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Ask your administrator if you think this is wrong. ====== Block 12 — Diode Applications ====== ===== Learning objectives ===== <callout> 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. * explain why LEDs and signal diodes need current limitation. * use Z-diodes for simple voltage limitation and voltage stabilization. * explain how a freewheeling diode protects a switching transistor or contact. * explain diode clamp circuits for sensitive microcontroller inputs. * distinguish half-wave, center-tap, and bridge rectifier circuits. * calculate the ideal average value \(U_{\rm di}\) of rectified sinusoidal voltages. * explain ripple voltage and ripple frequency. * estimate a smoothing capacitor for a simple diode rectifier power supply. </callout> ===== 90-minute plan ===== * **Warm-up (10 min):** * What happens if an LED is connected directly to \(24~{\rm V}\)? * Recall from [[block11|Block 11]]: diode polarity, forward voltage, reverse blocking. * Recall from [[block01|switching transients]]: inductor current cannot jump. * **Core concepts (55 min):** * LED operation with a series resistor. * Z-diode voltage limitation and stabilization. * Freewheeling diode for inductive loads. * Clamp diodes for sensitive inputs. * Diode rectifiers: M1, B2. * Capacitor smoothing and ripple. * **Practice (20 min):** * Calculate an LED series resistor. * Check Z-diode current limits. * Estimate switching overvoltages in an inductive load. * Calculate average rectifier voltages and smoothing capacitors. * **Wrap-up (5 min):** * Which diode application belongs to which engineering problem? * Preview: bipolar transistors as controlled switches and amplifiers. ===== Conceptual overview ===== <callout icon="fa fa-lightbulb-o" color="blue"> * A diode is useful because it is **nonlinear**: it behaves differently for the two voltage polarities. * In applications, a diode often has one of four jobs: * **conduct only one half-wave**: rectifier, * **limit a voltage**: Z-diode or clamp diode, * **provide a safe current path**: freewheeling diode, * **emit light**: LED. * A diode does not magically limit its own current. The circuit around it must do that. * Real diodes cause voltage drops and losses: \[ \begin{align*} P_{\rm D}=U_{\rm D}I_{\rm D}. \end{align*} \] * In mechatronics, diode circuits appear in power supplies, relay drivers, sensor inputs, motor-driver protection, status LEDs, and emergency signal paths. </callout> <panel type="info" title="Scope of this block"> This block uses the diode models from [[block11|Block 11]] and applies them to practical circuits. The focus is on **basic engineering estimates**, not yet on detailed datasheet design. </panel> ~~PAGEBREAK~~ ~~CLEARFIX~~ ===== Core content ===== ==== Operating point with a series resistor ==== A diode must usually be operated with a current-limiting element. <WRAP> <panel type="default"> <imgcaption op_point_circuit|Circuit of Diode with resistor.></imgcaption> {{drawio>op_point_circuit_v01.svg}} </panel> </WRAP> For the circuit in <imgref op_point_circuit> the loop equation is \[ \begin{align*} U_{\rm I} = U_R+U_{\rm D}. \end{align*} \] With the constant-voltage model, \[ \begin{align*} U_{\rm D}\approx U_{\rm TO}. \end{align*} \] Therefore \[ \begin{align*} I_{\rm D} \approx \frac{U_{\rm I}-U_{\rm TO}}{R}. \end{align*} \] <callout type="danger" icon="true"> Never connect a normal diode or LED directly to an ideal voltage source in forward direction. \\ The diode current must be limited. </callout> ==== LED (with series resistor) ==== An LED is operated in forward direction. It converts part of the electrical energy into light via electron-hole recombination. \\ The required forward voltage depends on the semiconductor material and therefore on the color. 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 \[ \begin{align*} \boxed{ R_{\rm V} = \frac{U_{\rm I}-U_{\rm F}}{I_{\rm F}} } \end{align*} \] <WRAP> {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=DwYwlgTgBAZgvAIgIwKgFwM6IAwDpsEEAsAnGeRZQGypgiJJ4BMBTA7GwMzYAcTVZNqhAAjRAFZsqAA5iERTqgBuECagC2mCQFMAtEhQA+AFBQowJVAAeDJEyhMmRKAftEeqeMhpQA7l5RYVQQqKSh1AEMrJQZcQJEwCKwEJlwSBAB6EzNgABN1KAA7dUQAL21C7QhdIk8cWmSSXE4mPQNUCGS8bCImVELETlwuVBEY+Uzs8wAZAFEAEWtEJnEqF1CHVYcnOoQw9QB7MoqqmuFgwJAAc3qoUVv1EXo9-BQs03NfJeQ7becVtaOWqwBiTD7AaA2FJbEiArZIUK7MIXAhgnJfKEA9bYP7YpFoz7fIHY3Gw-HvHIAZSJRBx7gctKgZJBe1QABswLc0AALZYE4AY2xuHgMukeFlSCnmKxEngi8QkKC9NYGHxeMIYTkpYFobSISm+MBoEDc2AHaAAHUKhp5UAyfiN3IOAFc0FbKdznYU0PyDlAKnyoBhpAw1YgrCwCO0oCHWVMctIoOMNc9uoQ2FQmCROCQiOJOHZxBouvhCEhKBWKOL1LHJfHzInxoog6n8CtizhXhpa-yMgcpsA-QHWUHYwjdhGZKD68BG7cMM8i1KE0nBqgF523gO+yZgBlwBATEA noborder}} </WRAP> For circuit design it is important the check the real resistor power with the absolute maximum ratings of the resistors \[ \begin{align*} P_R = (U_{\rm I}-U_{\rm F})I_{\rm F} = R_{\rm V}I_{\rm F}^2 \quad \overset{!}{<} \quad P_{R, \rm max} \end{align*} \] The LED power is approximately \[ \begin{align*} P_{\rm LED} = U_{\rm F}I_{\rm F}. \end{align*} \] <callout type="danger" icon="true"> Do not connect an LED directly to an ideal voltage source. The current must be limited, usually with a resistor or a current source. </callout> <tabcaption tab_led_values|Typical LED values for first estimates> ^ LED color ^ Typical forward voltage \(U_{\rm F}\) ^ Typical current ^ | infrared | \(\approx 1.3~{\rm V}\) | \(5\ldots 20~{\rm mA}\) | | red | \(\approx 1.6~{\rm V}\) | \(5\ldots 20~{\rm mA}\) | | yellow | \(\approx 1.7~{\rm V}\) | \(5\ldots 20~{\rm mA}\) | | green | \(\approx 1.8~{\rm V}\) | \(5\ldots 20~{\rm mA}\) | | blue / white | \(\approx 3.0\ldots 3.3~{\rm V}\) | \(5\ldots 20~{\rm mA}\) | </tabcaption> \\ <panel type="info" title="Engineering example"> A robot controller often uses a \(24~{\rm V}\) supply, but a status LED may need only \(10~{\rm mA}\) at about \(2~{\rm V}\). \\ Most of the voltage must therefore drop across the resistor, not across the LED. </panel> <callout type="warning" icon="true"> LEDs usually tolerate only small reverse voltages. Do not operate an LED in reverse direction unless the datasheet explicitly allows it. </callout> ~~PAGEBREAK~~ ~~CLEARFIX~~ ==== Z-Diodes ==== If the reverse voltage of a diode becomes too large, the diode enters **breakdown**. In this region, the reverse current rises strongly. Two physical effects can cause breakdown: * **avalanche breakdown:** charge carriers gain enough energy to free additional charge carriers by collisions. * **Zener breakdown:** in strongly doped pn junctions, charge carriers can cross the barrier by a quantum-mechanical effect. For ordinary diodes, breakdown is usually unwanted and can destroy the diode if the current is not limited. \\ **Z-diodes** are designed to operate safely in this reverse-breakdown region at a defined voltage \(U_{\rm Z}\). <WRAP> {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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-ccUi7ZCAAKocpCZFLTcgoLqiIMJmQQKAKe0CehEEBB6AAGggADUOQQ6NzFAR1BA1LGAD6eyjqJBQCiJbl0UsV6u23n0Gkl4h5iADtsczJ9ihiL1JGfxeNzxcdIR1PPdkuxgBan2OdfGJ-AECmwJhDWWHGa5-JIEf3QY1gAZjoAK5hVgAL2YfxmE0CpnwGapRHPOpb0keJEhfNIoJqOsCDAagkAwpFwSTaChBZRkyiZVl8kVaZcG6LY+DSaD9juXUXjOC5cCNJALXwwZoAIXBLWeN4Ph4bs+R2AhqDES8FRkXBoCgPFEIoyhfXoIQD1YcwrGYDkEkSODtTwWRA1jEMQFQ9QMGmPhRAzfc12zXT1FQ7UnQUAy13LKtNMgMc0gpPtsGstJ2xzPElKgZsu1XNJbObdpyBAKgYp7Q8OWndRcEUad2mbfyQECsRcAyOQ0G5MRa2s1oQEARMI504ktPKyAVys2cqqttMBZmYktpzylzmo7aYLgYJBqVqlBBRAd8LE0ewdE0ExjG0CJLH8OdkztIR6n68cxo8XQfD8QIjAWpa5wwAVRPINRwLG7QgmAowbEO5hFosZagA noborder}} </WRAP> The current must still be limited by the surrounding circuit. \\ In its operating range, the diode voltage is approximately constant: \[ \begin{align*} u_{\rm Z}\approx U_{\rm Z}. \end{align*} \] The piecewise-linear model is \[ \begin{align*} u_{\rm Z} \approx U_{\rm Z}+r_{\rm Z} \cdot i_{\rm Z}. \end{align*} \] <panel type="info" title="Z-diode"> * Z-diodes are useful for voltage limitation and voltage stabilization. * Z-diodes have a huge variety of breakdown voltages: $U_{\rm Z} \approx 1.0 ~\rm V... 400 ~ V $ \\ Z-diodes allow to get "knee voltages" above $0.7 ~\rm V$ * Z-diodes are still conventional diodes in the forward direction. </panel> \\ A typical application is a **Z-diode stabilizer** <panel type="info" title="Simulation: Z-diode voltage reference"> Use this simulation to observe how a Z-diode limits the output voltage. Things to try: * change the input voltage, * change the load resistance, * observe when the Z-diode current becomes too small for stabilization. <WRAP> {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=DwYwlgTgBAZgvAIgIwKgFwM6IAwDpsEEAsAnGeRZQGypgiJJ4BMBTA7GwMzYAcTVZNqhAAjRAFYiqAA5iE47KgBuECagC2mCQFMAtEhQA+AFBQowAEpQAHgyRMoSEg57ZHz1PASKoAdy8osKoIRD7qAIbWSgyoImDhWAhUuEIA9CZmwABeNoicnFRQTDw8UPmFLDyeMVDqAPaIACbaMOEArgA2aLpZ2gB22hAI6abmAOa5COVFJWUFZdhSsDjDGea+k9NIVG5ERKXbPl6KI5nQtiH7RbxQe6XFVcveqMGMhKujwBMXdzOlv9wlscPmdJk4HIc-o4dtVnlBggoTmtgNYweI2EV7ND7khxLCfBgwAwaOhtIgAKog8yoi7bHGFJBsCpOfGoQkMJioNBkhAAUSpKLBbBImIhsyYuNZUHZyBJ3MQiFO1M2LHcSDKTBFTiBK2lRNlXJ5iuRNLy2C1JHEZW27kewL1HMNiAAWgLTVNFu43NwDiROFKZdsnQgAJJuybFK24zhFEhERziEn2wNyo3hi7seO4jHFCEwp4E-X2YMANXTiH4WvEDnYMdxeILbKLqcQYaVgp+TCzPAxRGKXoDzeDbZNkz7FocRE4Uctg8dpMQABlyyFOFmSLsp+5-Y2HQaFwhjZ9pJNqwcdlAz39YTv1DyCfRkC8eW9CAQV4mrRL1ej4xKdXCKbBkemTuuIkiYlaibqhKdq6oGnIHq67ZgVu36XhB-5zvu8oIJS7YAILqHe3LnBWsyQls+YBBo95so+nKEcR2ikZs8yUfMlSwoEd7wQxAonrSezQm4SDCQ8N60XxNQQC+b7vO23wMOJFHCYCrKKae-YPJe-bqY27YbBcV6Qr8hwaaOPwXkwa63GwMGcEIu5AQeFgrkQ9lFJwIoeTmfbYcWB7LkxJGDGOVymVcXFPDxdHSvxKERn2jg8CKTAKO4iHJkWiG4QAyiuNnqkgPB-slJVOdlxLBm5hlgsJkLgiJFmfCIY4HFiTiFPkWUrIlFwCGUiaOFwQ0kAFjy4X0dRgBgACeUB5XUbQQCA2gCm1GY3AYDjed6+QtaBEbcENGIFGd1HwUWk08s6uiNGAdTNNKaDhHEHRgL0QztptiChAyWLiLM+Q7sC-V-Ta06FED3riONznXcGi51OEjRQAAFAAFoM2hQBY2iEhgaB1BAACUAp5fVWYXjtzW7tINQfbqaCYxWApWLSnX9q4jjFNxJL+LqERRIgaTIkZdgQg1XNwc8dWcxCziOJ1TKHcqtLYBCnolQy2CVVdDA3YgbQYLjGC+GAaAgJjAp1FA-QVg2GAM1MvUINYLAEIFLs+BcIZ9LoJZ1F04RjOtayZCe0SAY+jAaIkPEu-omsCqkdTInbDtPtKLucG7HsyDEEfmFH0nIEioyR1A0c7hgj54Cg7Zpxn9t9H9bK5-nnvYN7bOTAA8m03RByHYcIMXwClzHDBhAnGhJ4wjET1Ptf17gPGJHg2ANuoPup+nnyowAVvb7f2-qPhZ-jhOvX0a0Y3UmPqBg5O1G3HrvrUxlvmyaCIBYAB9RcGNkao3Ju2Y+p8QiIW0BfVAWcR6vTHm-NQX8Z6-0QIg0O60RjAFSOACAJggA noborder}} </WRAP> </panel> \[ \begin{align*} {\color{blue }{I_{\rm V}}} &= \frac{U_{\rm I}-U_{\rm Z}}{R_{\rm V}} &&\text{current supplied through the series resistor}, \\ {\color{green}{I_{\rm L}}} &= \frac{U_{\rm Z}}{R_{\rm L}} &&\text{useful load current}, \\ {\color{red}{I_{\rm Z}}} &= {\color{blue}{I_{\rm V}}} - {\color{green}{I_{\rm L}}} &&\text{remaining current through the Z-diode}. \end{align*} \] The Z-diode can stabilize the voltage only if \({\color{red}{I_{\rm Z}}}\) remains inside the allowed range: \[ \begin{align*} I_{\rm Z,min} \leq {\color{red}{I_{\rm Z}}} \leq I_{\rm Z,max}. \end{align*} \] The power limit is \[ \begin{align*} P_{\rm Z} = U_{\rm Z}\cdot {\color{red}{I_{\rm Z}}} \leq P_{\rm tot}. \end{align*} \] A Z-diode stabilizer is simple, but not efficient for large load currents. \\ It is useful for voltage limitation, small reference voltages, and robust simple circuits. ~~PAGEBREAK~~ ~~CLEARFIX~~ ==== Freewheeling diode for inductive loads ==== Inductors resist a sudden change of current: \[ \begin{align*} u_L=L\frac{{\rm d}i_L}{{\rm d}t}. \end{align*} \] If a relay coil, solenoid, or small motor is switched off, the current tries to continue flowing. Without a safe current path, the voltage can become very large. When the switch is opened, the freewheeling diode becomes forward-biased. The inductor current circulates through the diode and the coil. <panel type="info" title="Physical interpretation"> The coil is like a flywheel for current. * A mechanical flywheel cannot stop instantly. * An inductor current cannot stop instantly. * The freewheeling diode gives the current a safe path while the stored magnetic energy is dissipated. </panel> The magnetic energy stored in the inductance is \[ \begin{align*} W_L = \frac{1}{2}LI_0^2. \end{align*} \] With a freewheeling diode, the switch voltage is limited to a safe value. \\ The disadvantage is that the current decays more slowly, so a relay or solenoid may release more slowly. <callout type="info" icon="true"> For fast turn-off, additional components such as a Z-diode, TVS diode, or resistor-diode network can be used. The basic principle remains the same: provide a controlled path for the inductive current. </callout> <panel type="info" title="Simulation: inductive kickback protection"> Use this simulation to observe the overvoltage when switching an inductive load like a motor, and how a diode limits it. Things to try: * open and close the switch, * compare the circuit with and without the protection diode, * observe the voltage across the switch. <WRAP> {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=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 noborder}} </WRAP> </panel> ~~PAGEBREAK~~ ~~CLEARFIX~~ ==== Half-wave rectifier (M1) ==== 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> {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=DwYwlgTgBAZgvAIgIwKgFwM6IAwDpsEECsqYIiSAzPgJwDsALEkZQExLYAcSnAbClBAAjRERJQADiIRFsqAG4RRqALaZRAUwC0SFAD4AUFCjB5UAB4UkrKG2xRdNmr1TwEcqAHc3AmEoQMHioAhubyygD0hsbA0JYBSLxQzlBMSXauOKj+vIEIUUYmACYWVk7plEnOme6qAPaIRRowwQCuADZo+dEmnqXI1ras9pSVQx5ucgUxGP2jVRVJiRNZ0yYgc2PLtmMZsFmCOPiENKdn5xc0pOHuxwIY-h7yRYgMuIFEDCysDLm8jEROCQ1sA+vF5uNUolxjUpj1QZsFlCFrDuoVgHUoBoAHYHDASRC5GrmSioAm1HoxCRQG4eDDkZBwwpUmmIUlQelHVhomIROrwzE4iiofGEhjE9nk-TMkzU2kihnAymy1kIdmc27YaxK9F8wzACLgCCGIA noborder}} </WRAP> </panel> Assumptions for the basic formulas: * sinusoidal input voltage, * RMS value \(U_\sim\), * ohmic load, * ideal diode. For a half-wave rectifier: \[ \begin{align*} \boxed{ U_{\rm di} = \frac{\sqrt{2}}{\pi}U_\sim } \end{align*} \] The ripple frequency is \[ \begin{align*} f_\sigma=f. \end{align*} \] The output voltage can be split into an average DC value and an AC ripple component: \[ \begin{align*} u_{\rm out}(t) = U_{\rm di} + u_\sigma(t). \end{align*} \] Here * \(U_{\rm di}\) is the average value, i.e. the DC component, * \(u_\sigma(t)\) is the time-dependent ripple component, * \(U_\sigma\) is the RMS value of this ripple component. \[ \begin{align*} U_\sigma = \sqrt{ \frac{1}{T} \int_0^T u_\sigma^2(t)\,{\rm d}t }. \end{align*} \] The ripple factor for the ideal circuit is \[ \begin{align*} w_U = \frac{U_\sigma}{U_{\rm di}} \approx 1.21. \end{align*} \] <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> ==== Bridge rectifier B2 ==== A full-wave rectifier uses both half-waves. <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> For the bridge rectifier B2: \[ \begin{align*} \boxed{ U_{\rm di} = \frac{2\sqrt{2}}{\pi}U_\sim } \end{align*} \] The ripple frequency is \[ \begin{align*} f_\sigma=2f. \end{align*} \] The ideal ripple factor is \[ \begin{align*} w_U\approx 0.48. \end{align*} \] <panel type="info" title="Real diode voltage drops"> In a bridge rectifier, two diodes conduct at the same time. \\ Therefore, for silicon diodes, the output voltage is roughly reduced by \[ \begin{align*} 2U_{\rm TO}\approx 1.4~{\rm V}. \end{align*} \] This matters especially for low-voltage supplies. </panel> <tabcaption tab_rectifier_summary|Comparison of simple rectifier circuits> ^ Circuit ^ Uses half-waves ^ Ideal average voltage \(U_{\rm di}\) ^ Ripple frequency ^ | M1 half-wave | one half-wave | \(\frac{\sqrt{2}}{\pi}U_\sim\) | \(f\) | | B2 bridge | both half-waves | \(\frac{2\sqrt{2}}{\pi}U_\sim\) | \(2f\) | </tabcaption> ~~PAGEBREAK~~ ~~CLEARFIX~~ ==== Capacitor smoothing ==== A rectifier output is not constant. \\ A smoothing capacitor stores charge near the voltage maximum and supplies the load between maxima. <panel> <WRAP> {{url>https://www.falstad.com/circuit/circuitjs.html?running=false&ctz=DwYwlgTgBAZgvAIgIwKgFwM6IAwDpsEECsqYIiSAzPgJwDsALEkZQExLYAcSnAbClBAAjRERJQADiIRFsqAG4RRqALaZRAUwC0SFAD4AUFCjB5UAB4VWnKK1YMoSa495zYOVAHd4yVDCUIDG4qAIbm8soA9IbGwNCWgUi8UDTJTMnWnH4eUAG8QQjRRiYAJhaIrFwu2LZVqdkIwQD2iCUaMCEArgA2aIUxJhjlCJSUyalQo8lJbj5yRbEgw1PVk2O29g1u5I34hDQHh0fHNKQRuwQCGAFu8iWI1LycNPa8vKx070R0lFkLJp5lutMlB0rZOFl3I1+sVgICEisJmD6lD5gM4cNKjYZrUbHYGFsYbEygkseC8VV8YSoCoWgg2h0en1-hiEk5sa5cdVCSySRU6KxybYBRsCajVHSGV1ekSAZiRTiPoKqeLefLBYqRSi5hLWu1pcz0fCrHj7MLlZtVUagRkIebyTzraSRSCyZlHbDjQgyRMlSleB7Yl6-RNEQGrZ6baK1rbITqWU0oBoAHY5DASRD5BrmSioDPQgaxCRQc5uDA7Diyoslh6ocs4XCsWXASJNdGJlMUOv5rNQnN5ruFkzF0t1nYkf7V865qD1i7YL7N1uGFvgCCGIA noborder}} </WRAP> </panel> For a bridge rectifier with a sufficiently large smoothing capacitor, the DC voltage is approximately \[ \begin{align*} U_{\rm di} \approx \sqrt{2}U_\sim \end{align*} \] for ideal diodes and small ripple. With real silicon diodes in a bridge rectifier: \[ \begin{align*} U_{\rm di} \approx \sqrt{2}U_\sim - 2U_{\rm TO} - \frac{\Delta U}{2}. \end{align*} \] Here \(\Delta U\) is the approximate peak-to-peak ripple voltage. A simple estimate for the smoothing capacitor is \[ \begin{align*} \boxed{ C \approx \frac{I_{\rm d}}{f_\sigma\Delta U} } \end{align*} \] with * \(I_{\rm d}\): load current, * \(f_\sigma\): ripple frequency, * \(\Delta U\): allowed peak-to-peak ripple voltage. <panel type="info" title="Course approximation with RMS ripple"> If \(U_\sigma\) is used as the RMS value of the ripple voltage, a practical estimate is \[ \begin{align*} C \approx k\frac{I_{\rm d}}{f_\sigma U_\sigma}. \end{align*} \] Typical factors: \[ \begin{align*} k&=0.25 &&\text{for one-pulse rectification},\\ k&=0.20 &&\text{for two-pulse rectification}. \end{align*} \] </panel> <callout type="warning" icon="true"> 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. </callout> ===== 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@# #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1 Quick check: LED series resistor for a robot status LED #@TaskText_HTML@# A robot controller provides \[ \begin{align*} U_{\rm I}=24~{\rm V}. \end{align*} \] A green LED shall operate at \[ \begin{align*} U_{\rm F}=1.8~{\rm V}, \qquad I_{\rm F}=10~{\rm mA}. \end{align*} \] * Calculate the required series resistor \(R_{\rm V}\). * Choose a nearby standard value. * Calculate the resistor power for your calculated value. #@ResultBegin_HTML~ExerciseLEDResistor~@# The resistor value is \[ \begin{align*} R_{\rm V} &= \frac{U_{\rm I}-U_{\rm F}}{I_{\rm F}} \\ &= \frac{24~{\rm V}-1.8~{\rm V}}{10~{\rm mA}} \\ &= 2.22~{\rm k}\Omega. \end{align*} \] A suitable standard value is, for example, \[ \begin{align*} R_{\rm V}=2.2~{\rm k}\Omega. \end{align*} \] The resistor power is approximately \[ \begin{align*} P_R &= (U_{\rm I}-U_{\rm F})I_{\rm F} \\ &= 22.2~{\rm V}\cdot 10~{\rm mA} \\ &= 222~{\rm mW}. \end{align*} \] A \(0.25~{\rm W}\) resistor is very close to the limit. A \(0.5~{\rm W}\) resistor gives more margin. #@ResultEnd_HTML@# #@TaskEnd_HTML@# #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1 Quick check: Z-diode stabilizer #@TaskText_HTML@# A simple Z-diode stabilizer shall generate approximately \[ \begin{align*} U_{\rm Z}=5.1~{\rm V} \end{align*} \] from \[ \begin{align*} U_{\rm I}=12~{\rm V}. \end{align*} \] The series resistor is \[ \begin{align*} R_{\rm V}=470~\Omega. \end{align*} \] The load resistor is \[ \begin{align*} R_{\rm L}=1.0~{\rm k}\Omega. \end{align*} \] * Calculate \(I_{\rm V}\). * Calculate \(I_{\rm L}\). * Calculate \(I_{\rm Z}\). * Calculate the Z-diode power \(P_{\rm Z}\). #@ResultBegin_HTML~ExerciseZDiode~@# The current through the series resistor is \[ \begin{align*} I_{\rm V} &= \frac{U_{\rm I}-U_{\rm Z}}{R_{\rm V}} \\ &= \frac{12~{\rm V}-5.1~{\rm V}}{470~\Omega} \\ &= 14.7~{\rm mA}. \end{align*} \] The load current is \[ \begin{align*} I_{\rm L} = \frac{U_{\rm Z}}{R_{\rm L}} = \frac{5.1~{\rm V}}{1.0~{\rm k}\Omega} = 5.1~{\rm mA}. \end{align*} \] The Z-diode current is \[ \begin{align*} I_{\rm Z} = I_{\rm V}-I_{\rm L} = 14.7~{\rm mA}-5.1~{\rm mA} = 9.6~{\rm mA}. \end{align*} \] The Z-diode power is \[ \begin{align*} P_{\rm Z} = U_{\rm Z}I_{\rm Z} = 5.1~{\rm V}\cdot 9.6~{\rm mA} = 49~{\rm mW}. \end{align*} \] This is acceptable only if the datasheet permits this current and power. #@ResultEnd_HTML@# #@TaskEnd_HTML@# #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1 Quick check: freewheeling diode energy #@TaskText_HTML@# A relay coil has \[ \begin{align*} L=80~{\rm mH} \end{align*} \] and carries \[ \begin{align*} I_0=200~{\rm mA} \end{align*} \] just before switch-off. * Calculate the magnetic energy stored in the coil. * Explain why a freewheeling diode is useful. * State one disadvantage of a simple freewheeling diode. #@ResultBegin_HTML~ExerciseFreewheelEnergy~@# The stored magnetic energy is \[ \begin{align*} W_L = \frac{1}{2}LI_0^2 = \frac{1}{2}\cdot 80~{\rm mH}\cdot (200~{\rm mA})^2. \end{align*} \] Insert SI units: \[ \begin{align*} W_L = 0.5\cdot 0.080~{\rm H}\cdot (0.200~{\rm A})^2 = 1.6~{\rm mJ}. \end{align*} \] When the switch opens, this energy must go somewhere. The freewheeling diode provides a safe path for the coil current and limits the overvoltage. A disadvantage is that the coil current decays more slowly. Therefore, a relay or solenoid may release more slowly. #@ResultEnd_HTML@# #@TaskEnd_HTML@# #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1 Quick check: bridge rectifier average voltage #@TaskText_HTML@# A bridge rectifier B2 is supplied by a sinusoidal AC voltage with \[ \begin{align*} U_\sim=12~{\rm V} \end{align*} \] at \[ \begin{align*} f=50~{\rm Hz}. \end{align*} \] Assume an ohmic load and ideal diodes. * Calculate the ideal average rectified voltage \(U_{\rm di}\). * State the ripple frequency \(f_\sigma\). * Compare this with a half-wave rectifier M1 using the same \(U_\sim\). #@ResultBegin_HTML~ExerciseBridgeAverage~@# For the bridge rectifier: \[ \begin{align*} U_{\rm di,B2} = \frac{2\sqrt{2}}{\pi}U_\sim = \frac{2\sqrt{2}}{\pi}\cdot 12~{\rm V} = 10.8~{\rm V}. \end{align*} \] The ripple frequency is \[ \begin{align*} f_\sigma=2f=100~{\rm Hz}. \end{align*} \] For the half-wave rectifier: \[ \begin{align*} U_{\rm di,M1} = \frac{\sqrt{2}}{\pi}U_\sim = \frac{\sqrt{2}}{\pi}\cdot 12~{\rm V} = 5.4~{\rm V}. \end{align*} \] The bridge rectifier uses both half-waves. Therefore, the average voltage is twice as large and the ripple frequency is doubled. #@ResultEnd_HTML@# #@TaskEnd_HTML@# #@TaskTitle_HTML@##@Lvl_HTML@#~~#@ee2_taskctr#~~.1 Longer exercise: small DC supply with bridge rectifier and smoothing capacitor #@TaskText_HTML@# A \(12~{\rm V}\) RMS transformer secondary feeds a bridge rectifier with a smoothing capacitor. The mains frequency is \[ \begin{align*} f=50~{\rm Hz}. \end{align*} \] The load current is \[ \begin{align*} I_{\rm d}=250~{\rm mA}. \end{align*} \] The allowed peak-to-peak ripple voltage is \[ \begin{align*} \Delta U=1.0~{\rm V}. \end{align*} \] Assume silicon diodes with \[ \begin{align*} U_{\rm TO}=0.7~{\rm V}. \end{align*} \] * Calculate the peak value of the transformer secondary voltage. * Estimate the ripple frequency \(f_\sigma\). * Estimate the required capacitor \(C\). * Estimate the average DC output voltage with ripple and diode drops. * Explain why the transformer and diodes must tolerate current pulses. #@ResultBegin_HTML~ExerciseBridgeCapacitorSupply~@# The peak value of the secondary voltage is \[ \begin{align*} \hat{U}_\sim = \sqrt{2}U_\sim = \sqrt{2}\cdot 12~{\rm V} = 17.0~{\rm V}. \end{align*} \] For a bridge rectifier, \[ \begin{align*} f_\sigma=2f=100~{\rm Hz}. \end{align*} \] Using \[ \begin{align*} C\approx \frac{I_{\rm d}}{f_\sigma\Delta U}, \end{align*} \] we get \[ \begin{align*} C &\approx \frac{250~{\rm mA}}{100~{\rm Hz}\cdot 1.0~{\rm V}} \\ &= \frac{0.250~{\rm A}}{100~{\rm s}^{-1}\cdot 1.0~{\rm V}} \\ &= 2.5\cdot 10^{-3}~{\rm F} = 2500~\mu{\rm F}. \end{align*} \] A nearby practical value would be, for example, \[ \begin{align*} C=2200~\mu{\rm F} \quad \text{or} \quad C=3300~\mu{\rm F}, \end{align*} \] depending on the allowed ripple. In a bridge rectifier, two diodes conduct at the same time, so the diode drop is approximately \[ \begin{align*} 2U_{\rm TO}=1.4~{\rm V}. \end{align*} \] The average DC output voltage can be estimated as \[ \begin{align*} U_{\rm d} &\approx \hat{U}_\sim - 2U_{\rm TO} - \frac{\Delta U}{2} \\ &= 17.0~{\rm V} - 1.4~{\rm V} - 0.5~{\rm V} \\ &= 15.1~{\rm V}. \end{align*} \] The capacitor is recharged only near the peaks of the AC voltage. Therefore the diode current is not a smooth \(250~{\rm mA}\), but occurs in short charging pulses. The diodes, transformer, and capacitor must tolerate these pulse currents. #@ResultEnd_HTML@# #@TaskEnd_HTML@# ===== Common pitfalls ===== * **Connecting LEDs without current limitation:** The LED current can become destructive. * **Forgetting resistor power:** In \(24~{\rm V}\) control cabinets, LED resistors can dissipate noticeable heat. * **Using a Z-diode without load-current check:** The Z-current must remain between \(I_{\rm Z,min}\) and \(I_{\rm Z,max}\). * **Using clamp diodes without a series resistor:** The clamp current must be limited. * **Thinking the freewheeling diode removes energy instantly:** It gives the current a safe path, but turn-off may become slower. * **Ignoring diode drops in bridge rectifiers:** Two diodes conduct at the same time. * **Confusing RMS and peak values:** A \(12~{\rm V}\) RMS sine has a peak value of about \(17~{\rm V}\). * **Assuming a smoothing capacitor creates perfect DC:** The output still has ripple and charging-current pulses. * **Using capacitor formulas without checking ratings:** Check voltage rating, ripple current, polarity, and inrush current. ===== Embedded resources ===== <callout> The Falstad simulations are embedded directly in the relevant chapters above. </callout> ~~PAGEBREAK~~ ~~CLEARFIX~~ CKG Edit