On a 250V capacitor or a CRT, is the negative terminal -250V?

[Starglider]

Hi, I'm doing a little explanation of how capacitors work, leading to how CRTs are basically big capacitors.

Is it correct that a 250V capacitor or a CRT stores (for the sake of argument):
• +250V on the positive plate/terminal
• -250V on the negative

Or is it 0V on the negative?

What I'm trying to do is explain why you short circuit a cap or a CRT to discharge it.

Mathematically:

+250
-250
=
0

But if it isn't -250V on the negative then that explanation doesn't work [emoji848]

Thanks!

Sent from my SM-N950U1 using Tapatalk

[Brumby]

You're understanding and associated maths needs some work - but I won't get too deep into that right now.

The very first thing you need to understand is that there is no such thing as absolute voltage.  Voltage is RELATIVE.

What this means is that you must select a reference point before you can measure any voltage.  Typically, this reference point is where you would - in practice - connect the black lead of your DMM.  In the case of your capacitor, this is most commonly the "negative" terminal.  When you then place the red lead to the same point, the reading will (obviously) be 0V.  This is not silliness!  This is the definition of the reference point.  Leaving the black lead where it is, if we now move the red lead to the "positive" of your charged capacitor, it will now read 250V.  That is to say, the potential on the "positive" connection of the capacitor is a positive 250V with respect to the negative terminal.

Here is where it is appropriate to clarify what the +ve and -ve ends of any polarised capacitor really mean.  Very simply - the positive terminal must have a voltage which is positive with respect to the negative terminal and the negative terminal must have a voltage which is negative with respect to the positive terminal.

You can never say there is +250V on one terminal and -250V on the other terminal - as the difference between the two is now a mathematical 500V.  This is just plain wrong.

[Starglider]

Well, that's why I'm here.

That makes a lot of sense, thank you.

So it would be correct instead to say this?:

"The negative plate stores -250V relative to the positive, and the positive plate stores +250V relative to the negative."

If so, can you put into terms a ten year old could understand, why shorting the cap leads to it holding a charge of 0V instead of 250V?

I liked the equation idea that connecting -250 +250 leaves 0. But apparently that's not right.

Sent from my SM-N950U1 using Tapatalk

[Brumby]

In the case where you have a circuit with a number of components in it, you can easily have a capacitor (or any other component for that matter) which is not connected directly to the 0V connection - as specified in a schematic, for example.

It is quite possible that you could have the negative terminal of your 250V capacitor sitting at a voltage of +50V with respect to the circuit's 0V reference.  With the capacitor still charged to 250V, if you measure the voltage on its positive terminal with respect to the circuit's 0V reference, you will get a reading of 300V.

Similarly, you could have another circuit where the negative terminal of your 250V capacitor sitting at a voltage of -50V with respect to the circuit's 0V reference.  With the capacitor still charged to 250V, if you measure the voltage on its positive terminal with respect to the circuit's 0V reference, you will get a reading of 200V.

In all cases I have given, the capacitor has the same 250 volts and contains the same amount of energy.  The only thing that has changed is the reference point.

[Starglider]

Well, resetting the conversation slightly then, I simply want to explain why short circuiting a capacitor discharges it.

Can you explain it better than my equation idea please? Imagine you're explaining it to a 10 year old child you know, as that's who I'm explaining it to.

[Brumby]

I liked the equation idea that connecting -250 +250 leaves 0. But apparently that's not right.

No.  It is very wrong.

When you are working with voltages, you might have heard of the phrase "potential difference".  That is your clue to the maths involved!

To get the potential difference, you have to take the potential (voltage) at one point* and the voltage at another point* and subtract the two.  Thus +250V and -250V would give you 500V - which is clearly wrong.

* These two measurements must be taken with respect to the same reference point.  That is, put your black lead somewhere on the circuit and leave it there for the two measurements.

[Brumby]

Can you explain it better than my equation idea please? Imagine you're explaining it to a 10 year old child you know, as that's who I'm explaining it to.

That can be tricky - but here are a couple of ideas you might like to consider.  BE VERY AWARE though, that capacitors can be used for a variety of functions and those may not seem to fit these simplistic ideas....

 1. Consider a capacitor like a rechargeable battery.  You can put electricity into it and you can take electricity out of it.  When all the electricity has been taken out of the capacitor, it is like a dead battery.  How quickly the capacitor is discharged (or the battery goes dead) depends on how much you take electricity out at a time.  Capacitors can do this a lot faster than a battery but they can't hold the same amount of electricity as a battery of the same size - which means they go flat (discharged) a lot more quickly.

 2. Consider a capacitor is like a bucket.  Charging the capacitor is like filling the bucket with water.  Discharging it is like letting the water out.

[Starglider]

2 is very good.

Next the ten year old asks:

"if the capacitor is like a bucket, when the bucket is full why does touching metal across both handles make the bucket empty?"

[Arjunan M R]

Hi, I'm doing a little explanation of how capacitors work, leading to how CRTs are basically big capacitors.

Is it correct that a 250V capacitor or a CRT stores (for the sake of argument):
• +250V on the positive plate/terminal
• -250V on the negative

Or is it 0V on the negative?

What I'm trying to do is explain why you short circuit a cap or a CRT to discharge it.

Mathematically:

+250
-250
=
0

But if it isn't -250V on the negative then that explanation doesn't work [emoji848]

Thanks!

Sent from my SM-N950U1 using Tapatalk

+250-
-250
=
500 not 0
its not +250 -250.
+250-250 = +250-+250.
You are subtracting the same numbers.
its +250--250.

but, a 250V cap cant' hold 500V.
what is your reference point?

[Brumby]

Next the ten year old asks:

"if the capacitor is like a bucket, when the bucket is full why does touching metal across both handles make the bucket empty?"

OK.  Now things start to get "interesting".....

Building on from the bucket example, a capacitor is like two buckets sitting on a bench, each of them half filled with water.  When you charge up a capacitor, it is like taking water out of one of these buckets and putting it into the other.  Connecting a wire is like making a path between them for the water to flow.  A hose filled with water is like a wire and if you put both each end into one of the buckets (and keep them below the water line), then water will flow from the one with a higher level into the one with the lower level.

Notes:
 * A hose without water inside it will be like the insulation around a wire (without the wire).  It won't let anything flow.
 * When the buckets are at the same level (half full) there is no energy stored - as the energy can only come from a difference in their levels.
 * When considering this water analogy, please understand that all analogies will have difficulties explaining everything and they usually come with "conditions".  This bucket one requires that each capacitor has its own two buckets that must stay together and you can't have a single connection to another pair of "capacitor" buckets and have any flow.

[Starglider]

Thanks Arjunan. I don't really want to get off track of the point of this which is explaining how discharging a CRT or capacitor Works in terms a 10 year old child could understand. They wouldn't need to ask what the reference point is.

So let's rewind and reset.

Really the key is I want to explain WHY short circuiting a CRT does not require any connection to ground.

I had thought that I understood that when you short circuit the positive 250 and negative 250 of the cap it equals 0 volts. But apparently that is not what is happening. So if the positive terminal of the CRT is storing (for the sake of argument) +250 volts and the negative is 0 volts, why does connecting them together still equal 0 volts?

Mathematically that doesn't make sense to a ten year old because 250 plus 0 still equals 250. And 250 volts is dangerous.

Do you kind of get the point I'm trying to illustrate? The ten-year-old would ask where does the 250 volts go to if there is no ground wire connected? "It can't escape." That really is all I need to create a really basic analogy of. Where does that 250 volts disappear to?

[Starglider]

OK.  Now things start to get "interesting".....

Building on from the bucket example, a capacitor is like two buckets sitting on a bench, each of them half filled with water.  When you charge up a capacitor, it is like taking water out of one of these buckets and putting it into the other.  Connecting a wire is like making a path between them for the water to flow.  A hose filled with water is like a wire and if you put both each end into one of the buckets (and keep them below the water line), then water will flow from the one with a higher level into the one with the lower level.

Notes:
 * A hose without water inside it will be like the insulation around a wire.  It won't let anything flow.
 * When the buckets are at the same level (half full) there is no energy stored - as the energy can only come from a difference in their levels.
 * When considering this water analogy, please understand that all analogies will have difficulties explaining everything and they usually come with "conditions".  This bucket one requires that each capacitor has its own two buckets that must stay together and you can't have a single connection to another pair of "capacitor" buckets and have any flow.

This is great.

So let me just put this in my own words and you can tell me if it is still a correct understanding.

When I buy a brand new 250V capacitor from digi-key and open the package, could it be said that it is equivalent to 2 buckets, each containing 125 liters of water? We can call on bucket Mister negative can the other bucket Mister positive.

Then when I solder the capacitor to a circuit and turn it on, it is like pouring the water out of one bucket into the other bucket. Now the negative bucket is empty and the positive bucket has 250 liters of water?

When I turn the power off the water gets trapped in the full bucket.

But if I connect a metal hose the water shoots out of the full bucket until the negative bucket is half full and both buckets again have 125 liters inside?

[Brumby]

Yes, you've got it.

... and the thicker the "metal hose" as you put it is, the more current can flow and the quicker the discharge will happen.

[Starglider]

Many thanks!

[Brumby]

I can see more installments in this series in the not-too-distant future....   

[Brumby]

If they want to ask about that 125 litres of water, then you can tell them that's like all the electrons that are just sitting around doing nothing - waiting for something to move them.

[Starglider]

Well they might ask: Why is it safe to touch when there are 125 in both buckets but not when there are 250?

[Ysjoelfir]

(Noticed that you guys were much faster than me.... but hey, never the less this could be another way to see it )

If you short circuit any charge holding device, let it be a capacitor, a battery or your CRT, you basically introduce a second element into the equation: A resistor.

If you use a screwdriver to short out the energy container, like it is often done with CRTs, or if you just use a plain piece of wire, this resistor is very, very small.
Never the less it is there and closes the circuit.
Since - as you/the 10-year-old already noted - energy can't escape (=can't evaporate into nothing), we have to find a way for the energy to "get out" of the energy container and to the surrounding world, leaving the container empty.

This is where said resistor comes into the game. Energy can't be destroyed but it can be changed into another form of energy, in this example the energy is converted by the resistor into heat. If you use a smaller resistor you can absolutely feel it heating up.
If you use the screwdriver or a jumper wire you can't feel it - just because the thermal mass of those "resistors" are so massively huge that the converted energy just doesn't affect your "resistor" in any noticeable way - except to the points where you touched the capacitor: you will most likely see and hear some sparks while touching (because of the high resistance of the insulating air, potential dirt, grease,... on the "resistor" in the moment very very close to the point of it making contact) and see some black burnmarks = produced heat. And that is, where your stored energy went

[Starglider]

(Noticed that you guys were much faster than me.... but hey, never the less this could be another way to see it )

If you short circuit any charge holding device, let it be a capacitor, a battery or your CRT, you basically introduce a second element into the equation: A resistor.

If you use a screwdriver to short out the energy container, like it is often done with CRTs, or if you just use a plain piece of wire, this resistor is very, very small.
Never the less it is there and closes the circuit.
Since - as you/the 10-year-old already noted - energy can't escape (=can't evaporate into nothing), we have to find a way for the energy to "get out" of the energy container and to the surrounding world, leaving the container empty.

This is where said resistor comes into the game. Energy can't be destroyed but it can be changed into another form of energy, in this example the energy is converted by the resistor into heat. If you use a smaller resistor you can absolutely feel it heating up.
If you use the screwdriver or a jumper wire you can't feel it - just because the thermal mass of those "resistors" are so massively huge that the converted energy just doesn't affect your "resistor" in any noticeable way - except to the points where you touched the capacitor: you will most likely see and hear some sparks while touching (because of the high resistance of the insulating air, potential dirt, grease,... on the "resistor" in the moment very very close to the point of it making contact) and see some black burnmarks = produced heat. And that is, where your stored energy went

Funny, I asked someone else that and they said that the Sparks were just a side effects. I specifically asked if the 250 volts was being converted into heat and light and dissipating into The ether and was told no. So I guess that is correct after all? does the full 250 volts disappear as a spark?

This assumes that there was 250 volts sitting inside the CRT to begin with, which I know is unlikely after you hit the power especially if it's there is a discharge resistor on the flyback transformer.

[Brumby]

Well they might ask: Why is it safe to touch when there are 125 in both buckets but not when there are 250?

Because it's only dangerous when there is a difference If you've got the same in both buckets, then there's no reason for any water to move.

It's the same sort of thing as why things at 27ºC don't feel hot or cold - even though they are at 300ºC above absolute zero.  It's because they are the same temperature that we feel comfortable.

[Brumby]

(Noticed that you guys were much faster than me.... but hey, never the less this could be another way to see it )

If you short circuit any charge holding device, let it be a capacitor, a battery or your CRT, you basically introduce a second element into the equation: A resistor.

If you use a screwdriver to short out the energy container, like it is often done with CRTs, or if you just use a plain piece of wire, this resistor is very, very small.
Never the less it is there and closes the circuit.
Since - as you/the 10-year-old already noted - energy can't escape (=can't evaporate into nothing), we have to find a way for the energy to "get out" of the energy container and to the surrounding world, leaving the container empty.

This is where said resistor comes into the game. Energy can't be destroyed but it can be changed into another form of energy, in this example the energy is converted by the resistor into heat. If you use a smaller resistor you can absolutely feel it heating up.
If you use the screwdriver or a jumper wire you can't feel it - just because the thermal mass of those "resistors" are so massively huge that the converted energy just doesn't affect your "resistor" in any noticeable way - except to the points where you touched the capacitor: you will most likely see and hear some sparks while touching (because of the high resistance of the insulating air, potential dirt, grease,... on the "resistor" in the moment very very close to the point of it making contact) and see some black burnmarks = produced heat. And that is, where your stored energy went

Funny, I asked someone else that and they said that the Sparks were just a side effects. I specifically asked if the 250 volts was being converted into heat and light and dissipating into The ether and was told no. So I guess that is correct after all? does the full 250 volts disappear as a spark?

This assumes that there was 250 volts sitting inside the CRT to begin with, which I know is unlikely after you hit the power especially if it's there is a discharge resistor on the flyback transformer.

The energy held by that charge will get dissipated into heat energy, some electromagnetic radiation - which includes the light from the spark - and some mechanical deformation which is the pit created by the spark.  The end result of all this will eventually turn into heat.

[Starglider]

Alright guys that's all great. Let the lesson begin!

[radiolistener]

• +250V on the positive plate/terminal
• -250V on the negative

Or is it 0V on the negative?

Voltage exists between two leads. This is electric potential difference.
So, if there is 250V on the capacitor, the right way to say 250V.

In order to say 0V, there is need two leads and electric potential difference 0V between these leads.

[radiolistener]

"if the capacitor is like a bucket, when the bucket is full why does touching metal across both handles make the bucket empty?"

water will be a bad analogy because it almost cannot be compressed.

Technicaly electricity is like gas (for example air). The difference between gas and electricity is that gas consists of molecules and electricity consists of charged particles (for example electrons). This electron-gas can circulate inside metal, but almost cannot leave it.

In order to leave metal, there is need very high pressure for electron-gas, because air or empty space are isolators - they like glass for usual gas. When electron gas leaves metal you can see it as electric lightning.

Then you can imagine two glass vessels connected by a hose. When air pressure in both vessels are equals, there is no gas movement in the hose and this is equivalent for 0V voltage between two leads.

Then attach the pump into the hose break and pump air from one vessel to another. You will have pressure difference between two vessels. This is equivalent for electric potential between two leads. Higher pressure difference between vessels is equivalent for higher voltage between leads.

Now your capacitor equivalent is charged. You can replace pump on the hose with air valve. If you open valve, gas will starts to move from vessel with high pressure into vessel with low pressure. This is equivalent for electric current flow.

Electric current in the metals is represented by free electrons (negative charged particle). These electrons moving from "-" lead to "+" lead. It happens because there is excess electrons on the "-" lead and lack of electrons on the "+" lead. When count of electrons on "-" lead and "+" lead will be equal, the electric current will be stopped. It means that your capacitor is discharged and now has 0V potential difference between leads.

Vessels volume is equivalent for capacitor capacitance. Higher volume vessels can accumulate more air molecules. Higher capacitance capacitor can accumulate more electrons.

This is how electricity works in classic metal. In salt water electric current flow works in a little different way, it consist of positive particles - ions. Ion is an atom or molecule with missing electron. Since it consists of proton which has positive charge and this proton charge is not balanced with electron pair, it has total positive charge.

The difference between metal and salt water is that metal has free electrons, which can flow between atoms, but cannot leave metal. And water has ions (proton charge is NOT balanced with electrons), which can flow between normal atoms (proton charge is balanced with electrons), but cannot leave water.

When electrons moving in the metal, it can randomly crash into atom, so the atom starts to moving, it leads to thermal loss. This is why metal is heating when electric current flows through it. The same thing happens in the air hose when it has some kind of air obstruction, when high pressure air flows through this air obstruction, it heating it.

In some cases it is possible to make condition for ordered motion of electrons with no collide with atoms. This is known as superconductor effect. There is no thermal loss in such case.

AC electric current is equivalent to rapid pumping of air from one vessel to another and back with high speed. It leads to cyclic air flow through hose back and forth. This air flow can be used for some useful work.

[Starglider]

Would you guys say the simplified layman's explanation given at 4 minutes into this video is a good one for giving a basic understanding to non engineers? https://youtu.be/a6FMpvs71pc The video owns that it is a simplified explanation.

The energy being converted into the light and heat of a spark, so going into the atmosphere and therefore not needing to go into a ground cable? Makes sense but is it correct?