ECE305#Homework#SOLUTIONS:#Week#1492# - nanoHUB

Mark Lundstrom 12/5/2014

ECE-‐305 Fall 2014 1

ECE 305 Homework SOLUTIONS: Week 14-‐2 Mark Lundstrom Purdue University

(12.13.14)

1) The sketch below shows an NPN BJT. Assume that it is in the forward active region with

IC = 1.23 mA and IE = 1.27 mA . Answer the following questions.

1a) What is the common emitter current gain, βdc ?

Solution:

βdc ≡

IC

IB

=IC

IE − IC

= 1.231.27 −1.23

= 1.230.04

= 1.230.04

= 31

βdc = 31

1b) What is the common base current gain, α dc ?

Solution:

α dc ≡

IC

IE

= 1.231.27

= 0.97 α dc = 0.97

1c) What is the base transport factor, αT ?

Solution: The base transport factor is a quantity that describes internal current components. Without additional information, it cannot be determined from the terminal currents.

1d) What is the emitter injection efficiency, γ ?

Solution: The emitter injection efficiency is a quantity that describes internal current components. Without additional information, it cannot be determined from the terminal currents.


ECE-‐305 Fall 2014 2

HW Week 14-‐2 Solutions (continued) 1e) What is the emitter injection efficiency assuming that there is no recombination in

the base?

Solution: If there is no recombination in the base, then all of the base current is due to holes that are back-‐injected into the emitter, IEp . Also, since there is no base

recombination, the electron current at the end of the base ( ICn ) is equal to the current injected into the base from the emitter, IEn . Accordingly,

IEp = IB = IE − IC = 0.04 mA

IEn = ICn = IC = 1.23 mA

γ ≡

IEn

IEn + IEp

= 1.271.27 + 0.04

= 0.97

γ = 0.97 Note that when there is no recombination in the base, the emitter injection efficiency is equal to the common base current gain, γ =α dc

2) The sketch below shows an NPN BJT biased in the forward active region with the four

current components indicated. Assume:

IEn = 1.000 mA

IEp = 0.005 mA

ICn = 0.995 mA

ICp ≈ 0

and answer the questions below.


ECE-‐305 Fall 2014 3

HW Week 14-‐2 Solutions (continued)

2a) What is the emitter injection efficiency? Solution: We want the emitter current to consist almost entirely of electrons injected into the base, but we also get hole injected from the base to the emitter. The emitter injection efficiency, γ , is the ratio of the current we want to the total current across the forward emitter-‐base junction.

γ ≡

IEn

IEn + IEp

= 1.0001.000+ 0.005

= 0.995 γ = 0.995

2b) What is the base transport factor?

Solution: The base transport factor, αT , is the ratio of the electron current that leaves at the collector-‐base junction, ICn , to the electron current that was injected at emitter-‐base junction, IEn .

αT ≡

ICn

IEn

= 0.9951.000

= 0.995 αT = 0.995

2c) What is the common emitter current gain?

Solution:

βdc ≡

IC

IB

=ICn

IEp + IEn − ICn( ) =0.995

0.005+ 0.005= 99.5

The bas current consists of two terms, holes that are injected into the emitter (

IEp ) and holes that recombine with electrons in the base, IEn − ICn( ) .

βdc = 99.5

2d) What is the common base current gain?

Solution:

α dc ≡

IC

IE

=ICn

IEn + IEp

= 0.9951.000+ 0.005

= 0.990


ECE-‐305 Fall 2014 4


It is worth noting that we can write:

α dc =

ICn

IEn + IEp

=IEn

IEn + IEp

×ICn

IEn

= γ αT = 0.995× 0.995= 0.990

α dc ≡

IC

IE

= γ αT = 0.990

3) The four current components for the two diodes in a PNP transistor are defined below.

Assume that the base width is short (i.e. WB << Ln ) and that the transistor is biased in the forward active region with VCB = 0 . The common emitter current gain is βdc = 100 . Answer the following questions assuming that

ICp = 10 mA . Note that VCB = 0 places the

transistor on the borderline of the active and saturation regions. The behavior is continuous from one region to the next, so we can call this the active or saturation region. It is better to consider it to be the active region, because it takes a moderate forward bias on the base-‐collector junction to really get into the saturation region and see the collector current drop.

3a) What is IEp ?

Solution: Since the base is short, the base transport factor, αT , is very close to 1. Since,

ICp =αT IEp , we conclude that IEp = ICp

IEp = 10 mA

3b) What is IEn ?

Solution: In the active region,

IC ≈ ICp = 10 mA

In the active region, the base current is IB = IC βdc = 0.1 mA


ECE-‐305 Fall 2014 5

HW Week 14-‐2 Solutions (continued) Since the base is short, IB = IEn , so we conclude

IEn = 0.1mA

3c) What is ICn ?

Solution: Since the base-‐collector junction is zero-‐biased,

ICn = 0 mA

3d) What is the common base current gain, α dc ?

Solution:

α dc ≡

IC

IE

=IC

IC + IB

=IC

IC + IC βdc

= 11+1 βdc

=βdc

βdc +1

α dc =

βdc

βdc +1= 100

101= 0.99

α dc = 0.99

4) Consider a bipolar junction transistor. All three regions (emitter, base and collector) are

composed of the same semiconductor material (except for doping type/density). The excess minority carrier concentrations for a specific bias point are shown in the figure below – note that the scale is linear (although the concentrations near the CB junction are exaggerated for clarity). Assume that T = 300 K and that the base is doped at NB = 1.0 ×10

17 cm−3 (NB is either NA or ND you will need to figure out which.) The diffusion coefficients for electrons and holes are Dn = 20 cm

2 /s and Dp = 10 cm2 /s

respectively. Answer the following questions.

5)


ECE-‐305 Fall 2014 6

HW Week 14-‐2 Solutions (continued) 4a) What type of transistor is this? NPN or PNP? Explain how you know.

Solution PNP. The plot shows the excess minority carrier densities in each region, so the majority carriers are Emitter: holes, Base: electrons, Collector: holes. Hence, it is a PNP transistor.

4b) What bias regime is illustrated in the figure?

Solution: The emitter base junction is forward biased (because we have a positive number of excess carriers on each side of the junction). The base-‐collector junction is reverse-‐biased because we have a negative excess carrier density on each side. FB emitter-‐base, RB base collector, means that the bias region is Forward active region

4c) What is the doping density in the emitter (recall that NB is given)?

Solution: The Law of the Junction gives the excess hole concentration injected into the base as

Δp x = 0( ) = ni

2

N DB

eqVBE kBT −1( ) and the excess electron concentration injected into the emitter as

Δn ′x = 0( ) = ni

2

N AE

eqVBE kBT −1( ) Dividing the first equation by the second:

Δp x = 0( )Δn ′x = 0( ) =

N AE

N DB From the plot Δp x = 0( ) = 1×1012 cm-3

and

Δn ′x = 0( ) = 1×1011 cm-3 , so

Δp x = 0( )Δn ′x = 0( ) = 10 =

N AE

1017

so we conclude:

N AE = 1018 cm-3


ECE-‐305 Fall 2014 7

HW Week 14-‐2 Solutions (continued) 4d) Which of the following best describes the base region? Explain your answer.

i) the base width is much larger than a minority carrier diffusion length ii) recombination is the dominant process for minority carrier flow through the base iii) punch-‐through has occurred at the bias point illustrated in the figure iv) all of the above v) none of the above

Solution:

The base width is much shorter than a minority carrier diffusion length

There is a clear, linear, minority carrier diffusion profile, which indicates that recombination is negligible, so the minority carrier diffusion length is much longer than the base width and recombination is negligible.

4e) What is the emitter injection efficiency (numerical answer)? Solution: The hole current injected from the emitter to the base is

JEp = q

Dp

WB

ni2

N DB

eqVBE /kBT −1( ) (The base is short, so we use the base width, not the diffusion length.) The electron current injected from the base to the emitter is

JEn = q

Dn

Ln

ni2

N AE

eqVBE /kBT −1( ) (The emitter is long, so we use the electron diffusion length, not the emitter width.) For later use, we note:

JEn

JE p

=Dn

Dp

Ln

WB

N DB

N AE

The emitter injection efficiency is

γ =

JE p

JE p + JEn

= 11+ JEn JE p

= 1+Dn

Dp

Ln

WB

N DB

N AE

⎛

⎝⎜

⎞

⎠⎟

−1

Putting in numbers:

γ = 1+

Dn

Dp

Ln

WB

N DB

N AE

⎛

⎝⎜

⎞

⎠⎟

−1

= 1+ 20 cm2 s10 cm2 s

1µm0.2 µm

1017 cm-3

1018 cm-3

⎛

⎝⎜⎞

⎠⎟

−1

= 1+1( )−1= 0.5

γ = 0.5


ECE-‐305 Fall 2014 8


4f) What is βdc for this transistor?

Solution: Because there is no recombination in the base,

JEp = JCp = JC = q

Dp

WB

ni2

N DB

eqVBE /kBT −1( ) Because there is no recombination in the base:

JEn = JB = q

Dn

Ln

ni2

N AE

eqVBE /kBT −1( )

βdc ≡

JC

JB

=Dn

Dp

Ln

WB

N DB

N AE

= 1 βdc = 1

Note that beta is very low for this transistor because the emitter injection efficiency is low. We should increase the emitter doping and/or decrease the base width to increase the emitter injection efficiency.

5) Consider the impact of changing some important BJT device parameters on some key figures of merit. For each of the four changes below, explain whether the change increases, decreases, or has no effect on: i) the base transport factor, αT , ii) the emitter injection efficiency,γ , and iii) the current gain, βdc .

5a) Increase the base width. Solution: i) the base transport factor, αT DECREASES ii) the emitter injection efficiency,γ DECREASES iii) the current gain, βdc . DECREASES

5b) Increase the base lifetime. Solution: i) the base transport factor, αT INCREASES ii) the emitter injection efficiency,γ NO EFFECT (Assuming that we are talking

about a good transistor, there is very little base recombination. Increasing the lifetime and making base recombination even less, will not have much effect.)

iii) the current gain, βDC . INCREASES (Whether it increases a little or a lot depends

on the emitter injection efficiency.)


ECE-‐305 Fall 2014 9


5c) Increase the base doping (assume that mobility, lifetime, and base width do not change).

Solution: i) the base transport factor, αT NO EFFECT ii) the emitter injection efficiency,γ DECREASES iii) the current gain, βdc . DECREASES

5d) Increase the emitter doping. (Assume that mobility and lifetime do not change.)

Solution: i) the base transport factor, αT NO EFFECT ii) the emitter injection efficiency,γ INCREASES iii) the current gain, βDC . INCREASES

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ECE305#Homework#SOLUTIONS:#Week#1492# - nanoHUB