I see some ultrasonic sensors have legs with different length, such as below. Does it means they have polarity?
Are there any differences taking different leg as the signal pin?
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1\$\begingroup\$ For small signals piezos are effectively AC devices with polarity mattering mostly for phasing if you have an array. But some offering higher voltage support are polarized and could be damaged by getting that wrong (or at least require time with a suitable voltage source to "re-poll" them). If it's a receiver or a cheap surplus part operated as a transmitter at only a few volts you might not worry about it, otherwise as with most things, read the data sheet. \$\endgroup\$– Chris StrattonCommented Mar 8, 2019 at 3:09
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\$\begingroup\$ So, may I infer that it is meaning for 'transmitter' only? For a receiver, it has no difference? \$\endgroup\$– divergerCommented Mar 8, 2019 at 3:19
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1\$\begingroup\$ In the abstract it shouldn't matter to a receiver, no. It would matter if you were trying to measure phase, for example in a phased array of multiple receivers or a very sophisticated reflection system. Or if you wanted to impose a DC bias on your receiver. \$\endgroup\$– Chris StrattonCommented Mar 8, 2019 at 3:21
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3 Answers
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As people have already indicated in the comments, from the perspective of the actual ultrasound signal (transmitted and/or received) it doesn't make any difference at all. But my intuition tells me -- and small features in the drawing seem to indicate -- that one of these terminals is probably tied to the outer shield.
Without an actual part number and/or datasheet we have no way to be certain. But what we know for sure is that this part has a shield, and the shield needs to be connected to something to be useful. If this is a 2-pin device, then yes it's a safe bet that one of these terminals is connected to shield and the other isn't, so yes polarity matters.
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\$\begingroup\$ Yes, the short leg is indeed connected to the shield, see the diagram carefully, the short leg has two small arc connected to the base, the base connected to the shield then. But what strange is, when we toggle the pin, that is, take different legs as the signal pin, the received signal is different in amplitude. It's more stronger when take the short leg as signal pin, why? \$\endgroup\$– divergerCommented Mar 8, 2019 at 9:38
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\$\begingroup\$ Yep I saw those little arcs I just wasn't completely certain what to make of them. Regarding your other question (why does rx amplitude change), well that is puzzling. From a pure piezo-acoustic perspective, it should make no difference at all. My best guess is that it would have something to do with the shield again. In my experience ultrasound receivers also make great RF receivers. Perhaps the transducer is also acting a little like an antenna, and some of your received signal is getting amplified or reduced by RF coupling. The shield connection would definitely influence that... \$\endgroup\$ Commented Mar 8, 2019 at 17:32
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\$\begingroup\$ I think just like you. When use the long leg as signal, the short (shield) connected to the board ground. But when toggle that, the shield is the signal terminal, and no shielding exist actually. \$\endgroup\$– divergerCommented Mar 9, 2019 at 2:36
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Other answer mention that it matter if transducer have shield, but it also matter if it doesnt. Difference is that when terminal one has higher (lower) voltage than terminal two then piezo element will flex outward (inward) or other way around. When you use one transducer it usually doesnt matter which way you use it, but if you want to use more of them in array it does matter that same voltage will result in same acoustic wave and not inverted one.
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Regarding why amplitude could be different for different polarity, it can also depend on the geometry of the transducer. Though as an AC driven device it is intended to have symmetric (+/-) response, it may not be exactly the case. So pushing in one direction (+) may meet greater mechanical resistance/impedance than going in the other direction (-).
