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Capacitor DC charging question.

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lucky6969b

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In this circuit during the first cycle, the capacitor is charged up, but there isn't any DC component in the source.
How come will there be smoothing out effects on the load?
Thanks
Jack
 
Last edited:

Consider the AC generator to be single discrete DC values at one moment in time and evaluate the effect of that voltage value on the circuit
 
The capacitor stores juice received during the positive waveform. Then it releases it during the gap.

The capacitor must have a sufficiently high farad value, in order to power the load during the gap. That is smoothing action.

If it is too small a value, then it discharges too quickly, and there is little smoothing action.

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The diode blocks current from going in reverse. Therefore it prevents the capacitor from discharging current backwards into the source during the negative waveform.
 
lucky6969b

In this circuit during the first cycle, the capacitor is charged up, but there isn't any DC component in the source.
How come will there be smoothing out effects on the load?
Thanks
Jack

That question is answered extensively in the technical literature both in texts and on the net. You should not have to ask that question here unless you have a specific inquiry regarding half-wave rectification.

Ratch
 
the point is this is a half wave rectifier(diode) with a filter circuit (cap)

the diode allows only the positive half cycle thereby there is a pulsating dc op from the diode then the capacitor gets charged from this positive cycle till it reaches the maximum voltage

in the negative cycle no current flows from the diode and so the cap starts to discharge and there is almost a near constant dc op with a very small ripple in the resistor (load)
 
During the positive cycle, the capacitor will get charged. However, ac current will flow through the cap rather than establishing potential difference between the plates. All the energy should be "dissipated" during that time. That's what I don't understand.
 

lucky6969b,

During the positive cycle, the capacitor will get charged.

What does that mean? The charge accumulated on one plate of the cap will be the same as the charge depleted on the opposite plate. There will be a charge imbalance between the plates, but the net charge of a capacitor never changes whether the voltage across its plates is 0 volts or 100 volts. Therefore a capacitor is never "charged", it is energized. The energy stored in a capacitor is proportion to the square of the voltage.

However, ac current will flow through the cap rather than establishing potential difference between the plates

Current is charge flow, therefore "current flow" means "charge flow flow". Is that what you want to say? Current exists, but charge does not flow twice.

Charges never flow through a capacitor unless it is leaky. That is true no matter if the charges are AC, DC, or anything in-between.

So what is happening during half-wave rectification with a capacitor? During the first quarter-cycle, the capacitor is energized to the same voltage applied across the load. During the rest of the cycle, when the input voltage decreases, the capacitor supplies energy to the load, which causes the voltage of the capacitor to drop. Sometime at the beginning of the next input cycle, the capacitor will again be energized to the peak input voltage, and the process starts anew. Always remember that a capacitor is an electrical energy storage device whose energy level is determined by the voltage across its plates. Anything that decreases that voltage depletes its stored energy.

Ratch
 
During the positive cycle, the capacitor will get charged. However, ac current will flow through the cap rather than establishing potential difference between the plates. All the energy should be "dissipated" during that time. That's what I don't understand.

The waveform goes high and low cyclically. So it resembles alternating current on the face of it.

However notice that the waveform stays entirely in one polarity only. This is varying DC.

To visualize AC, picture electrons going back and forth in the wire. If something stops flow in either direction (such as the diode), then we don't have AC.

Even though it's not the same as alternating current, nevertheless it could be called AC riding a DC component.

The capacitor filters a portion of the AC component, by passing it to ground. There is that way of looking at it.

The result is that a smoother DC waveform goes to the load.
 
capacitor time constant is more compared to the time period of the input signal hence it responds slowly
therefore smoothing effect takes place if capacitor value is less then no smoothing effect takes place
 
you have misconceived the op from the diode as ac current it is not so that form is called pulsating DC and never AC

that should put you out of lots of misery
 
I just read my physics book today, something is related to the capacitor -
The displacement current. How does it relate to the operation of a capacitor?
Thanks
Jack
 

i mean it will give same pulsating dc if capacitor is of less value
 
Maybe a link to read?
I think it's better to think of it in physics perspective.
 
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pulsating dc means it contains dc value as well as ac components. if you want know the dc and ac components expand it using fourier series then you will get amount of dc in the output
 
We can't say a sinusoidal waveform has a fundamental frequency of 0Hz because the lowest frequency is 0, it's DC? Right?
Thanks
Jack
 

just see this pic this is ac and dc

sine can never be dc and can not have fundamental period of 0

ac dc.JPG
 

Positive varying voltages/currents in the first half of the cycle and non-varying second half are also called DC?
That's what it is called pulsating?
Thanks
Jack
 

just think i dono whether i am correct and all but

DC in general has no negative VOLTAGE COMPONENT

and AC alters between POSITIVE AND NEGATIVE VOLTAGE in a period of time
 

dc maybe -ve or +ve but it must not vary
 

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