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I'm working on a proximity sensor using a piezoelectric ultrasonic transceiver. While looking at driver ICs, I came across the following example in the datasheet for TI's PGA460-Q1:

Figure 8-1. Transformer-Driven Method Schematic

On the righthand side you can see the transceiver (XDCR) connected to the output pins through a step-up transformer. When transmitting a burst, Vpp at the transceiver is 100 V.

The transceiver is also connected to the input pins through some capacitors so the IC can receive echoes after sending out a burst. Echo signals are mV to uV depending on how far away the target is.

My question is, how does this circuit not overload IN when it generates a burst on OUT? The Vpp of the waveform at the transceiver during a burst is 100 V, but the absolute maximum voltage on the input pins is only 2 V. Rinp is described as optional, so it doesn't appear to simply be a matter of an RC time constant slowing down the rise time (and when I simulate this in LTspice, Rinp doesn't help anyway).

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    \$\begingroup\$ Just guessing, but I imagine they switch the inputs 'off' long enough to avoid deep saturation of their sensitive circuitry. So there likely is a short period during which they are deaf. If so, the datasheet may call out that dead-period somewhere. If not, my guess doesn't apply here. \$\endgroup\$
    – periblepsis
    Commented Jan 7 at 0:48

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As @periblespis mentioned, it is indeed that the pin only starts measuring after 50µs. Page 19. Added screenshot of the text.enter image description here

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