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TXB0108: txb0108 power up

Francis KImpel
Prodigy 50 points
Part Number: TXB0108

I have a question about a TI level translator PN: TXB0108RGYR I want to use in my design

 

I am working in an application where it is important to keep all A and B outputs of the part in Hi-Z until VCCA and VCCB are up and stable

 

I have an active low signal that will remain low until both voltages are stable and it is connected to the OE pin. Also have a pulldown resistor on OE pin.

 

 

 

The Data sheet indicates that  VCCA <= VCCB during normal operation after power up.

 

During Power up the order of turn on for VCCA and VCCB is not critical(so VCCA<VCCB if VCCB=0V) since there is a sensing circuit that keeps all outputs in HiZ as long as the OE pin is held low.

 

I cut and paste section 10 of the data sheet that appears to support these assumptions:

 

 

10 Power Supply Recommendations
During operation, ensure that VCCA ≤ VCCB at all times. During power-up sequencing, VCCA ≥ VCCB does not
damage the device, so any power supply can be ramped up first. The TXB0108 has circuitry that disables all
output ports when either VCC is switched off (VCCA/B = 0 V).
The output-enable (OE) input circuit is designed so that it is supplied by VCCA and when the (OE) input is low, all
outputs are placed in the high-impedance state. To ensure the high-impedance state of the outputs during
power-up or power-down, the OE input pin must be tied to GND through a pulldown resistor and must not be
enabled until VCCA and VCCB are fully ramped and stable. The minimum value of the pulldown resistor to ground
is determined by the current-sourcing capability of the driver.

 

 

Question???

 

1. Will the A-B I/O remain in HIZ through states 1-5???

2. According to the datasheet, the outputs are supposed to be in tristate as long as VCCA or VCCB are at ground or OE = logic 0. 

Assuming that OE is pulled high, At what voltage level for VCCA, VCCB during turn on do the outputs leave tristate?

 

State 1:  VCCA = 0V      VCCB = 0V           OE =LOGIC 0      A-B OUTPUTS = HIZ

State 2:  VCCA = 0V      VCCB = RAMP     OE =LOGIC 0      A-B OUTPUTS = HIZ

State 3:  VCCA = 0V      VCCB = 3,3V        OE =LOGIC 0      A-B OUTPUTS = HIZ

State 4:  VCCA =RAMP  VCCB = 3.3V       OE =LOGIC 0      A-B OUTPUTS = HIZ

State 5:  VCCA = 1.8V    VCCB = 3.3V       OE =LOGIC 0      A-B OUTPUTS = HIZ

State 6: VCCA = 1.8V     VCCB=  3.3V       OE =LOGIC 1      A-B OUTPUTS = LOGIC LEVEL 

  • 0
    TI__Guru* 97046 points
    Hi Frances,
    The outputs will remain Hi-Z until state 6.

    Ioff (partial power down) typically operates until from 0V to 0.5V. It is highly recommended to use the OE pin to tri-state the outputs (as discussed in your first question) rather than counting on Ioff to maintain high impedance. It is really intended as a protection against back-powering when a subcircuit is turned off.
  • 0 in reply to Emrys Maier
    Prodigy 50 points

    Makes sense to me but I am still concerned about what happens for the case where VCCB turns on first and then VCCA starts turning up with OE = logic 0

    (i.e.)

    VCCA        VCCB       OE             Output

    0V              3.3V          logic 0       HIz

    Ramp        3.3V          logic 0       ?????   Since VCCB is ramping (0-1.8V) willl it remain in Hiz during ramp?  NOTE: The OE logic works off of the VCCA voltage

    Frank 

  • 0 in reply to Francis KImpel
    TI__Guru* 97046 points

    Hi Frank,

    From the datasheet, page 1:

    To ensure the high-impedance state during power-up or power-down, OE should be tied to GND through a pulldown resistor.

    The device is guaranteed to remain in the high-impedance state while you ramp the supply.