BJT

1. CHAPTER 3
Zener Diode
Voltage
Regulator The load resistor sees a constant voltage regardless of the current
VL = VZ
Range of power
supply
Assume: IZmin = 0.1 IZ(max)
VRi = VPS - VZ
II = IZ + IL

2. CHAPTER 3 Rectifier
Half Wave
Full Wave
Center-
tapped
Bridge
Vo = Vs - V
Vo = Vs - 2V
Vo = Vs - V
PIV = Vspeak - V
PIV = 2Vspeak - V
PIV = Vspeak
Duty
Cycle
Filter
Ripple Voltage, Vr
Capacitor
Discharge Vr =
VmT′
RC
VC = Vme – t / RC
If the ripple is very small, we
can approximate T’ = Tp where
Tp is the period of the cycle
Half Wave
Vr =
VmTP
RC
Full Wave
Vr =
VmTP
2RC
Vm is the peak value
of the output
voltage
Multiple Diode Circuit
Peak and RMS
Vrms =
Vpeak
2

3. Chapter 4
Bipolar Junction Transistor

4. REMEMBER THIS
Current flow in the opposite direction of the electrons flow;
same direction as holes
e e e
I
h h h

5. Transistor Structures
 The bipolar junction transistor (BJT) has three separately
doped regions and contains two pn junctions.
 Bipolar transistor is a 3-terminal device.
 Emitter (E)
 Base (B)
 Collector (C)
 The basic transistor principle is that the
voltage between two terminals controls the
current through the third terminal.
 Current in the transistor is due to the flow of
both electrons and holes, hence the name
bipolar.

6. Transistor Structures
 There are two types of bipolar junction transistor: npn and
pnp.
 The npn bipolar transistor contains a thin p-region between
two n-regions.
 The pnp bipolar transistor
contains a thin n-region
sandwiched between two p-
regions.

7.  Active
Operating range of the amplifier.
Base-Emitter Junction forward biased.
Collector-Base Junction reverse biased
 Cutoff
The amplifier is basically off. There is voltage but
little current.
Both junctions reverse biased
 Saturation
The amplifier is full on. There is little voltage but lots
of current.
Both junctions forward biased
3 Regions of Operation

8. OPERATIONS - npn
 The base-emitter (B-E) junction
is forward biased and the base-
collector (C-B) junction is
reverse-biased,.
 Since the B-E junction is
forward biased, electrons
from the emitter are injected
across the B-E junction into
the base  IE
 Once in the base region, the
electrons are quickly
accelerated through the base
due to the reverse-biased C-B
region  IC
ACTIVE MODE
Some electrons, in passing
through the base region, recombine
with majority carrier holes in the
base. This produces the current
IB
+
-
VBE
iB
VBE = V

9. C B E
TO ILLUSTRATE
-
VBE
+
•Imagine the marbles as electrons
•A flat base region with gaps where the
marbles may fall/trapped – recombine
•A sloping collector region represents
high electric field in the C-B region
•Hence, when enough energy is given to
the marbles, they will be accelerated
towards to base region with enough
momentum to pass the base and straight
‘fly’ to the collector

10. MATHEMATICAL EXPRESSIONS
+
-
VBE
IE
IC
IB
IE = IS [ e VBE / VT -1 ] = IS e VBE / VT Based on KCL: IE = IC + IB
No. of electrons crossing the base region and then directly into the
collector region is a constant factor  of the no. of electrons exiting the
base region
IC =  IB
No. of electrons reaching the collector region is directly proportional to
the no. of electrons injected or crossing the base region.
IC =  IE
Ideally  = 1, but in reality it is between 0.9
and 0.998.

11. Based on KCL: IE = IC + IB IC =  IB IC =  IE
IE =  IB + IB = IB( + 1)
 = [  /  + 1 ]
IE = IB( + 1)
Now
With IC =  IB  IB = IC / 
Hence,
IE = [ IC / ] ( + 1)
IC = IE [  /  + 1 ]
Comparing with IC =  IE

12. B
C
E
-
+
VEB
OPERATIONS - pnp
FORWARD ACTIVE MODE
 The emitter – base (E- B)
junction is forward biased and
the base-collector (B- C) junction
is reverse-biased,.
IE = IS [ e VEB / VT -1 ] = IS e VEB / VT
**Notice that it is VEB
IE
IC
IB
Based on KCL: IE = IC + IB
VEB = V

13. pnp Transistor- Active mode

14. SUMMARY: Circuit Symbols and
Conventions
npn bipolar transistor simple
block diagram and circuit symbol.
Arrow is on the emitter terminal
that indicates the direction of
emitter current
pnp bipolar transistor simple
block diagram and circuit symbol.
Arrow is on the emitter terminal
that indicates the direction of
emitter current
Based on KCL: IE = IC + IB

15. IE = IS [ e VBE / VT ]
Based on KCL: IE = IC + IB
IC =  IB
IC =  IE
IE = IB( + 1)
IE = IS [ e VEB / VT]
NPN PNP
𝜶 =
𝜷
𝟏 + 𝜷
𝜷 =
𝜶
𝟏 − 𝜶

16. EXAMPLE
Calculate the collector and emitter currents, given the base current and current gain.
Assume a common-base current gain,  = 0.97 and a base current of iB = 25 µA . Also
assume that the transistor is biased forward in the forward active mode.
Solution: The common-emitter current gain is
The collector current is
And the emitter current is

17. Examples
• EXAMPLE 1
• Given IB = 6.0A and IC = 510 A. Determine ,  and IE
• EXAMPLE 2
• NPN Transistor
• Reverse saturation current Is = 10-13A with current gain,  = 90.
• Based on VBE = 0.685V, determine IC , IB and IE
Answers:
 = 85
= 0.9884
IE = 516 A
Answers:
IE = 10-13 (e 0.685/0.026) = 0.0277 A
IC = (90/91)(0.0277) = 0.0274 A
IB = IE – IC = 0.3 mA

18. BJT: Current-Voltage
Characteristic
IC versus VCE

19.  The Emitter is common to both input
(base-emitter) and output (collector-
emitter).
 Since Emitter is grounded, VC = VCE
 With decreasing VC (VCE), the junction
B-C will become forward biased too.
 The current IC quickly drops to zero
because electrons are no longer
collected by the collector
Common-Emitter Configuration - npn
Node B
0V

20. Characteristics of Common-Emitter - npn
NOTE: VEC for PNP