Protecting pins doubth

MapacheRaper · Joined: Feb 15, 2018 Posts: 166 Location: Spain

Im building some arduino circuits and I want to protect the pins from the eurorack voltages. I see most circuits use a resistor and then a 5v1 zener connected to ground.

In other schematics I see 2 schottky diodes. One between pin and ground and another between +12 and pin.

Are all correct?. I guess the double schotkky gives more protection, isn´t?

Why 5v1 zeners and no less?.

I would like to give all the possible protection to my circuits, but if just one 5v1 zener is enoght, why to put 2 schottkis?

blue hell · Posted: Thu Nov 22, 2018 6:05 pm Post subject:

Often input pins are specified to be able to tolerate input voltages between Vee / Vss (usually ground) - 0.3V and Vcc / Vdd + 0.3V. Now when your Vcc / Vdd is 5V a 5.1 zener would handle positive overloads, but the dual schottkky would do it for all Vcc / Vdd values and for negative overloads too. Sometimes on 3V3 devices inputs may be 5V tolerant - so in that case the zener thingie could work.

One caveat .. the zener will protect against positive over voltage, but for reversed current it acts as a silicon diode, giving a .7V drop .. so the zener solution does not protect too well for negative overloads. This can be mitigated though with a current limiting resistor between the zener and the input.

It's less thinking to just use the schottkys .. as long as the tolerated over voltages are -0.3V / +0.3V. And usually the the zener solution when needing to protect against negative over voltage would need an extra resistor.

hope this is not too technical?

JovianPyx · Posted: Thu Nov 22, 2018 8:04 pm Post subject:

Schottky diodes are very fast switching diodes. When the module is not being abused by an over-range signal, both of the diodes are reverse biased (off) and do nothing. When the signal goes bad, one diode turns on and you want that to happen as fast as possible. The protection circuit is called a "diode clamp" which describes what it does. When the diode turns on, it will conduct whatever current is available from the source signal, so the diodes need to be able to conduct at least that much current. I like to make things like that 15% or more larger than expected demand. If the diode blows, it's no longer doing it's job.

Grumble · Posted: Fri Nov 23, 2018 1:25 am Post subject:

http://electro-music.com/forum/topic-68022.html

Long story short: as long as you use a resistor at the in/output of about 1k the internal schottky diodes will provide sufficient protection, no need for external diodes.

JovianPyx · Posted: Fri Nov 23, 2018 7:58 am Post subject:

Well, I've been able to blow internal CMOS ESD diodes. If that happens on a 40106, no big deal, but if it's a CPU, it's more expensive. Just saying that a few cents worth of external diodes can be more than worth both the monetary cost as well as the PITA factor of blown internal diodes.

MapacheRaper · Joined: Feb 15, 2018 Posts: 166 Location: Spain

Sorry for the delay in response, for some reason I haven´t received notifications of new posts in this thread.

That´s the perfect balance between technical and common sense... Thanks for the answers!

But I must confess that I still not sure of what is the correct way. (Does even exists a correct way??). I guess 1k plus a schottky would do the trick. Definitively I don´t want to blow the processors.

Im going to re read slowly till I grasp it. Thanks again, maestros!

Edit: Afetr re-read and going thru the "torturing the arduino pins with -17V plus 1K resistor" thread now I think I will sleep better. If the can handly it, for sure they will handle standard eurorack voltages (with a 1k resis) with no problem