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Variable Load

  1. Jun 30, 2014 #1
    Hi,

    I'm currently working on my end of the year project for my studies.
    I have a DC motor mechanically linked to a wind turbine PMA. My aim is to control the speed of this motor (using PWM) to simulate different types of winds.
    My problem lies on the generator side. I need a load to dissipate the output power, but considering the fact that the output will constantly change (eg. gust of wind corresponding to a steep rise in speed) I need it to be automatically variable. I thought that a JFET would do the trick, but unfortunately they only support around 1 A at most. My PMA will deliver a maximum power of 350W and the 3 phase output goes through a rectifier to convert it to single phase.
    I have no idea what I'm looking for and what kind of solution I could use.
    Please help

    Below you will find the specifications of my PMA.
    Power Rating: 300W
    Maximum Power: 350W
    Rated Voltage: 24V
    Rated Speed: 450 rpm

    You can also find the voltage curve in the link below:
    http://www.aliexpress.com/store/pro...rbine-DHL-Free-Shipping/807020_528715296.html

    Thank you for your help =)

    Maxstag
     
  2. jcsd
  3. Jun 30, 2014 #2

    donpacino

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    look up IGBTs
     
  4. Jun 30, 2014 #3
    Thank You for the quick response.
    My knowledge of IGBTs is very limited. I have never used one and am not able to understand how it could be used to dissipate power.
     
  5. Jun 30, 2014 #4

    donpacino

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    Gold Member

    an IGBT is an Isulated gate bipolar transistor.
    To put it VERY simply it is a high power BJT with an insulated gate like a fet.
    You can set up the variable load similar to the way you would with a FET.
     
  6. Jun 30, 2014 #5

    Baluncore

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    Notice that the current does not change much with RPM. You can simulate those load characteristics with a 3.85 amp constant current sink, rated at 250W, in parallel with a 39 ohm resistor rated at 100W.

    Reading the W and V data from the graphs, then computing I gives;
    Code (Text):
     RPM     W    V       I     I-3.85A   R
      50    20   5.000   4.000   0.150  33.333
     100    40  10.000   4.000   0.150  66.667
     150    65  15.000   4.333   0.483  31.034
     200   100  21.000   4.762   0.912  23.029
     250   140  30.000   4.667   0.817  36.735
     300   180  38.000   4.737   0.887  42.849
     350   235  47.000   5.000   1.150  40.870
     400   290  56.000   5.179   1.329  42.151
     450   340  64.000   5.313   1.463  43.761
    The last two columns are the remaining current I-3.85, and the ideal parallel resistor.
     
  7. Jul 1, 2014 #6
    In general you won't be able to use a transistor as a variable load in a high power application. There's a limit to how much heat a transistor can dissipate.

    When a transistor is off it dissipates almost no heat. When it's saturated (fully turned on), it still doesn't dissipate too much heat because it's on-resistance is designed to be low. In the triode region where it has significant resistance you will reach the transistors heat limit pretty quick. In general, power transistors are not really intended to operate in triode mode.

    You can either get an insanely oversized transistor, forced air cooling, a fancy heat sink, or some combination of each in order to manage the heat.

    IBGT's can have runaway currents. When they get hot their resistance goes down which raises the current which then raises heat dissipation more and so on.

    When you test large generators you typically use a load bank. A load bank is basically a big toaster oven. It's full of heating elements and fans that blow the heat away. The load is increased by simply turning on more heaters. Load banks can get big too, up to 1 MegaWatt and beyond.

    A load bank is probably a little too fancy for a capstone project but that's how it's done in the field. I think Baluncore is on the right track.
     
  8. Jul 1, 2014 #7

    jim hardy

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  9. Jul 1, 2014 #8

    Baluncore

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    Here is the earlier data, recalculated for the load resistance needed at different RPM.
    Code (Text):
     RPM     W    V       I       R
      50    20   5.000   4.000   1.250  
     100    40  10.000   4.000   2.500
     150    65  15.000   4.333   3.462
     200   100  21.000   4.762   4.410
     250   140  30.000   4.667   6.429
     300   180  38.000   4.737   8.022
     350   235  47.000   5.000   9.400
     400   290  56.000   5.179  10.814
     450   340  64.000   5.313  12.047
    The column R shows the load required for each RPM.

    You control the speed, so I see no reason why the load must be automatic. You can build yourself a load bank from a dozen 1 ohm 25 watt wire wound resistors. If you go through the graphs carefully you will find the speeds at which different combinations of the 1 ohm resistor make a good load.

    Since “low voltage” stops at 50V, you will need to be very careful with the 64 V output at the high RPM end.

    If you want an automatic load then you would need to build a 4 amp constant current sink. The best way to do that would be with a switching current regulator into a 350 watt load of between 16 and 20 ohm. Attached is a schematic of such a beast. It works by turning on the mosfet until the current through the inductor rises above 4 amp, the transistor then turns off and the current flows through the diode and 16 ohm load. When the current falls again below 4 amp, the transistor switches on, and the cycle repeats. The transistor needed is only a low cost 4 amp, 100V N-chan mosfet. The duty cycle is voltage dependent so the same resistive load can serve all RPM requirements.

    If you want more info about a switching current sink, ask more specific questions. But consider first a manually switched load bank, it is much easier.
     

    Attached Files:

  10. Jul 1, 2014 #9
    EDIT: I hadn't seen the following comments. I will look into this current sink system. For the incandescent lamp, I have considered it however since this project will later be used for teaching purposes I have been told it would be a problem if several units emitted 350W of light in a single lab (where solar panel measurements are also made).

    So using your approach, if I used a 250W rated resistor that would mean I need a resistance of 17 ohm (P = R*I^2).
    So I could use this, customized to 17 ohms:
    http://www.arcolresistors.com/resistors/hs250-aluminium-housed-resisto/

    and for the second resistor I could use:
    http://www.amazon.co.uk/Green-Aluminum-Shell-Housed-Resistor/dp/B008FZABVU

    Would that be right?
     
    Last edited by a moderator: May 6, 2017
  11. Jul 1, 2014 #10

    Baluncore

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    The 100 watt 39 ohm resistor seems OK. It may be worth doing a more accurate analysis of the graphs before purchasing.

    It is true that 3.85A will dissipate 250W in a 250/3.852 = 16.866 ohm resistor.
    Unfortunately, it is NOT relevant to this design.
    Your understanding of the 250 W switching load has not quite caught up yet.

    Remember that for an inductor, V = L * di/dt
    A reversing voltage difference will appear across the inductor, but the current stays close to the set value.
    When the transistor switch is on, inductor current accelerates, downhill to ground.
    When the transistor switch is off, inductor current decelerates, uphill through the diode and resistor.

    The switch duty cycle will vary but, if the transistor remains off, the current through the diode and 250W resistor must not exceed the regulated current of 3.85A at the maximum generator voltage of 64V. Therefore 64V / 3.85A = 16.623 ohms minimum. That ignores the voltage drop across the diode and inductor, which happen to be in our favour.

    There is also a maximum resistance value. The transistor and the diode have a maximum voltage specification, to keep the cost down I set it at 100V. The high voltage is developed across the resistor while the transistor is switched off. 100V / 3.85A = 25.97 ohms.

    The minimum is 16.7 ohm, the maximum is 26 ohm, the standard values of 18 or 22 ohms lie in that range so either will do the job.

    If the transistor and diode were rated at 600 V you could use 600V / 3.85A = 155.8 ohms maximum, so the standard value of 150 ohms would then do. However, the voltage across the inductor would be 6 times higher. Since V = L * di/dt, the current would change 6 times faster, and the mosfet would have to switch more often.
     
  12. Jul 2, 2014 #11
    Ok thank you for the greatly precise and clear explanation. I will definitely look into this system in detail. Should the additional resistor (theoretically 39 ohms/100W) then be added in parallel to this current sink system (I would guess before the MOSFET)?
     
  13. Jul 2, 2014 #12

    Baluncore

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    The + and – terminals on the left of my circuit diagram should be connected in parallel with the 39 ohm load resistor and the generator. That way the generator current is the sum of the 3.85A constant current and the voltage dependent current through the 39 ohm resistor.
     
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