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Consider the case where A is an input and B an output and, by an error, another voltage is applied to B (5V instead of the output 0V with A=0V). B will now be considered as an input because of the transition detected, and A will become an output of 5V, which will create another short circuit. Is this a bug in the chip? A vicious cycle?

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There will be no burning shorts circuits. The drive strenght of the device is quite low. The device is also mostly for open drain IO where two pins actively grounding the line isn't a problem. The internal weak pullup won't kill anything.

I have used the TXB part for SDIO. The TXB is made for push-pull applications. Again, this device isn't capable of destroying things. I had issues where the A and B supply rail were turning on/off out of sequence and this could occasionally lock up the TXB but not to a dead short.

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  • \$\begingroup\$ Thanks for your answer. But if there is no destruction, will it be a vicious circle which will lock up the TXB or TXS ? Example : A Input & B output => Forcing B as input (with the other voltage 0 if Vcc and Vcc if 0) will force A as output, and so on... If you have seen occasionally a lock up of the chip, it can be also because of this case ? \$\endgroup\$ – zian Nov 28 '20 at 9:38
  • \$\begingroup\$ Moreover, if TXB and TXS chips don't have drive strength as you mentioned, the "one shot accelerator" have drive strength, but only during short time of transition, so as you said, it's not possible to destroy something, but it's enough to create a transition from 0 to Vcc (or from Vcc to 0), so that a pin which was an output will become an input, and a vicious circle can be created, it's my wondering. \$\endgroup\$ – zian Nov 28 '20 at 9:47
  • \$\begingroup\$ In some really badly designed application this could be a problem. Say if two MCUs talks to eachother and on a bidirectional-IO line, and one of the MCUs wouldn't switch to listening mode after sending something. \$\endgroup\$ – Dejvid_no1 Nov 29 '20 at 19:29
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If we ignore the edge accelerators, the TXS is a passive switch. If opposite voltages are applied to the two sides, then the switch is on, and there is a short through the device, which might exceed the absolute maximum ratings.

With the edge accelerators, if the A side is driven low, and then the B side is driven high, the rising edge on B will result in the TXS driving A high for a short time. If the TXS is stronger than the other device (or if there is enough capacitance/inductance so that the other device is not fast enough), then the A voltage can rise completely, and when the edge accelerator times out, the other device will drive A low again, which can be detected as a falling edge, which will trigger the edge accelerator in the other direction. So it is possible to make the TXS oscillate.

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  • \$\begingroup\$ Thanks, I agree with the second part of your answer, and I will test it to see what will happen. \$\endgroup\$ – zian Nov 29 '20 at 19:50
  • \$\begingroup\$ Concerning the first part of your answer, I agree that it is a passive switch without the edge accelerators=> there are 2 serial R1 and R2 with the "Npass" transistor on the TXS0108E. R1 and R2 are 150Ω (typical). So with 5V-0V=5V / 300Ω = 17mA => Pd=83mW, it should be ok with the Absolute Maximum Ratings (50mA max in continuous). The matter will be with the external chip : able or not to work with these 17mA : it should be ok. So the last problem should be with the potential oscillation of the edge accelerators. Note : I'm impressed with the high technical level of the 2 answers here ! \$\endgroup\$ – zian Nov 29 '20 at 19:58
  • \$\begingroup\$ The total resistance between the two I/O pins is 150 Ω typical. (But that's still only 33 mA.) \$\endgroup\$ – CL. Nov 29 '20 at 20:03
  • \$\begingroup\$ In the PDF "A Guide to Voltage Translation With TXS-Type Translators", it's written : "The series resistance values of R1 and R2 are 150Ω (typical)". In the TXS0108E PDF, it's written : "a high-on-resistance N-channel pass-gate transistor (on the order of 300Ω to 500Ω)". So i've supposed that it is 300Ω typical for R1+R2. Maybe there is another information in another PDF ? If it's only 150Ω for R1+R2, I will need to add a serial resistor, to preserve the pin of my external chip. \$\endgroup\$ – zian Nov 29 '20 at 23:04
  • \$\begingroup\$ The slope of the voltage-vs-current lines in section 7 of the TXS datasheet shows that the typical total resistance indeed is 150 Ω. \$\endgroup\$ – CL. Nov 30 '20 at 9:33
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After testing with a breadboard, here are the results :

  • With Va=3.3V Vb=5V
  • If A=0.00V and B is forced to 5.00V => A=0.01V and B=4.97V, and the current is 11mA
  • If B=0.00V and A is forced to 3.30V => B=0.01V and A=3.29V, and the current is 9.8mA
  • Same results with the 2 tests forced to 0.00V

So 4.96/0.011=451ohms and 3.28/0.0098=335ohms, it's confirmation of the datasheet : "N-channel pass-gate transistor (on the order of 300 Ω to 500 Ω)"

My conclusion : There are no oscillation bugs in the cases I was wondering.

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