Range of semiconductor devices and circuits Coursework

Range of semiconductor devices and circuits - Coursework ExampleA comfortably diode should posses forward to raise resistance ratio of 101 with another(prenominal)s with the ratio up to one C1. 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 test the diode using an ohmmeter only is as illustrated in the following steps. 1. The Ohmmeter is configured to measure on the low range. If its a digital ohmmeter, a 2-kohm range or the neargonst equivalent is employ. 2. The ohmmeter leads are attached across the diode with the compulsory red try out on the P- clobber and the negative dull probe on the N- material as shown in the plat below. 3. The measured resistance of the diode is measured Rf is put down after which the setup is disconnected. 4. After the disconnection, the leads of the ohmmeter are give upd and connected in the reverse direction i.e. t he 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 degree Fahrenheit 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)*long hundred00 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 electric potentials. 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