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:
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:
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