GATE EC

The bipolar junction transistor as a device

A bipolar junction transistor (BJT) has three terminals: emitter, base and collector. It amplifies because a heavily doped emitter injects carriers into a thin base, and most of them survive to reach the collector. This topic covers the quantities that describe how efficiently that happens: emitter injection efficiency, base transport factor, α and β, and the leakage currents $I_{C B O}$ and $I_{C E O}$ . It also covers the Ebers–Moll model, with its sign conventions, which describes the device in every mode of operation. Finally, it covers the non-idealities that make a real device depart from an ideal current source: base-width modulation, the Early effect, punch-through and collector breakdown. The same quantities underlie biasing, small-signal models and amplifier stages, so getting the conventions and the operating-region conditions right here carries through to every later BJT topic.

3 previous-year questions · 2 solved · 6 practice questions

BJT Structure, Doping and Modes of Operation

A bipolar junction transistor (BJT) has three regions and two pn junctions. The middle region, the base, is shared by both junctions and is thin. For an npn device the layers are n (emitter), p (base), n (collector); for a pnp device they are p, n, p. The terminals are the emitter (E), base (B) and collector (C). The two junctions are the emitter-base junction (EBJ) and the collector-base junction (CBJ). Two separate diodes wired back to back do not behave as a transistor. Transistor action needs the base to be much thinner than the minority-carrier diffusion length, so that carriers injected at the EBJ can diffuse across it to the CBJ before they recombine.

Doping pattern of a typical BJT (npn shown)

Emitter: heavily doped (written $n^{+}$ for an npn). Making $N_{E} ≫ N_{B}$ is what brings the emitter injection efficiency close to 1.
Base: doped more lightly than the emitter, and physically thin ( $W_{B} ≪ L_{B}$ , the minority-carrier diffusion length in the base). The thin base is what brings the base transport factor close to 1.
Collector: in a device designed for a high collector-base breakdown voltage, the collector next to the CBJ is the most lightly doped region ( $N_{E} ≫ N_{B} > N_{C}$ ). The CBJ depletion region then lies mainly in the collector. That raises the breakdown voltage, reduces base-width modulation (the Early effect) and delays punch-through. A heavily doped $n^{+}$ region further from the junction (the sub-collector) keeps the collector's series resistance low.
Some introductory treatments rank the collector doping between the emitter and the base. The physical requirement that the CBJ depletion region extend mainly into the collector is what the lighter collector doping achieves.

Modes of operation of an npn BJT (forward bias positive; for a pnp every voltage and current reverses)

Mode	EBJ bias	CBJ bias	Behaviour
Active (forward-active)	Forward	Reverse	$I_{C} \approx β_{F} I_{B}$ ; the region used for amplification
Saturation	Forward	Forward	Switch 'on' state; $V_{C E}$ small; $I_{C} < β_{F} I_{B}$ (the forced $β = I_{C} / I_{B}$ is below $β_{F}$ )
Cut-off	Reverse	Reverse	Switch 'off' state; only leakage currents flow
Reverse-active	Reverse	Forward	Roles of E and C interchanged; $β_{R} ≪ β_{F}$ because the device is not symmetric

Current Components in Active Mode

Take an npn device in active mode. The forward-biased EBJ injects electrons from the emitter, where they are majority carriers, into the base, where they are minority carriers. Call this electron current $I_{E n}$ . At the same time the base injects holes back into the emitter, giving a current $I_{E p}$ . This current does not reach the collector. Of the electrons injected into the base, a fraction recombines in the base, and the remainder, $I_{C n}$ , reaches the CBJ and is swept into the collector. The base current supplies both losses: the back-injected holes and the holes lost to recombination. With recombination in the EBJ depletion region neglected, $I_{B} = I_{E p} + (I_{E n} - I_{C n}) - I_{C B O}$ , where $I_{C B O}$ is the reverse leakage of the CBJ. Two figures of merit describe these losses.

EMITTER INJECTION EFFICIENCY

γ = \frac{I _{E n}}{I _{E n} + I _{E p}} = \frac{I _{E n}}{I _{E}}

This is the fraction of the emitter current made up of carriers injected into the base. It assumes recombination in the EBJ depletion region is negligible. It is made close to 1 by doping the emitter much more heavily than the base.

INJECTION EFFICIENCY IN TERMS OF DESIGN PARAMETERS (NPN)

γ \approx \frac{1}{1 + \frac{D _{E} N _{B} W _{B}}{D _{B} N _{E} W _{E}}}

Here $D_{E}$ is the hole diffusion coefficient in the emitter, $D_{B}$ the electron diffusion coefficient in the base, and $W_{E}$ , $W_{B}$ the neutral emitter and base widths. The expression assumes uniform doping, low-level injection and short regions ( $W_{B} ≪ L_{B}$ , $W_{E} ≪ L_{E}$ ). For a long emitter, replace $W_{E}$ by $L_{E}$ .

BASE TRANSPORT FACTOR

α_{T} = \frac{I _{C n}}{I _{E n}} = sech (\frac{W _{B}}{L _{B}}) \approx 1 - \frac{W _{B}^{2}}{2 L _{B}^{2}}

This is the fraction of carriers injected into the base that survive recombination and reach the CBJ. The sech form assumes uniform base doping and low-level injection. The approximation holds for $W_{B} ≪ L_{B}$ , where $L_{B} = D_{B} τ_{B}$ and $τ_{B}$ is the minority-carrier lifetime in the base.

COMMON-BASE CURRENT GAIN

α = γ α_{T} = \frac{I _{C n}}{I _{E}}, I_{C} = α I_{E} + I_{C B O}

This applies in active mode with no avalanche multiplication at the CBJ (multiplication factor $M = 1$ ). When $I_{C B O}$ is negligible, $α = I_{C} / I_{E}$ . For a real device $α < 1$ , and it is close to 1.

COMMON-EMITTER CURRENT GAIN

β = \frac{α}{1 - α}, \frac{1}{β} \approx (1 - γ) + (1 - α_{T})

When $I_{C B O}$ is negligible, $β = I_{C} / I_{B}$ . The second form holds when $γ$ and $α_{T}$ are both close to 1, and it shows how the two losses add. Because $α$ is close to 1, $β$ is very sensitive to small changes in $α$ . Carry enough significant figures in $1 - α$ before dividing.

TERMINAL CURRENT RELATIONS (ACTIVE MODE, LEAKAGE NEGLECTED)

I_{E} = I_{C} + I_{B}, I_{C} = α I_{E} = β I_{B}, I_{E} = (β + 1) I_{B}, α = \frac{β}{β + 1}

The first relation is KCL for the whole device (currents taken into C and B and out of E for an npn). The others convert between $α$ and $β$ and give any one of $I_{E}$ , $I_{B}$ , $I_{C}$ from another. A useful identity is $β + 1 = \frac{1}{1 - α}$ .

LEAKAGE CURRENTS

I_{C} = β I_{B} + (β + 1) I_{C B O}, I_{C E O} = (β + 1) I_{C B O}

$I_{C B O}$ is the collector current with the emitter open and the CBJ reverse-biased. $I_{C E O}$ is the collector current with the base open. $I_{C E O}$ is larger by a factor of $(β + 1)$ because the leakage current, flowing into the base, is itself amplified. Both relations hold in active mode.

Ebers–Moll Model

The Ebers–Moll model treats a BJT as two coupled diodes, the EBJ and the CBJ. It applies in all modes: active, saturation, cut-off and reverse-active. The active-mode formulas of the previous section are a special case of it. Conventions used here (npn): $I_{C}$ flows into the collector terminal, $I_{B}$ flows into the base, and $I_{E}$ flows out of the emitter, so that $I_{E} = I_{C} + I_{B}$ . Junction voltages $V_{B E}$ and $V_{B C}$ are measured p-side minus n-side, so forward bias is positive. $V_{T} = k T / q \approx 25.9 mV$ at 300 K.

EBERS–MOLL EMITTER CURRENT (NPN)

I_{E} = I_{E S} (e^{V_{B E} / V_{T}} - 1) - α_{R} I_{C S} (e^{V_{B C} / V_{T}} - 1)

$I_{E S}$ is the EBJ saturation current measured with the collector shorted to the base. $α_{R}$ is the reverse (inverse) common-base current gain.

EBERS–MOLL COLLECTOR CURRENT (NPN)

I_{C} = α_{F} I_{E S} (e^{V_{B E} / V_{T}} - 1) - I_{C S} (e^{V_{B C} / V_{T}} - 1)

$α_{F}$ is the forward common-base gain, the $α$ used elsewhere in this topic. $I_{C S}$ is the CBJ saturation current measured with the emitter shorted to the base.

RECIPROCITY CONDITION

α_{F} I_{E S} = α_{R} I_{C S} \equiv I_{S}

This follows from the reciprocity of the device. It reduces the four Ebers–Moll parameters to three independent ones, and it defines the transport saturation current $I_{S}$ .

ACTIVE-MODE REDUCTION

I_{C} \approx I_{S} e^{V_{B E} / V_{T}}, I_{B} \approx \frac{I _{S}}{β _{F}} e^{V_{B E} / V_{T}}, β_{F} = \frac{α _{F}}{1 - α _{F}}

These hold when the EBJ is forward-biased with $V_{B E}$ several times $V_{T}$ , so the $- 1$ is negligible, and $V_{B C} \leq 0$ , so the CBJ diode term contributes at most a small leakage. They follow from the Ebers–Moll equations using $I_{B} = I_{E} - I_{C} = (1 - α_{F}) I_{E S} e^{V_{B E} / V_{T}}$ and $I_{S} = α_{F} I_{E S}$ . This is the exponential law used for the active-mode $I_{C}$ – $V_{B E}$ characteristic.

SATURATION CURRENT IN TERMS OF DEVICE PARAMETERS (NPN)

I_{S} = \frac{q A _{E} D _{B} n _{i}^{2}}{N _{B} W _{B}}

$A_{E}$ is the emitter area, $D_{B}$ the electron diffusion coefficient in the base, $N_{B}$ the uniform base doping, $W_{B}$ the neutral base width and $n_{i}$ the intrinsic concentration (about $1 0^{10} cm^{- 3}$ for Si at 300 K). It assumes a short base ( $W_{B} ≪ L_{B}$ ), so the minority-carrier profile in the base is linear, and low-level injection. Because $I_{S} \propto 1/ W_{B}$ , narrowing the neutral base raises $I_{C}$ . That is the Early effect.

BASE TRANSIT TIME AND ITS LINK TO Β

τ_{t} = \frac{W _{B}^{2}}{2 D _{B}}, β \approx \frac{τ _{B}}{τ _{t}}

$τ_{t}$ is the average time a minority carrier takes to diffuse across a short base. $τ_{B}$ is the minority-carrier lifetime in the base. The expression for $β$ holds when base recombination is the only significant loss ( $γ \approx 1$ ): then $1 - α_{T} \approx W_{B}^{2} / (2 L_{B}^{2}) = τ_{t} / τ_{B}$ .

EXAM TIP

Two rules of thumb for silicon at room temperature. At constant $I_{C}$ , $V_{B E}$ falls by roughly 2 mV for each 1 °C rise in temperature. $I_{C B O}$ roughly doubles for every 10 °C rise. Use them only for estimates.

Early Effect and Output Resistance

The active-mode formulas above make $I_{C}$ independent of $V_{C E}$ , which treats the BJT as an ideal current source. In a real device, $I_{C}$ rises slightly with $V_{C E}$ at fixed $V_{B E}$ . This is the Early effect. It is caused by base-width modulation (next section), and it gives a BJT a finite output resistance.

EARLY EFFECT: COLLECTOR CURRENT

I_{C} (V_{C E}) = I_{C 0} (1 + \frac{V _{C E}}{V _{A}})

This first-order model holds in active mode at constant $V_{B E}$ . $I_{C 0} = I_{S} e^{V_{B E} / V_{T}}$ is the current extrapolated to $V_{C E} = 0$ . $V_{A}$ is the Early voltage. Extending the straight-line $I_{C}$ – $V_{C E}$ characteristic back to the $V_{C E}$ axis gives a crossing at $V_{C E} = - V_{A}$ .

SMALL-SIGNAL OUTPUT RESISTANCE

r_{o} = \frac{\partial V _{C E}}{\partial I _{C}}_{V_{B E}} = \frac{V _{A} + V _{C E}}{I _{C}} \approx \frac{V _{A}}{I _{C}}

It is evaluated at the DC operating point, in active mode. The middle expression follows exactly from the model above. The approximation $V_{A} / I_{C}$ is the one normally used, and it holds when $V_{A} ≫ V_{C E}$ .

EXAM TIP

In this first-order model $V_{A}$ is a property of the transistor and does not depend on the bias point. At different bias points of the same transistor, $r_{o} \approx V_{A} / I_{C}$ therefore scales inversely with $I_{C}$ .

Base-Width Modulation, Punch-Through and Breakdown

The physical cause of the Early effect is base-width modulation. In active mode the CBJ is reverse-biased, with reverse bias $V_{C B} = V_{C E} - V_{B E}$ . Its depletion region extends into each side in inverse proportion to that side's doping: mostly into the lightly doped collector, but partly into the base. When $V_{C E}$ is raised at fixed $V_{B E}$ , $V_{C B}$ rises, the depletion region extends further into the base, and the neutral (quasi-neutral) base width $W_{B}$ shrinks.

NEUTRAL BASE WIDTH VS BIAS

W_{B} = W_{met} - x_{pE} - x_{pC} (V_{C B})

$W_{met}$ is the metallurgical base width, measured junction to junction. $x_{pE}$ and $x_{pC}$ are the parts of the EBJ and CBJ depletion widths that lie in the base. For a step junction, $x_{pC}$ grows as $V_{bi, C} + V_{C B}$ , where $V_{bi, C}$ is the CBJ built-in potential and $V_{C B} > 0$ is reverse bias. With $V_{B E}$ fixed, $x_{pE}$ does not change.

For a short base, $I_{C} \propto I_{S} \propto 1/ W_{B}$ . A smaller neutral base width therefore gives a steeper minority-carrier gradient and a larger $I_{C}$ , which is exactly the Early effect. Punch-through is the extreme limit of the same mechanism. If the reverse bias is raised far enough, the CBJ depletion region takes up the entire neutral base and merges with the EBJ depletion region. The barrier that controls injection from the emitter is then lost, and the collector current rises sharply. This is a breakdown-like effect caused by depletion reach-through, not by avalanche multiplication. A base doped more heavily than the collector keeps $x_{pC}$ small, which delays punch-through.

COLLECTOR-EMITTER BREAKDOWN WITH THE BASE OPEN

B V_{C E O} \approx \frac{B V _{C B O}}{β ^{1/ n}}

$B V_{C B O}$ is the avalanche breakdown voltage of the CBJ with the emitter open. $n$ is an empirical constant, typically between 3 and 6. With the base open, the avalanche current is amplified by the transistor's gain, so breakdown occurs at a lower voltage than for the isolated junction: $B V_{C E O} < B V_{C B O}$ .

EXAM TIP

Exam trap: punch-through and avalanche breakdown ( $B V_{C B O}$ , $B V_{C E O}$ ) both show a sharp rise in current at high collector voltage, but their causes differ. Punch-through is set by the base width and the base and collector doping: the depletion regions merge. Avalanche breakdown is carrier multiplication in the high field at the CBJ. Do not treat them as the same effect.

Worked Examples

WORKED EXAMPLE

A silicon npn BJT in active mode has $α = 0.99$ and $I_{E} = 2 mA$ . Neglecting $I_{C B O}$ , find $I_{C}$ , $I_{B}$ and $β$ .

Solution

$I_{C} = α I_{E} = 0.99 \times 2 mA = 1.98 mA$
$I_{B} = I_{E} - I_{C} = 2 - 1.98 = 0.02 mA = 20 μ A$
$β = \frac{α}{1 - α} = \frac{0.99}{0.01} = 99$ . Check: $I_{C} / I_{B} = 1.98/0.02 = 99$ , which agrees.

Answer: $I_{C} = 1.98 mA$ , $I_{B} = 20 μ A$ , $β = 99$ .

WORKED EXAMPLE

A BJT in active mode has Early voltage $V_{A} = 80 V$ . With $V_{B E}$ fixed and $V_{C E} = 1 V$ , the measured current is $I_{C} = 1 mA$ . Estimate $I_{C}$ at the same $V_{B E}$ when $V_{C E} = 6 V$ .

Solution

Use $I_{C} (V_{C E}) = I_{C 0} (1 + V_{C E} / V_{A})$ . $V_{B E}$ is the same, so $I_{C 0}$ is the same, and it cancels in the ratio: $\frac{I _{C} ( 6 )}{I _{C} ( 1 )} = \frac{1 + 6/80}{1 + 1/80} = \frac{1.075}{1.0125} \approx 1.0617$
$I_{C} (6) \approx 1 mA \times 1.0617 \approx 1.062 mA$
Cross-check with the output resistance: $r_{o} = (V_{A} + V_{C E}) / I_{C} = 81 V /1 mA = 81 k Ω$ , so $Δ I_{C} = 5 V /81 k Ω \approx 0.062 mA$ . This agrees.

Answer: $I_{C} \approx 1.06 mA$ . That is about a 6% rise in $I_{C}$ for a 5 V increase in $V_{C E}$ , which shows why $r_{o}$ is large but finite.

WORKED EXAMPLE

An npn BJT in active mode has a neutral base width $W_{B}$ equal to one-tenth of the electron diffusion length in the base ( $W_{B} / L_{B} = 0.1$ ) and an emitter injection efficiency $γ = 0.99$ . Estimate $α_{T}$ , $α$ and $β$ , neglecting leakage.

Solution

$α_{T} \approx 1 - \frac{W _{B}^{2}}{2 L _{B}^{2}} = 1 - \frac{0.01}{2} = 0.995$ . The exact value $sech (0.1) = 0.99502$ agrees.
$α = γ α_{T} = 0.99 \times 0.995 = 0.98505$
$β = \frac{0.98505}{1 - 0.98505} = \frac{0.98505}{0.01495} \approx 65.9$
Check with $1/ β \approx (1 - γ) + (1 - α_{T}) = 0.01 + 0.005 = 0.015$ , giving $β \approx 66.7$ . This is close; the small gap arises because the check is only a first-order approximation.

Answer: $α_{T} \approx 0.995$ , $α \approx 0.985$ , $β \approx 66$ . Here the injection efficiency, not base recombination, is the larger loss.

PREVIOUS-YEAR QUESTIONS

3 questions from the EC papers

Try each question first, then open the answer. One shows the official answer only. We placed each question under a topic ourselves; the exam paper does not label topics.

See these papers in full: GATE ECE 2026 · GATE ECE 2024 · GATE ECE 2020

GATE 2026 EC Q.29

GATE 2026 · ECQ.29 · 1 mark · MSQ

Figure shows the output characteristics of two different Bipolar Junction Transistors (BJT), BJT 1 with magnitude of Early voltage $∣ V_{A 1} ∣$ , and BJT 2 with magnitude of Early voltage $∣ V_{A 2} ∣$ .

Which of the following options is/are correct regarding the Early voltages?

GATE 2026 EC question 29: A plot of collector current I_C (vertical axis) versus collector-emitter voltage V_CE (horizontal axis), both axes… — Figure from the GATE 2026 EC paper. It is part of the question.

(A)
$∣ V_{A 1} ∣ > ∣ V_{A 2} ∣$
(B)
$∣ V_{A 1} ∣$ is infinitely large
(C)
$∣ V_{A 1} ∣ < ∣ V_{A 2} ∣$
(D)
$∣ V_{A 2} ∣$ is finite

Answer and solution

GATE 2026 EC Q.29: answer and solution

GATE 2024 EC Q.52

GATE 2024 · ECQ.52 · 2 marks · MSQ

Which of the following statements is/are true for a BJT with respect to its DC current gain $β$ ?

(A)
Under high-level injection condition in forward active mode, $β$ will decrease with increase in the magnitude of collector current.
(B)
Under low-level injection condition in forward active mode, where the current at the emitter-base junction is dominated by recombination-generation process, $β$ will decrease with increase in the magnitude of collector current.
(C)
$β$ will be lower when the BJT is in saturation region compared to when it is in active region.
(D)
A higher value of $β$ will lead to a lower value of the collector-to-emitter breakdown voltage.

Answer and solution

GATE 2024 EC Q.52: answer and solution

See the question in the paper

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GATE 2024 EC Q.52: answer and solution

GATE 2020 EC Q.33

GATE 2020 · ECQ.33 · 2 marks · MCQ

The base of an npn BJT T1 has a linear doping profile $N_{B} (x)$ as shown below. The base of another npn BJT T2 has a uniform doping $N_{B}$ of $1 0^{17} cm^{- 3}$ . All other parameters are identical for both the devices. Assuming that the hole density profile is the same as that of doping, the common-emitter current gain of T2 is

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Figure from the GATE 2020 EC paper. It is part of the question.

Figure description

Doping profile of the base of T1. Two vertical dashed lines mark x = 0 (the Emitter/Base boundary) and x = W (the Base/Collector boundary) on a horizontal axis marked 0 and W, with the regions labelled Emitter, Base and Collector. A straight line labelled N_B(x) falls linearly from 10^17 (cm^-3) at x = 0 to 10^14 (cm^-3) at x = W. There is no labelled vertical axis; only these two values are marked.

(A)
approximately 2.0 times that of T1.
(B)
approximately 0.3 times that of T1.
(C)
approximately 2.5 times that of T1.
(D)
approximately 0.7 times that of T1.

Official answer

Official answer: marks to all candidatesSource: Official answer key, GATE 2020 EC (IIT Delhi)

GATE 2020 EC Q.33: official answer

PRACTICE

6 questions to test yourself

PRACTICE 11 mark · MCQ

A silicon BJT operating in active mode has emitter injection efficiency $γ = 0.995$ and base transport factor $α_{T} = 0.998$ . Neglecting $I_{C B O}$ and avalanche multiplication, the common-emitter current gain $β$ is closest to:

(A)
71
(B)
100
(C)
142
(D)
200

Answer and solution

Answer: (C)

$α = γ α_{T} = 0.995 \times 0.998 = 0.99301$
$1 - α = 0.00699$
$β = \frac{α}{1 - α} = \frac{0.99301}{0.00699} \approx 142.1$ , so the answer is (C).
This shows why a small change in $α$ near 1 causes a large change in $β$ .

PRACTICE 21 mark · NAT

A BJT operating in active mode has emitter current $I_{E} = 5 mA$ and common-base current gain $α = 0.98$ . Neglecting $I_{C B O}$ , find the base current $I_{B}$ in mA.

Answer and solution

Answer: $0.1$ (accepted range $\pm 0.005$ )

$I_{C} = α I_{E} = 0.98 \times 5 = 4.9 mA$
KCL for the device: $I_{B} = I_{E} - I_{C} = 5 - 4.9 = 0.1 mA$

PRACTICE 32 marks · MSQ

For an npn BJT biased in active mode, which of the following statements about the Early effect are correct?

(A)
Increasing $V_{C E}$ decreases the effective neutral base width
(B)
Increasing $V_{C E}$ increases $I_{C}$ for a fixed $V_{B E}$
(C)
The Early effect is modelled by a finite output resistance, approximately $r_{o} = V_{A} / I_{C}$
(D)
The Early effect arises from modulation of the emitter-base depletion width by $V_{C E}$

Answer and solution

Answer: (A), (B), (C)

Increasing $V_{C E}$ with $V_{B E}$ fixed increases the CBJ reverse bias $V_{C B} = V_{C E} - V_{B E}$ . The CBJ depletion region widens into the base, and the neutral base width shrinks. (A is correct.)
A narrower neutral base steepens the minority-carrier gradient in the base, so $I_{C}$ increases for the same $V_{B E}$ . (B is correct.)
The finite slope of $I_{C}$ against $V_{C E}$ is modelled by an output resistance $r_{o} \approx V_{A} / I_{C}$ in the small-signal model. (C is correct.)
The Early effect comes from modulation of the collector-base depletion width. The emitter-base depletion width is set by $V_{B E}$ , which is held fixed, so D is false.

PRACTICE 41 mark · NAT

A BJT in active mode has Early voltage $V_{A} = 100 V$ . At an operating point with $V_{C E} = 2 V$ , the collector current is $I_{C} = 2 mA$ . Using the standard approximation $r_{o} \approx V_{A} / I_{C}$ , find the small-signal output resistance in $k Ω$ .

Answer and solution

Answer: $50$ (accepted range $\pm 1$ )

$r_{o} \approx \frac{V _{A}}{I _{C}} = \frac{100 V}{2 mA} = 50 k Ω$
The model gives exactly $r_{o} = (V_{A} + V_{C E}) / I_{C} = 102/2 = 51 k Ω$ . The approximation drops $V_{C E}$ because $V_{A} ≫ V_{C E}$ .

PRACTICE 51 mark · MCQ

Punch-through in an npn BJT occurs when:

(A)
The emitter-base depletion region reaches the collector
(B)
The collector-base depletion region extends completely across the neutral base and reaches the emitter-base depletion region
(C)
Avalanche multiplication occurs at the collector-base junction
(D)
The base transport factor becomes negative

Answer and solution

Answer: (B)

Increasing the collector-base reverse bias widens the CBJ depletion region into the thin base.
If the reverse bias is large enough, this depletion region extends across the whole neutral base and meets the EBJ depletion region. That is punch-through, so the answer is (B).
It is a depletion reach-through effect, distinct from avalanche multiplication, which is carrier multiplication in the high field at the junction. $α_{T}$ is a fraction between 0 and 1 and cannot become negative.

PRACTICE 61 mark · NAT

An npn BJT has $β = 99$ and collector-base leakage current $I_{C B O} = 10 nA$ . With the base terminal open and the transistor in active mode, find the collector-emitter leakage current $I_{C E O}$ in nA.

Answer and solution

Answer: $1000$ (accepted range $\pm 1$ )

With the base open, $I_{B} = 0$ , so $I_{C} = β I_{B} + (β + 1) I_{C B O} = (β + 1) I_{C B O}$ .
$I_{C E O} = (99 + 1) \times 10 nA = 1000 nA$

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