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zener circuit

Discussion in 'General Electronics Discussion' started by Sator, Aug 6, 2014.

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  1. Sator


    Aug 6, 2014
    I have a simple voltage divider circuit going into a microcontroller. If i change the ratio of the divider i end up loosing resolution of the AD converter. the problem with not being able to change the ratio is that i have voltages coming into the divider above the controller limit. since i don't care about the value of the higher voltage i tried a zener on the bottom leg of the divider, in this case the voltage divider would work until the zener kicked in and then just read the zener voltage. instead the 3.3v doesn't go above 1.5v even at full voltage input. help? thanks zener.jpg
  2. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    Welcome Sator to our forum.

    Unfortunately the datasheets I find do not show the reverse breakdown characteristic in detail. However, a zener diode does not abruptly break down at the zener voltage but has some leakage current already at smaller voltages. This leakage may be sufficient to drop enough voltage across the 120k resistor.

    As an alternative, remove the zener and connect a simple switching diode (e.g. 1N4148) from the tap to 3.3.V (anode to tap).

    As a second alternative check whether a protection diode is necessary at all. A typical chip input has integrated protection diodes and can tolerate a certain input current through these diodes to Vcc or GND. This is normally specified in the datasheet of the chip (microcontroller). The 120k resistor limits the input current to approx. Iin=(Vin-Vcc+0.6V)/120kΩ (0.6V for the integrated protection diodes). If this current is less than the max. input current for max. input voltage Vin, then you are safe without additional protection circuitry.
    KrisBlueNZ likes this.
  3. Sator


    Aug 6, 2014
    i'm not so worried about current into the micro as applying 4.3V (divided down 25.8Vin) to a 3.3V A to D pin of the micro, the pin is not one of the 5v tolerant pins. i still need to read the voltages from 0-19v
  4. LvW


    Apr 12, 2014
    The current must not be larger than (25.8-3.3)/120k=0.1875mA to allow a voltage of 3.3V across the zener..
    That means: I don´t think that you can reach the breakpoint at 3.3V with such a small current.
  5. Sator


    Aug 6, 2014
    i agree, since this circuit has an extra "leg" with the divider to ground compared to a typical zener circuit i wasn't quite sure how current flow worked. i think i calculated 184uA, was thinking for halving the 2 resistors which should put it over the minimum current of .25ma up to 368uA (i think). at 5v the current is only 35.7uA so this may not work anyways. i will try it
  6. Arouse1973

    Arouse1973 Adam

    Dec 18, 2013
    Looking at the data sheet the lowest current for 3.3 V out is 1 mA. Your circuit at 25.8 volts will give 3.69 Volts to the zener. With a BZV55C3V3 I got 2.77 Volts out. Changing the 120K to 5K I got 3.28 Volts out. So I think you are going to need to increase the current a bit to get 3.3 Volts out. Remember the device I used is different but shows what I mean.
  7. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    Jan 21, 2010
    Zener diodes with low voltages have a really soggy knee. In some cases (very low currents) the forward voltage differs little from the reverse voltage!

    I would recommend you place a diode between the output of your voltage divider and your 3v3 supply. This will limit the voltage at the ADC pin to Vcc + 0.6V. The voltage divider will already act to limit the current.

    If you want to, you can use a schottky diode which will have a much lower forward voltage. All of these diode solutions will add a little capacitance and may load the voltage due to leakage (which is what you're seeing in a big way with the zener).

    As Harald suggests, the protection may not be required. If you think there will be high voltage pulses at the input then you may have a reason to include the additional protection.

    Edited to refer to Harald, not Dave -- sorry
    Last edited: Aug 8, 2014
  8. Harald Kapp

    Harald Kapp Moderator Moderator

    Nov 17, 2011
    The pin won't "see" 4.3 V.It is internally clamped to 3.3V via the protection diode (internal to the chip). This is the reason why I stated that you may be able to get along without any additional protection as long as the current into the micro's pin is not stressed above the limit.
  9. BobK


    Jan 5, 2010
    Or for more safety, you could place a beefier schottky diode from the pin (anode) to Vdd, which would parallel the internal diode.

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