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from which we can calculate vo t1 = ( vo Vs ) (T t1 ) (1120)
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This results in an average output voltage given by the expression
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t1 T
Vs =
1 Vs Vs 1
(1121)
Since the duty cycle, , is always less than 1, the theoretical range of the supply is VS vo < (1122)
Tm m
Figure 1139 Step-up chopper used for regenerative braking
S1 + VS S2
D2 io + vo DC motor D1 io
II S1 D1
vo S2 D2 III IV
Figure 1140 Two-quadrant chopper
The waveforms for the boost converter are shown in Figure 1138 A step-up chopper can also be used to provide regenerative braking: if the supply voltage is the motor armature voltage and the output voltage is the xed DC supply (battery) voltage, then power can be made to ow from the motor to the DC supply (ie, recharging the battery) This con guration is shown in Figure 1139 Finally, the operation of the step-down and step-up choppers can be combined into a two-quadrant chopper, shown in Figure 1140 The circuit shown schematically in Figure 1140 can provide both regenerative braking and motoring operation in a DC motor When switch S2 is open, and switch S1 serves as a chopper, the circuit operates as a step-down chopper, precisely as was described earlier in this section (convince yourself of this by redrawing the circuit with S2 and D2 replaced by open circuits) Thus, the drive and motor operate in the rst quadrant (motoring operation) The output voltage, vo , will switch between VS and zero, as shown in Figure 1136, and the load current will ow in the direction indicated by the arrow in Figure 1140; diode D1 free-wheels whenever S1 is open Since both output voltage and current are positive, the system operates in the rst quadrant When switch S1 is open and switch S2 serves as a chopper, the circuit resembles a step-up chopper The source is the motor emf, Ea , and the load is the battery; this is the situation depicted in Figure 1139 The current will now be negative, since the sum of the motor emf and the voltage across the inductor (corresponding to the energy stored during the on cycle of S2 ) is greater than the battery voltage Thus, the drive operates in the fourth quadrant Examples 119, 1110 and 1111 illustrate the operation of choppers as DC motor supplies
Part II
Electronics
EXAMPLE 119 Operation of Step-Down Chopper (Buck Converter)
Problem
Simulate the step-down chopper of Figure 1135 and verify numerically that the average output voltage is given by equation 1117
Solution
Known Quantities: Source voltage; load resistance and inductance; motor
characteristics
Find:
vo ; La =
Schematics, Diagrams, Circuits, and Given Data: VS = 220 V; Ra = 03 s 15 mH; ka = 00167 Vrev ; N = 0 2,000 rev/min, Ia = 25 A
Focus on Computer-Aided Solutions: The analysis of this design has been conducted in
simulation, using Electronics WorkbenchTM The simulation of this circuit may be found in the accompanying CD-ROM
EXAMPLE 1110 Operation of Step-Up Chopper (Boost Converter)
Problem
Simulate the step-up chopper of Figure 1137 and verify numerically that the average output voltage is given by equation 1122
Solution
Known Quantities: Source voltage; source series inductance; load resistance and inductance; motor characteristics Find:
vo ;
Schematics, Diagrams, Circuits, and Given Data: VS = 220; LS = 1 H; Ra = 03 La = 15 mH; ka = 00167 v s ; N = 0 2,000 rev/min rev
Focus on Computer-Aided Solutions: The analysis of this design has been conducted in
simulation, using Electronics WorkbenchTM The simulation of this circuit may be found in the accompanying CD-ROM
EXAMPLE 1111 Two-Quadrant Chopper
Problem
1 Determine the turn-on time of the chopper of Figure 1140 in the motoring mode if n = 500 rev/min and io = 90 A Also determine the power absorbed by the motor
11
Power Electronics
armature winding; the power absorbed by the motor; and the power delivered by the source 2 Determine the turn-on time of the chopper in the regenerative mode if n = 380 rev/min and io = 90 A Also determine the power absorbed by the motor armature winding, the power absorbed by the motor, and the power delivered by the source
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