Hi,

i want to use the SN74AVC8T245
datasheet: http://www.ti.com/general/docs/lit/getliterature.tsp?genericPartNumber=sn74avc8t245&fileType=pdf

I don't fully understand the datasheet. It says: use a pull up resistor at the OE port connected to VCCA to ensure stability during power up.
When the OE port is high, all the ports are in high impedance mode. But when connecting to VCCA the OE port is always high and so the shifter will not work.

Is the stability really a problem at power-up? or can i easily ground the OE port?

 

It depends on what you are using 74xx245 for and why.
What else is connected to the A-bus and B-bus?

 

It depends on what you are using 74xx245 for and why.
What else is connected to the A-bus and B-bus?

thanks for your reply!
On one side there is an atMega(3.3V) on the other side a sensor(1V)

 

The datasheet is saying you need a pullup resistor.

The resistor must be low enough in value to meet the input current requirement for logic HIGH. The data sheet specifies a current of 1 µA maximum over the full industrial temperature range with the control inputs at VCCA. We don't have any additional info, so we'll assume the current is the same with the input at the logic HIGH threshold. We'll assume VCCA is 1.2 volts and HIGH is 2/3 of that (from datasheet), so the voltage across the resistor must be no more 1.2/3 or 0.4 V.
0.4 V / 1 µA = 400 000 ohms
That's the maximum value of resistor you would use as shown in Figure 11 of the datasheet.

If the input were then pulled to 0 V, the current would be 1.2 V / 400k = 3 µA - easily managed by any logic output.

I'd use a resistor of about 100k. It could be much lower e.g. 10k. The reality is that it could likely be much higher.

Passive pull-up or pull-down is something that is very common to establish the logic states when the signal is not being actively controlled by a device output at all times.

 

but when connecting it to VCCA it will always be high also with a pull up resistor and no shifting will take place? Sorry i don't understand it.

 

but when connecting it to VCCA it will always be high also with a pull up resistor and no shifting will take place? Sorry i don't understand it.

Obviously, there would be no point in connecting OE- directly to VCCA; although shifting would take place internally with the right clock signals, the outputs of the chip would all be tri-stated (i.e., turned off). Connecting OE- to VCCA through a pullup resistor, on the other hand, presumes that OE- will also be connected to some output from some logic element that you're using to control the chip's outputs' ON/OFF state. When power is first applied to the circuit, the state of all the logic elements in the circuit is indeterminate, meaning the outputs of your chip may start out turned on, which could lead to bus contention if there are multiple chips connected to the same bus lines. The pullup resistor alleviates this problem by dragging the OE- line high until your circuit's logic is functioning.

This assumes the SN74AVC8T245 is going to be used as a bus transceiver along with other chips of that type; if all you're doing with the chip is logic level shifting, and no bus arbitration, just ground the OE- pin so the outputs will always be ON.

 

I think he is confused how a low signal can pull the gate low while being held high with the resistor.