Picture attached.
The network consists of:
+V through a:
50C = 32400
Then though two resistors in parallel:
94B = 9310
95B = 9530
Then Ground
So the voltage in the center is:
1/(1/9310 + 1/9530) = 4709
4709/(4709+32400) = 4709/37109 = 0.1269
19.6 x 0.1269 = 2.487V
So if I put a 100Kohm resistor in parallel with the 50C(32400ohm) I get a factor of:
1/(1/100000+1/32400)=24471
4709/(4709+24471) = 0.1614
2.487/0.1614=15.4V which is exactly what I get when I measure it! And the regulation seems to be working fine. This should work great. It'll be difficult to get the 100kOHM resistor in place but I'm off to try that now. I might try to order a 0 to 100k pot that I can put in series with a 50k resistor and in parallel with the 50C SMT resistor. This would allow me to adjust the voltage for other projects. It's a good brick so it should provide lots of power!
Thanks all!! I'll let you know if anything goes awry.
Richard
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OK, I'm updating this same post...final modified picture of Dell power Supply attached. Works great, output voltage is 15.36 on this cheapie meter.
Thanks again!