HV Probe Suggestions

[PCBONEZ]

Re: HV Probe Suggestions

But you can have extreamly high current with very low voltage. Your car battery example is is just that. A wrench across the car battery will drop the voltage down to millivolts, but you'll have hundreds of amps flowing.

NO. WRONG. - A lot of people seem to have this same GCE lately.

The Voltage [aka -STORED- ENERGY] has to be present -first-.
Just before the wrench touches -all- the energy is -stored- in the battery.
After you touch the wrench the energy moves [aka current] through the wrench.
The energy is turned into heat [I2R losses] and makes the wrench hot.
Only AFTER energy moves out does the batteries voltage go down.
[The voltage goes back up when you remove the wrench due to a chemical reaction which for all intents and purposes is the same as charging it with an outside source so that is irrelevant here.]
.
.
The SAME thing happens when you load an actively producing power source.
The voltage doesn't dip until AFTER current flow goes up.
In this case it goes down because the out energy is greater than the in energy so the stored energy [aka instantaneous Voltage at the source] is reduced..
The voltage only returns to what it was if there is a regualtor that causes more energy to in.
.
.

No, but you can have thousands of amps, with millivolts of voltage.

No you can't.
That argument ignores the Laws of Conservation of Energy.
Energy doesn't just go poof. It gets moved or changes in form.
.
You can have a high current reduce a pre-existing [aka 'open circuit'] high voltage by rapidly moving energy away from where it's stored.
But you cannot achieve that same high current by initially starting with the low voltage as the 'open circuit' voltage at the source.
~
In otherwords:
- High voltage is a prerequisite to high current.
- High current moves a lot of energy from one place to another.
- Moving the stored energy elsewhere causes the voltage to go down.
.
It all happens in a faction of a second which faster than human senses or meters can respond or detect what happens first.
[And that's where the GCE comes from because meters and uninformed humans think it's simultaneous.]
It is NOT simultaneous and it happens in exactly that order EVERY TIME.
.
.

Fuses have millohms of resistance, so millohms of voltage can push multiamps of current through them, and not just blow them, but vaporize them, assuming you're voltage source can push that current, a CRT, can't. Assuming an non powered TV, the only source of that voltage is a few hundred PF of charged capacitance. Enough to ruin your day, but not enough to push any real current through a fuse.

There is no such thing as "millohms of voltage" and the rest of your thought is either too screwed up or too poorly explained for me to figure out what you mean.

If you have 10,000v on a cap and put 600 ohms across it you will instanateously achieve 16.7 amps through the resistance.
THEN, as the energy is moved out of the cap the voltage goes down.
THEN, the current goes down because the voltage is lower.
~repeat~ for each instant until there is no more energy stored in the cap.
If there's not much stored energy in the cap then all the instants occur in a fraction of a second which is too fast for human senses [or meters] to detect and it 'appears' that voltage went to zero instantaneously.
But it didn't.
.

[PCBONEZ]

Re: HV Probe Suggestions

I am an electrical engineering technology student if that counts... but I am only in my second semester and all the HV stuff is in classes I take next semester.

I will probably save up for a proper probe...

Well Mr. Student....
If you want to get what I'm saying walk through what happens to voltage and current when you discharge a cap with a wrench but do it in 1 millisecond intervals from open circuit to fully discharged.
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Well say the cap starts at 12v and stores 10 amp-hours of charge.
That works out to 3000 Joules [[ 1 Watt = 1 Joule / 1 sec ]]
And we will assume the wrench is 0.5 ohms.
.
Don't forget the for each next millisecond you have to reduce the number of Joules in [and voltage on] the cap by the number of watts dissipated in wrench over the previous millisecond.
.
If you can get through that then you should understand how it actually works.
[Yeah, I've done that one. Put on some coffee...]
.

[ratdude747]

Re: HV Probe Suggestions

You still don't understand things.


First, before you go callign me "unqualified", I had a college class last semester that was about 50% based on the theory behind this circuit. I got an 86% in that class with a 90% on the final exam (I didn't cram either). I also am in a successive class that also involves this theory, currently where I have a high A.

Lets go back to this equation:

P= V *I

P = power.

V= voltage

I= current

when either is zero, power is zero (effectively an impossible situation).

otherwise, for a given amount of power (watts), Voltage and current have an inverse relationship.

for instance, it is very possible to have a high voltage low current situation. Like said above, this is the case with CRTs and spark plugs. This is also the case with static discharges, hence why there is enough voltage to spark (in the thousands) but not enough power to hurt your finger.

likewise, car batteries are a high current low voltage situation. ever notice the CCA rating? that is in hundreds of Amps. That is a very large amount of current.

Another way one can think of it is as like fluid pressure and flow. In this case, voltage is like pressure and current is like flow:

-high pressures pipes are thick walled, much like high voltage cable is thickly insulated

-high flow pipes have a large internal diameter, much like how high amperage cable is of thicker gauges.

The bottom line is that while where there is current, there is voltage, it is possible to have a high amount of one and a low amount of the other.

[PCBONEZ]

Re: HV Probe Suggestions

You still don't understand things.


First, before you go callign me "unqualified", I had a college class last semester that was about 50% based on the theory behind this circuit. I got an 86% in that class with a 90% on the final exam (I didn't cram either). I also am in a successive class that also involves this theory, currently where I have a high A.

Lets go back to this equation:

P= V *I

P = power.

V= voltage

I= current

when either is zero, power is zero (effectively an impossible situation).

otherwise, for a given amount of power (watts), Voltage and current have an inverse relationship.

for instance, it is very possible to have a high voltage low current situation. Like said above, this is the case with CRTs and spark plugs. This is also the case with static discharges, hence why there is enough voltage to spark (in the thousands) but not enough power to hurt your finger.

likewise, car batteries are a high current low voltage situation. ever notice the CCA rating? that is in hundreds of Amps. That is a very large amount of current.

Another way one can think of it is as like fluid pressure and flow. In this case, voltage is like pressure and current is like flow:

-high pressures pipes are thick walled, much like high voltage cable is thickly insulated

-high flow pipes have a large internal diameter, much like how high amperage cable is of thicker gauges.

The bottom line is that while where there is current, there is voltage, it is possible to have a high amount of one and a low amount of the other.

I understand perfectly well.
I understand you are lost.
.
You are trying to use an instantaneous Voltage as Power and it isn't.
Voltage is Potential Energy, not Power.
.
You are also trying to apply instantaneous average values of Potential Energy to a transient that occurs too fast the measure it.
.

[PCBONEZ]

Re: HV Probe Suggestions

The bottom line is that while where there is current, there is voltage.

Yes because you can not have current until you have voltage.

it is possible to have a high amount of one and a low amount of the other.

No kidding?
Ya think that might be due to resistance?
Air gaps on spark plugs are a REALLY high resistance. And the voltage on them is in the kV range..
- How much resistance does a 5/16" copper rod have?
.
You are here trying to tell me you are going to get low amps through a copper rod across a cap with a high open circuit voltage.
-
And YES - that's EXACTLY what you and that other guy have been saying.
.
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When you put the bar across it you will have VERY high amps -until- the voltage goes down and -only AFTER that- will you have low amps.
.
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That will be entirely clear if you bother to do the problem I just gave you.
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[LLLlllou]

Re: HV Probe Suggestions

NO. WRONG. - A lot of people seem to have this same GCE lately.

The Voltage [aka -STORED- ENERGY] has to be present -first-.
Just before the wrench touches -all- the energy is -stored- in the battery.
After you touch the wrench the energy moves [aka current] through the wrench.
The energy is turned into heat [I2R losses] and makes the wrench hot.
Only AFTER energy moves out does the batteries voltage go down.

Not even taking the battery's chemistry into account, the battery's internal resistance will act as a voltage divider with the wrench's resistance.

Assuming a .05 ohm resistance in both the wrench and the battery, you would have no more than 6v across the wrench. If the wrench had .01 ohm, you'd have 2V across the wrench.

[The voltage goes back up when you remove the wrench due to a chemical reaction which for all intents and purposes is the same as charging it with an outside source so that is irrelevant here.]

Most of what you're saying is irrelevant here.

The SAME thing happens when you load an actively producing power source.
The voltage doesn't dip until AFTER current flow goes up.
In this case it goes down because the out energy is greater than the in energy so the stored energy [aka instantaneous Voltage at the source] is reduced..
The voltage only returns to what it was if there is a regualtor that causes more energy to in.

Again irrelevant arguing for the sake of arguing, a long running theme in most of your posts.

No you can't.
That argument ignores the Laws of Conservation of Energy.
Energy doesn't just go poof. It gets moved or changes in form.
.
You can have a high current reduce a pre-existing [aka 'open circuit'] high voltage by rapidly moving energy away from where it's stored.
But you cannot achieve that same high current by initially starting with the low voltage as the 'open circuit' voltage at the source.

No one said you would. You seem to be obsessed with the conservation of energy, and try to make every discussion an opportunity to champion the cause of COE.

~
In otherwords:
- High voltage is a prerequisite to high current.

HORSE SHEET!

Low resistance is a prerequisite to high current. A 1V source will push 10000 amps through a .0001 ohm load. A 10000V source will only push .001 amp through a 1Mohm load. It's called ohm's law.

- High current moves a lot of energy from one place to another.

It moves allot of electrons from one place to another, the amount of energy depends on both current and voltage.

- Moving the stored energy elsewhere causes the voltage to go down.

No, resistance causes the voltage to "go down". If you were moving the current through a superconductor, there would be no voltage drop, or power loss.

It all happens in a faction of a second which faster than human senses or meters can respond or detect what happens first.
[And that's where the GCE comes from because meters and uninformed humans think it's simultaneous.]
It is NOT simultaneous and it happens in exactly that order EVERY TIME.

Um, yeah, whatever. Good thing you're an informed non human, so you can educate us.

There is no such thing as "millohms of voltage"

Cause it wasn't painfully obvious that I meant millohms of resistance.

and the rest of your thought is either too screwed up or too poorly explained for me to figure out what you mean.

There seems to be allot of things you can't figure out.

If you have 10,000v on a cap and put 600 ohms across it you will instanateously achieve 16.7 amps through the resistance.

Let's forget about the speed of light and the propagation delay of electrons moving through a conductor. You still can't disregard the impedance of your magical instantaneous current pulse traveling though the conductor.

THEN, as the energy is moved out of the cap the voltage goes down.
THEN, the current goes down because the voltage is lower.
~repeat~ for each instant until there is no more energy stored in the cap.
If there's not much stored energy in the cap then all the instants occur in a fraction of a second which is too fast for human senses [or meters] to detect and it 'appears' that voltage went to zero instantaneously.
But it didn't.
.

You love to bring up conservation of energy . . . so assuming zero power lost through resistance, inductance, magnetic fields, etc. I defy you to blow a simple one amp fuse using only the energy stored in the capacitance of a picture tube. It just can't happen.

A fuse blows because enough current flows through it's resistance to generate the heat required to melt the metal spanning it's contacts. Under normal circumstances that's a constant 1 amp to blow a 1 amp fuse. If you pulse that current, as long as the average current stays well under 1 amp, the fuse element will not get hot enough to melt.

Even if this magical CRT of yours could dump all of it's energy through the fuse in a femtosecond, we're still talking about WAY less than half a joule of energy, for a femtosecond. That won't melt a snowflake, never-mind a fuse.

[PCBONEZ]

Re: HV Probe Suggestions

Not even taking the battery's chemistry into account, the battery's internal resistance will act as a voltage divider with the wrench's resistance.

No:
#1:
The internal resistance would be in series with the wrench.
That is NOT a voltage divider.
#2:
Lacking chemistry the batteries internal resistance terminal to terminal would be infinity.
#3
The voltage across the wrench is measured at the terminals so any drops inside the battery are irrelevant.
#4
You are clearly talking way beyond your knowledge level.

Assuming a .05 ohm resistance in both the wrench and the battery, you would have no more than 6v across the wrench. If the wrench had .01 ohm, you'd have 2V across the wrench.

See what I mean about knowledge level?
If you have 12v at the terminals then there is 12v across the wrench.
Any voltage drops internal to the battery happened before the terminals.
So with 12v at the terminals there is 12v across the wrench the instant you touch it to the terminals.

Most of what you're saying is irrelevant here.
Again irrelevant arguing for the sake of arguing, a long running theme in most of your posts.

Not irrelevant at all - to anyone that truly understands electricity.

You seem to be obsessed with the conservation of energy, and try to make every discussion an opportunity to champion the cause of COE.

Yes I am. Ohms Law won't work without COE.
So, every time someone's theory violates COE I bring it up.

Low resistance is a prerequisite to high current.

By that theory a cell phone without a battery would work fine.
That resistance is going to cause current flow and create a voltage.
~ That's basically what you keep saying over and over...
.
There's a drill bit here on the table.
Pretty low resistance - And it has no current at all.
- Why? - NO VOLTAGE. ~ DUH!

A 1V source will push 10000 amps through a .0001 ohm load.
A 10000V source will only push .001 amp through a 1Mohm load. It's called ohm's law.

EXACTLY - You -HAVE- to have a voltage -BEFORE- you can 'push' electrons.
[Hence the less common alternate definition of Voltage: *Electro* *Motive* *Force*.]
No voltage = No current. [And the resistance value won't make a lick of difference with no voltage.]
...

It moves allot of electrons from one place to another, the amount of energy depends on both current and voltage.

No, the amount of POWER depends on both voltage and current.
Despite erroneous common use Power and Energy are not the same thing.
TIME is not a factor in Energy.
Time IS a factor in Current.[It's a rate.]
Power is Energy per unit Time.
.
Volts is Potential Energy.
Watts is Power
Amps is a Rate.
.
Energy x Rate = Power
~ Notice how it matches right up with Ohm's Law....
.
You can have LOADS of Energy at a zero Rate.
But you can't have a Rate with zero Energy.

.

No, resistance causes the voltage to "go down".

So now you are saying that when we short a charged cap [lower the resistance] the voltage will magically GO UP!
YEAH! ~ RIGHT!
I'm not even going to bother......

I defy you to blow a simple one amp fuse using only the energy stored in the capacitance of a picture tube. It just can't happen.

I haven't talked about the tubes at all.
I've been talking about caps and nothing else.
Primarily because because I don't give'a rats ass about picture tubes.

A fuse blows because enough current flows through it's resistance to generate the heat required to melt the metal spanning it's contacts.

Yes, that's what I said. - But your description is incomplete.
R = E/I
Since R doesn't change the ONLY way to get more I is for E to go up. [E goes up first because that's what drives I.]
Heat = Watts = I²R [And remember R doesn't change.]
So the ROOT cause of heating the fuse enough to blow is too much voltage across it which then causes too much current through it.

Under normal circumstances that's a constant 1 amp to blow a 1 amp fuse. If you pulse that current, as long as the average current stays well under 1 amp, the fuse element will not get hot enough to melt.

Um, yeah. If you don't have enough constant amps to blow the fuse and you then reduce the amps by pulsing it still won't blow the fuse.
The point here was?

And I'm pretty sure you are oblivious to the fact that fuses have a surge blow amp rating in addition to the constant current rating marked on the side.
OR, that these amp/blow ratings generally have specified time periods. - something like 'x-many amps for x-many milli-seconds'.

AND BESIDES THAT THIS ENTIRE COMMENT IS OUT OF THE CONTEXT OF THE DISCUSSION.
What I said is that you aren't going to blow a meter fuse if you properly discharge the cap.
[Of course by your theories that isn't true because there will be a really high voltage on the cap after it's discharged.]

Even if this magical CRT of yours could dump all of it's energy through the fuse in a femtosecond, we're still talking about WAY less than half a joule of energy, for a femtosecond. That won't melt a snowflake, never-mind a fuse.

CRT ?of mine? - My discussion was entirely about discharging caps.
-
In other words.... ACCORDING TO YOU...
There is SO LITTLE Energy stored in an undischarged picture tube that you can just grab it with your hand and you won't even feel it.

That won't melt a snowflake

You go right ahead and do that. - Enjoy the learning experience.

-
-
We can continue as soon as you explain what you said about how SHORTING a cap will magically cause the voltage to GO UP..
.

[ratdude747]

Re: HV Probe Suggestions

Whatever. I have the high grades in my classes so thus, I am confident in my knowledge.

I am declaring further argument a

[LLLlllou]

Re: HV Probe Suggestions

No:
#1:
The internal resistance would be in series with the wrench.
That is NOT a voltage divider.
#2:
Lacking chemistry the batteries internal resistance terminal to terminal would be infinity.
#3
The voltage across the wrench is measured at the terminals so any drops inside the battery are irrelevant.
#4
You are clearly talking way beyond your knowledge level.

I'll let that silliness stand on it's own silliness. If the battery's internal resistance was irrelevant you could pull infinite current through it.

See what I mean about knowledge level?
If you have 12v at the terminals then there is 12v across the wrench.
Any voltage drops internal to the battery happened before the terminals.
So with 12v at the terminals there is 12v across the wrench the instant you touch it to the terminals.

Again, silliness and argument for the sake of arguing. Obviously there would be 12 volts if there was 12 volts, the problem is there wouldn't be 12 volts, there'd be 12 volts minus the drop caused by the battery's internal resistance (which is irrelevant in your world) times the current flow.

So, every time someone's theory violates COE I bring it up.

No, every time you need something new to argue about you claim COE is violated so you can champion it's cause.

By that theory a cell phone without a battery would work fine.
That resistance is going to cause current flow and create a voltage.
~ That's basically what you keep saying over and over...

Another example of your arguing for the sake of arguing.

There's a drill bit here on the table.
Pretty low resistance - And it has no current at all.
- Why? - NO VOLTAGE. ~ DUH!

Yes, DUH indeed.

EXACTLY - You -HAVE- to have a voltage -BEFORE- you can 'push' electrons.
[Hence the less common alternate definition of Voltage: *Electro* *Motive* *Force*.]
No voltage = No current. [And the resistance value won't make a lick of difference with no voltage.]

Thank you captain abvious!

No, the amount of POWER depends on both voltage and current.
Despite erroneous common use Power and Energy are not the same thing.
TIME is not a factor in Energy.
Time IS a factor in Current.[It's a rate.]
Power is Energy per unit Time.
.
Volts is Potential Energy.
Watts is Power
Amps is a Rate.
.
Energy x Rate = Power
~ Notice how it matches right up with Ohm's Law....
.
You can have LOADS of Energy at a zero Rate.
But you can't have a Rate with zero Energy.

.

So now you are saying that when we short a charged cap [lower the resistance] the voltage will magically GO UP!
YEAH! ~ RIGHT!
I'm not even going to bother......

Yes, that's exactly what I said . . . you have a very convenient way of putting words into people's mouths just so you can argue with those words, rather than with reality.

I haven't talked about the tubes at all.
I've been talking about caps and nothing else.

Because a picture tube isn't just a giant glass dielectric capacitor, right?

Primarily because because I don't give'a rats ass about picture tubes.

Well, since this entire thread is about discharging the capacitance in a CRT, you sure have allot to say about something you don't give a rat's ass about.

Yes, that's what I said. - But your description is incomplete.
R = E/I
Since R doesn't change the ONLY way to get more I is for E to go up. [E goes up first because that's what drives I.]
Heat = Watts = I²R [And remember R doesn't change.]
So the ROOT cause of heating the fuse enough to blow is too much voltage across it which then causes too much current through it.

You keep saying E goes up, I have to assume the fuse in your world somehow magically induces a higher voltage in your charged cap.

In the real world here on Earth. The voltage begins at it's peak open circuit voltage, the second the fuse is shorted across it and current flows, the voltage begins to drop until the charge on the capacitor reaches zero. At no point in time will the voltage ever go up.

Um, yeah. If you don't have enough constant amps to blow the fuse and you then reduce the amps by pulsing it still won't blow the fuse.
The point here was?

The point there was the same as the point in most of your arguments. IE., you say the opposite that the other person said, then claim they said it, and then argue with them about something you said, not them.

Argument for the sake of argument.

And I'm pretty sure you are oblivious to the fact that fuses have a surge blow amp rating in addition to the constant current rating marked on the side.
OR, that these amp/blow ratings generally have specified time periods. - something like 'x-many amps for x-many milli-seconds'.

Yes the whole world is oblivious to everything, only you have knowledge, please enlighten us poor retarded masses.

\AND BESIDES THAT THIS ENTIRE COMMENT IS OUT OF THE CONTEXT OF THE DISCUSSION.

Um yeah, it's ME that's keeps changing the context just to argue, uhuh.

What I said is that you aren't going to blow a meter fuse if you properly discharge the cap.

So all this arguing, and you now claim it's over a discharged capacitor's ability to deliver power, really? Even you have to be astonished by your own silliness. You somehow think we believe that a completely discharged capacitor will have a voltage? Holly crap, I wonder what color the sky is in your world.

[Of course by your theories that isn't true because there will be a really high voltage on the cap after it's discharged.]

Again, yes, that's EXACTLY what I said . . . in your world.

CRT ?of mine? - My discussion was entirely about discharging caps.

Reading comprehensive must be one of your many faults, cause the rest of us are talking about the energy stored in the capacitance of CRTs. Also I vaguely remember you mentioning some kind of magic battery that has zero internal impedance, can output infinite current, and has zero voltage drop . . . I must have just imagined that.

In other words.... ACCORDING TO YOU...
There is SO LITTLE Energy stored in an undischarged picture tube that you can just grab it with your hand and you won't even feel it.

Again, you putting imagined words into other's mouths so you can argue with something you made up yourself. I said you couldn't pop a 1A fuse with a charged CRT . . . PERIOD.

You go right ahead and do that. - Enjoy the learning experience.

Actually, I was going to recommend it to you, I hear electro shock therapy work wonders on the delusional.

We can continue as soon as you explain what you said about how SHORTING a cap will magically cause the voltage to GO UP..
.

Well, since YOU are the one who said that, I'll let you argue that fact with yourself. Welcome to my ignore list.

[PCBONEZ]

Re: HV Probe Suggestions

I'll let that silliness stand on it's own silliness. If the battery's internal resistance was irrelevant you could pull infinite current through it.




Again, silliness and argument for the sake of arguing. Obviously there would be 12 volts if there was 12 volts, the problem is there wouldn't be 12 volts, there'd be 12 volts minus the drop caused by the battery's internal resistance (which is irrelevant in your world) times the current flow.




No, every time you need something new to argue about you claim COE is violated so you can champion it's cause.



Another example of your arguing for the sake of arguing.



Yes, DUH indeed.




Thank you captain abvious!




Yes, that's exactly what I said . . . you have a very convenient way of putting words into people's mouths just so you can argue with those words, rather than with reality.



Because a picture tube isn't just a giant glass dielectric capacitor, right?



Well, since this entire thread is about discharging the capacitance in a CRT, you sure have allot to say about something you don't give a rat's ass about.



You keep saying E goes up, I have to assume the fuse in your world somehow magically induces a higher voltage in your charged cap.

In the real world here on Earth. The voltage begins at it's peak open circuit voltage, the second the fuse is shorted across it and current flows, the voltage begins to drop until the charge on the capacitor reaches zero. At no point in time will the voltage ever go up.



The point there was the same as the point in most of your arguments. IE., you say the opposite that the other person said, then claim they said it, and then argue with them about something you said, not them.

Argument for the sake of argument.



Yes the whole world is oblivious to everything, only you have knowledge, please enlighten us poor retarded masses.



Um yeah, it's ME that's keeps changing the context just to argue, uhuh.



So all this arguing, and you now claim it's over a discharged capacitor's ability to deliver power, really? Even you have to be astonished by your own silliness. You somehow think we believe that a completely discharged capacitor will have a voltage? Holly crap, I wonder what color the sky is in your world.



Again, yes, that's EXACTLY what I said . . . in your world.



Reading comprehensive must be one of your many faults, cause the rest of us are talking about the energy stored in the capacitance of CRTs. Also I vaguely remember you mentioning some kind of magic battery that has zero internal impedance, can output infinite current, and has zero voltage drop . . . I must have just imagined that.



Again, you putting imagined words into other's mouths so you can argue with something you made up yourself. I said you couldn't pop a 1A fuse with a charged CRT . . . PERIOD.



Actually, I was going to recommend it to you, I hear electro shock therapy work wonders on the delusional.



Well, since YOU are the one who said that, I'll let you argue that fact with yourself. Welcome to my ignore list.

You made no valid points at all in any of the above.
It's all opinion, no facts.

And...

We can continue as soon as you explain what you said about how SHORTING a cap will magically cause the voltage to GO UP..
Which you did in fact say....
All one has to do to see that is read back...
.

[PCBONEZ]

Re: HV Probe Suggestions

Whatever. I have the high grades in my classes so thus, I am confident in my knowledge.

I am declaring further argument a

Then why did you ask the questions?
.

[ratdude747]

Re: HV Probe Suggestions

Then why did you ask the questions?
.

umm, to verify a circuit design?

at 12v the design worked perfectly... but due to the voltage limits of the 100K resistor, it will not be used for actual discharging.

Its a moot point because I have decided to get a decent probe.

[PCBONEZ]

Re: HV Probe Suggestions

Well
If you do the little problem I suggested and then take the values you get for voltage at the different times and plot volts vs time on a graph you will come up with a voltage curve IDENTICAL to the classic capacitive discharge curves.
You know... The curves in your school books.
.
Hmmmm, why do you think that might be?
.

[ratdude747]

Re: HV Probe Suggestions

Well
If you do the little problem I suggested and then take the values you get for voltage at the different times and plot volts vs time on a graph you will come up with a voltage curve IDENTICAL to the classic capacitive discharge curves.
You know... The curves in your school books.
.
Hmmmm, why do you think that might be?
.

No thanks, I'll stick to my class work and fixing things. I have a LIFE.

[PCBONEZ]

Re: HV Probe Suggestions

Well, you'll have to learn what I'm trying to teach you in the next year or two of school anyway..
.

[Jack Crow]

Re: HV Probe Suggestions

R
Some thoughts for ya.

Since money is tight expect to find them at ham radio flea markets.
Show a care.

When it comes to used probes heads up.

Much of this probe stuff depends on the type of meter.
If your using a 20k ohms per volt 'simpson' meter, a probe that works well on that won't be the same unit that works on a 10 or 100M ohm DVM input.

There will be DC vs AC reading issues.

What the other guys was saying about the ground wire, lisiten to them please.

Your average 'radio shack' jumper wire will become the fuse with enough energy behind it. If that happens you will be the low z path to ground.

I learned this the hard way when we were doing electronic weapons.
We were not only measuring high voltages but high bandwidth as well.

The lab had a couple of this places units.
http://www.highvoltageprobes.com/

We ran into an accuracy issue with of all things the coaxial cable used for the reading. Seems the things were designed with 93 ohm cable, and one of our EE's used 50 ohm. Wiped out a lot of good data.

For what it's worth, I have used my gear about 10 times in the last fifteen years. Can't find the thing now. Made more use of my degaussing ring than the probe.

Hope this helps.
Jack Crow

[PeteS in CA]

Re: HV Probe Suggestions

As some one pointed out, you have to pay attention to the max voltage rating of the probe. CRT anode voltages get pretty high. I can use 15KV or 20KV probes because I'm measuring 10KV max (so far). FWIW, the Fluke, HP and Agilent DMMs I use at work are 1000V max.

One thing to pay attention to in home-brew HV divider probes is the total resistance of the resistor network. At 20KV or 30KV, a few milliamps is a lot of power! Another thing to watch is the max voltage rating of each resistor. Exceed that and thing get "interesting". Caddock sells high-resistance high-voltage-rated resistors, but quality isn't cheap. Another is the voltage rating of your wires. You can probably get away with 300V or 600V wire, if you don't use it at points with voltages higher than that. Then there's that last potentiometer in the divider ... what happens if it opens? Full HV to your meter? Place a zener across that resistor with a zener voltage higher than the resistor will see and you'll be safe without compromising your measurements. Don't forget the physical layout of your divider network. High voltage can arc across short spaces. Lay it out in a straight line, and cover it with a generous coating of RTV (or some other coating that is good for high voltages and will block moisture). Finally, how will you hold it? We don't want you plastered against the wall with smoking, frizzled, hair! Maybe coated generously with RTV, wrapped with Kapton tape and then epoxied or RTVed into a PVC pipe of reasonable wall thickness? That might do. This is why high voltage probes are pricey.

That recovery effect after discharging a capacitor is called dielectric absorption. Lytics can give surprises, but I'm not sure about the dielectric absorption of glass (the CRT is essentially a really large glass dielectric capacitor).

[PeteS in CA]

Re: HV Probe Suggestions

Bottom line, we don't want no KFR, dude!

[ratdude747]

Re: HV Probe Suggestions

FYI- the homebrew probe concept has been ditched. Right now the CRT projects that require electrical work have been shelved for now.

[tom66]

Re: HV Probe Suggestions

A battery has two types of internal resistance:
- Electrical resistance, the effects of which occur immediately
- Ionic resistance, which takes some amount of time to occur (typically several seconds after a load is applied)

The atoms inside the battery resist current flow because free electrons can hit these atoms. This produces heat and ultimately reduces the average energy the electrons have, reducing the apparent cell EMF. (This is a primitive view of electrical resistance, but it's close enough.)

Ionic resistance is the result of a chemical reaction inside of the battery - it's a lot more complicated to explain but this bulletin does well:

So if you take the spanner and put it across the car battery terminals the voltage across the spanner WILL be less than ~13.8V because of the ESR of the battery which includes the ionic resistance, which is a purely chemical effect. The cell's EMF has actually decreased.

Ionic resistance explains why if you short out a battery, it takes a couple of seconds before the voltage returns to nearly the original voltage. (ESR explains why it doesn't lose all of its energy instantly.) And that means the EMF of the cell has actually decreased because the chemical reactions have slowed down.

Just sayin'.