I am looking for a deep explanation on why a TVS diode suppresses voltage. All of the research I have done on Google just says 'it suppresses voltage by shunting current', but this doesn't make sense to me. I also couldn't find any information on this device in the Sedra and Smith textbook.

My understanding of a TVS diode is that at the Vbr voltage, it will start to conduct a lot of current. An IV model neglecting other factors (like temperature etc.) looks like this for a bidirectional TVS: enter image description here

According to this, it makes sense that it passes a lot of current; at a constant voltage beyond Vbr, it could be though of as a resistor with very low resistance. This is where my understanding breaks down.

In a circuit, a TVS is used in parallel with the load:


simulate this circuit – Schematic created using CircuitLab

According to the nodal understanding, the voltage across the TVS is always the same as the voltage across the load. The only explanation for it clamping the voltage which slightly makes sense is that it acts as a voltage divider with Zload. But that would make the clamping of the TVS dependent on the impedance of the source.

Also, I'm not sure if this model for how the TVS works would make sense for transient events which aren't on the battery, like inductive coupling of a high voltage.

Thanks for reading my question, I appreciate anyone who can correct or confirm my thoughts on this.

TLDR: Does a TVS diode really work by voltage division

  • \$\begingroup\$ The value of \$Z_{load}\$ depends on the voltage across the TVS. Add switch in series with \$Z_{load}\$. That switch is only closed when the voltage across the TVS exceeds +/- Vbr. So: \$Z_{load}\$ has a high value when \$V_{TVS} < V_{BR}\$ and \$Z_{load}\$ has a low value when \$V_{TVS} > V_{BR}\$. In reality \$Z_{load}\$ is of course part of the TVS itself. \$\endgroup\$ Commented Oct 10, 2019 at 13:01
  • \$\begingroup\$ In my model Zload is the load you're protecting, which is usually some sort of MCU/IC for me. I understand the IV relationship of a TVS diode. I just don't understand how that implies that it will clamp voltage across the load. \$\endgroup\$
    – Tom
    Commented Oct 10, 2019 at 13:07
  • 1
    \$\begingroup\$ The (high voltage) source will always have some series resistance. During an overvoltage event, the TVS will get a low resistance value. Then we have a (some resistance) / (low TVS resistance) voltage divider. Indeed when the voltage source is ideal the TVS cannot clamp. But ideal sources do not exist. \$\endgroup\$ Commented Oct 10, 2019 at 13:21

1 Answer 1


that would make the clamping of the TVS dependent on the impedance of the source

A TVS diode cannot clamp a perfect voltage source; it requires that the source (an indirect lightning strike/stroke for example) has a source impedance. For lightning surge testing to EN 61000-4-5, the basic source impedance is 2 ohms so, if the surge voltage is 2000 volts then there can be a peak surge current of close to 1,000 amps. Many TVS diodes are rated for this current and do the job of surge protection very well (and quickly compared to GDTs and usually with a significantly longer lifespan than MOVs): -

enter image description here

So, it's feasible that the BZW50-12 is capable of protecting equipment from an EN 61000-4-5 surge of 4,000 volts. That's a pretty good device in my opinion but, you need to understand the source impedance.

Does a TVS diode really work by voltage division

I think using the term "voltage division" is one level away from just saying it clamps at a certain voltage and that voltage depends on the surge current.

  • \$\begingroup\$ You saying a TVS can't clamp a perfect voltage source reassures me that I'm on the right track. But wouldn't this effectively cause a short on the source, if the source sees an almost zero-ohm resistance? Could that damage the source? \$\endgroup\$
    – Tom
    Commented Oct 10, 2019 at 13:13
  • \$\begingroup\$ If the source is lightning, then damage to the lightning is irrelevant. \$\endgroup\$
    – Andy aka
    Commented Oct 10, 2019 at 13:14
  • \$\begingroup\$ And if it's a PSU, is the philosophy that it's better to damage that than the load? \$\endgroup\$
    – Tom
    Commented Oct 10, 2019 at 13:19
  • 4
    \$\begingroup\$ If the PSU outputs so much voltage that the TVS triggers: I would say that the PSU is damaged already. Under normal circumstances the TVS should not do anything. \$\endgroup\$ Commented Oct 10, 2019 at 13:23
  • 2
    \$\begingroup\$ Surge protectors protect down-stream equipment. If the up-stream equipment is producing an inappropriate surge situation then maybe that equipment is already faulty. If that happens then you need to also use a fuse to protect the cabling infrastructure. \$\endgroup\$
    – Andy aka
    Commented Oct 10, 2019 at 13:23

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