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Session 12Session 12
Analysis of DC circuits with BJT (polarization)
Electronic Components and Circuits
José A. Garcia Souto
www.uc3m.es/portal/page/portal/dpto_tecnologia_electronica/Personal/JoseAntonioGarcia
DC circuits with BJT and
polarization of transistors
OBJETIVESOBJETIVES• Summarizing the operating regions and introducing the
amplification (active)
Active, Cut-off, Saturation
• Analyzing basic BJT circuits in DC. Static characteristics and Load line.
• Understanding the need of polarization circuits• Understanding the need of polarization circuits
• Knowing and analyzing typical polarization circuits of transistors. Bias (quiescent) point.
2UC3M 2010 ECC - Session 12
Amplification concept with BJT
60
50
IC (mA) SATURATES
Vo ≈ 0 V
0 0,2 0,4 0,6 0,8
50
40
30
20
10
VBE (V)
CUT-OFF
Vo = Vcc
AMPLIFIES
Vo = G·Vi
Small changes of Vi result in greater variation of
Vo, thus providing gain Vo/Vi
It should be around a bias point or quiescent
point VBE-Q, VCE-Q
3UC3M 2010 ECC - Session 12
0 0,2 0,4 0,6 0,8 VBE (V)
Superposition of DC Bias and Signal
VBB is a continuous source that
provides along with RB a bias
point (polarization) : 0=vI point (polarization) : 0=gvCI
EI
BI
BEVCEV
60
50
40
30
20
10
IB (µA)
IB
A variable signal is introduced
and it is added to VBB
gBB vV +
4UC3M 2010 ECC - Session 12
EI
VBEVBB
10
VBE (V)
Amplification focuses on Active region
• Datasheet VBE(on) VCE(sat) hFE VBE(sat)
Search BC547 in http://www.fairchildsemi.com/
Region Base-Emitter Junction Base-Collector Junction
Cut-off Reverse(OFF) Reverse
Active Forward (ON) Reverse (Transistor Effect)
Saturation Forward (ON) Forward (Saturated)
Region Conditions NPN Operation NPN
5
Region Conditions NPN Operation NPN
Cut-off VBE < VBE-ON o IB = 0 IB=0, IC=IE=0
Active VBE-ON y VCE > VCE-SAT IC ≈≈≈≈ hFE·IB [VBE=VBE-ON]
Saturation VBE-ON y VCE-SAT VCE =VCE-SAT [VBE=VBE-SAT]
UC3M 2010 ECC - Session 12
Graphical DC Analysis
Cancel the signal (superposition)
0=vIAnalyze the input
Device Equation
(Input Characteristic)
Circuit Equation
(Load line)
Analyze the output
Device Equation
EXAMPLE
RC = 100 Ω
RB = 10 kΩ
VCC = 10 V
VBB = 10 V
VBE(on) = 0,7 V
VCE(sat) = 0,2 V
0=gvCI
EI
BI
BEVCEV
Device Equation
(Output Characteristic)
Circuit Equation
(Load line)
hFE = 100
6UC3M 2010 ECC - Session 12
E
Input Characteristic and
Load Line (I)
Graphically
Characteristic:60
IB (µA)
Characteristic:
Load line:
Analytically
VBB > VBE(on)
B
BEBBB
R
vVi
−=
Vv =
( )CEBEBB vvii ,=50
40
30
20
10
IB
7
)(onBEBE Vv =
BBBEBB RivV ⋅+=
)(onBEBE VV ≈B
onBEBB
BR
VVI
)(−=
UC3M 2010 ECC - Session 12
VBEVBB VBE (V)
Output Characteristic and
Load Line (II)
Graphically
Characteristic:6
IC
(mA)
60 µA Characteristic:
Load line :
Analytically
VCE > VCE(sat)
( )BCECC ivii ,=
C
CECCC
R
vVi
−=
⋅=
5
4
3
2
1
IB
=40 µA
50 µA
30 µA
20 µA
10 µA
IC
8
BFEC ihi ⋅=
CCCECC RivV ⋅+=
CCCCCE RIVV ⋅−=BFEC IhI ⋅=
UC3M 2010 ECC - Session 12
0µA
0 2 4 6 8 10 12 14 16
1 10 µA
VCE
(V)
VCE VCC
EXAMPLE
RC = 100 Ω
DC equivalent circuit: Analytically
Analyze the input RC = 100 Ω
RB = …
VCC = 10 V
VBB = …
VBE(on) = 0,7 V
VCE(sat) = 0,2 V
hFE = 100
Case 3
CI
I
BI
BEVCEV
Analyze the input
Device Equation
Circuit Equation
Analyze the output
Device Equation
Circuit Equation
9
Case 3
VBB = 10 V
RB = 10 kΩ
UC3M 2010 ECC - Session 12
EI Circuit Equation
CI
Solution: DC Analysis Analyze the input
VBB > VBE(on)
Device Equation )(onBEBE Vv =CI
EI
BI
BEVCEV
Device Equation
Circuit Equation
Analyze the output
VCE > VCE(sat)
Device Equation
Circuit Equation
)(onBEBE Vv =
BBBEBB RivV ⋅+=
BFEC ihi ⋅=
RivV ⋅+=
10UC3M 2010 ECC - Session 12
Circuit Equation
)(onBEBE VV ≈B
onBEBB
BR
VVI
)(−=
CCCECC RivV ⋅+=
CCCCCE RIVV ⋅−=BFEC IhI ⋅=
Bias circuits
11
Set a stable operating point desensitized against the transistor
parameters.
Optimize the signal amplifier circuit.
Separate bias circuit if necessary (signal AC coupling).
UC3M 2010 ECC - Session 12
Example
Reasoning operating region
Calculate the voltage VE
Calculate the current IE Calculate the current IE
Derive the relationship between IE, IB
and IC
Calculate the curents IC and IB
Calculate the voltage VC
DATA
VCC = 12 V
RC = RE = 10 kΩ
12UC3M 2010 ECC - Session 12
RC = RE = 10 kΩ
VBE(on) = 0,7 V
VCE(sat) = 0,2 V
hFE = 100
Types of polarization circuits
Biasing with REPractical circuit
13
Lower sensitivity to the transistor current gain
Stabilizes the bias point (quiescent point) against VBE, hFE, etc.
UC3M 2010 ECC - Session 12
Case 1: Biasing with RB
DATA:
VBB = 5 V
VCC = 10 V
RE = 2 kΩ
VBE(on) = 0,7 VVCC = 10 V
RB = 50 kΩ
RC = 3 kΩ
Reasoning the operating region
Indicate the value of VBE
Calculate the currents IB, IC e IE
Calculate the voltage VCE
Check that is in active
Recalculate if you change hFE = 200
VCE(sat) = 0,2 V
hFE = 100
14UC3M 2010 ECC - Session 12
Case 2: Biasing with REDATA:
VBB = 5 V
VCC = 10 V
RB = 10 kΩ
RE = 2 kΩ
VBE(on) = 0,7 V
VCE(sat) = 0,2 V
Reasoning the operating region
Indicate the value of VBE RB = 10 kΩ
RC = 3 kΩ
VCE(sat) = 0,2 V
hFE = 100
Indicate the value of VBE
Calculate the current IE (you can
neglect the base current IB)
Calculate the currents IC, IB
Check that IB is negligible
Calculate the voltage VCE
Check that the BJT is in active
Recalculate if you change hFE = 200
15UC3M 2010 ECC - Session 12
Recalculate if you change hFE = 200
Generalize if not negligible IB (eg if
RB = 50 kΩ)
Biasing with current mirrors
16
refO II ≈1refrefO IR
RII ⋅== 10
3
22
UC3M 2010 ECC - Session 12
Proposed exercise: Current sources
Zener ideal:
VZ = 3,3 V
IZmín = 10 mA
• Obtain the ratio IL/Iref
IL
IZmín = 10 mA
Transistor:
VCEsat = 0,2 V
VBEon = 0,7 V
β = 270
• Calculate IE, IC, IB and VB, VC, VE, VCE, IZ.
Iref
17
• Obtain the ratio IL/Iref
• Reason if IB is negligible
• Obtain Iref
• There is the scale factor IC2 =
2·IC1 between both transistor.
Calculate Vo.
• Calculate IE, IC, IB and VB, VC, VE, VCE, IZ.
• For RL = 1 kΩ, calculate IC y VC
• For RL = 22 kΩ, calculate IC y VC
• Plot the load lines (three cases).
• Maximum RL as a current source?.
• In the case RL = 22 kΩ, calculate IB, IZ
UC3M 2010 ECC - Session 12