Dear Support Team,
There are two different input sources with independent (isolated grounds) that feed the VDD1 and VDD2 of ADUM3190.
I want to understand whether there would be any ADVERSE IMPACT on the IC, if only one of the input power supplies (in desired range) is turned ON, while the other input power supply remains OFF for at least few hundred seconds?
The Supply (In desired range) is given to Pin 16 (VDD2) of the IC, while the supply to its pin 1 (VDD1) remains OFF for at least some hundreds of seconds.
The Supply to Pin 16 (VDD2) of the IC remains OFF for at least hundreds of seconds even after the the supply (in desired range) was given to to its pin 1 (VDD1) of the IC.
We are trying to monitor the VDD2 source (dropping it using series resistance divider and then feeding to pin 12 or Pin 11 based on conditions). Two scenarios exist:
1. Range check
2. State change
We are feeding VDD1 through the power source that turns ON our microcontroller. We are feeding VDD2 through an external power source that does not feed or turn ON our microcontroller.
If I want to use the Pin 7 as an Interrupt trigger to my microcontroller, then what circuit / schematic should I use, assuming VVD2 could already be ON when VDD1 was turned ON.
In Range check scenario I want a trigger pin 7 for conditions:
1. Input voltage < desired VMIN
2. Input voltage > desired VMAX.
In State Change the pin 7 will be used as Input Supply (source for VDD2) ON and OFF detection.
1 when VDD1 is lost. the encoded signal can not be received. so EAout/EAout2 are both 0V. when VDD1 gets back, the EAout/EAout2 will follow input like attached waveform. firstly there is some delay time. secondly, there would be some overshoot on EAout1 , as Eaout2 is open drain, during this delay time, EAout2 is just follow VDD1.
2 when VDD2 is lost. as there is no input voltage signal is encoded and transmitted. EAout will be at it lowest output voltage.
With regards to below,
fox said:when VDD2 is lost. as there is no input voltage signal is encoded and transmitted
I assume you meant below (additional text highlighted / Bold)
"When VDD2 is lost, as there is no input voltage, no signal is encoded and transmitted."
Yes, it was my typo. Sorry for that.