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LM3410: Datasheet correction

Jonathan Navor
Genius 9475 points
Part Number: LM3410

Hi team,

Could you please help to check our datasheet for correction. Here it is. 

I am fairly sure there is an error in the datasheet for LM3410 ICs.

Only found out once I got the board and started hooking thinkings up.

Application examples page.

8.2.6, OVP protection

R3 is basically in series with the pin 3, feedback pin and the R1 resistor that determines current through the LED. In the LM3410, The feedback voltage is 200mV, meaning the values for R1(4 Ohm) is right to deliver 50mA. However R1 value is 100 ohms and will reduce the current to 200mV/(100+4) = 1.9mA

The R3 seems to be a current limiter resistor for the Zener diode. So a proper schematic seems to be one where it's in series with the Zener. So the feedback pin will go direct to R1, not go through R3 first.

Figure 14 too.

So I think the intentions are correct, because in 8.1.1.7 it says Zener diode D2 and resistor R3 is placed from V OUT in parallel with the string of LEDs.

Thank you so much.

Best regards,

  • 0
    TI__Mastermind 18890 points

    Hello Johnathan,

    The schematic is correct.  Your are missing that the zener voltage is above the LED voltage.  So the current to thru the zener and R3 is 0A when the output voltage is driving the LED it's lower than the zener voltage therefore no current flow thru R3 and zero voltage drop.

    Thanks Tuan

  • 0 in reply to Tuan Tran74
    TI__Genius 9475 points

    Hi Tuan,

    Thank you so much for the support.

    Although our customer still have a question. The feedback voltage of Pin 3 is what determines current through the LED. And it has to go through R3 before it can reach R1, which is the feedback resistor. You are saying when the Zener is off, there will be no current through Zener and R3, which I can understand.

    Page 15, Figure 14,
    Let's assume the Zener is off and open circuit. The Vfb still needs to go through R3, then R1. The Vfb needs to be on the right side of R3 in order for it to reach R1 directly, not left. In fact that was the modification I needed to do to have the circuit reach desired brightness. Initially the LED was quite dim at max brightness.

    Hopefully you can understand my original question a bit better now.

    Best regards,

    Jonathan

  • +1 in reply to Jonathan Navor
    TI__Mastermind 18890 points

    Johnathan,

    I do understand your question.  The key is that the current going into the FB pin max is only 1uA so the error due to voltage drop across the 100 Ohm is only 0.1 mV with feedback of 190mV nominal (Vfb)...That is only 0.05% error due to the 100 Ohm.  The error due to the FB voltage is way higher than that.  You are mis-understanding the fundamental of current regulator.  It's not a voltage divider as you are thinking.

    Thanks Tuan