Into capacitor, half wave rectifier diode conducts for?

In summary: So then the answer is 180 degrees.In summary, the time for which the diode conducts in a circuit with a capacitor and a sinusoidal input is dependent on the initial charge of the capacitor. If the capacitor is initially uncharged, the diode will conduct for the entire first quarter-cycle of the sinusoid. However, if there is a load in parallel with the capacitor, the diode will conduct for part of each cycle after the initial quarter-cycle. The amount of time the diode conducts after the initial quarter-cycle is typically 180 degrees.
  • #1
jaus tail
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Homework Statement


upload_2017-1-18_12-59-25.png

The time for which the diode conducts is

Homework Equations


Using integration and differentiation Vc = 1 / C integral (current)(dt)

The Attempt at a Solution


Vs = -Vc
Vm sin(wt) = -Vc = - (1/C) integral (Idt)
now differentiate both sides to get expression for I that is current.
Vm (coswt) * w = -(1/C) I
So I = Vm(costwt) * C * w (-1)

So now i draw current wave, and then how to proceed? How do you get an answer for this? Am i to find out when current goes negative?

There was a similar question for RL circuit that was


Vs = L di/dt
so integrate to get expression for current.
Vs sin(wt) = L di/dt
Vs sin (wt) dt = L di
integrating
Vs cos (wt) (-1) / w + k = L * I
I = Vs cos (wt) (-1) /(wL) + k' where k' = k/L equation 1

to find value of K' take initial condition. At t = 0, I = 0
so 0 = Vs (1) (-1) / (wL) + k' from equation 1

K' = Vs/(wL) equation 2

so expression for current becomes from equation 1 and 2

I = Vs cos(wt) (-1) / (wL) + Vs / (wL) = Vs/ (wL) * (1 - cos (wt))

Drawing this we get I is always positive, so it conducts for 360 degrees when input goes 1 entire cycle.
360 degrees was right for RL circuit but for RC answer is not given as the book is old and the page is torn.


i'm not sure what to do in case of given question of diode + C with charged as shown.
Expression is
I = Vm(costwt) * C * w (-1)

So what to now?
 

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  • #2
For the circuit shown, if the capacitor is initially uncharged then the diode will conduct for the entire first ¼-cycle of sinusoid V.sin ωt. After that, the diode will never conduct again.

BUT your thread title says "RC" so shouldn't you have included a resistance in parallel to the capacitor?
 
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  • #3
Thanks, Quarter means 45 degrees right?
I was unable to find how to change title but found now. Learnt 2 new things today.:smile:
 
  • #4
jaus tail said:
Quarter means 45 degrees right?
A full cycle is 360°, a half-cycle is 180°, so a quarter-cycle is ⬜❓
 
  • #5
I don't like the question now. Are you sure it's how you want it? Almost no mathematics required.
 
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  • #6
NascentOxygen said:
I don't like the question now. Are you sure it's how you want it? Almost no mathematics required.
Quarter cycle is 90 degrees. Darn i am so stupid.

We do need maths to find equation of current by integrating... Vc = 1/C (integration) (Idt)
I = Vm(costwt) * C * w (-1)
upload_2017-1-18_13-54-36.png

This is current waveform. So it'll flow from 0 to 90 degrees.

Thanks.
 
  • #7
NascentOxygen said:
I don't like the question now. Are you sure it's how you want it? Almost no mathematics required.

+1

Usually there is a load in parallel with the capacitor that causes it to discharge. This in turn causes the diode to conduct for part of each cycle (after the initial 1/4 cycle). In fact problems like this usually ignore the initial 1/4 cycle and only want you to calculate the time per cycle after that.
 
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Related to Into capacitor, half wave rectifier diode conducts for?

1. What is a capacitor?

A capacitor is an electronic component that stores electrical energy in an electric field. It is made of two conductive plates separated by an insulating material, and its capacity to store charge is measured in farads (F).

2. How does a half wave rectifier diode work?

A half wave rectifier diode is a semiconductor device that allows current to flow in only one direction. When an alternating current (AC) signal is applied to the diode, it conducts current during one half of the cycle and blocks it during the other half, resulting in a pulsating direct current (DC) output.

3. What is the purpose of a half wave rectifier diode in a circuit?

The purpose of a half wave rectifier diode is to convert an AC signal to a DC signal. It is commonly used in power supplies to convert the AC current from the wall outlet to a usable DC current for electronic devices.

4. How does a capacitor affect the output of a half wave rectifier diode?

The capacitor in a half wave rectifier circuit helps to smooth out the pulsating DC output. It charges up during the conducting phase of the diode and discharges during the non-conducting phase, resulting in a more constant DC voltage. The larger the capacitor, the smoother the output will be.

5. What are the advantages and disadvantages of using a half wave rectifier diode?

The main advantage of using a half wave rectifier diode is its simplicity and low cost. It also has a low voltage drop, meaning less energy is lost during the conversion process. However, it has a lower efficiency compared to other rectifier circuits, and the output voltage is not as smooth as a full wave rectifier. Additionally, it only conducts during one half of the AC cycle, resulting in a lower average output voltage.

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