At the bottom right of our spreading resistance reports, we sometimes print some cryptic information. The following is an attempt to explain what it means. We will use the profile in Figure 1 as our example. In this case, we wrote:
1N Dose = 1.9 x 10^{15} cm^{2}  Sheet = 35 ohms/sq 
Fourpoint probe sheet = 32  
We shall attempt to explain each line. "1N" denotes the 1^{st} layer and it's type "N". We count the layers from left to right. Sometimes you will see several layers listed so this helps identify which layer the data refers to.
DOSE
"Dose" refers to the dose calculated from the n layer of the spreading resistance profile. We break up the n layer into 19 sublayers (the number of measurements made within the n layer.) We approximate the dose (net charges per area) of the sublayer by assuming that the sublayer has a thickness equal to the depth increment and is of uniform concentration. We then sum the calculated doses from each sublayer to get the total dose. In this example, concentrations within the nlayer were determined to be:
DEPTH (um)  CONC (cm^{3})   DEPTH (um)  CONC (cm^{3}) 
.020  1.22 x 10^{19}   .409  3.48 x 10^{19} 
.058  1.83 x 10^{19}   .449  3.22 x 10^{19} 
.098  2.86 x 10^{19}   .487  2.82 x 10^{19} 
.137  4.09 x 10^{19}   .526  2.47 x 10^{19} 
.175  2.85 x 10^{19}   .566  2.21 x 10^{19} 
.214  3.75 x 10^{19}   .605  1.35 x 10^{19} 
.254  4.49 x 10^{19}   .643  3.30 x 10^{18} 
.292  4.13 x 10^{19}   .682  2.95 x 10^{17} 
.332  4.38 x 10^{19}   .722  6.04 x 10^{16} 
.370  3.84 x 10^{19}   
The dose of a sublayer is then the concentration times the thickness (depth increment) which for this profile is 0.0391um or 3.91 x 10^{6} cm. (1um = 10^{4} cm)
So, the dose of the first sublayer is approximated as:
1.22 x 10^{19} cm^{3} x 3.91 x 10^{6} cm = 4.77 x 10^{13} cm^{2
} And the dose of the second sublayer is:
1.83 x 10^{19} cm^{3} x 3.91 x cm^{6} cm = 7.155 x 10^{13} cm^{2
} Okay, so we just do this calculation for all the sublayers and sum them for the total dose.
CONC(cm^{3})  Conc. (cm^{3}) x depth increment (cm) 
1.22 x 10^{19}  4.77020 x 10^{13} cm^{2} 
1.83 x 10^{19}  7.15530 x 10^{13} cm^{2} 
2.86 x 10^{19}  1.11826 x 10^{14} cm^{2} 
4.09 x 10^{19}  1.59919 x 10^{14} cm^{2} 
2.85 x 10^{19}  1.11435 x 10^{14} cm^{2} 
3.75 x 10^{19}  1.46625 x 10^{14} cm^{2} 
4.49 x 10^{19}  1.75559 x 10^{14} cm^{2} 
4.13 x 10^{19}  1.61483 x 10^{14} cm^{2} 
4.38 x 10^{19}  1.71258 x 10^{14} cm^{2} 
3.84 x 10^{19}  1.50144 x 10^{14} cm^{2} 
3.48 x 10^{19}  1.36068 x 10^{14} cm^{2} 
3.22 x 10^{19}  1.25902 x 10^{14} cm^{2} 
2.82 x 10^{19}  1.10262 x 10^{14} cm^{2} 
2.47 x 10^{19}  9.65770 x 10^{13} cm^{2} 
2.21 x 10^{19}  8.64110 x 10^{13} cm^{2} 
1.35 x 10^{19}  5.27850 x 10^{13} cm^{2} 
3.30 x 10^{18}  1.29030 x 10^{13} cm^{2} 
2.95 x 10^{17}  1.15345 x 10^{12} cm^{2} 
6.04 x 10^{16}  2.36164 x 10^{11} cm^{2} 
T0TAL DOSE  1.92980 x 10^{15} cm^{2} 
SHEET
The calculation of the sheet resistance follows the same scheme as the dose except for a few twists. If you have resistors in parallel (Figure 2) the reciprocal of equivalent resistance is equal to the sum of the reciprocal of the resistance of each parallel leg.
Figure 2
If the resistors were squares of material connected in parallel, the relationship would remain the same:
Figure 3
Squares have a convenient electrical property  the resistance encountered between opposite sides is the same for a given material regardless of the size of the square. Because of this we talk of sheet resistance, ,in units of "ohms per square". We can then substitute for R in the above equation and we have:
In the case of a "sheet" of uniform resistivity, the sheet resistance is equal to the bulk resistivity divided by the thickness:
Let's approximate the sheet resistance of each sub layer as the resistivity determined by spreading resistance divided by the depth increment. In our example, the depth increment is 3.91 x 10^{6} cm. In the table below, we have used the resistivity associated with each point and divided it by the depth increment to calculate a sheet resistance for each sublayer. We then calculated the reciprocal of each sheet resistance and added them together. Then we calculated the reciprocal of the total to get the sheet resistance for the nlayer. (Note that we could have introduced the idea of sheet conductance and avoided much of this talk of reciprocals, and we would have arrived at exactly the same result.)
Determined by SRA  Calculated from SRA data 
Depth  (ohmcm)  = /3.91 x 10^{6} cm  1/ 
.020  .00507  1296.675 ohms/sq  7.712 x 10^{4} 
.058  .00362  925.831  1.080 x 10^{3} 
.098  .00239  611.253  1.635 x 10^{3} 
.137  .00169  432.225  2.314 x 10^{3} 
.175  .00239  611.253  1.634 x 10^{3} 
.214  .00184  470.588  2.125 x 10^{3} 
.254  .00154  393.862  2.539 x 10^{3} 
.292  .00167  427.110  2.341 x 10^{3} 
.332  .00158  404.092  2.475 x 10^{3} 
.370  .00179  457.800  2.184 x 10^{3} 
.409  .00197  503.836  1.985 x 10^{3} 
.449  .00213  544.757  1.836 x 10^{3} 
.487  .00241  616.368  1.622 x 10^{3} 
.526  .00275  703.325  1.422 x 10^{3} 
.566  .00304  777.494  1.286 x 10^{3} 
.605  .00466  1191.816  8.391 x 10^{4} 
.643  .0117  2992.327  3.342 x 10^{4} 
.682  .043  10997.44  9.093 x 10^{5} 
.722  .12  30690.54  3.258 x 10^{5} 
Total (or )  2.855 x 10^{2} 
Sheet =  35.2 ohms/sq  
FOUR POINT PROBE SHEET
Finally, "Fourpoint probe sheet" (4PP) refers to the sheet resistance that we measured on our fourpoint probe. (You had already guessed that, hadn't you?) If the sample is big enough & unpatterned and if there is a junction and if the sheet rho of the top layer isn't too high, we will measure it. If there's room, we will probe at five locations and report the average.
Nearly always, the sheet resistance obtained by the fourpoint probe is more reliable than that obtained from SRA so this provides you a benchmark for the SRA measurements. If "Sheet" and "4PP" track each other pretty well, then the SRA is more credible.
Best of all, this quick and dirty "4PP" is free. If you would like something a little more elegant, we will be happy to provide you a 120 point fourpoint contour map for a little money.
If you have read this technical note in its entirety, we would like to thank you. You paid us a real compliment by bearing through it. And, as always, if you want more information, please let us know.
