\[ \begin{align*} P_{\rm D}=U_{\rm D}I_{\rm D}. \end{align*} \]
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.
A diode must usually be operated with a current-limiting element.
For the circuit in figure 1 the loop equation is
\[ \begin{align*} U_{\rm E} = U_R+U_{\rm D}. \end{align*} \]
With the constant-voltage model,
\[ \begin{align*} U_{\rm D}\approx U_{\rm TO}. \end{align*} \]
Therefore
\[ \begin{align*} I_{\rm D} \approx \frac{U_{\rm E}-U_{\rm TO}}{R}. \end{align*} \]
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 \(U_{\rm E}\), an LED forward voltage \(U_{\rm F}\), and a desired LED current \(I_{\rm F}\), the series resistor is
\[ \begin{align*} \boxed{ R_{\rm V} = \frac{U_{\rm E}-U_{\rm F}}{I_{\rm F}} } \end{align*} \]
For circuit design it is important the check the real resistor power with the absolute maximum ratings of the resistors
\[ \begin{align*} P_R = (U_{\rm E}-U_{\rm F})I_{\rm F} = R_{\rm V}I_{\rm F}^2 \quad \overset{!}{<} \quad P_{R, \rm max} \end{align*} \]
The LED power is approximately
\[ \begin{align*} P_{\rm LED} = U_{\rm F}I_{\rm F}. \end{align*} \]
| 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.
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:
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 \(U_{\rm Z}\).
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
\[ \begin{align*} u_{\rm Z} \approx U_{\rm Z}+r_{\rm Z} \cdot i_{\rm Z}. \end{align*} \]
A typical application is a Z-diode stabilizer
Use this simulation to observe how a Z-diode limits the output voltage.
Things to try:
\[ \begin{align*} {\color{blue }{I_{\rm V}}} &= \frac{U_{\rm E}-U_{\rm Z}}{R_{\rm V}} &&\text{current supplied through the series resistor}, \\ {\color{green}{I_{\rm L}}} &= \frac{U_{\rm Z}}{R_{\rm L}} &&\text{useful load current}, \\ {\color{red}{I_{\rm Z}}} &= {\color{blue}{I_{\rm V}}} - {\color{green}{I_{\rm L}}} &&\text{remaining current through the Z-diode}. \end{align*} \]
The Z-diode can stabilize the voltage only if \({\color{red}{I_{\rm Z}}}\) remains inside the allowed range:
\[ \begin{align*} I_{\rm Z,min} \leq {\color{red}{I_{\rm Z}}} \leq I_{\rm Z,max}. \end{align*} \]
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.
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 like a motor, and how a diode limits it.
Things to try:
A rectifier converts an AC voltage into a unidirectional voltage.
Use this simulation to observe how one half-wave is removed by a diode.
Things to try:
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 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
\[ \begin{align*} U_\sigma = \sqrt{ \frac{1}{T} \int_0^T u_\sigma^2(t)\,{\rm d}t }. \end{align*} \]
The ripple factor for the ideal circuit is
\[ \begin{align*} w_U = \frac{U_\sigma}{U_{\rm di}} \approx 1.21. \end{align*} \]
A full-wave rectifier uses both half-waves.
Use this simulation to compare half-wave and full-wave rectification.
Things to try:
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.
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:
\[ \begin{align*} k&=0.25 &&\text{for one-pulse rectification},\\ k&=0.20 &&\text{for two-pulse rectification}. \end{align*} \]
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.
1. Determine which lamps light up brightly.
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.
\[ \begin{align*} L_1 \quad \text{and} \quad L_5 \end{align*} \]
light up brightly.
2. Determine which lamps remain dark.
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.
\[ \begin{align*} L_2,\;L_3,\;L_4 \end{align*} \]
remain dark or almost dark.
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
1. Calculate the current through \(R_1\).
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:
\[ \begin{align*} I_{R1} = \frac{U_{R1}}{R_1} = \frac{3.4~{\rm V}}{200~\Omega} = 17~{\rm mA}. \end{align*} \]
2. Calculate the current through \(R_2\).
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:
\[ \begin{align*} I_{R2} = \frac{U_{R2}}{R_2} = \frac{2.8~{\rm V}}{100~\Omega} = 28~{\rm mA}. \end{align*} \]
3. Calculate the current through \(D_1\).
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:
\[ \begin{align*} I_{D1} &= 17~{\rm mA}+28~{\rm mA} \\ &= 45~{\rm mA}. \end{align*} \]
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
depending on the switch state \(S\).
1. Calculate the currents for open switch \(S\).
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,
\[ \begin{align*} I_{D1}=I_{R1}, \qquad I_{D2}=0. \end{align*} \]
\[ \begin{align*} I_{R1}&=4.3~{\rm mA}, \\ I_{D1}&=4.3~{\rm mA}, \\ I_{D2}&=0. \end{align*} \]
2. Calculate the currents for closed switch \(S\).
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:
\[ \begin{align*} I_{D1} \approx I_{D2} \approx \frac{4.3~{\rm mA}}{2} = 2.15~{\rm mA}. \end{align*} \]
\[ \begin{align*} I_{R1}=4.3~{\rm mA} \end{align*} \]
and
\[ \begin{align*} I_{D1}+I_{D2}=4.3~{\rm mA}. \end{align*} \]
For approximately identical real diodes:
\[ \begin{align*} I_{D1} \approx I_{D2} \approx 2.15~{\rm mA}. \end{align*} \]
3. Explain why the current sharing is not unique in the simple model.
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.
\[ \begin{align*} I_{D1}+I_{D2}=4.3~{\rm mA}. \end{align*} \]
It does not uniquely determine \(I_{D1}\) and \(I_{D2}\) separately.
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.