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SN74AUP1G07 Input Loading

Prodigy 135 points

Replies: 5

Views: 564

I am using 2 SN74AUP1G07 O/D buffers connecter to each other for doing some voltage translation.  I want to confirm the amount of current needed to source the input of this chip to a high state and what the worse case loading on the input pin could be?  I currently have the O/D buffer of the first chip with a 100K pull-up and I am running at 600 baud.  I have some boards where the signal is getting cutting in half (should swing between 0-1.5V but on the failing boards swings at certain times form 0-0.75V) and don't quite understand why.  It almost like the loading is changing at certain times.

5 Replies

  • The AUP can drive 4ma when driving low. If your Vcc is 1.5V and you use a 100k resistor that means your high drive is only 15ua. the high drive depend on the resistor. Your resistor is too large.

  • In reply to Chris Cockrill:

    Chris,

    The resistor is only needed when the first device is not driving low to pull the output high.  This is on a battery powered device where low power is key this is why I am using a 100K pull-up.  The parts that are working good I can go up to a 500K resistor with no issues.  Am I missing something here?

    Thanks

  • In reply to Clock Work:

    This is an open drain part so it either drives low or floats. The only high drive comes from the resistor.

    You may want to switch to the SN74AUP1G34 or SN74AUP17 instead.

  • In reply to Chris Cockrill:

    Chris,

    Yes I understand that it is a open drain output as this is by design for level translation.  The first buffer is powered from a 3V supply and the second buffer is powered from a 1.5V supply with the 100K pull-up tied to the 1.5V supply.  My question is how much current is required from my pull-up resistor to source the signal high on the input of the second buffer?  Or what is the smallest pull-up that I can use at my 600 SLOW BAUD rate?  Datasheet specifies CI = 1.5pF is the number higher at 1.5V.  Also the Ii is 0.5uA. So my 15uA should be no problem?  I am wondering if I have defective parts or what could be causing my signal on certain parts at certain times to be cut I half.

     

    Thanks, 

  • In reply to Clock Work:

    Got it.   15ua should be enough to drive the input. It could be defective.

    If you are down translating you can just use a standard AUP buffer. with 1.5 Vcc the output will be 1.5V and no pullup is nessesary.

    The inputs are 3.3 volt tolerant and you would not have the current drain from the resistor. Icc is less than 1ua for AUP parts.

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