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DC characteristics question

Category: Datasheet/Specs
Product Number: ADuM162N

Hi,

I am reading this datasheet : https://www.analog.com/media/en/technical-documentation/data-sheets/ADuM160N-161N-162N-163N.pdf 

And at page 5 i simply don't understand the output currents.

Like i understand that there is a default low and high ordering option but i don't understand the meaning of those currents.

The way the results are depicted are hard to understand what is powered what is not and how the output is tied.

They are output current but in what conditions is it IO to Vcc/GND or some other load? 

I would like to know what is the output current in these states:

both sides are powered on and IO is:

high and tied to GND 

low and tied to Vcc

And same states but Vcc on the other side is not powered.

Best regards,

Robert

Parents
  • Hi Robert,

    With the OOK architecture, each channel has a transmitter on the input side and a receiver on the output side. The transmitter is either idle or active based on the state detected at the input. A transmitter will be idle with the input detecting the default state and active when detecting the non-default state. An active transmitter will consume more power. It doesn't matter if the receiver side is powered or not, when the input side is powered and presented with the non-default state at the input, the transmitter of that channel goes active. 

    The receiver side outputs the default state when no transmitter activity is detected. When the receiver detects the channel's transmitter is active, that channel's output will go to the non-default state. 

    With VDD1 and VDD2 powered, if you provide input high and direct short the output to GND, it will be a violation of the abs max table on the output side. The output has ~70ohm impedance and it will source/sink more than the -10mA current limit in the abs max table. Same goes with an input low and the output shorted to the VDD rail. 

    If you present the input (default low or high models) with a high signal and the input side VDD is unpowered, the high signal on the input will back feed to the input side VDD rail through the channel's ESD diode. When the input is unpowered and output side powered, the receiver will detect no transmitter signal(its unpowered) and set the output to the default state.

    Regards,

    Jason   

  • THX

    That is all i wanted to know i somehow missed the output impedance 

  • Apologies, I reread and realized I didn't clarify your first question. 

    The top 2 rows show power consumption with all transmitters idle. The bottom 2 rows show power consumption with all transmitters active. The ADuM162N has transmitters on both sides, and we can see both Idd1 and Idd2 increase when all channels simultaneously see the non-default state.

    If your application might actually have outputs shorted to the rail, I suggest series resistors to not violate the abs max table. You may also need to consider some ESD protection depending on what the signal traces are coming in contact with. The output pins are rated for only +/-4kV HBM.  

    FYI, the newest generation of OOK standard digital isolators (ADuM34xE, ADuM32xN...) have lower power consumption.

    Regards,

    Jason

Reply
  • Apologies, I reread and realized I didn't clarify your first question. 

    The top 2 rows show power consumption with all transmitters idle. The bottom 2 rows show power consumption with all transmitters active. The ADuM162N has transmitters on both sides, and we can see both Idd1 and Idd2 increase when all channels simultaneously see the non-default state.

    If your application might actually have outputs shorted to the rail, I suggest series resistors to not violate the abs max table. You may also need to consider some ESD protection depending on what the signal traces are coming in contact with. The output pins are rated for only +/-4kV HBM.  

    FYI, the newest generation of OOK standard digital isolators (ADuM34xE, ADuM32xN...) have lower power consumption.

    Regards,

    Jason

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