Control on/off state of IC with arduino

Discussion in 'The Projects Forum' started by jpborunda, Aug 22, 2014.

  1. jpborunda

    Thread Starter Member

    Apr 9, 2014
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    Hello everyone. I am testing a 5V-200VDC-DC converter IC. I tested the IC with a bench power supply, and I saw that it is drawing approximately with 5V: 450mA at start-up, then the current decreases exponentially until it stabilizes at aprox 120mA when connected to the Load.

    I would like to control when the ON/OFF state of the converter with an arduino, however, I am aware that the arduino digital pins can only source 20mA safely, so it cannot be done directly.

    To provide the needed current, I thought maybe I could use a battery, say a 9V battery with a decent amp-hr, a regulator or Linear Dropout Regulator to step down the voltage to 5V, and connect that output to the converter. But again, how could I control the on/off state of the battery using the arduino?

    Is this an okay approach? I'm open to suggestions of course, since I'm fairly new at this.

    Thanks, any help is appreciated.
     
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  2. shteii01

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    Feb 19, 2010
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    Does the ic chip has On/Off or Enable/Disable pin?
     
  3. Alec_t

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    You could use a logic-level MOSFET, or a relay, to switch the 5V supply to the converter.
     
  4. jpborunda

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    Apr 9, 2014
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    No it does not. It only has a +5 V input, and a Gnd input.

    Using a relay, I asume it would be something like this:
    http://img.bhs4.com/b1/6/b1642fbf2a12f5c4c1d4a3427518bafebeff8993_large.jpg
    Just a couple of questions to make sure Im picturing it right.
    1. Could I use the 9V battery for "supply voltage" (for the relay coil)?
    2. I need to connect the common terminal to the LDO_Input, then LDo_Output to the converter.
    3. The Normally open/closed pins to 9V and Ground respectively.
    Right?
     
  5. Alec_t

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    Q1) Yes. Make sure the relay coil rating is 9V. You will need to choose the transistor and base resistor value dependent on the coil current requirement.
    Q2) Yes (assuming you are referring to the common terminal of the relay contacts).
    Q3) Connect 'N.O' pin to 9V, leave 'N.C' pin unconnected.
     
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  6. MikeML

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    Oct 2, 2009
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    Here is how I would do it.

    If you place the PFET power switch between the output of the 5V regulator and the inverter, it is simple. If you want to place the PFET upsteam of the 5V regulator, it is doable, but a bit harder.
     
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  7. jpborunda

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    Apr 9, 2014
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    Thanks for the help guys. I've managed to simulate a circuit like MikeML suggested. (Attachment 1) . This will get the job done too. I've noticed that the converter signal will be active when the arduino output is low.

    Could I use an n channel fet to activate the converter when the arduino output is high? What would I have to change based on this configuration? Im thinking a NOT GATE could do that, but I imagine there must be a "cleaner" way by using the Nfet.
    Thanks again everyone!
     
    Last edited: Aug 24, 2014
  8. MikeML

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    One extra ! in the Arduino code will do it, or output a low when you would have output a high...

    I assume you control the Arduino code.
     
  9. ScottWang

    Moderator

    Aug 23, 2012
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    Chnage to N msofet and using the arduino output a high level voltage is a good way for the start time.

    Move R1 to the Vg and GND, move the converter to the Vd and the output of LT1083, the Vs connecting to GND.
     
  10. MikeML

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    Oct 2, 2009
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    Scott,

    If you mean what is shown in the attached schematic, then it probably will not work, because the converter minus input (bottom end of Rc) likely cannot be floated from ground.
     
  11. jpborunda

    Thread Starter Member

    Apr 9, 2014
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    Yes, I was trying to analyze what Scott proposed, but as you say the minus terminal cannot be floated from ground. Do you guys think my best bet is to just go with the PFet? I CAN change the arduino code.

    Thanks a ton!
     
  12. MikeML

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    If you own the code, then use the high-side switching with the PFET.

    How about using the PFET to switch the input of the voltage regulator? To do that, it would add an NPN transistor and a couple of resistors between the Arduino and the PFET. It would, however, add another inversion, so the Arduino port pin could use positive logic...
     
  13. jpborunda

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    Apr 9, 2014
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    Oh I see, thanks for the solutions. I'm trying to keep components to a minimum, so prob. the quickest way will be to just change the code. Thank you guys, this is so useful for us beginners!
     
    Last edited: Aug 24, 2014
  14. ScottWang

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    Aug 23, 2012
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    You mean the converter minus input already fixed to ground, if really did that, then we have to thinking some other ways.
     
  15. ScottWang

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    Aug 23, 2012
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    Do you have any infos about 5V-200VDC-DC converter IC?
     
  16. jpborunda

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    Apr 9, 2014
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  17. ScottWang

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  18. MikeML

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    Take an Ohmmeter. Measure between IN- and OUT-. If the resistance is ∞, then you could use a low-side switch...
     
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  19. ScottWang

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    Aug 23, 2012
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    The module you choose is a DC to DC, so you don't need the pulse to drive it, you just using a high to driver the n-ch mosfet, and the mosfet will be turn on, and drive the the 5V side, and then the output side will output a 200Vdc voltage.

    When you need to using the pulse to drive a mosfet, the load almost is a low voltage to high voltage transformer, not the DC to DC converter.
     
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  20. jpborunda

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    Apr 9, 2014
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    Ok guys, I will go ahead and check the resistance. The reason I said it couldn't be connected on the low side is that when I ordered it, they said just put "5V on one terminal and the other connected to ground", so I assumed that it HAD to stay that way. I dont own a multimeter, so I'll check the resistance sometime this day and report back, just to let you know.]

    Oh and I'm aware that I dont need to pulse drive it. It was just for the sake of simulation, so that I could see the "ON, OFF" states. Thank you very much!
     
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