Testing an Industrial AC and DC Variable Speed Drives

Discussion in 'General Electronics Chat' started by NiCeBoY, May 6, 2011.

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  1. NiCeBoY

    Thread Starter Active Member

    Aug 20, 2008
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    Hi all,
    i did this test in the lab. can anyone help me with some doubts that i have?

    below is how the system was setup.
    Power circuit of lab drive setup
    [​IMG]
    The shafts of motors M1 and M2 are coupled directly. The AC VSD (A1) is a two-quadrant drive, and hence cannot provide braking. Motor M1, therefore, operates always in the motoring mode. The DC VSD of motor M2 can provide operation in four quadrants. It is set for torque control (Fig. 2). If the set torque of the DC drive is of the opposite direction to the direction of rotation selected on the AC drive, the DC
    drive will operate in the braking mode. Consequently, M1 operates as a motor, and M2 – as a generator.
    [​IMG] fig2
    Torque/Speed Characteristics of AC Drive
    1. On the control panel of AC drive, select a frequency in the range of 15 Hz – 20 Hz. This is the parameter of the first characteristic to be taken. The selected direction of rotation is clockwise (positive).
    2. Start the AC motor by depressing the START pushbutton.
    3. On the control panel of DC drive, switch to local control (LOC) by depressing the LOC/REM pushbutton.
    4. Start the DC drive by depressing the START p/b.
    5. On the DC drive control panel, set the reference torque at a negative value corresponding to the braking of the AC motor. The torque is displayed as a percentage of the motor nominal value, and can be varied in the range of 0 to (-80%).
    6. In the table below, record the displayed values of frequency f (AC drive control panel), torque T and speed n (DC drive control panel).
    7. Record the drive speed for at least four other frequency values, selected to cover evenly the remaining range up to 60 Hz. The torque is maintained constant by the torque controller of the DC drive.
    8. Repeat steps 5, 6 and 7 for at least four other torque values, selected to obtain measurement points spread evenly in the range of 0 to (- 80 %). The set of frequency values at which the speed is measured remains the same for each torque value.
    [​IMG] tab1
    Torque/Speed Characteristics of a DC Generator
    1. The procedure is similar to that described in the previous section, except that the parameter is the dc machine terminal voltage Vt, displayed on the control panel of the DC drive. The frequency is adjusted to obtain the required Vt at a given value of T.
    2. Select at least four values of Vt spread evenly over the range of 0 – 380 V. These are the parameter values for which a family of characteristics will be taken.
    3. With the AC drive on, set the torque at a value within the available range of 0 to (-80%). Adjust the AC drive frequency until the selected value of Vt is obtained. For each value of Vt, record the displayed speed n.
    4. Repeat step 3 for at least three other values of T, making sure that all torque values are spread evenly over the full range of 0 to (- 80 %).
    [​IMG] tab2
    Output Voltage and Current Waveforms of a PWM VSD.
    1. On the oscilloscope, monitor the phase current and the phase voltage of the induction motor. Keep the motor loaded with a torque of at least (-50%) (as displayed on the DC drive).
    2. Sketch the waveforms for three widely spaced frequencies in the range of 25 – 60 Hz. Note the differences between the voltage waveforms. Note that, in your report, you are required to include the scale of the presented graphs.
    3. Measure the amplitude of pulses in the voltage waveforms.
    4. Note the response of the phase current to the voltage pulses.

    [​IMG]


    50Hz
    [​IMG]



    Anyone can tell my why the current waveforms are of the shape shown above and why the amplitude of the pulses in the output voltage is constant and how the output voltage magnitude is varied?

    Thanks
     
  2. kingdano

    Member

    Apr 14, 2010
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    I didnt read your post very thoroughly, but i think i can add some clarity here.


    In order to approximate the ideal drive waveform for motors, which is sinusoidal, the method is often to PWM the drive circuitry.

    What you see on the yellow (voltage) trace isnt constant voltage, it is PWM voltage, and it appears constant because of your time scale.

    Were you to zoom in your time scale i would imagine you would see square waves.

    The current waveforms are sinusoidal because thats what the motor windings need to operate smoothly - granted they arent the prettiest sine waves, but varying load and frequency will cause the induced current to not be necessarily ideal, but still effective.
     
  3. strantor

    AAC Fanatic!

    Oct 3, 2010
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    The voltage ouput of an AC drive isn't a pretty sine wave like you might expect. It's called Pulse Width Modulation. Full voltage ("full" as determined by a V/Hz table or other algorithm depending on control mode) is pulsed in the KHz range in the + or - direction. The motor windings "average out" these pulses. That is why the current graph looks like a sine wave, while the voltage graph looks like garbage. The Duty Cylce (time ON vs. time OFF, expressed in 0-100%) is varied produce peak "averaged" voltage at the peak of the wave form and half voltage in the middle of the waveform and everything in between. This pluse frequency is referred to as "carrier frequency" and this is what you will read if you measure the frequency of the drive output with a digital multimeter. In order to measure the 0-60Hz (or 0-400Hz or whatever) output of a drive accurately you need a DMM with a low pass filter, designed for the task such as Fluke 289. If you want to see a pretty voltage sine wave on your o-scope you may be able to use a lo-pass filter in the same manner, but I wouldn't be able to instruct you how to do this as I've never attempted it.
     
  4. NiCeBoY

    Thread Starter Active Member

    Aug 20, 2008
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    so is there a way i can change the magnitude of the output voltage?
     
  5. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    So you want to TRANSFORM one AC voltage to another?

    How oh, how shall you ever accomplish that?
    :)
     
  6. NiCeBoY

    Thread Starter Active Member

    Aug 20, 2008
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    I mean if you look at the waveforms @ 30, 40 and 50Hz, they voltage (Square shape) waveform is the same with the same magnitude.
    LEt say i want to increase or decrease it. How i can do that? i am thinking of reducing the DC bus voltage.

    Any idea?

    thanks
     
  7. gerty

    AAC Fanatic!

    Aug 30, 2007
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    Look up your drive specs, I believe what you need is the "volts per hertz" setting. You typically only have a few choices, maybe it'll do what you want
     
  8. NiCeBoY

    Thread Starter Active Member

    Aug 20, 2008
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    How about the current waveform(blue).
    Any Idea why the shaped is as recorded?

    Thanks
     
  9. strantor

    AAC Fanatic!

    Oct 3, 2010
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    Depending on what drive you have, read the manual. I would say probably not, because that voltage is calculated by the drive with factors of motor pole wiring, motor power factor, motor rated slip, motor rated current, motor thermal time constant, et.al. (and ultimately by the V/Hz table)

    As I was trying to explain, the motor windings "average out" (yes, I will continue to put quotes around that) the voltage pulses sent through them. Think of them as big inductors. since the current flowing through them is an average of pulses, the amperage graph shows a more-or-less nice sine wave.

    EDIT: I guess the "quick & dirty" way to change that voltage would be to change the nameplate motor rated voltage parameter in the drive. Should be safe to go down with this number, but don't enter a number higher than what's on the nameplate. Just remember that a whole lot of stuff is calulated from those few motor namplate data paramters, so changing them could drastically alter the performance. that specific voltage is being applied for a reason.
     
    Last edited: May 9, 2011
  10. strantor

    AAC Fanatic!

    Oct 3, 2010
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    please don't attempt to connect a transformer to the output of an inverter drive as alluded to in post #5.
     
  11. noman.akbar

    New Member

    Dec 8, 2013
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    Hi Bro,
    from basic circuit analysis, u know that for inductor, current lags the voltage, i.e. V=L*di/dt. The same case is here, voltage waveform leads and current lags because motor is inductive load with windings as inductors.
     
  12. noman.akbar

    New Member

    Dec 8, 2013
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    hi bro, motor is inductive load and current lags the voltage in inductor
     
  13. Meixner

    Member

    Sep 26, 2011
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    This thread is over 2 years old.
     
  14. bertus

    Administrator

    Apr 5, 2008
    15,645
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    Hello,

    The OP has not been online for over an year.
    He probably was satisfied with the help or stopped his project.

    Bertus
     
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