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LSF0204D: Controlling LED

Nobuhiro Yoshizawa
Prodigy 10 points
Part Number: LSF0204D

Tool/software:

Hello, I'm using LSF0204D to control LED.
But I have a trouble where the LED turns ON very weakly even when it should be completely OFF.
So, I'd like to know the cause of the issue.

In my design, Vref_A is 1.8V and Vref_B is 5V.
A control input signal (push-pull) is connected to A port, and the output signal from B port drives the LED. LED is connected to 5V through a pull-up resistor.

When the control signal is low, LSF0204D FET is ON and the LED turns ON correctly.
But when the control signal is high, LSF0204D FET is OFF and the LED still turns ON very weakly.
With the measured voltage of the B port pin, it seems that B port accepts about 50uA current even when the FET is OFF.

I'd like to know whether 50uA is reasonable value or not.
If the "IIH" in the datasheet is the definition of the leak current, it should be less than 5uA.
Is my understanding correct?

Thank you very much in advance.

  • 0
    TI__Mastermind 27011 points

    Hi Yoshiza-san,

    Is there a schematic that you can provide? In the event that the device is off (EN= LOW, or logic high is driven from A-side), the FET should also be off and the only current the LED should see is through the external pullup on 5V side. Can we disconnect this load and measure this current again? 

    Regards.

    Jack

  • 0
    Prodigy 10 points

    Thank you very much for your quick comment and suggestions.
    I really appreciate your support.

    Please refer to my schematics attached.

    "LEDR" on A-side is the control signal for LED(Red).
    "LED1" is the LED (Red and Green in one package) with external 1.2K ohm pullup to 5V.

    Red and Green are almost the same design, so I'm now considering Red at first.

    I measured the current at the pullup resistor R30. It is about 19uA.
    I said 50uA in the previous question, but probably it was not exact.
    It was a value I estimated with the measured voltage at B-side being about 3.45[V].
    I suppose the difference was caused by my estimation without the actual characteristics of LED when the current is below 1mA.

    Now I actually measured the current at R30, and it was about 19uA.
    When R30 was removed, the current was 0 and of course LED did not turn ON.
    It seems that 19uA is still large enough to turn the LED ON (though very weakly).
    Is it reasonable for LSF0204D to accept 19uA when FET is OFF?

  • 0 in reply to Nobuhiro Yoshizawa
    Guru 283130 points

    The LSF indeed allows a small current to flow when the voltages on both sides are not higher than Vref_A.

    See section 9.2.1.2 of the TCA9535 datasheet for a similar circuit where the forward voltage of the LED keeps the voltage too low (which leads to a different problem). A 100 kΩ pull-up, or a 47 kΩ resistor in parallel with the LED, will allow the voltage to go higher and then should shut off all current.

  • 0 in reply to Clemens Ladisch
    Prodigy 10 points

    Thank you very much for your suggestion.

    For trial, I added 100K ohm pullup and confirmed that the brightness of the LED was actually reduced.
    But adding 100K ohm requires to modify the circuit(PCB), and I don't want to modify it.

    I need to reduce the brightness of the LED to be invisible when FET is off.
    If you have any good solution, I would appreciate your advice.