0.1\,\text{mA}}$.
$\beta = 99$.

Answer: The collector current is $9.9\,\text{mA}$ and the current gain $\beta = 99$.

Worked Example

A transistor has a common-base current gain $\alpha = 0.98$. Find the common-emitter current gain $\beta$.

Solution

The two gains are related by $\beta = \dfrac{\alpha}{1-\alpha}$.
$1 - \alpha = 1 - 0.98 = 0.02$.
$\beta = \dfrac{0.98}{0.02}$.
$\beta = 49$.

Answer: The common-emitter current gain is $\beta = 49$.

Worked Example

A common-emitter amplifier has a current gain $\beta = 100$, an input resistance $R_{in} = 1\,\text{k}\Omega$ and a load resistance $R_L = 5\,\text{k}\Omega$. Find the voltage gain (magnitude).

Solution

Voltage gain in CE mode: $|A_V| = \beta\,\dfrac{R_L}{R_{in}}$.
$|A_V| = 100 \times \dfrac{5\,\text{k}\Omega}{1\,\text{k}\Omega}$.
$|A_V| = 100 \times 5$.
$|A_V| = 500$.

Answer: The voltage gain has magnitude $500$; the output is $180^\circ$ out of phase with the input.

Worked Example

Write the truth table for a two-input NAND gate and state its Boolean expression.

Solution

A NAND is an AND followed by a NOT: $Y = \overline{A \cdot B}$.
AND gives 1 only when both inputs are 1; NAND inverts this output.
So $Y = 0$ only when $A = 1$ and $B = 1$; otherwise $Y = 1$.
Truth table: $A,B = 00 \to 1$; $01 \to 1$; $10 \to 1$; $11 \to 0$.

Answer: $Y = \overline{A \cdot B}$; output is 0 only for inputs $(1,1)$ and 1 for all other input combinations.

Worked Example

A transistor amplifier produces an output voltage change of $4\,\text{V}$ for an input voltage change of $20\,\text{mV}$. Find the voltage gain.

Solution

Voltage gain is the ratio of output to input voltage change: $A_V = \dfrac{\Delta V_{out}}{\Delta V_{in}}$.
$\Delta V_{out} = 4\,\text{V}$ and $\Delta V_{in} = 20\,\text{mV} = 0.02\,\text{V}$.
$A_V = \dfrac{4}{0.02}$.
$A_V = 200$.

Answer: The amplifier has a voltage gain of $200$.

Worked Example

Using the universal property of NAND gates, show how a NOT gate is obtained from a single NAND gate.

Solution

A two-input NAND gives $Y = \overline{A \cdot B}$.
Tie both inputs together, so $B = A$: $Y = \overline{A \cdot A}$.
Since $A \cdot A = A$ (idempotent law), $Y = \overline{A}$.
This is exactly the NOT (inverter) function.

Answer: Connecting both inputs of a NAND gate together gives $Y = \overline{A}$, a NOT gate — which is why NAND is called a universal gate.

Key Points

A BJT (npn or pnp) has emitter, base and collector; in active mode the emitter-base junction is forward-biased and the collector-base junction reverse-biased, with $I_E = I_B + I_C$.
Current gains: $\alpha = \dfrac{I_C}{I_E} < 1$ and $\beta = \dfrac{I_C}{I_B}$ (50-300), related by $\beta = \dfrac{\alpha}{1-\alpha}$.
As a switch the transistor uses cut-off (OFF) and saturation (ON); as an amplifier it works in the active region with CE voltage gain $A_V = -\beta R_L/R_{in}$ (output $180^\circ$ out of phase).
Basic gates: OR ($Y=A+B$), AND ($Y=A\cdot B$), NOT ($Y=\bar{A}$); each has a defined truth table.
NAND ($Y=\overline{A\cdot B}$) and NOR ($Y=\overline{A+B}$) are universal — any logic circuit can be built from NAND alone or NOR alone.

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.In a transistor in the active region, the emitter-base and collector-base junctions are respectively:

Explanation: For amplification the emitter-base junction is forward-biased (to inject carriers) and the collector-base junction reverse-biased (to collect them).

Q2.If $\beta = 100$ and the base current is $20\,\mu\text{A}$, the collector current is:

Explanation: $I_C = \beta I_B = 100 \times 20\,\mu\text{A} = 2000\,\mu\text{A} = 2\,\text{mA}$.

Q3.The Boolean expression for a NOR gate is:

Explanation: A NOR gate is an OR followed by a NOT, so $Y = \overline{A + B}$; it gives 1 only when both inputs are 0.

Q4.A transistor used as a switch operates in:

Explanation: As a switch the transistor is driven between cut-off (OFF, no $I_C$) and saturation (ON, maximum $I_C$), avoiding the active region.

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