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I have Atmega328 MCU and MAX7219 connected to each other. However, they have diode separated power sources. There is a case, when MCU will get power, but MAX7219 will not.

Would it damage MAX7219? While in the application having the knowledge that only MCU will be powered on, I will try to pull down 3 MAX input pins, but there may be some short period during MCU bootup when logical 1 may reach MAX.

Is it safe for MAX to keep it that way, considering that this situation may happen occasionally?

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  • \$\begingroup\$ Add a schematic so that the connection between the chips is evident. \$\endgroup\$ – Chetan Bhargava Feb 6 '13 at 20:54
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    \$\begingroup\$ There is no need for schematics. Very simple. 3 output pins on MCU going to 3 input pins on MAX (data, clock, latch). \$\endgroup\$ – Pablo Feb 6 '13 at 20:58
  • \$\begingroup\$ Check the data sheet for the MAX7219. What does it say about voltages applied to pins with power removed? \$\endgroup\$ – Leon Heller Feb 6 '13 at 21:02
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The answer depends on why you have a diode separated power supply to begin with. You may not be achieving your original goal by tying the pins together if what you're going for isolation.

That said, in most cases a few microseconds of bad behavior on input pins of digital ICs is usually OK, in that the IC may survive it, but you may have some bad behavior when things are not within the specifications. You need to see if this bad behavior is OK for your application. It may involve, for instance, your LEDs turning on for a brief time when they aren't supposed to. If this also leads to a high current inrush through the digital inputs of your MAX, that could be a bad thing. There could be IC damage and set your uC in a reset loop if it's power supply in unable to sustain whatever the badly behaving MAX7219 demands.

This bad behavior can stay on as long and the power supplies put either device in a region where operation is not specified. The longer it keeps up, the more likely you are to cause actual damage. Its never a good idea to power things through digital lines, and that is precisely what can happen in this condition. i would recommend using a level translating buffer such as the SN74LVC1T45 and others of its family to do a better job dealing with connections across power supplies that aren't guaranteed to exist together. You simply put a resistor between the Vcc ad ground on both sides, something in the 10-100K range, and when one side loses power, the buffer shuts down and the line is effectively disconnected.

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  • \$\begingroup\$ The reason of separation is that during USB firmware update my device is powered from USB and it will power up also the LED's which will draw too much current from USB. I could also try to put MAX7219 into shutdown mode, but still it may cause current spike. \$\endgroup\$ – Pablo Feb 6 '13 at 23:20
  • \$\begingroup\$ USB may not like that much if the current draw is high. Most modern computers would shut off operation beyond 450mA, and laptops sometimes do that at lower points. Using a shutdown mode is probably the solution you would be better off with. If you are using an FT232, it has an output you can use to do this. Other USB serial ICs may have them too. Also, since this probably means the AVR is probably just coming out of reset at that point, the pins are all inputs and high impedance. Dont set any as output until your power supply is ready for it. \$\endgroup\$ – Chintalagiri Shashank Feb 6 '13 at 23:28
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The serial interface pins are high impedance digital inputs (think CMOS gates), I wouldn't worry about them seeing a voltage before the chip has power briefly. I would, however, pull down the LOAD pin to GND through a weak resistor (say 10k) as a good practice.

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