Where does constant voltage come from?

[Sanity1234]

http://www.edexcel.com/migrationdoc...um 2000/june2010-qp/6PH04_01_que_20100618.pdf

Hi, I'm preparing for my exams, but I can't understand something about the question Q16 (part b i).

The answer is:

Capacitor stores charge/charges up (1)
(If voltage is constant) capacitor doesn’t discharge (1)

But, where is constant voltage? Is it across a capacitor or in a power supply?

As far as I know, voltage across the capacitor matches the voltage of power supply, and because we basically have no resistance the time lag between increase/decrease in voltage of power supply and change in voltage across the capacitor is small.

So, if the voltage of the power supply changes, and the voltage across the capacitor changes (matches with the change in power supply) as well, where does constant voltage come from?

Thanks.

 

[Studiot]

Well I think you should have another look at 16 (a) which described the power supply.

Did you call the graph constant?

Come back if this hint is not enough.

 

[Sanity1234]

Well I think you should have another look at 16 (a) which described the power supply.

Did you call the graph constant?

Come back if this hint is not enough.

Not sure, if I understand your question. Values on a graph are not constant. But I suspect there's something wrong with my reasoning.

 

[Studiot]

Well both parts of the question were really about how 'constant' is a constant power supply.

Part (a) showed a graph having a nearly 6 volt variation. This had peaks and troughs, although it was unidirectional and could therefore be called direct voltage.

The troughs reached down nearly to zero so what would supply the current to any load during these periods of time in the waveform?
Say you wanted 5 volts across the load what happens when the graph is less than this?

A capacitor charges up to the peak voltage of the supply and stores the energy inside.
So in the when the waveform is less than 5volts ( in my example) the capacitor maintains the voltage across the load.

Obviously it can't do this indefinitely and its own voltage drops as it supplies current to the load.

However if it is recharged every peak it 'smoothes out' the variation shown in 16(a)

 

[Sanity1234]

Well both parts of the question were really about how 'constant' is a constant power supply.

Part (a) showed a graph having a nearly 6 volt variation. This had peaks and troughs, although it was unidirectional and could therefore be called direct voltage.

The troughs reached down nearly to zero so what would supply the current to any load during these periods of time in the waveform?
Say you wanted 5 volts across the load what happens when the graph is less than this?

A capacitor charges up to the peak voltage of the supply and stores the energy inside.
So in the when the waveform is less than 5volts ( in my example) the capacitor maintains the voltage across the load.

Obviously it can't do this indefinitely and its own voltage drops as it supplies current to the load.

However if it is recharged every peak it 'smoothes out' the variation shown in 16(a)

In part b i, no load is applied.

Is it right to say that voltage across the capacitor will match the voltage of power supply? So, it charges up to 5V, and then the voltage of power supply goes down, will the voltage across the capacitor go down as well?


I understand what you're saying, and it's one of the application of capacitors to smooth DC voltage after applying rectifiers, but it doesn't help me to see where constant voltage occurs.

 

[gneill]

In part b i, no load is applied.

Is it right to say that voltage across the capacitor will match the voltage of power supply? So, it charges up to 5V, and then the voltage of power supply goes down, will the voltage across the capacitor go down as well?

If the output of the power supply is via a rectifier, then no, the capacitor voltage will not drop after it is charged --- current cannot flow back into the power supply from the capacitor because the rectifier will block it. However, if some load is also connected then the capacitor voltage will dip as it supplies current to the load. The dip will be less for larger capacitance. The power supply will "top up" the charge on the capacitor whenever the supply voltage exceeds the capacitor voltage. This repeated dip and rise in the supplied voltage is called the "ripple" voltage, since it looks like ripples superimposed on a steady DC value.

I understand what you're saying, and it's one of the application of capacitors to smooth DC voltage after applying rectifiers, but it doesn't help me to see where constant voltage occurs.

By choosing a large enough capacitor the amount of ripple in the voltage supplied to the load can be made arbitrarily small. For a given load's requirements, beyond some value of "arbitrarily small" the output can be called constant for practical purposes.

 

[Sanity1234]

If the output of the power supply is via a rectifier, then no, the capacitor voltage will not drop after it is charged --- current cannot flow back into the power supply from the capacitor because the rectifier will block it. However, if some load is also connected then the capacitor voltage will dip as it supplies current to the load. The dip will be less for larger capacitance. The power supply will "top up" the charge on the capacitor whenever the supply voltage exceeds the capacitor voltage. This repeated dip and rise in the supplied voltage is called the "ripple" voltage, since it looks like ripples superimposed on a steady DC value.By choosing a large enough capacitor the amount of ripple in the voltage supplied to the load can be made arbitrarily small. For a given load's requirements, beyond some value of "arbitrarily small" the output can be called constant for practical purposes.

What if we don't have a rectifier? In this situation it's not mentioned (and they're not on a syllabus of the exam).

Is it right to say that voltage across the capacitor will match the voltage of power supply? So, it charges up to 5V, and then the voltage of power supply goes down, will the voltage across the capacitor go down as well in this case?

And then, where is a constant voltage?

 

[gneill]

What if we don't have a rectifier? In this situation it's not mentioned (and they're not on a syllabus of the exam).

A power supply (interpreted to be a device that takes a source of AC current and coverts it to a DC current) without a rectifier would be an oddity. But if such a thing were assumed then the output would behave like a variable battery and drive the voltage on the capacitor accordingly. This situation would NOT allow you to smooth the output with a capacitor.

Is it right to say that voltage across the capacitor will match the voltage of power supply? So, it charges up to 5V, and then the voltage of power supply goes down, will the voltage across the capacitor go down as well in this case?

In the oddball case where the power supply isn't rectified but acts like some kind of signal generator as you've described, then the capacitor voltage would follow it. NOTE that this is definitely NOT the usual case for a typical power supply. I'm certain that the course material authors are assuming a typical power supply with rectified output.

And then, where is a constant voltage?

Nowhere. If you create a situation where you can't smooth the output because you choose not to have a rectified output, then you take the consequences of your choice.

 

[MrWarlock616]

wow gneill, never would have thought of that.. your explanation was awesome.

 

[CWatters]

What if we don't have a rectifier? In this situation it's not mentioned (and they're not on a syllabus of the exam).

If rectifiers really aren't on the syllabus I consider this question unfair. Perhaps even worthy of a complaint. The question relies on you recognising that the waveform has most likely been produced by a rectifier from AC. The existence of the rectifier is key to understanding how a capacitor can convert it to DC. If you programmed an ideal voltage source (eg a signal generator) to produce that waveform and fed it directly to a capacitor you would get a different result.

Aside:

Many years ago I had a summer job working for an exam board. We were only meant to check the examiner had marked every page. One day I was checking physics exam papers and noticed that the examiner had incorrectly marked a students work. The question required the student to draw a 4 diode bridge rectifier. Most people had drawn the classic diamond with two diodes pointing up and two down like this...

http://forum.allaboutcircuits.com/cache.php?url=http://static.electro-tech-online.com/imgcache/11158-3523.jpg

however this student had drawn it a different, but equally correct, way. Electrically he had drawn exactly what was was asked for but because he hadn't drawn it that way the examiner had marked it as wrong! I brought it to the attention of my supervisor but never heard what happened. I wondered how many others had been marked incorrectly.

 

[NascentOxygen]

In part b i, no load is applied.

Is it right to say that voltage across the capacitor will match the voltage of power supply? So, it charges up to 5V, and then the voltage of power supply goes down, will the voltage across the capacitor go down as well?

That's a good point, and makes that part of the question a poor assessment tool (unless your class has studied power supply design). I can see how the examiner came to that wording, but the question is open to criticism.