BJT operation

YOU WILL BE ABLE TO

Describe the npn structure and the forward-active bias condition.
Relate emitter, base, and collector currents; define α and β.
State the exponential relationship between $V_{BE}$ and $I_C$ .
Identify cutoff, active, and saturation regions.

The MOSFET of ME-2 controls current with a voltage across an insulator. The bipolar junction transistor — invented first, in 1947, and still unbeaten at certain jobs — controls current with a current, through two coupled PN junctions. Its structure is three doped layers: for an npn device, a heavily doped n-type emitter, an extremely thin lightly doped p-type base, and an n-type collector. Two junctions, back to back, and the thinness of that base is the whole trick.

The npn transistor: base on the left, collector above, emitter below. The arrow on the emitter points outward — that is what makes it npn.

Bias the base-emitter junction FORWARD and the base-collector junction REVERSE, and the device enters its useful state: forward-active. Electrons pour from the emitter into the base, as they would in any forward-biased diode — but the base is so thin, and so lightly doped, that almost none of them find a hole to recombine with before they wander into the collector's reverse-biased field, which sweeps them across. The result is that a large emitter current arrives, a tiny fraction leaks out of the base, and the rest emerges from the collector.

I_E = I_B + I_C

KCL, applied to a transistor

That first equation is nothing but KCL from CA-3 — a transistor is a circuit node like any other. The interesting part is the RATIO. Define α as the fraction of emitter current reaching the collector (typically 0.98 to 0.998), and β as the current gain from base to collector:

\alpha = \frac{I_C}{I_E} \qquad\text{typically } 0.98\text{–}0.998

\beta = \frac{I_C}{I_B} = \frac{\alpha}{1 - \alpha}

α and β

A base current of $20$ µA commanding a collector current of $2$ mA is a current gain of 100 — and that amplification is the device's reason for existing. But note where β comes from: it is $\alpha/(1-\alpha)$ , a ratio dominated by a small difference of numbers near one. That makes β enormously variable — the same part number can arrive with β anywhere from 100 to 300, and β also drifts with temperature and with collector current. NEVER design a circuit whose behaviour depends on β. Lesson 2 is largely about how to obey that rule.

Underneath the current-gain description sits the real physics, and it is ME-1's Shockley equation again: the collector current is set exponentially by the base-emitter VOLTAGE, because the base-emitter junction is simply a forward-biased diode.

I_C = I_S\, e^{V_{BE}/V_T}

the exponential control law

Collector current against base-emitter voltage: the same exponential as ME-1's diode, because that junction is a diode.

Collector characteristics: each curve is one base current. The steep rise is saturation; the plateau is the active region.

The three regions, then. CUTOFF: base-emitter not forward biased, no current, an open switch. ACTIVE: B-E forward, B-C reverse — the plateaus above, where $I_C \approx \beta I_B$ and the device is an amplifier. SATURATION: both junctions forward biased, which happens when the collector voltage falls to within a couple of tenths of a volt of the emitter; the collector can no longer take all the current β would suggest, and $V_{CE}$ collapses to roughly $0.2$ V. That is the closed-switch state — and it is why a BJT switch drops about $0.2$ V while a good MOSFET switch drops millivolts.

Everything above describes the npn. The pnp is its mirror: p-type emitter and collector around an n-type base, all voltages and currents reversed, emitter arrow pointing inward on the symbol, and current flowing INTO the emitter and out of the base and collector. Every npn statement becomes a true pnp statement under sign reversal — the same duality discount ME-2's channel types offered.

WORKED EXAMPLE

α, β, and the three currents

A transistor has $\alpha = 0.99$ . Find β.
$\beta = \alpha/(1-\alpha) = 0.99/0.01 = 99$ .
If $I_B = 20$ µA, then $I_C = \beta I_B = 99 \times 20\ \mu\text{A} \approx 1.98$ mA.
And $I_E = I_B + I_C = 0.02 + 1.98 = 2.00$ mA — which checks against $I_C = \alpha I_E = 0.99 \times 2.00 = 1.98$ mA. ✓

\beta = 99

I_C = 1.98

mA,

I_E = 2.00

WORKED EXAMPLE

Which region?

A silicon npn has $V_B = 2.7$ V, $V_E = 2.0$ V, $V_C = 7.0$ V.
B-E: $V_{BE} = 0.7$ V — forward biased and conducting. ✓
B-C: $V_{BC} = 2.7 - 7.0 = -4.3$ V — reverse biased. ✓
Both conditions met: forward-active. $V_{CE} = 5.0$ V, comfortably above the $\approx 0.2$ V saturation floor.

Forward-active, with 4.8 V of margin before saturation

PRACTICE

A BJT has $I_B = 15$ µA and $\beta = 200$ . Find $I_C$ and $I_E$ .

A transistor's datasheet gives $\beta$ between 100 and 300. If a circuit's collector current is proportional to β, by what factor could the current vary between two supposedly identical parts?

An npn has $V_B = 3.0$ V, $V_E = 2.3$ V, $V_C = 2.4$ V. Which region is it in, and how do you know?

Contrast the BJT and the MOSFET on two points: what the control terminal draws in steady state, and what each drops when used as a closed switch.