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Exercise E1.1 Circuit with multiple diodes: which lamps light up?
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.
Simulation: multiple diodes and lamps
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.
3. Explain the role of the diodes.
A conducting diode can create a low-voltage bypass path.
That means:
\[ \begin{align*} \text{forward-biased diode} \quad \Rightarrow \quad \text{small voltage drop} \end{align*} \]
If this low-voltage path is parallel to a lamp or to part of the lamp chain, the lamp voltage becomes too small for bright operation.
Therefore, only the outer lamps receive enough voltage to light brightly.
Exercise E2.1 Circuit with multiple diodes II: current calculation
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
- \(D_1\),
- \(R_1\),
- \(R_2\).
Simulation: two diodes and two resistors
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:
\[ \begin{align*} I_{D1} &= 17~{\rm mA}+28~{\rm mA} \\ &= 45~{\rm mA}. \end{align*} \]
Exercise E3.1 Circuit with multiple diodes III: switch-dependent currents
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 \(S\).
Simulation: switch-dependent diode circuit
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.