SN74LVC1G125 OE pull up resistor

Yosy Shmuly1

Prodigy 10 points

can I connect the OE in directly to VCC or to an FPGA IO or a driver without resistor?

I am asking it because of this sentence in the data sheet:

"To ensure the high-impedance state during power up or power down, OE should be tied to VCC through a pullup

resistor; the minimum value of the resistor is determined by the current-sinking capability of the driver."

why a minimum resistor value is needed?

over 12 years ago

0 Chris Cockrill over 12 years ago

TI__Mastermind 46480 points

Al that statment means it that the pullup current should not over power what ever is drivng your OE pin.

0 John Westmoreland43 over 11 years ago in reply to Chris Cockrill

Guru 13915 points

Hello Chris,

Please have a look at the attached circuit - is this OK for an LED driver from a 1 pulse per second signal? I must admit even though the statement in the datasheet seems straightforward, it made me wonder about current flow in the '125. As if the OE* was open drain or something like that.

So, a weak pull-up is OK - correct? The entire device current is not going through the 4.7K-ohm resistor, correct?

The statement in the datasheet makes it seem the 4.7K-ohm resistor is setting the drive current for the '125.

Thanks,
John W.

0 Chris Cockrill over 11 years ago in reply to John Westmoreland43

TI__Mastermind 46480 points

Hi John

The circuit is Ok. The drive comes from Vcc. The only thing I waould watch is since your signal ia 1sec pulse be sure the rise and fall will meet spec.

0 John Westmoreland43 over 11 years ago in reply to Chris Cockrill

Guru 13915 points

Chris,

Yes - I have tested the circuit and it looks OK.

Thanks for your reply.

Best Regards,
johnw

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SN74LVC1G125 OE pull up resistor