Regardless of whether the transistor (Q2) is turned on and the output is forced to be turned off, the transistor cannot be used as a switch. What is the specific reason and solution for the leakage power of the transistor in this circuit?
Gisele Posted on March 2, 2021
Reverse saturation current (diffusion current caused by the difference in the distribution of minority carriers near the junction surface of the PN junction), or caused by electron-hole pairs generated in the depletion area of the PN junction due to a strong electric field when a higher reverse bias is applied The leakage current of the short-channel device, and the source-drain series current of the short-channel device. In addition, the heavily doped drain region will also produce band-to-band tunneling current. For MOSs below 5V, the band-to-band tunneling current even becomes the main source of leakage current due to the high source and drain doping concentration.
Tammy Posted on March 2, 2021
The voltage at the G terminal of your circuit is greater than -10. Of course, no matter whether the transistor is on or off, there is a positive grid voltage. Am I wrong?
Lynn Posted on March 2, 2021
Does your VBS-10 mean -10V? ? ? If it is -10V, then the lift diode in the NMOS tube directly turns on -10V. . . .
If VBS-10V refers to 10V, replace the above-mentioned NMOS tube with a PMOS tube, and then put the input voltage on the S terminal and the output voltage on the D terminal. . .
