2
\$\begingroup\$

TVS is a good practice to protect USB ports. A USB (at least 2.0) connection has the power path supplied with a 5 V source (Vbus=5V0) at the host side and the data path supplied with a 3.3 V sources (3V3) at both ends of the connection.

A TVS protecting the data lines (D+, D-), is connected like the following:

enter image description here

Should it be connected to 5V0 (Vbus) or 3V3? Why so? Arguments? References?

(This is especially interesting in the case of FTDI and similar ICs whose transceiver logic is supplied from the internal 3.3 V LDO.)

\$\endgroup\$
  • \$\begingroup\$ What's the clamping voltage of the TVS? \$\endgroup\$ – Matt Young Apr 19 '15 at 16:42
  • \$\begingroup\$ @MattYoung I think the TVS clamping voltage is a second question and depends on the connection scheme being selected. \$\endgroup\$ – asndre Apr 19 '15 at 16:49
2
\$\begingroup\$

This style of TVS is intended to be either connected to a power rail, or to be connected to ground only. In that second mode, a positive transient pulse will go through the top-side diode, then break down the TVS diode around 6V, clamping to ground through that device.

Here is a datasheet from Semtech that describes how to use one of their parts like this. See Figure 4 on page 7.

tl;dr, they connect the common cathode to \$V_{bus}\$, which is +5V.

This additionally serves as ESD protection for \$V_{bus}\$.

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

Theoretically speaking, hooking the TVS to 3.3V should give lower (better) level of signal clamping, by 1.7V. However, practically it might be harder to route a 3.3V rail in near vicinity of USB connector. More, when ESD (or EFT) strikes the wires, the peak signal will still be at 10-15V, due to finite dynamic impedance of diodes. So the 10%-15% gain may not be worth troubles of wasting PCB space.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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