1
$\begingroup$

I have been building my first simple circuits with just resistors, LED's wire, and alkaline batteries.

Now I would like to add capacitors to my components box. Read a couple of articles online on the matter and it seems pretty deep, so to get the next step in the learning process going, I wanted to use the same technique I did with resistors and LED's, which is to plug a capacitor or two up and make some examinations.

but first, I wanted to straighten a vague point in my notes on the concept of capacitors:

Between the plates of a capacitor is the dielectric, is the built up charge between the two plates eventually suppose to traverse through the dielectric?

Improve this question
$\endgroup$
0
2
$\begingroup$

Here is a good link on Capacitors. Capacitor is a energy storage device. Capacitor can act like a battery. Stored energy is discharged to a device like an LED. Take a look at this capacitor battery LED resistor animation.

enter image description here

Improve this answer
$\endgroup$
3
  • $\begingroup$ you really should add more text information here to have a good answer. Don't make folks follow links for fairly basic info. $\endgroup$ – agentp Sep 26 '17 at 14:24
  • 1
    $\begingroup$ @agentp Thank you for the feedback. I will add more text when I find free time. I have day job that needs full attention. :-) $\endgroup$ – user8055 Sep 26 '17 at 14:59
  • $\begingroup$ @user8055 So do we ! $\endgroup$ – Donald Gibson Sep 26 '17 at 20:40
0
$\begingroup$

First I want to address some fundamental capacitor properties for you to consider:

  1. Certain capacitors have specific polarity markings that you need to take into account and insert them correctly into your circuit. Not doing so can cause damage to your components. If you don't see any indication of polarity, you should be okay with either direction you place it.
  2. As briefly mentioned in another answer, capacitors are "energy storage devices", specifically charge. Current is simply the flow of charge carriers, so a capacitor will prevent this flow while it is charging and when it discharges, the current will flow.

So, think about the circuit implemented in user8055's answer. A switching circuit needs to be thought of as 2 pseudo-independent circuits

First: When the switch is initially creating the smaller circuit of the DC source and the capacitor, charge builds up in the capacitor. As long as the switch is in the first position and not connected to the resistor/LED branch, this charge will not go anywhere and current will not flow, as the capacitor is fully charged due to the source constantly supplying voltage to it, thus keeping it in an equilibrium state.

In this first state, think of the capacitor as a fully charged battery.

Second: As soon as the switch flips to the other branch, the capacitor is no longer being supplied with power, and it begins to discharge. This discharge creates a movement of charge carriers (current) that passes through the resistor and LED, lighting it up. Make sure you notice the DC source isn't actually powering anything at this time, as it isn't part of a complete circuit. The only source of energy is the charged capacitor. Eventually, the capacitor will fully discharge, and the current will stop flowing, thus the LED turning off, due to not having any power source.

Another thing to consider is how a capacitor will work when the circuit is driven by an AC source, instead of DC. Depending on the configuration of the resistor and capacitor in the circuit, this is how filters are created to restrict specific frequency ranges.

Improve this answer
$\endgroup$
4
  • $\begingroup$ What was bothering me is how a negative charge could build up on the negative terminal/plate of the capacitor, even though the dielectric between the plates pretty much creates an open circuit. But this is what I know now: 1.) the size of the space between the plates, and thus the dielectric itself, helps dictate the potential for capacitance of a capacitor, the smaller the volume of the dielectric, the greater the capacitance will be. I believe this is because there exists an electric field between the two plates, and it will be stronger if the opposite charges are closer to each other... $\endgroup$ – Iam Pyre Sep 28 '17 at 2:38
  • $\begingroup$ If you are specifically considering parallel plate capacitors, the capacitance is dependent on the area of the plates, the dielectric constant of the dielectric itself, as the distance between the plates. Other types of capacitors will have slightly different things to consider, but parallel plate caps are the most fundamental to consider. And, in regards to the field: the field within the cap will balance with the field generated by the source when it is fully charged. That's when it truly is acting as an open circuit. Up until this equilibrium point, there is a transient current flowing. $\endgroup$ – jthom Sep 28 '17 at 2:45
  • $\begingroup$ ...(phone app cut me off) It just wasn't making sense to me how a charge could build up without electron flow, but I believe I get that part now, which Q=CV pretty much wraps up. Im trying to build a simple circuit in which i can make observations about capacitance using an LED and a resistor(pretty hard because it's my first "multi-path" circuit), but from what I read, I don't have large enough capacitors to afford me a time constant that would give me anything measurable. I have two 220micro-farad caps, that's the best I can do until i get a shipment. $\endgroup$ – Iam Pyre Sep 28 '17 at 2:45
  • $\begingroup$ There is a transient current happening, this is what is causing the charge build up. Then, an induced charge builds up on the other side of the cap. $\endgroup$ – jthom Sep 28 '17 at 2:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.