Helwan University - Faculty of Engineering
Electronics and Communications Engineering Department
Electronics
Bipolar Junction Transistors
(BJTs)
Presented By:
Assistant Professor: Azza M. Anis
Bipolar Junction Transistor (BJT)
The term bipolar refers to the use of both holes and free elections as
current carriers.
BJT consists of three doped semiconductor regions:
[1] NPN BJT: has two n regions [2] PNP BJT: has two p regions
separated by p region. separated by n region.
E B C E B C
The three regions are called emitter (E), base (B), and collector (C).
The emitter region is heavily doped.
Collector Collector
N The collector region is moderately doped.
P
P Base Base N
N
P
The base region is lightly doped and very thin.
Emitter Emitter
NPN BJT PNP BJT
Each region is connected to a conductive terminal ( metallic lead).
These metallic leads are labeled E, B, and C for emitter, base, and
collector.
NPN BJT PNP BJT
Junctions of NPN BJT
NPN BJT has two pn junctions.
Base
Emitter Collector
BEJ BCJ
The pn junction joining the base region and the emitter region is called
the base-emitter junction (BEJ).
The pn junction joining the base region and the collector region is called
the base-collector junction (BCJ).
Modes (Operation Regions) of NPN BJT
Base (B)
Emitter Collector
(E) (C)
BEJ BCJ
Modes BEJ BCJ Applications
Active Forward Reverse Amplifier
Saturation Forward Forward Closed Switch Digital
Cutoff Reverse Reverse Open Switch Circuits
Symbol of NPN BJT
The arrow on the emitter terminal refers
to the flow direction of emitter current
when the emitter-base junction is
NPN N forward-biased.
Symbol
P
− 𝒗𝒆 Emitter
injects free
Emitter electrons
Current
Active Mode of NPN BJT (Amplifier) Collector
In active mode, external dc voltages are applied to set
BEJ is forward-biased and BCJ is reverse-biased.
Base
Forward-Biased Reverse-Biased
BEJ BCJ
Emitter
The forward biasing between The reverse biasing between
base and emitter narrows the base and collector widens the
base-emitter depletion region base-collector depletion region
Operation of Active Mode of NPN BJT
In npn BJT, the heavily doped n-type emitter region has a very high
density of free electrons.
− +
electrons
These free electrons easily diffuse (move) into the p-type base region.
Base
Most of the emitter electrons
Emitter diffuse into the collector region by
the attraction of the positive
collector voltage source.
+ 𝒗𝒆
Small percentage of the
emitter electrons recombine
with holes in the base region.
Collector + 𝒗𝒆
Just as in a forward-biased diode:
The flow of electrons from emitter into base results in electron current 𝑰𝒏𝑬
The flow of holes from base into emitter results in hole current 𝑰𝒑𝑬
The total emitter current is the sum of these two currents: 𝑰𝑬 = 𝑰𝒏𝑬 + 𝑰𝒑𝑬
The flow of the emitted electrons from emitter into collector results in electron
current 𝑰𝒏𝑪
The flow of minority holes from collector into base results in hole current 𝑰𝒑𝑪
𝑰𝑪𝑩
The other part of the collector current (𝑰𝑪𝑩 ) is due to the flow of the minority
electrons from base into collector.
The total collector current is the sum of these currents: 𝑰𝑪 = 𝑰𝒏𝑪 + 𝑰𝒑𝑪 + 𝑰𝑪𝑩
In base region, a small percentage of
emitter electrons recombine with holes.
These recombined electrons
move from one hole to the
next and finally flow out of
+ 𝒗𝒆 the base lead toward the
positive side of external
Base source forming the base
Current electron current (𝑰𝑩 ).
(𝑰𝑩 )
This recombination are less in number, because
of smaller area and light-doping of base.
Collector
Current (𝑰𝑪 )
𝑰𝑪
𝑰𝑩
Base 𝑰𝑬
Current
(𝑰𝑩 )
Emitter
Current (𝑰𝑬 )
The emitter current (𝑰𝑬 ) is the sum of the collector current (𝑰𝑪 ) and the
base current (𝑰𝑩 ).
𝑰𝑬 = 𝑰𝑪 + 𝑰𝑩
𝑰𝑪
The ratio of the collector current (𝑰𝑪 ) to the
base current (𝑰𝑩 ) is beta (𝜷).
𝑰𝑪
𝜷= 𝑰𝑩
𝑰𝑩 𝑰𝑬
The ratio of the collector current (𝑰𝑪 ) to the emitter current (𝑰𝑬 ) is
alpha (𝜶). 𝑰𝑪
𝜶=
𝑰𝑬
Related Expressions
∵ 𝑰𝑬 = 𝑰𝑪 + 𝑰 𝑩
𝑰𝑪
∵𝜷= 𝑰𝑪 = 𝜷 𝑰𝑩
𝑰𝑩
∴ 𝑰𝑬 = 𝜷 𝑰𝑩 + 𝑰𝑩
𝑰𝑬 = 𝜷 + 𝟏 𝑰𝑩
𝑰𝑪
∵𝜶=
𝑰𝑬
∴ 𝑰𝑪 = 𝜶 𝑰 𝑬 𝒂𝒏𝒅 ∵ 𝑰𝑬 = 𝑰𝑪 + 𝑰𝑩
𝑰𝑪 = 𝜶 (𝑰𝑪 + 𝑰𝑩 ) = 𝜶 𝑰𝑪 + 𝜶 𝑰𝑩
𝑰𝑪 − 𝜶 𝑰𝑪 = 𝜶 𝑰𝑩
𝑰𝑪 (𝟏 − 𝜶) = 𝜶 𝑰𝑩
𝜶 𝜶
𝑰𝑪 = 𝑰𝑩 𝒂𝒏𝒅 ∵ 𝑰𝑪 = 𝜷 𝑰𝑩 𝜷=
(𝟏 − 𝜶) (𝟏 − 𝜶)
𝜶
∵𝜷=
(𝟏 − 𝜶)
∴ 𝜷(𝟏 − 𝜶) = 𝜶
𝜷−𝜷𝜶=𝜶
𝜷=𝜶+𝜷𝜶
𝜷 = 𝜶 (𝟏 + 𝜷)
𝜷
=𝜶
(𝟏 + 𝜷)
Summary of Active-Mode Currents Relations
𝑰𝑬 = 𝑰𝑪 + 𝑰𝑩 𝑰𝑬 = 𝟏 + 𝜷 𝑰𝑩
𝑰𝑪
𝜷= 𝑰𝑪 = 𝜷 𝑰𝑩
𝑰𝑩
𝑰𝑪
𝜶= 𝑰𝑪 = 𝜶 𝑰𝑬
𝑰𝑬
𝜶 𝜷
𝜷= 𝜶=
(𝟏 − 𝜶) (𝟏 + 𝜷)
Application of Active Mode BJT is Amplifier
Amplifier is a device which is used to increase the amplitude of the signal
applied to its input.
𝑰𝒏𝒑𝒖𝒕 𝒐𝒖𝒕𝒑𝒖𝒕
𝒙𝒊 𝑺𝒊𝒈𝒏𝒂𝒍 𝑺𝒊𝒈𝒏𝒂𝒍 𝒙𝒐
𝑨𝒎𝒑𝒍𝒊𝒇𝒊𝒆𝒓
x: represents either a voltage or a current signal.
Amplifier Gain (A): is defined as the ratio of the output signal 𝒙𝒐
𝑨=
to the input signal. 𝒙𝒊
Small-Signal Amplifier vs. Large-Signal Amplifier
𝑫𝒄 𝑷𝒐𝒘𝒆𝒓
𝑺𝒖𝒑𝒑𝒍𝒚
𝑺𝒎𝒂𝒍𝒍 − 𝑺𝒊𝒈𝒏𝒂𝒍 𝑳𝒂𝒓𝒈𝒆 − 𝑺𝒊𝒈𝒏𝒂𝒍
𝒎𝑽 𝑨𝒎𝒑𝒍𝒊𝒇𝒊𝒆𝒓 𝑽 𝑨𝒎𝒑𝒍𝒊𝒇𝒊𝒆𝒓 Block Diagram of Audio Amplifier
❑ Microphone: converts sound to electrical signal in microvolts/millivolts
range.
❑ Small-Signal Amplifier: converts the millivolts signal to large output
signal in the volt-range and output power less than 1 watt.
❑ Large-Signal Amplifier: produces output power greater than 1 watt to the
load (speaker). ❑ Speaker: converts electrical signal to sound.
In general, an amplifier has triangle symbol indicates
the direction of signal flow from the input side to the
output side.
Rin : input resistance of the amplifier, is seen at the
input terminal by the signal source (vs). It is a ratio
of the input voltage to the input current.
𝑹𝒊𝒏 𝑹𝒐𝒖𝒕
Rout : output resistance of the amplifier, is seen at the output terminal by
the load (RL).
Non-Inverting Amplifier Amplified output signal
with 0o phase shift
between input and
output signal
Input Output Input Signal
Signal Signal
Input Signal
Inverting Amplifier Amplified output signal
with 180o phase shift
between input and
output signal
BJT Configurations
❑ Bipolar transistor can be connected in three different configurations.
❑ The three different configurations are common-emitter, common-
collector, and common-base configurations.
❑ In each of these configurations, one lead is connected to the input, a
second lead is connected to the output, and a third lead is common
(connected) to both input and output and is used as a circuit reference
point.
𝑪
𝑩
𝑪
𝑬
𝑩
𝑬
Cutoff Mode of NPN BJT
In cutoff mode, external dc voltages are applied to set BEJ and BCJ are
reverse-biased.
All the terminal currents are zero (when neglecting the reverse saturation
current across the junctions) and the BJT acts as an open switch.
𝐼𝑓 𝑖𝑛𝑝𝑢𝑡 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑖𝑠 𝑧𝑒𝑟𝑜 (𝑉𝑖𝑛 = 0)
+VCC
𝑅𝐶
RC IC = 0
𝑅𝐵 𝑰𝑪
𝑉𝑖𝑛 𝑰𝑩
+
- 𝑉𝑐𝑐 ≅ C
E
Open
Switch
𝑇ℎ𝑒 𝑏𝑎𝑠𝑒 − 𝑒𝑚𝑖𝑡𝑡𝑒𝑟 𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑟𝑒𝑣𝑒𝑟𝑠𝑒 − 𝑏𝑖𝑎𝑠𝑒𝑑
𝑇ℎ𝑒 𝑏𝑎𝑠𝑒 − 𝑐𝑜𝑙𝑙𝑒𝑐𝑡𝑜𝑟 𝑗𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑟𝑒𝑣𝑒𝑟𝑠𝑒 − 𝑏𝑖𝑎𝑠𝑒𝑑
∴ 𝑰𝑩 = 𝑰𝑪 = 𝑰𝑬 = 𝟎
Saturation Mode of NPN BJT
In saturation mode, external dc voltages are applied to set BEJ and BCJ
are forward-biased.
In saturation mode, BJT acts as a closed switch.
In saturation, the collector current (𝑰𝑪 ) consists of two components:
The first component is due to the diffusion of electrons from emitter
into collector.
The other is due to the forward current across the base-collector
junction.
The two components are opposite, 𝑰𝑪
so collector current decreases in
saturation than in active mode.
𝐼𝐶 𝑖𝑛 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛 < 𝐼𝐶 (𝑖𝑛 𝑎𝑐𝑡𝑖𝑣𝑒)
𝑰𝑪 𝑰𝑪
𝒊𝒏 𝒔𝒂𝒕𝒖𝒓𝒂𝒕𝒊𝒐𝒏 < (𝒊𝒏 𝒂𝒄𝒕𝒊𝒗𝒆)
𝑰𝑩 𝑰𝑩
When the base-emitter junction becomes forward-biased, the base
current is increased (𝐼𝐵 ), the collector current also increases (𝐼𝐶 = β 𝐼𝐵 ),
and 𝑉𝐶𝐸 decreases as a result of more voltage drop across the collector
resistor (𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶 ).
When 𝑉𝐶𝐸 reaches its saturation value, 𝑽𝑪𝑬(𝒔𝒂𝒕) = 𝟎. 𝟐𝑽,
the base-collector junction becomes forward-biased.
+VCC
RC IC (sat)
≅ C
𝑽𝑪𝑬(𝒔𝒂𝒕) = 𝟎. 𝟐𝑽
Closed E
Switch
Example 1:
Determine IB , IC , IE , VBE , VCE , VCB in the circuit if β is 150.
IC
RC= 100 Ω
VCB
RB= 10KΩ VCC =10V
VCE
IB
VBB = 5V VBE
IE
Solution:
For the npn BJT :
❑ Base is connected to VBB (5V) 𝒏
through a resistor 10KΩ and 𝒑
emitter is connected to ground
(0V). VB > VE , base–emitter 𝒏
junction will be forward biased.
❑ Base is connected to VBB (5V) through a resistor 10KΩ and
collector is connected to VCC (10V) through a resistor 100Ω.
VB ≤ VC , base–collector junction is reverse biased.
❑ Assume transistor is in active mode and VBE=0.7V
In active mode: 𝑉𝐵𝐸 = 0.7𝑉
Applying KVL in the base-emitter circuit:
𝑉𝐵𝐵 − 𝐼𝐵 𝑅𝐵 − 𝑉𝐵𝐸 = 0
𝑉𝐵𝐵 − 𝑉𝐵𝐸 5𝑉 − 0.7𝑉
𝐼𝐵 = = = 430𝜇A
𝑅𝐵 10𝐾Ω
𝐼𝐶 = 𝛽 𝐼𝐵 = (150)(430𝜇A) = 64.5mA
𝐼𝐸 = 𝐼𝐶 + 𝐼𝐵 = 64.5mA + 430𝜇A = 64.9mA
Applying KVL in the collector-emitter circuit:
𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶 − 𝑉𝐶𝐸 = 0 +
𝑉𝐶𝐸 = 𝑉𝐶𝐶 − 𝐼𝐶 𝑉𝐶 𝑉𝐶𝐸
-
𝑉𝐶𝐸 = 10V − (64.5mA)(100Ω)
𝑉𝐶𝐸 = 10V − 6.45V = 3.55V
∵ 𝑉𝐶𝐸 ≫ 0.2𝑉
∴ 𝑇𝑟𝑎𝑛𝑠𝑖𝑠𝑡𝑜𝑟 𝑖𝑠 𝑖𝑛 𝑎𝑐𝑡𝑖𝑣𝑒 𝑚𝑜𝑑𝑒
The voltage across the base-collector junction (𝑉𝐶𝐵 ) is given by a KVL
around transistor terminals:
𝑉𝐶𝐸 − 𝑉𝐶𝐵 − 𝑉𝐵𝐸 = 0
𝑉𝐶𝐵 +
𝑉𝐶𝐵 = 𝑉𝐶𝐸 − 𝑉𝐵𝐸 𝑉𝐶𝐸
-
𝑉𝐵𝐸
𝑉𝐶𝐵 = 3.55V−0.7V = 2.85V
∵ 𝑉𝐶𝐵 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑣𝑎𝑙𝑢𝑒 ∴ 𝑇𝑟𝑎𝑛𝑠𝑖𝑠𝑡𝑜𝑟 𝑖𝑠 𝑖𝑛 𝑎𝑐𝑡𝑖𝑣𝑒 𝑚𝑜𝑑𝑒
Example 2:
Assume transistor in active mode with α=0.97 and VBE=0.7V,
find R in the circuit shown to yield IE=2mA.
RC=2KΩ, RE=100Ω, R1=25KΩ, VCC=10V.
𝐼2
Solution:
𝑉𝐶
𝐼𝐶 = 𝛼 𝐼𝐸 𝐼𝑅
𝑉𝐵 𝐼𝐶
𝐼𝐵 = 𝐼𝐸 − 𝐼𝐶 𝐼𝐵
𝐼1
𝐼𝐸
𝑉𝐶 − 𝑉𝐵
𝑅=
𝐼𝑅
𝑉𝐵 − 𝑉𝐵𝐸 − 𝑅𝐸 𝐼𝐸 = 0 𝑉𝐵 = 𝑉𝐵𝐸 + 𝑅𝐸 𝐼𝐸
𝑉𝐵 −
𝐼1 =
𝑅1
𝑽𝑪
𝐼2
𝐼𝑅 = 𝐼1 + 𝐼𝐵 +
𝐼𝑅
𝐼2 = 𝐼𝑅 + 𝐼𝐶 + 𝐼𝐶
𝑉𝐵𝐸
−𝑉𝐶 + 𝑉𝐶𝐶 − 𝐼2 𝑅𝐶 = 0 𝐼1 𝑉𝐵
𝐼𝐸
𝑉𝐶 = 𝑉𝐶𝐶 − 𝐼2 𝑅𝐶 −
Example 3:
A transistor given in the figure has 𝜷 = 𝟏𝟎𝟎.
If 𝑽𝑪𝑪 = 𝟏𝟎𝑽, 𝑹𝑪 = 𝟐. 𝟕𝑲𝛀, 𝑹𝑩 = 𝟏𝟖𝟎𝑲𝛀.
Determine the value of 𝑽𝑪𝑬 and 𝑰𝑪 .
Solution: Assume transistor in active mode
Applying KVL in the base-emitter circuit:
𝐼𝐸
𝑉𝐶𝐶 − 2.7𝐾 𝐼𝐸 − 180𝑘 𝐼𝐵 − 𝑉𝐵𝐸 = 0
𝑉𝐶𝐶 − 2.7𝐾 1 + 𝛽 𝐼𝐵 − 180𝑘 𝐼𝐵 − 0.7 = 0
𝐼𝐵 = 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒
𝐼𝐸 = 1 + 𝛽 𝐼𝐵
𝐼𝐶 = 𝐼𝐸 − 𝐼𝐵
Applying KVL in the collector-emitter circuit:
𝑉𝐶𝐶 − 2.7𝐾 𝐼𝐸 − 𝑉𝐶𝐸 = 0
+
𝑉𝐶𝐸 ≫ 0.2V 𝑉𝐶𝐸
−
∴ 𝑇𝑟𝑎𝑛𝑠𝑖𝑠𝑡𝑜𝑟 𝑖𝑠 𝑖𝑛 𝑎𝑐𝑡𝑖𝑣𝑒 𝑚𝑜𝑑𝑒
Example 4:
Determine the node voltages VB , VE , VC and the currents IB , IC , IE .
Solution:
For the npn BJT :
❑ Base is connected to 𝑉𝐼𝑁 (0V) through a resistor 1KΩ
and emitter is connected to ground (0V). 𝑉𝐵 ≤ 𝑉𝐸 ,
base-emitter junction will be reverse biased.
❑ Base is connected to 𝑉𝐼𝑁 (0V) through a resistor 1KΩ and
collector is connected to 𝑉𝐶𝐶 (10V) through a resistor 1KΩ.
𝑉𝐵 ≤ 𝑉𝐶 , base-collector junction will be reverse biased.
❑ Assume transistor is in cutoff mode.
In cutoff mode: 𝐼𝐵 = 𝐼𝐶 = 𝐼𝐸 = 0
𝑉𝐵 = 𝑉𝐼𝑁 − 𝑅𝐵 𝐼𝐵 = 0𝑉
𝑉𝐸 = 0𝑉 𝑰𝑪
𝑰𝑩 𝑽𝑪
𝑉𝐶 = 𝑉𝐶𝐶 − 𝑅𝐶 𝐼𝐶 = 10𝑉 𝑽𝑩
𝑽𝑬
𝐼𝑓 𝑉𝐵 ≤ 𝑉𝐸 → ∴ base−emitter junction is reverse biased.
𝐼𝑓 𝑉𝐵 ≤ 𝑉𝐶 → ∴ base−collector junction is reverse biased.
∴ 𝑇𝑟𝑎𝑛𝑠𝑖𝑠𝑡𝑜𝑟 𝑖𝑠 𝑖𝑛 𝑐𝑢𝑡𝑜𝑓𝑓 𝑚𝑜𝑑𝑒.
Example 5:
If β=100, determine whether or not a transistor is in saturation and
find 𝐼𝐵 , and 𝐼𝐶 . Repeat with the 2KΩ emitter resistance is added.
Solution: Assume transistor is in saturation mode:
𝑉𝐵𝐸 = 0.7𝑉, 𝑉𝐶𝐸 = 0.2𝑉, 𝐼𝐸 = 𝐼𝐵 + 𝐼𝐶
Apply KVL in the base-emitter circuit: 𝑰𝐼𝐵𝑩
𝑉𝐵𝐵 − 𝐼𝐵 𝑅𝐵 − 𝑉𝐵𝐸 = 0
𝑽𝑩𝑬
𝑉𝐵𝐸
𝑉𝐵𝐵 − 𝑉𝐵𝐸
𝐼𝐵 = = 86𝜇A
𝑅𝐵
Apply KVL in the collector-emitter circuit:
𝑉𝐶𝐶 − 𝑉𝐶𝐸
𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶 − 𝑉𝐶𝐸 = 0 𝐼𝐶 = = 3.26𝑚A
𝑅𝐶
𝐼𝐶 3.26𝑥10−3
𝑖𝑛 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛 = −5
= 38
𝐼𝐵 8.6𝑥10
𝑰𝑪
𝐼𝐶 𝐼𝐶
𝑖𝑛 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛 < (𝛽 = = 100 𝑖𝑛 𝑎𝑐𝑡𝑖𝑣𝑒) +
𝐼𝐵 𝐼𝐵
𝑽𝑪𝑬
-
∴ 𝑇𝑟𝑎𝑛𝑠𝑖𝑠𝑡𝑜𝑟 𝑖𝑠 𝑖𝑛 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑚𝑜𝑑𝑒.
❑ Repeat with the 2KΩ emitter resistance is added.
Assume transistor is in saturation mode:
𝑉𝐵𝐸 = 0.7𝑉, 𝑉𝐶𝐸 = 0.2𝑉, 𝐼𝐸 = 𝐼𝐵 + 𝐼𝐶
Apply KVL in the base-emitter circuit:
𝑉𝐵𝐵 − 𝐼𝐵 𝑅𝐵 − 𝑉𝐵𝐸 − 𝐼𝐸 𝑅𝐸 = 0 𝑰𝑩
𝑉𝐵𝐵 − 𝐼𝐵 𝑅𝐵 − 𝑉𝐵𝐸 − (𝐼𝐵 + 𝐼𝐶 )𝑅𝐸 = 0
𝑰𝑬
(1)
Apply KVL in the collector-emitter circuit:
𝑰𝑪
𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶 − 𝑉𝐶𝐸 − 𝐼𝐸 𝑅𝐸 = 0
𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶 − 𝑉𝐶𝐸 − (𝐼𝐵 + 𝐼𝐶 )𝑅𝐸 = 0 (2)
+
𝑽𝑪𝑬
-
From (1) and (2):
𝐼𝐵 = 6.25x10−6 A
𝐼𝐶 = 1.94x10−3 A 𝑰𝑬
𝐼𝐶 1.94x10−3 Transistor is not
= −6
= 310.4 > (𝛽 = 100 𝑖𝑛 𝑎𝑐𝑡𝑖𝑣𝑒) in saturation
𝐼𝐵 6.25x10
Assume again transistor is in active mode: 𝑉𝐵𝐸 = 0.7𝑉
Apply KVL in the base-emitter circuit:
𝑉𝐵𝐵 − 𝐼𝐵 𝑅𝐵 − 𝑉𝐵𝐸 − 𝐼𝐸 𝑅𝐸 = 0
𝑉𝐵𝐵 − 𝐼𝐵 𝑅𝐵 − 𝑉𝐵𝐸 − (1 + 𝛽) 𝐼𝐵 𝑅𝐸 = 0
𝐼𝐵 = 0.017𝑚𝐴 𝑰𝑩
𝐼𝐸 = (1 + 𝛽) 𝐼𝐵 = 1.72𝑚𝐴
𝐼𝐶 = 𝛽 𝐼𝐵 = 1.70𝑚𝐴 𝑰𝑬
𝑰𝑪
Apply KVL in the collector-emitter circuit:
𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶 − 𝑉𝐶𝐸 − 𝐼𝐸 𝑅𝐸 = 0 +
𝑽𝑪𝑬
-
𝑉𝐶𝐶 − 𝐼𝐶 𝑅𝐶 − 𝐼𝐸 𝑅𝐸 = 𝑉𝐶𝐸
𝑉𝐶𝐸 = 1.46𝑉
𝑰𝑬
∵ 𝑉𝐶𝐸 > 0.2𝑉 ∴ 𝑇𝑟𝑎𝑛𝑠𝑖𝑠𝑡𝑜𝑟 𝑖𝑠 𝑖𝑛 𝑎𝑐𝑡𝑖𝑣𝑒 𝑚𝑜𝑑𝑒.
∴ 𝑇ℎ𝑒 𝑐𝑜𝑟𝑟𝑒𝑐𝑡 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑐𝑢𝑟𝑟𝑒𝑛𝑡𝑠 𝑎𝑟𝑒
𝐼𝐵 = 0.017𝑚𝐴
𝐼𝐶 = 1.70𝑚𝐴
𝐼𝐸 = 1.72𝑚𝐴
