Related Concepts: 01 Digital Systems & Signal Propagation | 04 NPN Transistor Basics & DTL Logic | 05 TTL Logic Families & Open Collector Bus Systems

4.03 BJT as NOT Gate & Inverter Operation

BJT Transistor Switch Concept

In digital logic, a Bipolar Junction Transistor (BJT) operates strictly as a digital switch driven between two extreme operating states:

1. Cut-off Region (OFF Switch): Base-Emitter junction reverse-biased $⟹I_B=0,I_C=0,V_{out}=V_{CC}$ (Logic HIGH).
2. Saturation Region (ON Switch): Base-Emitter junction forward-biased $⟹I_B\ge I_{B(sat)},V_{out}=V_{CE(sat)}\approx 0.2\text{V}$ (Logic LOW).

graph TD
    Input[Input Voltage Vin] --> Check{Vin Level}
    
    Check -->|Vin = LOW e.g. 0.2V| Cutoff[Cut-off Region: BE Junction OFF]
    Cutoff --> ICZero[Collector Current IC = 0]
    ICZero --> OutHigh[Output Vout = VCC = HIGH Logic 1]
    
    Check -->|Vin = HIGH e.g. 5V| Saturation[Saturation Region: BE Junction ON]
    Saturation --> ICSat[Max Collector Current IC-sat]
    ICSat --> OutLow[Output Vout = VCE-sat = 0.2V = LOW Logic 0]

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1. Mathematical Saturation & Cut-off Proof

Major Exam Numerical Problem (2016, 2019 - 13 Marks)

Given Circuit Parameters:

• $V_{CC}=5\text{V}$, $V_{in(HIGH)}=5\text{V}$, $V_{in(LOW)}=0.2\text{V}$
• Base Resistor $R_B=10\text{k}\Omega$, Collector Resistor $R_C=1\text{k}\Omega$
• Minimum Current Gain $h_{FE}(\beta )=30$
• Transistor Constants: $V_{BE(sat)}=0.7\text{V}$, $V_{CE(sat)}=0.2\text{V}$

Part A: Operation when $V_{in}=0.2\text{V}$ (Logic LOW)

1. Base-Emitter voltage $V_{BE}=0.2\text{V}<0.7\text{V}$ {below cut-in threshold}.
2. Transistor is in Cut-off. Base current $I_B=0$, Collector current $I_C=0$.
3. Output voltage: $V_{out}=V_{CC}-(I_C\cdot R_C)=5\text{V}-(0\cdot 1\text{k}\Omega )=5\text{V}(\text{Logic HIGH})$

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Part B: Operation when $V_{in}=5\text{V}$ (Logic HIGH)

1. Calculate actual base current ($I_B$): $I_B=\frac{V_{in}-V_{BE(sat)}}{R_B}=\frac{5\text{V}-0.7\text{V}}{10\text{k}\Omega }=\frac{4.3\text{V}}{10\text{k}\Omega }=0.43\text{ mA}$

2. Calculate saturation collector current ($I_{C(sat)}$): $I_{C(sat)}=\frac{V_{CC}-V_{CE(sat)}}{R_C}=\frac{5\text{V}-0.2\text{V}}{1\text{k}\Omega }=\frac{4.8\text{V}}{1\text{k}\Omega }=4.8\text{ mA}$

3. Calculate minimum base current required for saturation ($I_{B(min)}$): $I_{B(min)}=\frac{I_{C(sat)}}{h_{FE}}=\frac{4.8\text{ mA}}{30}=0.16\text{ mA}$

4. Saturation Condition Check: $I_B(0.43\text{ mA})\ge I_{B(min)}(0.16\text{ mA})$ Since $I_B>I_{B(min)}$, the transistor is driven deep into saturation.

5. Output voltage drops to saturation voltage: $V_{out}=V_{CE(sat)}=0.2\text{V}(\text{Logic LOW})■$

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Past Year Questions (PYQs)

• [PYQ 2016, 2019]: BJT inverter operation: Mathematical proof of saturation/cut-off given circuit parameters. (13 Marks)