Why is not the $g$-field of the sun considered in the GPS-settings?

I have read this SE-Physics question and its answers, however, I could not find anything regarding the effect of Sun’s $g$-field on the GPS clocks. When a GPS clock orbits earth, depending on whether it is in superior or inferior conjunction, the $g$-potential of the sun changes. In the traditional approach to the GPS-settings, which possibly proves GR, it is claimed that two kinds of time rates are involved:

1- Due to SR by factor $sqrt{1-v_{Sat}^2/c^2}$, and

2- Due to GR by factor $sqrt{1-2GM_E/r_{Sat}c^2}=sqrt{1-Deltaphi/c^2}approx 1-Deltaphi/2c^2$.

The key term in the above equations is the difference in the $g$-potentials having the value:

$$Deltaphi=frac{2GM_E}{r_{Sat}}=frac{2×6.67×10^{-11}×6×10^{24}}{2×10^7}=4×10^7$$

However, if we consider the gravitational field of the sun, and if we consider the difference in the $g$-potentials, for when the satellite is between the sun and earth and when the earth is between the satellite and sun, we reach:

$$Deltaphi=GM_S left (frac{1}{R_{SE}-r_{Sat}}-frac{1}{R_{SE}+r_{Sat}} right)approx GM_Sfrac{2r_{Sat}}{R_{SE}^2}$$

$$=6.67×10^{-11}×2×10^{30}frac{2×2×10^7}{(1.5×10^{11})^2}$$

$$=3×10^5,$$

where $R_{SE}$ is the distance between the sun and earth. Although the recent value is nearly one hundredth the previous one, we can make them of the same order by assuming that the GPS clocks orbit the earth at slightly greater distances, say, $r_{Sat}=2×10^8space m$ (ten times greater than the orbital height of today’s GPS devices). Does this mean that if we want to send a GPS device at larger distances from the earth, we need to consider the gravitational potential of the sun as well? Is the obtained value for $g$-potential of the sun small enough that can be neglected as the GPS error?