The tunnel field-effect transistor (TFET) has emerged as a promising device for biosensing applications due to band-to-band tunneling (BTBT) operation mechanism and a steep subthreshold swing. In this paper, an electrically doped cavity on source junctionless tunnel field-effect transistor (ED-CS-JLTFET)-based biosensor is proposed for label-free detection of biomolecules. In the proposed model, the electrically doped concept is enabled to reduce fabrication complexity and cost. In order to create a nano-cavity at the source region, some portion of the dielectric oxide of the polarity gate terminal is etched away. To perceive the presence of biomolecules, two important properties of biomolecules, such as dielectric constant and charge density, are incorporated throughout the simulation. The sensing performance of the proposed ED-CS-JLTFET-based biosensor has been analyzed in terms of transfer characteristics, threshold voltage and subthreshold swing. In addition, the sensitivity of the proposed biosensor has also been analyzed with respect to different fill factors (FFs), varying nano-cavity dimension and work-function of the control gate. It is found from the simulated results that the proposed ED-CS-JLTFET-based biosensor offers higher current sensitivities with neutral, positively charged and negatively charged biomolecules of $3.01\times 10^{11}$ (at k$=12$), $9.51\times 10^7$ (at $k=6$ and $\rho =2\times 10^{12}$ C$\cdot$cm${}^{-2}$) and $4.04\times 10^{10}$ (at k$=6$ and $\rho =-2\times 10^{12}$ C$\cdot$cm${}^{-2}$), respectively.

由于带间隧穿 (BTBT) 操作机制和陡峭的亚阈值摆幅，隧道场效应晶体管 (TFET) 已成为一种有前途的生物传感应用器件。在本文中，提出了一种基于源极无结隧道场效应晶体管（ED-CS-JLTFET）的电掺杂腔生物传感器，用于生物分子的无标记检测。在所提出的模型中，电掺杂概念能够降低制造复杂性和成本。为了在源极区形成纳米腔，蚀刻掉极性栅极端子的电介质氧化物的一部分。为了感知生物分子的存在，在整个模拟过程中结合了生物分子的两个重要特性，例如介电常数和电荷密度。所提出的基于 ED-CS-JLTFET 的生物传感器的传感性能已在传输特性、阈值电压和亚阈值摆幅方面进行了分析。此外，还针对不同的填充因子 (FF)、不同的纳米腔尺寸和控制栅极的功函数分析了所提出的生物传感器的灵敏度。从模拟结果中发现，所提出的基于 ED-CS-JLTFET 的生物传感器对中性、带正电和带负电的生物分子具有更高的电流灵敏度$3.01\times 10^{11}$（在k$=12$),$9.51\times 10^7$（在$ķ=6$和$\rho =2\times 10^{12}$C$\cdot$厘米${}^{-2}$） 和$4.04\times 10^{10}$（在k$=6$和$\rho =-2\times 10^{12}$C$\cdot$厘米${}^{-2}$）， 分别。