How can clamping voltage be lower than breakdown voltage. I reviewed the datasheet as well.
Can somebody shed some light please.
https://www.digikey.com/en/products/detail/nexperia-usa-inc/PESD5V0X1BCSFYL/5981345
Breakdown voltage: the voltage at which some specified and significant conduction occurs, for example catastrophic damage.
Clamping voltage: the voltage at which the device will conduct its fully rated current (hundreds to thousands of amperes.
The typical voltage breakdown for this part is 10V, but its in the reverse direction. I know, this is a bi-directional component, but each internal diode has its forward and reverse specifications.
Interesting that on page 10 here, it says:
The device is not designed to be used onlines connected to a DC supply
Back to the clamping and breakdown voltage, this brilliant application note (page 16) will answer your question.
I agree it is peculiarly specified. It seems to be missing a specification for minimum stand off voltage (or similar).
As for the clamping voltage being lower than Vrwm, note one is a max the other is typical.
If you were considering using this part and plan to hand solder, you should look at a 1:1 representation. This part is tiny.
Which is why I asked how can it be conductive before breakdown.
Yeah I agree. Was looking for some parts...but thiswas bugging me.
Thanks
Sorry, I really don't see your problem. Avoid "breakdown" conditions, by limiting the current though the device to below the absolute maximum value, and you will be fine.
Typical/max standoff voltage is the important spec to look at.
It must be the same as the rail voltage you want to protect.
Then look if clamping/breakdown voltage is not going to destroy your circuit.
I hope you're not thinking of protecting input pins with TVS diodes.
Leo..
Breakdown in diodes does not mean catastrophic breakdown. Just means silicon becomes conductive.
Tvs diodes are designed for transient / surge / spike protection.
Question still stands how can clamping voltage be at a lower voltage than breakdown.
Nope just for induced noise and transients etc.
Supposed to be. But if you have a clamping voltage lower than breakdown . You are going to have a problem. If breakdown is at 5.2v and clamp voltage is at 4.5v what happens at 5v?
I thought it was an error. But I do see quite a few TVS diodes with clamping voltage lower than min. Breakdown voltage
In that condition:
Isn't it almost impossible to reach the A-K voltage to 5 V as long as it is used properly within the rating?
It will be clamped at 4.5 V.
Even so, if a extreme terrible power surge occurs and the A-K voltage exceeds 5.2V, the diode will be destroyed.
follow #15
IMO the datasheet isn't all that mysterious. Look at figure 3 combined with table 6, which gives a good impression of how the device behaves. Vbr = 6-10V, Vrwm = 5V, suggesting that a logic level of 5V would only involve its specified leakage current of 1-50nA. It also shows that the device gives no protection against voltage 'peaks' that are below its Vbr, so it really should only be used for actual ESD protection where the spikes are short, but very high voltage (several kV).
As koraks has pointed out, this happens near rated peak pulse current (IPPM), where clamping voltage (VCL) is lower than breakdown voltage (VBR).
EDIT: Its because IPP < IPPM, therefore VCL < VBR
Guh, I understand. The I posted above is wrong.
The "breakdown voltage" represents the voltage at which a specified current flows by gradually applying a voltage.
The "clamp voltage" represents the voltage that remains between A-K when the specified pulse surge is applied.
It is that?
Perhaps also helpful in this context: https://www.ti.com/lit/an/slvae37/slvae37.pdf
Great help. Thanks
I use tvs diodes to prevent arcing when opening relay that is switching high inductive current
I'm not sure if they work particularly well for that due to the higher energy of an inductor coil than a typical ESD pulse.
Prevents welding of relay latch as well as inductive power surge to the IC if in close proximity