* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Mathcad - HW3_ECE427_soln
Survey
Document related concepts
Cavity magnetron wikipedia , lookup
Integrating ADC wikipedia , lookup
Josephson voltage standard wikipedia , lookup
Nanofluidic circuitry wikipedia , lookup
Wilson current mirror wikipedia , lookup
Operational amplifier wikipedia , lookup
Charlieplexing wikipedia , lookup
Schmitt trigger wikipedia , lookup
Valve RF amplifier wikipedia , lookup
Resistive opto-isolator wikipedia , lookup
Power electronics wikipedia , lookup
Switched-mode power supply wikipedia , lookup
Voltage regulator wikipedia , lookup
Current source wikipedia , lookup
Power MOSFET wikipedia , lookup
Surge protector wikipedia , lookup
Current mirror wikipedia , lookup
Transcript
ECE 427 Homework #3 Solutions 1. Question 12.6 on page 632. What is a static VAR compensator (SVC)? An answer is on pages 612-613. Important points to address are the circuit diagram, the combination of a TSC and a TCR, what it does for the power system (stabilize voltage levels in a fast and continuously controllable fashion), and the role of reactive power in its operation. 2. Problem 2.1 on page 73. 2.1 The reverse recoery time of a diode is trr=5μs, and the rate of fall of the diode current is di/dt=80A/μs. If the softness factor is SF=0.5, determine a. the storage charge QRR and b. the peak reverse recovery current IRR. A trr 5 μs rate 80 SF 0.5 For SF=0, 1 2 QRR rate trr 1000 μC 2 μs But SF=0.5. Deriving a value for ta , tb = SF ta ta tb = trr ta SF t a = trr ta trr 1 SF 3.333 μs Solving for IRR from the di/dt value given, IRR ta rate 266.667 A QRR is the area under the discharge curve, 1 QRR IRR trr 666.667 μC 2 IRR trr rate 400 A 3. Problem 2.6 on page 74. Two diodes are connected in series as shown in Figure 2.11 and the voltage across each diode is maintained the same by connecting a voltage-sharing resistor, such that V D1 =VD2 =2000V and R1 =100kOhms. The v-i characteristics of the diodes are shown in Figure P2.6. determine the leakage currents of each diode and the resistance R2 across diode D2 . VD1 2 kV VD2 2 kV R1 100 kΩ From Figure P2.6, at 2kV, Is1 8 mA Is2 16 mA From Figure 2.11, VD1 IR1 20 mA R1 By KCL, we find the current in R2 , IR2 IR1 Is1 Is2 12 mA R2 VD2 IR2 R2 166.667 kΩ It is possible to interpret the problem as having Is1 16 mA Is2 8 mA From Figure 2.11, VD1 IR1 20 mA R1 By KCL, we find the current in R2 , IR2 IR1 Is1 Is2 28 mA R2 VD2 IR2 R2 71.429 kΩ The direction of currents shown in Figure 2.11 can make this problem tedious. Be careful. 4. Problem 2.26 on page 76 A diode circuit is shown in Figure P2.26 with R=10Ω, L=5mH, and Vs=220V. If a load current of 10A is flowing thriough freewheeling diode Dm and switch S1 is closed at t=0, determine the expression for the current i through the switch. R5 10 L5 0.005 Vs 220 IDm 10 With the switch open initially, the inductor current is equal to the diode current. IL0 IDm 10 Closing the switch, where u(t) is a unit step function, d L5 iL R5 iL = Vs u ( t) dt Take a LaPlace transform of the differential equation, Vs L5 s R5 IL( s) L5 IL0 = s Partial fraction expansion, Vs L5 IL0 s IL( s) = s L5 s R5 Vs L5 IL0 s s L5 s R5 K2 K1 IL( s) = L5 s R 5 s = K1 L5 s R5 K2 s s L5 s R5 Vs K1 = R5 Vs = K1 R5 L5 IL0 = K1 L5 K2 L5 IL0 = Write the solution , Vs R5 IL( s) = s Vs R IL0 5 s R5 L5 Vs L5 R5 K2 K2 = L5 IL0 Vs L5 R5 Back to time domain, t L5 Vs Vs R5 iL( t) IL0 e R5 L5 4 τ5 5 10 R5 R5 Substituting into the expression for iL (t) Vs R5 22 1 IL0 10 τ5 3 2 10 iL( t) 22 12 e A 2000 t 20 iL( t) 10 0 0 4 5 10 3 1 10 3 1.5 10 3 2 10 t 6. Question 9.7 on page 500. 9.7 What are the means of turning on thyristors? An answer is on page 449 of the textbook. This explains the five important ways to turn on a thyristor: a. Gate pulse b. Light pulse c. Excessive forward voltage d. Excessive dv/dt between anode and cathode e. Too high temperature 6. How does a thyristor turn off? Show a diagram revealing charge concentrations to explain your answer. To turn off, we must evaucate the gate region of excess charge carriers. We can do this by natural commutation, reverse bias turnoff, or gate turnoff. In a sequence of snapshots of the excess charge carrier concentrations, np pn pn np Anode Cathode np pn pn np Anode Cathode np pn pn np Anode Cathode np pn pn Anode np Cathode 7. Questions 9.16, 9.18, and 9.21 on page 500. 9.16 What is the difference between an SC and a TRIAC? The SCR has a unidirectional current flow. The TRIAC is bidirectional and can be turned on with a gate signal in either polarity. 9.18 What are the advantages and disadvantages of a GTO Thyristors? The main advantage is the ability to turn off. The main disadvantage is the low reverse current gain during turnoff. Other advantages are the same advantages as an SCR. Other disadvantages are the long tail curent and the minimum sustaining voltage and current. The GTO is more expensive and does not have quite the peak voltage and current range overall as the SCR. 9.21 What are the advantages and disadvantages of an LASCR Thyristor? The main advantage is the optical triggerring and the isolation capability that is brings. The main disadvantage is the speed and lack of an ability to turn off optically. The other characteristics are the same as an SCR. 7. Question 9.28 on page 500. 9.28 What are the common techniques for voltage sharing of series connected thyristors? An answer is on pages 475 and 476. The resistors provide steady state voltage balancing when the thyristor block while the interaction of resistors and capacitors provide dynamic voltage balancing. Another problem not in the homework set provided for your study examples. Problem 9.8 on page 502. Two thyristors are connected in parallel to share a total load current of IL =500A. The on-state voltage drop of one thyristor is VT1 =1.0V at 300A and that of the other thyristor is VT2 =1.5V at 300A. Determine the values of series resistances to force current sharing with 10% difference. Total voltage v=2.5V. VT1 1.0 V VT2 1.5 V IT 500 A IT share ITmax 1 262.5 A 2 2 v 2.5 V share 10 % IT share ITmin 1 237.5 A 2 2 If the thyristors have the give currents and the given voltages, we want to make the sum of voltage across them to be equal. We use a model of the thyristor as a constant forward voltage. Therefore, series resistors much make up the difference in voltage to get an overallvoltage of 2.5 Volts. 3 mΩ 10 Ω VT1 ITmax RT1 = v RT1 v VT1 ITmax VT2 ITmax RT2 = v 5.714 mΩ RT2 v VT2 ITmax 3.81 mΩ