ADUM1251/ADUM1250 Isolation

I am investigating the ADUM1250/1251 devices for an I2C isolation application.

My concern is that it appears both VDD pins are considered to be supply pins because the datasheet lists a sink current for both VDD1 and VDD2.  If both pins are sinking current and powering the device, will the device still be able to provide isolation on the power rail?

Also, we plan to use the chip to prevent latchup when hot plugging a non-powered device into a powered device.  Will the ADUM1251 need to place on the non-powered device or the powered device?

Thanks in advance.

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    •  Analog Employees 
    on May 25, 2012 5:26 PM

    Hello Vaughn,

    If 5V is applied to the GND1 pin and ground is connected to VDD1 this has a high likelyhood of damaging the part.  The first thing that will happen is the ESD protection diodes at each input will be forward biased and a large current will flow.  These diodes are not meant to work in standard forward bias mode, and they can't dissipate a lot of continuous power, so they will burn out.  Then there are other circuits internal to the chip that will start to conduct and suffer a similar fate.

    I have hooked parts up backwards before, and if the current is limited, and you turn off the supply quickly, they can survive.  The situation you don't want is to have burned out the protection diodes and left the rest of the part functional, that is a latent failure mechanism for ESD.

    To protect against this, you can use a Zener diode between Vdd and Gnd and a current limiting resistor in the VDD line.

    Best regards,

    MSCantrell

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  • 0
    •  Analog Employees 
    on May 25, 2012 5:26 PM

    Hello Vaughn,

    If 5V is applied to the GND1 pin and ground is connected to VDD1 this has a high likelyhood of damaging the part.  The first thing that will happen is the ESD protection diodes at each input will be forward biased and a large current will flow.  These diodes are not meant to work in standard forward bias mode, and they can't dissipate a lot of continuous power, so they will burn out.  Then there are other circuits internal to the chip that will start to conduct and suffer a similar fate.

    I have hooked parts up backwards before, and if the current is limited, and you turn off the supply quickly, they can survive.  The situation you don't want is to have burned out the protection diodes and left the rest of the part functional, that is a latent failure mechanism for ESD.

    To protect against this, you can use a Zener diode between Vdd and Gnd and a current limiting resistor in the VDD line.

    Best regards,

    MSCantrell

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