# 🎛️ BJT and MOSFET Switching Basics
## Learning Objectives
By the end of this lesson you should be able to:
- Name the three operating regions of a BJT and of a MOSFET, and state the terminal-voltage conditions that define each one.
- Apply the saturated-switch design routine: compute load current, size the base resistor, then verify with a consistency check.
- Decide when a MOSFET is fully enhanced and why insufficient gate drive causes overheating.
- Compare BJT and MOSFET switches on drive current, on-state voltage drop, and power loss.
## Quick Review
Last lesson you built a routine you will reuse today: replace the device with a model, test the bias, solve with KVL, and confirm the answer is consistent with the model you assumed. For a forward-biased silicon junction that model is a fixed {{c1::0.7 V}} drop. Keep the routine. Only the device changes.
## Theory
### The BJT as a switch
An NPN transistor has three terminals (base, collector, emitter) and two junctions. The base-emitter junction behaves like a diode: below roughly 0.7 V it does nothing, above it conducts. That single fact drives everything.
Three regions matter:
| Region | Condition | Behaviour |
| --- | --- | --- |
| Cutoff | $V_{BE} < 0.7\ \text{V}$ | No collector current. Switch OFF. |
| Active | $V_{BE} \approx 0.7\ \text{V}$, $V_{CE} > 0.3\ \text{V}$ | $I_C = \beta I_B$. Used for amplifiers. |
| Saturation | $V_{BE} \approx 0.7\ \text{V}$, $V_{CE} \approx 0.2\ \text{V}$ | Fully ON, current set by the load. |
The active-region relation is $I_C = \beta I_B$, and therefore $I_E = (\beta+1) I_B$. In saturation that relation no longer holds: the load sets the current, and the transistor simply looks like a small voltage source of about $V_{CE(sat)} = 0.2\ \text{V}$.
> [!WARNING]
> The word "saturation" is a trap. In switching it means fully ON. In amplifier design it also means the transistor is pinned against the rail and clipped. Always read it from context: today, saturation = switch closed.
To switch a load, put the load in the collector path and drive the base through a resistor. The load current when ON is
$$
I_{C(sat)} = \frac{V_{CC} - V_{CE(sat)}}{R_C}
$$
You need $I_B \geq I_{C(sat)}/\beta$. In practice you supply 3 to 10 times that, called the **overdrive factor**, because $\beta$ varies wildly between parts and with temperature, and because a deeply saturated transistor drops less voltage and runs cooler.
### The MOSFET as a switch
The MOSFET is voltage controlled. The gate is separated from the channel by an oxide, so the gate draws essentially no DC current, only a brief charging current each time it switches.
- **Cutoff**: $V_{GS} < V_{TH}$. OFF.
- **Triode (ohmic)**: $V_{GS} > V_{TH}$ and $V_{DS} < V_{GS} - V_{TH}$. Behaves like a resistor $R_{DS(on)}$. This is the ON state for switching.
- **Saturation**: $V_{GS} > V_{TH}$ and $V_{DS} \geq V_{GS} - V_{TH}$. Current is roughly constant. This is the amplifier region.
For switching you want the triode region, which means driving the gate well above $V_{TH}$, typically 4 to 10 V above. On-state drop is $I_D R_{DS(on)}$, often only tens of millivolts.
### Side-by-side
| Item | BJT (NPN) | N-MOSFET |
| --- | --- | --- |
| Controlled by | Base current $I_B$ | Gate-source voltage $V_{GS}$ |
| ON condition | $V_{BE} \approx 0.7\ \text{V}$, saturated | $V_{GS}$ well above $V_{TH}$ |
| ON-state drop | $V_{CE(sat)} \approx 0.2\ \text{V}$ | $I_D \times R_{DS(on)}$ |
| Drive current | Milliamps, continuous | Nanoamps, plus gate charge spikes |
| OFF condition | $V_{BE} < 0.7\ \text{V}$ | $V_{GS} < V_{TH}$ |
## Worked Examples
### Example 1: Sizing the base resistor for an NPN switch
A relay coil of $R_C = 120\ \Omega$ sits between $+12\ \text{V}$ and the collector. A 3.3 V logic output drives the base. Given $\beta_{min} = 100$, $V_{BE} = 0.7\ \text{V}$, $V_{CE(sat)} = 0.2\ \text{V}$. Design $R_B$ with an overdrive factor of 5.
```steps
1. Load current when ON: $I_{C(sat)} = (12 - 0.2)/120 = 98.3\ \text{mA}$
2. Minimum base current: $I_B = 98.3/100 = 0.983\ \text{mA}$
3. Apply overdrive: $I_B = 5 \times 0.983 = 4.92\ \text{mA}$
4. Resistor value: $R_B = (3.3 - 0.7)/4.92\ \text{mA} = 528\ \Omega$, pick the standard 470 Ω
5. Verify: $I_B = 2.6/470 = 5.53\ \text{mA}$, and $\beta I_B = 553\ \text{mA}$ which is far above 98.3 mA, so the transistor is firmly saturated
6. Transistor dissipation: $P = 0.2 \times 98.3\ \text{mA} = 19.7\ \text{mW}$
```
### Example 2: A MOSFET that looks fine but is not
An N-MOSFET with $V_{TH} = 2\ \text{V}$ and $R_{DS(on)} = 50\ \text{m}\Omega$ at $V_{GS} = 10\ \text{V}$ drives a 12 V, 2 A load from a 3.3 V logic pin.
$V_{GS} = 3.3\ \text{V}$ clears $V_{TH}$ by only 1.3 V, so the channel is weakly formed. The resistor is not 50 mΩ here, it might be several tenths of an ohm. At a quarter ohm the dissipation is $I^2R = 4 \times 0.25 = 1\ \text{W}$ in a part sized for 0.2 W. Two fixes: choose a **logic-level** MOSFET whose $R_{DS(on)}$ is specified at 3.3 V, or add a gate driver that pulls the gate to 10 V. With proper drive the loss is $4 \times 0.05 = 0.2\ \text{W}$.
## Guided Practice
Work each one on paper before opening the answer.
**Exercise 1: Classify the region.** An NPN transistor with $V_{BE} = 0.68\ \text{V}$ measures (a) $V_{CE} = 0.15\ \text{V}$, (b) $V_{CE} = 6\ \text{V}$, (c) $V_{BE} = 0.4\ \text{V}$.
Click to reveal answer
(a) Saturation: base-emitter forward biased and collector pinned low. (b) Active: forward biased base-emitter with a large collector-emitter voltage, current set by $\beta I_B$. (c) Cutoff: base-emitter not conducting.
Click to reveal answer
$I_{C(sat)} = (5 - 0.2)/470 = 10.2\ \text{mA}$. Minimum base current is $10.2/50 = 0.204\ \text{mA}$; with overdrive, $2.04\ \text{mA}$. Then $R_B = (5 - 0.7)/2.04\ \text{mA} = 2.1\ \text{k}\Omega$. Sanity check: with 2 kΩ you get 2.15 mA and $\beta I_B = 107\ \text{mA} > 10.2\ \text{mA}$, saturated.
Click to reveal answer
(a) Cutoff, gate below threshold. (b) Triode, since $V_{DS} = 0.1 < V_{GS} - V_{TH} = 8.5\ \text{V}$; this is the switching state. (c) Saturation, since $V_{DS} = 6 \geq 4 - 1.5 = 2.5\ \text{V}$.
Click to reveal answer
BJT: $P = 0.25 \times 3 = 0.75\ \text{W}$. MOSFET: $P = 9 \times 0.02 = 0.18\ \text{W}$. The MOSFET runs cooler, but only if the gate is driven properly; gate charge must be supplied at every switching edge, so high switching frequency needs a real driver.