Transistors and Logic Gates: Quick Revision

A brief note on Transistors and Logic Gates. Read it before a test. Try each question at the end before you open its answer.

The transistor

• Regions: the emitter is heavily doped, the base is thin and lightly doped, and the collector is larger and moderately doped. The arrow of the symbol is on the emitter.
• Currents: $I_E = I_B + I_C$, $\alpha = \dfrac{I_C}{I_E}$, $\beta = \dfrac{I_C}{I_B}$, $\beta = \dfrac{\alpha}{1 - \alpha}$.
• Regions of work: active (emitter forward, collector reverse, $I_C \approx \beta I_B$), saturation (both forward, $V_{CE} \approx 0.2$ V), cut-off (no base current).
• Circuit: $V_{CC} = I_C R_C + V_{CE}$ and $V_{BB} = I_B R_B + V_{BE}$, with $V_{BE} = 0.7$ V for silicon.
• Resistances: $r_i = \left(\dfrac{\Delta V_{BE}}{\Delta I_B}\right)_{V_{CE}}$ and $r_o = \left(\dfrac{\Delta V_{CE}}{\Delta I_C}\right)_{I_B}$.

Amplifier and switch

• Voltage gain: $A_V = -\dfrac{\beta R_C}{r_i}$. The output is $180^\circ$ out of phase. $A_I = \beta$ and $|A_P| = \beta |A_V|$.
• Clipping: the output stays between $0.2$ V and $V_{CC}$. The Q point is near $V_{CC}/2$.
• Switch: cut-off gives $V_o = V_{CC}$ and saturation gives $V_o \approx 0.2$ V. Saturation needs $I_B \ge \dfrac{V_{CC} - 0.2\ \text{V}}{\beta R_C}$. The switch is a NOT gate.
• Oscillator: an amplifier with positive feedback and a tuned circuit, $f = \dfrac{1}{2\pi\sqrt{LC}}$.

Logic gates

• Gates: NOT $\overline{A}$, OR $A + B$, AND $A \cdot B$, NAND $\overline{A \cdot B}$, NOR $\overline{A + B}$, XOR $A \oplus B$.
• Remember: NAND is $0$ only for $1, 1$. NOR is $1$ only for $0, 0$. XOR is $1$ when the inputs differ. $n$ inputs give $2^n$ rows.
• De Morgan: $\overline{A + B} = \overline{A} \cdot \overline{B}$ and $\overline{A \cdot B} = \overline{A} + \overline{B}$.
• Universal gates: NAND alone, or NOR alone, can make every gate. NOT is one gate, AND is two, OR is three.

A worked example

A transistor has $\beta = 100$, $V_{CC} = 9$ V, $R_B = 300\ \text{k}\Omega$ and $R_C = 1.0\ \text{k}\Omega$. Find $V_{CE}$.

1. The base current. $I_B = \dfrac{9\ \text{V} - 0.7\ \text{V}}{300\ \text{k}\Omega} = 27.7\ \mu\text{A}$.
2. The collector current. $I_C = 100 \times 27.7\ \mu\text{A} = 2.77$ mA.
3. The voltage. $V_{CE} = 9\ \text{V} - 2.77\ \text{mA} \times 1.0\ \text{k}\Omega = 6.23$ V. The transistor is active.

Practise with new numbers

Check yourself

1. Which gate is made by tying the two inputs of a NAND gate together?

2. The input of a transistor inverter changes from $0$ V to $5$ V. What happens to the output?

3. An amplifier has a voltage gain of $-150$. What are the size and the phase of the output for an input of $4.0$ mV?

Sources

• NCERT Physics Class 12, Semiconductor Electronics: Materials, Devices and Simple Circuits.
• Horowitz and Hill, The Art of Electronics, third edition.