\[ \begin{align*} P_{\rm D}=U_{\rm D}I_{\rm D}. \end{align*} \]
This block uses the diode models from Block 11 and applies them to practical circuits.
The focus is on basic engineering estimates, not yet on detailed datasheet design.
An LED is operated in forward direction. It converts part of the electrical energy into light.
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
\[ \begin{align*} P_{\rm LED} = U_{\rm F}I_{\rm F}. \end{align*} \]
<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 | \(\approx 3.0\ldots 3.3~{\rm V}\) | \(5\ldots 20~{\rm mA}\) |
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.
LEDs tolerate only small reverse voltages. Therefore, operation directly at AC voltage needs protection.
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.
A Z-diode is operated in reverse breakdown. In its working range, the voltage is approximately constant:
\[ \begin{align*} u_{\rm Z}\approx U_{\rm Z}. \end{align*} \]
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
\[ \begin{align*} P_{\rm Z} = U_{\rm Z}I_{\rm Z} \leq P_{\rm tot}. \end{align*} \]
\[ \begin{align*} {\color{blue}{I_{\rm V}}} &= \frac{U_{\rm E}-U_{\rm Z}}{R_{\rm V}} &&\text{current supplied through the series resistor}, \\ {\color{green}{I_{\rm L}}} &= \frac{U_{\rm Z}}{R_{\rm L}} &&\text{useful load current}, \\ {\color{red}{I_{\rm Z}}} &= {\color{blue}{I_{\rm V}}} - {\color{green}{I_{\rm L}}} &&\text{remaining current through the Z-diode}. \end{align*} \]
The Z-diode can stabilize the voltage only if \({\color{red}{I_{\rm Z}}}\) remains inside the allowed range.
Use this simulation to observe how a Z-diode limits the output voltage.
Things to try:
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.
The coil is like a flywheel for current.
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.
Use this simulation to observe the overvoltage when switching an inductive load, and how a diode limits it.
Things to try:
Microcontroller and sensor inputs tolerate only a limited voltage range. Clamp diodes can conduct disturbances away from the sensitive input.
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:
\[ \begin{align*} I_{\rm clamp} \approx \frac{U_{\rm disturb}-U_{\rm clamp}}{R_{\rm V}}. \end{align*} \]
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.
A rectifier converts an AC voltage into a unidirectional voltage.
Assumptions for the basic formulas:
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
\[ \begin{align*} w_U = \frac{U_\sigma}{U_{\rm di}} \approx 1.21. \end{align*} \]
Use this simulation to observe how one half-wave is removed by a diode.
Things to try:
A full-wave rectifier uses both half-waves.
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
\[ \begin{align*} w_U\approx 0.48. \end{align*} \]
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.
<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 | \(\frac{2\sqrt{2}}{\pi}U_\sim\) | \(2f\) |
Use this simulation to compare half-wave and full-wave rectification.
Things to try:
A rectifier output is not constant. A smoothing capacitor stores charge near the voltage maximum and supplies the load between maxima.
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
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:
\[ \begin{align*} k&=0.25 &&\text{for one-pulse rectification},\\ k&=0.20 &&\text{for two-pulse rectification}. \end{align*} \]
<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? |
A robot controller provides
\[ \begin{align*} U_{\rm E}=24~{\rm V}. \end{align*} \]
A green LED shall operate at
\[ \begin{align*} U_{\rm F}=1.8~{\rm V}, \qquad I_{\rm F}=10~{\rm mA}. \end{align*} \]
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.
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
\[ \begin{align*} R_{\rm L}=1.0~{\rm k}\Omega. \end{align*} \]
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.
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.
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.
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.
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.
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
\[ \begin{align*} U_{\rm TO}=0.7~{\rm V}. \end{align*} \]
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.