Hard to drive MOSFET to work harder?

Discussion in 'General Electronics Chat' started by oookey, Mar 19, 2011.

  1. oookey

    Thread Starter Member

    May 24, 2010
    62
    0
    Hi everybody :),

    Please ref to the attached file for the circuit schematic.
    I need advice where went wrong to this circuit, it could not fully drive the MOSFET to let desired current to flow :(!
    I expecting 1.8A to flow through but only 1.33A, the LEDs (10W) are well attached with heatsink (individually), the temperature measured 43°C~45°C. The MOSFET flange temperature 30°C, I believe overheating is not an issue here.
    The power supply is 24V 2.65A (SMPS adapter).
    Thanks:).
     
  2. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    According to your circuit, point A voltage is 24x27/(2200+27) = 291mV.

    Therefore the constant current value through the MOSFET with the source 0.1Ω sensing resistor will be:

    I = V/R = 0.291 / 0.1 = 2.91A, not the 1.8A you would have wanted.

    If you cannot get this value, something is not quite right with the circuit.

    To locate the problem, can you measure the two voltages as shown when you are getting 1.33A and post back?

    [​IMG]
     
  3. oookey

    Thread Starter Member

    May 24, 2010
    62
    0
    Thanks eblc1388,

    please refer the attached for voltage readings.;)
     
  4. eblc1388

    Senior Member

    Nov 28, 2008
    1,542
    102
    From the result it is clear that the current is limited by the LEDs themselves and not by the constant current circuit.

    The MOSFET has been fully turned ON.

    The LEDs will only take 1.33A at 24V. It is their power characteristic. There is no way you can increase their current beyond that, besides raising the supply voltage.

    [​IMG]
     
  5. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    If your current is 1.33A then the voltage across R5 should be 0.133v.
    You're indicating 21.9v on the output of the LM358, so approximately the same voltage would be on the gate.
    21.9v - 0.133v = 21.767v for Vgs. Most International Rectifier power MOSFETs have an absolute maximum limit of +/-20v for Vgs. If you exceed that value, the MOSFET will be destroyed.

    You are operating LEDs in parallel without current limiting in each series string. This is not a good idea, as it is likely that one string will have a lower Vf than the other, and thus receive more current, which causes more power dissipation as heat, which lowers the Vf, which causes yet more current to flow, etc.
     
  6. shortbus

    AAC Fanatic!

    Sep 30, 2009
    4,015
    1,531
    I may be totally off base here because I'm still learning, but. Doesn't having R5 between the mosfet source and ground make the mosfet a HIGH side device? That would make the gate voltage not high enough to completely turn on the mosfet. By moving R5 to the drain side of the mosfet(between the drain and the LED connection) it would still limit the current but make the mosfet a LOW side device allowing it to turn on completely.

    Please correct me if wrong :)
     
  7. SgtWookie

    Expert

    Jul 17, 2007
    22,182
    1,728
    Shortbus,
    R5 is simply a current sense resistor. The voltage across R5 will be relatively low, even with 3A current it would only measure 3a*0.1 Ohms= 0.3 Volts, or 300mV. This is not very significant compared to the range available for gate control.

    One big problem is that the maximum Vgs of the MOSFET device is being exceeded, which usually results in the destruction of the MOSFET. However, the MOSFET seems to be conducting normally, so something is wrong in the area of the LEDs; they are not performing according to their ratings.
     
  8. oookey

    Thread Starter Member

    May 24, 2010
    62
    0
    thanks sgtwookie,

    yah, i'll add series R along each LED path, but for the value of R3, how would i choose the correct Ω value? there isn't specific gate current in the datasheet, pls advice.
    :confused:
     
  9. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    Why not jack up R3 from 47 ohms to 2.7k or so?
    This will protect the MOSFET's gate from exceeding 20v yet still allow enough voltage to drive it at lower levels.
     
  10. oookey

    Thread Starter Member

    May 24, 2010
    62
    0
    but i think there should have some kind of calculation to workout a specific value, rather try an error , right? may be 2.7k would slow down the charging of Ciss, how about 1k or 500...:confused:
     
  11. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Oookey, what kind of MOSFET are you really using? Your schematic doesn't match the datasheet.
    And what is the part number of the LEDs?
     
  12. oookey

    Thread Starter Member

    May 24, 2010
    62
    0
    Ron H,

    the datasheets of the component that i used. the MOSFET is attached in the 1st thread.
    :)
     
  13. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    If the max output of the op-amp goes to +24v, and if you do the voltage divider math, Vg will never exceed ~19 volts with 2.7k. For lower values of max op-amp output voltage (which is very likely), the resistor will be proportionately less.
     
    Last edited: Mar 20, 2011
  14. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    Are you planning to modulate the LED current?
     
  15. someonesdad

    Senior Member

    Jul 7, 2009
    1,585
    141
    Ookey, you're in good hands with RonH and SgtWookie. However, one thing I would suggest if you haven't done it is to verify that your power supply can indeed supply its rated current at 24 volts. I've found myself more than once troubleshooting something in a circuit that's not working right only to notice that my power supply is in constant current mode and the current knob is adjusted too low! Once you've verified the power supply, then you can use it along with appropriate resistors to verify the operating characteristics of each of your LEDs.

    In other words, verify the operation of each of the components. Because you may have damaged the FET, you should also check it.

    When you're troubleshooting, measure the voltage drop across each high-current component from the power supply to the sense resistor. That's valuable to help you (and us) diagnose what's going on.
     
  16. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    Look at page 19 on the bottom left for how you can do this with 2 parts:

    http://www.national.com/ds/LM/LM117.pdf

    Make sure the resistor is at least 5 watts and make sure to give the regulator 3 volts or more for itself.
     
    Last edited: Mar 20, 2011
  17. someonesdad

    Senior Member

    Jul 7, 2009
    1,585
    141
    Excellent suggestion, Jaguarjoe! At the moment, I'm using the two-part 1 A current source on the previous page and it works nicely. I'm using a TO-3 package and it requires a heat sink at 1 amp (I like to be able to touch things with my fingers and not get burned), so if you use this method oookey, be sure to provide a good heat sink.
     
  18. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    That's one reason why I asked if he was planning to modulate the current. Easy with his circuit, tough or impossible with the LM117.
     
  19. Jaguarjoe

    Active Member

    Apr 7, 2010
    770
    90
    OP never mentioned modulation, I just showed an easy way to light him up. I think right now he'd just be happy see something work. This will do modulation though, if it must be linear:

    http://www.modulatedlight.org/optical_comms/LED_linear_modulator.html

    The basics of this circuit and his are identical, the 2 big differences are choice of Rs value and the inclusion of Rl.
     
    Last edited: Mar 20, 2011
  20. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
    657
    The other reason I asked about modulation is that oscillation may occur, or frequency response can be pretty bad (peaking in the frequency domain, overshoot or ringing in the time domain) if the compensation network (R8, R11, C4) is omitted. In fact, his original circuit may oscillate if no compensation is included, because the MOSFET input capacitance adds another pole in the feedback loop.
    I have to admit that I have not breadboarded these circuits, but simulations indicate that the stability is marginal without compensation.
     
Loading...