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.
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.
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
- \(D_1\),
- \(R_1\),
- \(R_2\).
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 \(S\).
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.