Sol The thyristor will conduct when the instantaneous value of ac voltage is

Sol the thyristor will conduct when the instantaneous

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Sol: The thyristor will conduct when the instantaneous value of ac voltage is more than 50V or 100sin ω t = 50 Or ω t = π /6 and 5 π/6 i = = 10sinωt -5 Average current = = |- 10cosωt - 5ωt| π/6 5π/6 = (-10cos + 10cos - 5x + 5x ) = 1.09A 7. If a half-wave controlled rectifier has a purely resistive load of R and the delay angle is α = π /3. Determine (a) Rectification efficiency, (b) Form factor, (c) Ripple factor, (d) Transformer utilization factor and (e) Peak inverse voltage for SCR Solution: (a) Rectification efficiency, η = where, Pdc = dc load power = V 2 dc/R and Pac = rms load power = V 2 rms/R.
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47 Vdc = (1+cos α ) , since, α = π /3 therefore Vdc = 0.239Vm Also Vrms = Vm[ + ] 0.5 For firing angle α = π /3 , Vrms = 0.485Vm Therefore rectification efficiency η = (0.239Vm) 2 /(0.485Vm) 2 = 0 .2428 = 24.28% (b) Form factor, (ff) = Vrms/Vdc = 0.485Vm/0.239Vm = 2.033 = 203.3% (c) Ripple factor (Rf) = (ff 2 1) = 1.77 = 177% (d) Trasformer utilization factor (TUF) = where Vs and Is are the rms secondary voltage and current respectively. Now Vs = Vm/ 2 = 0.707Vm and Is = rms load current = Vrms/R = 0.485Vm/R Therefore TUF =(V 2 dc/R)/Vs.Is = (0.239Vm) 2 /R/(0.707Vm x 0.485Vm/R) = 0.166 or 16.6% (e) Peak inverse voltage = Vm 8. An SCR is used to control the power of 1kW, 230V, 50Hz heater. Determine the heater power for firing angles of 45° and 90°. Solution : The heater resistance = R, and the rms current is the heat producing component of load current. Vrms = Vm[ + ] 0.5 (i) At α = π /4, Vrms = 155V, Therefore heat power W = V 2 rms/R R = 230 2 /1kW = 52.90 Therefore W = (155) 2 /52.90 = 454.15 watts (ii) At α = π /2, Vrms = 115V, Therefore W = (115) 2 /52.90 = 250watts UJT Triggered SCR Phase Control In SCR power control circuit a control circuit is needed to vary the firing or the phase angle at which the SCR is triggered. The value of phase angle controls the power delivered to the load. If we want to deliver the maximum power to the load, then the SCR must be triggered as soon as the a.c. voltage across it goes positive i.e., the firing angle must be zero. Therefore, the SCR will conduct for 180°after triggering. If the power requirement is below the maximum value, the firing angle must be between 0° and 180°. In order to adjust the firing angle between 0° and 180°, we need a control circuit. We have already seen that a 180° half-wave rectifier circuit can be used to accomplish the control. But this circuit is affected by loading and supply voltage variation. Therefore, this circuit does not give a better phase control. However, a better control cab be achieved by using a unijunction transistor (UJT) relaxation oscillator circuit.
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48 This circuit shows a UJT oscillator controlling a S CR circuit. The SCR is in it ‘OFF’ state until the UJT fires. When the UJT fires, the capacitor will discharge quickly through the resistor R4. This positive pulse will turn ‘ON’ the SCR. The SCR will remain ‘ON’ until the SCR line voltage approaches z ero. At this point, the SCR will turn ‘OFF’ throughout the entire negative circle. A similar circuit can also be made for triac phase control.
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