open collector constant-current driver -> MOSFET gate

Discussion in 'General Electronics Chat' started by damiannz, Apr 1, 2010.

  1. damiannz

    Thread Starter New Member

    Apr 1, 2010
    hi everyone,

    this is my first post here so i hope i'm doing everything correctly!

    i have a Texas Instruments TLC5940 constant-current PWM LED driver which has open collector output pins. normally, these allow an adjustable constant current to flow through an LED connected to a power source up to 18V. but i need 350mA, which the TLC5940 can't deliver; so rather than using the TLC5940 current directly i'd like to use it to open the gate of the MOSFET in this circuit.

    this is the first time i've encountered the 'open collector' concept and i think i understand it, but just to make sure i have two questions:

    1. since the pins on the TLC5940 are rated up to 18V, do i need that Zener diode between the output pin and ground in the MOSFET circuit?

    2. the TLC5940 includes error-detection circuitry. my initial tests with a multimeter and pull-up resistor indicate that it will not complain, but could i be doing damage to the device in the long-term?

    thanks in advance for any advice!
  2. JDT

    Well-Known Member

    Feb 12, 2009
    I think your main problem is that when the TLC5940 switches on a LED attached to one of its output pins it pulls that pin down (connects it to 0V). This will switch off the MOSFET in your circuit. So your LED will be off when it should be on.

    Possible circuit attached. Best to have V+ as low as possible as otherwise excess power will be dissipated in the transistor.

    EDIT: the small LED is there just to produce a useful drive voltage to the base of the transistor but could be a useful on-board indicator as well!
  3. Bychon


    Mar 12, 2010
    The zener thing is there to protect both the mosfet's gate and the brain chip. You haven't named the voltage you will be using, so I can't tell if anything is in danger. The circuit from JDT allows the full voltage to get to the brain chip and this still leaves open the question of whether you still need a zener to protect the chip, even with the JDT circuit.

    Give us specific information and you will get specific answers.
  4. kkazem

    Active Member

    Jul 23, 2009
    Another possible fix is to parallel 2 or more of these TI chip outputs. But you should carefully review the datasheet first. If this isn't what you want, there are lots of constant current circuits with more than 350mA capability. For example, look on the TI website as they have numerous op-amp and other constant current circuits that may be better suited to your needs than the TI chip you selected. Also, TI has live application engineers to help you that you can call or email for help. In fact, not just TI, but all IC manufacturers have app notes, white papers, and app engineers.
    Good luck,
    Kamran Kazem
  5. damiannz

    Thread Starter New Member

    Apr 1, 2010
    thanks for the replies!

    @JDT: i had noticed that there was some inversion going on; that clears up why. i'm not sure where you're getting the value of that resistor from, though; more importantly i want to steer away from using a resistor-in-series circuit here because of changes due to heat buildup.

    @Bychon: apologies. the supply voltage for the LEDS is 12V. so i take it i wouldn't need it.

    @kkazem: yes, thanks, i'd heard about that; but that is using up too many channels of the TLC5940 (yes, i really need all 16 of them - in fact for the project in mind i need about 50, not all high-current LEDs though).

    a current-mirror circuit has been suggested elsewhere as a way of getting a constant current i could then pass through a resistor to obtain a +ve voltage when current is flowing.. i will try that out.
  6. JDT

    Well-Known Member

    Feb 12, 2009
    The resistor sets the 350mA current through the big LED.
    It will have about 0.6V across it. This is the forward voltage of the small red LED (about 1.2V) minus the base-emitter voltage of the transistor (0.6V) giving about 0.6V across the resistor. Then Ohm's law.

    This is just a simple way of making a constant-current. Most of the voltage will be dropped in the transistor. Therefore most of the power will be lost in the transistor.