I am looking to select an appropriate Zener diode from the ZMM range for MCU input pin protection.

They are specified at a Zener current I_ZT = 5 mA having a maximum impedance of Z_ZT = 90 ohm.

They also specify a maximum impedance (at working current threshold?) Z_ZK = 600 ohm @ I_ZK = 1 mA.

I am assuming absolute maximum input pin voltage of 5V.

My analysis is as follows:

  1. The Zener diode requires at least 1mA to break down: 5V / 1mA = 5kOhm current limiting resistor. Below 1mA the Zener is not guaranteed to break down and my input pin could see the full voltage.

  2. At 1mA the Zener impedance is 600 Ohm leading to a voltage drop of 0.6V.

  3. 5V permissible input voltage - 0.6V zener voltage drop: --> require zener diode with a maximum zener voltage of 4.4V.

  4. From the ZMM range, the next best part would be the ZMM3V9 with a V_Z(max) = 4.1V @ I_ZT = 5mA.

My issue is, that steps 1-3 of my analysis were based on I_Z = 1mA but the Zener voltages are specified at 5mA.

Assuming we calculate an 'junction voltage' without the resistance voltage drop:

V_j = V_Z - 90ohm * 5mA

Does V_j vary with respect to the current I_Z?

Is it acceptable to calculate a Zener voltage at lower current as follows?

V_Z @ 1mA = V_j + 600 ohm * 1 mA

I started my analysis at the working threshold current, because a selection at 5mA gives insufficient over-voltage protection:

Assuming 0.25 W current limiting resistors:

V_input_max = 35V with 5kOhm (1mA analysis)

V_input_max = 15V with 1kOhm (equivalent 5mA analysis)

I am looking for protection from sustained voltages of 30V and an expected normal operating voltage of 15V.

  • \$\begingroup\$ The edit helps. \$\endgroup\$
    – tcrosley
    Commented Feb 23, 2017 at 16:39

3 Answers 3


You should base your analysis on \$I_{ZT}\$ instead of \$I_{ZK}\$. The 5mA current is the one that guarantees that the voltage is going to be under regulation.

Does the zener voltage depends on the current?

Yes, that's why they give you an impedance (e.g 89 Ohms max @\$I_{ZT}\$) when working as a regulator. So if the current changes, so does the zener voltage.

\$I_{ZK}\$ and \$Z_{ZK}\$ are the current and impedance they get near the breakdown region (the knee in the IV curve).

In order to be certain that the zener will work as a voltage regulator, you have to ensure there is always 5mA (the test current they use) running through it.

Your minimum sustained voltage is 15V, your zener voltage is 5V, and you want 5mA flowing to the zener at all times to keep the voltage regulated.


$$ R_s = \dfrac{V_s-V_z}{I_t}$$

Where \$I_t\$ is the total current from the power supply (zener current + current drawn by the pin or load). So if the pin draws an additional 10mA for exaple, your total current is 15mA.

$$ R_s = \dfrac{15\text{V}-5\text{V}}{15\text{mA}}=670\Omega$$

\$R_s\$ is a resistor between your source and the zener diode.

If the voltage goes any higher, say 30V as you mentioned, then there will be more current going through the zener, but as long as you don't go over the power dissipation limit of the diode, you should be ok. So if the voltage increases to 30V, the total current will be around 37mA with the chosen \$R_s\$. If you had no load, the zener will dissipate a max power of \$P_z\$=(37mA)(5V) = 0.187Watts, so choose accordingly.

  • \$\begingroup\$ Ok, I was unclear. I was asking the following really: Assuming V_1 = V_Z - 90 ohm * 5 mA, does V_1 vary with respect to I_Z? V_Z does vary, obviously. \$\endgroup\$
    – ARF
    Commented Feb 23, 2017 at 15:54
  • \$\begingroup\$ @ARF yes, model the zener diode as a voltage source in series with an impedance, which is what you are given. The voltage source has a nominal value, call it \$V_n\$, and and impedance \$R_z\$. So the actual zener voltage is \$V_z\$=\$V_n\$+\$I_zR_z\$, which depends on the current. That's why the \$V_z\$ isn't constant but should remain reasonable close to a nominal value. \$\endgroup\$
    – Big6
    Commented Feb 23, 2017 at 16:37
  • \$\begingroup\$ Thanks. I am now having doubts thoug. E.g. looking at the 3.3V diode in the datasheet referenced in George's answer, I have $V_n = 3.3V - 28 Ohm * 20 mA = 2.74 V$. The problem is, at 0.1 mA the figure 1 shows about 1.8V - which is seems inconsistent with the voltage source + series impedance model. Something more complicated seems to be happening. At least the Zener voltage @ 0.1 mA is below that specified at 20mA. I am wondering whether this will always be the case. \$\endgroup\$
    – ARF
    Commented Feb 23, 2017 at 17:29
  • \$\begingroup\$ @ARF First, 0.1mA is well below the test current for regulation (20mA). And even the current near breakdown is higher as per the datasheet (0.25mA), at that current of 0.1mA, you need to use a different Vn and use Zzk instead of Zzt. Those are two different slopes. Now, near the 20mA current, for 3.3V zener, you can use a linear approximation. That's the point (20mA) where the specified impedance is found and the 28 Ohms is a max value, so it could be lower. In other words, the approx is valid near the operating point of (3.3V, 20mA) \$\endgroup\$
    – Big6
    Commented Feb 23, 2017 at 18:21

I'm not sure I understand your question. But a good zener data sheet will show you the I-V curve for the various diodes. Here's one...not the best but the first I found. http://www.mccsemi.com/up_pdf/1N5221-1N5267(DO-35).pdf


Although the Shockley equation defines the diode voltage well up to saturation, beyond this it fails. This is because the bulk resistance affects and determines the ESR or Zt of the diode. Although the Zt will drop with rising current above rated current, the rate is low enough to assume it is constant even if it is -25%. The variations are actually process quality related where best case Zt=ESR ~ 1/4Pd (high power high efficacy, high quality) , and typical is ESR=1/Pd for a package power rating of Pd at 85'C . This is not in any textbook, so I call it ;) drum roll... ;)

Stewart's law . Zt=1/Pd (+/-50%) for Vf = Vth + Zt *If

  • and Vth ~=90% of Vf rating best case ( higher indicates sorted by lower quality higher ESR) thus Vf=0.9Vf + Zt*If this would imply Zt=0.1Vf/If which does not prove as accurate as the datasheet intercept. so Zt=1/Pd typical is my Rule of Thumb.

You can try this out on 100V zeners or 100A power diodes or 15mW zener regulators

Note: Zeners are rated at lower Izt currents than max so Zt is much higher at the regulated voltage and will drop in Zt value significantly at max current.

Therefore for currents near the rated current for If=5mA, your analysis is correct. as @SiztoCabrera writes \$V_z=V_n+I_zR_z\$ where \$V_n\$ is around 90% of \$V_z\$. ...where the threshold \$Vth(=V_n)\$ For example this 3V white LED is rated at 100 to 300mA meaning the package can handle 1/3 to 1W depending on heatsink and bin sorting of quality at factory. So we expect ESR to range from 1 to 3Ω.

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back to the solution

Ultimately for protection you want a Schottky diode to supply rail with a current limiting R just as they do with 2 Sch Diodes in all CMOS inputs or a TVS diode rated for 5V.

  • \$\begingroup\$ Thanks for this detailed answer. Two issues though: 1. I have positive rail @ 3.3V and maximum input voltage at 30V. Is seems imprudent to put voltage anywhere near the positive rail. Hence my question on shunting the excess to ground with Zener diodes. 2. To what extent does your analysis apply to Zener diodes? After all the operating regime of a Zener is reverse bias while Shottky diodes to supply rail would be forward biased. \$\endgroup\$
    – ARF
    Commented Feb 23, 2017 at 17:41

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