Motor Drive Topology

Discussion in 'General Electronics Chat' started by pilko, Oct 20, 2011.

  1. pilko

    Thread Starter Senior Member

    Dec 8, 2008
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    20
    Hi all,
    I am in the process of designing a PWM single direction Motor driver for
    my HVAC Fan.
    The motor I will be using is a Leeson 1/2 HP; 1800 RPM; 90 Volt; 5 Amp;
    PM DC Motor.
    I will be dealing with overcurrent and overspeed protection seperately.
    I havn't made a final decision on the components yet, but to keep it simple the topology I would like to use is attached.
    So -- my question is, "is this topology suitable?" It looks too simple.
     
  2. strantor

    AAC Fanatic!

    Oct 3, 2010
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    your diode is backwards. your circuit looks too simple because you haven't included current limiting (yes, I read what you said) or power supplies (where's your 5, 12, & 90V coming from?) and all the rectifiers, caps, regulators, et. al. that goes along with that. You haven't included bypass caps in your circuit or pullup/pulldown resistors, diodes, snubbers, and capacitors assosciated with your driver (see figures 18-24). and theres' a lot more than 3 traces going to a PIC. I would call that circuit a block diagram.
     
  3. pilko

    Thread Starter Senior Member

    Dec 8, 2008
    213
    20
    High Strantor,
    Thanks for your reply.
    1 "Diode reversed" oops --- dragged symbol into place and forgot to rotate it.
    2 I agree with all of your comments and I realise all of the requirements,however at this stage I am only interested in the topology.I was hoping I could deal with the details later (I will probably be seeking more help).

    regards
    pilko
     
    Last edited: Oct 20, 2011
  4. strantor

    AAC Fanatic!

    Oct 3, 2010
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    sure then yeah that looks like a good motor control circuit, minus all the finer details
     
  5. SgtWookie

    Expert

    Jul 17, 2007
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    If you're going to use resistive current sensing, it'll be a lot easier if you do so on the low side of the load, which means that you'll need to switch the current on the high side of the load; which also means using a high-side driver.

    You could use a diode for the flyback, but 100v is cutting it rather close. Consider using a diode rated for at least 110v.

    Better yet, consider using a MOSFET switch as an "ideal diode".

    I see you're planning on using a PICAXE, but that's probably going to cripple your efficiency. You won't be able to use PWM at a frequency high enough to minimize losses in the motor.

    Consider using something like a synchronous buck regulator, and just use the PICAXE to monitor overall performance, adjust the current, check to ensure the motor is rotating at a speed that's reasonable for the current being supplied & measured airflow, and issue a warning if not, or shut the power down if the current is far in excess of what it should be - etc.

    Something like this regulator:
    http://www.national.com/pf/LM/LM5116WG.html#Overview
     
  6. pilko

    Thread Starter Senior Member

    Dec 8, 2008
    213
    20
    @ SgtWookie,

    Thanks for your response and recomendations.

    - I am planning on using a Hall current sensor (I have a couple spare)

    - I have an air flow monitor in my air duct (previous Picaxe project)

    - The diodes on the schematic are rated at 200V, -- did I miss something?

    - The Picaxe can PWM out up to 10MHz.

    - I need to get my head around your Sync Buck Reg idea.

    - Use Mosfet as ideal diode -- I need to study that one also.

    Regards

    pilko
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    The MBR20100CT are rated for 20A, 100v, according to ONsemi's datasheet.

    You don't say which PICAXE you're using. However, PIXAXE uC's are Microchip PIC uC's with bootloaders. I find the 10MHz statement rather optimistic, as that would be simply on/off at 50% duty cycle with no provisions for adjustment - IF the PIC being used had a clock frequency of ~40MHz.

    The synchronous buck concept isn't really terribly difficult, once you realize that the MOSFET that takes the place of the diode is simply switched on when the current is flowing in the proper direction - basically, when the voltage at the source terminal is higher (more positive) than the voltage on the drain.

    [eta]
    If you have Java installed, click on this link:
    http://www.falstad.com/circuit/#$+1...r+384+224+384+256+0+20.0 w+336+224+384+224+0
    Click the RESET button (upper right corner of the simulation) after it's loaded.
    This is a simplified simulation of a synchronous buck, where the output is grounded. The 1H inductor represents your motor, the 20 Ohm resistor represents a combination of things, but mainly the resistance provided by the work that your motor is doing (turning the fan which moves air).
    The upper MOSFET is on 10% of the time, supplying current to the "motor". The upper MOSFET turns OFF, and after a brief "dead-time" delay (where you can see a short period of current flow through the diode; representing the MOSFET body diode) the lower MOSFET turns ON, and all of the current flows through that MOSFET back to the inductor.

    As the current through the inductor wanes, the lower MOSFET turns off (more dead-time) and you can see a brief pulse of current through the body diode; then the upper MOSFET turns on and the cycle repeats.

    Hopefully this will help you to understand the basics a bit easier.

    You CAN adjust the duty cycles of the 40Hz supplies, but you need to get the phase delay on the lower supply correct, and the duty cycles should add up to somewhat less than 100% - as some "dead time" is required.

    If you increase the duty cycle for the upper MOSFET, you must increase the resistance, or the simulation will complain of a convergence problem and stop.
     
    Last edited: Oct 21, 2011
  8. SgtWookie

    Expert

    Jul 17, 2007
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    I suggest that you also monitor the RPM of the fan. Then you can compare the RPM to the airflow, and if the airflow is lower than it should be, you can have the PICAXE alert you that perhaps the filter is dirty/blocked, fan needs cleaning, or other airflow problem.
     
  9. pilko

    Thread Starter Senior Member

    Dec 8, 2008
    213
    20
    Thanks Wookie --- your right,that diode is 100V,its the other one on the schematic that's 200V.
    Also thanks for the other info (need to do some studying)--I thought when I passed 65 I had done studying -- seems like Iam just getting started.

    pilko
     
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