Understanding TVS Diodes: A Comprehensive Guide

A TVS diode’s peak pulse power (PPP) is the maximum rate at which the electrical energy is transferred during a pulse without the TVS being damaged. Expressed in watts, peak pulse power is critical when selecting a TVS diode, and it indicates the diode’s ability to handle spikes and energy levels effectively to prevent damage to connected components and systems.

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Peak pulse power is not to be confused with the peak pulse current (IPP). Peak pulse current is measured in amperes and represents the maximum amount of electrical current that flows during a pulse. A selected TVS should have a peak pulse current that exceeds the anticipated transient current to ensure the component can safely absorb the surge.

1. Is one diode (either on Vcc or GND) is enough? Which one to choose?

One diode will do the job either on Vcc or GND. But a Diode to GND will shift the GND voltage so if you have another equipment powered with the same battery connected to the arduino that voltage shift can lead to some problem

2. What type of diode to choose (voltage etc ratings)?

Voltage : the battery voltage
Current : the current you deliver to your board
Power : the current time the voltage drop across the diode (wich is quite low)

3. I have battery powered device with very low power consuming (most of the time MCU is in "Power down mode" -- around 1uA or less) will diode take some power?

You should take into account the voltage drop across the diode. This voltage drop can be an issue with battery powered equipment. You should choose the diode in the Schottky family as the voltage drop is much lower than in standard diode.

Thanks for the detailed response.

My battery is 3*AAA batteries (around 4.5V) and I need no lower than this for 16x2 LCD (4.5V is lower value for it according datasheet).

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What is typical voltage drop of Schottky diode?

Sorry for being stupid, but I'd just want to make sure I get it right. If voltage drop is, say, 0.5V and MCU current in sleep mode is 1uA, so I can easilly compute dissipation by multiplication 0.5 * 1*10^-6, is this correct? Diode is not linear...

so I can easilly compute dissipation by multiplication 0.5 * 1*10^-6, is this correct?

Yes that is right.

Diode is not linear.

A diode's linearity is the way voltage varies with current. It is because the diode is not linear (in the way it is) that you can use it with a constant voltage drop no matter what current you put through it.