TVS diodes are used for transient overvoltage protection. You should not use a TVS for clamping a voltage continously. It is only for protection against voltage transients,
A TVS diode is placed in parallel (not in series) with the power input that you want to protect, so, the 15A that you mention will not cross the TVS.
You can see a TVS as a big resistor (one lead of the resistor connected to the voltage and the other one connected to GND) which, when the voltage that is being monitored is normal, it puts an infinite resistence value, and when the voltage reaches its breakdown voltage threshold, the TVS changes its internal resistence to near 0Ohm achieving with this to drive the overvoltage to GND.
Breakdown voltage: This is a threshold. If the voltage that is being monitored reaches this value, the TVS will turn on (decreasing its resistence) and driving to GND the overvoltage detected.
Clamping voltage: When the breakdown voltage is reached, the TVS goes to its on state and it will try to clamp the voltaje to this value (clamping voltage). While the TVS is trying to do this, it will be responsible of absorving (drive to GND) all the current that is generated due to the difference of voltages [it's something like (Vtransient - Vclamp)/Rtvs].
Maximum Peak Pulse Current: This is the maximun current capacity that the TVS can drive before failing due to excessive heat dissipasion.
If you want to clamp the voltage to 5.5V continously, you should use other element and not a TVS. For example a zenner diode with its proper resistor.
If you are plannig to use the TVS only for protection against transients, so you should take into account these steps to choose the right one:
Choose a breakdown voltage higher than the maximum normal operation voltage. For example, if you voltage range is up to 5V, you could use a TVS whose minimum breakdown voltage is 5.5V. If you voltage range is up to 5.5V, you could use a TVS with breakdown of 6V.
After choosing the breakdown voltage, you will see that the TVS diodes that are available for your app is reduced.
Choose a TVS with a high heat dissipation capacity (Watts). This will prevent your TVS diode from being damaged when activated. This feature is important because the price and the size of the TVS depending on it.
Choose the TVS whose clamping voltage is the lower (from the TVSs that you will have to choose at this point).
Follow the previous steps in that specify order.
Also, consider the idea of adding a PTC resetteable fuse to your design. The pair TVS+PTC fuse is perfect. First, you should put the fuse in series with the input and then the TVS. The PTC fuse will reduce the amount of current that the TVS must dissipate.
I hope this helps.
Both solutions are not fully correct. First of all, look at the datasheet of 1.5KE series. It claims that 1.5KE39 has nominal breakdown voltage \$V_{br}\$ of 39V, but don't forget about accuracy and possible breakdown voltage in 37V to 42V range (7% accuracy). In practice breakdown voltage of TVS never equals to its nominal, I've checked that many times.
First diode (Z2033) has nominal \$V_{br}\$ of 33V (29.7 to 36.3V), so less than Your original one. The best choice from VRD series is Z2039 with 39V nominal breakdown voltage.
If you combine two 1.5KE18 (nominal \$V_{br}=18V\$) in series, theoretically you will have 36V breakdown voltage. But due to accuracy of every diode, the breakdown voltage will be somewhere between 34V and 38V - it's still too little. You should combine two 1.5KE20, then range of the breakdown voltage will be 39V to 42V.
Best Answer
Reverse Standoff Voltage is the voltage at which no noticeable current flows, uA or nA scale usually, for the diode you have listed it is 1uA. This means that below the standoff voltage the diode will not conduct more than 1uA.
Breakdown Voltage is the voltage at which the diode begins to conduct, at the breakdown voltage for your diode the current is 1mA. When 1mA is flowing through the diode, the voltage will be 40V-44.2V.
Clamping Voltage is the voltage across the diode where the power limit is reached, which for your diode is 58.V @ 6.9A. The pulse rating for the diode is 400W and 58.1V * 6.9A = 400.89W
The current flowing through the diode is more of a curve(not discrete steps) and these voltages/currents just help give a better understanding of the current that flows through the diode as the voltage increases. If your circuit continually operates at 38V you will have between 1uA-1mA of leakage through the diode. your actual power dissipation at that level will be no greater than 38mW, so this diode should work for your design, depending on your power budget and the spike levels you want to protect against.
You may want to use a circuit like the following:
simulate this circuit – Schematic created using CircuitLab
The resistor and capacitor filter the repetitive spikes and the diode is for infrequent large or sharp spikes. The actual values should be determined based on the amount of power going through the filter and acceptable noise.