microphoneresistorsultrasound
I have seen a few circuits where a resistor is connected across the positive and negative (grounded) terminals of an ultrasonic receiver. It seems to "stabilize" the received signal. I have attached an example of one of these circuits, but I am using a simpler circuit with just a comparator.
What does this resistor do, and how does it work?
The high voltage signal (+/- 50 Volts, i.e. 100 Volts peak to peak) can be attenuated and DC biased using a filter like this:
Note that all the components used are standard values. The output will be a signal of 3.242 Volts peak to peak, with a 1.65 Volts DC bias added. Thus, the resultant output to the ADC pin will not have any negative values, which addresses the concern mentioned in the question.
Using high precision (low tolerance %, e.g. 1%) tantalum capacitors and resistors, and keeping all traces / leads short, will prevent much variance in values and little stray inductance / capacitance.
Why this works:
. Output impedance of this filter is around 1.14 KOhms, and input impedance is around 36.36 kOhms.
I guess they mean load resistor, not loaded resistor.
A load resistor is like a dummy. You can use it to test a power supply, antenna tuner, .... Instead of the device you'd connect normally to the device you want to test, you connect a dummy load or load resistor to it, so that if something goes wrong, you mess up the dummy and not the real device.
About your datasheet: the manufacturer did some tests with his device and a dummy load or load resistor. The diagrams give explanation on that.
Best Answer
It's a pull down resistor. It's job is to eliminate small noise, spurious energy that can capacitively develop. Kind of a filter. The value of the resistor matters, as well as the capactance specs for the receiver.