How Does a Diode Affect Capacitor Voltage in an AC-DC Conversion Circuit?

[mtr]

Homework Statement

There is a circuit which consists of a capacitor with capacity C, a diode and a source of changing EMF such that during time $$\frac{1}{2}T$$ EMF equals E and during another half of T it is -E (current flows in the opposite direction). What will be the voltage on the capacitor after a long time?

Homework Equations

$$C=\frac{Q}{U}$$

The Attempt at a Solution

First of all, I don't understand what is the role of the diode in this circuit. Does it only influence on the direction of the current? And in fact I do not have any reasonable idea how to solve this whole problem.

 

[Irid]

A diode is a device which allows current in one direction, but not the other. Thus, the capacitor can be charged, but not discharged.

 

[mtr]

So after a long time the capacitor is charged completely, but still how can I find the voltage? Does it equal EMF? Or maybe I should find Q somehow and use $$C=\frac{Q}{U}$$?

 

[Defennder]

You didn't provide the angular frequency of the voltage source. What 1/2 T here? The period of the sinusoidal AC voltage source? Is the source sinusoidal in the first place, or is it a square wave?

Note that the voltage across the capacitor and the source is always the same at all times because it is connected across the capacitor.

 

[mtr]

You didn't provide the angular frequency of the voltage source. What 1/2 T here? The period of the sinusoidal AC voltage source? Is the source sinusoidal in the first place, or is it a square wave?

Note that the voltage across the capacitor and the source is always the same at all times because it is connected across the capacitor.

Actually, the EMF is produced by a pendulum in a magnetic field, with period 2T.

Isn't there any drop of voltage that is connected with the diode?

 

[Irid]

Isn't there any drop of voltage that is connected with the diode?

In an ideal diode there is no voltage drop. In a realistic case you should know the internal resistance of the diode, but it shouldn't be much greater than that of connecting wires, etc.

 

[mtr]

So it seems, that I have to find the maximum value of the EMF and it will be the voltage on the capacitor. Am I right?
On the other hand what is the capacity C for? It is among task's data, but it hasn't been used yet.

 

[Irid]

I came to the same conclusion myself.

 

[mtr]

What about C?

 

[Irid]

Strange problem... Anyway, you can use dimensional analysis arguments to show that you can't use C (or, more likely, $$1/\omega C$$) and E to obtain a voltage, because capacity introduces another dimension of ohms, which doesn't cancel with anything. Probably, if you had a resistor somewhere, you could and should incorporate C somehow.

 

[mtr]

Strange problem... Anyway, you can use dimensional analysis arguments to show that you can't use C (or, more likely, $$1/\omega C$$) and E to obtain a voltage, because capacity introduces another dimension of ohms, which doesn't cancel with anything. Probably, if you had a resistor somewhere, you could and should incorporate C somehow.

I don't exactly understand what you mean. How can we obtain a voltage using C? And is E the EMF or the stored energy in the capacitor?

 

[Redbelly98]

You didn't provide the angular frequency of the voltage source. What 1/2 T here? The period of the sinusoidal AC voltage source? Is the source sinusoidal in the first place, or is it a square wave?

Note that the voltage across the capacitor and the source is always the same at all times because it is connected across the capacitor.

No circuit diagram was provided, but I assume the diode and capacitor are connected in series. The EMF is applied to that series combination, not directly to the capacitor. This is the typical way to introduce the concept of a rectifier.

The forward voltage drop of a real diode is typically 0.6 to 0.8 V in practice. It's not clear if they are assuming an ideal (Vf = 0) or a real diode though.

 

[mtr]

No circuit diagram was provided, but I assume the diode and capacitor are connected in series. The EMF is applied to that series combination, not directly to the capacitor. This is the typical way to introduce the concept of a rectifier.

The forward voltage drop of a real diode is typically 0.6 to 0.8 V in practice. It's not clear if they are assuming an ideal (Vf = 0) or a real diode though.

That's right, the capacitor and the diode are connected in series. I suppose they are considering here and ideal diode, otherwise there should be something about its properties.

 

[Redbelly98]

So it seems, that I have to find the maximum value of the EMF and it will be the voltage on the capacitor. Am I right?

Yes.

At least that's right for an "ideal diode" that has no voltage drop across it.

For a real diode, the capacitor voltage would be about 0.6 to 0.8V less than E.

E = Vdiode + Vcap
Vcap = E - Vdiode

Ideal diode:
Vcap = E - 0 = E

Real diode:
Vcap = E - (about 0.6 to 0.8 V)

 

[mtr]

Yes.

At least that's right for an "ideal diode" that has no voltage drop across it.

For a real diode, the capacitor voltage would be about 0.6 to 0.8V less than E.

E = Vdiode + Vcap
Vcap = E - Vdiode

Ideal diode:
Vcap = E - 0 = E

Real diode:
Vcap = E - (about 0.6 to 0.8 V)

That's true, but still there is the capacity which wasn't used. I think that there might be something about impedance, as Irid suggested.

 

[Irid]

I don't exactly understand what you mean. How can we obtain a voltage using C? And is E the EMF or the stored energy in the capacitor?

That's the point, we can't. Look at it this way. You have some physical quantities (E, C, ...) which are measured in some units (V, F, ...). Now you must combine these quantities in some way to obtain the voltage across capacitor - U. How can you do it? It's trivial to see that in this problem the only possible way is to make

$$U=E$$

up to a dimensionless constant. You can't plug in C anywhere here, because it will ruin the dimensions. Volts must equal Volts, and unless the Farads have some other Farads to cancel with, they have no place in the answer.
If you still have doubts, search the web for dimensional analysis.

 

[Redbelly98]

What this circuit does is, given an AC voltage source, a DC voltage is created which can be used to run DC electronic devices.

For this homework assignment, it only matters that C > 0.

The value of C comes into play once you connect something which will draw current. Drawing a current will reduced the charge, and hence the voltage (=Q/C), during the half-cycle where the EMF is negative. It's part of electronic design to choose a large enough C so that the drop in voltage is acceptably small.