In therapeutic ultrasound, why is the ERA smaller than the ultrasound head area?

• A. The crystal does not vibrate uniformly which causes the ERA to be smaller than area of the ultrasound head
• B. ERA equals head area
• C. ERA is larger than head area
• D. ERA is always zero

Answer: (A)

The main idea is that only part of the crystal actually radiates ultrasound into the tissue. The crystal doesn’t vibrate uniformly across its entire surface; due to the backing, design, and the physics of piezoelectric vibration, only a portion of the crystal surface moves in a way that effectively emits energy in phase with the wave. The rest of the surface either vibrates out of phase or is damped, contributing little to the radiated field. That portion defines the effective radiating area, which is smaller than the physical head area. This matters because treatment intensity is based on this ERA; for a given device output, a smaller ERA yields a higher energy density in the tissue. The other statements don’t fit with how transducers actually behave.