---
title: "Transistor as a switch"
subject: "Electronics Devices and Circuit"
chapter: 2
key: "semester_3/electronics_devices_and_circuit:bjt-switch-cutoff-saturation"
source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/bjt-switch-cutoff-saturation
site: BCT Engineering
---

# Transistor as a switch

Electronics Devices and Circuit · Chapter 2: The Bipolar Junction Transistor

- Asked in: 2081 Baisakh, 4 marks
- Video: [Electronic Basics #22: Transistor (BJT) as a Switch](https://www.youtube.com/watch?v=WRm2oUw4owE) — GreatScott!
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/bjt-switch-cutoff-saturation
- Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-2

## Question

Explain Transistor as a Switch in cutoff and saturation region.

## Short answer

In cutoff, VBE<0.7V, IB=0, VCE≈VCC, acting as open switch. In saturation, both junctions forward biased, IB > IC(sat)/β, VCE≈0.2V, acting as closed switch. Example: for VCC=5V, RC=1kΩ, IC(sat)=4.8mA, IB(min)=48μA.

## Answer

A BJT used as a switch operates in the **cut-off** region (OFF) and the **saturation** region (ON); it passes quickly through the active region.

```text
          +VCC
            |
           RC
            |
 Vin --RB--B C----o Vout
            E
            |
           GND
```

#### Cut-off region (switch OFF)
- Base-emitter junction is not forward biased: Vin < 0.7 V (usually 0 V), so IB = 0.
- IC ≈ 0 (only leakage ICEO), so no drop across RC.
- VCE ≈ VCC, hence Vout = VCC (high).
- The collector-emitter path acts as an open switch.

#### Saturation region (switch ON)
- Both junctions are forward biased when IB is large enough:
```text
IC(sat) = (VCC - VCE(sat))/RC ≈ VCC/RC ,    IB > IC(sat)/β
IB = (Vin - 0.7)/RB
```
- VCE(sat) ≈ 0.2 V, so Vout ≈ 0.2 V (low). 
- The collector-emitter path acts as a closed switch of small resistance.

#### Example
VCC = 5 V, RC = 1 kΩ, β = 100: IC(sat) = 4.8 mA, IB(min) = 48 µA. For Vin = 5 V choose RB = 4.3 V/0.1 mA = 43 kΩ (about twice the minimum for hard saturation).

Because Vout is high when Vin is low and vice versa, the circuit is also an inverter (NOT gate). Uses: relay/LED drivers, digital logic.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. In cut-off (switch OFF) what happens?
   A. IC = beta IB with VCE = VCC/2
   B. IB = 0, IC ≈ 0, VCE ≈ VCC so Vout is high
   C. Both junctions forward biased
   D. IB is large and Vout ≈ 0.2 V
   Answer: B. With no forward bias on the BE junction no current flows, no drop on RC, so Vout = VCC. (Key point: Cut-off state)

2. What condition ensures saturation?
   A. Vin < 0.7 V
   B. IB > IC(sat)/beta, with IC(sat) ≈ VCC/RC
   C. IB < IC(sat)/beta
   D. IB = 0
   Answer: B. The base current must exceed what is needed for IC(sat) so both junctions become forward biased. (Key point: Saturation condition)

3. What is Vout in saturation and what does the C-E path act like?
   A. VCC; an open switch
   B. About 0.2 V; a closed switch of small resistance
   C. About 0.7 V; an open switch
   D. VCC/2; a resistor
   Answer: B. VCE(sat) ≈ 0.2 V so the C-E path is like a closed switch. (Key point: Output in saturation)

4. Why is the circuit also an inverter?
   A. It amplifies Vin by beta
   B. Vout equals Vin
   C. Vout is always VCC
   D. Vout is high when Vin is low and low when Vin is high
   Answer: D. Low input gives cut-off (Vout high) and high input gives saturation (Vout low), acting as a NOT gate. (Key point: Inverter action)

## Related in Electronics Devices and Circuit

- [MOSFET ID, VGS, VDS and saturation check](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Amosfet-saturation-check-10m-6m8) — Electronics Devices and Circuit, chapter 3

## Related in other subjects

- [Transistor as static switch and base drive](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Atransistor-static-switch-base-drive) — Power Electronics, chapter 1, asked 4×
- [Power transistor V-I and static switch](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Apower-transistor-vi-static-switch) — Power Electronics, chapter 1, asked 2×
- [TTL NAND gate; transistor switch](https://www.bctengineer.com/question.md?key=semester_3%2Fdigital_logic%3Attl-nand-transistor-switch) — Digital Logic, chapter 10
- [Transistor switch with opto-coupler base drive](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Atransistor-switch-opto-base-signal) — Power Electronics, chapter 1