ECE 305 Spring 2015 ECE 305 Homework SOLUTIONS: Week 14 Mark Lundstrom Purdue University 1)
(5.6.15) For MOSFETs, we focus on understanding drain current. The gate current is taken to be negligibly small. For BJTs, we focus first on the collector current. For good BJTs, the base current is small, but not negligible. Before we discuss the base current, we focus on the collector current. We can understand the most important features of the common emitter collector current characteristic, I C (VBE ,VCE ) or the common base collector current characteristic, I C (VBE ,VCB ) , by examining the minority carrier concentration in the base. Answer the following questions for an NPN BJT by providing a sketch of !n ( x ) in the base. 1a) Sketch !n ( x ) vs. x for the forward active region of operation. 1b) Sketch !n ( x ) vs. x for the saturation region of operation. 1c) Sketch !n ( x ) vs. x for the reverse active region of operation. 1d) Sketch !n ( x ) vs. x for the cut-­‐off region of operation. HINT: Begin with the Law of the Junction and assume a good transistor, which means the base is short compared to a minority carrier diffusion length. ECE-­‐305 1 Spring 2015 ECE 305 Spring 2015 HW Week 14 Solutions (continued) 1a) Sketch !n ( x ) vs. x for the forward active region of operation. Solution: The Law of the Junction for an NP junction says that the excess minority electron concentration injected in the neutral P-­‐region (which begins at the end of the depletion region, is ni2 qVA kBT
!n ( x = 0 ) =
e
" 1 NA
(
)
where x = 0 is the beginning of the neutral base and V A is the forward bias across the junction. In a BJT, we have two junctions. At x = 0 , the emitter-­‐base junction controls !n ( x = 0 ) , so (
ni2 qVBE
!n ( x = 0 ) =
e
N AB
)
k BT
" 1 At the end of the neutral base, x = WB , the base-­‐collector junction controls !n ( x = WB ) , so (
)
ni2 qVBC kBT
!n ( x = WB ) =
e
" 1 N AB
So if we know the bias across each junction, then we know !n at each end of the base. Since the base is assumed to be short, we simply connect the two end points with a straight line. Forward active region means: The emitter base junction is forward biased ( VBE > 0 ) and the base collector junction is reverse-­‐biased VBC < 0 . (
ni2 qVBE
!n x = 0 =
e
N AB
(
)
(
)
!n x = WB =
2
i
(
k BT
n
qV
e BC
N AB
)
"1
k BT
) "1
For a moderate forward bias on the emitter-­‐base junction, the exponential in the first equation dominates, so to a very good approximation: n2 qV k T
n2
!n x = 0 " i e BE B >> i
N AB
N AB (
)
For a moderate reverse bias on the base-­‐collector junction, the exponential in the second equation above approaches zero, so to a very good approximation: ECE-­‐305 2 Spring 2015 ECE 305 Spring 2015 HW Week 14 Solutions (continued) ni2
!n x = WB " #
<< !n x = 0
N AB
Because !n x = WB is so small, we can usually assume that !n ( x = WB ) " 0 . In the (
)
(
(
)
)
(
)
figure below, we have exaggerated !n x = WB so that it can be seen on the plot. Additional note: We could also ask about the total electron density, not just the excess electron density. In that case, we would have: n2
n2 qV k T
n2
n x = 0 = !n x = 0 + i = i e BE B >> i N AB N AB
N AB
(
(
)
(
)
)
(
)
n x = WB = !n x = WB +
ni2
n2 qV
= i e BC
N AB N AB
k BT
" 0 1b) Sketch !n ( x ) vs. x for the saturation region of operation. Solution: Saturation region means: The emitter base junction is forward biased ( VBE > 0 ) and the base collector junction is forward-­‐biased VBC > 0 . n2
n2
n2
!n ( x = 0 ) = i eqVBE kBT " 1 = i eqVBE kBT >> i
!n ( x = 0 ) >> 0
N AB
N AB
N AB (
)
(
ni2 qVBC
!n ( x = WB ) =
e
N AB
ECE-­‐305 k BT
)
ni2 qVBC
"1 =
e
N AB
3 k BT
ni2
>>
N AB 0 << !n ( x = WB ) < !n ( x = 0 )
Spring 2015 ECE 305 Spring 2015 HW Week 14 Solutions (continued) 1c) Sketch !n ( x ) vs. x for the reverse active region of operation. Reverse active region means: The emitter base junction is reverse biased ( VBE << 0 ) and the base collector junction is forward-­‐biased VBC > 0 . n2
n2
!n ( x = 0 ) = i eqVBE kBT " 1 = " i
!n ( x = 0 ) " 0
N AB
N AB (
!n ( x = WB ) =
)
(
ni2 qVBC
e
N AB
k BT
)
"1 =
ni2 qVBC
e
N AB
k BT
>>
ni2
N AB !n ( x = WB ) >> 0
Again, the small negative excess carrier concentration is exaggerated for clarity. ECE-­‐305 4 Spring 2015 ECE 305 Spring 2015 HW Week 14 Solutions (continued) 1d) Sketch !n ( x ) vs. x for the cut-­‐off region of operation. Reverse active region means: The emitter base junction is reverse biased ( VBE << 0 ) and the base collector junction is reverse-­‐biased VBC << 0 . n2 qV k T
n2
!n ( x = 0 ) = i e BE B " 1 = " i
!n ( x = 0 ) " 0
N AB
N AB (
)
(
ni2 qVBC
!n ( x = WB ) =
e
N AB
k BT
)
ni2
"1 = "
N AB !n ( x = WB ) " 0
2)
Again, the small negative excess carrier concentrations are exaggerated for clarity. The sketch below shows an NPN BJT. Assume that it is in the forward active region with I C = 1.23 mA and I E = 1.27 mA . Answer the following questions. ECE-­‐305 5 Spring 2015 ECE 305 Spring 2015 HW Week 14 Solutions (continued) 2a) What is the common emitter current gain, ! dc ? Solution: I
IC
1.23
1.23 1.23
! dc " C =
=
=
=
= 31 I B I E # I C 1.27 # 1.23 0.04 0.04
! dc = 31 2b) What is the common base current gain, ! dc ? Solution: I
1.23
! dc = 0.97 ! dc " C =
= 0.97 I E 1.27
2c) What is the base transport factor, ! T ? Solution: The base transport factor is a quantity that describes internal current components. Without additional information, it cannot be determined from the terminal currents. 2d) What is the emitter injection efficiency, ! ? Solution: The emitter injection efficiency is a quantity that describes internal current components. Without additional information, it cannot be determined from the terminal currents. 2e) 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, I Ep . Also, since there is no base recombination, the electron current at the end of the base ( I Cn ) is equal to the current injected into the base from the emitter, I En . Accordingly, ECE-­‐305 6 Spring 2015 ECE 305 Spring 2015 HW Week 14 Solutions (continued) I Ep = I B = I E ! I C = 0.04 mA I En = I Cn = I C = 1.23 mA γ ≡
I En
1.27
=
= 0.97 I En + I Ep 1.23 + 0.04
! = 0.97
Note that when there is no recombination in the base, the emitter injection efficiency is equal to the common base current gain, ! = " dc 3)
The sketch below shows an NPN BJT biased in the forward active region with the four current components indicated. Assume: I En = 1.000 mA I Ep = 0.005 mA I Cn = 0.995 mA I Cp ! 0 and answer the questions below. 3a) 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, ! , is the ratio of the current we want to the total current across the forward emitter-­‐base junction. I En
1.000
! "
=
= 0.995 ! = 0.995 I En + I Ep 1.000 + 0.005
ECE-­‐305 7 Spring 2015 ECE 305 Spring 2015 HW Week 14 Solutions (continued) 3b) What is the base transport factor? Solution: The base transport factor, ! T , is the ratio of the electron current that leaves at the collector-­‐base junction, I Cn , to the electron current that was injected at emitter-­‐base junction, I En . !T "
I Cn 0.995
=
= 0.995 I En 1.000
! T = 0.995 3c) What is the common emitter current gain? Solution: I
I Cn
0.995
! dc " C =
=
= 99.5 I B I Ep + ( I En # I Cn ) 0.005 + 0.005
The bas current consists of two terms, holes that are injected into the emitter ( I Ep ) and holes that recombine with electrons in the base, ( I En ! I Cn ) . ! dc = 99.5 3d) What is the common base current gain? Solution: I
I Cn
0.995
! dc " C =
=
= 0.990 I E I En + I Ep 1.000 + 0.005
It is worth noting that we can write: I Cn
I En
I
! dc =
=
" Cn = # ! T = 0.995 " 0.995 = 0.990 I En + I Ep I En + I Ep I En
! dc "
IC
= # ! T = 0.990 IE
ECE-­‐305 8 Spring 2015 ECE 305 Spring 2015 HW Week 14 Solutions (continued) 4) 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 ! dc = 100 . Answer the following questions assuming that I Cp = 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. 4a) What is I Ep ? Solution: Since the base is short, the base transport factor, ! T , is very close to 1. Since, I Cp = ! T I Ep , we conclude that I Ep = I Cp I Ep = 10 mA 4b) What is I En ? Solution: In the active region, I C ! I Cp = 10 mA In the active region, the base current is I B = I C ! dc = 0.1 mA Since the base is short, I B = I En , so we conclude I En = 0.1 mA 4c) What is I Cn ? Solution: Since the base-­‐collector junction is zero-­‐biased, I Cn = 0 mA ECE-­‐305 9 Spring 2015 ECE 305 Spring 2015 HW Week 14 Solutions (continued) 4d) What is the common base current gain, ! dc ? Solution: I
IC
IC
# dc
1
! dc " C =
=
=
=
I E I C + I B I C + I C # dc 1+ 1 # dc # dc + 1
! dc =
" dc
100
=
= 0.99 " dc + 1 101
! dc = 0.99 5)
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 N B = 1.0 ! 1017 cm "3 ( N B is either N A or N D you will need to figure out which.) The diffusion coefficients for electrons and holes are Dn = 20 cm 2 /s and D p = 10 cm 2 /s respectively. Answer the following questions. 5a) 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. ECE-­‐305 10 Spring 2015 ECE 305 Spring 2015 HW Week 14 Solutions (continued) 5b) 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 5c) What is the doping density in the emitter (recall that N B is given)? Solution: The Law of the Junction gives the excess hole concentration injected into the base as ni2 qVBE kBT
!p ( x = 0 ) =
e
" 1 N DB
and the excess electron concentration injected into the emitter as ni2 qVBE kBT
!n ( x " = 0 ) =
e
# 1 N AE
Dividing the first equation by the second: !p ( x = 0 ) N AE
=
!n ( x " = 0 ) N DB (
)
(
)
From the plot !p ( x = 0 ) = 1" 1012 cm -3 and !n ( x " = 0 ) = 1# 1011 cm -3 , so !p ( x = 0 )
N AE
= 10 = 17
10 !n ( x " = 0 )
so we conclude: N AE = 1018 cm -3 5d) 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 ECE-­‐305 11 Spring 2015 ECE 305 Spring 2015 HW Week 14 Solutions (continued) 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. 5e) What is the emitter injection efficiency (numerical answer)? Solution: The hole current injected from the emitter to the base is D n2
J Ep = q p i eqVEB /kBT ! 1 WB N DB
(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 D n2
J En = q n i eqVEB /kBT ! 1 Ln N AE
(The emitter is long, so we use the electron diffusion length, not the emitter width.) For later use, we note: J En Dn WB N DB
=
J E p D p Ln N AE
(
)
(
)
The emitter injection efficiency is ! =
JE p
J E p + J En
=
(1
1
1+ J En J E p
"
D W N %
= $ 1+ n B DB ' D p Ln N AE &
#
Putting in numbers: "
D W N %
! = $ 1+ n B DB '
D p Ln N AE &
#
! = 0.96
ECE-­‐305 (1
"
20 cm 2 s 0.2 µm 1017 cm -3 %
= $ 1+
)
)
'
# 10 cm 2 s 1.0 µm 1018 cm -3 &
(1
(
= 1+ 0.04
)
(1
= 1.04 12 Spring 2015 ECE 305 Spring 2015 HW Week 14 Solutions (continued) 5f) What is ! dc for this transistor? Solution: Because there is no recombination in the base, D p ni2 qV /k T
J Ep = J Cp = J C = q
e EB B ! 1 WB N DB
Because there is no recombination in the base: D n2 qV /k T
J En = J B = q n i e EB B ! 1 Ln N AE
(
(
! dc "
)
)
J C D p Ln N DB
=
= 25 J B Dn WB N AE
! dc = 25
A quicker way to get this answer is to use "
0.96
! dc =
=
= 24 1# " 1# 0.96
(Difference is due to round-­‐off error.) ECE-­‐305 13 Spring 2015