Connect with us

Zener diodes for input protection

Discussion in 'General Electronics Discussion' started by adamp524, Apr 28, 2010.

  1. adamp524

    adamp524

    4
    0
    Apr 28, 2010
    Hi,

    I'm trying to use zener diodes to protect the inputs to an IO driver chip (Microchip MCP23017)
    The IO chip runs off 3.3V and cannot accept any inputs higher than this; however, all of my inputs are at 5V.

    Attached is a schematic for my protection circuit arrangement. Pins 1-8 on CONN1 are where 5V inputs will be introduced and I1-I8 is the connection to the IO driver chip. As can be seen, each input has a 3.3V zener diode and 10K resistor placed in parallel with it.

    My problem is that this arrangement does not seem to work! The IO driver chip still gets 5V at its inputs (I1-I8)

    Does anyone have any ideas as to what I'm doing wrong?

    Thanks


    The datasheet for the 3.3V zeners I'm using is located here:
    ---docs-europe.origin.electrocomponents.com/webdocs/0dad/0900766b80dad746.pdf---
     

    Attached Files:

  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,186
    2,692
    Jan 21, 2010
    Firstly, what is doing the driving of these inputs?

    Can you indicate why a simple resistive divider would not work (that would be my first choice)?

    A second option is a series resistor and a diode clamp (using a schottky diode to act before the input protection) to the 3.3V rail.

    Another option may simply be to use a series resistor (and limit the current through the input's protection diodes)

    Another option is a 3.3V buffer with 5V tolerant inputs.

    The Zener option "correctly" would be a series resistor and the zener connected to ground. However be aware that these low voltage zeners have a very soft knee. As a colleague found out, at low currents they can easily clamp the voltage to a level far lower than you expect. (take a look at figure 8 and figure 9 in those specs -- the 3.3V zeners are a real mess)

    Your circuit is bad for several reasons, the major one being that there is no limiting of the current through the 3.3V zener. This circuit simply loads the output of whatever is driving it. Depending on what is driving it, you *could* turn the zener, or the chip driving it into a charred mess.

    Have you tested the zeners to make sure that you actually have 3.3V parts? Is it possible that your supplier accidentally supplied 33V (for example).
     
  3. adamp524

    adamp524

    4
    0
    Apr 28, 2010
    The inputs will be driven from a range of voltages from 5V to 12V.

    As far as I'm aware, a potential divider will not work because of the fluctuating input voltage.

    If I were to use a series resistor with a Schottky diode, what sort of resistor value would I use when the input voltage is ranging from 5 - 12?
    Would the resistor just be to limit the current going through the diode?
    Also, what are the advantages of using a Schottky diode vs using the zener diode I suggested?

    Thanks,

    Adam
     
  4. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    Select the series resistor low enough to drive the parasitic capacitance on the input depending on your signal speed, and high enough to limit the maximum current through the diode. Your zener solution should also work, but using a Schottky diode is the standard practice. In fact, for analog signal processing the input is usually protected through two Schottky diodes, reversed biased to the positive and negative power rails. A Schottky diode has faster response and less leakage than a zener diode. A good reference would be this application note:
    http://www.cirrus.com/en/pubs/appNote/an20.pdf
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,186
    2,692
    Jan 21, 2010
    Check what the knee on 3.3V zeners looks like. The reverse voltage varies so much with current that they're barely better than a resistor (ok, maybe a little more than barely, but you get the idea)
     
  6. Laplace

    Laplace

    1,252
    184
    Apr 4, 2010
    OK, (*steve*) now you are speaking like a design engineer. I won't argue with that except to observe that sometimes, to an experimenter, a solution that barely works is indistinguishable from one that works very well.
     
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,186
    2,692
    Jan 21, 2010
    OK, as a practical example, a colleague of mine used a low voltage zener in a manner similar to this and found the input was clamped well below the "advertised" zener voltage.

    As an experimenter, he had all sorts of problems until he discovered this.

    It highlights the fact that reading spec sheets is really important. *I* was surprised at how soft these things are. Their relatively high leakage probably doesn't help either.
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-