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1、Figure 1Figure 1 shows the circuit diagram for a simple d.c. power supply. Identify the type of rectifier circuit represented in figure 1 and explain the operation of the circuit with reference to the function of each component within the circuit.This is a bridge rectifier circuit.The mains voltage
2、is applied to the primary winding of the transformer T1. This typically produces a reduced amplitude voltage on the secondary winding closer to the desired d.c. output voltage than the original mains voltage. The transformer also provides electrical isolation between the mains supply and the load.Th
3、e diodes D1 to D4 perform rectification. On positive half cycles of the secondary voltage, D2 and D4 are forward biased and conduct connecting positive half cycles across RLoad, with the top of RLoad positive.On negative half cycles of the secondary voltage, D1 and D3 are forward biased and conduct
4、connecting negative half cycles across RLoad, with the top of RLoad still positive.RLoad represents the circuit to which it is desired to supply d.c. electricity.Sketch the voltage across RLoad as a function of time showing its relationship to the secondary voltage from the transformer.The rectifier
5、 circuit shown in figure 1 requires the addition of a filter to produce a near constant d.c. voltage across RLoad. Redraw figure 1 showing where a smoothing capacitor should be connected.Explain how the smoothing capacitor sustains a d.c. voltage across RLoad, despite the pulsating nature of the rec
6、tifier output.The capacitor charges rapidly from the transformer secondary voltage via the diodes in the bridge when the a.c. rises towards its peak voltage. When the a.c. has reached its peak and starts to drop again, the capacitor holds on to its voltage and the diodes in the bridge become reverse
7、 biased. Current can only now be delivered to the load by discharge of the capacitor. The capacitor voltage will drop gradually under these circumstances at a rate dependent on the capacitor value and the load resistance value. After a short time, the a.c. starts to rise back towards its peak and wi
8、ll forward bias the bridge diodes when it exceeds the voltage to which the capacitor has dropped. The capacitor charge will be topped-up ready for the next half cycle of a.c. Provided that the capacitor is large enough, its terminal voltage will not drop substantially between peaks of the a.c. Thus,
9、 the load voltage is held almost constant.Calculate a value for the smoothing capacitor in order to keep the percentage ripple voltage across RLoad below 5%. Assume a value of 500 for RLoad and a mains frequency of 50 Hz.Peak-peak ripple voltage is given by Where:VLOAD NOM = nominal output voltage f
10、rom PSUC = the value of the smoothing capacitorf = the frequency of the a.c. supplyfor 5% ripple, = =The power supply shown in figure 1 is said to be unregulated. Explain the meaning of this term and show how a three terminal regulator chip may be used to provide a regulated output voltage.An unregu
11、lated power supply is one where the output voltage may change substantially from the stated nominal output voltage under specific operating conditions. In particular, change in output voltage may arise from Fluctuations in the mains supply voltage Change in the amount of current drawn by the loadA t
12、hree terminal regulator chip is an integrated circuit designed to prevent changes in the output voltage occurring within design limits. It is used as shown in the circuit belowFigure 2Figure 2 shows a bipolar junction transistor (BJT) used to switch a lamp (L1) on and off in response to the output v
13、oltage from a programmable logic controller (PLC). VS switches between 0 V and 5 V as the PLC output changes state. The BJT is specified with DC = 150 and BVCEO = 50 V. The lamp is rated at 0.5 W when supplied from a 20 V source.Calculate the base current in the BJT when The PLC output is 0 V.The PL
14、C output is 5 V.The base current may be calculated from when S1 is closed when S1 is openedSketch to scale an approximate collector characteristic curve for the BJT used in this circuit when the PLC output switches to 5 V. Identify on the characteristic the active region, the saturation region and t
15、he breakdown region.Calculate the collector current needed to turn the lamp on fully. Hence determine if the PLC output will be powerful enough to turn the lamp on fully using this interface circuit.With reference to typical values where necessary, estimate the power dissipation in the BJT when the
16、PLC output is at 5V.Typical VCE when the BJT operates in the saturation region might be about 0.3 volts.The collector current when the lamp is on should be Thus power dissipated in the BJT will be Explain the consequences for the BJT if a large amount of power is dissipated and describe a technique
17、that may be used to minimise the effect.Excessive power dissipation in a BJT will lead to the device heating up to a point where the temperature of the device exceeds the maximum rated operating temperature. In this case the device may fail or at least the operating lifetime of the device may be sho
18、rtened.To reduce the heating, the device may be fitted with a heatsink to conduct heat away from the device and radiate this heat to the surrounding air.Figure 3 Identify the type of electronic component represented by each of the symbols shown in figure 3 above and state the function of the circuit
19、.R1 is a resistorR2 is a potentiometerC1 is a capacitorL1 is a lampT1 is a SCRThe function of the circuit is a Lamp Dimmer control6 marks out of 33Briefly describe the principle of operation of the device T1 including an explanation of how the device is made to turn on and off.T1 will not conduct wh
20、en reverse biased i.e. on negative half cycles of the 220 V 50 Hz supply.When T1 is forward biased it will not commence conduction until a sufficient voltage is applied on the gate terminal. Once triggered into conduction, the SCR will continue to conduct until it becomes reverse biased at which poi
21、nt it will turn off.6 marks out of 33Sketch the typical shape of the voltage waveform that would be measured across L1 in this circuit given that R2 is set to approximately half of its maximum value.7 marks out of 33Explain how the circuit operates to reduce the r.m.s. voltage across L1.The rms volt
22、age is related to the area underneath the LAMP waveform above. Keeping the SCR turned off until a point somewhere after zero crossing of the AC waveform has the effect of reducing this area below the area given by a complete sine half cycle. Thus in the diagram above, each positive half cycle is onl
23、y half used giving half the rms voltage.7 marks out of 33State the likely effect of removing component C1 from the circuit and explain the reason why the behaviour of the circuit is modified.C1 is used to delay the instant at which the voltage on the SCR gate reaches the voltage necessary to turn th
24、e SCR on. This allows triggering to occur almost anywhere between 0 and 180 degrees. Removing C1 will remove this delay and triggering will only be possible between 0 and 90 degrees.7 marks out of 33Figure 3Write out the equation for Ohms law showing the relationship between voltage, current and res
25、istance in an electrical circuit.3 marks out of 33Given a resistance of 220 connected to an e.m.f. of 10 VCalculate the expected current flow through the resistor.Explain the likely consequences of increasing the value of the resistor to 270 .Increasing R to 270 will cause a reduction in current flo
26、wing in the resistor and a consequent reduction in power dissipationThe resistor is replaced by one with a value of 22 . Calculate the new value of current flow and the required power rating for the resistor.6 marks out of 33With reference to figure 1:Calculate the current flow in resistor R1 Calcul
27、ate the voltage drop across resistor R2Calculate the power dissipation in resistor R1Calculate the power dissipation in resistor R210 marks out of 33Comment on the consequences of using a 10 resistor with a power rating of 5 watts as R2 in figure 1The power dissipation for R2 has been calculated as
28、6.56W. A device with a rated power of 5W will heat up excessively and experience a shortened lifetime relative to a higher wattage device.4 marks out of 33State what a heatsink is and explain why R2 may need one if a 10 10 watt component were used in the construction of this circuit.A heatsink is a
29、mechanism designed to conduct heat away from a heat sensitive device so that the device does not operate at an excessive temperature that would contribute to premature failure of the device. It is usually made of metal for its high thermal conductivity. The surface area will be maximised using fins
30、to permit heat to escape to the surrounding air.5 marks out of 33Explain what the term de-rating means and describe how it may be used in order to improve the long-term reliability of an electronic component.De-rating is a technique used to prolong the service lifetime of a device. A device operated
31、 near to its maximum specified operating limits will wear out faster than a device operated well below its maximum limits. Derating calls for the use of devices with highe limits than are strictly needed in an application so that they will provide longer service. 5 marks out of 33我的大学爱情观1、什么是大学爱情：大学
32、是一个相对宽松，时间自由，自己支配的环境，也正因为这样，培植爱情之花最肥沃的土地。大学生恋爱一直是大学校园的热门话题，恋爱和学业也就自然成为了大学生在校期间面对的两个主要问题。恋爱关系处理得好、正确，健康，可以成为学习和事业的催化剂，使人学习努力、成绩上升；恋爱关系处理的不当，不健康，可能分散精力、浪费时间、情绪波动、成绩下降。因此，大学生的恋爱观必须树立在健康之上，并且树立正确的恋爱观是十分有必要的。因此我从下面几方面谈谈自己的对大学爱情观。2、什么是健康的爱情：1) 尊重对方，不显示对爱情的占有欲，不把爱情放第一位，不痴情过分；2) 理解对方，互相关心，互相支持，互相鼓励，并以对方的幸福为
33、自己的满足; 3) 是彼此独立的前提下结合；3、什么是不健康的爱情：1)盲目的约会，忽视了学业;2)过于痴情，一味地要求对方表露爱的情怀，这种爱情常有病态的夸张;3)缺乏体贴怜爱之心，只表现自己强烈的占有欲;4)偏重于外表的追求;4、大学生处理两人的在爱情观需要三思：1. 不影响学习：大学恋爱可以说是一种必要的经历，学习是大学的基本和主要任务，这两者之间有错综复杂的关系，有的学生因为爱情，过分的忽视了学习，把感情放在第一位；学习的时候就认真的去学，不要去想爱情中的事，谈恋爱的时候用心去谈，也可以交流下学习，互相鼓励，共同进步。2. 有足够的精力：大学生活，说忙也会很忙，但说轻松也是相对会轻松的
34、！大学生恋爱必须合理安排自身的精力，忙于学习的同时不能因为感情的事情分心，不能在学习期间，放弃学习而去谈感情，把握合理的精力，分配好学习和感情。3、 有合理的时间；大学时间可以分为学习和生活时间，合理把握好学习时间和生活时间的“度”很重要；学习的时候，不能分配学习时间去安排两人的在一起的事情，应该以学习为第一；生活时间，两人可以相互谈谈恋爱，用心去谈，也可以交流下学习，互相鼓励，共同进步。5、大学生对爱情需要认识与理解，主要涉及到以下几个方面：（1） 明确学生的主要任务“放弃时间的人，时间也会放弃他。”大学时代是吸纳知识、增长才干的时期。作为当代大学生，要认识到现在的任务是学习学习做人、学习知
35、识、学习为人民服务的本领。在校大学生要集中精力，投入到学习和社会实践中，而不是因把过多的精力、时间用于谈情说爱浪费宝贵的青春年华。因此，明确自己的目标，规划自己的学习道路，合理分配好学习和恋爱的地位。（2） 树林正确的恋爱观提倡志同道合、有默契、相互喜欢的爱情：在恋人的选择上最重要的条件应该是志同道合，思想品德、事业理想和生活情趣等大体一致。摆正爱情与学习、事业的关系：大学生应该把学习、事业放在首位，摆正爱情与学习、事业的关系，不能把宝贵的大学时间，锻炼自身的时间都用于谈情说有爱而放松了学习。 相互理解、相互信任，是一份责任和奉献。爱情是奉献而不时索取，是拥有而不是占有。身边的人与事时刻为我们
36、敲响警钟，不再让悲剧重演。生命只有一次，不会重来，大学生一定要树立正确的爱情观。（3） 发展健康的恋爱行为 在当今大学校园，情侣成双入对已司空见惯。抑制大学生恋爱是不实际的，大学生一定要发展健康的恋爱行为。与恋人多谈谈学习与工作，把恋爱行为限制在社会规范内，不致越轨，要使爱情沿着健康的道路发展。正如马克思所说：“在我看来，真正的爱情是表现在恋人对他的偶像采取含蓄、谦恭甚至羞涩的态度，而绝不是表现在随意流露热情和过早的亲昵。”（4） 爱情不是一件跟风的事儿。很多大学生的爱情实际上是跟风的结果，是看到别人有了爱情，看到别人幸福的样子（注意，只是看上去很美），产生了羊群心理，也就花了大把的时间和精力
37、去寻找爱情（5） 距离才是保持爱情之花常开不败的法宝。爱情到底需要花多少时间，这是一个很大的问题。有的大学生爱情失败，不是因为男女双方在一起的时间太少，而是因为他们在一起的时间太多。相反，很多大学生恋爱成功，不是因为男女双方在一起的时间太少，而是因为他们准确地把握了在一起的时间的多少程度。（6） 爱情不是自我封闭的二人世界。很多人过分的活在两人世界，对身边的同学，身边好友渐渐的失去联系，失去了对话，生活中只有彼此两人；班级活动也不参加，社外活动也不参加，每天除了对方还是对方，这样不利于大学生健康发展，不仅影响学习，影响了自身交际和合作能力。总结：男女之间面对恋爱，首先要摆正好自己的心态，树立自尊、自爱、自强、自重应有的品格，千万不要盲目地追求爱，也不宜过急追求爱，要分清自己的条件是否成熟。要树立正确的恋爱观，明确大学的目的，以学习为第一；规划好大学计划，在不影响学习的条件下，要对恋爱认真，专一，相互鼓励，相互学习，共同进步；认真对待恋爱观，做健康的恋爱；总之，我们大学生要树立正确的恋爱观念，让大学的爱情成为青春记忆里最美的风景，而不是终身的遗憾！