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Range of semiconductor devices and circuits - Coursework Example

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Date: April 10, 2015 Question 1 The internal circuitry of an ohmmeter has a voltage source which is connected to its leads when it is being used to measure resistance. The existence of this voltage can be utilized to bias a diode without connecting it to a circuit…
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Range of semiconductor devices and circuits
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Download file to see previous pages A good diode should posses forward to reverse resistance ratio of 10:1 with others with the ratio up to 100:1. If a diode is open, it will show infinite resistance in both directions. On the other hand, a shorted resistance will show low resistance in both directions. The procedure for testing the diode using an ohmmeter only is as illustrated in the following steps. 1. The Ohmmeter is configured to measure on the lowest range. If it’s a digital ohmmeter, a 2-kohm range or the nearest equivalent is used. 2. The ohmmeter leads are connected across the diode with the positive red probe on the P-material and the negative black probe on the N- material as shown in the diagram below. 3. The measured resistance of the diode is measured; Rf is recorded after which the setup is disconnected. 4. After the disconnection, the leads of the ohmmeter are reversed and connected in the reverse direction i.e. the positive red probe is connected to the N – material on the diode and the negative black probe is connected to the P – material on the diode as indicated in the figure below. 5. The measured reverse resistance RR (reverse resistance is recorded) 6. The forward and the reverse resistance ratio is calculated as RR/RF 7. If the reverse resistance ratio is equal to one or very close to one, then the diode is faulty. In the same way the polarity of the diode can be identified. When the diode shows infinite resistance the material connecting to the positive red probe of the ohmmeter is the P – material of the diode. On the contrary, the part connected to the Negative black probe of the ohmmeter is the N – material of the Diode. Question3 VRMS = Vm*0.7071 = 100*0.7071 =70.71 Question 4 a) Ripple factor = Vrms/Vdc Capacitance = 50micro Farad 50*80*10-6 =4000*10-6 4*10-3 (4*10-3)*2500 10 10*2 = 20 20* v3 = 34.64 34.64*4 = 138.56 1/138.56 = 0.0072 Ripple factor = 0.0072*100% = 0.72% b) 50*100*10-6 5*10-3 (5*10-3)*12000 60,000*10-3 =60 60*2 = 120 120* v3= 207.84 207.84*4 = 831.38 1/831.38 = 0.0012 Ripple factor = 0.0012*100% = 0.12% Question 5 Block diagram of a complete power supply Analyze the operation of different types of amplifier Summing up amplifier This type of amplifier is used to add two voltages. The output voltage is equivalent to the sum of the input voltages. With summing up amplifiers, the output voltage cannot be greater than the power supply in case of exceeding the power supply, the amplifier is damaged. Summing up amplifiers are of twp types, Non – inverting and inverting amplifiers. Non – inverting amplifiers does not have an effect on the polarity of t ...Download file to see next pagesRead More
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