For the circuit shown in Figure 1, calculate: 1- Voc, Isc, RTH, and Pmax from load side, i.e., between points 'a' and 'b' R₁-10 kn R₁ = 1 k R₂-33 kn E₁-10V R₁-22 kn VL R₁ =47 kn E₂-5V b Figure 1
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- Design of a DC-DC Converter. The Circuit Below shows a basic Buck-Boost Converter with a highly inductive Load. With a Deliberate mistake in the configuration of the Circuit Connection. The Value of L1 is 22mH, C1 is 10uF, R1 is 220 and L2 is 10mH. The circuit operates with a switching frequency of 20KHZ. The Duty cycle of the Clock signal V2 is given by the function below: ( 17 ) х0.2 Duty Cycle, k = 0. 55 + 100 Your Unique Duty Cycle = %3D Buck-Boost Converter Q1 IRG4BC10U D1 R1 222 1N4009 C1 LV1 12V L1 22mH 10µF L2 10mH a) Find the mistake in the configuration of the circuit and correct the connection error.Draw a timing diagram for the circuit in Figure 2.24a. Show the waveforms that can be observed on all wires in the circuit. X2 f X3 X1 (a) A minimal sum-of-products realization Repeat Problem 2.8 for the circuit in Figure 2.24b. X1 X3 X2 (b) A minimal product-of-sums realizationPropose 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?
- 1.28 a b c dQ1. a) Figure Q1 is a single phase 2-level voltage source converter (VSC) with a total DC voltage V-1000V. A PWM scheme, by comparing a reference value with a 1kHz triangular waveform, is used to control the switches S₁ and S2, and output 340v at Vo. With the aid of graphs explain how pulse signals for the switches are generated, sketch the output voltage Vo, determine the duty cycle and the modulation index. Va + 2 it Va d + TS₁ V {1+5₂ = Figure Q1 A single phase VSCQ3) A. A step down de chopper is connected to an R=52 and L=10 mH. The de supply voltage is 100 V. The chopper is switching at a frequency of 1 kHz with a duty cycle of 50%. Determine the load current, lo and the peak to peak ripple current as an absolute value and as percentage of de value.
- In the circuit of the following figure, the input voltage Vs is 15 volts rms with a frequency of 60 Hz, R equals 150 Ohms and C equals 100,000 Pico Farads. The diodes are Germanium (Vd = 0.2 volts) and the Zener diode is 12 volts. a) The magnitude of the ripple voltage at Cb) The Magnitude of the Peak Inverse Voltage (PIV) for D1 and D2.solar cells short circuit current, Isc-5A and open circuit voltage, vo.c.-D0.6. * The power delivered by a solar cell is Between 2.295 & 2.565 Between 2.565 & 2.6 ) Between 2.04 & 2.565Can someone please provide a step by step solution? Thank you! The full wave rectifier in the circuit below is to deliver 0.25A and 15V (peak) to a load. The ripple is to be no larger than 0.4 V peak-to-peak. The input signal is 120 V (rms) at 60 Hz. Assume V(lambda) = 0.7 V . Determine the required turns ratio, the filter capacitance, and the diode PIV rating.
- Helli. I only need the last part. Plotting power vs voltage if used as a solar cell and give the definition of fill factor. TnxFAIRCHILD Discrete POWER & Signal Technologies SEMICONDUCTOR ru 1N4001 - 1N4007 Features • Low torward voltage drop. 10 a14 * High aurge eurrent cepablity. 0.160 4.06) DO 41 COLOR BAND DGNOTEs CAT-Cos 1.0 Ampere General Purpose Rectifiers Absolute Maximum Ratings T-26*Cuness atnerwioe rated Symbol Parameter Value Units Average Recttied Current 1.0 375" lead length a TA - 75°C Tsargei Peak Forward Surge Current 8.3 ms single halr-sine-wave Superimposed on rated load JEDEC method) 30 A Pa Total Device Dissipetion 2.5 20 Derste above 25°C Ra Tag Thermal Resistence, Junction to Amblent 5D Storage Temperature Range 55 to +175 -55 to +150 Operating Junetion Temperature PC "These rarings are imithg valuee above whien the serviceatity or any semiconductor device may te impaired. Electrical Characteristics T-20'Cunieas ofherwise roted Parameter Device Units 4001 4002 4003 4004 4005 4006 4007 Peak Repetitive Reverse Vellage Maximum RME votage DC Reverse Voltage Maximum Reverse Current @ rated VR…1) AC power in a load can be controlled by using a. Two SCR's in parallel opposition b. Two SCR's in series c. Three SCR's in series d. Four SCR's in series 2) The advantage of using free - wheeling diode in half controlled bridge converter is that a. There is always a path for the ac current independent of the ac line b. There is always a path for the dc current independent of the ac line c. There is always a path for the dc current dependent of the ac line d. There is always a path for the ac current dependent of the ac line 3) Silicon controlled rectifier can be turned on a. By applying a gate pulse and turned off only when current becomes zero b. And turned off by applying gate pulse c. By applying a gate pulse and turned off by removing the gate pulse d. By making current non zero and turned off by making current zero