# Why do capacitors need both leads?

#### jaygatsby

Joined Nov 23, 2011
182
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

#### praondevou

Joined Jul 9, 2011
2,942
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?
What do you mean by "air has 0 potential"?

The bypass capacitor needs to be connected from + to - with low resistance (via the leads). Air doesn't have low resistance. If you leave the GND pin open it is the same as if you put a VERY high resistance value in series with the capacitor.

If what you said was true, you wouldn't even need to connect the negative pole of your battery to your circuit. See the difference?

#### MrChips

Joined Oct 2, 2009
23,245
Electronics is all about current (and electron flow). If the other wire is not connected, electrons have nowhere to flow. You must complete the circuit (loop).

Joined Aug 6, 2010
65
the capacitor must be connected to 0v because current follows the path of least resistance. without the other lead connected, the current would skip past the capacitor like it wasnt there.
Air has a very large resistance, it takes a high potential difference to pass a current through it.

#### strantor

Joined Oct 3, 2010
5,537
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
Dirt is conductive, despite what your meter says. Especially when it is moist. The deeper you go in the ground, the more likely you are to find moisture, which is why grounding rods need to be pounded like 4ft in the ground. Also, because the dirt is 3dimensional, unlike a pcb trace, the resistivity plays a big part. There is dirt contacting the grounding rod on all sides for 4ft which makes an almost nil resistance. I think (not 100% sure, someone correct me) that the further you go down, the resistance would decay exponentialy due to the 3dimensional resistivity.
Air, however has an extremely high resistance up until you hit is breakdown voltage,, and then it ionizes the air and a spark gap happens.

#### thatoneguy

Joined Feb 19, 2009
6,359
Air, however has an extremely high resistance up until you hit is breakdown voltage,, and then it ionizes the air and a spark gap happens.
Air breakdown to plasma gradient is approximately 2,000Volts/Inch (Rule of Thumb)

Ground resistance can be measured, in dry dirt, if your meter supports the higher resistance ranges, actually conduction ranges, in Siemens.

Also think about air being conductive. Before applying power, one would need to conformal coat both sides of the board so it wasn't shorted out by the air.

#### JMac3108

Joined Aug 16, 2010
348
Jay,

Forget about ground for a second. Your capacitor (along with all your other components) needs to be part of a complete path so that current can flow. Both leads must be connected to complete the circuit.

The concept of ground is not so complicated really. You have a circuit that is a closed loop. You specify a single point on the circuit that you call ground. This is your reference point that all other voltages are measured with respec to, and it is assigned the value 0V. Since voltage measurements are potential "differences" you have to measure them at two points. Assigning ground simply gives you reference point to measure to.

Now it happens that for safety reasons we tie our circuit ground to the earth and call it earth ground. Don't let this confuse you.

Another example is a hand-held battery-powered device that does not plug into the AC mains. We assign ground to the negative side of the battery. Again, this is just a reference point.

Joined Dec 26, 2010
2,148
@ jaygatsby: If what you said was true, every wire in contact with the air would be at zero potential, which is obviously wrong.

A disconnected wire may appear to be at zero potential if measured with a practical meter of only moderate sensitivity, but this is by no means the same thing as saying that it will remain so if connected to something else, like a bypass capacitor for instance. In fact, with sufficiently sensitive measurement a wire in the air will normally be seen to have potentials induced on it by electromagnetic fields - this is how radio and TV antennas work.

Have you not seen circuits in which a device such as a lamp or a relay is switched on and off by a device in its common return - if a wire in the air always held zero potential, this could not work, the lamps etc would remain on.

If the capacitor's return connection to common were not necessary, neither would be the return connections of any other components, and we would have a situation where any device could be powered merely by connecting one wire to it. As many of us may have discovered when playing with batteries and bulbs as youngsters, this is not so.

This is electrical theory at its most basic level: the concept of a complete circuit, which can be stated more more formally in terms of Kirchhoff's Laws. You would do well to get this clearly sorted out in your mind.

#### joeyd999

Joined Jun 6, 2011
4,477
...current follows the path of least resistance...
Where does this misconception come from??? And why is it propagated?

By this logic, if I have a 1K resistor in parallel with a 5K resistor, the current will "choose" to follow the 1K path, and ignore the 5K path. Ludicrous!

#### Potato Pudding

Joined Jun 11, 2010
688
The open lead would technically form a capacitance of extremely small value in series with the bypass capacitor and if you look up the rule for capacitors in series you will see that it no longer matters how large the bypass capacitor is the small capacitance at the open lead will be the dominant factor in determining the series capacitance.

You could leave the bypass cap out entirely instead of leaving an open lead. That part of the circuit won't work either way.

The series capacitors stop conducting when their total charges equal the applied voltage and that small capacitance at the open lead will drop all of the voltage almost immediately. The amount of conduction to charge a stray capacitance will be very small and very brief. So small and fast that we would probably say there was no conduction at all.

#### Potato Pudding

Joined Jun 11, 2010
688
Where does this misconception come from??? And why is it propagated?

By this logic, if I have a 1K resistor in parallel with a 5K resistor, the current will "choose" to follow the 1K path, and ignore the 5K path. Ludicrous!
It should probably be said instead that current prefers the path of least resistance.

#### K7GUH

Joined Jan 28, 2011
190
Originally Posted by joeyd999
Where does this misconception come from??? And why is it propagated?

Perhaps some budding scientist conducted an elementary experiment and wrote down his conclusions. When all else fails, try following a good example.

#### thatoneguy

Joined Feb 19, 2009
6,359
It should probably be said instead that current prefers the path of least resistance.
Still not true, and confusing to people new to electronics.

If they were shown a 1Ω resistor in parallel with a 1MΩ resistor, and asked how much current would flow through the 1MΩ, most would say "none".

Joined Aug 6, 2010
65
easy there, sorry if im being confusing but that is a fairly common saying in electronics (at least in ireland anyway) and i should have realised that it would be confusing for a beginner.
it would be more correct to say it prefers as someone said earlier. my bad

#### joeyd999

Joined Jun 6, 2011
4,477
easy there, sorry if im being confusing but that is a fairly common saying in electronics (at least in ireland anyway) and i should have realised that it would be confusing for a beginner.
it would be more correct to say it prefers as someone said earlier. my bad
I don't think electrons have preferences...if so, they wouldn't follow very explicit rules (except by choice???). Then again, I've never bothered to ask

#### Potato Pudding

Joined Jun 11, 2010
688
More than 9 out of 10 electrons agree that given parallel paths across the same potential, a 10k resistor is a better way to drop than a 100k resistor!

#### Lundwall_Paul

Joined Oct 18, 2011
228
joey i like your comment. "if I have a 1K resistor in parallel with a 5K resistor, the current will "choose" to follow the 1K path, and ignore the 5K". This line can be used somewhere not sure where. This is too funny.

Joined Dec 26, 2010
2,148
@jaygatsby: If you think about it, can you see that if a wire in the air was as good as a ground, all wires in the universe would be effectively be grounded. This would make it impossible to use electricity at all, as the whole universal system would just be one enormous short-circuit?

It would be interesting to hear if any of our replies have helped to clear this matter up, or if further help is needed. If the contradiction explained above is not clear to you, please come back and explain why - you may be heading for difficulties if these ideas are not understood.

If you work as a technician, it would really help you to understand grounding, both as applied to signals, and in protective systems. This may affect your own and other peoples' safety.

#### jaygatsby

Joined Nov 23, 2011
182
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?

#### praondevou

Joined Jul 9, 2011
2,942
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?
A capacitor is an "open circuit" for DC, not for AC. There is no sense in having a capacitor in your circuit if the voltage applied to it or the current into or out of it would never change.
A bypass capacitor does not have pure DC applied to it, thatoneguys article described that quite well. You put it in because there are load changes. You don't want to draw any peak currents from your power supply but from your bypass capacitors. So, if the current in and out of the cap changes that means it is not DC, it is AC.

A capacitor has an impedance that depends on the frequency of this AC, the higher the frequency the lower its impedance.

Since almost every practical circuit has a dynamic behaviour you should start looking at capacitors like a frequency dependend component, same as wires and traces that act as inductors.

Everything else is pure theory and simplification.