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I'm working with an MCU that sees a 0-5V input range. I have a possibility where they may accidentally see 12V (user error).

My problem is this analog input using a 5V Zener has a knee that isn't as sharp as I'd want and can load the sensor reading giving a bad measurement.

How can I get a sharper knee for around 5-5.5V or so?

enter image description here

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  • \$\begingroup\$ The shape of the "knee" is what it is, it is impossible to get a zener to "start zenering abruptly" at 5.00 V. Explain why a normal zener's knee isn't good enough, "not sharp enough" isn't a properl explanation. Explain what (bad things) happen due to that knee not being sharp enough. \$\endgroup\$ – Bimpelrekkie Apr 6 at 15:17
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    \$\begingroup\$ Have seen 5.1V zener diodes (these exhibit avalanche reverse breakdown) with very sharp knee. Some even show negative resistance, once they start conducting. However, capacitance is somewhat larger than a regular small-signal diode. This isn't all bad - it assists anti-alias filtering. \$\endgroup\$ – glen_geek Apr 6 at 16:10
  • \$\begingroup\$ @glen_geek could you find a part number I can look on for that behavior? \$\endgroup\$ – DonP Apr 8 at 0:23
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Here is one method:

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

If the errant user leaves the +12 on there the resistor will see almost half a watt and the transistor a few hundred mW. You could increase R1 to a few hundred ohms to reduce both those numbers. Note that the input current is diverted to ground through Q1 rather than into the +5V supply. The latter can cause unwanted side effects.

This is still dependent on the characteristics of a PN junction so the clamp voltage will change with temperature and there will be some leakage as you approach the clamping voltage. You can play with R2/R3 ratio to change the clamping voltage and even try to make it temperature dependent with a thermistor or diode to a higher voltage source but it's far from perfect.

If you want something more precise, especially over temperature, you'd probably have to combine an active op-amp clamp with something like the above to handle fast transients.

Edit: As Jack Creasey mentioned in a comment, in this case R2 can be replaced with a diode or a transistor with collect tied to base (ideally in a dual NPN-PNP transistor array) to give better temperature compensation. You can add a small resistor in series to give a bit more headroom if you want to be able to go closer to the supply rail.

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  • \$\begingroup\$ The TL431 "super zener" also comes to mind if you need a really sharp zener knee. \$\endgroup\$ – Hearth Apr 6 at 16:17
  • \$\begingroup\$ @Hearth The OP's problem is that the protection is loading the input signal, that will also happen with the voltage divider to a TL431's sense terminal, though at least it will be linear. \$\endgroup\$ – Neil_UK Apr 6 at 16:37
  • \$\begingroup\$ @Neil_UK Fair point! \$\endgroup\$ – Hearth Apr 6 at 16:38
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    \$\begingroup\$ IMO you should replace R2 with a diode. \$\endgroup\$ – Jack Creasey Apr 6 at 17:37
  • \$\begingroup\$ @JackCreasey Yes, that will work and improve temperature compensation. Added. \$\endgroup\$ – Spehro Pefhany Apr 6 at 17:41
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For input protection of MCUs, they can usually tolerate a small over-voltage. A diode to the positive rail is usually good enough. You can choose between a low leakage silicon device that clamps around 0.7v, or a higher leakage schottky that clamps at around 0.3v.

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    \$\begingroup\$ The problem with a diode to the supply is that most regulators cannot sink current so you end up with a supply rail that approaches the input voltage minus a diode drop, which destroys chips attached to the rail that can’t handle the voktage. Unless the sum of all the loads on the rail will always exceed the fault current into the rail. \$\endgroup\$ – Spehro Pefhany Apr 6 at 16:56
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    \$\begingroup\$ This won't work where the potential is at least one 70mA clamp, you could easily destroy an MCU ...if you have the potential for multiple faults (as students well might) then you need to sink potentially several amps total. \$\endgroup\$ – Jack Creasey Apr 6 at 18:20
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How can I get a sharper knee for around 5-5.5v or so for MCU port signal clamp?

No Zener diode will be effective in this type of application, the knee voltage curve is just too broad.
Your definition of the clamp voltage as 5-5.5V may be too broad as well. Most MCU's have internal conduction increasing rapidly at VDD + 0.3V and VSS - 0.3V (intrinsic diodes), so you really need to clamp at LESS then 0.3V above VDD or below VSS.
If your 'user error' includes +12V as a connection possibility, then perhaps you should also include -12V as a possibility.

I assume that you have multiple inputs (both digital and A/D inputs) on your MCU to protect, so the following may be effective for you:

schematic

simulate this circuit – Schematic created using CircuitLab

Note:

  1. The negative clamp is simply the diode Vf so varies considerably with temp.
  2. The positive clamp can be adjusted to clamp just above VDD but before the internal intrinsic diode conducts.
  3. If there is the possibility of multiple inputs being wired to 12V (if this is a student environment this may well happen), then you have to account for 70mA for EACH clamp. This adds up, so you may need a small heatsink for the TIP42 which may need to sink several amps in total.

Note: This clamp circuit could drive about 30 inputs or more depending on the number of simultaneous errors to 12V you want to allow.

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  • \$\begingroup\$ If i'm reading this diagram correctly, this is accepting up to 3 inputs right? \$\endgroup\$ – DonP Apr 8 at 0:15
  • \$\begingroup\$ The clamp driver would be suitable for up to about 30 inputs at 70mA per input. \$\endgroup\$ – Jack Creasey Apr 8 at 4:00
  • \$\begingroup\$ @DonP ….you can drive lots of clamped inputs with the schematic I showed ….compare this to Spehro's answer which needs two transistors (or one transistor and one diode) per input protected. \$\endgroup\$ – Jack Creasey Apr 8 at 4:45

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