- #1
What happens in a circuit that has a capacitor in parallel with a diode?
In summary, the circuit will shift the input up, preserving its shape (**PROVIDED** the load is very light), until the lowest voltage of the output is -0.6V.
- #2
DragonPetter
- 830
- 1
Storm Butler said:
Can you give some numbers for the waveform? Amplitude, DC bias/offset? Also, that diode is in series, not parallel.
- #3
Dickfore
- 2,987
- 5
Model the diode as a resistor in the forward bias, and an open circuit in the reversed bias. Suppose [itex]v(t)[/itex] is the potential, and [itex]i(t)[/itex] is the current through the external circuit, and let [itex]\varepsilon(t)[/itex] be the e.m.f. of the source. Then, the circuit equations become:
[tex]
i(t) - \frac{v(t)}{R_d} = C \, \left( \dot{\varepsilon}(t) - \dot{v}(t) \right), \ v(t) < 0
[/tex]
[tex]
i(t) = C \, \left( \dot{\varepsilon}(t) - \dot{v}(t) \right), \ v(t) > 0
[/tex]
You also need to supply a connection between [itex]i(t)[/itex], and [itex]v(t)[/itex]. These equations are non-linear, and require further analysis.
[tex]
i(t) - \frac{v(t)}{R_d} = C \, \left( \dot{\varepsilon}(t) - \dot{v}(t) \right), \ v(t) < 0
[/tex]
[tex]
i(t) = C \, \left( \dot{\varepsilon}(t) - \dot{v}(t) \right), \ v(t) > 0
[/tex]
You also need to supply a connection between [itex]i(t)[/itex], and [itex]v(t)[/itex]. These equations are non-linear, and require further analysis.
- #4
fbs7
- 345
- 37
The circuit in the diagram is called "clamper" or "DC restoration".
Say the input varies from +5V to -5V. Initially the capacitor has no voltage (Vc = 0). Say that the input input is -5V; as the capacitor has no charge, for a very brief moment the diode gets 5V, so it is forward polarized and there is a big surge of current to the capacitor, which charges until it has 4.4V, so the output is -0.6V. At that point the diode cuts, as it can only conduct while it has 0.6V or 0.7V.
A little time later the input changes to +5V; as the capacitor has 4.4V, the other terminal of the capacitor rises to 9.4V. The diode is reverse polarized, so it doesn't conduct anything. There will be a current to the load, but as long as the load has a large resistance (say 100k or more), the current will be small and the capacitor will keep the 4.4V charge.
Therefore, if the input of the circuit is a square wave of -Vo to +Vo, the output will be a square wave of -0.6V to 2Vo - 0.6V. That is, the circuit will shift the input up, preserving its shape (**PROVIDED** the load is very light), until the lowest voltage of the output is -0.6V.
ps: if the load is significant (say 1kΩ) or the frequency of the input is high, then the output will be distorted and the bias will not be enough to reach DC.
Say the input varies from +5V to -5V. Initially the capacitor has no voltage (Vc = 0). Say that the input input is -5V; as the capacitor has no charge, for a very brief moment the diode gets 5V, so it is forward polarized and there is a big surge of current to the capacitor, which charges until it has 4.4V, so the output is -0.6V. At that point the diode cuts, as it can only conduct while it has 0.6V or 0.7V.
A little time later the input changes to +5V; as the capacitor has 4.4V, the other terminal of the capacitor rises to 9.4V. The diode is reverse polarized, so it doesn't conduct anything. There will be a current to the load, but as long as the load has a large resistance (say 100k or more), the current will be small and the capacitor will keep the 4.4V charge.
Therefore, if the input of the circuit is a square wave of -Vo to +Vo, the output will be a square wave of -0.6V to 2Vo - 0.6V. That is, the circuit will shift the input up, preserving its shape (**PROVIDED** the load is very light), until the lowest voltage of the output is -0.6V.
ps: if the load is significant (say 1kΩ) or the frequency of the input is high, then the output will be distorted and the bias will not be enough to reach DC.
Last edited:
- #5
Related to What happens in a circuit that has a capacitor in parallel with a diode?
1. How does a capacitor affect the flow of current in a circuit with a diode?
A capacitor in parallel with a diode acts as a temporary storage unit for electrical energy. When the diode is forward biased, the capacitor charges up to the same voltage as the power source, and then discharges when the diode is reverse biased. This causes a fluctuation in the current flow through the circuit.
2. Can a capacitor and diode in parallel cause a short circuit?
No, a capacitor in parallel with a diode does not cause a short circuit. The diode only allows current to flow in one direction, while the capacitor stores energy in the opposite direction. This allows for a controlled flow of current through the circuit.
3. How does the capacitance value affect the behavior of a circuit with a capacitor and diode in parallel?
The capacitance value of the capacitor determines the amount of energy it can store. A higher capacitance value will result in a larger charge and discharge of the capacitor, causing a larger fluctuation in current flow through the circuit. A lower capacitance value will result in a smaller fluctuation in current.
4. What happens to the voltage across the diode when a capacitor is added in parallel?
Adding a capacitor in parallel with a diode does not significantly affect the voltage across the diode. The diode will still have a forward voltage drop when conducting current, and the voltage across the capacitor will fluctuate depending on its charge and discharge. However, the overall voltage across the circuit may change due to the added component.
5. Can a circuit with a capacitor in parallel with a diode be used for rectification?
Yes, a circuit with a capacitor in parallel with a diode can be used for rectification, but it is not as efficient as a traditional diode rectifier. The capacitor will only partially smooth out the AC waveform, resulting in a pulsating DC output. This can be useful in some applications, but a diode rectifier would be a more effective choice.
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