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The value of an a.c. voltage is defined by v(t) = 5.sin(314.t). Determine the frequency and peak amplitude of this voltage waveform and hence sketch the voltage waveform to scale. A sinusoidal voltage is described mathematically by an expression of the form v(t) = VPEAK.sin(2.π.f.t) where: VPEAK represents the peak amplitude of the sinusoid f represents the frequency of the sinusoid Thus by comparison, VPEAK = 5 V and 314 2. . f f 50 Hz The voltage waveform looks like – 5 4 3 2 1 0 0 10 20 30 40 50 -1 -2 -3 -4 -5 The vertical axis is calibrated in units of Volts and the horizontal axis is calibrated in units of mS. The voltage defined in section (a) is applied across a 10 Ω resistor in series with a 1.6 μF capacitor - Calculate the impedance of this series R-C circuit Impedance Z = R + XC 1 j.2. . f .C 1 Z 10 10 j.20 j.2. .50.1.6 106 Z R In polar form: 20 Z 10 2 20 2 tan 1 22.4 63.4 10 Calculate the peak magnitude of the current flowing in the series R-C circuit V 5314.t 0.22(314.t 63.4 ) A Z 22.4 63.4 Current magnitude = 0.22 A I Calculate the phase shift between the applied voltage and the current From above, the phase shift has current leading voltage by 63.4˚. A circuit comprises an inductor of 10 H in series with a 10 μF capacitor. The wire used to wind the inductor has resistance of 2 Ω. Explain the circumstances under which resonance occurs in the circuit described above. Resonance occurs at the frequency which gives capacitive and inductive reactances of equal magnitude. These two components of the overall impedance have opposite polarity and thus cancel out at resonance. This leaves the impedance equal to the resistance in the circuit when operating at the resonant frequency. Calculate the resonant frequency of the circuit. The resonant frequency is determined from the equation – 1 1 fo 16 Hz 2. . L.C 2. . 10.10 10 6 Calculate the Q-Factor for the inductor used in this circuit The Q factor for an inductor is defined as – X 2. .16.10 Q L 500 R 2 State how the inductor should be modified in order to improve the QFactor and explain why the modification leads to an improvement. The inductor could be wound from a thicker gauge wire in order to reduce the self-resistance of the coil. This might require extra turns and the reduction in resistance might be compromised by the extra length – but some improvement may be obtained. T1 D1 D2 D4 D3 Vmains RLoad Figure 1 Figure 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 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. The 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 – 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. Secondary Voltage Load Voltage The load voltage reaches a peak value that is approximately 1.2 V lower than the peak value of the secondary voltage. This is due to the forward voltage drop across two silicon diodes used in the bridge rectifier at any time. The rectifier 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. T1 D1 D2 D4 D3 Vmains RLoad + C The smoothing capacitor, C, should be included us shown above. Explain how the smoothing capacitor sustains a d.c. voltage across RLoad, despite the pulsating nature of the rectifier 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 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 will 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, 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 Ω for RLoad and a mains frequency of 50 Hz. Peak-peak ripple voltage is given by – VLOAD NOM VRIPPLE 2 RLOAD C f Where: VLOAD NOM = nominal output voltage from PSU C = the value of the smoothing capacitor f = the frequency of the a.c. supply VRIPPLE 1 0.05 VLOADNOM 2 RLOAD C f 1 C 400F 2 500 0.05 50 for 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 unregulated 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 load A three 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 below T1 D1 D2 U1 In Vmains D3 + D4 C Out Gnd RLoad D1 S1 M R2 Q1 V1 5.6k Vcc 24V 5V Figure 2 Figure 2 shows a schematic diagram of a bipolar junction transistor (BJT) used to switch a motor on and off. The BJT is specified with βDC = 100 and BVCEO = 40 V. The motor presents a load resistance of 300 Ω when running. Calculate the base current in the BJT when S1 is closed The base current may be calculated from – V V 5V 0.6V I B Control BE 786A RB 5.6k Calculate the base current in the BJT when S1 is opened The base current will be zero Sketch to scale an approximation to the collector characteristic curves for the BJT used in this circuit when the switch is open and closed and identify on the characteristic the active region, the saturation region and the breakdown region. Collector Characteristics 100 80 IC 60 40 20 0 -5 5 15 25 35 45 VCE Char S1 Closed Char S1 Opened Load Line 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 equation is – VCC VCE RLoad This may be plotted by finding two points and joining up – see above. IC The load current is got by finding the intersection of the load line with the characteristic curve – i.e. approximately 80 mA. With reference to typical values where necessary, estimate the power dissipation in the BJT when switched on. Typically we might expect about 0.3 volts between collector and emitter when turned fully on. Given 80 mA of load current, power is – P 0.3V 80mA 24mW State the circumstances under which the power dissipation in the BJT is at a maximum and hence calculate the maximum instantaneous power dissipation in the BJT. Maximum power dissipation occurs in the BJT as it switches from fully on to fully off. Specifically, there will be an instant during the switching action where VCE = VCC/2 and load current is one half of the fully on current. The value of this maximum instantaneous power is V I PPEAK CC MAX 12V 40mA 0.48W 2 2 Explain the consequences for the BJT if a large amount of power is dissipated and describe a technique 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 shortened. To reduce the heating, the device may be fitted with a heatsink to conduct and radiate heat away from the device. Explain the reason for including the diode D1 in the circuit. The motor is an inductive load that the transistor is required to switch on and off. Switching the current in an inductor will cause it to generate a large voltage in an attempt to prevent the current changing. This inductive voltage could potentially be large enough to damage the BJT. D1 is connected so as to present a low impedance to the induced voltage and hence protecting the BJT. The diode is reverse biased under static operating conditions and does not affect normal operation of the circuit. D1 is often called a flywheel diode. A Darlington circuit is sometimes used in place of the BJT in switching applications. Sketch a Darlington switching circuit and explain the advantages associated with the circuit compared to a single BJT. M S1 R2 Vcc V1 The Darlington circuit uses two BJTs. The first one provides base current to the second one from its emitter. Thus the collector current of the second BJT is related to the base current of the first BJT by – I C 2 1 2 I B1 I.e. the effective current gain of the combined BJTs is equal to the product of the individual current gains of the BJTs. The result is a device with a much higher current gain than is possible with a single BJT. This allows even smaller control currents to control larger load currents.