• State Resolved
  • Locked Locked
  • Replies 2 replies
  • Subscribers 34 subscribers
  • Views 525 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

AM26LV32E: Input leakage to VCC and outputs when unpowered?

Andrew Morley
Prodigy 40 points
Part Number: AM26LV32E

The AM26LV32E (we're using AM26LV32EIPWR) is intended for use in applications when its inputs are connected to an active (driven) line even though it - and the electronics connected to its outputs (a microcontroller) - are not powered. Thus it is specified to tolerate +/-14v on those inputs even with no power, and the outputs are still guaranteed to output no more than +/-100µA. Correct me if I'm wrong in my interpretation of the data sheet.

However this raises a few questions:

  • Is Ioff the total output current for all four outputs or is it per output?
  • What is the leakage path from the inputs to the supply rail, ie what current should we expect to see driven onto the (unpowered) Vcc from a powered line.  (I am aware of the figure in section 8.2).

In short, I want to avoid the differential lines back-powering the rest of our system when it remains connected but powered-off. What amount of current to I need to expect back-powering from the AM26LV32E?

If this question looks a bit familiar it's because I asked it earlier, but named the wrong (but closely-related) part, here. kindly responded to that. But I meant to ask about this. And in this case there is some DC path (see section 8.2) - but is that all?

Thanks (Again!)

Andrew

  • 0
    TI__Mastermind 25705 points

    Andrew,

    The schematic in section 8.2 represents the device's structure. With Vcc, the input resistance is about 17kOhm (ri) and the input current (Ii) flows from the pin to ground or from Vcc to the pin. With Vcc=0V, the input voltage is blocked by the reversely biased diode. However at A pin, there is a DC path with 200kOhm resistance.

    I hope it makes sense to you.

    Regards,

    Hao 

  • 0 in reply to Hao L
    Prodigy 40 points

    Many thanks - that's very clear. It seem to me, then, that voltages on the inputs may leak (via 200k) into the power rail and may then find their way to the output, and that's the leakage path. But it's a very small current in practice.