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I came through this schematic for powering a 3g module (SIM800L). enter image description here

What is this diode connected with the bypass capacitors? By googling 5.1V 500mW I came to conclusion that it is probably a Zener diode(however the symbol does not seem like a zener diode) Is that true? What is the use of it?

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    \$\begingroup\$ Quite simply overvoltage protection. If the input power goes above 5.1V the zener (you are correct) will clamp. Without any current limiting it might not survive very long, though. \$\endgroup\$
    – Peter Smith
    Feb 3, 2021 at 14:30
  • \$\begingroup\$ so if Voltage increases above 5.1V, then current will not flow to what is connected to this circuit (the sim800L in that case) but it will be grounded through the zener diode. is that correct? like zener will "suck" everthing to the ground \$\endgroup\$
    – Kris
    Feb 3, 2021 at 14:35
  • \$\begingroup\$ @Not quite grounded. "Grounded" implies near 0V. More like the zener will break down and conduct in reverse causing it to be 5.1V across it which keeps everything else in parallel at 5.1V...as long as the zener survives. \$\endgroup\$
    – DKNguyen
    Feb 3, 2021 at 14:37
  • \$\begingroup\$ so zener will "stabilise" the Voltage to 5.1 V incase that goes above 5.1V \$\endgroup\$
    – Kris
    Feb 3, 2021 at 14:38
  • \$\begingroup\$ Yeah. Or clamp, or limit. \$\endgroup\$
    – DKNguyen
    Feb 3, 2021 at 14:39

1 Answer 1

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This is a Zener diode for over-voltage protection. Ordinarily, a diode conducts only in one direction. However, every diode has a "reverse breakdown" voltage, where the reverse bias voltage across the diode (or, the electric field across the PN junction) becomes great enough for the diode to start to conduct in the opposite direction.

Zener Diodes have finely tuned reverse breakdown voltages, and they can then be used to protect circuits from over-voltage by taking advantage of this reverse breakdown. If too much voltage is present, the Zener conducts in reverse bias. If designed properly, this will either (a) remove harmful voltage transients and allow the circuit to continue operating normally or (b) conduct enough current from the overvoltage source that a fuse blows, causing the system to fail safe.

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  • \$\begingroup\$ isn't every diode working the same way? If I connect a regular diode reversed instead of zener with similar voltage 5.1V, wouldn't it work the same way? \$\endgroup\$
    – Kris
    Feb 3, 2021 at 14:37
  • \$\begingroup\$ Theoretically yes, but regular diodes are not optimized to work that way: imprecise breakdown, not designed to handle all the heat, far higher and fewer breakdown voltage selection etc \$\endgroup\$
    – DKNguyen
    Feb 3, 2021 at 14:40
  • \$\begingroup\$ A general purpose diode usually has a reverse breakdown voltage that is very large, like 1kV, and that will not protect your circuit. You might be able to use any diode with a reverse breakdown of 5.1V, but there are more things to consider, such as the clamping voltage, peak and DC power rating, and failure mode. If you are not designing for any particular over-voltage case, you should get away with any Zener or TVS with 5.1V reverse breakdown voltage. \$\endgroup\$
    – Dylan Klepps
    Feb 3, 2021 at 14:46
  • \$\begingroup\$ copy that, so principle is the same/similar, zener is the best option for this tho due to its charasteristics (low reverse breakdown etc) \$\endgroup\$
    – Kris
    Feb 3, 2021 at 15:24
  • \$\begingroup\$ Doing some research I found out that there should be a resistor in series with the Zener Diode. Why in this schematic the designer of the schematic did not use one? Not always required? \$\endgroup\$
    – Kris
    Feb 3, 2021 at 18:27

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