What Kind of Diode to Use with ADC Inputs

Diode protection of inputs is well covered by answers here plus many web references.Less obvious is that the spec sheet will have "absolute maximum" specs and "operating specs". The latter may be termed "Recommended operating conditions". This should be understood to mean "recommended if you want your system to work OK conditions". Specs which allow the pins to be taken outside the supply rails are almost always in the "absolute maximum" specs and NOT in the "operating" specs. "Operating" specs almost always specify voltages that NEVER are taken outside the supply rails. If you want your design to ALWAYS work then you must NEVER allow it's input pins to be driven outside operating specification range when in use. Maximum specs are system destruction limits. There is a good possibility that if Vin exceeds operating specs but not absolute maximum specs then the device may functionally misbehave. This can be a one-off problem at the time, or a problem which starts when input is out of spec and continues while input remains out of spec, or may start when Vin is out of spec and continue until the system is fully powered down. An obvious example of the latter is a full IC "crash" where the IC goes gaga and will not recover until powered off. Out of spec input on an ADC pin may affect readings on other ADC channels - perhaps an electrically adjacent one or all ADC inputs. This may occur while Vin is out of spec but (less commonly but possible) remain in this mode until repowered.Worst case, an input that meets Abs max spec but fails operating spec MAY result in IC malfunction even after the IC is powered completely down, then again powered up and reset. I have seen this happen. The IC may remain non compus mentis for many minutes after being wholly powered down. This is caused by charge being injected via the "protetcion" diodes into nodes that are not normally driven. I have seen this happen in ICs used in commercial designs (including mine :-( ) such that special anti-glitching circuitry is needed to prevent or recover the condition. Note: Many people will tell you that taking pins to maximum limits during normal operations is acceptable. This is not true if you never want uncontrollable outcomes. If it does not bother you if your system faults, locks up or walks funny then by all means follow the advice of people who say its OK to violate operating specs

1. How to reduce voltage without reducing the current using zener diode?

For something like that, I think you should rather use a 5V voltage regulator

2. Why does a power diode have a much larger p-n junction area than the regular signal diode?

A power diode has a much larger p-n junction area because all solids are ultimately limited to a maximum current density. Beyond this maximum they will heat up so much that they melt. Power diodes are designed to handle lots of power, P = IV, which means that for a fixed potential, V, the current must be large. Current density is current per unit area. If you want a particular maximum current, all of the cross sectional areas of the diode must have a minimum area so that the maximum current divided by the area never exceeds the maximum current density.In most solids, this number is between one tenth of a million and one million amps per square centimeter, 10^6A/cm^2. It sounds like a big number but most junctions are quite small. A typical diode for small signal applications will have an area about one micrometer square, mu m^2, which is 10^-8 cm^2. Such a diode can only handle 0.01 Amps or 10 milliAmps (10^6 * 10^-8). To handle 1 Amp the diode area must be 100 times larger 10x10 mu m^2 . Power diodes need to handle tens of amps even for household power. Some household diodes rated at 6 amps have surge current capacity of 400 amps. That makes the area of such diodes at least 200 x 200 mu m^2. This is huge for integrated circuit technology.Why does a power diode have a much larger p-n junction area than the regular signal diode?

3. What diode should I use that doesn't drop power?

If you put a string of diodes between a power supply and a resistive load, the current through the load will be Iload=(Vps-n Vd)/Rload. The load power will be Iload^2 Rload

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