Help for Microcontroller isolation!

Discussion in 'The Projects Forum' started by elborgy, Apr 24, 2015.

  1. elborgy

    Thread Starter New Member

    Nov 20, 2012
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    I'm working on a project that controls 12V 5A heater with pic Microcontroller, I need to control it without relay. So I designed a circuit using optocoupler (PC817) and two power sources (Vcc 5V for PIC, Vpp 12V for heater) for isolation, I need to know if this circuit will work good with this heater or not and if there is another idea for isolation?

    Thanks in advance,
    Borgy
     
  2. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    Just out of curiosity, why do you need isolation? A low Rds FET can do this job quite well and inexpensively.
     
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  3. crutschow

    Expert

    Mar 14, 2008
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    It's unlikely that you need isolation. Just run a separate ground back to the power supply for the 12V heater current (transistor emitter) and tie the two grounds together at the power supplies.

    Note that you need to provide 500mA to the base of the transistor to properly switch a 5A load. The opto isolator or the PIC cannot provide that.
    It might be better to use a MOSFET instead of a BJT to switch the load since they can be controlled with a voltage.
     
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  4. Reloadron

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    Jan 15, 2015
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    I also see no need for isolation. However, you may want to reflect in your drawing that the transistor MJ11016 is in fact a Darlington and not a standard NPN transistor as drawn.

    Ron
     
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  5. NorthGuy

    Active Member

    Jun 28, 2014
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    Why without relay? Using relay would save at least 5-7% of energy and would eleminate heat sinks. Not to mention, it would automatically provide isolation, so that bad swings on the heater's ground wouldn't affect your MCU.
     
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  6. elborgy

    Thread Starter New Member

    Nov 20, 2012
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    Thanks for answers ,,
    Ok, there is no need for isolation, MSFET is enough!

    Which one of those circuits?
     
  7. joeyd999

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    Jun 6, 2011
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    The first.
     
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  8. Reloadron

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    Jan 15, 2015
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    If you run with the first drawing which will work just make sure the MOSFET used is a logic level MOSFET because the gate voltage will be low. For that reason I would use the second circuit allowing the 5 volts to drive the transistor to turn on the MOSFET. I would use a common 2N3904 or 2N3906 transistor.

    Ron
     
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  9. MaxHeadRoom

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    Jul 18, 2013
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    I have used the IRL540 direct driven from Picmicro OK.
    Max.
     
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  10. Reloadron

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    Jan 15, 2015
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    That would be a real good choice for the first drawing as it uses a logic level gate.

    Ron
     
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  11. elborgy

    Thread Starter New Member

    Nov 20, 2012
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    Thanks all for fast reply,,
    I'll do it with IRF634
     
  12. MaxHeadRoom

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    I would be more inclined to go with a logic gate version, the V(th) is 2v -4v which is getting close to the switching voltage being used.
    Max.
     
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  13. lmarklar

    New Member

    Apr 23, 2015
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    The IRF634 won't work in the first drawing. as MaxHeadRoom said, it's Vgs(th) is 2 - 4v and actual Vgs is 10v. You would need to use the second drawing for that guy as it will provide enough power to ensure you are fully switched. (I just went through the same thing with the wrong switching power because I misread the datasheet)
     
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  14. SgtWookie

    Expert

    Jul 17, 2007
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    The IRF634 is a poor choice. It's not a logic-level MOSFET, which complicates the gate drive requirements. It's rated for far higher Vds (voltage from drain to source) than required, and it's Rds(on) is significantly higher than it could be.

    Take a look at something like an IRLU7807:
    http://www.digikey.com/product-detail/en/IRLU7807ZPBF/IRLU7807ZPBF-ND/811176
    Logic level, Vdss=30, rated for 43A continuous (if you can keep it's junction temp down), Rds(on) is 13.8m Ohms when Vds is 10v, max of 18.2m Ohms (0.0182 Ohms) when Vds = 4.5v and Id=12A.
     
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  15. elborgy

    Thread Starter New Member

    Nov 20, 2012
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    Thanks all,,
    I found in my lab IRL640,, i think it 'll work,, i'll try it
     
  16. SgtWookie

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    Jul 17, 2007
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    With Vds=5v, Rds(on) = 0.18 Ohms; so with a drain current of 5A, you will have 5A x 0.18 Ohms = 0.9V dropped across the MOSFET, with a resulting 4.5 Watts of power dissipation in the MOSFET. You will definitely need a heatsink on it, and probably a fan blowing air on the heatsink.

    On the other hand, if you used the MOSFET that I suggested, you would have 69mW power dissipation in the MOSFET, and no heatsink required.
     
    Last edited: Apr 27, 2015
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  17. elborgy

    Thread Starter New Member

    Nov 20, 2012
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    So what can i do I didn't find IRLU7807 , i need something available to switch, can you please help me SgtWookie for that,, i tried it and yes it need cooling,,
     
  18. joeyd999

    AAC Fanatic!

    Jun 6, 2011
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    For $1.41 in singles at Digikey, you can get a IRLB8743PBF with a .0042 ohm Rds,on at 4.5V Vgs. This will dissipate about 100mW at 5A, for a temperature rise of about 6.5C without a heat sink.
     
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  19. SgtWookie

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    Jul 17, 2007
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    I already helped you with that - I gave you an example part# and where you could buy it - if you are in the States, that is. I have no idea what else you might have available where you are, or even where in the world you are, as you have not mentioned that as of yet.

    If you must use what you have available wherever you are, then you will need to do research on what you have. Make a list of your available power MOSFETs, and then go online and find their datasheets. Find out what their Rds(on) will be when Vgs is 4.5v to 5v, and multiply that by 5 to find out how much power the MOSFET will be dissipating. Hopefully, you can find something that will dissipate less than 1/4 Watt of power. If not, you'll need to use a heat sink, and possibly a fan to keep the MOSFET from burning up.
     
  20. joeyd999

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    Of course, you meant 25, right?
     
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