TVS for 5V power supply ? (2024)

Dear Members,

What breakdown voltage do I need for protecting 5V power supply with TVS diode ?
Is 6.8V breakdown too low ? or good enough ?

Thanks

Hi,

It's not clear to me what you want to achieve.

A 5V supply output should be 5V...so where do you expect a higher voltage come from?
Is the high voltage DC .... in the correct direction only .... or is it AC?

Usually the 5V is connected to a load. Maybe the more sensitive device. If you want to protect the load against overvoltage, then you need to know it's specifications.

And what current do you talk about? What time? And is there a current limiting device or a fuse that blows?

Generally the TVS comes with a datasheet. There you can find it's V-I characteristic.

Klaus

It's DC power supply, just want to protect the circuit supplied with it,
Yes I have put fuse but I want to put extra TVS as protector and experiment.

The breakdown of the TVS is 6.8 V or I need higher abit ?
I'm looking at P6KE6.8 thru P6KE540A

Thanks

What breakdown voltage do I need for protecting 5V power supply...

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Are you really trying to protect the 5V power supply? Or, more likely, stuff that are fed with the 5V supply?

If you are trying to protect the 5V power supply from the input side, we need to know more. But (more likely) you are trying to protect the output side (due to some fault, the output may go above 5V).

It is better to try to protect the downstream systems (components) - things that are connected to 5V. (else you may save the power supply but rest will be cooked)

If your 5V power supply is reasonably well designed, do not worry. Most electronic parts rated for 5V will work fine at 5+/- 10%; 4.5 to 5.5V

Hi,

We know nothing about the limits of voltage and currents in your application, thus it's impossible to help.

The only thing we can say for now is, that the TVS need to be rated for a continous operating voltage of 5V (you need to include upper supply voltage tolerance here).

But we can not say anything about breakdown voltage. Additionalky the breakdown voltage is defined with breakdown current. Some are defined in the uA range, others may be defined in the 10A range...

Klaus

more likely, stuff that are fed with the 5V supply?

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Yes I want to protect the output side...

So with breakdown 6.8V, good enough, the breakdown is as close as to 5V ? or opposite way ?

Hi,

Please specify the tolerance of 5V voltage the device can withstand...

You have to select the spec "clamp voltage" which is the maximum voltage allowable on that rail. Breakdown voltage is below than that..

Please go through the link..

https://www.semtech.com/uploads/documents/tvs_diode_selection.pdf

**broken link removed**

- - - Updated - - -

Your device have absolute ratings, Vc (Clamp voltage) should not exceed this...

bianchi77 said:

It's DC power supply, just want to protect the circuit supplied with it,

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bianchi77 said:

5.25V maximum for the device

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First of all, the DC power supply has DC ripple which you said nothing about it. We assume it is less than 0.25V ?
Second, whenever a load step happens, there is an overvoltage/undervoltage which the amplitude depends on the load step change. You must check that the maximum load step can happen is lower than 0.25V. Otherwise, you must re-design the supply.

That being said, a well designed SMPS does not need TVS.

I can see from datasheet Vclamp is 10V,
Good enough I reckon for protecting my board, I combine it with fast blow fuse.
It's an experiment, I want to know the response.

What TVS do you suggest ? Vishay or ON？ I'm looking for through hole model.
谢谢 你们

I reckon for protecting my board, I combine it with fast blow fuse...

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Fuses are excellent devices that can prevent a fire and save lives. Are you sure your electronic components can be saved by the fuse?

In my humble experience, electronic items often protect the fuse.

My suggestion is to get a decent 5V power supply and forget about the TVS diode. Use small ceramic capacitors generously (to filter spikes and provide surge currents) close to all sensitive components...

I can see from datasheet Vclamp is 10V. Good enough I reckon for protecting my board.

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That's fine, but rather unlikely. Many 5V circuits have absolute maximum voltage of 7V, being often destroyed when the level is exceeded.

There's however a difference between short transients and enduring overvoltage, e.g. due to defective power supply. As said, it's absolutely unrealistic to expect protection against the latter from TVS diodes. Notice that the TVS behavior for continuous overload isn't defined at all. Except for the lucky case, that the overvoltage is high enough to blow the upstream fuse, it's well possible that the TVS overheats and fails open circuit.

bianchi77 said:

I can see from datasheet Vclamp is 10V,
Good enough I reckon for protecting my board, I combine it with fast blow fuse.
It's an experiment, I want to know the response.

What TVS do you suggest ? Vishay or ON？ I'm looking for through hole model.
谢谢 你们

Click to expand...

I'm wondering how you come to the conclusion that a Vclamp of 10V is good enough for devices rated at 5.25V....

TVS diodes aren't appropriate for clamping sensitive nodes (like I/Os and antennas) from huge and fast transient events which with high effective source impedance (ESD). For power supplies, your bypass capacitors will generally swallow ESD and other very fast surges. And usually the clamping voltage is still above the tolerance of the device it's protection, in which case you need a secondary clamping circuit (like a small series resistance and schottky diodes to the supply rail).

But for low-impedance transient sources with relatively long durations (like overvoltage or overshoot on a power supply), a simple diode can't really help you. You need a proper crowbar circuit or a disconnect switch.

Hi,

How about a precision clamp, if it will work? Another option is to use a comparator, an NMOS switch (as inverter) and a PMOS pass device to clamp on/off at a reasonably precise voltage that way, despite downside of additional components.