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LSF0108: Removal of Rpu

Ella KIM
Expert 4212 points
Part Number: LSF0108

Hi Team, 

I have a question on the removal of Rpu in LSF0108 translating down application. From the video link below, RB1 can be removed only if the leakage current into the receiver is <1uA. Can you please let me know why this condition is required? If it's relevant to IC's spec, can I find the reason in the datasheet?

https://training.ti.com/down-translation-lsf-family?context=1134826-1139264-1134793

"There are special circumstances that allow the removal of one or both of the pull-up resistors. If the signal is always going to be down translated from a push-pull transmitter, then the resistor RB1 can be removed.

If the leakage current into the receiver is less than 1 microamp, then the resistor RA1 can also be removed. Recall that CB1 and CA1 are parasitic capacitances, and are shown here only as a reminder."

Thanks a lot!

  • +1
    Guru 317280 points

    The LSF is a passive device and has no drive strength. Furthermore, the current that can flow through the switch for high-level signals is extremely small. If a device tries to draw more than 1 µA from the output, the output voltage will drop.

  • 0 in reply to Clemens Ladisch
    TI__Expert 4212 points

    Hi Clemens,

    I don't think I fully understand the 1uA condition. Could you help advise?

    I target to remove Rpu 4.7kohm on 1.8V side. The reason why I try to remove Rpu is : 

    1) pcb space save

    2) lower VOL while 3.3V-->1.8V translation down

    When transmitting LOW, VOL becomes lower than before, because there's no I_A flowing thru FET in the capture below.

    When transmitting HIGH, even though there's no Rpu (R_A1) on the output side, the push pull input driver on the B side can provide enough current to form the output voltage on the A side. For my system, input driver can source upto 8mA, so 1uA of leakage on the receiver side cannot be a problem.

    Those are my logic so far. Could you help advise if any misunderstanding on my side?

  • 0 in reply to Ella KIM
    Guru 317280 points

    What is the direction? A to B, B to A, or both?

  • 0 in reply to Clemens Ladisch
    TI__Expert 4212 points

    In the system there will be both direction.

  • 0 in reply to Ella KIM
    TI__Guru* 97046 points

    Hi Ella,

    In the high state, the output of the LSF generally depends on the pull-up resistor to maintain the output voltage.

    When you remove the resistor RA1, then the only source of voltage is from the channel MOSFET, which shuts off when the output is at ~1.8V (in your case).

    Technically in this case, the MOSFET never really turns off -- it must continue to provide a very small amount of current (less than 1uA) to the input in order to offset any leakage current. This isn't a problem since the MOSFET can supply that much current in this state, however if you start to draw more current, the output voltage will suffer -- ie in order to turn the MOSFET on enough to provide more current, the channel will require additional voltage (stronger field).

    I built a quick simulation to show this on a 'nominal' LSF device:

    Here you can see that as the output current is increased (I_leakage), the output voltage (V_A1) decreases.  In this case the simulation shows a relatively small decrease (about 80mV), however this is dependent on many variables and we can't guarantee that it will be this small in all cases (process, voltage, and temperature variations). This is why we recommend a 1uA max input leakage for removing that pull-up resistor.

    If the device has a relative large input leakage current specification, I would recommend adding a weak pull-up resistor (typically 10kohm is sufficient) for the low-voltage side). This just helps keep the output in the HIGH state by having the resistor provide some of the current instead of the MOSFET.