2
\$\begingroup\$

I was reading about ESD diode arrays and protection devices and saw this question over here.

My question is:

In this diagram below, can someone help me to draw/identify the path of current flow through the ESD diode (RClamp part) when an ESD strike occurs?

enter image description here

Also, I saw the datasheet of the Rclamp part over here.

enter image description here

I think that the arrangement of the diodes of the Rclamp part is different from the one in the question and the one in the datasheet. Can someone clarify which one is better?

Also, I would be grateful if the current path for the datasheet part is also helped with. Thank you.

\$\endgroup\$
6
  • 1
    \$\begingroup\$ It would help to draw a diagram how the current flows if you gave the starting point by specifying where would this ESD strike. \$\endgroup\$
    – Justme
    Commented Apr 17, 2023 at 10:24
  • \$\begingroup\$ Where did the first diagram come from? \$\endgroup\$
    – Andy aka
    Commented Apr 17, 2023 at 10:39
  • \$\begingroup\$ @Justme, suppose the ESD strikes on pin 1 of the RJ45 connector, any idea how the current would flow through the ESD diode? \$\endgroup\$
    – Freshman
    Commented Apr 17, 2023 at 12:00
  • \$\begingroup\$ @Andyaka, the first diagram comes from the question linked in my question. \$\endgroup\$
    – Freshman
    Commented Apr 17, 2023 at 12:00
  • \$\begingroup\$ @Andyaka I found a possible source for it; edited the linked question. Given the coloring, likely it comes from Semtech themselves but I didn't see anything on the product page at least. \$\endgroup\$ Commented Apr 17, 2023 at 13:15

3 Answers 3

1
\$\begingroup\$

As per your example of what happens if ESD strikes on pin 1 of RJ45 connector.

As Ethernet is a transformer coupled interface, it would mean that ESD current going into pin 1 would have to come out from pin 2, so it just applies a pulse on the transformer coil. It would generally not jump over through the transformer directly.

The pulse would couple from transformer primary to secondary, and would try to apply differential voltage of the data pair, pushing one wire high above supply voltage and the other wire low below ground voltage.

The ESD protection device would then clamp the voltage that tries to go below GND to GND by conducting current, and also clamp the voltage that tries to rise above the TVS voltage to ground by conducting current.

\$\endgroup\$
1
\$\begingroup\$

Current flow path

In the long tradition of crudely drawn lines -- just follow the diodes. The tilted lines on the zener (huh, kinda like a 'Z', innit?) means current flows through it given enough voltage (which is given here).

Take your pick of which line(s) to apply to. In this case, either pin 1 went positive, and/or pin 10 went negative. Current may flow between pairs, or from one or both lines to/from ground, depending on intensity and bias voltage.

\$\endgroup\$
4
  • \$\begingroup\$ Thank you for the answer. So, for the positive ESD spike, the voltage on the line will be the = ( ESD Voltage - 0.7V (forward diode drop) - TVS Clamping voltage ) ? And for the negative ESD spike line voltage will be = ESD Spike + Forward diode drop? Am I correct? \$\endgroup\$
    – Freshman
    Commented Apr 18, 2023 at 3:54
  • \$\begingroup\$ For the negative voltage, it would imply that there's no clamping if the spike doesn't pass through the TVS diode, right? \$\endgroup\$
    – Freshman
    Commented Apr 18, 2023 at 3:55
  • \$\begingroup\$ I'm not sure what you're trying to express there. There is a path from negative pulse to GND if connected, and/or through zener to positive. Follow the diodes. \$\endgroup\$ Commented Apr 18, 2023 at 4:14
  • \$\begingroup\$ @Newbie Excuse me? \$\endgroup\$ Commented Apr 19, 2023 at 14:25
0
\$\begingroup\$

The diode arrangements are exactly the same.

The first diagram does not show the important components, i.e., the ground connection and the power supply's decoupling capacitors. But that is where the current flows.

Once the current spike is on the ground line, the decoupling capacitors eat the charge. (ESD strikes do not have much energy, so the capacitors' voltage does not change much.) If you have a multilayer board, the power/ground planes can act as a capacitor.

In theory, you could put a capacitor directly on the signal line, but this would filter out high-frequency signals.

\$\endgroup\$
2
  • \$\begingroup\$ But there are no capacitors, and you don't explain where this capacitor would be, or how any ESD current would end up to the capacitor. As a hint, it takes two terminals to charge a capacitor, and the ESD protector has only a ground terminal. \$\endgroup\$
    – Justme
    Commented Apr 17, 2023 at 10:38
  • \$\begingroup\$ @CL, can you clarify you answer a bit more? A little diagram showing the current path would help me greatly. \$\endgroup\$
    – Freshman
    Commented Apr 17, 2023 at 12:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.