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schematic

simulate this circuit – Schematic created using CircuitLab

I'm trying to hook up a Zener diode in order to add some measure of protection (fused "crowbar") to a circuit that includes a loop powered oxygen sensor with output read by a GHI Cobra ADC/Analog In. The idea is that if there was a short in the sensor that caused 12V on its output, the Zener with 4.7 V breakdown between the sensor output (and the ADC input) and ground would keep the 12V current away from the ADC long enough for the fuse to blow as the limit on the Cobra pin is 5V.

However, I find that when I put in the Zener diode it drops the voltage being read by the ADC (volt meter right now), e.g., from 4.38 volts to 3.98 volts. When I replace the Zener with a regular diode just to test, the voltage does not drop. What gives? Can a Zener be used for a "crowbar" so it doesn't change the voltage on the sensor output side? The Zener is installed with the ring towards the ADC side. The electronics store said it has a 4.7V breakdown, but I'm not sure how to test that.

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  • \$\begingroup\$ Regular diodes don't usually display so good Zener break-down as Zener diodes do (that's why have special Zener diodes), so it's normal that a regular diode does nothing. Zener diodes may leak a small amount of current and in that way affect the measurement. What kind of test equipment do yo have available? Can you get a transient voltage suppression diode (TVS diode)? They should have lower leakage current than a Zener diode. \$\endgroup\$
    – AndrejaKo
    Commented Sep 1, 2013 at 19:39
  • \$\begingroup\$ Kinda tough to get a TVS diode right now (on small island.) The voltage drop is 10%--seems high for leakage. No measurable leakage with the regular diode. \$\endgroup\$
    – Matt
    Commented Sep 1, 2013 at 20:12
  • \$\begingroup\$ That is not a crowbar, it's a clamp. A crowbar shorts the input on overvoltage conditions; a clamp merely limits it to the specified voltage. \$\endgroup\$
    – marcelm
    Commented Nov 8, 2016 at 18:40

2 Answers 2

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The reverse voltage of a zener diode does not show a sharp knee with increasing current, so the zener will conduct several milliamperes at voltages well below its rated voltage. There's no need for a fuse in this case assuming that the input impedance of your microcontroller is high. By the way, what is a "GHI Cobra"?

I suggest replacing the fuse with a 4.7k resistor and using a 5.1V or 5.6V zener. It would be best if you could reduce the normal operating range of the ADC input so that the maximum expected voltage is around 4V...perhaps by changing the 250 ohm resistor to 200 ohm. Then the soft knee of the zener will not affect your measurements and you are giving the zener a little room to start conducting.

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  • \$\begingroup\$ GHI Cobra is a microcontroller (think Arduino) only with faster clock and 10x the RAM. The GHI and Netduino controllers are .NET based and use MS Visual Studio to develop in. A small list in their site: goo.gl/xDU7m5 \$\endgroup\$
    – Krista K
    Commented Sep 1, 2013 at 22:04
  • \$\begingroup\$ If the zener leaks "several milliamps", then, yes, that would certainly be the problem. I'm surprised there is such leakage. Don't understand replacing the fuse with a 4.7k resistor. If i stick w/ the diagram, then the ADC wouldn't be reading the voltage right, correct? \$\endgroup\$
    – Matt
    Commented Sep 1, 2013 at 23:17
  • \$\begingroup\$ Actually, i would be probably use a 165 ohm resistor to drop the max voltage to 3.3V, the limit on the Cobra's (Cobra II, actually) ADC. That would be safe even though i plan to keep the oxygen concentration that the sensor would be exposed to below 5%, thus the sensor would output low current even with a large resistor. \$\endgroup\$
    – Matt
    Commented Sep 1, 2013 at 23:22
  • \$\begingroup\$ So if the leakage is caused by a non-sharp knee at the breakdown voltage, then leakage may be insignificant when the sensor output is low which is where it will operate in my case. E.g., if the oxygen concentration is kept below 5%, the sensor output will be (roughly) below 7mA. Hmm. Will test for degree of accuracy with and without zener at such concentration/sensor output. \$\endgroup\$
    – Matt
    Commented Sep 2, 2013 at 2:17
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    \$\begingroup\$ Regarding the fuse: the zener you specified is a 1W device so its maximum current is about 200mA. It is difficult to find an inexpensive, very fast fuse that will blow at this current level. The more likely scenario is that the zener will be destroyed and the ADC input subjected to the full 12V output of your sensor. Using a resistor (say 4.7k) limits the current that will pass through the zener so you don't destroy it. Yes, the goal is for normal operation to be at a voltage where the zener leakage is negligible. \$\endgroup\$
    – Joe Hass
    Commented Sep 2, 2013 at 2:37
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Though it seems there isn't overvoltage on ADC pin according the Ohm's law, but, what case we need to avoid?

TVS diode is a good choice. every type MCU has a maxiam supply voltage, the best active voltage is lower then maxiam voltage and higher than work voltage. but if we consider the polarity of the surge voltage, two TVS diode series In the opposite direction.

There is another method to do this, you can use two diode to protect the ADC port, for example 1N4148.one diode connect betweeen VCC and ADC pin, another connect ADC pin and GND. it is best to use a resistor connect between ADC pin adn junction of the two diode. in fact, this is a common method to protect port inside MCU, It is a extensibility enhancements.

If it is acceptable, use optocoupler isolation is best choice. but it is very complex.

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  • \$\begingroup\$ Thanks, DreamCat. Actually, using a regular diode between VCC and the sensor output is my default protection scheme, i was trying to understand how i could use a zener diode and if there are any advantages. I hadn't thought of attempting to protect for a reverse polarity. Don't understand what you mean by, "it is best to use a resistor connect between ADC pin and junction of the two diodes." A TVS diode (new to me) sounds ideal. \$\endgroup\$
    – Matt
    Commented Sep 2, 2013 at 16:43

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