Why do capacitors need both leads?

Adjuster

Joined Dec 26, 2010
2,148
Thanks all for the great info. BUT this brings up another question. Some of you have said that there must be a closed loop.... this is not the case with a capacitor! There is no loop through it, but a difference of potential on the plates which allows an electric field to exist inside the cap. So... difference in potential, not the path, is key here right? So I could keep the + lead in the circuit, put the - lead to any ground, even earth ground, and still get the field in the cap. Right? Wrong?
There is a lot of confusion going on here. I have put a statement you have made in bold, because I think it is important to get this straight. The fact that a capacitor blocks DC does not in the least alter the requirement to have a loop. AC circuits require to be complete just as much as DC ones do.

Actually, the function of a bypass capacitor frequently requires it to be placed as physically close as possible to the point in the circuit where it is required, with both the "hot" and ground leads kept as short as possible to keep down parasitic inductance and resistance.

Edit: Connecting a bypass capacitor to a physically remote ground might get the same DC field in the capacitor, but generally it might not fulfil the purpose of providing a low-impedance local path for supply current fluctuations. The result might for instance be the creation of ground loop problems, possibly resulting in interference pickup or instability.
 
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MrChips

Joined Oct 2, 2009
23,227
Let's say I use a bypass cap, as outlined in thatoneguy's excellent post about their usage. Why can't I just put the positive lead on the Vcc breadboard power rail, and let the ground lead hang? Air has 0 potential, the same as ground, right? So what's the difference?

This is a general issue I have with understanding ground, actually. For instance, I don't understand why dirt, which has 0 potential and conductivity if I put my meter into it, can attract current or act as 0 reference, whereas air cannot, which according to my meter has the same properties!

Thank you
There are three flaws in this original post.

1) The potential of air is NOT 0V.

2) The potential of the other lead, unconnected (floating) lead of the capacitor is NOT at 0V.

3) Hence as such, the capacitor is not acting as a bypass capacitor and would serve ZERO function in an electronic circuit.
 

Adjuster

Joined Dec 26, 2010
2,148
It may be worth pointing out that these ideas are particularly important for RF work, where parasitic inductances may become critical. As frequencies increase, it becomes more and more important to adhere to the rules for good grounding and bypassing. It eventually becomes impossible to think in terms of a single ground potential everywhere, as transmission line effects become significant even on a scale of a short PCB connection.

One particularly poisonous example is where a common impedance - maybe that of just a short PCB track - is shared between the input and output ground connections of a high-frequency amplifier. With enough gain at a high enough frequency, just millimetres too much can be a recipe for poor frequency responses or even oscillations.
 

Thread Starter

jaygatsby

Joined Nov 23, 2011
182
There are three flaws in this original post.

1) The potential of air is NOT 0V.

2) The potential of the other lead, unconnected (floating) lead of the capacitor is NOT at 0V.

3) Hence as such, the capacitor is not acting as a bypass capacitor and would serve ZERO function in an electronic circuit.
If the potential of air is not 0V, then it must be something other than 0V. That something is probably a different potential than the positive rail that I will plug the capacitor + lead into. So there is a difference of potential?

I understand that my theory is very wrong here, but I'm showing you my point of view so that I can keep getting all this great information.... eventually I'll understand this.
 

praondevou

Joined Jul 9, 2011
2,942
If the potential of air is not 0V, then it must be something other than 0V.
For all practical beginner projects just forget air, it has a negligble effect on your circuit. Just see it as non-existant.

If you really want to then look at it as a very very very big resistance although I don't know what that would be good for.

Considering "air potential" or "air resistance" in simple practical low voltage circuits is absolutely useless.

Or are you trying to build a laser induced plasma channel gun???
 
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Adjuster

Joined Dec 26, 2010
2,148
This is really a very fundamental misunderstanding that you have here. I feel that it is likely to give you an awful lot of trouble if you do not get past it.

Try imagining an AC mains lamp, connected to 120V live on one side, and back to the neutral via a large capacitor: the lamp lights up. Now disconnect the capacitor from the neutral: the lamp goes out.
 

MrChips

Joined Oct 2, 2009
23,227
Every metal has the capacity to hold a charge (electrons).

The end of the capacitor that is not connected has some capacity, but in this case it is very small. The earth has a huge capacity to hold electrons.

If I were to add one extra electron to the unconnected lead of the capacitor I would alter its potential, analogous to placing one drop of water into an drinking glass. Adding one extra electron to the earth is akin to adding one drop of water to the ocean.
 

strantor

Joined Oct 3, 2010
5,534
Guys I don't think he's considering air as a current path in the circuit. I think the problem here is a variation of the old "measuring voltage on a point". Voltage can only exist between 2 points in the same circuit. As you have said, but I'm not convinced you understand, potential. if you have an isolated 12VDC power supply and you place your meter leads, one on the + terminal, and one on the grounding rod of your house (earth), what will you read? Nothing! because they are not in the same circuit; the power supply is isolated from ground. If you take that same power supply, and you put one lead on the positive terminal, and one lead in the air, what will you read? nothing as well. Because the air is not in the same circuit just like the earth ground wasn't in the same circuit. Now, you take your lead and you place one on the positive terminal, and one on the negative terminal, you will read 12V, because those 2 terminals are in the same circuit and a voltage exists between them; a difference in potential. a difference in potential must exist between 2 points within the same circuit. So, a capacitor acts like your meter. if you take one lead of the cap and place it on the positive terminal of the p/s and one terminal on earth ground, there will still be no voltage across it, no charge inside it, and it won't do anything. same thing if you put one cap lead on the positive terminal and one cap in the air; the 2 leads do not exist within the same circuit, so no charge is present, and no voltage across it. now, if you take the 2 cap leads, put one on + and one on -, now the 2 leads of the cap exist in the same circuit, and the cap builds a charge. having only one lead of the cap existing in the circuit, means that the cap functionally does not exist at all in the circuit, because like voltage itself, the cap is only a function of the potential difference between 2 points, and air is not a point. having only one lead on the positive is about as effective as sticking a toothpick in your +V rail.
 

Thread Starter

jaygatsby

Joined Nov 23, 2011
182
Guys I don't think he's considering air as a current path in the circuit. I think the problem here is a variation of the old "measuring voltage on a point". Voltage can only exist between 2 points in the same circuit. As you have said, but I'm not convinced you understand, potential. if you have an isolated 12VDC power supply and you place your meter leads, one on the + terminal, and one on the grounding rod of your house (earth), what will you read? Nothing! because they are not in the same circuit; the power supply is isolated from ground. If you take that same power supply, and you put one lead on the positive terminal, and one lead in the air, what will you read? nothing as well. Because the air is not in the same circuit just like the earth ground wasn't in the same circuit. Now, you take your lead and you place one on the positive terminal, and one on the negative terminal, you will read 12V, because those 2 terminals are in the same circuit and a voltage exists between them; a difference in potential. a difference in potential must exist between 2 points within the same circuit. So, a capacitor acts like your meter. if you take one lead of the cap and place it on the positive terminal of the p/s and one terminal on earth ground, there will still be no voltage across it, no charge inside it, and it won't do anything. same thing if you put one cap lead on the positive terminal and one cap in the air; the 2 leads do not exist within the same circuit, so no charge is present, and no voltage across it. now, if you take the 2 cap leads, put one on + and one on -, now the 2 leads of the cap exist in the same circuit, and the cap builds a charge. having only one lead of the cap existing in the circuit, means that the cap functionally does not exist at all in the circuit, because like voltage itself, the cap is only a function of the potential difference between 2 points, and air is not a point. having only one lead on the positive is about as effective as sticking a toothpick in your +V rail.
This is the second time that you've given me an answer that really has helped me out. Your ability to educate is in welcome but stark contrast to your avatar, sir.
 

t_n_k

Joined Mar 6, 2009
5,455
Some other thoughts.

It might be possible to put some numbers to a hypothetical 'thought experiment'.

Suppose one lead of a 1nF capacitor is connected to an earthed 100V DC supply. Suppose also that the external metal lead (including the unconnected capacitor 'plate') of the capacitor is physically positioned such that its capacitance to earth is 1 pico Farad - an extreme overestimate - but that doesn't matter for the purposes of the thought experiment.

Effectively the situation is equivalent to two capacitors in series between the 100V supply and earth - a capacitive divider in fact. The equivalent series capacitance from the connected end of the capacitor to ground would be the order of 0.999pF. The voltage at the unconnected end w.r.t ground would be close to 99.9V with a drop of about 100mV across the 1nF capacitor.

Could you measure the voltage at the unconnected end of the capacitor with respect to ground? Yes with an ideal voltmeter. Trouble is an ideal voltmeter doesn't exist. The moment you tried to measure the voltage with a real voltmeter the voltmeter internal resistance would discharge the 1pF capacitance to ground.
 
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Adjuster

Joined Dec 26, 2010
2,148
Edit: This was made in answer to a question about whether voltages would "different" circuits in neighbouring houses, but I cannot find this now.

It would be better to say that they are different parts of a larger network. Potentials in mains circuits are customarily tied to Earth, connections being made between neutral wires and ground at certain points as specified by the wiring codes applicable to the region concerned. It is therefore likely that the voltages in your house wiring and that of your neighbour will be related in a predictable way.

On the other hand, you could not say the same about the potentials in the wiring say of a helicopter hovering above your house.
 
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