EE 330 Lecture 12 evices in Semiconuctor Processes ioes
Review from Last Lecture http://www.ayah.com/perioic/mages/perioic%20table.png
Review from Last Lecture
Review from Last Lecture Silicon opants in Semiconuctor Processes B (Boron) wiely use a opant for creating p-type regions P (Phosphorus) wiely use a opant for creating n-type regions (bulk oping, iffuses fast) As (Arsenic) wiely use a opant for creating n-type regions (Active region oping, iffuses slower)
ioes (pn junctions) epletion region create that is ionize but voi of carriers
pn Junctions Physical Bounary Separating n-type an p-type regions f oping levels ientical, epletion region extens equally into n-type an p-type regions
pn Junctions Physical Bounary Separating n-type an p-type regions Extens farther into p-type region if p-oping lower than n-oping
pn Junctions Physical Bounary Separating n-type an p-type regions Extens farther into n-type region if n-oping lower than p-oping
pn Junctions Positive voltages across the p to n junction are referre to forwar bias Negative voltages across the p to n junction are referre to reverse bias As forwar bias increases, epletion region thins an current starts to flow Current grows very rapily as forwar bias increases Current is very small uner revere bias
pn Junctions Anoe Anoe Cathoe Cathoe Circuit Symbol
pn Junctions As forwar bias increases, epletion region thins an current starts to flow Current grows very rapily as forwar bias increases Anoe Cathoe Simple ioe Moel: =0 >0 =0 <0 Simple moel often referre to as the eal ioe moel
pn Junctions Simple ioe Moel: pn junction serves as a rectifier passing current in one irection an blocking it n the other irection
Rectifier Application: 1 OUT Simple ioe Moel: N 1K N = M sinωt M N t OUT M t
- characteristics of pn junction mprove ioe Moel: (signal or rectifier ioe) S in the 10fA to 100fA range kt = t q ioe Equation t e 1 S What is t at room temp? t is about 26m at room temp k= 1.380 6504(24) 10 23 JK -1 q = 1.602176487(40) 10 19 C k/q=8.62 10 5 K -1 ioe equation ue to William Schockley, inventor of BJT n 1919, William Henry Eccles coine the term ioe n 1940, Russell Ohl stumble upon the p-n junction ioe
(amps) - characteristics of pn junction mprove ioe Moel: (signal or rectifier ioe) ioe Characteristics 0.01 0.008 0.006 0.004 ioe Equation Uner reverse bias ( <0), Uner forwar bias ( >0), t e 1 S Simplification of ioe Equation: S Se t 0.002 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (volts) S in the 10fA to 100fA range kt = t q k= 1.380 6504(24) 10 23 JK -1 q = 1.602176487(40) 10 19 C k/q=8.62 10 5 K -1 t is about 26m at room temp Simplification essentially ientical moel except for very close to 0 ioe Equation or forwar bias simplification is unwiely to work with analytically
- characteristics of pn junction mprove ioe Moel: (signal or rectifier ioe) ioe Equation Simplification of ioe Equation: Uner reverse bias, Uner forwar bias, t e 1 S S S e t S often in the 10fA to 100fA range S proportional to junction area t is about 26m at room temp How much error is introuce using the simplification for > 0.5? t S e 1 Se t e 1 S t 1 05. 4. 4 10. 026 e How much error is introuce using the simplification for < - 0.5? 05. 026 e. 4. 4 10 Simplification almost never introuces any significant error 9 9
Will you impress your colleagues or your boss if you use the more exact ioe equation when < -0.5 or > +0.5? Will your colleagues or your boss be unimpresse if you use the more exact ioe equation when < -0.5 or > +0.5?
pn Junctions Anoe Cathoe ioe Equation: (goo enough for most applications) JSAe 0 n T 0 0 Note: S =J s A J S = Sat Current ensity (in the 1aA/u 2 to 1fA/u 2 range) A= Junction Cross Section Area T =kt/q (k/q=1.381x10-23 C/ K/1.6x10-19 C=8.62x10-5 / K) n is approximately 1
pn Junctions ioe Equation: J 0 S Ae n T 0 0 Anoe J S is strongly temperature epenent With n=1, for >0, Cathoe - G0 (T) J T m e Ae t t SX Typical values for key parameters: J SX =0.5A/μ 2, G0 =1.17, m=2.3
Example: pn Junctions - G0 (T) J m t T e Ae SX t What percent change in S will occur for a 1 C change in temperature at room temperature? - -G0 - - - G0 G0 G0 G0 (T ) t T1 (T ) (T ) m m m m t T1 t 2 t t 2 t 2 J T e Ae - J T e Ae T e - T e SX T SX T T T @ 1 2 1 S S -G0 - - G0 G0 m t T1 (T ) m t T1 t 2 J T e Ae T e SX T T 1 1 15 15 1 240x10 1 025x10 15 1. 025x10 S. -. 100% 21% S
pn Junctions Anoe Cathoe ioe Equation: (goo enough for most applications) JSAe 0 n T 0 0 S =J s A Simple ioe Moel: Often terme the conucting or ON state Often terme the nonconucting or OFF state
Consier again the basic rectifier circuit OUT N R Previously consiere sinusoial excitation Previously gave qualitative analysis Rigorous analysis metho is essential? O U T
Analysis of Nonlinear Circuits (Circuits with one or more nonlinear evices) What analysis tools or methos can be use? KCL? KL? Superposition? Noal Analysis Mesh Analysis Two-Port Subcircuits oltage ivier? Current ivier? Thevenin an Norton Equivalent Circuits?
Consier again the basic rectifier circuit OUT N R N OUT R R t e 1 S OUT S R e N t O U T 1 Even the simplest ioe circuit oes not have a close-form solution when ioe equation is use to moel the ioe!! ue to the nonlinear nature of the ioe equation Simplifications are essential if analytical results are to be obtaine
(amps) Lets stuy the ioe equation a little further t S e 1 ioe Characteristics 10000 8000 6000 4000 2000 0 0 0.2 0.4 0.6 0.8 1 Linear-Linear Axis (volts) Power issipation Becomes estructive if > 0.85 (actually less)
(amps) Lets stuy the ioe equation a little further t e 1 S ioe Characteristics 10000 100 1 0.01 0.0001 1E-06 1E-08 1E-10 1E-12 Linear-Log Axis 0 0.2 0.4 0.6 0.8 1 (volts) For two ecaes of current change, is close to 0.6 This is the most useful current range for many applications
(amps) Lets stuy the ioe equation a little further t e 1 S ioe Characteristics 0.01 0.008 0.006 0.004 0.002 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (volts) For two ecaes of current change, is close to 0.6 This is the most useful current range when conucting for many applications
(amps) Lets stuy the ioe equation a little further t e 1 S ioe Characteristics 0 0.6 0.6 0 0.01 0.008 0.006 0.004 0.002 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (volts) Wiely Use Piecewise Linear Moel
(amps) Lets stuy the ioe equation a little further t e 1 S ioe Characteristics 0.01 0.008 0.006 0.004 0.002 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (volts) Better moel in ON state though often not neee nclues ioe ON resistance
Lets stuy the ioe equation a little further t e 1 S Piecewise Linear Moel with ioe Resistance 0 if 0.6 R 0.6 0 (R is rather small: often in the 20Ώ to 100Ώ range): if Equivalent Circuit A C A C Off State A C 0.6 R On State
The eal ioe 0 if 0 0 if 0
The eal ioe 0 if 0 0 if 0 OFF ON ON OFF ali for >0 0
(amps) (amps) ioe Moels (amps) ioe Characteristics 0.01 0.008 0.006 0.004 0.002 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (volts) ioe Characteristics 0.01 0.008 0.006 0.004 0.002 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (volts) ioe Characteristics 0.01 0.008 0.006 0.004 0.002 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (volts) Which moel shoul be use? The simplest moel that will give acceptable results in the analysis of a circuit
(amps) (amps) (amps) ioe Moels ioe Characteristics ioe Equation t e 1 S 0.01 0.008 0.006 0.004 0.002 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (volts) ioe Characteristics 0 0.6 if if 0.6 0 0.01 0.008 0.006 0.004 0.002 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (volts) ioe Characteristics Piecewise Linear Moels 0 0.6 R if if 0.6 0 0.01 0.008 0.006 0.004 0.002 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 (volts) 0 if 0 0 if 0
0 0 0 if if 0 ioe Moels ioe Equation 1 e t S Piecewise Linear Moels 0 0.6 0.6 0 0 R 0.6 0.6 0 When are the piecewise-linear moels aequate? When it oesn t make much ifference whether =0.6 or =0.7 is use When is the ieal PWL moel aequate? When it oesn t make much ifference whether =0 or =0.7 is use
Example: etermine OUT for the following circuit 10K OUT 12 1 Solution: Strategy: 1. Assume PWL moel with =0.6, R =0 2. Guess state of ioe (ON) 3. Analyze circuit with moel 4. aliate state of guess in step 2 (verify the if conition in moel) 5. Assume PWL with =0.7 6. Guess state of ioe (ON) 7. Analyze circuit with moel 8. aliate state of guess in step 6 (verify the if conition in moel) 9. Show ifference between results using these two moels is small 10. f ifference is not small, must use a ifferent moel
Solution: 1. Assume PWL moel with =0.6, R =0 2. Guess state of ioe (ON) 10K OUT 12 0.6 3. Analyze circuit with moel 12-0.6 = 1. 14mA OUT 10K 4. aliate state of guess in step 2 To valiate state, must show >0 = OUT =1.14mA>0
Solution: 5. Assume PWL moel with =0.7, R =0 6. Guess state of ioe (ON) 10K OUT 12 0.7 7. Analyze circuit with moel 12-0.7 = 1. 13mA OUT 10K 8. aliate state of guess in step 6 To valiate state, must show >0 = OUT =1.13mA>0
Solution: 9. Show ifference between results using these two moels is small =1.14mA an =1.13 ma are close OUT OUT Thus, can conclue OUT 1.14mA
Example: etermine OUT for the following circuit 10K 0.8 OUT 1 Solution: Strategy: 1. Assume PWL moel with =0.6, R =0 2. Guess state of ioe (ON) 3. Analyze circuit with moel 4. aliate state of guess in step 2 5. Assume PWL with =0.7 6. Guess state of ioe (ON) 7. Analyze circuit with moel 8. aliate state of guess in step 6 9. Show ifference between results using these two moels is small 10. f ifference is not small, must use a ifferent moel
Solution: 1. Assume PWL moel with =0.6, R =0 2. Guess state of ioe (ON) 10K 0.8 OUT 0.6 3. Analyze circuit with moel 0.8-0.6 = OUT 10K 20 A 4. aliate state of guess in step 2 To valiate state, must show >0 = =20A>0 OUT
Solution: 5. Assume PWL moel with =0.7, R =0 6. Guess state of ioe (ON) 10K 0.8 OUT 0.7 7. Analyze circuit with moel 0.8-0.7 = OUT 10K 10 A 8. aliate state of guess in step 6 To valiate state, must show >0 = =10A>0 OUT
Solution: 9. Show ifference between results using these two moels is small =10A an =20A are not close OUT OUT 10. f ifference is not small, must use a ifferent moel Thus must use ioe equation to moel the evice OUT OUT 0.8- = 10K = e S t 0.8 10K OUT 0.6 Solve simultaneously, assume t =25m, S =1fA Solving these two equations by iteration, obtain = 0.6148 an OUT =18.60μA
En of Lecture 12