RECALL LECTURE 9
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Introduction to BJT
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3 modes of operation
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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
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DC analysis of BJT
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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
turn-on voltage VBE (on). Unless given , always
assume VBE = 0.7V
Common-Emitter Circuit
 The base current: KVL at B-E loop
 Implicitly assuming that VBB > VBE (on), which means
that IB > 0. When VBB < VBE (on), the transistor is cut
off and IB = 0.
Common-Emitter Circuit
 In the C-E loop of the circuit, we can use:
and
 Implicitly assuming that the transistor is biased in the
forward-active mode.
Examples
BJT DC Analysis
Common-Emitter Circuit
Example
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.
BJT Circuits at DC
 = 0.99
KVL at BE loop: 0.7 + IERE – 4 = 0
IE = 3.3 / 3.3 = 1 mA
Hence, 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
Find IB, IC, IE and RC such that VEC =
½ VCC for a common-emitter circuit .
Consider:
VBB = 1.5 V, RB = 580 Ω, VCC = 5 V,
VEB (on) = 0.6 V, β = 100.
EXAMPLE
Given  = 75 and VEC = 6V. Find the
values of the labelled parameters, RC
and IE,
IE