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DC Motor Voltage Ratings
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jbartos
Electrical
- Jan 15, 2000
- 6,440
No. The 48VAC sine peak to peak has an amplitude or Vmax=48VAC/2=24 VAC
Vrms=Vmax/2**0.5=24/2**0.5=16.97VAC ~ 17VAC
Vdc=Vaverage-of-fully-rectified-17VAC=Vmax x 2/(PI)=Vrms x 2**0.5 x 2/3.1416=15.3VDC, where
PI=3.1416
2**0.5 is square root of 2
The dc motor rated at 24VDC should have 24VDC across its terminals. If it happen to comply with NEMA standards then it may have some small percentage over, usually 10% at most. This means that the motor shall not have more than 24VDC + 2.4VDC=26.4VDC, which is however the very risky voltage to have across the motor terminals.
Vrms=Vmax/2**0.5=24/2**0.5=16.97VAC ~ 17VAC
Vdc=Vaverage-of-fully-rectified-17VAC=Vmax x 2/(PI)=Vrms x 2**0.5 x 2/3.1416=15.3VDC, where
PI=3.1416
2**0.5 is square root of 2
The dc motor rated at 24VDC should have 24VDC across its terminals. If it happen to comply with NEMA standards then it may have some small percentage over, usually 10% at most. This means that the motor shall not have more than 24VDC + 2.4VDC=26.4VDC, which is however the very risky voltage to have across the motor terminals.
Looks like we need to clear up some things first.
DC voltage from a battery a fixed voltage, ie. the amplitude is constant and Vpeak equals Vavg. When DC is supplied by a generator or a rectifier, it is not pure and actually will contain a small portion of ripple such that (depending on the specifics of the source) the value of Vpeak will be higher than Vavg or Vdc. Generators supply very little ripple and single phase half wave rectifiers supply the most.
With respect to Jbartos' post, it would appear that he mathematically proved that Vrms of a 48VAC source would be equal to 17VAC....NOT true. Any source rated 48VAC has a rms value of 48VAC by definition. The instantaneous peak value Vpeak of a 48VAC source is equal to 48 x 2**0.5= 67.8V. The DC value or average value of 48VAC is 0V.
Back to the ripple...DC motor nameplates and or specifications will include a "Power Supply Code" describing the acceptable power supplies for the motor. The idea is that ripple causes heating and heating is a primary factor for motor ratings. In other words, the value Vpeak may be somewhat higher than Vdc or Vavg by an amount allowed due to the power code. An incomplete listing of Codes is :
A: battery, generator, or 12 pulse, 6 phase, fully controlled rectifier
C: 6 pulse, 3 phase, fully controlled rectifier
K: 2 pulse, single phase full wave rectifier
Finally, the amount of AC voltage you need to produce 24VDC depends on the number of AC phases and the type of rectifier used. Also, as implied by the power codes, the amount of ripple will be determined by the same factors.
I hope this helps.
DC voltage from a battery a fixed voltage, ie. the amplitude is constant and Vpeak equals Vavg. When DC is supplied by a generator or a rectifier, it is not pure and actually will contain a small portion of ripple such that (depending on the specifics of the source) the value of Vpeak will be higher than Vavg or Vdc. Generators supply very little ripple and single phase half wave rectifiers supply the most.
With respect to Jbartos' post, it would appear that he mathematically proved that Vrms of a 48VAC source would be equal to 17VAC....NOT true. Any source rated 48VAC has a rms value of 48VAC by definition. The instantaneous peak value Vpeak of a 48VAC source is equal to 48 x 2**0.5= 67.8V. The DC value or average value of 48VAC is 0V.
Back to the ripple...DC motor nameplates and or specifications will include a "Power Supply Code" describing the acceptable power supplies for the motor. The idea is that ripple causes heating and heating is a primary factor for motor ratings. In other words, the value Vpeak may be somewhat higher than Vdc or Vavg by an amount allowed due to the power code. An incomplete listing of Codes is :
A: battery, generator, or 12 pulse, 6 phase, fully controlled rectifier
C: 6 pulse, 3 phase, fully controlled rectifier
K: 2 pulse, single phase full wave rectifier
Finally, the amount of AC voltage you need to produce 24VDC depends on the number of AC phases and the type of rectifier used. Also, as implied by the power codes, the amount of ripple will be determined by the same factors.
I hope this helps.
jbartos
Electrical
- Jan 15, 2000
- 6,440
Suggestion/Comment on rhatcher (Electrical) Apr 27, 2001
1. There is a big difference between 48V(peak-peak) sine wave appearing in the original posting and 48VAC sine wave appearing in the previous posting. Namely, 48V peak to peak means that one considers voltage from the minimum, i.e. -Vmax=48V(peak-peak)/2 to the maximum +Vmax=48V(peak-peak)/2, or 48V(peak-peak)= Vmax + (-Vmax) = 48V(peak-peak)/2 + 48V(peak-peak).
48VAC sinusoidal rms has 2**0.5 x 48VAC sinusoidal rms x 2 = 135.76V(peak-peak)
The AC voltage, when rectified is called a rectified sinusoidal dc voltage, either half-way of full-wave. Half-wave rectified sinusoidal dc pulses would give mathematically one half of full wave rectified sinusoidal voltage average value. The full-wave DC sinusoidal voltage, which has a ripple, has its average equivalent to the battery dc, which is difficult or impossible to obtain in practice since the filtering is used, which somewhat smoothes the ripple; however, some ripple is preserved in case that the power supply is sufficiently loaded. Obviously, the three-phase ac power supply dc equivalent is different and has a different ripple. I provided some posting in this forum in the past covering the three phase six pulse rectifier dc equivalent.
Reference:
1. Dr. M. Foigel, Director, "The Electric Circuits Problem Solver," Research and Education Association, 61 Ethel Road West, Piscataway, New Jersey 08854, Published in 1992
Problem 8-9 on page 380
1. There is a big difference between 48V(peak-peak) sine wave appearing in the original posting and 48VAC sine wave appearing in the previous posting. Namely, 48V peak to peak means that one considers voltage from the minimum, i.e. -Vmax=48V(peak-peak)/2 to the maximum +Vmax=48V(peak-peak)/2, or 48V(peak-peak)= Vmax + (-Vmax) = 48V(peak-peak)/2 + 48V(peak-peak).
48VAC sinusoidal rms has 2**0.5 x 48VAC sinusoidal rms x 2 = 135.76V(peak-peak)
The AC voltage, when rectified is called a rectified sinusoidal dc voltage, either half-way of full-wave. Half-wave rectified sinusoidal dc pulses would give mathematically one half of full wave rectified sinusoidal voltage average value. The full-wave DC sinusoidal voltage, which has a ripple, has its average equivalent to the battery dc, which is difficult or impossible to obtain in practice since the filtering is used, which somewhat smoothes the ripple; however, some ripple is preserved in case that the power supply is sufficiently loaded. Obviously, the three-phase ac power supply dc equivalent is different and has a different ripple. I provided some posting in this forum in the past covering the three phase six pulse rectifier dc equivalent.
Reference:
1. Dr. M. Foigel, Director, "The Electric Circuits Problem Solver," Research and Education Association, 61 Ethel Road West, Piscataway, New Jersey 08854, Published in 1992
Problem 8-9 on page 380
raghavan100
Electrical
the d.c votage vs ac voltage depends on the number of phases and type of connection ie.,halfwave and full wave
Ripple factor again governs the final dc voltage
sixpulse rectifier can give a higher dc voltage than its rms value if it is with proper ripple factor correction and these basic data of ac to dc votage is avilable in all handbooks for different connections
K.S.RAGHAVAN
K.S.RAGHAVAN
Ripple factor again governs the final dc voltage
sixpulse rectifier can give a higher dc voltage than its rms value if it is with proper ripple factor correction and these basic data of ac to dc votage is avilable in all handbooks for different connections
K.S.RAGHAVAN
K.S.RAGHAVAN
jbartos
Electrical
- Jan 15, 2000
- 6,440
Comment: The 24VDC motors are usually small HP motors supplied from small AC-DC converters single phase converters. Three-phase AC-DC converter would be more appropriate for 125VDC or higher voltage and higher horsepower rated DC motors. However, the previous posting is a very valuable supplement.
The full wave rectification is mentioned in my Apr. 26, 2001 posting:
Vdc=Vaverage-of-fully-rectified-17VAC=Vmax x 2/(PI)=Vrms x 2**0.5 x 2/3.1416=15.3VDC
The full wave rectification is mentioned in my Apr. 26, 2001 posting:
Vdc=Vaverage-of-fully-rectified-17VAC=Vmax x 2/(PI)=Vrms x 2**0.5 x 2/3.1416=15.3VDC
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