TXS0108E High Speed Full Duplex 8 Channel Way Logic Level Converter Module

Im using one of these between a nano and a MFRC522 module, so stepping down the voltage from 5v top 3.3v. I'm stuck on where the OE connection needs to be connected too, i've read the data sheet.Could someone explain exactly in stupid lingo.
Data sheet

Regards

Page 20.

"The TXS0108E has an OE pin input that is used to disable the device by setting the OE pin low"

"To disable the device set the OE pin input low, which places all I/Os in a high impedance state.
Setting the OE pin input high enables the device."

Leo..

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Wawa:
Page 20.

"The TXS0108E has an OE pin input that is used to disable the device by setting the OE pin low"

"To disable the device set the OE pin input low, which places all I/Os in a high impedance state.
Setting the OE pin input high enables the device."

Leo..

Yes thank you leo I seen this in the link I provided,, but do I attach the OE to 3.3+ ?

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The datasheet mentions "OE to A or B", so I assume the 3.3volt VCC is ok.
Leo..

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Brilliant thanks leo.

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OE must be ground or VCCA, not VCCB. From deep within the datasheet under power supply considerations:

During operation, ensure that VCCA ≤ VCCB at all times. The sequencing of each power supply will not damage the device during the power up operation, so either power supply can be ramped up first. 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 pull-down resistor and must not be enabled until VCCA and VCCB are fully ramped and stable. The minimum value of the pull-down resistor to ground is determined by the current- sourcing capability of the driver.

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avr_fred:
OE must be ground or VCCA, not VCCB. From deep within the datasheet under power supply considerations:
"the OE input pin must be tied to GND through a pull-down resistor"

This pull up resistor is it included in the board chip? If not can you explain in layman's terms, aka noobie style.

Regards

The official Adafruit board has the resistor. If you bought from Adafruit or one of their distributors, you’re good to go.

I don’t know about Chinese’s knock-offs. A 10k resistor would typically be marked 103. If there is a resistor, verify connections to OE with an ohmmeter.

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avr_fred:
The official Adafruit board has the resistor. If you bought from Adafruit or one of their distributors, you’re good to go.

I don’t know about Chinese’s knock-offs. A 10k resistor would typically be marked 103. If there is a resistor, verify connections to OE with an ohmmeter.

Thank you, as i am really special(simple) can you whip this into paint and draw how this goes, i’m really confused. I’ve attached a jpg just need to tie in my ends.

Regards

Connect OE to GND with a 10K resistor. This is a pull-down which keeps OE low with no connection. OE should also be connected to VCCA so when the supply is high, the outputs are enabled by the logic high. From the TI application notes referenced on the Adafruit website:

If the application does not require output enable control, The OE pin should be tied to the VCCA supply. One should never leave the OE pin floating in an indeterminate state as this can cause undesirable quiescent current to flow in the device, which subsequently increases its overall power dissipation.

Edit: the 10k pull-down is optional so long as OE is not left floating. So long as OE has a valid logic level, the chip is happy.The 10k guarantees OE is always low with no connection and a floating input is something that can happen with breadboards. The Adafruit part adds a 1 cent resistor to make sure OE is always low unless pulled high. The Chinese clone you showed saved a fraction of a yuan and skipped the resistor.

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avr_fred:
Connect OE to GND with a 10K resistor. This is a pull-down which keeps OE low with no connection. OE should also be connected to VCCA

This is the bit that confuses me, so i will place a jumper from OE to GND with a 10k resistor in between, but also have a jumper going from VCCA to OE?? Sorry, or should it be like this, see attached
Regards

Your attached schematic is correct. The only reason to have the resistor is so the pin doesn’t float if the supply is disconnected.

avr_fred:
Your attached schematic is correct. The only reason to have the resistor is so the pin doesn’t float if the supply is disconnected.

Got ya, looked into the crazy interference from other electrical devices changing it!! Thank you

Hi,
Similar problem solved.

In fact search txs0108e arduino in YouTube.

Tom… :slight_smile:

TomGeorge:
Hi,
Similar problem solved.

NodeMCU + I2C LCD + Arduino IDE + TXS0108E logic level converter - YouTube

In fact search txs0108e arduino in YouTube.

Tom... :slight_smile:

Thanks Tom, there was only him and the Russian guy on there, it was that YouTube video which made me question it, but he ran a resistor between OE and Vcca. When I prototyped I used 2x 4 Ch LLC's to save on board space I thoughts a 8 Ch would be better, but me being me didn't think that it might need to be wired in differently, I started then questioning my code, my modules and soldering

Ok, trying to understand this more, i am putting a pull down resistor on GND and OE but also attaching to VCCA to enable high-impedance Hi-Z

Quote “Hi-Z (or High-Z or high impedance) refers to an output signal state in which the signal is not being driven. The signal is left open, so that another output pin (e.g. elsewhere on a bus) can drive the signal or the signal level can be determined by a passive device (typically, a pull-up resistor).”

This is to leave the comms open right? So why didnt the other guy in the youtube clip do this??

Regards

When OE is high, the outputs are enabled/active. When you disconnect the voltage from OE, the resistor ensures the OE pin is low which forces the outputs to a hi-z state.

This behavior was covered back in post #4.

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