Capacitor Charging and Discharging

[mark_3094]

Hi Everyone,

I have some more notes that I've been working on.
This is supposed to provide a basic walk-through of capacitor charging and discharging.
I've used a very basic circuit that simply charges and discharges a capacitor through a resistor and LED

I welcome comments on it's accuracy, and hopefully I've avoided being misleading

I look forward to hearing your opinions!

 

[tresca]

I dont think the Resistor and Capacitor section is correct.
This capacitor voltage is time dependent, you cannot say that I = V/R

First order circuits are pretty easy. There are 2 ways, one uses differential equations, the other is alot easier.

x(t) = x(infinite) + [ x(0) - x(infinite)] exp(-t/tau) <--- general form

A capacitor acts as an open circuit at t=infinite. So the voltage across it would be your source voltage. In this case, 9V. At t=0, we will assume that power was not applied, and at t=0 you just hit the switch so that the battery is connected in the charging circuit. Since a capacitors voltage cannot change instantaneously, at t=0, even though the battery is connected is still 0V. Apply these values into the equation above, and thats how you get your equation for voltage.

The discharging is the same thing except that v(infinite)=0 (since we are discharging) and v(0)=9V (a fully charged capacitor)

For a single order circuit like this...the charge time for C1 would be approx
v(t) = 9-9exp(-t/R1C1).
This accounts for all values of time.

Discharging would be
v(t)= 9exp(-t/R1C1)

And so because the voltage is time dependent, current also becomes time dependent. So yes, Ohms Law is used, but I = v(t) / R instead of I = V/R

Hope that makes sense.

 

[mark_3094]

Do you mean the 5xRxC part, or using the different R and C values to get the same result part?

 

[tresca]

No that I = V/R is not entirely correct.
If you connect the battery to the circuit, at t=0 you will get 1.6mA of current flowing through the LED (but only at t=0 will that be true) However, as time increases, or your time constants, the current in your circuit will begin to decrease until all current stops flowing since the capacitor will act as an open circuit.

 

[transistance]

Basically, your current will start trickling as time passes so you can only say I=V/R at t=0 - as soon as the switch activates. I guess you could say I = V/R given that V across your power supplies will decrease as well in which case you should calculate the V in terms of time.

 

[colin55]

I have never heard the expression "the capacitor acts as an open circuit."
When the capacitor is fully charged it produces a voltage equal to the charging voltage but of the opposite polarity. The net result is this: there is no voltage available to charge the capacitor and thus no current flows.
Your description of initial current flow (when the capacitor is uncharged) is correct, and is good for the beginner, but you should also point out that the current decreases as observed by the LED dimming.

Don't go into any complex discussion on capacitor-charging as you will lose your reader. Keep it simple as it is.

 

[transistance]

I have never heard the expression "the capacitor acts as an open circuit."

This is a very common expression used in ECE101 & ECE102 courses. Well, at least I remember Purdue University professors using it frequently in beginner courses.

 

[ericgibbs]

I
When the capacitor is fully charged it produces a voltage equal to the charging voltage but of the opposite polarity.

hi,
I think this statement is misleading, it not 'opposite polarity', its the same polarity as the charging voltage.

 

[colin55]

The voltage produced by the charged capacitor is opposing the charging voltage in this case.
If you put all the components in a straight line, you will see the voltage produced by the capacitor is opposite to the battery. Replace the capacitor with a battery and you will see why no current flows.


I don't like the expression "acts as an open circuit" because the capacitor in fact acts a very heavy dampener to any change in voltage and it is ready to do this when the voltage changes.

 

[ericgibbs]

The voltage produced by the charged capacitor is opposing the charging voltage in this case.

This is not what your previous post stated, 'opposite polarity', now you are saying opposing which is correct, ie: at the same polarity.

Ive heard the 'acts an open circuit' reference to caps, which I also think is misleading.
I would think 'appears' an open circuit to dc, is a better choice.

 

[colin55]

The "polarity is opposing" and "it is of opposite polarity". Both are the same thing.

What do you mean by: "at the same polarity."

The capacitor is not: "at the same polarity" when it is fully charged.

 

[ericgibbs]

The "polarity is opposing" and "it is of opposite polarity". Both are the same thing.

What do you mean by: "at the same polarity."

The capacitor is not: "at the same polarity" when it is fully charged.

I would suggest you post a sketch of what you are trying to say.

 

[flat5]

I'm often wrong, but what I visualize is the + of the supply is connected to the + of the cap. so not opposite in the sense that the circuit has (say) a left side + and a right side - when charging.
When the cap discharges, the current flows in the wires in the opposite direction.

 

[colin55]

The circuit is in the "Charging and Discharging" document and I am simply replying to the request to read the document and rectify the mistakes.

 

[ericgibbs]

The circuit is in the "Charging and Discharging" document and I am simply replying to the request to read the document and rectify the mistakes.

hi,
Im sure that you realise that when you make a statement on an open Forum that your peers will expect you to explain
any ambiguity or error in your posts.

Its not personal or nit picking, its just that many 'learners' read these forums and IMHO the content should be correct.

 

[transistance]

i feel heat...

I think Eric's change on the words would make the article more intuitive though.

 

[ericgibbs]

i feel heat...

I think Eric's change on the words would make the article more intuitive though.

hi t,
No heat.

When I've made incorrect or misleading comments, I have had my 'nuts' tweaked by my Forum peers and rightly so.!

After all that what an open forum is, an exchange of ideas and opinions

 

[flat5]

Sorry, I did not read the doc file because graphics were in the file and Wordpad can't deal with that. I did not want to use Open Office. So I was just following the discussion.

 

[mark_3094]

Thanks Colin55. You have been a great source of assistance (in multiple threads) in clarifying what I'm learning.
That goes for everyone else too!

Thanks