Re: Answers: Successful HP-11C speedup Message #12 Posted by Gerson W. Barbosa on 21 Sept 2005, 3:28 p.m., in response to message #11 by don wallace
What if just replacing the capacitor instead of adding another in series? That might be a better option for earlier boards. On newer boards, it may be easier just to solder another inductor in parallel with the existing one. In this case, the terminals should not be overheated as the existing weld may melt under the other side of the board. I have to open up the calculator to see if there's room for another inductor.
In the table below (DECIMAL POINT IS COMMA) are listed both options (I am not sure if all capacitors are easily available). Notice that the mutual inductance between inductors have not been taken into account in the calculations. The table quite matches one of Ken's while the other is slightly different, which makes me wonder he may have soldered the new inductor below the existing one in the older board, thus minimizing the influence of the mutual inductance, and side by side on the newer board. Just a guess though. Please correct me if I am wrong.
Gerson.
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L (uH) C (pF) f (kHz)* speedup
factor
original 180 180 884,19 1,00
LC circuit
180 180 1250,44 1,41
100 180 1479,54 1,67
82 180 1580,49 1,79
inductor 47 180 1943,17 2,20
in 33 180 2246,37 2,54
parallel 22 180 2679,24 3,03
15 180 3188,01 3,61
12 180 3536,78 4,00
10 180 3854,11 4,36
8,2 180 4235,95 4,79
180 100 1186,27 1,34
180 82 1310,02 1,48
capacitor 180 47 1730,35 1,96
replacement 180 33 2065,03 2,34
180 22 2529,14 2,86
180 15 3062,94 3,46
180 12 3424,47 3,87
180 10 3751,32 4,24
* not actual clock frequencies
Edited: 21 Sept 2005, 3:35 p.m.
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