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dummy8 [2026/06/02 03:43] – mexleadmindummy8 [2026/06/02 03:50] (current) – mexleadmin
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   * 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+ymqzIdzYOYJLU7EAzsbbk83gIBhFxaT1HbCf646EdEl7NhAA 700,500 noborder}} +
-</panel>+
  
-#@ResultBegin_HTML~ExerciseMultipleDiodesLamps~@# 
  
-**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 37: Line 38:
 After closing the switch, check the voltage across each lamp in the simulation. After closing the switch, check the voltage across each lamp in the simulation.
  
-The important idea is:+A lamp is assumed to light brightly if
  
-  * a forward-biased diode behaves approximately like a low-voltage path, +\[ 
-  * if a diode is connected in parallel to a lamp or a group of lamps, it can bypass this part of the circuit, +\begin{align*} 
-  * a bypassed lamp has too little voltage across it and therefore remains dark.+U_{\rm lamp}\geq 5~{\rm V}. 
 +\end{align*} 
 +\]
  
-The middle lamps are bypassed by forward-biased diode paths.   +The simulation shows that the outer lamps have a sufficiently large voltage across them, while the inner lamps are bypassed by conducting diodes. 
-Therefore, their lamp voltage is much smaller than the required value of approximately \(5~{\rm V}\).+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP>
  
-The outer lamps are not bypassed in the same way.   +<WRAP half column> 
-They have a sufficiently large voltage across them and light up brightly. +#@ResultBegin_HTML~12001~@# 
- +The lamps
-**Result**+
  
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +L_1 
-L_1 \text{ and } L_5 \text{ light up brightly.} +\quad \text{and} \quad 
-}+L_5
 \end{align*} \end{align*}
 \] \]
 +
 +light up brightly.
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +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
  
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +U_{\rm lamp}<5~{\rm V},
-L_2,\;L_3,\;L_4 \text{ remain dark or almost dark.} +
-}+
 \end{align*} \end{align*}
 \] \]
  
-The reason is not that the dark lamps are defective.   +the lamp does not light brightly. 
-They are bypassed by conducting diodes.+</WRAP> 
 +#@PathEnd_HTML@# 
 +</WRAP>
  
 +<WRAP half column>
 +#@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 111: 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 \(R_1\).
  
-**Solution path**+<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~12004~@# 
 +<WRAP leftalign> 
 +The current through \(R_1\) passes through one forward-biased diode.
  
-First decide which diodes conduct.   +Therefore the voltage across \(R_1\) is
-With the given polarity, both diodes are forward-biased. +
- +
-The current through \(R_1\) flows after one forward-biased diode.   +
-Therefore, the voltage across \(R_1\) is+
  
 \[ \[
Line 138: Line 171:
 \begin{align*} \begin{align*}
 U_{R1} U_{R1}
-=+&=
 4.0~{\rm V}-0.6~{\rm V} 4.0~{\rm V}-0.6~{\rm V}
-=+\\ 
 +&=
 3.4~{\rm V}. 3.4~{\rm V}.
 \end{align*} \end{align*}
 \] \]
  
-Thus+Now apply Ohm's law:
  
 \[ \[
Line 158: Line 192:
 \end{align*} \end{align*}
 \] \]
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
  
-The current through \(R_2\) flows after two forward-biased diodes.   +<WRAP half column> 
-Therefore, the voltage across \(R_2\) is+#@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
  
 \[ \[
Line 175: Line 230:
 \begin{align*} \begin{align*}
 U_{R2} U_{R2}
-=+&=
 4.0~{\rm V}-2\cdot 0.6~{\rm V} 4.0~{\rm V}-2\cdot 0.6~{\rm V}
-=+\\ 
 +&=
 2.8~{\rm V}. 2.8~{\rm V}.
 \end{align*} \end{align*}
 \] \]
  
-Thus+Now apply Ohm's law:
  
 \[ \[
Line 195: Line 251:
 \end{align*} \end{align*}
 \] \]
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
  
-The diode \(D_1\) supplies both branches.   +<WRAP half column> 
-Therefore, Kirchhoff's current law gives +#@ResultBegin_HTML~12005~@#
 \[ \[
 \begin{align*} \begin{align*}
-I_{D1} +I_{R2}=28~{\rm mA}
-= +
-I_{R1}+I_{R2}.+
 \end{align*} \end{align*}
 \] \]
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
  
-So+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,
  
 \[ \[
Line 213: Line 280:
 I_{D1} I_{D1}
 = =
-17~{\rm mA}+28~{\rm mA} +I_{R1}+I_{R2}.
-= +
-45~{\rm mA}.+
 \end{align*} \end{align*}
 \] \]
  
-**Result**+Insert the values:
  
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +I_{D1} 
-I_{R1}=17~{\rm mA} +&= 
-}+17~{\rm mA}+28~{\rm mA} 
 +\\ 
 +&= 
 +45~{\rm mA}.
 \end{align*} \end{align*}
 \] \]
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
  
 +<WRAP half column>
 +#@ResultBegin_HTML~12006~@#
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ 
-I_{R2}=28~{\rm mA} 
-} 
-\end{align*} 
-\] 
- 
-\[ 
-\begin{align*} 
-\boxed{ 
 I_{D1}=45~{\rm mA} I_{D1}=45~{\rm mA}
-} 
 \end{align*} \end{align*}
 \] \]
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
  
-#@ResultEnd_HTML@# 
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#
  
Line 286: 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 \(S\).
  
-**Solution path**+<WRAP group> 
 +<WRAP half column rightalign> 
 +#@PathBegin_HTML~12007~@# 
 +<WRAP leftalign> 
 +With the switch open, only \(D_1\) is connected to the resistor path.
  
-First consider the switch open. +The conducting diode clamps the node voltage to approximately
- +
-With the switch open, only \(D_1\) is connected to the resistor path.   +
-The conducting diode clamps the node voltage approximately to+
  
 \[ \[
 \begin{align*} \begin{align*}
-U_{\rm node} +U_{\rm node}\approx U_{\rm F}=0.7~{\rm V}.
-\approx +
-U_{\rm F} +
-= +
-0.7~{\rm V}.+
 \end{align*} \end{align*}
 \] \]
Line 314: Line 374:
 \begin{align*} \begin{align*}
 I_{R1} I_{R1}
-= +&= 
-\frac{U_0-U_{\rm F}}{R_1}. +\frac{U_0-U_{\rm F}}{R_1} 
-\end{align*} +\\ 
-\] +&=
- +
-Insert the values: +
- +
-\[ +
-\begin{align*} +
-I_{R1} +
-=+
 \frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega} \frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega}
-=+\\ 
 +&=
 4.3~{\rm mA}. 4.3~{\rm mA}.
 \end{align*} \end{align*}
Line 340: Line 394:
 \end{align*} \end{align*}
 \] \]
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
  
-Now consider the switch closed. +<WRAP half column> 
- +#@ResultBegin_HTML~12007~@# 
-With the switch closed, \(D_1\) and \(D_2\) are connected in parallel.   +For open switch:
-The node voltage is still approximately clamped to+
  
 \[ \[
 \begin{align*} \begin{align*}
-U_{\rm node} +I_{R1}&=4.3~{\rm mA}, 
-\approx +\\ 
-0.7~{\rm V}.+I_{D1}&=4.3~{\rm mA}, 
 +\\ 
 +I_{D2}&=0.
 \end{align*} \end{align*}
 \] \]
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
  
-Therefore, the total resistor current is still+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 \(R_1\):
  
 \[ \[
 \begin{align*} \begin{align*}
 I_{R1} I_{R1}
-=+&= 
 +\frac{U_0-U_{\rm F}}{R_1} 
 +\\ 
 +&=
 \frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega} \frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega}
-=+\\ 
 +&=
 4.3~{\rm mA}. 4.3~{\rm mA}.
 \end{align*} \end{align*}
 \] \]
  
-This current now splits between the two parallel diodes:+Kirchhoff's current law gives
  
 \[ \[
Line 374: Line 447:
 \] \]
  
-With the ideal constant-voltage model, the individual diode currents are not uniquely determined.   +With the ideal constant-voltage diode model, the individual currents through two parallel diodes are not uniquely determined.
-The model can calculate the total current, but not how two idealized parallel diodes share this current.+
  
-If the two real diodes are approximately identical, the current is approximately shared equally:+If both real diodes are approximately identical, the current splits approximately equally:
  
 \[ \[
Line 385: Line 457:
 I_{D2} I_{D2}
 \approx \approx
-\frac{I_{R1}}{2}.+\frac{4.3~{\rm mA}}{2} 
 += 
 +2.15~{\rm mA}.
 \end{align*} \end{align*}
 \] \]
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
  
-**Result** +<WRAP half column> 
- +#@ResultBegin_HTML~12008~@# 
-For open switch:+For closed switch:
  
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +I_{R1}=4.3~{\rm mA}
-S \text{ open: } +
-I_{R1}=4.3~{\rm mA}, +
-\quad +
-I_{D1}=4.3~{\rm mA}, +
-\quad +
-I_{D2}=0 +
-}+
 \end{align*} \end{align*}
 \] \]
  
-For closed switch, the total current is+and
  
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ 
-S \text{ closed: } 
-I_{R1}=4.3~{\rm mA}, 
-\quad 
 I_{D1}+I_{D2}=4.3~{\rm mA}. I_{D1}+I_{D2}=4.3~{\rm mA}.
-} 
 \end{align*} \end{align*}
 \] \]
Line 423: Line 488:
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +I_{D1} 
-I_{D1}\approx I_{D2}\approx 2.15~{\rm mA}. +\approx 
-}+I_{D2} 
 +\approx 
 +2.15~{\rm mA}.
 \end{align*} \end{align*}
 \] \]
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +3. Explain why the current sharing is not unique in the simple model.
 +
 +<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:
 +
 +\[
 +\begin{align*}
 +U_{\rm D}=U_{\rm F}.
 +\end{align*}
 +\]
 +
 +For two parallel diodes, this condition is true for many possible current distributions.
 +
 +Therefore, the model only determines the sum
 +
 +\[
 +\begin{align*}
 +I_{D1}+I_{D2},
 +\end{align*}
 +\]
 +
 +not the individual diode currents.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +
 +<WRAP half column>
 +#@ResultBegin_HTML~12009~@#
 +The constant-voltage diode model determines only
 +
 +\[
 +\begin{align*}
 +I_{D1}+I_{D2}=4.3~{\rm mA}.
 +\end{align*}
 +\]
 +
 +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>
- 
 #@ResultEnd_HTML@# #@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 #@TaskEnd_HTML@# #@TaskEnd_HTML@#