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Im designing a simple usb application and I used this USBLC6-2 for ESD protection and voltage surge on the Vusb. On the datasheet it says that the breakdown voltage is min 6V, however I did some test for this if this is for real. I input a 10V on the Vusb and I am expecting that the zener diode should clamp the voltage to 6V however I can measure 10V.

Can anybody explain why the input voltage was not clamped down to 6V as it is supposed to be clamp according to the datasheet? With regards with my PCB layout i just followed what the datasheet says schematic diagram

Dave Tweed
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jasp
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2 Answers2

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Well, actually those Diode that you are looking at are not zener diode but rather TVS diode, TVS diodes do not have as sharp of a curve as a zener diode does are intended to take fast power surges, while a zener diode is intended to create a sharp curve at the zener voltage. If you see the data sheet, it states two voltages that are Vbr = Breakdown voltage and Vcl= clamping voltage. The breakdown voltage is the voltage at which the TVS will begin to do minimum conduction, specify in the datasheet as 1mA. And the Vcl is specified at 12V with a 1A and and 17V at 5(remember that its ESD protection and during and ESD event 1A and 5A for a few pico seconds is not out of the question). This Vcl is the actual voltage at which the TVS diode will clamp.

Kvegaoro
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  • Hi Kvegaoro, Thanks for your reply. If that is the case then I need to put a zener diode of at least 5V6 Breakdown voltage for this protection. I know its for an ESD protection but I thought with this zener diode on the Vbus will clamp to 6V. – jasp Jan 30 '13 at 01:25
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It's unclear what exactly you did, but it sounds like you put a 10 V source accross the 6 V zener. If this was a typical small zener rated for a few 100 mW, then it is now dead. The 10 V supply was able to provide enough current to fry the zener, and is probably failed open. It basically acted like a fuse. Poof.

To measure the zener voltage, put some resistance in series with the 10 V source. 1 kΩ would work here. That would cause 4 mA when the zener is working and clamping at 6 V. 4mA x 6V = 24mW, which will be well within the power dissipation capabilities of even a small zener.

Olin Lathrop
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