Download BJT DC Analysis

Recall Lecture 10

Introduction to BJT

3 modes of operation




Cut-off
Active
Saturation
Active mode operation of NPN
NPN
PNP
IE = IS [ e VBE / VT ]
IE = IS [ e VEB / VT]
IC =  IB
IC =  IE
IE = IB(  + 1)
=[ /+1]
 = [ / 1 -  ]
Based on KCL: IE = IC + IB
DC Analysis of BJT Circuit
and Load Lines

DC analysis of BJT



BE Loop (EB Loop) – VBE for npn and VEB for pnp
CE Loop (EC Loop) - VCE for npn and VEC for pnp
When node voltages are known, branch current
equations can be used.
Common-Emitter Circuit
 The figures below is showing a common-emitter circuit
with an npn transistor and the DC equivalent circuit.
 Assume that the B-E junction is forward biased, so the
voltage drop across that junction is the cut-in or turnon voltage VBE (on).
Unless given , always
assume VBE = 0.7V
 KVL at B-E loop
You can calculate
IC using IC = βIB
Calculate IE using
IE = IB + IC
Input Load Line
IB versus VBE
Input Load Line – IB versus VBE
Derived using B-E loop

The input load line is obtained from
Kirchhoff’s voltage law equation
around the B-E loop, written as
follows:
VBE – VBB + IBRB = 0
y = mx + c

Both the load line and the quiescent base current change as either or
both VBB and RB change.
For example;
 = 200
 The input load line is essentially
the same as the load line
characteristics for diode circuits.
VBE – 4 + IB(220k) = 0
IB = -VBE + 4
220k 220k
y = mx + c
 IBQ = 15 μA
 KVL at C-E loop of the circuit
You have calculated the
value of IC from the value
of IB
Output Load Line
IC versus VCE
Output Load Line – IC versus VCE
Derived using C-E loop

For the C-E portion of the circuit, the
load line is found by writing
Kirchhoff’s voltage law around the
C-E loop. We obtain:
y = mx + c
For example;
IC = -VCE + 10
2k
2k
y = mx + c
 = 200
Q-point is the intersection of the load line with
the iC vs vCE curve, corresponding to the
appropriate base current

To find the intersection
points setting IC = 0,
VCE = VCC = 10 V

setting VCE = 0
IC = VCC / RC = 5 mA
Examples
BJT DC Analysis
Common-Emitter Circuit
Example 1
Calculate the base, collector and emitter currents
and the C-E voltage for a common-emitter
circuit by considering VBB = 4 V, RB = 220kΩ, RC
= 2 kΩ, VCC = 10 V, VBE (on) = 0.7 V and β =
200.
Example 2
 = 0.99
KVL at BE loop:
Hence,
0.7 + IERE – 4 = 0
IE = 3.3 / 3.3 = 1 mA
IC =  IE = 0.99 mA
IB = IE – IC = 0.01 mA
KVL at CE loop:
ICRC + VCE + IERE – 10 = 0
VCE = 10 – 3.3 – 4.653 = 2.047 V
Common-Emitter Circuit - PNP
Example 3
Find IB, IC, IE and RC such that VEC = 2.5 V for
a common-emitter circuit .
Consider:
VBB = 1.5 V, RB = 580 kΩ, VCC = 5 V, VEB (on)
= 0.6 V,  = 100.
Example 4: DC Analysis and Load Line
Calculate the characteristics of a circuit containing an emitter resistor
and plot the output load line.
For the circuit, let VBE (on) = 0.7 V and β = 75.
Output Load Line
Use KVL at B-E loop to find the value of IB
Use KVL at C-E loop – to obtain the linear equation
IC RC + VCE + IE RE – 12 = 0
IC RC + VCE + (IC / )RE – 12 = 0
VCE = 12 – IC (1.01)
IC = - VCE + 12
1.01
IB = 75.1 A
VCE = 12 – 5.63 (1.01)
VCE = 6.31 V
Example: DC Analysis – Common Base (npn)
E
C
B
C
B
E
 = 100
5k
BE Loop
0.7 + IERE – 5 = 0
15V
IE = 0.43 mA
CE Loop
10 k
VCE + ICRC + IERE – 5 – 15 = 0
VCE = 20 – 4.3 – 2.13
5V
VCE = 13.57 V