DC-DC Converter Drives
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Principle of Power Control
Principle of Regenerative Brake Control
Principle of Rheostatic Brake Control
Combined Regenerative and Rheostatic
Brake Control
• Two and Four Quadrant DC – DC
Converter Drives
ECE 442 Power Electronics
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Converter-fed DC Drive for a
Separately-Excited Motor
ECE 442 Power Electronics
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Waveform Summary
Highly Inductive Load
Ripple-free Armature Current
ECE 442 Power Electronics
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Principle of Power Control
• The average armature voltage is
Va  kVs
• The power supplied to the motor is
Po  Va I a  kVs I a
ECE 442 Power Electronics
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Principle of Power Control (continued)
• The average value of the input current is
I s  kIa
• The equivalent input resistance seen by the
source is
Vs Vs
Req  
I s kIa
Control Power Flow by adjusting the duty cycle
ECE 442 Power Electronics
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Principle of Power Control (continued)
• To find the maximum peak-to-peak ripple
current
I max
Vs
Rm

tanh
Rm
4 fLm
ECE 442 Power Electronics
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Example 15.7
A dc separately excited motor is powered by a dc-dc converter, as shown, from a 600V dc source. The armature
resistance is Ra = 0.05Ω. The back emf constant is Kv = 1.527V/A rad/s. The average armature current is Ia = 250A. The
field current is If = 2.5A. The armature current is continuous and has negligible ripple. If the duty cycle of the dc-dc
converter is 60%, determine:
ECE 442 Power Electronics
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Example 15.7
Solution
Vs  600V
I a  250 A
k  0.6
Rm  Ra  0.05
the input power from the source
Po  Va I a  kVs I a
Po  (0.6)(600V )(250 A)  90kW
the equivalent input resistance of the dc-dc converter drive
Req 
Vs Vs 1

Is Ia k
Req 
600V
 4
(250 A)(0.6)
ECE 442 Power Electronics
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the motor speed
E g  K v I f
Eg  Va  Rm I m
Va  kVs  (0.6)(600V )  360V
Eg  360V  (0.05)(250 A)  347.5V

Eg
Kv I f

347.5V
 91.03rad / s
(1.527V / Arad / s )(2.5 A)
 30 
  869.3rpm
 
  91.03 
the developed torque
Td  Kt I f I a  Kv I f I a
Td  (1.527V / Arad / s)(250 A)(2.5 A)  954.38 N  m
ECE 442 Power Electronics
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Application of a DC – DC Converter
in Regenerative Braking
ECE 442 Power Electronics
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Waveform Summary
Armature Current Continuous
and Ripple-Free
ECE 442 Power Electronics
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Regenerative Braking
• Begin with the motor
turning by kinetic
energy of the vehicle
• Armature current flows
as shown
• Turn the transistor on
• Armature current rises
• Turn the transistor off
• Diode turns on, current
flows into the supply
ECE 442 Power Electronics
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Principle of Regenerative Braking
• The average voltage across the transistor is
Vch  (1  k )Vs
• The regenerated power can be found from
Pg  I aVs (1  k )
ECE 442 Power Electronics
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Principle of Regenerative Braking
(continued)
• The voltage generated by the motor acting as a
generator is
Eg  K v I f 
Eg  Vch  Rm I a  (1  k )Vs  Rm I a
• The equivalent load resistance of the motor
acting as a generator is
Control Power by changing k
Eg
Vs
Req 
 (1  k )  Rm
Ia Ia
ECE 442 Power Electronics
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0  ( E g  Rm I a )  Vs
E g  K v min I f  Rm I a
 min
Rm I a

Kv I f
Minimum Braking Speed
   min
K v max I f  Rm I a  Vs
 max
Vs
Rm I a


Kv I f Kv I f
Maximum Braking
Speed
   max
ECE 442 Power Electronics
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Example 15.8
A dc-dc converter is used in regenerative braking of a dc series motor similar to the arrangement shown below. The dc
supply voltage is 600V. The armature resistance is Ra = 0.02Ω and the field resistance is Rf = 0.03Ω. The back emf
constant is Kv = 15.27mV/A rad/s. The average armature current is maintained constant at Ia = 250A. The armature
current is continuous and has negligible ripple. If the duty cycle of the dc-dc converter is 60%, determine the following:
For this example, the field and
armature need to be in series
ECE 442 Power Electronics
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Example 15.8
Solution
Vs  600V
I a  250 A
K v  0.01527V / Arad / s
k  0.6
Rm  Ra  R f
Determine the average voltage across the converter.
Vch  (1  k )Vs
Vch  (1  0.6)(600V )  240V
Determine the power regenerated to the dc supply
Pg  I aVs (1  k )
Pg  (250 A)(600V )(1  0.6)  60kW
Determine the equivalent resistance of the motor acting as a generator
Req 
Eg
Ia

Vs
(1  k )  Rm
Ia
Rm  Ra  R f  0.02  0.03  0.05
Req 
600V
(1  0.6)  0.05  1.01
250 A
ECE 442 Power Electronics
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Determine the minimum permissible braking speed ωmin
min 
Rm I a
0.05
250 A

 3.274rad / s
K v I f 0.01527V / Arad / s 250 A
min  3.274rad / s
1rev 60s
 31.26rpm
2 rad 1min
Determine the maximum permissible braking speed ωmax
max 
Vs
R I
 m a
Kv I f Kv I f
600V
0.05

(0.01527V / Arad / s)(250 A) 0.01527V / Arad / s
 160.445rad / s
max 
max
 30 
  1532.14rpm
 
max  160.445 
Determine the motor speed
Eg  K v I f    
Eg
Kv I f
Eg  (1  k )Vs  Rm I a  240V  (0.05)(250 A)  252.5V

252.5V
 66.14rad / s
(0.01527V / Arad / s)(250 A)
 30 
  631.6rpm
 
  66.14 
ECE 442 Power Electronics
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Rheostatic Brake Control
Dynamic Braking
ECE 442 Power Electronics
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Waveform Summary
ECE 442 Power Electronics
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Principle of Rheostatic Brake Control
• The average current in the braking resistor is
I b  I a (1  k )
• The average voltage across the braking
resistor is
Vb  Rb I a (1  k )
ECE 442 Power Electronics
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Principle of Rheostatic Brake Control
(continued)
• The equivalent load resistance of the generator
Vb
Req 
 Rb (1  k )  Rm
Ia
• The power dissipated in the resistor Rb is
Pb  I a2 Rb (1  k )
ECE 442 Power Electronics
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Example 15.9
A dc-dc converter is used in rheostatic braking of a dc separately excited motor as shown below. The armature
resistance is Ra = 0.05Ω. The braking resistor is Rb = 5Ω. The back emf constant is Kv = 1.527V/A rad/s. The average
armature current is maintained constant at Ia = 150A. The armature current is continuous and has negligible ripple. The
field current is If = 1.5A. If the duty cycle of the dc-dc converter is 40%, determine:
ECE 442 Power Electronics
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Example 15.9
Solution
I a  150 A
K v  1.527V / Arad / s
k  0.4
Rm  Ra  0.05
the average voltage across the dc-dc converter.
Vch  Vb  Rb I a (1  k )
Vch  (5)(150 A)(1  0.4)  450V
the power dissipated in the braking resistor
Pb  I a2 Rb (1  k )
Pb  (150 A)2 (5)(1  0.4)  67.5kW
the equivalent resistance of the motor acting as a generator
Req 
Vb
 Rb (1  k )  Rm
Ia
Req  (5)(1  0.4)  0.05  3.05
ECE 442 Power Electronics
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the motor speed ω
Eg  K v I f 

Eg
Kv I f

457.5V
 199.74rad / s
(1.527V / Arad / s)(1.5 A)
 30 
  1907.4rpm
 
  199.74 
the peak dc converter voltage
Vp  I a Rb
Vp  (150 A)(5)  750V
ECE 442 Power Electronics
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Combined Regenerative and
Rheostatic Brake Control
ECE 442 Power Electronics
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Combined Regenerative and
Rheostatic Brake Control (continued)
• Used when the supply
is partly “receptive”
• Remove regenerative
braking if line voltage
is too high
–
–
–
–
Turn thyristor TR on
Divert current to RB
Apply rheostatic braking
TR is “self-commutated”
ECE 442 Power Electronics
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Two-Quadrant DC–DC Converter Drive
ECE 442 Power Electronics
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Quadrant Operation Summary
Regenerative Braking Control
Power Control
ECE 442 Power Electronics
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Power Control
• Q1 and D2 operate
• Q1 ON, Vs applied to
the motor
• Q1 turned OFF, D2
“free-wheels”
• Armature current
decays
ECE 442 Power Electronics
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Regenerative Control
• Q2 and D1 operate
• Q2 turned ON, motor
acts as a generator,
and the armature
current rises
• Q2 turned OFF, motor
returns energy to the
supply via D1 “freewheeling”
ECE 442 Power Electronics
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Four Quadrant DC-DC Converter Drive
ECE 442 Power Electronics
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Quadrant Operation Summary
Forward Regeneration
Reverse Power Control
Forward Power Control
Reverse Regeneration
ECE 442 Power Electronics
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Forward Power Control
• Q1 and Q2 turned ON
• Supply voltage appears
across the motor
• Armature current rises
• Q1 and Q2 turned OFF
• Armature current
decays via D3 and D4
ECE 442 Power Electronics
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Forward Regeneration
• Q1, Q2, and Q3 turned
OFF
• Turn Q4 ON
• Armature current rises
and flows through Q4, D2
• Q4 turned OFF, motor
acts as a generator,
returns energy back to
the supply via D1, D2
ia reverses
ECE 442 Power Electronics
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Reverse Power Control
• Q3 and Q4 turned ON
• Supply voltage appears
in the reverse direction
across the motor
• Armature current rises
and flows in the
reverse direction
• Q3 and Q4 turned OFF
• Armature current
decays via D1 and D2
ECE 442 Power Electronics
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Reverse Regeneration
• Q1, Q3, Q4 turned OFF
• Q2 turned ON
• Armature current rises
through Q2 and D4
• Q2 turned OFF
• Armature current falls
and returns energy via
D3 and D4
ECE 442 Power Electronics
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