DW EditShow pageOld revisionsBacklinksAdd to bookExport to PDFFold/unfold allBack to top This page is read only. You can view the source, but not change it. Ask your administrator if you think this is wrong. ====== Block 12 — Diode Applications ====== ===== Learning objectives ===== <callout> After this 90-minute block, you can * 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, M2, 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 ===== ==== LED with series resistor ==== An LED is operated in forward direction. It converts part of the electrical energy into light. <WRAP> <panel type="default"> <imgcaption fig_led_series_resistor|LED with current-limiting series resistor.></imgcaption> {{drawio>block12_led_series_resistor.svg}} </panel> </WRAP> 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 \[ \begin{align*} \boxed{ R_{\rm V} = \frac{U_{\rm E}-U_{\rm F}}{I_{\rm F}} } \end{align*} \] The resistor power is \[ \begin{align*} P_R = (U_{\rm E}-U_{\rm F})I_{\rm F} = R_{\rm V}I_{\rm F}^2. \end{align*} \] 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 | \(\approx 3.0\ldots 3.3~{\rm V}\) | \(5\ldots 20~{\rm mA}\) | <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> ==== LED operation with AC voltage ==== LEDs tolerate only small reverse voltages. Therefore, operation directly at AC voltage needs protection. <WRAP> <panel type="default"> <imgcaption fig_led_ac_protection|LED operation with AC voltage and reverse-voltage protection.></imgcaption> {{drawio>block12_led_ac_protection.svg}} </panel> </WRAP> A second diode can be placed antiparallel to the LED. Then, during the reverse half-wave, the normal diode conducts and limits the reverse voltage across the LED. <callout> For low-frequency indicator circuits, a visible flicker may occur if only one half-wave is used. For higher quality indicators, rectification or dedicated LED drivers are used. </callout> ==== Z-diode voltage limitation and stabilization ==== A Z-diode is operated in reverse breakdown. In its working range, the voltage is approximately constant: \[ \begin{align*} u_{\rm Z}\approx U_{\rm Z}. \end{align*} \] <WRAP> <panel type="default"> <imgcaption fig_z_diode_stabilizer|Simple Z-diode voltage stabilizer with load resistor.></imgcaption> {{drawio>block12_z_diode_stabilizer.svg}} </panel> </WRAP> The input current through the series resistor is \[ \begin{align*} I_{\rm V} = \frac{U_{\rm E}-U_{\rm Z}}{R_{\rm V}}. \end{align*} \] The load current is \[ \begin{align*} I_{\rm L} = \frac{U_{\rm Z}}{R_{\rm L}}. \end{align*} \] The Z-diode current is \[ \begin{align*} \boxed{ I_{\rm Z} = I_{\rm V}-I_{\rm L} } \end{align*} \] and must stay in the allowed operating range: \[ \begin{align*} I_{\rm Z,min} \leq I_{\rm Z} \leq I_{\rm Z,max}. \end{align*} \] The power limit is \[ \begin{align*} P_{\rm Z} = U_{\rm Z}I_{\rm Z} \leq P_{\rm tot}. \end{align*} \] <panel type="info" title="Color scheme for the Z-diode stabilizer"> \[ \begin{align*} {\color{blue}{I_{\rm V}}} &= \frac{U_{\rm E}-U_{\rm Z}}{R_{\rm V}} &&\text{current supplied through the series resistor}, \\ {\color{green}{I_{\rm L}}} &= \frac{U_{\rm Z}}{R_{\rm L}} &&\text{useful load current}, \\ {\color{red}{I_{\rm Z}}} &= {\color{blue}{I_{\rm V}}} - {\color{green}{I_{\rm L}}} &&\text{remaining current through the Z-diode}. \end{align*} \] The Z-diode can stabilize the voltage only if \({\color{red}{I_{\rm Z}}}\) remains inside the allowed range. </panel> <callout type="warning" icon="true"> 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. </callout> <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. {{url>https://www.falstad.com/circuit/e-zenerref.html 700,500 noborder}} </panel> ~~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. <WRAP group> <WRAP column half> <panel type="default"> <imgcaption fig_inductive_switch_without_diode|Switching an inductive load without freewheeling diode: dangerous overvoltage can occur.></imgcaption> {{drawio>block12_inductive_load_without_diode.svg}} </panel> </WRAP> <WRAP column half> <panel type="default"> <imgcaption fig_inductive_switch_with_diode|Switching an inductive load with freewheeling diode: the current has a safe path.></imgcaption> {{drawio>block12_inductive_load_with_freewheel_diode.svg}} </panel> </WRAP> </WRAP> 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, 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. {{url>https://www.falstad.com/circuit/e-inductkick-block.html 700,500 noborder}} </panel> ==== Clamp diodes for sensitive inputs ==== Microcontroller and sensor inputs tolerate only a limited voltage range. Clamp diodes can conduct disturbances away from the sensitive input. <WRAP> <panel type="default"> <imgcaption fig_input_clamp_diodes|Clamp diodes protecting a sensitive input against overvoltage and undervoltage.></imgcaption> {{drawio>block12_input_clamp_diodes.svg}} </panel> </WRAP> For a \(5~{\rm V}\) input, the input node is often clamped approximately to \[ \begin{align*} -0.7~{\rm V} \lesssim u_{\rm in} \lesssim 5.7~{\rm V}. \end{align*} \] The resistor \(R_{\rm V}\) limits the clamp current: \[ \begin{align*} I_{\rm clamp} \approx \frac{U_{\rm disturb}-U_{\rm clamp}}{R_{\rm V}}. \end{align*} \] <callout type="danger" icon="true"> Clamp diodes are not a substitute for proper EMC design. For external connectors, use suitable protection components and check the datasheets. </callout> <panel type="info" title="Mechatronics example"> A sensor cable near a motor cable can pick up short disturbance pulses. Clamp diodes can prevent the input voltage from exceeding the allowed range, while the series resistor limits the injected current. </panel> ~~PAGEBREAK~~ ~~CLEARFIX~~ ==== Half-wave rectifier M1 ==== A rectifier converts an AC voltage into a unidirectional voltage. <WRAP> <panel type="default"> <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> 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 ripple factor for the ideal M1 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. </callout> <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, * compare input and output voltage. {{url>https://www.falstad.com/circuit/e-rectify.html 700,500 noborder}} </panel> ==== Center-tap rectifier M2 and bridge rectifier B2 ==== A full-wave rectifier uses both half-waves. <WRAP group> <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> <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: \[ \begin{align*} U_{\rm di} = \frac{2\sqrt{2}}{\pi}U_{1{\rm N}} = \frac{\sqrt{2}}{\pi}U_{\rm S}. \end{align*} \] 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. 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 | \(\frac{\sqrt{2}}{\pi}U_\sim\) | \(f\) | | M2 center-tap | both half-waves | \(\frac{2\sqrt{2}}{\pi}U_{1{\rm N}}\) | \(2f\) | | B2 bridge | both half-waves | \(\frac{2\sqrt{2}}{\pi}U_\sim\) | \(2f\) | <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~~ ==== Capacitor smoothing ==== A rectifier output is not constant. A smoothing capacitor stores charge near the voltage maximum and supplies the load between maxima. <WRAP> <panel type="default"> <imgcaption fig_bridge_rectifier_with_capacitor|Bridge rectifier with smoothing capacitor.></imgcaption> {{drawio>block12_bridge_rectifier_with_capacitor.svg}} </panel> </WRAP> 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> ==== Application overview ==== <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 #@TaskText_HTML@# A robot controller provides \[ \begin{align*} U_{\rm E}=24~{\rm V}. \end{align*} \] 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 \[ \begin{align*} R_{\rm V} &= \frac{U_{\rm E}-U_{\rm F}}{I_{\rm F}} \\ &= \frac{24~{\rm V}-1.8~{\rm V}}{10~{\rm mA}} \\ &= 2.22~{\rm k}\Omega. \end{align*} \] A suitable standard value is, for example, \[ \begin{align*} R_{\rm V}=2.2~{\rm k}\Omega. \end{align*} \] The resistor power is approximately \[ \begin{align*} P_R &= (U_{\rm E}-U_{\rm F})I_{\rm F} \\ &= 22.2~{\rm V}\cdot 10~{\rm mA} \\ &= 222~{\rm mW}. \end{align*} \] 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 \[ \begin{align*} U_{\rm E}=12~{\rm V}. \end{align*} \] 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 \[ \begin{align*} I_{\rm V} &= \frac{U_{\rm E}-U_{\rm Z}}{R_{\rm V}} \\ &= \frac{12~{\rm V}-5.1~{\rm V}}{470~\Omega} \\ &= 14.7~{\rm mA}. \end{align*} \] 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