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Original question:

LSF0204: Higher push pull resistor

LSF0204: Continue the question thread : Higher push pull resistor

Masayuki Uchihara
Intellectual 1676 points
Part Number: LSF0204

Hi Team, 

Let me continue the question. https://e2e.ti.com/support/logic/f/151/t/861992

1. About 3, adding resistor. (Please refer attached e2e thread)

- Could you explain in detail the impact the low level output voltage of the device?

2.  About 4, the RC circuit (Please refer attached e2e thread)

- It should be get rid of the capacitor of the RC circuit right? because the speed is too slow.

3.  If too slow the speed, could you explain detail failure mode?

Thanks,

Uchihara

 

  • 0
    TI__Guru* 96947 points

    Masayuki Uchihara said:

    1. About 3, adding resistor. (Please refer attached e2e thread)

    - Could you explain in detail the impact the low level output voltage of the device?

    Having a series resistor on a wire that is carrying current will result in a voltage drop across the resistor as described by Ohm's Law: V = I*R

    When driving LOW into one side of an LSF device, current sinks through the device into the driver. There is a good depiction of this in the video "Down Translation with the LSF Family" starting at ~1:09 time.

    Here I added a blue series resistor. You can see that some current must flow through that resistor in order to operate the device, and thus the voltage at the input to the next stage will be higher than expected.

    Masayuki Uchihara said:

    2.  About 4, the RC circuit (Please refer attached e2e thread)

    - It should be get rid of the capacitor of the RC circuit right? because the speed is too slow.

    Yes, remove the capacitor.

    Masayuki Uchihara said:

    3.  If too slow the speed, could you explain detail failure mode?

    I'm not 100% sure what you mean. If you're talking about slow edges, the issue is with oscillation and shoot-through current, as explained here: 

    https://e2e.ti.com/support/logic/f/151/t/737694?tisearch=e2e-sitesearch&keymatch=faq%2520slow

  • 0 in reply to Emrys Maier
    TI__Intellectual 1676 points

    Hi Emrys,

    Could you please review the schematic which I attached again?

    The voltage source of 3.3V  need 1uF output capacitance. Therefore it seems the 1uF capacitor add to VrefA. How do you think about it? Is it no problem?

    Thank you for your support.

    Uchihara

  • 0 in reply to Masayuki Uchihara
    TI__Guru* 96947 points

    The capacitance at the VrefA pin won't really affect the operation of the LSF0204. What is more of a concern is the current flowing from the LSF0204 into the IC(3.3V) that is supplying the voltage. If this is a linear regulator, and it is only supplying the LSF0204, the reverse current could cause disregulation.

  • 0 in reply to Emrys Maier
    TI__Intellectual 1676 points

    Hi Emrys,

    Should put on a diode to avoid the reverse current?

    Thanks,

    Uchihara

  • 0 in reply to Masayuki Uchihara
    TI__Guru* 96947 points

    Hi Uchihara-san,

    No, blocking the current will stop the LSF0204 from working.  If necessary, a resistor can be added to ground to sink the current. 

  • 0 in reply to Emrys Maier
    TI__Intellectual 1676 points

    Hi Emrys,

    All I have to add the resistor between Vref A to Grand?

    Thanks,

    Uchihara

  • 0 in reply to Masayuki Uchihara
    TI__Guru* 96947 points

    It depends on your system design. Is the linear regulator driving anything else?  How much current do those devices require?

    Your goal should be to prevent current from going back into the regulator while still providing a path for current to source from the VrefA pin of the LSF0204. The bias circuit for the LSF0204 requires some current to operate ( ~= (5V - 3.3V - 0.7V) / 200kohm = 5uA -- details available in this video: LSF Bias Video )  If there are no other devices in the circuit, then a resistor would work.

    The current through resistors is governed by Ohm's Law (V = I * R), so if you want to have at least 5uA of current flowing through the resistor at 3.3V, then the largest resistor value you can use is: R = V/I = 3.3V/6uA = 660kohm. I would recommend using a smaller resistor just for a safety margin - 330kohm should work fine (330kohm will sink 10uA at 3.3V).