EE 330 Lecture 15 Devices in Semiconductor Processes • Diodes • Capacitors • MOSFETs Review from Last Lecture Basic Devices and Device Models • Resistor • Diode • Capacitor • MOSFET • BJT Review from Last Lecture Analysis of Nonlinear Circuits (Circuits with one or more nonlinear devices) What analysis tools or methods can be used? KCL ? Nodal Analysis KVL? Mesh Analysis Superposition? Two-Port Subcircuits Voltage Divider ? Current Divider? Thevenin and Norton Equivalent Circuits? Review from Last Lecture Diode Models Diode Characteristics Diode Characteristics 0.01 0.008 0.01 0.008 0.006 0.006 0.004 Id (amps) 0.004 Id (amps) 0.002 Id (amps) 0.002 0 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Vd (volts) Vd (volts) ID VD Which model should be used? The simplest model that will give acceptable results in the analysis of a circuit Review from Last Lecture Diode Model Summary Piecewise Linear Models ID Idd = 0 if V < 0 V Vdd =0 if I > 0 D Diode Characteristics 0.01 0.008 I = 0 if V < 0.6V 0.006 0.004 Id (amps) dd Id (amps) 0.002 0 V =0.6V if I > 0 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 dd Vd (volts) Idd = 0 if V < 0.6 Vd =0.6+I d R d if I d > 0 Diode Equation Vd I = I eVt -1 dS Review from Last Lecture Diode Model Summary Piecewise Linear Models Idd = 0 if V < 0 Vdd =0 if I > 0 Idd = 0 if V < 0.6V Vdd =0.6V if I > 0 Diode EquationIdd = 0 if V < 0.6 Vd =0.6+I d R d if I d > 0 When is the ideal model adequate? Vd V WhenI it = doesn’t I e t make -1 much difference whether Vd=0V or Vd=0.6V dS When is the second piecewise-linear model adequate? When it doesn’t make much difference whether Vd=0.6V or Vd=0.7V Review from Last Lecture Example: Determine IOUT for the following circuit 10K IOUT 12V D1 Solution: Strategy: 1. Assume PWL model with VD=0.6V, RD=0 2. Guess state of diode (ON) 3. Analyze circuit with model Select Select Model 4. Validate state of guess in step 2 (verify the “if” condition in model) 5. Assume PWL with VD=0.7V 6. Guess state of diode (ON) 7. Analyze circuit with model Model 8. Validate state of guess in step 6 (verify the “if” condition in model) Validate 9. Show difference between results using these two models is small 10. If difference is not small, must use a different model Review from Last Lecture Solution: 1. Assume PWL model with VD=0.6V, RD=0 2. Guess state of diode (ON) 10K IOUT 12V 0.6V 3. Analyze circuit with model 12V-0.6V I = 1 . 14mA OUT 10K 4. Validate state of guess in step 2 To validate state, must show ID>0 ID =I OUT =1.14mA>0 Review from Last Lecture Solution: 5. Assume PWL model with VD=0.7V, RD=0 6. Guess state of diode (ON) 10K IOUT 12V 0.7V 7. Analyze circuit with model 12V-0.7V I = 1 . 13mA OUT 10K 8. Validate state of guess in step 6 To validate state, must show ID>0 ID =I OUT =1.13mA>0 Review from Last Lecture Solution: 9. Show difference between results using these two models is small are close IOUT =1.14mA and I OUT =1.13 mA Thus, can conclude IOUT 1.14mA Example: Determine IOUT for the following circuit 10K IOUT 0.8V D1 Solution: Strategy: 1. Assume PWL model with VD=0.6V, RD=0 2. Guess state of diode (ON) 3. Analyze circuit with model 4. Validate state of guess in step 2 5. Assume PWL with VD=0.7V 6. Guess state of diode (ON) 7. Analyze circuit with model 8. Validate state of guess in step 6 9. Show difference between results using these two models is small 10. If difference is not small, must use a different model Solution: 1. Assume PWL model with VD=0.6V, RD=0 2. Guess state of diode (ON) 10K IOUT 0.8V 0.6V 3. Analyze circuit with model 0.8 - 0.6V I= 20 A OUT 10K 4. Validate state of guess in step 2 To validate state, must show ID>0 ID =I OUT =20 A>0 Solution: 5. Assume PWL model with VD=0.7V, RD=0 6. Guess state of diode (ON) 10K IOUT 0.8V 0.7V 7. Analyze circuit with model 0.8V-0.7V I= 10 A OUT 10K 8. Validate state of guess in step 6 To validate state, must show ID>0 ID =I OUT =10 A>0 Solution: 9. Show difference between results using these two models is small IOUT =10 A and I OUT =20 A are not close 10. If difference is not small, must use a different model Thus must use diode equation to model the device 10K VD 0.8-VD IOUT I=OUT 10K 0.8V VD 0.6V Vt IOUT =I S e Solve simultaneously, assume Vt=25mV, IS=1fA Solving these two equations by iteration, obtain VD= 0.6148V and IOUT=18.60μA Use of Piecewise Models for Nonlinear Devices when Analyzing Electronic Circuits Process: 1. Guess state of the device 2. Analyze circuit 3. Verify State 4. Repeat steps 1 to 3 if verification fails 5. Verify model (if necessary) Observations: o Analysis generally simplified dramatically (particularly if piecewise model is linear) o Approach applicable to wide variety of nonlinear devices o Closed-form solutions give insight into performance of circuit o Usually much faster than solving the nonlinear circuit directly o Wrong guesses in the state of the device do not compromise solution (verification will fail) o Helps to guess right the first time o Detailed model is often not necessary with most nonlinear devices o For practical circuits, the simplified approach usually applies Key Concept For Analyzing Circuits with Nonlinear Devices Use of Piecewise Models for Nonlinear Devices when Analyzing Electronic Circuits Process: 1. Guess state of the device 2. Analyze circuit 3. Verify State 4. Repeat steps 1 to 3 if verification fails 5. Verify model (if necessary) What about nonlinear circuits (using piecewise models) with time-varying inputs? 1K Vout D1 80Vsin500t 1K Same process except state verification (step 3) may include a range where solution is valid Example: Determine V for V =80sin500t OUT IN 1K Vout VIN D1 80Vsin500t 1K Guess D1 ON (will use ideal diode model) 1K ID Vout D1 VIN 1K VOUT=VIN=80sin(500t) VIN Valid for I >0 ID D 1K Thus valid for VIN > 0 Example: Determine V for V =80sin500t OUT IN 1K Vout VIN D1 80Vsin500t 1K Guess D1 OFF (will use ideal diode model) 1K VD Vout D1 VIN 1K VOUT=VIN/2=40sin(500t) V Valid for V <0 V IN D D 2 Thus valid for VIN < 0 Example: Determine V for V =80sin500t OUT IN 1K Vout VIN D1 80Vsin500t 1K Thus overall solution VIN 80V t 80sin 500t for VIN 0 VOUT 40sin 500t forVIN 0 VOUT 80V t -40V Use of Piecewise Models for Nonlinear Devices when Analyzing Electronic Circuits Process: 1. Guess state of the device 2. Analyze circuit 3. Verify State 4. Repeat steps 1 to 3 if verification fails 5. Verify model (if necessary) What about circuits (using piecewise models) with multiple nonlinear devices? 1K 20V Vout 80V D1 D2 4K Guess state for each device (multiple combinations possible) Example: Obtain VOUT 1K 20V Vout 80V D1 D2 4K Example: Obtain VOUT Guess D1 and D2 on 1K 20V Vout 80V ID1 I D1 D2 D2 1K 20V Vout 4K 80V D VOUT=-20V 1 D2 Valid for ID1>0 and ID2>0 20V 4K I 50 mA D2 4K Since validates, solution is valid 80V I I 85 mA 0 DD121K Use of Piecewise Models for Nonlinear Devices when Analyzing Electronic Circuits Single Nonlinear Device Process: 1. Guess state of the device 2. Analyze circuit 3. Verify State 4. Repeat steps 1 to 3 if verification fails 5. Verify model (if necessary) Process: Multiple Nonlinear Devices 1. Guess state of each device (may be multiple combinations) 2. Analyze circuit 3. Verify State 4. Repeat steps 1 to 3 if verification fails 5. Verify models (if necessary) Analytical solutions of circuits with multiple nonlinear devices are often impossible to obtain if detailed non-piecewise nonlinear models are used Types of Diodes pn junction diodes Id I Id Id d Id Id Vd Vd Vd Vd Vd Vd Pin or Light Emitting Signal or Zener Varactor or Rectifier Photo LED Varicap Laser Diode Metal-semiconductor junction diodes Id Id Id Vd Vd Vd Schottky Barrier Basic Devices and Device Models • Resistor • Diode • Capacitor • MOSFET • BJT Capacitors • Types – Parallel Plate – Fringe – Junction Parallel Plate Capacitors A2 C A1 cond1 cond2 d insulator A = area of intersection of A1 & A2 One (top) plate intentionally sized smaller to determine C A C d Parallel Plate Capacitors Cap If C d unit area ε A C d C CdA where ε C d d Fringe Capacitors C d ε A C d A is the area where the two plates are parallel Only a single layer is needed to make fringe capacitors Fringe Capacitors C Capacitance Junction Capacitor C V p D d d n depletion A region C d Note: d is voltage dependent -capacitance is voltage dependent CjoA φB C n for VFB -usually parasitic caps V 2 D -varicaps or varactor diodes exploit 1 φB voltage dep.