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Known Quantities: Supply voltage; motor parameters; chopping frequency armature resistance and inductance Find: For each of the two cases: t1 ; Pa ; Pm ; PS Schematics, Diagrams, Circuits, and Given Data:
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1 VS = 120 V; Ea = 01n; Ra = 02 2 VS = 120 V; Ea = 01n; Ra = 02 300 Hz
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; 1/T = chopping frequency = 300 Hz ; LS ; 1/T = chopping frequency =
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Assumptions: The switches in the chopper of Figure 1140 act as ideal switches Assume that the motor inductance is suf ciently small to be neglected in the calculations (ie, assume a short circuit) Analysis:
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1 Analysis of motoring operation To analyze motoring operation of the chopper, we refer to Figure 1135 and apply KVL to the motor side: vo = Ra Ia + Ea = Ra io + 01n = 02 90 + 01 500 = 68 V From equation 1117 we can then compute the duty cycle of the chopper, : = vo 68 t1 = = 0567 = T VS 120 1 T = = 189 ms 300 0567
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Since the chopping frequency is 300 Hz, we can compute t1 : t1 =
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The power absorbed by the armature is:
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2 P a = R a Ia = R a io
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= 02 902 = 162 kW
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The power absorbed by the motor is: Pm = Ea Ia = 01n io = 01 500 90 = 45 kW The power delivered by the voltage supply is: PS = VS io = 0567 120 90 = 612 kW 2 Analysis of regenerative operation To analyze regenerative operation of the chopper, we refer to Figure 1137 and apply KVL to the motor side, noting that now the current is owing in the reverse direction: vo = Ra Ia + Ea = Ra io + Ea = 90 02 + 01 380 = 20 V We now turn to equation 1122, and observe that in this equation the motor acts as the source, and the supply voltage as the load, thus: VS = vo 1 tT1 1 or 120 = 1 20 1 300t1
Part II
Electronics
We can then compute t1 = 28 ms from the above equation The duty cycle for the step-up chopper is now: 5 t1 = = 0833 T 6 The power absorbed by the armature is: =
2 P a = R a Ia = R a io 2
= 02 ( 90)2 = 162 kW
The power absorbed by the motor will now be negative, since current is owing in the reverse direction; the motor is in fact generating power: Pm = Ea Ia = 01n io = 01 380 ( 90) = 342 kW The power delivered by the voltage supply is: PS = VS io = 0567 120 ( 90) = 18 kW This power is negative because the supply is actually absorbing power, not delivering it
Comments:
1 Note that the sum of the motor and armature power losses is equal to the power delivered by the source; this is to be expected, since we have assumed ideal (lossless) switches In a practical chopper, the chopping circuit would actually absorb power; heat dissipation is therefore an important issue in the design of choppers This chopper operates in quadrant I (see Figure 1140) 2 Note that in the regenerative case the equivalent duty cycle is greater than 1 Note also that now the power absorbed by the motor is a negative quantity; that is, the motor delivers power to the rest of the circuit However, the power absorbed by the armature resistance is still a positive quantity because the armature resistance dissipates power regardless of the direction of the current ow through it Here VS might, for example, represent a battery pack in an electric vehicle, which would be recharged at the rate of 18 kW The source of energy capable of producing this power is the inertial energy stored in the vehicle: when the vehicle decelerates, this mechanical energy causes the electric motor to act as a generator (see 17), producing the 90-ampere current in the reverse direction This chopper operates in quadrant IV (see Figure 1140)
Inverters (DC-AC Converters) As will be explained in 17, variable-speed drives for AC motors require a multiphase variable-frequency, variable-voltage supply Such drives are called DC-AC converters, or inverters Inverter circuits can be quite complex, so the objective of this section is to present a brief introduction to the subject, with the aim of illustrating the basic principles A voltage source inverter (VSI) converts the output of a xed DC supply (eg, a battery) to a variable-frequency AC supply Figure 1141 depicts a half-bridge VSI; once again, the switches can be either bipolar or MOS transistors, or thyristors The operation of this circuit is as follows When switch S1 is turned on, the output voltage is in the positive half-cycle, and vo = VS /2 To generate the negative half-cycle, switch S2 is turned on, and
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