# Pmos or Nmos? Which has the biggest contribution to flicker noise?

I would like to ask if this sentence below is correct.I had find it, when I was reading a paper.

"PMOS transistors have lower flicker noise than similarly-sized nMOS transistors in most CMOS processes."

I am a little confused why PMOS transistors have lower flicker noise.I didn't find a good explanation for this.

• As flicker noise is directly related to the underlying chemistry /topology of the device and different devices are different chemistries /topologies? Google "flicker noise" - there's a page... – Paul Uszak Apr 30 '17 at 12:40
• I think PFET channels are deeper in the substrate than are NFETS. Thus surface traps, releasing and refilling the tiny charge buckets, affect NFETS more. – analogsystemsrf Apr 30 '17 at 13:01
• thank you analogsystemsrf!!This sentence refer to MOSFET. I know that this question is little difficult for someone to answer, but I think is very important for someone to understand the reason. – elecV1 Apr 30 '17 at 13:10
• PFET and NFET in @analogsystemsrf's comment refer to p-channel and n-channel MOSFETs. – Hearth Apr 30 '17 at 13:28
• @analogsystemsrf I think PFET channels are deeper in the substrate than are NFETS. I doubt that, both NMOS and PMOS are very much surface devices. A more likely cause for the difference in noise between N- and P-type MOS becomes clear when you look at the formulas describing 1/f noise: the K-parameter is in there. K is directly related to electron/hole mobility and higher for NMOS (electron mobility) than for PMOS (hole mobility) resulting in higher K for the NMOS resulting in higer noise for the NMOS. – Bimpelrekkie Apr 30 '17 at 14:50

The answer lies in the mean time between collisions and this is based on mobility. Whichever device has a higher mobility, $\mu$, will have higher collisions because you have a greater probability of have a collision. This generally means that nFETs have higher collisions and therefore higher flicker noise, but on undoped channels, you will see similar mobilities. I have 10nm fins on my bench that show higher $\mu_p$.
Due to Brownian motion, you have movement whenever you have heat, and skipping a bunch of physics, you end up with the average net velocity for drift to be $v_{dn}= -\mu_nE$ and $v_{dp}= \mu_pE$ respectively. The mobility $\mu_x$ has the "mean free time between collisions" term of $\tau_c$, as $\mu_{n,p}=\frac{q\tau_c}{2m_{n,p}}$.
Once you calculate $\tau_c$, what this tells you is that for a field, $E$, for similar devices you will have more collisions just due to higher mobility. To actually calculate $\tau_c$, you will need to pull out a device physics book and look at the density of states at a temperature under field conditions. This is one of those things that we just empirically measure. The math says it's proportional to $T^{\frac{1}{2}}$, but on the bench you see $T^{\frac{3}{2}}$ as your channel changes.