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electrical_engineering_and_electronics_2:block12 [2026/06/02 02:41] – mexleadminelectrical_engineering_and_electronics_2:block12 [2026/06/10 03:06] (current) – mexleadmin
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 After this 90-minute block, you can After this 90-minute block, you can
  
 +  * identify basic diode types such as universal diodes, Z-diodes, and LEDs.
 +  * calculate simple diode operating points with a series resistor.
   * design a simple LED circuit with a series resistor.   * design a simple LED circuit with a series resistor.
   * explain why LEDs and signal diodes need current limitation.   * explain why LEDs and signal diodes need current limitation.
Line 28: Line 30:
     * Freewheeling diode for inductive loads.     * Freewheeling diode for inductive loads.
     * Clamp diodes for sensitive inputs.     * Clamp diodes for sensitive inputs.
-    * Diode rectifiers: M1, M2, B2.+    * Diode rectifiers: M1, B2.
     * Capacitor smoothing and ripple.     * Capacitor smoothing and ripple.
  
Line 75: Line 77:
 <WRAP> <WRAP>
 <panel type="default"> <panel type="default">
-<imgcaption op_point_circuit|Circuit of Diode with Shunt resistor.></imgcaption>+<imgcaption op_point_circuit|Circuit of Diode with resistor.></imgcaption>
 {{drawio>op_point_circuit_v01.svg}} {{drawio>op_point_circuit_v01.svg}}
 </panel> </panel>
Line 84: Line 86:
 \[ \[
 \begin{align*} \begin{align*}
-U_{\rm E}+U_{\rm I}
 = =
 U_R+U_{\rm D}. U_R+U_{\rm D}.
Line 104: Line 106:
 I_{\rm D} I_{\rm D}
 \approx \approx
-\frac{U_{\rm E}-U_{\rm TO}}{R}.+\frac{U_{\rm I}-U_{\rm TO}}{R}.
 \end{align*} \end{align*}
 \] \]
Line 110: Line 112:
 <callout type="danger" icon="true"> <callout type="danger" icon="true">
 Never connect a normal diode or LED directly to an ideal voltage source in forward direction.  \\ Never connect a normal diode or LED directly to an ideal voltage source in forward direction.  \\
-The diode current must be limited. \\ +The diode current must be limited.
-The used resistor is often called **shunt resistor**.+
 </callout> </callout>
  
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 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 138: 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 239: 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 326: Line 327:
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
  
-==== Half-wave rectifier M1 ====+==== Half-wave rectifier (M1) ====
  
 A rectifier converts an AC voltage into a unidirectional voltage. A rectifier converts an AC voltage into a unidirectional voltage.
 +
 +<panel type="info" title="Simulation: half-wave rectifier">
 +Use this simulation to observe how one half-wave is removed by a diode.
 +
 +Things to try:
 +
 +  * reverse the diode direction,
 +  * change the load resistance,
 +  * change capacitor,
 +  * compare input and output voltage.
  
 <WRAP> <WRAP>
-<panel type="default"> +{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=DwYwlgTgBAZgvAIgIwKgFwM6IAwDpsEECsqYIiSAzPgJwDsALEkZQExLYAcSnAbClBAAjRERJQADiIRFsqAG4RRqALaZRAUwC0SFAD4AUFCjB5UAB4UkrKG2xRdNmr1TwEcqAHc3AmEoQMHioAhubyygD0hsbA0JYBSLxQzlBMSXauOKj+vIEIUUYmACYWVk7plEnOme6qAPaIRRowwQCuADZo+dEmnqXI1ras9pSVQx5ucgUxGP2jVRVJiRNZ0yYgc2PLtmMZsFmCOPiENKdn5xc0pOHuxwIY-h7yRYgMuIFEDCysDLm8jEROCQ1sA+vF5uNUolxjUpj1QZsFlCFrDuoVgHUoBoAHYHDASRC5GrmSioAm1HoxCRQG4eDDkZBwwpUmmIUlQelHVhomIROrwzE4iiofGEhjE9nk-TMkzU2kihnAymy1kIdmc27YaxK9F8wzACLgCCGIA noborder}} 
-<imgcaption fig_half_wave_rectifier|Half-wave rectifier M1 with ideal diode and resistive load.></imgcaption> +
-{{drawio>block12_half_wave_rectifier_m1.svg}} +
-</panel>+
 </WRAP> </WRAP>
 +</panel>
  
 Assumptions for the basic formulas: Assumptions for the basic formulas:
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 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +\boxed{ U_{\rm di} = \frac{\sqrt{2}}{\pi}U_\sim }
-U_{\rm di} +
-= +
-\frac{\sqrt{2}}{\pi}U_\sim +
-}+
 \end{align*} \end{align*}
 \] \]
Line 364: Line 369:
 \] \]
  
-The ripple factor for the ideal M1 circuit is+The output voltage can be split into an average DC value and an AC ripple component:
  
 \[ \[
 \begin{align*} \begin{align*}
-w_U +u_{\rm out}(t) = U_{\rm di} + u_\sigma(t).
-= +
-\frac{U_\sigma}{U_{\rm di}} +
-\approx +
-1.21.+
 \end{align*} \end{align*}
 \] \]
  
-<callout> +Here
-The half-wave rectifier is simple, but it uses only one half-wave.   +
-Therefore the ripple is large and the transformer is used poorly. +
-</callout>+
  
-<panel type="info" title="Simulation: half-wave rectifier"> +  * \(U_{\rm di}\) is the average value, i.e. the DC component, 
-Use this simulation to observe how one half-wave is removed by a diode.+  * \(u_\sigma(t)\) is the time-dependent ripple component, 
 +  * \(U_\sigma\) is the RMS value of this ripple component.
  
-Things to try:+\[ 
 +\begin{align*} 
 +U_\sigma = \sqrt{ \frac{1}{T} \int_0^T u_\sigma^2(t)\,{\rm d}t }. 
 +\end{align*} 
 +\]
  
-  * reverse the diode direction, +The ripple factor for the ideal circuit is
-  * change the load resistance, +
-  * compare input and output voltage.+
  
-{{url>https://www.falstad.com/circuit/e-rectify.html 700,500 noborder}} +\[ 
-</panel>+\begin{align*} 
 +w_U = \frac{U_\sigma}{U_{\rm di}} \approx 1.21. 
 +\end{align*} 
 +\]
  
-==== Center-tap rectifier M2 and bridge rectifier B2 ====+<callout> 
 +The half-wave rectifier is simple, but it uses only one half-wave. \\   
 +Therefore the ripple is large and the transformer is used poorly. \\ 
 +Damping capacitors must be relatively large. 
 +</callout>
  
-A full-wave rectifier uses both half-waves. 
  
-<WRAP group> +==== Bridge rectifier B2 ====
-<WRAP column half> +
-<panel type="default"> +
-<imgcaption fig_center_tap_rectifier|Center-tap rectifier M2.></imgcaption> +
-{{drawio>block12_center_tap_rectifier_m2.svg}} +
-</panel> +
-</WRAP>+
  
-<WRAP column half> +A full-wave rectifier uses both half-waves.
-<panel type="default"> +
-<imgcaption fig_bridge_rectifier|Bridge rectifier B2.></imgcaption> +
-{{drawio>block12_bridge_rectifier_b2.svg}} +
-</panel> +
-</WRAP> +
-</WRAP>+
  
-For the center-tap rectifier M2:+<panel type="info" title="Simulation: bridge rectifier"> 
 +Use this simulation to compare half-wave and full-wave rectification.
  
-\[ +Things to try:
-\begin{align*} +
-U_{\rm di} +
-= +
-\frac{2\sqrt{2}}{\pi}U_{1{\rm N}} +
-= +
-\frac{\sqrt{2}}{\pi}U_{\rm S}. +
-\end{align*} +
-\]+
  
-Here \(U_{1{\rm N}}\) is the RMS voltage of one half of the secondary winding and \(U_{\rm S}\) is the RMS voltage of the full secondary winding.+  * observe which two diodes conduct in each half-wave, 
 +  * compare input and output voltage, 
 +  * add or remove smoothing if available in the simulation. 
 + 
 +{{url>https://www.falstad.com/circuit/e-fullrect.html 700,500 noborder}} 
 +</panel>
  
 For the bridge rectifier B2: For the bridge rectifier B2:
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 \begin{align*} \begin{align*}
 \boxed{ \boxed{
-U_{\rm di} +U_{\rm di} = \frac{2\sqrt{2}}{\pi}U_\sim
-= +
-\frac{2\sqrt{2}}{\pi}U_\sim+
 } }
 \end{align*} \end{align*}
Line 456: Line 447:
  
 <panel type="info" title="Real diode voltage drops"> <panel type="info" title="Real diode voltage drops">
-In a bridge rectifier, two diodes conduct at the same time.  +In a bridge rectifier, two diodes conduct at the same time.  \\ 
 Therefore, for silicon diodes, the output voltage is roughly reduced by Therefore, for silicon diodes, the output voltage is roughly reduced by
  
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 ^ Circuit ^ Uses half-waves ^ Ideal average voltage \(U_{\rm di}\) ^ Ripple frequency ^ ^ Circuit ^ Uses half-waves ^ Ideal average voltage \(U_{\rm di}\) ^ Ripple frequency ^
-| M1 half-wave | one half-wave | \(\frac{\sqrt{2}}{\pi}U_\sim\) | \(f\) | +| M1 half-wave  | one half-wave    | \(\frac{\sqrt{2}}{\pi}U_\sim\)   | \(f\)   | 
-| M2 center-tap | both half-waves | \(\frac{2\sqrt{2}}{\pi}U_{1{\rm N}}\) | \(2f\) | +| B2 bridge     | both half-waves  | \(\frac{2\sqrt{2}}{\pi}U_\sim\)  | \(2f\)  | 
-| B2 bridge | both half-waves | \(\frac{2\sqrt{2}}{\pi}U_\sim\) | \(2f\) |+</tabcaption>
  
-<panel type="info" title="Simulation: bridge rectifier"> 
-Use this simulation to compare half-wave and full-wave rectification. 
- 
-Things to try: 
- 
-  * observe which two diodes conduct in each half-wave, 
-  * compare input and output voltage, 
-  * add or remove smoothing if available in the simulation. 
- 
-{{url>https://www.falstad.com/circuit/e-fullrect.html 700,500 noborder}} 
-</panel> 
  
 ~~PAGEBREAK~~ ~~CLEARFIX~~ ~~PAGEBREAK~~ ~~CLEARFIX~~
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 ==== Capacitor smoothing ==== ==== Capacitor smoothing ====
  
-A rectifier output is not constant. A smoothing capacitor stores charge near the voltage maximum and supplies the load between maxima.+A rectifier output is not constant. \\  
 +A smoothing capacitor stores charge near the voltage maximum and supplies the load between maxima.
  
 +<panel>
 <WRAP> <WRAP>
-<panel type="default"> +{{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=DwYwlgTgBAZgvAIgIwKgFwM6IAwDpsEECsqYIiSAzPgJwDsALEkZQExLYAcSnAbClBAAjRERJQADiIRFsqAG4RRqALaZRAUwC0SFAD4AUFCjB5UAB4VWnKK1YMoSa495zYOVAHd4yVDCUIDG4qAIbm8soA9IbGwNCWgUi8UDTJTMnWnH4eUAG8QQjRRiYAJhaIrFwu2LZVqdkIwQD2iCUaMCEArgA2aIUxJhjlCJSUyalQo8lJbj5yRbEgw1PVk2O29g1u5I34hDQHh0fHNKQRuwQCGAFu8iWI1LycNPa8vKx070R0lFkLJp5lutMlB0rZOFl3I1+sVgICEisJmD6lD5gM4cNKjYZrUbHYGFsYbEygkseC8VV8YSoCoWgg2h0en1-hiEk5sa5cdVCSySRU6KxybYBRsCajVHSGV1ekSAZiRTiPoKqeLefLBYqRSi5hLWu1pcz0fCrHj7MLlZtVUagRkIebyTzraSRSCyZlHbDjQgyRMlSleB7Yl6-RNEQGrZ6baK1rbITqWU0oBoAHY5DASRD5BrmSioDPQgaxCRQc5uDA7Diyoslh6ocs4XCsWXASJNdGJlMUOv5rNQnN5ruFkzF0t1nYkf7V865qD1i7YL7N1uGFvgCCGIA noborder}}  
-<imgcaption fig_bridge_rectifier_with_capacitor|Bridge rectifier with smoothing capacitor.></imgcaption> +</WRAP>
-{{drawio>block12_bridge_rectifier_with_capacitor.svg}}+
 </panel> </panel>
-</WRAP> 
  
 For a bridge rectifier with a sufficiently large smoothing capacitor, the DC voltage is approximately For a bridge rectifier with a sufficiently large smoothing capacitor, the DC voltage is approximately
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 \[ \[
 \begin{align*} \begin{align*}
-U_{\rm di} +U_{\rm di} \approx \sqrt{2}U_\sim \end{align*}
-\approx +
-\sqrt{2}U_\sim +
-\end{align*}+
 \] \]
  
Line 516: Line 493:
 \[ \[
 \begin{align*} \begin{align*}
-U_{\rm di} +U_{\rm di} \approx \sqrt{2}U_\sim - 2U_{\rm TO} - \frac{\Delta U}{2}.
-\approx +
-\sqrt{2}U_\sim +
-- +
-2U_{\rm TO} +
-- +
-\frac{\Delta U}{2}.+
 \end{align*} \end{align*}
 \] \]
Line 532: Line 503:
 \[ \[
 \begin{align*} \begin{align*}
-\boxed{ +\boxed{ C \approx \frac{I_{\rm d}}{f_\sigma\Delta U} }
-C +
-\approx +
-\frac{I_{\rm d}}{f_\sigma\Delta U} +
-}+
 \end{align*} \end{align*}
 \] \]
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 \[ \[
 \begin{align*} \begin{align*}
-C +C \approx k\frac{I_{\rm d}}{f_\sigma U_\sigma}.
-\approx +
-k\frac{I_{\rm d}}{f_\sigma U_\sigma}.+
 \end{align*} \end{align*}
 \] \]
Line 568: Line 533:
  
 <callout type="warning" icon="true"> <callout type="warning" icon="true">
-A larger capacitor reduces ripple, but it also creates short high charging-current pulses through the diodes and transformer.  +A larger capacitor reduces ripple, but it also creates short high charging-current pulses through the diodes and transformer.  \\
 For power supplies, check diode peak current, transformer rating, capacitor ripple current, and inrush current. For power supplies, check diode peak current, transformer rating, capacitor ripple current, and inrush current.
 </callout> </callout>
  
-==== Application overview ====+===== 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@#
  
-<tabcaption tab_diode_applications|Typical diode applications in mechatronics and robotics> 
  
-^ Problem ^ Diode application ^ Main design question ^ 
-| Status indication | LED with resistor | Which current and resistor value? | 
-| Small reference voltage | Z-diode stabilizer | Is \(I_{\rm Z}\) inside the allowed range? | 
-| Relay or solenoid switch-off | freewheeling diode | Where can the inductor current flow? | 
-| Sensor input disturbance | clamp diodes | Is the clamp current limited? | 
-| AC to DC conversion | rectifier | M1, M2, or B2? | 
-| DC supply with lower ripple | smoothing capacitor | Which ripple voltage is acceptable? | 
  
-===== Exercises ===== 
  
 #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Quick check: LED series resistor for a robot status LED #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1  Quick check: LED series resistor for a robot status LED
Line 593: Line 1100:
 \[ \[
 \begin{align*} \begin{align*}
-U_{\rm E}=24~{\rm V}.+U_{\rm I}=24~{\rm V}.
 \end{align*} \end{align*}
 \] \]
Line 619: 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 643: 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 674: Line 1181:
 \[ \[
 \begin{align*} \begin{align*}
-U_{\rm E}=12~{\rm V}.+U_{\rm I}=12~{\rm V}.
 \end{align*} \end{align*}
 \] \]
Line 707: 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}}
 \\ \\
 &= &=