====== 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>

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