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<p>rezistory su odmerane. Prvy stlpec je poradove cislo rezistora z
pasky, druhy hodnota<br>
</p>
<p>
</p>
<table style="border-collapse:
collapse;width:220pt" width="293" cellspacing="0" cellpadding="0"
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<colgroup><col
style="mso-width-source:userset;mso-width-alt:4224;width:99pt"
width="132"> <col
style="mso-width-source:userset;mso-width-alt:5162;width:121pt"
width="161"> </colgroup><tbody>
<tr style="height:16.0pt" height="21">
<td class="xl63" style="height:16.0pt;width:99pt" width="132"
height="21">10k nominal</td>
<td class="xl63" style="width:121pt" width="161">Value (k<font
class="font6">Ohm</font><font class="font5">)</font></td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl64" style="height:16.0pt" height="21">9</td>
<td class="xl65">10,000164</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl64" style="height:16.0pt" height="21">1</td>
<td class="xl65">10,000362</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl64" style="height:16.0pt" height="21">6</td>
<td class="xl65">10,000424</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl64" style="height:16.0pt" height="21">4</td>
<td class="xl65">10,000433</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl64" style="height:16.0pt" height="21">5</td>
<td class="xl65">10,000451</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl64" style="height:16.0pt" height="21">10</td>
<td class="xl65">10,000473</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl64" style="height:16.0pt" height="21">15</td>
<td class="xl65">10,000503</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl64" style="height:16.0pt" height="21">7</td>
<td class="xl65">10,000592</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl64" style="height:16.0pt" height="21">2</td>
<td class="xl65">10,000620</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl64" style="height:16.0pt" height="21">11</td>
<td class="xl65">10,000630</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl64" style="height:16.0pt" height="21">3</td>
<td class="xl65">10,000660</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl64" style="height:16.0pt" height="21">14</td>
<td class="xl65">10,000715</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl64" style="height:16.0pt" height="21">12</td>
<td class="xl65">10,000749</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl64" style="height:16.0pt" height="21">13</td>
<td class="xl65">10,000789</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl64" style="height:16.0pt" height="21">8</td>
<td class="xl65">10,000814</td>
</tr>
</tbody>
</table>
<p><br>
</p>
<p>Vezmeme 12 z nich a zapojime ich do serie. Pomocou skratovacej
listy sa daju zapojit paralelne. Podla teorie [1] by mala byt
tolerancia seriovej kombinacie a paralelnej kombinacie druha
mocnina tolerancie jednotlivych rezistorov, t.j. 0.01% x 0.01% =
0.01 ppm. Excelovy vypocet to nevylucuje:<br>
</p>
<p>
</p>
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collapse;width:240pt" width="320" cellspacing="0" cellpadding="0"
border="0">
<colgroup><col
style="mso-width-source:userset;mso-width-alt:3882;width:91pt"
width="121"> <col
style="mso-width-source:userset;mso-width-alt:3413;width:80pt"
width="107"> <col
style="mso-width-source:userset;mso-width-alt:2944;width:69pt"
width="92"> </colgroup><tbody>
<tr style="height:16.0pt" height="21">
<td class="xl67" style="height:16.0pt;width:91pt" width="121"
height="21">Hodnoty kOhm</td>
<td class="xl67" style="width:80pt" width="107"> Deliaci
pomer</td>
<td class="xl67" style="width:69pt" width="92"> Chyba pomeru
ppm</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl66" style="height:16.0pt" height="21">10,000424</td>
<td class="xl67">1</td>
<td class="xl67">0</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl66" style="height:16.0pt" height="21">10,000789</td>
<td class="xl67">0,91666822</td>
<td class="xl67">1,690806</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl66" style="height:16.0pt" height="21">10,000451</td>
<td class="xl67">0,83333339</td>
<td class="xl67">0,0699957</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl66" style="height:16.0pt" height="21">10,000715</td>
<td class="xl67">0,75000138</td>
<td class="xl67">1,8443319</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl66" style="height:16.0pt" height="21">10,000503</td>
<td class="xl67">0,66666717</td>
<td class="xl67">0,7624535</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl66" style="height:16.0pt" height="21">10,000814</td>
<td class="xl67">0,58333473</td>
<td class="xl67">2,3998536</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl66" style="height:16.0pt" height="21">10,000620</td>
<td class="xl67">0,4999997</td>
<td class="xl67">-0,599963</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl66" style="height:16.0pt" height="21">10,000592</td>
<td class="xl67">0,41666628</td>
<td class="xl67">-0,919944</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl66" style="height:16.0pt" height="21">10,000660</td>
<td class="xl67">0,3333331</td>
<td class="xl67">-0,699957</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl66" style="height:16.0pt" height="21">10,000473</td>
<td class="xl67">0,24999935</td>
<td class="xl67">-2,599841</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl66" style="height:16.0pt" height="21">10,000649</td>
<td class="xl67">0,16666716</td>
<td class="xl67">2,9498201</td>
</tr>
<tr style="height:16.0pt" height="21">
<td class="xl66" style="height:16.0pt" height="21">10,000630</td>
<td class="xl67">0,0833335</td>
<td class="xl67">1,999878</td>
</tr>
</tbody>
</table>
<p>
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<p>stredna hodnota 10,000610 kOhm</p>
<p>vsetky seriovo 120,007320 kOhm, vydelene strednou hodnotou
12,0000000</p>
<p>vsetky paralelne 0,833384167 kOhm, vydelene strednou hodnotou
12,0000000</p>
<p>dokonca aj pomer ser/par je 144,0000000<br>
</p>
<p>Tato cast zda sa funguje vyborne, stary Hamon to vymyslel
perfektne. </p>
<p>A teraz kombinatoricka otazka. Dany geret chceme pouzivat nie len
na transfer absolutnej hodnoty odporu a jej presne
nasobenie/delenie, ale aj ako velmi presny napatovy delic. Ako
vybrat hodnoty tak, aby bola celkova chyba delica co najmensia?
T.j. aby kazdy z 12 vystupov bol co najblizsie idealnym 1/12
prirastkom.<br>
</p>
<p>Hruba sila asi nebude uplne optimalna, 15! je velmi vela operacii
na to aby sa vyskusala kazda kombinacia :-)</p>
<p>Intuitivne som vyskusal prestriedat maly/velky/maly/velky a dava
to pouzitelne vysledky. Ale ako by islo najst optimalne riesenie?
Len pre zaujimavost. <br>
</p>
<p>b.</p>
<p><br>
</p>
<p>[1] Hamon, B. V. "A 1-100 Ω build-up resistor for the calibration
of standard resistors." Journal of Scientific Instruments 31.12
(1954): 450
<a class="moz-txt-link-freetext" href="https://iopscience.iop.org/article/10.1088/0950-7671/31/12/307/pdf">https://iopscience.iop.org/article/10.1088/0950-7671/31/12/307/pdf</a></p>
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<div class="moz-cite-prefix">On 14/11/2023 17:31, balu wrote:<br>
</div>
<blockquote type="cite"
cite="mid:C69CA420-641C-4E03-8640-0EBDC066DB5F@k-net.fr">
<meta http-equiv="content-type" content="text/html; charset=UTF-8">
<div dir="ltr">Asi som to vyriesil :) udaj sa meni na 7. az 8.
mieste, podla toho ako na to dycham. To mi staci. </div>
<div dir="ltr"><br>
</div>
<div dir="ltr"><br>
</div>
<div dir="ltr"><br>
<blockquote type="cite">On 14 Nov 2023, at 16:00, balu
<a class="moz-txt-link-rfc2396E" href="mailto:balu@k-net.fr"><balu@k-net.fr></a> wrote:<br>
<br>
</blockquote>
</div>
<blockquote type="cite">
<div dir="ltr">
<meta http-equiv="content-type"
content="text/html; charset=UTF-8">
<div dir="ltr">Su to metal foil 0.01% 2512. Presne na tuto
otazku budem mat odpoved hned po zaspajkovani :) </div>
<div dir="ltr">Ucelom nie je presna absolutna hodnota, ale
najst z kotuca take aby ked sa z toho urobi 12-odporovy
delic aby bol co najviac monotonny. A ak bude viditelny
rozdiel medzi 0.01% hodnotami najst take aby boli co
najrovnakejsie. <br>
</div>
</div>
</blockquote>
</blockquote>
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