Help with my first Opto Isolator / Mosfet circuit

Discussion in 'The Projects Forum' started by nchoop, Jul 10, 2015.

  1. nchoop

    Thread Starter New Member

    Jul 1, 2015
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    Following on from project on the following thread

    http://forum.allaboutcircuits.com/threads/help-with-adding-a-led-to-my-circuit.113051/#post-877304

    I've been talking to the manufactures of the module I'm using to drive this circuit and they have suggested that I use MOSFETs to switch my load together with an Opto Isolator to keep everything separated.

    The only problem is that I knew nothing about either until I started doing some research and I'm not sure I know much more now !!!

    I think my needs are simple but I may be wrong.

    In my original design the relays are used to switch the +12V line of some led strips. I already do this successfully using a different type of module from the same manufacturer that has built in relays but it is quite an expensive way of doing it.

    I have created a schematic which I have uploaded.

    The input form the module (controller) is 3.3V @ 10ma so I think that I've got that bit covered.

    When the controller output goes high I need the MOSFET to turn on, I think I've achieved that but I'm not sure as there are so many different circuit examples out there which appear to contradict themselves; at least in my mind.

    This is not going to need to switch these lights on and off either quickly or repeatedly.

    I've attached the Datasheets for the two components. The opto isolator is a four channel one but I've only shown one channel in my schematic.

    The maximum that I need to switch is 12V @ 10A.

    Can the circuit be this simple or am I totally missing the mark ?

    - Neil
     
  2. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    This can work, but you're not quite there yet.

    1. Easy part: even though one end of a diode is called the cathode, the abbreviation is K, not C.

    2. You show the pull-down resistor going to -12. Is this really a negative voltage with respect to ground, or is this what you are calling the return leg of a +12V power source. If you don't have two separage voltages, +12 V and -12 V, then rename this GND.

    3. The connections to the MOSFET are incorrect. U2 pin 3 goes to the Gate.

    4. If you are going to stay with an N-channel MOSFET, then the connections to the LED strip have to change. The + end of the LED strip goes directly to the LED driver, the - end of the strip goes to the Drain, and the Source goes to ground, the return leg to the LED driver. The LED driver return and the controller circuit ground do not have to be connected together, but they can be if necessary.

    Attached is a redrawing of your schematic using parts in my design library. Your parts should be fine for this project, but at 10 A you can't use the surface mount version of the MOSFET because it will overheat. At 1K, R2 will turn off the MOSFET quickly, but it will sink 12 mA when the MOSFET is on. Make sure your optocoupler has a minimum CTR of 100%, or increase R2 to something between 2.2K and 4.7K.

    ak
     
  3. nchoop

    Thread Starter New Member

    Jul 1, 2015
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    ak,

    Thank you for your reply.

    I'm not sure that the circuit that you have provided for me will work, not that I don't appreciate your time and effort but please tell me if I'm wrong.

    The driver for the led strip is http://neonica.eu/rgb-amplifier-12-24v-max-12a.html

    At the moment I am able to switch the + output of this via one of the relays in this module. http://www.idratek.com/public/docs/datasheets/QRI002_DS.pdf

    The only problem with this module is that the relays can only handle 4A. I could use the relays on this to operate more relays of a larger capacity but this involves adding more moving parts.

    I was hoping that by using a MOSFET I could just interrupt the + output as I do already.

    - Neil
     
  4. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    There is a way to modify the circuit I posted for switching the +12 side of the LED strips. But first, please post a schematic sketch of the way these modules are wired together.

    ak
     
  5. nchoop

    Thread Starter New Member

    Jul 1, 2015
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    ak
    I've created a schematic of the way that the system is setup at the moment.

    The module that is to be replaced is the Idratek Relay Module.

    I hope this makes sense now.

    - Neil
     
  6. nchoop

    Thread Starter New Member

    Jul 1, 2015
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  7. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Yup.

    ak
    LED-Control-2-c.gif
     
  8. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    Is the signal coming from the controller a simple on/off command, or is it PWM for dimming?

    ak
     
  9. nchoop

    Thread Starter New Member

    Jul 1, 2015
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    Thanks ak,

    The signal from the controller is a simple on/off command.

    The LT-393-5A carries out the dimming by PWM.

    The only thing I don't understand is where I connect U1 pin4 to as in my mind the +12V is coming from the LED driver, through Q? and on to the LED strip +12V i.e. replacing the relay contact in my existing setup. The strip has 3 returns being an RGB strip. My understanding is that the dimming is carried out for each channel by PWM which I think is turning the -ve for each channel off and on at a very fast rate.

    Does this mean that I need to connect U1 pin4 to the LT-393-5A -ve / power supply terminal.

    Thanks.

    - Neil
     
  10. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    There is no -ve connection on you schematic, but I think I know what you mean, what I call the Led Driver Return.

    Top right corner, Power Supply for LEDs, the - terminal; that is the reference voltage for the entire LED system, everything to the right of the relay module, including the circuit that energizes the gate.
     
  11. nchoop

    Thread Starter New Member

    Jul 1, 2015
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    ak

    Thank you for your reply.

    My mistake I thought I had labelled those two wires.

    I'll get the parts ordered which should be here on Tuesday, get it tested and report back.

    - Neil
     
  12. nchoop

    Thread Starter New Member

    Jul 1, 2015
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    ak,

    Could I get your opinion on something.

    Could I substitute the P type MOSFET with an N type to switch -ve and use the same circuit and could it also be used to switch AC with a suitable MOSFET? I'm thinking that as there are going to be 8 of these on one board I could possibly mix and match components as needed.

    - Neil
     
  13. AnalogKid

    Distinguished Member

    Aug 1, 2013
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    1. The circuit in post #2 is an N-channel MOSFET switching the ground leg of the load. Note that the parts are what is sitting in my design library. There are a zillion MOSFETs that will work in this application, and the resistor can be 5% or 10% tolerance.

    2. "switch AC" - Low voltage AC like the secondary of a power transformer is different from what we have discussed so far, and AC powerline switching is *very* different. Post a sketch of what you are considering.

    ak
     
  14. nchoop

    Thread Starter New Member

    Jul 1, 2015
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    ak,

    Thank you for all your help on this. After some consideration I'm not going to do any AC switching. The modules that this project are designed to replace are more than capable of switching AC voltage and are primarily designed for that purpose. No point in re-inventing the wheel.

    All of the parts arrived today and I assembled a test circuit and I can happily report that it works. Although I never doubted that it would.

    The only one small problem that I have is that when the LED strip is turned on in the software it's actually off and vice versa. I'm waiting to hear from the manufacturers because they may be able to reverse the logic in the module firmware, they're quite good at that sort of thing. However, if it turns out that they can't, how can I modify / improve the circuit to achieve this ?

    - Neil
     
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