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dummy8 [2026/06/02 03:41] – mexleadmindummy8 [2026/06/02 03:50] (current) – mexleadmin
Line 19: Line 19:
   * Explain the result using the idea of diode bypass paths.   * Explain the result using the idea of diode bypass paths.
  
-<panel type="info" title="Simulation: multiple diodes and lamps"> +{{url>http://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxG3KQBZsQEBTAWjDACgBzEa4wkTFN14U0UKGwAmQvgPAYZGQYIkMAZgEMArgBsALmwBK4MILDFBeSOHNir1K7ltRoCSf3vuHhPJ-4gVGjr6AO4U3r7Y4WaCkGyhKB4JVknWMWxgeBC0pjY8fNFiuFYQSDBwEGWQ7KFg8vyKcvk2saG41KkUkO0mPi2NHbX5KL1ubeDD-T1+AVp6o1ET2eM+ymqzIdzYOYJLU7EAzsbbk83gIBra+wzpmWE+BbunRWelsFXO5TcQKQVjBQ5wF4fd6VdgZCB-GyRe5PQElYEVN5g26DDo-WHFegIhFxaT1HbCf646EdEl7NhAA noborder}}
-{{url>http://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxG3KQBZsQEBTAWjDACgBzEa4wkTFN14U0UKGwAmQvgPAYZGQYIkMAZgEMArgBsALmwBK4MILDFBeSOHNir1K7ltRoCSf3vuHhPJ-4gVGjr6AO4U3r7Y4WaCkGyhKB4JVknWMWxgeBC0pjY8fNFiuFYQSDBwEGWQ7KFg8vyKcvk2saG41KkUkO0mPi2NHbX5KL1ubeDD-T1+AVp6o1ET2eM+ymqIdzYOYJLU7EAzsbbk83gIBhFxaT1HbCf646EdEl7NhAA 700,500 noborder}} +
-</panel>+
  
-#@ResultBegin_HTML~ExerciseMultipleDiodesLamps~@# 
  
-<panel type="info" title="Solution path">+1. Determine which lamps light up brightly. 
 + 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~12001~@# 
 +<WRAP leftalign>
 Number the lamps from left to right: Number the lamps from left to right:
  
Line 34: Line 36:
 \] \]
  
-After the switch is closed, check which diode paths become forward-biased.+After closing the switch, check the voltage across each lamp in the simulation.
  
-A forward-biased diode behaves approximately like a low-voltage path.   +A lamp is assumed to light brightly if
-Therefore, if a diode is connected in parallel to a lamp or to a group of lamps, it can bypass this part of the circuit.+
  
-In this circuit: +\[ 
- +\begin{align*} 
-  * one diode bypasses the middle part of the lamp chain, +U_{\rm lamp}\geq 5~{\rm V}. 
-  * another diode bypasses one of the inner lamps, +\end{align*} 
-  * therefore the voltage across the bypassed lamps is too small to make them light brightly.+\]
  
-The outer lamps are not bypassed in the same way. They receive a sufficiently large voltage. +The simulation shows that the outer lamps have a sufficiently large voltage across them, while the inner lamps are bypassed by conducting diodes. 
-</panel>+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP>
  
-<panel type="success" title="Result">+<WRAP half column> 
 +#@ResultBegin_HTML~12001~@#
 The lamps The lamps
  
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{L_1 \text{ and } L_5}+L_1 
 +\quad \text{and} \quad 
 +L_5
 \end{align*} \end{align*}
 \] \]
  
 light up brightly. light up brightly.
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
  
-The lamps+2. Determine which lamps remain dark. 
 + 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~12002~@# 
 +<WRAP leftalign> 
 +The inner lamps are connected in parts of the circuit that are bypassed by forward-biased diodes. 
 + 
 +A forward-biased diode has only a small voltage drop. Therefore, a lamp in parallel with such a diode path receives only a small voltage. 
 + 
 +If
  
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{L_2,\;L_3,\;L_4}+U_{\rm lamp}<5~{\rm V},
 \end{align*} \end{align*}
 \] \]
  
-remain dark or almost dark.+the lamp does not light brightly. 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP>
  
-The reason is not that the lamps are broken, but that the conducting diodes create bypass paths with a much smaller voltage drop. +<WRAP half column> 
-</panel>+#@ResultBegin_HTML~12002~@# 
 +The lamps
  
 +\[
 +\begin{align*}
 +L_2,\;L_3,\;L_4
 +\end{align*}
 +\]
 +
 +remain dark or almost dark.
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
  
Line 113: Line 146:
   * \(R_2\).   * \(R_2\).
  
-<panel type="info" title="Simulation: two diodes and two resistors"> +{{url>https://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxABZykLsQEBTAWjDACgA3EFFC7iyN17gUeKOIGVxgmAjYAnENjQixywbxnc4CpXj5hRevpvFgdAd2P9B6m1DZW7pnicmQ2AD24IU4c9wE-nRuIAAi7N7Y2EistoSxYCH84SheSmTgGH4UmFk0KQBKaVG4WXTYhIJgGIRSwoWRQmI1dCgILbX1fACqHgAmQgZGdoZifv0MAGYAhgCuADYALmyDoyP6qtwgk7OLK0A noborder}}
-{{url>https://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxABZykLsQEBTAWjDACgA3EFFC7iyN17gUeKOIGVxgmAjYAnENjQixywbxnc4CpXj5hRevpvFgdAd2P9B6m1DZW7pnicmQ2AD24IU4c9wE-nRuIAAi7N7Y2EistoSxYCH84SheSmTgGH4UmFk0KQBKaVG4WXTYhIJgGIRSwoWRQmI1dCgILbX1fACqHgAmQgZGdoZifv0MAGYAhgCuADYALmyDoyP6qtwgk7OLK0A 700,500 noborder}} +
-</panel>+
  
-#@ResultBegin_HTML~ExerciseMultipleDiodesII~@#+1. Calculate the current through \(R_1\).
  
-<panel type="info" title="Solution path"> +<WRAP group> 
-Both diodes are forward-biased.+<WRAP half column rightalign> 
 +#@PathBegin_HTML~12004~@# 
 +<WRAP leftalign> 
 +The current through \(R_1\) passes through one forward-biased diode.
  
-First consider the branch with \(R_1\).   +Therefore the voltage across \(R_1\) is
-There is one forward-biased diode in series with \(R_1\), so the resistor voltage is+
  
 \[ \[
Line 129: Line 162:
 U_{R1} U_{R1}
 = =
-U_0-U_{\rm F} +U_0-U_{\rm F}. 
-=+\end{align*} 
 +\] 
 + 
 +Insert the values: 
 + 
 +\[ 
 +\begin{align*} 
 +U_{R1} 
 +&=
 4.0~{\rm V}-0.6~{\rm V} 4.0~{\rm V}-0.6~{\rm V}
-=+\\ 
 +&=
 3.4~{\rm V}. 3.4~{\rm V}.
 \end{align*} \end{align*}
 \] \]
  
-Thus+Now apply Ohm's law:
  
 \[ \[
Line 150: Line 192:
 \end{align*} \end{align*}
 \] \]
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
  
-Now consider the branch with \(R_2\).   +<WRAP half column> 
-Here the current path includes two forward-biased diode drops, so+#@ResultBegin_HTML~12004~@# 
 +\[ 
 +\begin{align*} 
 +I_{R1}=17~{\rm mA} 
 +\end{align*} 
 +\] 
 +#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +2. Calculate the current through \(R_2\). 
 + 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~12005~@# 
 +<WRAP leftalign> 
 +The current through \(R_2\) passes through two forward-biased diodes. 
 + 
 +Therefore the voltage across \(R_2\) is
  
 \[ \[
Line 158: Line 221:
 U_{R2} U_{R2}
 = =
-U_0-2U_{\rm F} +U_0-2U_{\rm F}. 
-=+\end{align*} 
 +\] 
 + 
 +Insert the values: 
 + 
 +\[ 
 +\begin{align*} 
 +U_{R2} 
 +&=
 4.0~{\rm V}-2\cdot 0.6~{\rm V} 4.0~{\rm V}-2\cdot 0.6~{\rm V}
-=+\\ 
 +&=
 2.8~{\rm V}. 2.8~{\rm V}.
 \end{align*} \end{align*}
 \] \]
  
-Thus+Now apply Ohm's law:
  
 \[ \[
Line 179: Line 251:
 \end{align*} \end{align*}
 \] \]
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
  
-The diode \(D_1\) carries the current feeding both branches: +<WRAP half column> 
 +#@ResultBegin_HTML~12005~@#
 \[ \[
 \begin{align*} \begin{align*}
-I_{D1} +I_{R2}=28~{\rm mA}
-= +
-I_{R1}+I_{R2}.+
 \end{align*} \end{align*}
 \] \]
-</panel>+#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP> 
 + 
 +3. Calculate the current through \(D_1\). 
 + 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~12006~@# 
 +<WRAP leftalign> 
 +The diode \(D_1\) supplies both current paths. 
 + 
 +Therefore, by Kirchhoff's current law,
  
-<panel type="success" title="Result"> 
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +I_{D1} 
-I_{R1}=17~{\rm mA} += 
-}+I_{R1}+I_{R2}.
 \end{align*} \end{align*}
 \] \]
 +
 +Insert the values:
  
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +I_{D1} 
-I_{R2}=28~{\rm mA} +&= 
-}+17~{\rm mA}+28~{\rm mA} 
 +\\ 
 +&= 
 +45~{\rm mA}.
 \end{align*} \end{align*}
 \] \]
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
  
 +<WRAP half column>
 +#@ResultBegin_HTML~12006~@#
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +I_{D1}=45~{\rm mA}
-I_{D1}=I_{R1}+I_{R2}=45~{\rm mA} +
-}+
 \end{align*} \end{align*}
 \] \]
-</panel> 
- 
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
  
Line 258: Line 351:
 depending on the switch state \(S\). depending on the switch state \(S\).
  
-<panel type="info" title="Simulation: switch-dependent diode circuit"> +{{url>https://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWKswDZ0A4BMYDMAWfAdiIE59UFcRVIQl9qEBTAWjDACgA3ELLfH3x1+gsFgxQpw+lLowEnAE5C64ybkhiJUsPE4B3EIyyqQuVJIHzD5y2dFnInACZ3J69w-AA5fGHwMTgAPcwxqMCJBfwiSYyEQABEuUKwENUhSPgw1cXiBEAAlFL4dSOo0jyJUfMEAVWc3E3AdZus+X39AkONicCjjOMiiWqSsTgBnL09mz3kQADMAQwAbCeZOXCI6TW0NCxaPOVtHT3a521wDzwsPHWdQ3HJwTOM9cDzBAoBlTiA noborder}}
-{{url>https://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWKswDZ0A4BMYDMAWfAdiIE59UFcRVIQl9qEBTAWjDACgA3ELLfH3x1+gsFgxQpw+lLowEnAE5C64ybkhiJUsPE4B3EIyyqQuVJIHzD5y2dFnInACZ3J69w-AA5fGHwMTgAPcwxqMCJBfwiSYyEQABEuUKwENUhSPgw1cXiBEAAlFL4dSOo0jyJUfMEAVWc3E3AdZus+X39AkONicCjjOMiiWqSsTgBnL09mz3kQADMAQwAbCeZOXCI6TW0NCxaPOVtHT3a521wDzwsPHWdQ3HJwTOM9cDzBAoBlTiA 700,500 noborder}} +
-</panel>+
  
-#@ResultBegin_HTML~ExerciseMultipleDiodesIII~@#+1. Calculate the currents for open switch \(S\).
  
-<panel type="info" title="Solution path: switch open">+<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~12007~@# 
 +<WRAP leftalign>
 With the switch open, only \(D_1\) is connected to the resistor path. With the switch open, only \(D_1\) is connected to the resistor path.
  
-The node voltage is clamped to approximately one forward voltage:+The conducting diode clamps the node voltage to approximately
  
 \[ \[
 \begin{align*} \begin{align*}
-U_{\rm node} +U_{\rm node}\approx U_{\rm F}=0.7~{\rm V}.
-\approx +
-U_{\rm F} +
-= +
-0.7~{\rm V}.+
 \end{align*} \end{align*}
 \] \]
Line 304: Line 394:
 \end{align*} \end{align*}
 \] \]
-</panel>+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 + 
 +<WRAP half column> 
 +#@ResultBegin_HTML~12007~@# 
 +For open switch:
  
-<panel type="success" title="Result: switch open"> 
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +I_{R1}&=4.3~{\rm mA}, 
-I_{R1}=4.3~{\rm mA} +\\ 
-}+I_{D1}&=4.3~{\rm mA}, 
 +\\ 
 +I_{D2}&=0.
 \end{align*} \end{align*}
 \] \]
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +2. Calculate the currents for closed switch \(S\).
 +
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~12008~@#
 +<WRAP leftalign>
 +With the switch closed, \(D_1\) and \(D_2\) are connected in parallel.
 +
 +The resistor current is still determined by the source voltage, the forward diode voltage, and \(R_1\):
  
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +I_{R1} 
-I_{D1}=4.3~{\rm mA} +&= 
-}+\frac{U_0-U_{\rm F}}{R_1} 
 +\\ 
 +&= 
 +\frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega} 
 +\\ 
 +&= 
 +4.3~{\rm mA}.
 \end{align*} \end{align*}
 \] \]
 +
 +Kirchhoff's current law gives
  
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +I_{D1}+I_{D2}=I_{R1}.
-I_{D2}=0 +
-}+
 \end{align*} \end{align*}
 \] \]
-</panel> 
  
-<panel type="info" title="Solution path: switch closed"> +With the ideal constant-voltage diode model, the individual currents through two parallel diodes are not uniquely determined.
-With the switch closed, \(D_1\) and \(D_2\) are connected in parallel.+
  
-The node voltage is still clamped to approximately+If both real diodes are approximately identical, the current splits approximately equally:
  
 \[ \[
 \begin{align*} \begin{align*}
-U_{\rm node}+I_{D1}
 \approx \approx
-U_{\rm F}+I_{D2} 
 +\approx 
 +\frac{4.3~{\rm mA}}{2}
 = =
-0.7~{\rm V}.+2.15~{\rm mA}.
 \end{align*} \end{align*}
 \] \]
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
  
-Therefore the resistor current remains+<WRAP half column> 
 +#@ResultBegin_HTML~12008~@# 
 +For closed switch:
  
 \[ \[
 \begin{align*} \begin{align*}
-I_{R1} +I_{R1}=4.3~{\rm mA}
-= +
-\frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega} +
-= +
-4.3~{\rm mA}.+
 \end{align*} \end{align*}
 \] \]
  
-This current now splits between \(D_1\) and \(D_2\):+and
  
 \[ \[
 \begin{align*} \begin{align*}
-I_{D1}+I_{D2}=I_{R1}.+I_{D1}+I_{D2}=4.3~{\rm mA}.
 \end{align*} \end{align*}
 \] \]
  
-With the simple constant-voltage diode model, the individual diode currents are not uniquely determined.   +For approximately identical real diodes:
-The model only fixes the total current. +
- +
-If both real diodes are assumed to be identical, the current approximately splits equally:+
  
 \[ \[
Line 378: Line 492:
 I_{D2} I_{D2}
 \approx \approx
-\frac{I_{R1}}{2}.+2.15~{\rm mA}.
 \end{align*} \end{align*}
 \] \]
-</panel>+#@ResultEnd_HTML@# 
 +</WRAP> 
 +</WRAP>
  
-<panel type="success" title="Result: switch closed"> +3. Explain why the current sharing is not unique in the simple model. 
-The total current is+ 
 +<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~12009~@# 
 +<WRAP leftalign> 
 +The constant-voltage diode model assumes that each conducting diode has exactly the same voltage drop:
  
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +U_{\rm D}=U_{\rm F}.
-I_{R1}=4.3~{\rm mA} +
-}+
 \end{align*} \end{align*}
 \] \]
  
-and+For two parallel diodes, this condition is true for many possible current distributions. 
 + 
 +Therefore, the model only determines the sum
  
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +I_{D1}+I_{D2},
-I_{D1}+I_{D2}=4.3~{\rm mA}. +
-}+
 \end{align*} \end{align*}
 \] \]
  
-With identical real diodes, approximately+not the individual diode currents. 
 +</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP> 
 + 
 +<WRAP half column> 
 +#@ResultBegin_HTML~12009~@# 
 +The constant-voltage diode model determines only
  
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +I_{D1}+I_{D2}=4.3~{\rm mA}.
-I_{D1}\approx I_{D2}\approx 2.15~{\rm mA}. +
-}+
 \end{align*} \end{align*}
 \] \]
 +
 +It does not uniquely determine \(I_{D1}\) and \(I_{D2}\) separately.
  
 <callout type="warning" icon="true"> <callout type="warning" icon="true">
-This exercise shows a limitation of the constant-voltage diode model.   +Parallel diodes are sensitive to small differences in real diode characteristics.   
-For parallel diodes, the total current can be calculated, but the current sharing between idealized identical diodes is not uniquely determined.+Current sharing should not be assumed to be perfect without checking the design.
 </callout> </callout>
-</panel> 
- 
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#