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I have a Zener diode of the type BZX84C3V6LT1G.

enter image description here

It has a leakage current of 5 uA at 1 volt. Even though it has a leakage current, I need to increase my current source to 24-25 mA, until I get get a 3.2V voltage drop over the Zener diode.

24mA over 160 ohm is 3.84V over the shunt resistor. If I have 20 mA as current source, then I will have like 2.8V over the Zener diode, but 3.2V over the shunt resistor. Why?

Is it because the 160 resistor to the left is stopping the current from reaching the Zener diode and the Zener diode then has a chance to leak out the current?

enter image description here

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  • \$\begingroup\$ What is the purpose of R2 (the vertical resistor)? \$\endgroup\$
    – Transistor
    Jun 23 at 18:21
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    \$\begingroup\$ @Transistor Covert 4-20mA to 0.6-3.2V. That's because my STM32 processor has its ADC reference to 3.3V and 160 ohm resistor is a standard resistor. \$\endgroup\$
    – DanM
    Jun 23 at 18:28
  • \$\begingroup\$ @user287001 Well, I could remove the 160 ohm resistor to the left. But that resistor also works as a protection resistor for the zener diode. \$\endgroup\$
    – DanM
    Jun 23 at 18:36
  • \$\begingroup\$ @user287001 The zener diode has a maximum power consumtion of 250 mW. \$\endgroup\$
    – DanM
    Jun 23 at 18:37
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    \$\begingroup\$ So you are trying to design a protection circuit for a 4 - 20 mA ADC input? You should explain that in the question. \$\endgroup\$
    – Transistor
    Jun 23 at 18:46

5 Answers 5

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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. The circuit redrawn with voltmeters and ammeters added.

24mA over 160 ohm is 3.84V over the shunt resistor. If I have 20 mA as current source, then I will have like 2.8V over the Zener diode, but 3.2V over the shunt resistor.

We measure voltage across a resistor and current through a resistor. (It may be different in other languages.)

In Figure 1 we can see that most of the 20 mA returns to the current source through R2. The 19.61 mA through 160 Ω causes the voltage to rise to only 3.138 V and this isn't enough to turn on the Zener diode properly so very little current flows through it.

You can see from the datasheet that the current varies rather a lot at different voltages. They are not precision devices.

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  • \$\begingroup\$ If little current flows throug it, then it means I will have no or very little voltage drop over the 160 ohm resistor. But it seems that the zener diode leak more than 5 uA at 1 voltage across the zener diode. \$\endgroup\$
    – DanM
    Jun 23 at 19:07
  • \$\begingroup\$ I have a 3.6V zener diode. But the zener seems to open way to early. \$\endgroup\$
    – DanM
    Jun 23 at 19:08
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Zener diodes, like other components, are not ideal. They do not simply switch sharply on at their Zener voltage, but slowly near the Zener voltage, and low voltage Zeners turn on very slowly, they have a very soft "knee" in the curve.

As your circuit has already 160 ohms over 20mA source, it has a 3.2V output. There is virtually no voltage drop over the Zener bias resistor and thus virtually no current flows through it.

So if you increase current, the output voltage is limited by the 160 ohms parallel with it, which defines the output voltage. Voltage needs to be high enough to get the rated Zener current to flow through the Zener in order to have the rated Zener voltage over it.

The Zener marked in red requires 5mA through it to have 3.6V over it.

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  • \$\begingroup\$ But if there was no current through the zener, that means no voltage dropp over 160 ohm resistor. right? Well, in this case, it's a voltage drop over the 160 ohm resistor. First I used 10 kOhm, but then when I had 3.2V drop over shut resistor, then I had 2.2V over the zener diode. So that means that the left 160 ohm resistor blocking much current, so the zener diode have a posibility to leak out current? Right? \$\endgroup\$
    – DanM
    Jun 23 at 18:04
  • \$\begingroup\$ @Transistor I mean that the zener diode let more current to pass by, compared to the 160 ohm resistor to the left. Therefore, the voltage is less after the 160 ohm resistor compared to the 160 ohm resistor to the right. \$\endgroup\$
    – DanM
    Jun 23 at 18:25
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Low-voltage zeners have worse characteristics than higher voltage devices, as you have observed. Perhaps consider using a higher voltage device and then reduce this voltage using a divider (two resistors). Zeners of about 6V2 to 6V8 are desirable because the temperature coefficient is close to zero.

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Do not use a zener diode. Precision voltage reference-comparators (a.k.a. adjustable voltage references and precision zeners and shunt regulators) are generally used in power supplies to replace far too unsharply working low voltage zener diodes. Check this circuit:

enter image description here

If VR1 for example is a 2,5V type the ADC gets the voltage that sensor output current makes to resistance R2+R3. As soon as the voltage over R3 reaches 2,5V VR1 starts to suck current as much as it's needed to keep the voltage over R3 =2,5V.

Reference-comparators can sink as much as 100mA. This circuit protects your ADC input if R1 can be so high that 100 mA sink is enough to keep ADC input voltage equal or lower than the mentioned limit 3,2V.

You must calculate the voltage divider R2,R3 so that you get the wanted voltage and the treshold 2,5V is reached just when ADC input voltage is in its maximum. If you can scale ACD output numbers with software you can well get working system with standard resistors without a need to connect resistors in parallel. I skip those calculations.

A link to LM341 datasheet. That's one possible VR1. https://www.ti.com/lit/ds/symlink/lm431.pdf?ts=1655992465902&ref_url=https%253A%252F%252Fwww.google.com%252F

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Why does a Zener diode leak so much current?

enter image description here

On a log scale, 5 uA is NOT EXCESSIVE and is consistent with the exponential behavior of all current rectifier diodes with a straight line except near the knee current where the bulk resistance dominates the curve as a linear slope.

The value of the linear slope is what defines the value of Rs and tolerance of Vf above 1 mA . This is the incremental resistance (from bulk resistance) or slope in V/I. "onsemi" calls this Rs , the Zzt value or rated threshold resistance. (Zzk is the lower current knee threshold) The rated current, Rs = Zzt are inversely proportioned to the size and power rating of the diode Rs=<=1/Pmax and for this part ranges in the 50-ohm range so its Vf rises significantly above 5mA as such. 0.75V/15mA = 50 Ohms (results from 20-5=15mA)

The linear plot below on the right Y-axis shows the Z value 1 to 5 mA and drops down from 5 to 20 mA. enter image description here

As a matter of comparison a 5 mm White LED has a threshold voltage of 2.85V with a nominal Vf of 3.1V @ 20 mA with a supplier depending tolerance, thus the Rs or Zzt or bulk resistance is around 12 Ohms much lower than this Zener but makes a cheap regulator and indicator and is more consistent between suppliers if kept in the 2 to 3 mA range at 2.9V. Poor suppliers in the past have rated their 5mm LEDs up to 3.6V at 20 mA. The chip itself is much smaller than 5mm, which is just the lens.

This is an XY question from the comments and lack of clarity in the query text.

  • the above correctly answers the question in the title.
  • The perceived question is how to improve a 4-20mA to voltage converter with adequate protection for uC input, such as over-voltage or over-current with Vdd= 3.3V +/- TBD tolerance.
    • Since the ADC has a high Zin and the load is low 160 ohms, all you need is a >=10 k series R from the 160 and then optional Schottky diode clamps to Vdd,Vss or rely on ESD diode pairs internally. If there is still common mode noise, then a differential Op-Amp performs better than a single-ended input to local ground. This often occurs due to SMPS CM noise with floating supplies.
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