I'm looking at incorporating a LSF0108 level shifter into a design to convert bidirectional GPIO signals from a 1.8V CPLD (Vcca) to an off-board device via ribbon cable. The header contains a voltage reference at one of several known voltages (1.8V, 2.5V, 3.3V, and 5V possible), which I have connected to Vccb.

The issue I'm running into is the datasheet calls for the enable pin to be connected via pull-up resistor to Vccb. Because there is an offboard connector, Vccb may not always be present when power is applied to Vcca. The suggested layout also calls for a .1uF decoupling capacitor on both the Vcca and Vccb rails. The datasheet also mentions that if either Vcca or Vccb are grounded, then this will also disable translation.

On to my question: how can I ensure that Vcca or Vccb are effectively grounded when a cable isn't attached? Using a pull-down doesn't seem right especially since the enable pin is pulled up when Vccb is present.

(Apologies in advance, I'm a hobbyist trying to step up his game and I'll gladly read any reference you can put in front of me!)


That EN pin connection is needed due to the internal structure of the device (see figure 1 of the LSF application report).

You could use a mux or analog switch to connect these pins to either ground or the cable. At the input of that switch, you would be able to use a pulldown resistor.

Alternatively, consider another device where the OE pin is referenced to VCCA, like the TXS0108E, or, if you do not have any open-collector signals, TXB0108.

  • \$\begingroup\$ Interesting. I was looking at the TXS0108E as a replacement (I'm dealing with SPI, and I'm worried that any pullups on the board could affect the drive of the TXB). It looks like as long as there is 0v applied to Vccb, then the output drivers will be disabled, which is what I want. Is it possible that since this pin would effectively float, that this could be a problem? How can I ensure that Vccb is at 0v when nothing is attached to the voltage reference? \$\endgroup\$ – Steven Stallion Feb 14 '17 at 21:46
  • \$\begingroup\$ "Floating" is not the same as "grounded". But what about a pulldown of about 200 kΩ? \$\endgroup\$ – CL. Feb 15 '17 at 7:30
  • \$\begingroup\$ Out of curiosity, is there an effect on current draw if a pull down is present on a rail when voltage is applied? What are the guidelines for choosing the resistor value? \$\endgroup\$ – Steven Stallion Feb 15 '17 at 15:15
  • \$\begingroup\$ The current through the resistor can be computed with Ohm's law (5 V / 200 kΩ = 25 µA). Larger resistors react slower to changes (because parasitic capacitances form an RC filter), but that probably does not matter on a supply line. \$\endgroup\$ – CL. Feb 15 '17 at 16:11
  • \$\begingroup\$ Gotcha. That makes total sense - thanks for the help. In retrospect, I think this is the first time I've had to deal with a potentially disconnected supply net. Glad to see the usual rules apply here! \$\endgroup\$ – Steven Stallion Feb 15 '17 at 17:00

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