330 Lecture 20 ipolar Device Modeling
xam 2 Friday March 9 xam 3 Friday April 13
Review from Last Lecture ipolar Transistors npn stack pnp stack ipolar Devices Show asic Symmetry lectrical Properties not Symmetric Designation of and critical npn transistor pnp transistor With proper doping and device sizing these form ipolar Transistors
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 ipolar Operation onsider npn transistor Forward Active Operation npn stack F 1 So, what will happen? F 2 Some will recombine with holes and contribute to base current and some will be attracted across junction and contribute to collector
Review from Last Lecture ipolar Operation onsider npn transistor Forward Active Operation β is typically very large α 1 defn 1 often 50<β<999
Review from Last Lecture ipolar Operation onsider npn transistor Forward Active Operation β is typically very large ipolar transistor can be thought of a current amplifier with a large current gain n contrast, MOS transistor is inherently a tramsconductance amplifier urrent flow in base is governed by the diode equation ollector current thus varies exponentially with ~ S e ~ S e t t
Review from Last Lecture ipolar Operation onsider npn transistor Forward Active Operation β is typically very large ollector current thus varies exponentially with ~ S e t This exponential relationship (in contrast to the square-law relationship for the MOSFT) provides a very large gain for the JT and this property is very useful for many applications!!
Review from Last Lecture Simple dc model npn transistor Forward Active Operation t ~ S e ~ S kt q e t t J J t kt q S S A β A e e t t J S is termed the saturation current density Process Parameters : J S,β Design Parameters: A nvironmental parameters and physical constants: k,t,q At room temperature, t is around 26m J S very small around.25fa/u 2
Review from Last Lecture Simple dc model npn transistor Forward Active Operation Output haracteristics 300 250 200 d 150 100 50 or 0 0 1 2 3 4 5 ds J S A e t
Review from Last Lecture Simple dc model Typical Output haracteristics 300 250 200 Saturation Forward Active d 150 or 100 50 0 0 1 2 3 4 5 ds utoff Forward Active region of JT is analogous to Saturation region of MOSFT Saturation region of JT is analogous to Triode region of MOSFT
etter Model of Output haracteristics or
JT and MOSFT omparison D or GS DS Same general characteristics Spacings a bit different (xponetial vs square law) Slope steeper for small compared to DS
etter Model of Output haracteristics or With scaled axis, transition in saturation very steep
JT and MOSFT omparison D or GS DS Same general characteristics Spacings a bit different (xponetial vs square law) Slope steeper for small compared to DS
Simple dc model Typical Output haracteristics 300 250 200 d 150 or 100 50 0 0 1 2 3 4 5 ds Projections of these tangential lines all intercept the axis at the same place and this is termed the arly voltage, AF (actually AF is intercept) Typical values of AF are in the 100 range an multiply expression for by term 1+ AF to account for slope
Simple dc model mproved Model 300 250 200 d 150 100 or 50 0 0 1 2 3 4 5 ds JA S = =J A e S t e t 1+ AF alid only in Forward Active Region
Simple JT dc model Typical Output haracteristics or
Simple JT dc model Typical Output haracteristics Saturation Forward Active or utoff
Simple JT dc model Typical Output haracteristics D Saturation Forward Active or Triode Saturation GS utoff utoff DS Forward Active region of JT is analogous to Saturation region of MOSFT Saturation region of JT is analogous to Triode region of MOSFT
mproved dc model 300 250 d 200 150 or t kt q 100 50 0 0 1 2 3 4 5 ds J A S t t e 1 J A 1 S e F J t SA t J 1 1 SA e e R alid in All regions of operation AF effects can be added Not mathematically easy to work with Note dependent variables changes Termed bers-moll model Reduces to previous model in FA region
mproved dc model Forward Active ncreasing (or ) ncreasing (or ) t kt q Reverse Active J A S t t e 1 J A 1 S e F J t SA t J 1 1 SA e e R alid in All regions of operation AF effects can be added Not mathematically easy to work with Note dependent variables changes Termed bers-moll model Reduces to previous model in FA region
Simplified Multi-Region Model 5 5 4 4 3 3 2 2 1 1 bers-moll Model SAT Simplified Multi-Region Model Observe around 0.2 when saturated around 0.6 when saturated n most applications, exact and voltage in saturation not critical Simplified model in saturation: =0.7 =0.2 Saturation
Simplified Multi-Region Model J S A JSA β e e t 1 t AF Forward Active t kt q =0.7 =0.2 Saturation = =0 utoff
Simplified Multi-Region Model J S A JSA β e e t 1 t AF >0.4 <0 Forward Active t kt q =0.7 =0.2 <β Saturation = =0 <0 <0 utoff A small portion of the operating region is missed with this model but seldom operate in the missing region
Simplified Multi-Region Model Alternate equivalent model β JSA β 1 e t AF >0.4 <0 Forward Active t kt q =0.7 =0.2 <β Saturation = =0 <0 <0 utoff A small portion of the operating region is missed with this model but seldom operate in the missing region
Further Simplified Multi-Region dc Model Forward Active β β 0.6 β Adequate when it makes little difference whether =0.6 or =0.7
Simplified Multi-Region dc Model Forward Active β 0.6 β Mathematically =0.6 =β Or, if want to show slope in - characteristics =0.6 =β (1+ / AF ) 0.6 β R DS R DS = β AF Q R DS highly nonlinear
Further Simplified Multi-Region dc Model quivalent Further Simplified Multi-Region Model β 0.6 t kt q >0.4 <0 Forward Active =0.7 =0.2 <β Saturation = =0 <0 <0 utoff A small portion of the operating region is missed with this model but seldom operate in the missing region
onditions for Regions of Operation in Multi-Region Model >0.4 <0 Forward Active <β Saturation <0 <0 utoff Note: One condition is on dependent variables! Observe that in saturation, <β an t condition on independent variables in saturation because they are fixed in the model
Regions of Operation in ndependent Parameter Domain used n multi-region models Saturation Forward Active 0.4 0.4 utoff Reverse Active Seldom operate in regions excluded in this picture
xcessive Power Dissipation if either junction has large forward bias Saturation Melt Down!! Forward Active Saturation 0.4 0.4 utoff Reverse Active
Safe regions of operation Simplified Forward Saturation Saturation Melt Down!! Forward Active Saturation 0.4 0.4 utoff Reverse Active
Simplified Forward Saturation Saturation Melt Down!! Forward Active Saturation 0.4 0.4 utoff Reverse Active 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 Simplified Forward Saturation Saturation Forward Active 0.4 0.4 utoff Reverse Active
xample: Determine and OUT 12 500K 4K OUT A =100u 2 J =10 s β 100 A/μ -16 2
xample: Determine and OUT 500K 12 4K Solution: 1. Guess Forward Active Region 2. Solve ircuit with Guess 12 OUT A =100u 2 J =10 s β 100 A/μ -16 2 4K 500K OUT OUT 0.6 100 12 0.6 500K 12 0.6 100 2.28mA 500K 12 4K 2.88 OUT 3. erify FA Region 0.6 0.4 0.6 2.88 2.28 0 erify Passes so solution is valid 2.28mA OUT 2.88 >0.4 and <0
xample: Determine and OUT, 12 50K 4K OUT A =100u 2 J =10 s β 100 A/μ -16 2
50K xample: Determine and OUT. 12 4K Solution: 1. Guess Forward Active Region 2. Solve ircuit with Guess 12 50K 4K OUT A =100u 2 OUT 0.6 100 OUT J =10 s β 100 A/μ -16 2 12 0.6 50K 12 0.6 100 22.8mA 50K 12 4K 79.2 OUT 3. erify FA Region >0.4 and <0 0.6 0.4 0.6 79.2 79.8 0 erify Fails so solution is not valid
50K xample: Determine and OUT 12 4K Solution: 4. Guess Saturation 5. Solve ircuit with Guess 12 OUT A =100u 2 50K OUT 0.6 0.2 4K OUT J =10 s β 100 A/μ -16 2 OUT 12 0.6 228 A 50K 12 0.2 2.95mA 4K 0.2 6. erify SAT Region <β 100 228A 22.8mA 2.95mA 2.95mA 22.8mA erify Passes so solution is valid 2.95mA 0.2 OUT
xample: Determine and OUT. Assume is large and N is very small. 12 500K 4K OUT A =100u 2 N J =10 s β 100 A/μ -16 2
xample: Determine and OUT. Assume is large and N is very small. 12 Solution: N 500K 4K OUT A =100u 2 J =10 s β 100 A/μ -16 2 Assume N =0, then no current flows through so circuit is identical to circuit of previous-previous example so 2.28mA 2.88 OUT Note: Since N is coupled directly to base, will amplify N if it is a small time varying signal and gain will be very large
nd of Lecture 20