--- title: "Power transistor switch and opto-coupler isolation" subject: "Power Electronics" chapter: 1 key: "semester_7/power_electronics:power-transistor-switch-opto" source: https://www.bctengineer.com/q/semester_7/power_electronics/power-transistor-switch-opto site: BCT Engineering --- # Power transistor switch and opto-coupler isolation Power Electronics · Chapter 1: Characteristics and specification of power electronic devices - Asked in: 2069 Chaitra, 8 marks - Video: [What is Optocoupler ? How Optocoupler Works ? The Optocoupler Explained](https://www.youtube.com/watch?v=eHevtwPn27w) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/power-transistor-switch-opto - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-1 ## Question Explain how a power transistor can be used as a switch in the electric circuit. How an opto-coupler can be used to isolate the gate signal generator and power circuit? ## Short answer A power transistor switches by operating in saturation (base current above Ics/beta_min) for ON, and cut-off (zero base current) for OFF, so it acts as closed/open switch with low loss. An opto-coupler isolates by using an LED on the control side to turn on a phototransistor on the power side, which supplies base current from an auxiliary supply. ## Answer A power transistor is used as a switch by driving it into **saturation** (ON) or **cut-off** (OFF) with its base current, so it behaves like a closed or open switch with very small power loss. #### Power transistor as a 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 make sure the transistor saturates, 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 the storage time). - The active region is avoided because both $V_{CE}$ and $I_C$ are large there, giving a high power loss $V_{CE} I_C$. In switching mode the loss is small: in cut-off the current is nearly zero, and in saturation the voltage is nearly zero. ```text iC ^ saturation | | IB4 | |---------------------- | | IB3 | |---------------------- active | | IB2 region | |---------------------- | | IB1 | |---------------------- |/_______________________ IB = 0 +----------------------------> vCE cut-off region ``` #### Isolation of the drive circuit using an opto-coupler **Opto-coupler isolation:** the gate (base) signal generator (logic level, grounded to the control circuit) and the power circuit (hundreds of volts, emitter floating) must be electrically separated. An opto-coupler does this using light. ```text control side | power side | + Vaux R1 | | o--/\/\--+ | R2 pulse | | | LED ~~~~> | photo-transistor | light | (C top, E below) o--------+ | | GND | +------> B (base of power transistor) | | | R3 | | | Vaux(-) +------> E (emitter of power transistor) ``` 1. The control pulse drives a current through the LED via resistor $R_1$. 2. The LED emits infrared light, which falls on the phototransistor (or photo-darlington) inside the same package. 3. The phototransistor turns ON and connects the auxiliary supply $V_{aux}$ (referenced to the emitter of the power transistor) through $R_2$ to its base, supplying base current. $R_3$ prevents false turn-on by noise. 4. When the pulse ends, the LED goes dark, the phototransistor turns OFF and the base current stops. There is no electrical path between the two sides, so the isolation can withstand a few kV, and noise from the power circuit does not reach the logic circuit. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which states is a power transistor driven into when used as a switch? A. Saturation (ON) and cut-off (OFF) B. Active region and cut-off C. Active region and saturation D. Only the active region Answer: A. In saturation V_CE is nearly zero and in cut-off I_C is nearly zero, so power loss is small. (Key point: Switch operating states) 2. Why is the active region avoided while switching? A. The base current is zero there B. Both V_CE and I_C are large there, giving high power loss C. Both junctions are reverse biased there D. The transistor acts as an open switch there Answer: B. Loss = V_CE I_C is large in the active region, whereas it is small at the two extremes. (Key point: Avoiding the active region) 3. How is the base current chosen to guarantee saturation? A. I_B < I_CS/beta_min B. I_B = 0 for the ON state C. I_B > I_BS = I_CS/beta_min, with overdrive factor usually 1.5 to 3 D. I_B as large as possible, since overdrive has no drawback Answer: C. Overdrive ensures saturation, but too much overdrive increases storage time. (Key point: Saturation base current) 4. In the opto-coupler drive, what makes the gate/base current flow? A. The LED current flows directly to the base B. A transformer induces voltage in the base circuit C. R3 supplies the gate current from the logic supply D. Light from the LED turns on the phototransistor, connecting Vaux through R2 to the gate Answer: D. The phototransistor connects the isolated auxiliary supply (referenced to the emitter) to the gate; R3 prevents noise triggering. (Key point: Opto-coupler operation) ## Related in Power Electronics - [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 - [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× - [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× - [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× - [BJT as a switch in power circuit](https://www.bctengineer.com/question.md?key=semester_7%2Fpower_electronics%3Abjt-as-power-switch) — Power Electronics, chapter 1, asked 2× ## Related in other subjects - [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 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