https://fweb.wallawalla.edu/class-wiki/index.php?action=history&feed=atom&title=Electronics_QuestionsElectronics Questions - Revision history2026-04-04T16:38:25ZRevision history for this page on the wikiMediaWiki 1.43.0https://fweb.wallawalla.edu/class-wiki/index.php?title=Electronics_Questions&diff=8818&oldid=prevFonggr at 23:05, 28 January 20102010-01-28T23:05:24Z<p></p>
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<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*Capacitor_L acts like an open circuit for DC and a short circuit for AC</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*Capacitor_L acts like an open circuit for DC and a short circuit for AC</div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*AC coupling goes through a blocking capacitor, while DC coupling doesn't</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*AC coupling goes through a blocking capacitor, while DC coupling doesn't</div></td></tr>
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</table>Fonggrhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Electronics_Questions&diff=8817&oldid=prevFonggr at 22:59, 28 January 20102010-01-28T22:59:51Z<p></p>
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<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*1/22/2010,3: Explain the protection diode so that the transistor doesn't arc over.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*1/22/2010,3: Explain the protection diode so that the transistor doesn't arc over.</div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*Capacitor_L acts like an open circuit for DC and a short circuit for AC</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*Capacitor_L acts like an open circuit for DC and a short circuit for AC</div></td></tr>
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</table>Fonggrhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Electronics_Questions&diff=8814&oldid=prevFonggr at 17:51, 28 January 20102010-01-28T17:51:06Z<p></p>
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<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*If <math>V_{BE}</math> is 0, and <math>V_{CE}</math> is very large, will you get any current flow?</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*If <math>V_{BE}</math> is 0, and <math>V_{CE}</math> is very large, will you get any current flow?</div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*1/22/2010,3: Explain the protection diode so that the transistor doesn't arc over.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*1/22/2010,3: Explain the protection diode so that the transistor doesn't arc over.</div></td></tr>
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</table>Fonggrhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Electronics_Questions&diff=8778&oldid=prevFonggr at 20:02, 27 January 20102010-01-27T20:02:20Z<p></p>
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<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>:* The resistor is there because it is a Norton equivalent. Loading effects will play a part, but if you are designing this section of the amplifier, your design can take this into account to provide minimum losses.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>:* The resistor is there because it is a Norton equivalent. Loading effects will play a part, but if you are designing this section of the amplifier, your design can take this into account to provide minimum losses.</div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*If <math>V_{BE}</math> is 0, and <math>V_{CE}</math> is very large, will you get any current flow?</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*If <math>V_{BE}</math> is 0, and <math>V_{CE}</math> is very large, will you get any current flow?</div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">*1/22/2010,3: Explain the protection diode so that the transistor doesn't arc over.</ins></div></td></tr>
</table>Fonggrhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Electronics_Questions&diff=8777&oldid=prevFonggr at 19:28, 27 January 20102010-01-27T19:28:59Z<p></p>
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<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*31, F1.25: Why are the output resistances placed where they are? It seems there would be a voltage drop over the voltage amplifier and extra current not being accounted for in the current amplifier?</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*31, F1.25: Why are the output resistances placed where they are? It seems there would be a voltage drop over the voltage amplifier and extra current not being accounted for in the current amplifier?</div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>:* The resistor is there because it is a Norton equivalent. Loading effects will play a part, but if you are designing this section of the amplifier, your design can take this into account to provide minimum losses.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>:* The resistor is there because it is a Norton equivalent. Loading effects will play a part, but if you are designing this section of the amplifier, your design can take this into account to provide minimum losses.</div></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>*If <math><del style="font-weight: bold; text-decoration: none;">V_BE</del></math> is 0, and <math><del style="font-weight: bold; text-decoration: none;">V_CE</del></math> is very large, will you get any current flow?</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>*If <math><ins style="font-weight: bold; text-decoration: none;">V_{BE}</ins></math> is 0, and <math><ins style="font-weight: bold; text-decoration: none;">V_{CE}</ins></math> is very large, will you get any current flow?</div></td></tr>
</table>Fonggrhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Electronics_Questions&diff=8776&oldid=prevFonggr at 19:28, 27 January 20102010-01-27T19:28:44Z<p></p>
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<td colspan="2" style="background-color: #fff; color: #202122; text-align: center;">Revision as of 12:28, 27 January 2010</td>
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<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*31, F1.25: Why are the output resistances placed where they are? It seems there would be a voltage drop over the voltage amplifier and extra current not being accounted for in the current amplifier?</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>*31, F1.25: Why are the output resistances placed where they are? It seems there would be a voltage drop over the voltage amplifier and extra current not being accounted for in the current amplifier?</div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>:* The resistor is there because it is a Norton equivalent. Loading effects will play a part, but if you are designing this section of the amplifier, your design can take this into account to provide minimum losses.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>:* The resistor is there because it is a Norton equivalent. Loading effects will play a part, but if you are designing this section of the amplifier, your design can take this into account to provide minimum losses.</div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">*If <math>V_BE</math> is 0, and <math>V_CE</math> is very large, will you get any current flow?</ins></div></td></tr>
</table>Fonggrhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Electronics_Questions&diff=7209&oldid=prevFonggr at 19:15, 8 January 20102010-01-08T19:15:26Z<p></p>
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<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>:* Loading effects are the impedances that reduce the voltage or current to the intended source. For F1.18, these are <math> R_s</math> and <math>R_o</math>, for F1.28, it is <math>R_o</math></div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>:* Loading effects are the impedances that reduce the voltage or current to the intended source. For F1.18, these are <math> R_s</math> and <math>R_o</math>, for F1.28, it is <math>R_o</math></div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>:* Yes! It depends on what part of the project you get to design. If you get to work on the pre-amplifier section, you can have small resistances (for a Thevenin equivalent) or high resistances (for a Norton equivalent). If you are designing the amplifier section, you can stick in a buffer amplifier in so that there is no voltage drop (due to no current) flowing into the amplifier. This ensures the voltage drop is completely across the load/input resistor.</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>:* Yes! It depends on what part of the project you get to design. If you get to work on the pre-amplifier section, you can have small resistances (for a Thevenin equivalent) or high resistances (for a Norton equivalent). If you are designing the amplifier section, you can stick in a buffer amplifier in so that there is no voltage drop (due to no current) flowing into the amplifier. This ensures the voltage drop is completely across the load/input resistor.</div></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>*31,<del style="font-weight: bold; text-decoration: none;">1,5</del>: Why are the output resistances placed where they are? It seems there would be a voltage drop over the voltage amplifier and extra current not being accounted for in the current amplifier?</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>*31, <ins style="font-weight: bold; text-decoration: none;">F1.25</ins>: Why are the output resistances placed where they are? It seems there would be a voltage drop over the voltage amplifier and extra current not being accounted for in the current amplifier?</div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">:* The resistor is there because it is a Norton equivalent. Loading effects will play a part, but if you are designing this section of the amplifier, your design can take this into account to provide minimum losses.</ins></div></td></tr>
</table>Fonggrhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Electronics_Questions&diff=7208&oldid=prevFonggr at 19:11, 8 January 20102010-01-08T19:11:52Z<p></p>
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<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><br></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>'''Chapter 1'''</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>'''Chapter 1'''</div></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">#</del>20,1,3: Shouldn't the output voltage get larger as <math>R_L</math> gets smaller?</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">*</ins>20,1,3: Shouldn't the output voltage get larger as <math>R_L</math> gets smaller?</div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>:* No.<math>R_O</math> is in series with <math>R_L</math>, so if <math>R_L</math> gets smaller, there will be a smaller voltage drop over <math>R_L</math> and a larger voltage drop over <math>R_O</math></div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>:* No.<math>R_O</math> is in series with <math>R_L</math>, so if <math>R_L</math> gets smaller, there will be a smaller voltage drop over <math>R_L</math> and a larger voltage drop over <math>R_O</math></div></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">#</del>21, F1.18: Why does it have to have an source impedance? How big is this impedance? Loading effects? Could we get rid of this if we had superconducting wires?</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">*</ins>21, F1.18: Why does it have to have an source impedance? How big is this impedance? Loading effects? Could we get rid of this if we had superconducting wires?</div></td></tr>
<tr><td class="diff-marker" data-marker="−"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del style="font-weight: bold; text-decoration: none;">#</del>31,1,5: Why are the output resistances placed where they are? It seems there would be a voltage drop over the voltage amplifier and extra current not being accounted for in the current amplifier?</div></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">:* The source impedance is simply to reflect the real world internal losses of the system. If you want to use these models, you will need to make a Thevenin or Norton equivalent.</ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">:* When you make the Thevnin or Norton equivalent of a circuit, you will find out how big these impedances are.</ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">:* Loading effects are the impedances that reduce the voltage or current to the intended source. For F1.18, these are <math> R_s</math> and <math>R_o</math>, for F1.28, it is <math>R_o</math></ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">:* Yes! It depends on what part of the project you get to design. If you get to work on the pre-amplifier section, you can have small resistances (for a Thevenin equivalent) or high resistances (for a Norton equivalent). If you are designing the amplifier section, you can stick in a buffer amplifier in so that there is no voltage drop (due to no current) flowing into the amplifier. This ensures the voltage drop is completely across the load/input resistor.</ins></div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">*</ins>31,1,5: Why are the output resistances placed where they are? It seems there would be a voltage drop over the voltage amplifier and extra current not being accounted for in the current amplifier?</div></td></tr>
</table>Fonggrhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Electronics_Questions&diff=7207&oldid=prevFonggr at 18:48, 8 January 20102010-01-08T18:48:37Z<p></p>
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<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>'''Chapter 1'''</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>'''Chapter 1'''</div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>#20,1,3: Shouldn't the output voltage get larger as <math>R_L</math> gets smaller?</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>#20,1,3: Shouldn't the output voltage get larger as <math>R_L</math> gets smaller?</div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">:* No.<math>R_O</math> is in series with <math>R_L</math>, so if <math>R_L</math> gets smaller, there will be a smaller voltage drop over <math>R_L</math> and a larger voltage drop over <math>R_O</math></ins></div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>#21, F1.18: Why does it have to have an source impedance? How big is this impedance? Loading effects? Could we get rid of this if we had superconducting wires?</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>#21, F1.18: Why does it have to have an source impedance? How big is this impedance? Loading effects? Could we get rid of this if we had superconducting wires?</div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>#31,1,5: Why are the output resistances placed where they are? It seems there would be a voltage drop over the voltage amplifier and extra current not being accounted for in the current amplifier?</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>#31,1,5: Why are the output resistances placed where they are? It seems there would be a voltage drop over the voltage amplifier and extra current not being accounted for in the current amplifier?</div></td></tr>
</table>Fonggrhttps://fweb.wallawalla.edu/class-wiki/index.php?title=Electronics_Questions&diff=6845&oldid=prevFonggr at 00:40, 6 January 20102010-01-06T00:40:58Z<p></p>
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<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>#20,1,3: Shouldn't the output voltage get larger as <math>R_L</math> gets smaller?</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>#20,1,3: Shouldn't the output voltage get larger as <math>R_L</math> gets smaller?</div></td></tr>
<tr><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>#21, F1.18: Why does it have to have an source impedance? How big is this impedance? Loading effects? Could we get rid of this if we had superconducting wires?</div></td><td class="diff-marker"></td><td style="background-color: #f8f9fa; color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>#21, F1.18: Why does it have to have an source impedance? How big is this impedance? Loading effects? Could we get rid of this if we had superconducting wires?</div></td></tr>
<tr><td colspan="2" class="diff-side-deleted"></td><td class="diff-marker" data-marker="+"></td><td style="color: #202122; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins style="font-weight: bold; text-decoration: none;">#31,1,5: Why are the output resistances placed where they are? It seems there would be a voltage drop over the voltage amplifier and extra current not being accounted for in the current amplifier?</ins></div></td></tr>
</table>Fonggr