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Basics of DC Motors
HIMANSU JHA
Induced Magnetic Field
(Due to current)
Fixed Magnetic Field
Force
A Conductor in a Fixed
Magnetic Field
A Current Carrying Conductor
in a Fixed Magnetic Field
Electromagnetic drag
Fleming's Left Hand (Motor) Rule
N
S
Direction of Rotation
Fixed Magnetic Field Direction
Conventional
Current Direction
N
S
N
S Use the Left Hand Rule to
Determine the Rotation
Direction of the Armatures
in A and B
Notice that when the
current through the
armature is reversed,
it moves (Rotates) in
the opposite direction
A
B
Hint: You will have to turn
your left hand upside down
for example A
Which position requires
least force to turn the
crowbar?
SIGNIFICANCE OF THE BACK E.M.F.:
• Back emf Eb = (ΦZN/60) x (P/A)
• VOLTAGE EQUATION:
• V = Eb + IaRa
• Power equation
• V Ia = Eb Ia + Ia2Ra
• Eb Ia = Electrical equivalent of mechanical power
Pm developed in the armature
• Ia2Ra = Copper loss in the armature.
TORQUE The turning twisting moment of force about an axis
• Let Force is F Newton
• r = Radius of shaft in meters
• Then Torque T = F x r (Nm)
• Work done = Force x distance
• Distance travelled by the shaft in 1 revolution = perimeter =
2 x  x r
• Work done by shaft in 1 rev = F x 2 x  x r
As F x r = T,
Work done = F x 2 x  x r = 2  T
Speed = N rpm or
𝑁
60
rps. Or
60
𝑁
second for 1 revolution
Power developed by shaft in 1 second = 2  T x
𝑁
60
J/s or
Watts
Horse power developed =
2NT
60 𝑥 746
HP
Power developed by shaft
Torques in DC Motors
• Gross torque / armature torque
• Shaft torque
• Lost torque
• Power equation of a shaft =
2NT
60
J/s or Watts
• Power equation of a shaft =
2NT
60 𝑥 746
HP
Armature torque(Ta) / Gross torque
• Power equation of a shaft =
2NT
60
J/s or Watts
• Power developed in armature = Eb Ia =
2NTa
60
Armature Torque Ta =
Eb Ia x 60
2N
=
9.55 Eb Ia
N
N-m
Shaft torque(Tsh) / Output torque
• Power equation of a shaft =
2NT
60 𝑥 746
BHP
BHP =
2NTsh
60 𝑥 746
•Shaft torque Tsh =
𝐵𝐻𝑃 𝑥 60 𝑥 746
2N
Speed and Torque DC Motors
• Ta =
Eb Ia x 60
2N
• Eb =
ZNx 𝑃
60 𝑋 𝑎
• Ta =
ZNx 𝑃
60 𝑋 𝑎
𝑋
Ia x 60
2N
• Ta = K .  Ia or
•Ta   Ia
• Eb =
ZNx 𝑃
60 𝑋 𝑎
• Eb = K x N
•
Eb

= K x N or
•N 
Eb

Speeds ,back emf and fluxes
• Let N1 , 1, Eb1 be
Speed, flux per pole and back emf at No Load
• Let N2 , 2, Eb2 be
Speed, flux per pole and back emf at Load
N1 =
𝑘 𝐸𝑏1
1
----eq.(1) and N2 =
𝑘 𝐸𝑏2
2
----eq.(2)
Dividing eq(1) by eq(ii)
𝑁1
𝑁2
=
𝐸𝑏1
1
x
2
𝐸𝑏2
𝑁1
𝑁2
=
𝐸𝑏1
𝐸𝑏2
x
2
1
Flux and torques
𝑇𝑎1
𝑇𝑎2
=
𝐼𝑎1
𝐼𝑎2
x
1
2
Basics of dc motors

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Basics of dc motors

  • 1. Basics of DC Motors HIMANSU JHA
  • 2. Induced Magnetic Field (Due to current) Fixed Magnetic Field Force A Conductor in a Fixed Magnetic Field A Current Carrying Conductor in a Fixed Magnetic Field Electromagnetic drag
  • 3.
  • 4.
  • 5. Fleming's Left Hand (Motor) Rule N S Direction of Rotation Fixed Magnetic Field Direction Conventional Current Direction
  • 6. N S N S Use the Left Hand Rule to Determine the Rotation Direction of the Armatures in A and B Notice that when the current through the armature is reversed, it moves (Rotates) in the opposite direction A B Hint: You will have to turn your left hand upside down for example A
  • 7. Which position requires least force to turn the crowbar?
  • 8. SIGNIFICANCE OF THE BACK E.M.F.: • Back emf Eb = (ΦZN/60) x (P/A) • VOLTAGE EQUATION: • V = Eb + IaRa • Power equation • V Ia = Eb Ia + Ia2Ra • Eb Ia = Electrical equivalent of mechanical power Pm developed in the armature • Ia2Ra = Copper loss in the armature.
  • 9. TORQUE The turning twisting moment of force about an axis • Let Force is F Newton • r = Radius of shaft in meters • Then Torque T = F x r (Nm)
  • 10. • Work done = Force x distance • Distance travelled by the shaft in 1 revolution = perimeter = 2 x  x r • Work done by shaft in 1 rev = F x 2 x  x r As F x r = T, Work done = F x 2 x  x r = 2  T Speed = N rpm or 𝑁 60 rps. Or 60 𝑁 second for 1 revolution Power developed by shaft in 1 second = 2  T x 𝑁 60 J/s or Watts Horse power developed = 2NT 60 𝑥 746 HP Power developed by shaft
  • 11. Torques in DC Motors • Gross torque / armature torque • Shaft torque • Lost torque • Power equation of a shaft = 2NT 60 J/s or Watts • Power equation of a shaft = 2NT 60 𝑥 746 HP
  • 12. Armature torque(Ta) / Gross torque • Power equation of a shaft = 2NT 60 J/s or Watts • Power developed in armature = Eb Ia = 2NTa 60 Armature Torque Ta = Eb Ia x 60 2N = 9.55 Eb Ia N N-m
  • 13. Shaft torque(Tsh) / Output torque • Power equation of a shaft = 2NT 60 𝑥 746 BHP BHP = 2NTsh 60 𝑥 746 •Shaft torque Tsh = 𝐵𝐻𝑃 𝑥 60 𝑥 746 2N
  • 14. Speed and Torque DC Motors • Ta = Eb Ia x 60 2N • Eb = ZNx 𝑃 60 𝑋 𝑎 • Ta = ZNx 𝑃 60 𝑋 𝑎 𝑋 Ia x 60 2N • Ta = K .  Ia or •Ta   Ia • Eb = ZNx 𝑃 60 𝑋 𝑎 • Eb = K x N • Eb  = K x N or •N  Eb 
  • 15. Speeds ,back emf and fluxes • Let N1 , 1, Eb1 be Speed, flux per pole and back emf at No Load • Let N2 , 2, Eb2 be Speed, flux per pole and back emf at Load N1 = 𝑘 𝐸𝑏1 1 ----eq.(1) and N2 = 𝑘 𝐸𝑏2 2 ----eq.(2) Dividing eq(1) by eq(ii) 𝑁1 𝑁2 = 𝐸𝑏1 1 x 2 𝐸𝑏2 𝑁1 𝑁2 = 𝐸𝑏1 𝐸𝑏2 x 2 1