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electrical_engineering_and_electronics_2:block12 [2026/06/02 03:48] – mexleadminelectrical_engineering_and_electronics_2:block12 [2026/06/10 03:06] (current) – mexleadmin
Line 86: Line 86:
 \[ \[
 \begin{align*} \begin{align*}
-U_{\rm E}+U_{\rm I}
 = =
 U_R+U_{\rm D}. U_R+U_{\rm D}.
Line 106: Line 106:
 I_{\rm D} I_{\rm D}
 \approx \approx
-\frac{U_{\rm E}-U_{\rm TO}}{R}.+\frac{U_{\rm I}-U_{\rm TO}}{R}.
 \end{align*} \end{align*}
 \] \]
Line 120: Line 120:
 The required forward voltage depends on the semiconductor material and therefore on the color. The required forward voltage depends on the semiconductor material and therefore on the color.
  
-For a supply voltage \(U_{\rm E}\), an LED forward voltage \(U_{\rm F}\), and a desired LED current \(I_{\rm F}\), the series resistor is+For a supply voltage \(U_{\rm I}\), an LED forward voltage \(U_{\rm F}\), and a desired LED current \(I_{\rm F}\), the series resistor is
  
 \[ \[
 \begin{align*} \begin{align*}
 \boxed{ \boxed{
-R_{\rm V} = \frac{U_{\rm E}-U_{\rm F}}{I_{\rm F}}+R_{\rm V} = \frac{U_{\rm I}-U_{\rm F}}{I_{\rm F}}
 } }
 \end{align*} \end{align*}
Line 139: Line 139:
 \[ \[
 \begin{align*} \begin{align*}
-P_R = (U_{\rm E}-U_{\rm F})I_{\rm F} = R_{\rm V}I_{\rm F}^2 \quad \overset{!}{<} \quad P_{R, \rm max}+P_R = (U_{\rm I}-U_{\rm F})I_{\rm F} = R_{\rm V}I_{\rm F}^2 \quad \overset{!}{<} \quad P_{R, \rm max}
 \end{align*} \end{align*}
 \] \]
Line 240: Line 240:
 \[ \[
 \begin{align*} \begin{align*}
-{\color{blue }{I_{\rm V}}} &= \frac{U_{\rm E}-U_{\rm Z}}{R_{\rm V}} &&\text{current supplied through the series resistor},+{\color{blue }{I_{\rm V}}} &= \frac{U_{\rm I}-U_{\rm Z}}{R_{\rm V}} &&\text{current supplied through the series resistor},
 \\ \\
 {\color{green}{I_{\rm L}}} &= \frac{U_{\rm Z}}{R_{\rm L}}           &&\text{useful load current}, {\color{green}{I_{\rm L}}} &= \frac{U_{\rm Z}}{R_{\rm L}}           &&\text{useful load current},
Line 538: Line 538:
  
 ===== Exercises ===== ===== Exercises =====
 +
 +#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Circuit with multiple diodes: which lamps light up?
 +#@TaskText_HTML@#
 +
 +The following simulation includes multiple diodes and several lamps.  
 +A lamp lights brightly when a voltage of approximately
 +
 +\[
 +\begin{align*}
 +U_{\rm lamp}\geq 5~{\rm V}
 +\end{align*}
 +\]
 +
 +drops across it.
 +
 +Close the switch in the simulation.
 +
 +  * Which lamps light up brightly?
 +  * Which lamps remain dark?
 +  * Explain the result using the idea of diode bypass paths.
 +
 +{{url>http://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxG3KQBZsQEBTAWjDACgBzEa4wkTFN14U0UKGwAmQvgPAYZGQYIkMAZgEMArgBsALmwBK4MILDFBeSOHNir1K7ltRoCSf3vuHhPJ-4gVGjr6AO4U3r7Y4WaCkGyhKB4JVknWMWxgeBC0pjY8fNFiuFYQSDBwEGWQ7KFg8vyKcvk2saG41KkUkO0mPi2NHbX5KL1ubeDD-T1+AVp6o1ET2eM+ymqzIdzYOYJLU7EAzsbbk83gIBra+wzpmWE+BbunRWelsFXO5TcQKQVjBQ5wF4fd6VdgZCB-GyRe5PQElYEVN5g26DDo-WHFegIhFxaT1HbCf646EdEl7NhAA noborder}}
 +
 +
 +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:
 +
 +\[
 +\begin{align*}
 +L_1,\;L_2,\;L_3,\;L_4,\;L_5.
 +\end{align*}
 +\]
 +
 +After closing the switch, check the voltage across each lamp in the simulation.
 +
 +A lamp is assumed to light brightly if
 +
 +\[
 +\begin{align*}
 +U_{\rm lamp}\geq 5~{\rm V}.
 +\end{align*}
 +\]
 +
 +The simulation shows that the outer lamps have a sufficiently large voltage across them, while the inner lamps are bypassed by conducting diodes.
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +
 +<WRAP half column>
 +#@ResultBegin_HTML~12001~@#
 +The lamps
 +
 +\[
 +\begin{align*}
 +L_1
 +\quad \text{and} \quad
 +L_5
 +\end{align*}
 +\]
 +
 +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
 +
 +\[
 +\begin{align*}
 +U_{\rm lamp}<5~{\rm V},
 +\end{align*}
 +\]
 +
 +the lamp does not light brightly.
 +</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@#
 +</WRAP>
 +</WRAP>
 +
 +#@TaskEnd_HTML@#
 +
 +
 +#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Circuit with multiple diodes II: current calculation
 +#@TaskText_HTML@#
 +
 +The following simulation includes two diodes and two resistors.
 +
 +Assume a simple constant-voltage diode model:
 +
 +\[
 +\begin{align*}
 +U_{\rm F}=0.6~{\rm V}.
 +\end{align*}
 +\]
 +
 +The source voltage is
 +
 +\[
 +\begin{align*}
 +U_0=4.0~{\rm V}.
 +\end{align*}
 +\]
 +
 +The resistors are
 +
 +\[
 +\begin{align*}
 +R_1=200~\Omega,
 +\qquad
 +R_2=100~\Omega.
 +\end{align*}
 +\]
 +
 +Calculate the currents through
 +
 +  * \(D_1\),
 +  * \(R_1\),
 +  * \(R_2\).
 +
 +{{url>https://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWcMBMcUHYMGZIA4UA2ATmIxABZykLsQEBTAWjDACgA3EFFC7iyN17gUeKOIGVxgmAjYAnENjQixywbxnc4CpXj5hRevpvFgdAd2P9B6m1DZW7pnicmQ2AD24IU4c9wE-nRuIAAi7N7Y2EistoSxYCH84SheSmTgGH4UmFk0KQBKaVG4WXTYhIJgGIRSwoWRQmI1dCgILbX1fACqHgAmQgZGdoZifv0MAGYAhgCuADYALmyDoyP6qtwgk7OLK0A noborder}}
 +
 +1. Calculate the current through \(R_1\).
 +
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~12004~@#
 +<WRAP leftalign>
 +The current through \(R_1\) passes through one forward-biased diode.
 +
 +Therefore the voltage across \(R_1\) is
 +
 +\[
 +\begin{align*}
 +U_{R1}
 +=
 +U_0-U_{\rm F}.
 +\end{align*}
 +\]
 +
 +Insert the values:
 +
 +\[
 +\begin{align*}
 +U_{R1}
 +&=
 +4.0~{\rm V}-0.6~{\rm V}
 +\\
 +&=
 +3.4~{\rm V}.
 +\end{align*}
 +\]
 +
 +Now apply Ohm's law:
 +
 +\[
 +\begin{align*}
 +I_{R1}
 +=
 +\frac{U_{R1}}{R_1}
 +=
 +\frac{3.4~{\rm V}}{200~\Omega}
 +=
 +17~{\rm mA}.
 +\end{align*}
 +\]
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +
 +<WRAP half column>
 +#@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
 +
 +\[
 +\begin{align*}
 +U_{R2}
 +=
 +U_0-2U_{\rm F}.
 +\end{align*}
 +\]
 +
 +Insert the values:
 +
 +\[
 +\begin{align*}
 +U_{R2}
 +&=
 +4.0~{\rm V}-2\cdot 0.6~{\rm V}
 +\\
 +&=
 +2.8~{\rm V}.
 +\end{align*}
 +\]
 +
 +Now apply Ohm's law:
 +
 +\[
 +\begin{align*}
 +I_{R2}
 +=
 +\frac{U_{R2}}{R_2}
 +=
 +\frac{2.8~{\rm V}}{100~\Omega}
 +=
 +28~{\rm mA}.
 +\end{align*}
 +\]
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +
 +<WRAP half column>
 +#@ResultBegin_HTML~12005~@#
 +\[
 +\begin{align*}
 +I_{R2}=28~{\rm mA}
 +\end{align*}
 +\]
 +#@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,
 +
 +\[
 +\begin{align*}
 +I_{D1}
 +=
 +I_{R1}+I_{R2}.
 +\end{align*}
 +\]
 +
 +Insert the values:
 +
 +\[
 +\begin{align*}
 +I_{D1}
 +&=
 +17~{\rm mA}+28~{\rm mA}
 +\\
 +&=
 +45~{\rm mA}.
 +\end{align*}
 +\]
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +
 +<WRAP half column>
 +#@ResultBegin_HTML~12006~@#
 +\[
 +\begin{align*}
 +I_{D1}=45~{\rm mA}
 +\end{align*}
 +\]
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +#@TaskEnd_HTML@#
 +
 +
 +#@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Circuit with multiple diodes III: switch-dependent currents
 +#@TaskText_HTML@#
 +
 +The following simulation includes two diodes and a switch.
 +
 +Assume a simple constant-voltage diode model:
 +
 +\[
 +\begin{align*}
 +U_{\rm F}=0.7~{\rm V}.
 +\end{align*}
 +\]
 +
 +The source voltage is
 +
 +\[
 +\begin{align*}
 +U_0=5.0~{\rm V}.
 +\end{align*}
 +\]
 +
 +The resistor is
 +
 +\[
 +\begin{align*}
 +R_1=1.0~{\rm k}\Omega.
 +\end{align*}
 +\]
 +
 +Calculate the currents through
 +
 +  * \(R_1\),
 +  * \(D_1\),
 +  * \(D_2\),
 +
 +depending on the switch state \(S\).
 +
 +{{url>https://www.falstad.com/circuit/circuitjs.html?hideSidebar=true&running=false&ctz=CQAgjCAMB0l3BWKswDZ0A4BMYDMAWfAdiIE59UFcRVIQl9qEBTAWjDACgA3ELLfH3x1+gsFgxQpw+lLowEnAE5C64ybkhiJUsPE4B3EIyyqQuVJIHzD5y2dFnInACZ3J69w-AA5fGHwMTgAPcwxqMCJBfwiSYyEQABEuUKwENUhSPgw1cXiBEAAlFL4dSOo0jyJUfMEAVWc3E3AdZus+X39AkONicCjjOMiiWqSsTgBnL09mz3kQADMAQwAbCeZOXCI6TW0NCxaPOVtHT3a521wDzwsPHWdQ3HJwTOM9cDzBAoBlTiA noborder}}
 +
 +1. Calculate the currents for open switch \(S\).
 +
 +<WRAP group>
 +<WRAP half column rightalign>
 +#@PathBegin_HTML~12007~@#
 +<WRAP leftalign>
 +With the switch open, only \(D_1\) is connected to the resistor path.
 +
 +The conducting diode clamps the node voltage to approximately
 +
 +\[
 +\begin{align*}
 +U_{\rm node}\approx U_{\rm F}=0.7~{\rm V}.
 +\end{align*}
 +\]
 +
 +The resistor current is therefore
 +
 +\[
 +\begin{align*}
 +I_{R1}
 +&=
 +\frac{U_0-U_{\rm F}}{R_1}
 +\\
 +&=
 +\frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega}
 +\\
 +&=
 +4.3~{\rm mA}.
 +\end{align*}
 +\]
 +
 +Since only \(D_1\) conducts,
 +
 +\[
 +\begin{align*}
 +I_{D1}=I_{R1},
 +\qquad
 +I_{D2}=0.
 +\end{align*}
 +\]
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +
 +<WRAP half column>
 +#@ResultBegin_HTML~12007~@#
 +For open switch:
 +
 +\[
 +\begin{align*}
 +I_{R1}&=4.3~{\rm mA},
 +\\
 +I_{D1}&=4.3~{\rm mA},
 +\\
 +I_{D2}&=0.
 +\end{align*}
 +\]
 +#@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 \(R_1\):
 +
 +\[
 +\begin{align*}
 +I_{R1}
 +&=
 +\frac{U_0-U_{\rm F}}{R_1}
 +\\
 +&=
 +\frac{5.0~{\rm V}-0.7~{\rm V}}{1.0~{\rm k}\Omega}
 +\\
 +&=
 +4.3~{\rm mA}.
 +\end{align*}
 +\]
 +
 +Kirchhoff's current law gives
 +
 +\[
 +\begin{align*}
 +I_{D1}+I_{D2}=I_{R1}.
 +\end{align*}
 +\]
 +
 +With the ideal constant-voltage diode model, the individual currents through two parallel diodes are not uniquely determined.
 +
 +If both real diodes are approximately identical, the current splits approximately equally:
 +
 +\[
 +\begin{align*}
 +I_{D1}
 +\approx
 +I_{D2}
 +\approx
 +\frac{4.3~{\rm mA}}{2}
 +=
 +2.15~{\rm mA}.
 +\end{align*}
 +\]
 +</WRAP>
 +#@PathEnd_HTML@#
 +</WRAP>
 +
 +<WRAP half column>
 +#@ResultBegin_HTML~12008~@#
 +For closed switch:
 +
 +\[
 +\begin{align*}
 +I_{R1}=4.3~{\rm mA}
 +\end{align*}
 +\]
 +
 +and
 +
 +\[
 +\begin{align*}
 +I_{D1}+I_{D2}=4.3~{\rm mA}.
 +\end{align*}
 +\]
 +
 +For approximately identical real diodes:
 +
 +\[
 +\begin{align*}
 +I_{D1}
 +\approx
 +I_{D2}
 +\approx
 +2.15~{\rm mA}.
 +\end{align*}
 +\]
 +#@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">
 +Parallel diodes are sensitive to small differences in real diode characteristics.  
 +Current sharing should not be assumed to be perfect without checking the design.
 +</callout>
 +#@ResultEnd_HTML@#
 +</WRAP>
 +</WRAP>
 +
 +#@TaskEnd_HTML@#
  
  
Line 549: Line 1100:
 \[ \[
 \begin{align*} \begin{align*}
-U_{\rm E}=24~{\rm V}.+U_{\rm I}=24~{\rm V}.
 \end{align*} \end{align*}
 \] \]
Line 575: Line 1126:
 R_{\rm V} R_{\rm V}
 &= &=
-\frac{U_{\rm E}-U_{\rm F}}{I_{\rm F}}+\frac{U_{\rm I}-U_{\rm F}}{I_{\rm F}}
 \\ \\
 &= &=
Line 599: Line 1150:
 P_R P_R
 &= &=
-(U_{\rm E}-U_{\rm F})I_{\rm F}+(U_{\rm I}-U_{\rm F})I_{\rm F}
 \\ \\
 &= &=
Line 630: Line 1181:
 \[ \[
 \begin{align*} \begin{align*}
-U_{\rm E}=12~{\rm V}.+U_{\rm I}=12~{\rm V}.
 \end{align*} \end{align*}
 \] \]
Line 663: Line 1214:
 I_{\rm V} I_{\rm V}
 &= &=
-\frac{U_{\rm E}-U_{\rm Z}}{R_{\rm V}}+\frac{U_{\rm I}-U_{\rm Z}}{R_{\rm V}}
 \\ \\
 &= &=