BJT Simulator

Two tools in one: sweep V_in to explore cutoff / active / saturation operating regions, then tune β and temperature to see how different biasing schemes maintain — or lose — the Q-point.

BJT as a Switch

Drive an NPN BJT from cutoff through active to saturation by sweeping V_in. Watch V_CE, I_C, and the LED brightness respond in real time.

CUTOFF

ACTIVE

SATURATION

$V_{BE}$ below ~0.5 V — base–emitter junction not yet forward-biased. $I_B \approx 0$ , $I_C \approx 0$ , LED off.

V_{BE}

0.000 V

I_B

\beta \cdot I_B

(would-be)0

I_C

(actual)0

V_{CE}

3.20 V

V_{in}

: 0.00 V

0 V5 V

Fixed: $\beta$ = 100, $V_{BE(\text{on})}$ = 0.7 V, $V_{CE(\text{sat})}$ = 0.2 V, $V_{LED}$ ≈ 1.8 V. Saturation kicks in when $\beta \cdot I_B$ exceeds $(V_{CC} - V_{LED} - V_{CE(\text{sat})}) / R_C$ ≈ 3.00 mA.

Q-point Stability (Biasing)

Compare voltage-divider bias (VDB) against fixed bias. Sweep β or temperature and watch the Q-point dot move — or stay put.

Voltage-Divider Bias (stable)

Fixed Bias (β-sensitive)

\beta

: 150

50300

Q-point drift across β = 50 → 300

VDB

Fixed

I_C

@ min

3.61 mA

1.71 mA

I_C

@ max

5.66 mA

10.27 mA

drift

44.3 %

142.9 %

Voltage-divider bias is stable because $R_E$ provides DC negative feedback: if $I_C$ tries to rise, the emitter voltage rises, $V_{BE}$ drops, and the rise is opposed. Fixed bias has no such feedback — $I_C = \beta \cdot I_B$ tracks $\beta$ directly.

BJT fundamentals

A Bipolar Junction Transistor (BJT) is a current-controlled device: a small base current I_B controls a much larger collector current I_C. The ratio β (hFE) = I_C / I_B is typically 50–300 but varies with temperature and operating point — never design a circuit that depends on an exact β value.

Three operating regions

Cutoff: V_BE < 0.6 V — transistor off, I_C ≈ 0.
Active: V_BE ≈ 0.7 V, V_CE > V_CE(sat) — amplifier region, I_C = β × I_B.
Saturation: both junctions forward-biased — transistor fully on, V_CE(sat) ≈ 0.2 V.

Voltage-divider bias

Voltage-divider bias (VDB) sets the Q-point using a resistor divider at the base, independent of β. If the divider current is ≥ 10 × I_B, the Q-point is stable across transistor variations and temperature — the standard approach for analog amplifier stages.

Learn more → BJT Amplifier — Learn

Quick experiments

Walk through all three regions. Sweep base current from zero. At 0 the transistor is in cutoff; a small current puts it in the active region where Ic = β·Ib; push further and Vce collapses to a few hundred millivolts — saturation.
Design a saturated switch. For 100 mA of collector current, use a forced β of 10 and drive 10 mA into the base. From 5 V that needs (5 − 0.7)/10 mA ≈ 430 Ω. Use 390 Ω and it saturates with margin.
Watch β betray you. Change β from 100 to 300 with the bias fixed by a single base resistor. The Q-point moves dramatically. Add an emitter resistor and the same β change barely shifts it.
Measure the cost of saturation. At Vce(sat) ≈ 0.2 V and 100 mA, the transistor burns 20 mW. Scale to 1 A and it is 200 mW — the point where a MOSFET with milliohm on-resistance becomes the better switch.
Push it toward thermal runaway. Raise the temperature with no emitter resistor: Vbe falls about 2 mV/°C, so Ic rises, which heats it further. The emitter resistor is what breaks that loop.

Formula reference

Current gain in the active region: $I_C = \beta I_B$
Only valid in the active region — not in saturation.
Emitter current: $I_E = I_C + I_B = (\beta + 1) I_B$
Emitter carries both collector and base current.
Base resistor for a saturated switch: $R_B = \frac{V_{in} - V_{BE}}{I_C / \beta_{forced}}$
Use a forced β of about 10, never the datasheet β.
Power dissipated in the transistor: $P = V_{CE} \, I_C$
Worst in the active region; small in cutoff and saturation.

Symbol	Meaning	Unit
$\beta$	DC current gain (hFE)	—
$V_{BE}$	Base-emitter drop, about 0.7 V when conducting	V
$V_{CE(sat)}$	Collector-emitter drop in saturation	V
$I_C$	Collector current	A

Common mistakes

Sizing the base resistor with the datasheet β.
Datasheet β is a typical active-region figure that varies 3:1. For a switch use a forced β of about 10 so the transistor saturates even on a low-gain sample.
Applying Ic = β·Ib in saturation.
Once saturated, collector current is set by the load, not by β. Extra base current changes nothing except making it harder to turn off.
Leaving out the emitter resistor in an amplifier.
Without emitter degeneration the Q-point depends on β and temperature, so it drifts between samples and warms into distortion. An emitter resistor sets the operating point with resistor ratios instead.
Driving an inductive load without a flyback diode.
Interrupting coil current forces the collector voltage to spike far beyond the supply, breaking down the transistor. A reverse diode across the coil clamps it.
Forgetting the 0.7 V base-emitter drop.
Base current is (Vin − 0.7) divided by the base resistor, not Vin divided by it. Driving from 3.3 V, ignoring the drop overestimates base current by more than 20 %.

Frequently asked questions

What are the three operating regions of a BJT?

Cutoff, where both junctions are reverse biased and no current flows; active, where the base-emitter junction is forward biased and collector current is beta times base current; and saturation, where both junctions conduct and the transistor behaves like a closed switch.

How do I choose a base resistor to saturate a transistor?

Work out the collector current the load needs, divide by a forced beta of about 10 to get base current, then size the resistor as (Vin − 0.7) divided by that base current. Using the datasheet beta instead leaves the transistor in the active region where it dissipates heat.

Why is Vce not zero in saturation?

A saturated transistor still drops a saturation voltage, typically 0.1 to 0.3 V for small-signal parts. That drop times the collector current is the switching loss, which is why MOSFETs with milliohm on-resistance are preferred for heavy loads.

Why does beta vary so much between transistors?

Beta depends on manufacturing spread, collector current and temperature, and can vary three to one within the same part number. Reliable designs use emitter degeneration or feedback so the operating point is set by resistors, not by beta.

What is thermal runaway in a BJT?

As a transistor heats, its base-emitter voltage falls and collector current rises, which produces more heat. Without an emitter resistor or thermal feedback the cycle escalates until the device fails.

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