Q.In the figure, the $V_{BB}$ supply can be varied from 0 V to 5.0 V. The transistor has $\beta_{dc} = 250$ and $R_B = 100\,\Omega$, $R_C = 1\,k\Omega$, $V_{CC} = 5.0\,V$. Assume that when the transistor is saturated, $V_{CE} = 0\,V$, and $V_{BE} = 0.8\,V$. Calculate the minimum base current and input voltage for which the transistor will reach saturation.

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:

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:

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:

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 ($\beta \approx 100$, $V_{CE(sat)} \approx 0.2\ \text{V}$).

1. Load current: $I_C = 100\ \text{mA}$.
2. Base current for saturation: Use $\beta_{forced} = 10$. Then $I_B = 100\ \text{mA} / 10 = 10\ \text{mA}$.
3. 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.

Why This Works: The Physics in One Sentence …