What happens in a circuit that has a capacitor in parallel with a diode?

[Storm Butler]

I attached a PNG of the simple schematic. I know that the cuircuit should shift the waveform that's coming from the source. What i don't understand is why or by how much it will be shifted up.
Any help in explaining this is appreciated.

 

[DragonPetter]

I attached a PNG of the simple schematic. I know that the cuircuit should shift the waveform that's coming from the source. What i don't understand is why or by how much it will be shifted up.
Any help in explaining this is appreciated.

Can you give some numbers for the waveform? Amplitude, DC bias/offset? Also, that diode is in series, not parallel.

 

[Dickfore]

Model the diode as a resistor in the forward bias, and an open circuit in the reversed bias. Suppose $v(t)$ is the potential, and $i(t)$ is the current through the external circuit, and let $\varepsilon(t)$ be the e.m.f. of the source. Then, the circuit equations become:

$$i(t) - \frac{v(t)}{R_d} = C \, \left( \dot{\varepsilon}(t) - \dot{v}(t) \right), \ v(t) < 0$$

$$i(t) = C \, \left( \dot{\varepsilon}(t) - \dot{v}(t) \right), \ v(t) > 0$$

You also need to supply a connection between $i(t)$, and $v(t)$. These equations are non-linear, and require further analysis.

 

[fbs7]

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.

 

[Jony130]

This imagine will help you understand how the circuit works