--- title: "BJT as a switch in power circuit" subject: "Power Electronics" chapter: 1 key: "semester_7/power_electronics:bjt-as-power-switch" source: https://www.bctengineer.com/q/semester_7/power_electronics/bjt-as-power-switch site: BCT Engineering --- # BJT as a switch in power circuit Power Electronics · Chapter 1: Characteristics and specification of power electronic devices - Asked in: 2074 Asoj, 8 marks; 2073 Shrawan, 8 marks (asked 2 times) - 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_7/power_electronics/bjt-as-power-switch - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-1 ## Question Explain how a transistor (BJT) can be used as a switch in power circuit. ## Short answer A BJT acts as a switch by operating only in cut-off for OFF and saturation for ON, avoiding the active region. In cut-off IC is nearly zero; in saturation VCE(sat) is about 0.2 to 1 V, so power loss is small. Base current must exceed ICS/beta_min, with overdrive factor 1.5 to 3. ## Answer A BJT is used as a switch in power circuits by driving it either fully OFF (**cut-off**) or fully ON (**saturation**) with its base current, never in the active region. In this way it behaves like an ideal ON/OFF switch with very small power loss. #### Circuit and operation ```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. ```text iC ^ saturation | | IB4 | |---------------------- | | IB3 | |---------------------- active | | IB2 region | |---------------------- | | IB1 | |---------------------- |/_______________________ IB = 0 +----------------------------> vCE cut-off region ``` #### Switching waveforms and losses ```text vB ^ ____ ____ || | | | |+ +--------+ +---- vCE ^ ______ ______ | | | | Vcc |----' |------' ~0.2 | '__ON__ iC ^ ___ ___ || | | | |' '------' '-----> t ``` - Conduction loss $= V_{CE(sat)} I_C \times$ duty ratio (small because $V_{CE(sat)}$ is small). - Switching loss occurs during the short turn-on and turn-off intervals; it rises with frequency. #### Example For $V_{CC} = 200$ V, $R_L = 10\ \Omega$, $V_{CE(sat)} = 1$ V and $\beta_{min} = 20$: $I_{CS} = (200 - 1)/10 = 19.9$ A, $I_{BS} = 19.9/20 = 0.995$ A. With ODF = 2, the base drive is designed for about 2 A. #### Points to note - A freewheeling diode is connected across inductive loads to protect the BJT from voltage spikes at turn-off. - A negative base current at turn-off reduces storage time. - The BJT is a current-controlled device, so its driver must supply a continuous base current; for higher frequency a MOSFET or IGBT is often preferred. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. When the BJT switch is in the OFF state, what appears across collector-emitter? A. About 0.2 V B. Zero volts C. VCC/2 D. The full supply voltage VCC Answer: D. In cut-off IC is about zero, so there is no drop across RL and the full VCC appears across C-E, like an open switch. (Key point: Switch states) 2. With VCC = 200 V, RL = 10 ohm, VCE(sat) = 1 V and beta_min = 20, what is the minimum base current IBS? A. 0.995 A B. 19.9 A C. 2 A D. 10 A Answer: A. ICS = (200-1)/10 = 19.9 A, so IBS = 19.9/20 = 0.995 A. With ODF = 2 the drive is about 2 A. (Key point: Saturation current) 3. Which statement about BJT switch losses is correct? A. Conduction loss is large because VCE(sat) is large B. Switching loss occurs at turn-on and turn-off and rises with frequency C. Switching loss is independent of frequency D. There is no loss in a BJT switch Answer: B. Conduction loss is VCE(sat) x IC x duty (small), while switching loss occurs in the short transitions and increases with frequency. (Key point: Losses) 4. Why is a freewheeling diode connected across an inductive load of a BJT switch? A. To increase the base current B. To reduce the ODF C. To protect the BJT from voltage spikes at turn-off D. To keep the BJT in the active region Answer: C. The inductive load current continues at turn-off; the diode provides a path and prevents a large voltage spike across the BJT. 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