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The load current through the 74- resistor is: 2 200 sin (2 250t) 75 iL (t) = 0 and the triggering voltage is: vt (t) = Vt sin [(2 250t) ]
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The triggering voltage will go positive at the desired ring angle, , thus injecting a current into the gate of the thyristor, turning it on Thus, the requirement that the average load current be equal to 1 A is equivalent to requiring that 1 2 200 iL (t) = sin 2 t d t = 1 A 1 75 Performing the integration, we determine that the requirement is 2 200 (1 + cos( )) = 1 2 75 Solving for , we nd = 4823 Now, to determine the value of R, we observe that the AC source voltage appears across the RC circuit; thus, vt can be computed from an impedance voltage divider by
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using phasor methods: 1 VS j C VS (j ) = arctan ( RC) Vt (j ) = 1 1 + 2 R 2 C 2 R+ j C We then observe that the phase of Vt (j ) is the ring angle , and we can therefore determine by setting arctan( RC) = = 4823 R= tan( ) tan(4823) = 713 = C 2 250 10 6
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Focus on Computer-Aided Solutions: An Electronics WorkbenchTM simulation of the
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circuit analyzed in this example is supplied in the accompanying CD-ROM You may wish to experiment with changing the value of R to see its effect on the average load current
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Check Your Understanding
112 Using the approximation given in equation 118, nd the DC and AC load currents for the circuit of Figure 1117 if R = 10 , L = 03 H, A = 170 V, and = 377 rad/s 113 Calculate the load voltage in Figure 1126 for A = 100, = /3 114 For the circuit in Example 116, the input AC voltage is 240 V Find the rms value
of the load voltage and the power at the ring angle = /4
i II Reverse braking (generation) Motoring (forward) v Forward braking (generation) IV I
ELECTRIC MOTOR DRIVES
Motoring (reverse) III
Figure 1133 The four quadrants of an electric drive
+ Ia Va m Tm
The advent of high-power semiconductor devices has made it possible to design effective and relatively low-cost electronic supplies that take full advantage of the capabilities of the devices introduced in this chapter Electronic power supplies for DC and AC motors have become one of the major elds of application of power electronic devices The last section of this chapter is devoted to an introduction to two families of power supplies, or electric drives: choppers, or DC-DC converters; and inverters, or DC-AC converters These circuits nd widespread use in the control of AC and DC motors in a variety of applications and power ranges Before we delve into the discussion of the electronic supplies, it will be helpful to introduce the concept of quadrants of operation of a drive Depending on the direction of current ow, and on the polarity of the voltage, an electronic drive can operate in one of four possible modes, as indicated in Figure 1133 Choppers (DC-DC Converters) As the name suggests, a DC-DC converter is capable of converting a xed DC supply to a variable DC supply This feature is particularly useful in the control of the speed of a DC motor (described in greater detail in 17) In a DC motor, shown schematically in Figure 1134, the developed torque, Tm , is proportional
Figure 1134 DC motor
Part II
Electronics
to the current supplied to the motor armature, Ia , while the electromotive force (emf), Ea , which is the voltage developed across the armature, is proportional to the speed of rotation of the motor, m A DC motor is an electromechanical energy-conversion system; that is, it converts electrical to mechanical energy (or vice versa if it is used as a generator) If we recall that the product of torque and speed is equal to power in the mechanical domain, and that current times voltage is equal to power in the electrical domain, we conclude that in the ideal case of complete energy conversion, we have Ea Ia = Tm m (1116)
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