EE 330 Lecture 13 evices in Semiconuctor Processes ioes Capacitors Transistors
Review from Last Lecture pn Junctions Physical Bounary Separating n-type an p-type regions Extens farther into n-type region if n-oping lower than p-oping
Review from Last Lecture pn Junctions Anoe Anoe Cathoe Cathoe Circuit Symbol
Review from Last Lecture 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
Review from Last Lecture - 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
Review from Last Lecture 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?
Review from Last Lecture 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
Lets stuy the ioe equation a little further t S e 1 ioe Characteristics 10000 8000 (amps) 6000 4000 2000 0 0 0.2 0.4 0.6 0.8 1 (volts) Power issipation Becomes estructive if > 0.85 (actually less)
Lets stuy the ioe equation a little further t e 1 S ioe Characteristics (amps) 10000 100 1 0.01 0.0001 1E-06 1E-08 1E-10 1E-12 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
Lets stuy the ioe equation a little further t e 1 S ioe Characteristics 0.01 (amps) 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 for many applications
Lets stuy the ioe equation a little further t e 1 S ioe Characteristics 0 0.6 0.6 0 0.01 0.008 (amps) 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
Lets stuy the ioe equation a little further t e 1 S ioe Characteristics 0.01 0.008 (amps) 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 0.6 R 0.6 0 (R is rather small: often in the 20Ώ to 100Ώ range): 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
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) (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 (amps) 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
ioe Moels ioe Characteristics ioe Equation t e 1 S (amps) 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 0.6 0 (amps) 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 0.6 0 (amps) 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 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 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
Use of Piecewise Moels for Nonlinear evices when Analyzing Electronic Circuits Process: Observations: 1. Guess state of the evice 2. Analyze circuit 3. erify State 4. Repeat steps 1 to 3 if verification fails o Analysis generally simplifie ramatically (particularly if piecewise moel is linear) o Approach applicable to wie variety of nonlinear evices o Close-form solutions give insight into performance of circuit o Usually much faster than solving the nonlinear circuit irectly o Wrong guesses in the state of the evice o not compromise solution (verification will fail) o Helps to Guess Right the first time
Types of ioes pn junction ioes Signal or Rectifier Pin or Photo Light Emitting LE Laser ioe Metal-semiconuctor junction ioes Zener aractor or aricap Schottky Barrier
Basic evices an evice Moels Resistor ioe Capacitor MOSFET BJT
Capacitors Types Parallel Plate Fringe Junction
Parallel Plate Capacitors A 2 con 2 con 1 A 1 C insulator A = area of intersection of A 1 & A 2 One (top) plate intentionally C A size smaller to etermine C
Parallel Plate Capacitors f C Cap unit area where C C C ε A C ε A
Fringe Capacitors C C ε A A is the area where the two plates are parallel Only a single layer is neee to make fringe capacitors
Fringe Capacitors C
Junction Capacitor Capacitance C C jo 1 φ p C A A B n n for FB φ 2 φ B 0.6 n ; 0.5 B C epletion region Note: is voltage epenent -capacitance is voltage epenent -usually parasitic caps -varicaps or varactor ioes exploit voltage ep. of C
Capacitance Junction Capacitor 1.6 1.4 1.2 C Cj0A 1 0.8 0.6 0.4-4 -3-2 -1 0 1 0.2 0 C C jo 1 φ A B n for FB φ 2 φ B 0.6 n ; 0.5 B oltage epenence is substantial
En of Lecture 13