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I've been told that these are disconnect diodes. However, I am unsure as to what purpose they serve. They are used on the line that sends a signal to a piezoelectric, the return signal from the piezoelectric returns on the same path but has a RX/TX switch that doesn't allow the initial signal in.

schematic

simulate this circuit – Schematic created using CircuitLab

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In this application, the forward voltage drop of the diodes means that they have a relatively high impedance for voltages less than about ±0.65V, which effectively disconnects the source impedance of the pulse driver from the transducer when receiving, preventing the driver from "loading down" the high impedance of the sensor and reducing the overall sensitivity.

When transmitting, the drop is negligible.

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  • \$\begingroup\$ @Dave_Tweed what impedance value does the initial high voltage signal to the piezoelectric see due to the diodes, once the initial signal makes it past the first forward biased diode and incurs the 0.7V drop isn't the signal high enough to travel backwards the now forward (previously reverse) facing diode? \$\endgroup\$ – user1084113 Sep 11 '14 at 19:02
  • \$\begingroup\$ Huh? No, in the pair of back-to-back diodes, only one of them conducts at a time. Whenever either one is forward-biased, the other one is reverse-biased (by 0.65 V). \$\endgroup\$ – Dave Tweed Sep 11 '14 at 20:04

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