The collector current Ic (see the circuit below) is:
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A: Given G1(w,x,y,z) = Σm(0,1,2,3,7,8,10) + Σd(5,6,11,15). Derive the POS and SOP.
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- What is CEMF?Repeat example 35 for FWD and firing angle Fa) 60°. 215 l65 D1985 Example 35: A full wave rectifier used 220V, with firing angle 10° ,total resistance load 5 KQ, inductance 2.34 H,and frequency 60HZ, Draw and calculate: (a) VD.c and Ipc (b) VD.C(Max) (c) Vn (d) Vorms-Q3: The figure below includes a bridge full-wave rectifier circuit with a diode type is1N4007. The input voltage is Vin V2 120 sin 2n60t 5:1 D3 D1 120 V RL Vp(out) D2 DA 10 kN a) Draw the output voltage waveform for the circuit in the figure and include thevoltage values. b) What is the peak inverse voltage (PIV) across each diode? c) Determine the rms voltage, current and power delivered to RL d) Determine the average voltage, current, and power delivered to RL e) What is the ratio of Po(de) to Po(ac)? lll
- Consider the circuit shown in the figure below with Ry=13 ko and R=9 kO. The Zener diode voltage is Vz = 3.5 V, then the currents Iz and , respectively, are equal to: R1 ww 20 V O a. 1.280 mA, 0.789 mA O b. 0.680 mA, 0.189 mA O c. 0.880 mA 0.389 mA O d. 1.080 mA, 0.589 mAIn a single phase half-wave controlled rectifier has 400 sinwt as the input voltage and R as the load.For the firing angle of 60 degrees for the SCR,the average output voltage is a. 300/pi b. 200/pi c. 240/pi d. 400/piDesign the power supply using diode components. This power supply sholud have 2 outputs. 12Vdc and 5Vdc. Input 240 Vac Supply i. Draw the circuit diagram ii. Bill of material/components that will be used iii. Teh output waveform for each stage
- The gate current of an SCR full-wave rectifier is adjusted to 1.2mA and the forward breakdown voltage of SCR corresponding to this gate current is 150V. The applied voltage is a sinusoidal voltage of 300V peak, the load resistance is 100 ohms and the holding current is zero. Determine the a) firing angle b) conduction angle c) average output voltage d) average current and e) average power (Our lesson is POWER ELECTRONIC DEVICES, in the lesson of INDUSTRIAL ELECTRONIC)In the circuit shown below. Let Vm=37 V and i,=21 mA and Voa -0.7 V: iD is C0.8 ko 3 ko Vmcos(wt)V For t = Os, the current in the diode equals: Oa. 2 mA Ob. 0 mA Oc. -2 mA Od. 4 mA Ift = T/4, then the current in the diode equals: Оа. 4.24 mA Оb. 3.39 mA Oc. 5.3 mA Od. 2.71 mAConsider the circuit shown in the figure below with R1=16 ko and R=8 klQ. The Zener diode voltage is Vz = 3.5 V, then the currents Iz and lu respectively, are equal to: Ri of ww RL 20 V O a. 0.594 mA, 0.438 mA Ob. 0.394 mA 0.237 mA Oc 0.794 mA, 0.637 mA O d. 0.994 mA, 0.838 mA
- FAIRCHILD 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…In the circuit shown below. Determine the following a. Vs (at the secondary) b. Vout (across RL) c. Vrip (ripple votage) d. VDC e. PIV (Peak Inverse Voltage) 10:1 Output 115 V mis 60 Hz RL 2.2 k) D, 50μF- All diodes are IN4001. Tund AlMannai EENG261 Page 5/11A 60 Hz sinusoidal voltage is applied to an SCR full-wave rectifier. If the DC output voltage is 74 V and the firing angle of the SCR is 60 degrees, the RMS voltage at the input would be: O A. 220 V O B.99 V O C. None of the other choices are correct OD. 110 V O E.256 V