TXS0108E: Question about input signal waveform

Johnny Guo

Part Number: TXS0108E

Hi, team,

My customer has some questions about TXS0108E input signal may need your confirmation.

In customer's design, VCCA=1.8V, VCCB=3.3V. In TXS0108E datasheet, when setting A-port power voltage to 1.8V, VIH range is 1.6~1.8V, VIL range is 0~0.15V.

On customer's board, there is a step voltage in rising and falling edge as shown below.

My questions are below:

1. Will below voltage step bring risk to TXS0108E port A signal detection?

2. If no risk on below signal, is it because of short step time duration? How long of time will the step voltage bring risk?

Thanks.

Johnny

over 7 years ago

0 ShreyasRao over 7 years ago

TI__Guru 63920 points

Hi Johnny,

Is there an output scopeshot you can provide to see the response?
The one shot is designed to trigger around Vcc/2 and then the pullup resistor will hold the output signal at VccB level (for B ports).
I believe that the plateau will not cause issues at the output as the input signal transition rate is within the recommended datasheet spec.

Is there a reason why the input signal behavior is like this and can this be fixed? It could be impedance mismatch causing this behavior?

0 Johnny Guo over 7 years ago in reply to ShreyasRao

TI__Expert 4910 points

Hi, Shreyas,

The output waveform is good without any voltage step. Customer just needs to check whether this input behavior will cause any risk to TXS0108E.

The input voltage is caused by impedance mismatch, and this result is already the best waveform.

Thanks.

Johnny

0 ShreyasRao over 7 years ago in reply to Johnny Guo

TI__Guru 63920 points

Johnny,
assuming that rise and fall time is going to be fast it shouldn't cause issues on the output with the step well above and below the threshold.

0 Johnny Guo over 7 years ago in reply to ShreyasRao

TI__Expert 4910 points

Okay, Thanks Shreyas.

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TXS0108E: Question about input signal waveform