- #1
What happens in a circuit that has a capacitor in parallel with a diode?
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Discussion Overview
The discussion centers around the behavior of a circuit that includes a capacitor in parallel with a diode, particularly focusing on how the waveform from the source is affected. Participants explore the shifting of the waveform and the implications of circuit components on the output signal.
Discussion Character
- Exploratory
- Technical explanation
- Debate/contested
Main Points Raised
- Some participants express uncertainty about the extent and reason for the waveform shift caused by the circuit configuration.
- One participant suggests modeling the diode as a resistor in forward bias and an open circuit in reverse bias, presenting equations that describe the circuit behavior under different conditions.
- Another participant describes the circuit as a "clamper" or "DC restoration" and provides a detailed example of how the capacitor charges and affects the output voltage based on the input waveform.
- Concerns are raised about the impact of load resistance on the output, indicating that a significant load may distort the output signal and affect the DC bias.
Areas of Agreement / Disagreement
Participants have not reached a consensus on the specifics of the waveform shift or the exact behavior of the circuit under different conditions. Multiple competing views and interpretations remain present in the discussion.
Contextual Notes
Participants mention various assumptions regarding the load resistance and input waveform characteristics, which may influence the circuit's behavior. The discussion includes non-linear equations that require further analysis, indicating potential complexities in the circuit's response.
- #2
DragonPetter
- 831
- 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
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