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Question 1 DT002 Power Conditioning January 2007 – Answers Page 1 of 10 R1 27 Ω E 30 V R2 10 Ω Figure 1 Write out the equation for Ohms law showing the relationship between voltage, current and resistance in an electrical circuit. Voltage Current Re sis tan ce 3 marks out of 33 Given a resistance of 220 Ω connected to an e.m.f. of 10 V Calculate the expected current flow through the resistor. I V R 10V 45.5mA 220 Explain the likely consequences of increasing the value of the resistor to 270 Ω. Increasing R to 270 Ω will cause a reduction in current flowing in the resistor and a consequent reduction in power dissipation The resistor is replaced by one with a value of 22 Ω. Calculate the new value of current flow and the required power rating for the resistor. V 10V 455mA R 22 Power I 2 R 0.455 A 2 22 I 4.55W 6 marks out of 33 With reference to figure 1: Calculate the current flow in resistor R1 Question 1 DT002 Power Conditioning January 2007 – Answers I E R1 R 2 Page 2 of 10 30V 0.81A 37 Calculate the voltage drop across resistor R2 V I R 0.81A 10 8.1V Calculate the power dissipation in resistor R1 Power I 2 R1 0.81A 2 27 17.7W Calculate the power dissipation in resistor R2 Power I 2 R 2 0.81A 2 10 6.56W 10 marks out of 33 Comment on the consequences of using a 10 Ω resistor with a power rating of 5 watts as R2 in figure 1 The power dissipation for R2 has been calculated as 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 33 State what a heatsink is and explain why R2 may need one if a 10 Ω 10 watt component were used in the construction of this circuit. A heatsink is a 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 to permit heat to escape to the surrounding air. 5 marks out of 33 Explain 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 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 Question 2 DT002 Power Conditioning January 2007 – Answers Page 3 of 10 T 220 Vrms 50 Hz B 10:1 C R Figure 2 Figure 2 shows the circuit diagram for a simple d.c. power supply. Identify the type of rectifier circuit represented in figure 2 and explain the operation of the circuit with reference to the function of each component within the circuit. This is a bridge rectifier circuit. Transformer T provides a reduction in the amplitude of the mains voltage to a value near to that of the required dc output voltage. It also provides electrical isolation between the mains and the load R. The diodes in the bridge circuit B conduct on alternate half cycles to provide only positive ‘pulses’ to the capacitor and load. C is a capacitor included to ‘smooth’ the raw d.c. from the rectifier diodes. It accumulates electric charge as the a.c. from the transformer approaches its peak value. After the peak, the capacitor delivers the stored charge to the load resistor, preventing the load voltage from dropping in between peaks in the a.c. voltage. R represents the resistance of the load being powered from the supply. 10 marks out of 33 Sketch to scale the voltage across R as a function of time showing its relationship to the secondary voltage from the transformer. Unsmoothed Smoothed ac Question 2 DT002 Power Conditioning January 2007 – Answers Page 4 of 10 6 marks out of 33 Calculate a value for the capacitor, C in order to keep the percentage ripple voltage across R below 5%. Assume a value of 500 Ω for R and a mains frequency of 50 Hz. 100% 2 f RC 100% 5% 2 50 500 C 100 C 400 F 5 2 50 500 %ripple voltage 5 marks out of 33 The power supply unit shown in figure 2 is said to be unregulated Explain the meaning of the term unregulated as used in relation to power supplies. An unregulated power supply is one where the output voltage is not protected from fluctuation in the event that – The mains input voltage changes (within permitted limits) The current drawn by the load changes. Either of these occurrences will result in a change in output voltage amplitude from an unregulated power supply. Give one example where an unregulated power supply may be acceptable and one example where a regulated power supply is required. An unregulated power supply example is a battery charger – many chargers provide unregulated output voltages. A regulated power supply is necessary for powering up digital and microprocessor circuits. Show how a three terminal regulator chip may be used to provide a regulated output voltage and explain how a regulated output voltage of 9 V may be obtained. T1 7809 U1 In Vmains + C1 The 9 volt output is obtained using a dedicated 9 volt regulator chip. Out Gnd RLoad Question 2 DT002 Power Conditioning January 2007 – Answers Page 5 of 10 In relation to power supply units, explain the meaning of the terms percentage load regulation and percentage line regulation. Percentage line regulation is a figure of merit that describes the extent of change in the output voltage from a power supply unit as a consequence of a change in the mains amplitude. % Line Re gulation VLoad Max VLoad Min VLoad No min al 100% Where: VLoadMax = Maximum output voltage as mains voltage increases above nominal value VLoadMin = Minimum output voltage as mains voltage decreases below nominal value Percentage load regulation is a figure of merit that describes the extent of change in the output voltage from a power supply unit as a consequence of a change in the load current. % Line Re gulation VLoad Max VLoad Min VLoad No min al 100% Where: VLoadMax = Maximum output voltage as load current decreases below nominal value VLoadMin = Minimum output voltage as load current increases above nominal value 12 marks out of 33 Question 3 DT002 Power Conditioning January 2007 – Answers R1 Page 6 of 10 L1 220 V rms 50 Hz T1 R2 C1 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. R1 is a resistor R2 is a potentiometer C1 is a capacitor L1 is a lamp T1 is a SCR The function of the circuit is a Lamp Dimmer control 6 marks out of 33 Briefly 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 when 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 point it will turn off. 6 marks out of 33 Question 3 DT002 Power Conditioning January 2007 – Answers Page 7 of 10 Sketch 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. 400 300 200 100 AC 0 0 10 20 30 40 50 LAMP -100 -200 -300 -400 7 marks out of 33 Explain how the circuit operates to reduce the r.m.s. voltage across L1. The rms voltage 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 only half used giving half the rms voltage. 7 marks out of 33 State 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 the 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 33 Question 4 DT002 Power Conditioning January 2007 – Answers M D1 S1 R2 Q1 5.6k V1 Page 8 of 10 Vcc 24V 5V Figure 4 Figure 4 shows a bipolar junction transistor (BJT) used to switch a small motor on and off in response to switch S1 closing and opening. The BJT is specified with β DC = 100 and BVCEO = 40 V. The motor draws 1 Amp from a 24 volt supply when running. Calculate the base current in the BJT when S1 is closed. With S1 closed, we can write Kirchoffs Voltage Law in the V1 loop as V1 I B R2 VBE 0 Thus: IB V1 VBE 5V 0.6V 10.7mA R2 410 S1 is opened. S1 opened prevents base current flowing in Q1 IB 0 6 marks out of 33 Question 4 DT002 Power Conditioning January 2007 – Answers Page 9 of 10 Sketch to scale approximate collector characteristic curves for the BJT used in this circuit when S1 is open and S1 is closed Identify on the characteristics the active region, the saturation region and the breakdown region. 10 marks out of 33 Set out the load-line equation for the circuit and plot this on the same graph as the collector curve. Hence or otherwise determine the load current flowing when the BJT is switched on. The load line intersects the characteristic curve at collector current of 1 Amp approx 8 marks out of 33 With reference to typical values where necessary, estimate the power dissipation in the BJT when switched on. Typically, the voltage between the collector and emitter when saturated is about 0.3 volts. Thus power dissipation is of the order of – P 0.3V 1A 0.3W 5 marks out of 33 State the reason for including device D1 in this circuit. Question 4 DT002 Power Conditioning January 2007 – Answers Page 10 of 10 As the BJT turns on it attempts to change the current in the windings of the motor suddenly from zero up to about 1 amp. The self inductance of the motor windings will resist this change in current by creating a large reverse e.m.f. that may damage the BJT. The diode D1 is included to provide a discharge path for this e.m.f so that it never becomes large enough to damage the BJT. 4 marks out of 33