Problem 1 Rp Vt = 1.2 V K = 0.02 mA/V² Voo VDD = 8 V Rp =2K
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- Use only simple circuit techniques to solve for Vout and IoutPropose Snubber circuits that will take care of dv/dt, di/dt protection and provide Trapped energy recovery and save against the inductive kick from the load?is the simplest way of producing a source of lower EME from a source of higher EMF.
- For the circuit in the figurea) Determine the time constant.b) Write the mathematical expression for IL, VL and VR, after the switch is closed.c) Determine IL, VL for one, three and five time constants.d) Draw the waveforms of IL, VL and VR.Vth=.....2..v. Rh =... IL=Vth / (Rth + RL) PL = IL? * RL %3D Practical Simulation Theoretical, RL(0) IL PL IL PL IL PL 96.4 77.7m 65.5 m 56.7m 49.8 m 25 50 75 100 125 44.5m 40.33m 36.6 m 33.7m 150 175 200Known Vsat for LM741 ± 14V for source voltage Vs ± 15V. If Vi1 = 3 V and Vi2 = 4 V, Ri1 = Ri2 = Ro2 = 1 K Ohms. Define what is the value of Vo? how much
- You are required to design a two types of load cells. A Cantilever beam load cell and a Hollow cylinder load cell. What equations would you use to measure strain and voltage.A step-up chopper has input voltage of 220 V and output voltage of 660 V. If the conducting time of the IGBT based chopper is 100 micro- sec, then duty cycle will be a. -0.66 b. 0.7 c. 1.66 d. 0.66Which of the following statement is true? a. Angle between load voltage and load current in a pure capacitive load is 45 degree. b. Angle between load voltage and load current in a pure capacitive load is is 180 degree. c. Angle between load voltage and load current in a pure capacitive load is is zero. d. Angle between load voltage and load current in a pure capacitive load is is 90 degree.
- The solution of (3y2²e3=y – 1)dx + (2yešzy + 3ry²e*#v)dy = 0, y(0) = 1 is %3D Select one: a. y'e3ry – y = 4 O b. y?e3ry – y = 1 O . y?e3ry – x = 4 O d. y?e3ry - x = 1Since R1=18.10Kohm, R2=67.58Kohm, RC= 1.08Kohm, RE= 3.23Kohm, VCC=14.00V, Beta=241.00 in the circuit given in the figure, calculate the IC current by doing a complete analysis. When performing your operations, 2 steps will be taken after the point. choose the closest one from the stylish ones according to the +/-10% margin of error. There is only 1 correct answer to the question.We want to investigate how the field strength will be with air as dielectric and with steatite. A plate capacitor is placed in a 24 kV network between phase and ground. The relative permittivity of air is &=1 and for steatite &-6. The plate capacitor has area = 1m². The distance between the electrodes is 2 mm. Ep = 8.854-10-12 F/m A=1m² d=2mm a) Calculate the maximum field strength in the plate capacitor. (Answer: 9.8kV/mm) b) What is the capacitance of the capacitor if we use steatite? (Answer: C = 26.6nF)