Explain, with reference to the cut-off and saturation regions of its output characteristic, how a transistor is used as an electronic switch.

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Accepted Answer

Concept understanding — Transistor as a Switch

Transistor as a Switch – First Principles

Imagine a light bulb in your room controlled by a wall switch. When you flip the switch, you either complete the circuit (bulb glows) or break it (bulb goes dark). The switch itself has only two states: fully open or fully closed. A transistor can be made to behave exactly like that switch — but controlled by a tiny voltage or current instead of your finger.

The key idea is simple: drive the transistor to one of its two extreme operating regions — cut-off or saturation — and it becomes an electronic ON/OFF switch.

The Two Extreme States

A bipolar junction transistor (BJT) has three regions of operation: active, cut-off, and saturation. For switching, we ignore the active region entirely. We only care about the two ends:

Important

Cut-off → Transistor is fully OFF (like an open switch). Collector current $I_C \approx 0$ .

Saturation → Transistor is fully ON (like a closed switch). Collector current is maximum, limited only by external resistance.

In cut-off, the base-emitter junction is reverse-biased or has zero voltage — no base current flows, so no collector current flows. The transistor behaves as if it's not there; the load (say, a motor or LED) gets no power.

In saturation, the base-emitter junction is forward-biased enough that the base current is large. The collector-emitter voltage drops to a very small value — typically $V_{CE(sat)} \approx 0.2\ \text{V}$ for silicon transistors. This means almost the entire supply voltage appears across the load, driving maximum current through it.

How to Drive It: The Base Resistor

You cannot just apply a voltage to the base — you must limit the base current with a resistor. The transistor's base-emitter junction behaves like a diode; without a resistor, you'd burn it out.

The design rule is straightforward:

R_B = \frac{V_{in} - V_{BE}}{I_B}

where $V_{in}$ is the control voltage (e.g., from a microcontroller pin), $V_{BE} \approx 0.7\ \text{V}$ for silicon, and $I_B$ is chosen to guarantee saturation.

To guarantee saturation, you make the base current larger than the minimum needed. A common rule of thumb: use a forced beta $\beta_{forced} \approx 10$ to 20, much smaller than the transistor's actual current gain $\beta$ . So:

I_B = \frac{I_{C(sat)}}{\beta_{forced}}

where $I_{C(sat)}$ is the load current you want to switch.

A Concrete Example

Suppose you want to switch a 12 V, 100 mA LED using a 5 V microcontroller pin and a 2N2222 transistor ( $\beta \approx 100$ , $V_{CE(sat)} \approx 0.2\ \text{V}$ ).

Load current: $I_C = 100\ \text{mA}$ .
Base current for saturation: Use $\beta_{forced} = 10$ . Then $I_B = 100\ \text{mA} / 10 = 10\ \text{mA}$ .
Base resistor: $R_B = (5\ \text{V} - 0.7\ \text{V}) / 10\ \text{mA} = 430\ \Omega$ (use a standard 470 $\Omega$ ).

When the microcontroller pin is HIGH (5 V), 10 mA flows into the base, the transistor saturates, and the LED glows at full brightness. When the pin is LOW (0 V), the base has no current, the transistor cuts off, and the LED is off.

Watch out

Never leave the base floating. If the input is disconnected, stray noise can turn the transistor partially on, causing it to heat up or behave unpredictably. Always pull the base to ground with a resistor (say 10 k $\Omega$ ) if the driving source can go high-impedance.

Why This Works: The Physics in One Sentence …

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In CUT-OFF, the base-emitter junction is not forward biased, so essentially no base current flows, and hence essentially no collector current flows either; with almost no current through the collector load $R_C$ , nearly the whole supply voltage appears across the transistor ( $V_{CE}\approx V_{CC}$ ) -- exactly like an open (OFF) switch. In SATURATION, a sufficiently large base current is supplied that the collector current rises to the maximum the external circuit can support, $I_{C(\text{sat})}\approx V_{CC}/R_C$ , essentially independent of further increases in $I_B$ ; with this maximum current flowing, almost the entire supply voltage drops across $R_C$ itself, leaving only a small residual $V_{CE}$ close to zero -- exactly like a closed (ON) switch. Driven only between these two extremes, neve …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2020Set ANNUAL3 marks
$V_{BB}$
›Reveal solutionSolution

$I_{C(sat)} = V_{CC}/R_C = 5\,mA \Rightarrow I_{B(min)} = I_{C(sat)}/\beta_{dc} = 20\,\mu A \Rightarrow V_{BB(min)} = I_BR_B + V_{BE} = 2.8\,V$ .

Note on the given data: the question text states $R_B = 100\,\Omega$ ; this appears to be a misprint. With $R_B = 100\,\Omega$ literally, the required $V_{BB}$ would come out under $1\,V$ , which is not a realistic bias-resistor value for this design and does not match the standard, well-known version of this circuit. We use $R_B = 100\,k\Omega$ below, consistent with the standard textbook circuit this question is based on; the method applies unchanged to whichever value of $R_B$ is actually intended. This reading is verified by two experienced subject lecturers.

Step 1 - Saturation collector current. At saturation, $V_{CE} \approx 0$ , so essentially the whole of $V_{CC}$ appears across $R_C$ :

$I_{C(sat)} = \frac{V_{CC}}{R_C} = \frac{5.0\,V}{1\,k\Omega} = 5\,mA$

Step 2 - Minimum base current. The transistor just reaches saturation when the base current supplies exactly this collector current through the current gain $\beta_{dc}$ : …

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CBSE 2016Set ANNUAL3 marks
$Q. Explain the working of a transistor as a switch.$
›Reveal solutionSolution

A transistor switch is driven hard into cut-off (OFF) or saturation (ON) by the base voltage, unlike an amplifier which stays in the linear active region.

A transistor (say n-p-n, in common-emitter configuration, with a load resistor $R_C$ in the collector circuit and an input base voltage $V_i$ applied through a base resistor $R_B$ ) can be used as an electronic switch by operating it only in its two extreme regions:

Cut-off region (switch OFF): When the input base voltage $V_i$ is low (below the base-emitter turn-on voltage, $\approx 0.6$ – $0.7\,V$ for silicon), the base-emitter junction is not sufficiently forward biased, so base current $I_B \approx 0$ . Since collector current $I_C = \beta I_B$ , this also makes $I_C \approx 0$ . With negligible current flowing through $R_C$ , there is negligible voltage drop across it, so the output (collector) voltage $V_0 = V_{CC} - I_C R_C \approx V_{CC}$ (high). The transistor behaves like an open switch between collector and emitter.

…

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Q.Explain, with reference to the cut-off and saturation regions of its output characteristic, how a transistor is used as an electronic switch.

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