330 EE Lecture 20
Bipolar Device Modeling Bipolar Process
Review from Last Lecture
Basic Devices and Device Models
Resistor Diode Capacitor MOSFET BJT
Review from Last Lecture
Bipolar Junction Transistors
Operation and Modeling
Review from Last Lecture
Bipolar Transistors
E B
npn stack
E B
pnp stack
C
Bipolar Devices Show Basic Symmetry Electrical Properties not
C B
C
B
E C
Symmetric
Designation of C and E critical
E
npn transistor
pnp transistor
With proper doping and device sizing these form Bipolar Transistors
Review from Last Lecture
Bipolar Operation
Consider npn transistor
E
B
npn stack
C
So, what will happen?
F1 F2
Some will recombine with holes and contribute to base current and some will be attracted across BC junction and contribute to collector Size and thickness of base region and relative doping levels will play key role in percent of minority carriers injected into base contributing to collector current
Review from Last Lecture
Bipolar Operation
Consider npn transistor
E
B
npn stack
C
Under forward BE bias and reverse BC bias current flows into base region Efficiency at which minority carriers injected into base region and contribute to collector current is termed is always less than 1 but for a good transistor, it is very close to 1 For good transistors .99 < < .999 Making the base region very thin makes large
Review from Last Lecture
Bipolar Operation
Consider npn transistor
E
B
npn stack
C
In contrast to MOS devices where current flow in channel is by majority carriers, current flow in the critical base region of bipolar transistors is by minority carriers
Review from Last Lecture
Bipolar Operation
E B
IB IE IC
C
IC IB
is typically very large
Bipolar transistor can be thought of a current amplifier with a large current gain In contrast, MOS transistor is inherently a tramsconductance amplifier Current flow in base is governed by the diode equation Collector current thus varies exponentially with VBE
VBE
~ IB I S e Vt
~ Vt IC I S e
VBE
Review from Last Lecture
Simple dc model
Summary:
~ IB I S e
VBE Vt
IC IB B VBE
C VCE
~ IC I S e
kT Vt q
VBE Vt
E
This has the properties we are looking for but the variables we used in introducing these relationships are not standard It can be shown that
~ IS
is proportional to the emitter area AE
~ Define I S 1JS AEand substitute this into the above equations
Review from Last Lecture
Simple dc model
~ IB I S e
VBE Vt
~ IC I S e
kT Vt q
VBE Vt
JS A E IB e
VBE Vt
IC JS A Ee
kT Vt q
VBE Vt
JS is termed the saturation current density Process Parameters : JS,
Design Parameters: AE
Environmental parameters and physical constants: k,T,q At room temperature, Vt is around 26mV JS very small around .25fA/u2
Review from Last Lecture
Simple dc model
Typical Output Characteristics Saturation
300 250 200 150 100 50 0 0 1 2 Vds 3 4 5
IC
Forward Active
VBE or IB
Id
VCE
Cutoff
Forward Active region of BJT is analogous to Saturation region of MOSFET Saturation region of BJT is analogous to Triode region of MOSFET
Simple dc model
Typical Output Characteristics
300 250 200 150 100 50 0 0 1 2 Vds 3 4 5
IC
VBE or IB
Id
VCE
Projections of these tangential lines all intercept the VCE axis at the same place and this is termed the Early voltage, VAF (actually VAF is intercept) Typical values of VAF are in the 100V range
Simple dc model
Improved Model
300 250 200 150 100 50 0 0 1 2 Vds 3 4 5
IC
VBE or IB
Id
VCE
JS A E IB e
IC JSe
VBE Vt
VBE Vt
V 1 CE VAF
Valid only in Forward Active Region
Simple dc model
Improved Model
300 250 200 150 100 50
IC
VBE or IB
Vt
kT q
VBE Vt
Id
0 0 1 2 3 Vds 4 5
VCE
JS AE e IE F
J A e 1 S E
VBC Vt
1
Valid in All regions of operation VAF effects can be added Not mathematically easy to work with Note dependent variables changes Termed Ebers-Moll model Reduces to previous model in FA region
BC BE VV J A VV e t 1 S E e t 1 IC JS AE R
Simple dc model
Ebers-Moll model
VBE BC VV JS AE Vt e 1 J A e t 1 IE S E F
BC BE VV J A VV S E t e t 1 IC JS AE e 1 R
Vt
kT q
Process Parameters: {JS, F, R} Design Parameters: {AE} F is the parameter discussed earlier R is termed the "reverse "
= 1-
F F
F
= 1-
R R
F =
R
F 1+F
R =
R 1+R
Typical values for process parameters: JS ~10-16A/2 F~100, R~0.4
Simple dc model
Ebers-Moll model
VBE BC VV JS AE e Vt 1 J A e t 1 IE S E F
BC BE VV J A VV kT S E t e t 1 IC JS AE e 1 Vt R q With typical values for process parameters in forward active region (VBE~0.6V, VBC~-3V), with Vt=26mV and if AE=1002:
JS ~10-16A/2
F~100, R~0.4
10 I 10 1.05x10 1 7.7x10 1 .28 Completely
14 10 -51 C
BC BE VV J A VV S E t e t 1 IC JS AE e 1 R -14
Makes no sense to keep anything other than I J A e in forward active
VBE Vt C S E
dominant!
Simple dc model
Ebes-Moll model
VBE BC VV JS AE Vt e 1 J A e t 1 IE S E F
BC BE VV J A VV S E t e t 1 IC JS AE e 1 R
Vt
kT q
Alternate equivalent expressions for dependent variables {IC, IB} defined earlier for Ebers-Moll equations in terms of independent variables {VBE, VCE} after dropping the "-1" terms
1+ I J A e 1 e 1 1 I JA e - e
VBE Vt R -VCE Vt C S E R
VBE Vt -VCE Vt B S E F R
No more useful than previous equation but in form consistent with notation Introduced earlier
Simple dc model
Simplified Multi-Region Model
IC
VBE5 VBE4 VBE3 VBE2 VBE1
IC
VBE5 VBE4 VBE3 VBE2 VBE1
Ebers-Moll Model
VCE
VCESAT
VCE
Simplified Multi-Region Model
Observe VCE around 0.2V when saturated VBE around 0.6V when saturated In most applications, exact VCE and VBE voltage in saturation not critical
VBE=0.7V VCE=0.2V
Saturation
Simple dc model
Simplified Multi-Region Model
IC
VBE5 VBE4 VBE3 VBE2 VBE1
IC
VBE5 VBE4 VBE3 VBE2 VBE1
Ebers-Moll Model
VCE
VCESAT
VCE
Simplified Multi-Region Model
1+ I J A e 1 e
VBE Vt R C S E R
-VCE Vt
V 1+ V
CE AF
IC JS A E e
IB
VBE Vt
1 1 I JA e - e
VBE Vt B S E F R
-VCE Vt
JS A E Vt e
VCE 1 V AF VBE
Forward Active
VBE=0.7V VCE=0.2V IC=IB=0
Saturation
Cutoff
Simple dc model
Simplified Multi-Region Model
VBE Vt
IC JS A E e
VCE 1 V AF
Forward Active
J A IB S E e Vt
Vt kT q
VBE
VBE=0.7V VCE=0.2V IC=IB=0
Saturation
Cutoff
Simple dc model
Simplified Multi-Region Model
IC JS A E e
VBE Vt
VCE 1 V AF
VBE>0.4V VBC<0
J A IB S E e Vt
Vt kT q
VBE
Forward Active
VBE=0.7V VCE=0.2V
IC=IB=0
IC<IB
Saturation
VBE<0
Cutoff
VBC<0
A small portion of the operating region is missed with this model but seldom operate in the missing region
Simple dc model
Equivalent Simplified Multi-Region Model
V IC IB 1 CE V AF
VBE>0.4V VBC<0
J A IB S E e Vt
Vt kT q
VBE
Forward Active
VBE=0.7V VCE=0.2V
IC=IB=0
IC<IB
Saturation
VBE<0
Cutoff
VBC<0
A small portion of the operating region is missed with this model but seldom operate in the missing region
Simplified dc model
Forward Active
C B E B IB IB E C
B
IB IB
C
B 0.6V
IB IB
C
E
E
Adequate when it makes little difference whether VBE=0.6V or VBE=0.7V
Simplified dc model
Forward Active
B C IB E E B 0.6V C IB IB IB
Mathematically VBE=0.6V IC=IB Or, if want to show slope in IC-VCE characteristics
VBE=0.6V IC=IB(1+VCE/VAF)
IB B 0.6V
IC IB
C
RDS =
RDS
VAF ICQ
RDS highly nonlinear
E
Simplified dc model
Equivalent Simplified Multi-Region Model
IC IB
VBE 0.6V
Vt kT q
VBE>0.4V VBC<0
Forward Active
VBE=0.7V VCE=0.2V
IC=IB=0
IC<IB
Saturation
VBE<0
Cutoff
VBC<0
A small portion of the operating region is missed with this model but seldom operate in the missing region
Conditions for Regions of Operation in Multi-Region Model
VBE>0.4V
VBC<0 IC<IB VBE<0 VBC<0 Note: One condition is on dependent variables !
Forward Active
Saturation
Cutoff
Observe that in saturation, IC<IB Can't condition on independent variables in saturation because they are fixed in the model
Regions of Operation in Independent Parameter Domain used In multi-region models
Seldom operate in regions excluded in this picture
Excessive Power Dissipation if either junction has large forward bias
Safe regions of operation
Actually cutoff, forward active, and reverse active regions can be extended modestly as shown and multi-region models still are quite good
Sufficient regions of operation for most applications
Example: Determine IC and VOUT
12V
4K 500K IC VOUT AE=100u2
J s =10-16 A/ 2 100
Example: Determine IC and VOUT , assume C is large and VIN is very small.
12V
4K 50K IC VOUT AE=100u2
J s =10-16 A/ 2 100
Example: Determine IC and VOUT.
12V
Assume C is large and VIN is very small.
4K 500K IC C VOUT AE=100u2 VIN
J s =10-16 A/ 2 100
End of Lecture 20