---
title: "Transistor switch with opto-coupler base drive"
subject: "Power Electronics"
chapter: 1
key: "semester_7/power_electronics:transistor-switch-opto-base-signal"
source: https://www.bctengineer.com/q/semester_7/power_electronics/transistor-switch-opto-base-signal
site: BCT Engineering
---

# Transistor switch with opto-coupler base drive

Power Electronics · Chapter 1: Characteristics and specification of power electronic devices

- Asked in: 2076 Chaitra, 8 marks
- Video: [Electronic Basics #32: Relays & Optocouplers](https://www.youtube.com/watch?v=2BdevOmN-Zk) — GreatScott!
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/power_electronics/transistor-switch-opto-base-signal
- Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-1

## Question

Explain how a transistor can be used as a static switch. Describe a base current signal generating circuit using an opto-coupler.

## Short answer

A transistor operates as a static switch by alternating between cutoff (OFF) and saturation (ON) with an overdrive factor ensuring low loss. An opto-coupler circuit isolates the PWM control signal, while a totem-pole driver on the power side supplies forward base current for saturation and reverse current for fast turn-off.

## Answer

A transistor acts as a **static switch** (no moving parts) when it is operated only in **cut-off** (OFF) and **saturation** (ON), switched by its base current.

#### Transistor as a static switch

```text
                 +Vcc
                  |
                 RL (load)
                  |
          Rb     C|  iC
   vB o--/\/\/---B|/
                  |\  NPN
                 E|
                  |
                 GND
```

- **OFF state (cut-off):** when $v_B \le 0$, $I_B = 0$, both junctions are reverse biased and $I_C \approx 0$ (only leakage). The transistor acts as an open switch and the full supply voltage $V_{CC}$ appears across C-E.
- **ON state (saturation):** when enough base current is supplied, both junctions are forward biased. $V_{CE(sat)}$ is only about 0.2 to 1 V, so the transistor acts as a closed switch and $I_C = \frac{V_{CC} - V_{CE(sat)}}{R_L}$.
- To be sure of saturation, the base current is made larger than the minimum value:
$$
I_B > I_{BS} = \frac{I_{CS}}{\beta_{min}}, \qquad \text{ODF} = \frac{I_B}{I_{BS}}
$$
where ODF (overdrive factor) is usually 1.5 to 3 (too much overdrive increases storage time).
- The active region is avoided because both $V_{CE}$ and $I_C$ are large there, giving high power loss $V_{CE} I_C$. In the switching mode the loss is small: in cut-off the current is nearly zero, and in saturation the voltage is nearly zero.

#### Base current signal generating circuit using an opto-coupler

```text
 control side      |          power side
 +--------+        |         +--------------+
 | pulse  |  LED   | photo-  | totem pole   |-Rb-> B
 | (PWM)  |--|>~~~~~~> tran- | Q1 to +Vaux  |   (power
 +--------+        |  sistor | Q2 to -Vaux  |    BJT)
                   |         +--------------+---> E
     only light crosses the isolation barrier
```

1. The control pulse (PWM from a 555 timer or microcontroller) lights the LED of the opto-coupler through a current-limiting resistor $R_1$.
2. The phototransistor turns ON and drives the totem-pole stage $Q_1$-$Q_2$, which is powered by an isolated auxiliary supply referenced to the emitter of the power BJT.
3. When the pulse is high, $Q_1$ conducts and supplies positive base current $I_{B1}$ through $R_b$; the power transistor saturates.
4. When the pulse is low, $Q_2$ conducts and connects the base to $-V_{aux}$, giving a negative base current that removes stored charge and turns the power BJT OFF quickly.
5. Since the only link between the two sides is light, the logic circuit is fully isolated from the high-voltage power circuit.

## Check yourself

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

1. In which regions does a transistor operate as a static switch, and what is VCE(sat) roughly?
   A. Active region; about 10 V
   B. Cut-off and saturation; about 0.2 to 1 V in saturation
   C. Active and cut-off; about 5 V
   D. Saturation only; exactly 0 V
   Answer: B. The switch is OFF in cut-off (IC ~ 0) and ON in saturation with only about 0.2 to 1 V across CE. (Key point: OFF and ON states)

2. Why is the active region avoided in switching?
   A. Both VCE and IC are large, giving high power loss VCE*IC
   B. The gate cannot be controlled
   C. The transistor cannot conduct there
   D. Leakage current is zero
   Answer: A. In cut-off the current is nearly zero and in saturation the voltage is nearly zero, so the loss is small only in these states. (Key point: Avoiding active region)

3. How is saturation ensured, and what is a limit on overdrive?
   A. IB < ICS/beta; ODF below 1
   B. IB > IBS = ICS/beta_min; ODF usually 1.5 to 3, since too much overdrive increases storage time
   C. IB = 0; ODF of 10
   D. IB > ICS; ODF of 100
   Answer: B. Base current must exceed IBS = ICS/beta_min. ODF = IB/IBS is kept at 1.5 to 3 because heavy overdrive lengthens storage time. (Key point: Overdrive factor)

4. In the opto-coupler base drive circuit, what happens when the control pulse is low?
   A. Q1 conducts and supplies +IB1
   B. Q2 connects the base to -Vaux, giving a negative base current that turns the BJT OFF quickly
   C. The LED lights up
   D. The isolation barrier is bypassed
   Answer: B. With the pulse low, Q2 conducts and the negative base current removes stored charge. (Key point: Opto-coupler turn-off)

## Related in Power Electronics

- [Power transistor switch and opto-coupler isolation](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Apower-transistor-switch-opto) — Power Electronics, chapter 1
- [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 switching and base drive](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Apower-transistor-switching-base-drive) — Power Electronics, chapter 1
- [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×
- [Thyristor V-I and opto-coupler isolation](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athyristor-vi-opto-isolation) — Power Electronics, chapter 1, asked 2×
- [Thyristor V-I, application and opto-isolation](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Athyristor-vi-application-opto) — Power Electronics, chapter 1, asked 2×

## Related in other subjects

- [Transistor as a switch](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Abjt-switch-cutoff-saturation) — Electronics Devices and Circuit, chapter 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