How do BJTs and MOSFETs work as electronic switches between cut-off and saturation?

In one line

A transistor works as a switch by being driven fully OFF (cut-off, like an open switch) or fully ON (saturation for a BJT, low-resistance ohmic region for a MOSFET, like a closed switch), with no stay in the active region.

Explained simply

A wall switch is either on or off, never half-way. A transistor can be used the same way, but it is switched by a small electrical signal and can switch thousands or millions of times per second with no moving parts. A small signal from a microcontroller can thus turn on a relay, a motor or a bulb that needs much more current.

Technically, for a BJT: cut-off when \( V_{BE} \) is below about 0.5 V (\( I_C \approx 0 \), \( V_{CE} \approx V_{CC} \)); saturation when base current is large enough that \( I_C \) reaches its maximum \( V_{CC}/R_C \) (\( V_{CE(sat)} \approx 0.2 \) V). For an enhancement MOSFET: OFF when \( V_{GS} < V_T \); ON when \( V_{GS} \) is well above \( V_T \), giving a small \( R_{DS(on)} \).

How it works

1. Load (LED, relay coil, lamp) is connected between \( V_{CC} \) and the collector (or drain).
2. Input LOW: no base current (or \( V_{GS} < V_T \)); transistor OFF; load has no current.
3. Input HIGH: base current \( I_B \) larger than \( I_{C(sat)}/\beta \) (usually 2 to 10 times larger, to be sure); transistor saturates; load gets nearly full \( V_{CC} \).
4. Power loss in the transistor is small in both states (either current or voltage is nearly zero); most loss occurs during the brief transitions.
5. For inductive loads like relays and motors, a freewheeling diode across the load absorbs the voltage spike at switch-off.
6. MOSFETs need almost no gate current and switch fast, so they dominate high-power and high-frequency switching (SMPS, inverters, motor drives); BJTs are common for simple low-power switching.

Key formula

\( I_{C(sat)} \approx \dfrac{V_{CC} - V_{CE(sat)}}{R_C}, \quad I_{B} > \dfrac{I_{C(sat)}}{\beta_{min}}, \quad R_B = \dfrac{V_{in} - V_{BE}}{I_B} \)

\( V_{CC} \) = supply (V), \( V_{CE(sat)} \approx 0.2 \) V, \( R_C \) = load resistance (\( \Omega \)), \( \beta_{min} \) = minimum current gain, \( V_{in} \) = input HIGH voltage (V), \( V_{BE} \approx 0.7 \) V, \( R_B \) = base resistor (\( \Omega \)).

Quick example

A 12 V relay with a 200 \( \Omega \) coil needs about 60 mA. With \( \beta_{min} = 100 \), \( I_B \) must exceed 0.6 mA; choosing 1.2 mA from a 5 V logic output gives \( R_B = (5 - 0.7)/0.0012 \approx 3.6 \) kΩ. A diode across the coil protects the transistor. (Illustration only.)

Common mistakes

• Operating the 'switch' in the active region, where it heats up because both \( V_{CE} \) and \( I_C \) are large.
• Omitting the base resistor and connecting the base straight to a supply.
• Forgetting the freewheeling diode with a relay or motor.

How MU asks it

Numericals are not expected in this module (MU syllabus), so questions are theory with diagrams. Typical: "Explain the transistor as a switch" (5 marks), sometimes "Explain FET/MOSFET as a switch". For full marks draw the CE switch circuit with load and base resistor, mark the cut-off and saturation points on the output characteristics/load line, write conditions for ON and OFF with \( V_{CE} \) values, and give a real application.