MOSFET speed for driving high current into LEDs

Discussion in 'General Electronics Chat' started by ray242, Jul 14, 2017.

  1. ray242

    Thread Starter Member

    Nov 27, 2016
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    Hi,

    I've been trying to use a MOSFET (model PSMN8R2-80YS) to drive 50A current at 50us pulse width into 7 parallel LED load, but I realized the current can only rise at a rather slow speed (about 10us to get 40A).

    What can be the cause of this and how can I improve it? The wire loop length was about 10cm so I don't think inductance is the problem problem.

    I attached the circuit diagram. My power supply can not generate 50A so a big capacitor was used and the voltage was adjusted for 50A.

    In the oscillograph, the yellow line is current (10A/1V), the blue line is the drain voltage and the red line is the gate voltage.

    Thank you very much for your help(^^)
     
  2. crutschow

    Expert

    Mar 14, 2008
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    What was the voltage you adjusted the power supply to?
    Is the capacitor connected with as short leads as possible between the LED's anodes and the MOSFET source?
    At that current level, even small wire inductance can have an effect.
     
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  3. ray242

    Thread Starter Member

    Nov 27, 2016
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    The voltage was 6.5V.
    Yes, I used very short wires, an inductance calculator tells me the inductance is around 150nH, which corresponds to di/dt of about 35A/us for 6.5V.

    Thank you.
     
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  4. crutschow

    Expert

    Mar 14, 2008
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    How much voltage does the LEDs drop?
    The voltage for the di/dt calculation (that which appears across the parasitic inductance) is the difference between the supply voltage and the LED drop, not the supply voltage.
    You can add a resistor in series so you can use a higher supply voltage to reduce the di/dt value.
     
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  5. ray242

    Thread Starter Member

    Nov 27, 2016
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    I looked at the datasheet and it's about 2.6V, does that mean there's a drop of 3.9V across the wire? But that still doesn't give me a slow enough value to account for the speed.

    Do you mean increase the di/dt value? Because I want to increase the speed but not reduce it...

    Thank you
     
  6. ray242

    Thread Starter Member

    Nov 27, 2016
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    Sorry I didn't notice that the unit is mA in the datasheet. It only provides a diagram for current up to 1.2A. So can't know how much the voltage drop is.
     
  7. ian field

    AAC Fanatic!

    Oct 27, 2012
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    First place to look is the time it takes to charge the gate capacitance.

    Not exactly the same thing - but a lot of fast SMPSU designers drive the gate with a complementary emitter follower pair. As switching speed rises, the transition losses can make things get hot.

    Also look for proprietary technology low capacitance gates. One manufacturer boasts a mosaic gate structure that mimics multiple parallel MOSFETs.
     
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  8. ray242

    Thread Starter Member

    Nov 27, 2016
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    If you look at the gragh I posted, the gate voltage rises rather sharply compared to the current, so I don't think the gate capacitance is the problem. Is that correct?

    Thank you
     
  9. crutschow

    Expert

    Mar 14, 2008
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    Yes, I got it backwards. :oops:
    Well, if you have no resistor in series with the LEDs, then they must be dropping most of the voltage.

    How high a voltage can your power supply provide?
    I would try a fraction of an ohm resistor in series with the LED
    Experiment until you find a value that will give the desired 50A pulse with the full power supply voltage.
    A short piece of nichrome wire should work.
    See how much that improves the risetime.
    Correct.
     
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  10. ian field

    AAC Fanatic!

    Oct 27, 2012
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    With very narrow pulses - you may only need to worry about abs-max peak current.
     
  11. crutschow

    Expert

    Mar 14, 2008
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    He presently adjusts the voltage to give the peak current he wants.
    If he wants a faster rise-time he needs to increase the voltage.
    To control the peak current with the increased voltage he will likely need to add a small amount of resistance in series with the LEDs.
     
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  12. ray242

    Thread Starter Member

    Nov 27, 2016
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    Hi Crutschow,

    I won't be able to try that before going back to uni on Monday, but can I ask why would that work? If you add a small resistor and increase the voltage, wouldn't the voltage across the inductance remain the same? Because the increased voltage would drop across the resistance?

    Thank you.
     
  13. crutschow

    Expert

    Mar 14, 2008
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    No.
    At the instant the transistor turns on, all the voltage appears across the inductance, since there is no current.
    In effect it gives a "kick-start" to the inductive current.
    The voltage across the resistor does increase with time but that's after the initial turn-on which is now faster.
    The higher the resistance compared to the inductance, the more the risetime will look like that of a pure resistance.

    Here's an LTspice simulation that shows the difference.
    Both inductors are the same value but one has a 1Ω resistor in series and the other has a 5Ω.
    The 1Ω is driven by a 1V pulse and the 5Ω is driven by a 5V pulse so they both end at the same 1A of current.
    As you can see the inductor with the higher resistance and the higher voltage has a much (5 times) faster rise-time.

    Make sense now?

    upload_2017-7-15_20-22-9.png
     
    Last edited: Jul 15, 2017
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  14. ray242

    Thread Starter Member

    Nov 27, 2016
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    Yes!! Thank you so much.
     
  15. crutschow

    Expert

    Mar 14, 2008
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    I forgot to include a plot of the inductor voltages, which clearly shows why the L2 current has a faster rise than L1.

    upload_2017-7-15_22-26-30.png
     
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  16. ray242

    Thread Starter Member

    Nov 27, 2016
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    The only thing that worries me is the power dissipation. If I want to use 20V, after turning on, the voltage drop across the resistor would be 14V, so the power would be 14*50=700W, whereas the power dissipated by the LEDs is only 6*50=300W. So how can I deal with the power dissipation?

    Thank you.
     
  17. crutschow

    Expert

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    You said the pulse is only 50μs wide, so the total energy dissipated is low.
    It's the energy dissipated, not the peak power that's important.
    The resistor would only dissipate 700W*50μs = 35mJ of energy (equivalent to 35mW for 1 second).
     
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  18. ebeowulf17

    Active Member

    Aug 12, 2014
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    A related question would be whether or not the pulse repeats. If so, how often and at what intervals? If it's many times and at short intervals, perhaps power dissipation becomes an issue again?
     
  19. crutschow

    Expert

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    It could.
    The average power dissipation is the average pulse power times the pulse width times the pulse rep rate.
     
  20. Sensacell

    Senior Member

    Jun 19, 2012
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    If you are going for really short, high current pulses- pay attention to the physical layout.

    build the circuit with the shortest possible leads, with large conductors, use a ground plane.
    Place your gate driver as close as possible to the FET.

    Build it like and RF amplifier- because it is.
     
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