ECN Forum Forums General Topics of the Trade General Discussion Area Megger Death

I own a small 500/1000 volt megger made by AVO It is a model BM2222. It is powered by six AA batterys.It has a built in powered low resistance meter. It works excellent for troubleshooting motors and such. I would reccomend having one on every electricians truck.

One more thing, A few years ago I was testing a Large AC motor 6000 Hp. I had it apart and in a rewind shop. The reccomended Hi pot voltage was twice the voltage plus a 1000...I hipotted the motor to 10k volts and got distracted and went to lunch. Needless to say when I did discharge it packed quite a punch. Meggers are an excellent tool and just need to be used properly,and discharged after to ground after.

And:


Sounds like a Van De Graaff (sp???) Generator.

This device creates high potential Static Charges, which may easily exceed 100K Volts between Charged body and opposite potential (typically Earth).


If done with a Van De Graaff Genny, then it's neither AC or DC - only Static.
If done with a Tesla coil, or similar plasma tossing Induction Animal, then this would be AC.


This sounds more like a Tesla coil, but a Van De Graaff Genny could do the same, especially if the air was dusty.


It's a trick!

Plasmas from a Tesla coil will flow around and sometimes through wood, but rarely would this only cause the wood to burn and not ignite or shokken ze sh*@zenhouzen out of the person holding the wood!

Static charge transfers would unlikely carry enough energy to ignite wood alone, but if the wood was soaked with a highly flammable material - such as Gasoline - a few sparks will get the flames going!

Man, I would hate to have that guy's job!

Scott35

Pauluk, thanks again, your answers are great. I sat and thought about that until I rose to a new level, referring back and forth from your explanation to the cover of my Ugly's Electrical References book until I got it. I know how to figure volts, amps and watts calculations, but haven't done much with ohms yet. This thread is taking me headlong into the land of Ohm. Thanks!!! It's going to be great to take a resistance reading and calculate how many amps I'm looking at. That was a hole in my knowledge that had to be filled.

Can I can test the resistance through a battery?

Is there a way to test resistance while a wire is hot, or does it always have to be dead?

Scott35, I believe it was a Tesla coil. Thanks to you, too, for the enlightening answer!

What do people normally do with a Tesla coil or a Van De Graaf generator?

Here's a cheap new megger I found, has anybody used one of these?
http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=2548791771&category=4678#ebayphotohosting

$55 isn't too bad for a new megger, is it worth it? Seems like a megger's job is only to crank out some voltage to see if there's a short...

I tried to copy and paste the picture alone, it didn't work, so I had to post the url.

Are new meggers better than the old antique ones? I like the looks of the old ones. They have that Ben Franklin look to them. Maybe there's something I need to know about them that somebody here can tell me. Thanks for the help, guys, this message board is a godsend.

mlk682, how long did the AC motor hold the charge before it discharged? Was that a 4160v motor?

Here's a shot of the Van de Graf generator(s) at the Museum of Science in Boston. They give a great show there. A bit loud (and scary) for small children when the lightning bolts start flying around!



(The person you see is in something resembling a large Bird Cage)

Bill

[This message has been edited by Bill Addiss (edited 08-02-2003).]

Amazing.

The coil I saw was a pedestal about 5 feet tall, around 18" to 24" in diameter. The guy stood on top of it, holding the 2x4 with foil wrapped around one end for a handle. That kept the gasoline off his hands, as Scott35 told us.

It must have been scary to be the first one to stand on that thing and say, "Switch."

I wouldn't want to grab the leads of a megger and get surprised by some physics law I'm not aware of.

So a megger can't shock me very badly, right? It's not going to knock me out, stop my heart or anything, even if I'm standing in water? I haven't tested the feeling of various amounts of amperage through my body, so I have no idea of how much it would hurt to get hit by a megger, even with 5mA. Maybe somebody can tell me, like George Corron who did it in class, or somebody who accidentally came in contact with one. I want to know what a megger can do to me, worst case scenario. Is it dangerous if used improperly, or does all the built-in resistance make it as safe as a GFI that won't allow me to take a lethal shock? Will I just feel a little tickle? If the unit gets wet, is there a danger of getting more than I bargained for if the water causes the power to get re-routed around the resistors?

[This message has been edited by Spark Master Flash (edited 08-02-2003).]

Don’t forget that with insulation-resistance testing you are charging the natural capacitor whose two “plates” are the normally grounded and normally ungrounded {er, “phase” conductors.} That stored energy can be discharged through you, and should not be dismissed as always harmless.

Another mode of stored energy might be the inductance in a winding should one mistakenly do a test where the instrument is paralleled not with a capacitor, but an inductor.

What you have to always keep in mind is that volts, amps, watts, and ohms are so interlinked that you can't change one without changing at least one of the others. When you get into the realm of AC, then reactance also plays a big part, but I think you'll find it best to get a thorough understanding of the DC basics before trying to tackle that.

Not directly with an ohmmeter, because the meter supplies power from its own batteries and the EMF from the battery you're trying to test would interfere with the reading (that's the best case scenario; the worst is that you'd burn out your meter!).

You can calculate the internal resistance of the battery indirectly, however. What you need to do is measure the open-circuit voltage at the battery terminals using a voltmeter of fairly high resistance.

Next, you connect a load across the battery (e.g. a lamp), and measure both the current flowing and the voltage now appearing at the battery's terminals. You can then use Ohm's Law to calculate the internal resistance of the battery.

An example:
You get an off-load voltage of 12.6V. When you connect a load which draws 2A the voltage across the battery drops to 11.8V.

The voltage being lost across the internal resistance of the battery is therefore

12.6 - 11.8 = 0.8V

You know that 2A is flowing through the battery, so the internal resistance by Ohm's Law is then

R = E / I = 0.8 / 2 = 0.4 ohm.


Again, you can't use your ohmmeter because the circuit is energized, but you can apply Ohm's Law again if you know the current and voltage.

Measure the voltage between the ends of the wire you want to test, check the current flowing through it, then the resistance is R = E / I.

That's for DC or non-reactive AC. It gets more complex when inductance and capacitance are involved.

Just be a little careful how you take that assumption.

A GFI certainly provides a very high level of protection, but a shock at just below 6mA could still be a little risky in some situations.

The other point is that the GFI trips only if the shock you receive is from line to ground. If you get yourself across hot and neutral (and there is no appreciable current flowing to ground at the same time), then it won't help you one bit.



[This message has been edited by pauluk (edited 08-03-2003).]

Pauluk, a million thanks.

So...in the case of the DC battery showing 12.8 volts, can I assume that the battery is capable of providing up to 31.5 amps? I=E/R = 12.8/.4 = 31.5 amps.

Maybe the voltage will drop and change all the numbers as the amp load is applied. Do I work from the full, unladen voltage to calculate amps and ohms, or do I have to let the amp load bring the voltage down before I'll have good numbers to work with?

--------------------------------------------
"The voltage being lost across the internal resistance of the battery is therefore

12.6 - 11.8 = 0.8V

You know that 2A is flowing through the battery, so the internal resistance by Ohm's Law is then

R = E / I = 0.8 / 2 = 0.4 ohm."
-------------------------------------------

So it looks like you're saying that we're not working off the full EMF when we calculate the resistance through a battery, but instead, we work off the voltage drop and the full amp reading. Seems like the formula would be this:

R=EMF lost/I

It would be easy to make the mistake of working off the full voltage divided by the amps drawn, which would be 11.8 volts/2 amps = 5.9 ohms.

Please correct me if I'm wrong on any of this, I appreciate it.


[This message has been edited by Spark Master Flash (edited 08-03-2003).]

Correct -- Although as you need a dead short across the terminals to achieve that much current, you couldn't put it to any productive use. There's also the point that such a small battery wouldn't supply this much power (12.8V x 31.5A = 403.2W) for more than a very brief period. The battery would be fully exhausted very quickly and the EMF (and therefore the current also) would quickly taper off.

Ohm's Law applies equally to the entire circuit, or to any portion of the circuit. You just have to make sure you match up the values correctly.

Let's take the last part of your post and apply it to the example in hand:

What you've done there is to calculate the resistance of the load. We know that 2A is flowing through the whole circuit, and 11.8V is the voltage which appears across the load. That's applying Ohm's Law to the load portion of the circuit.

Let's extend it to the whole circuit. We know from the previous measurement that the actual EMF of the battery is 12.6V. Now, applying Ohm's Law to the entire circuit we get:

R = E / I = 12.6 / 2 = 6.3 ohms.

That's the resistance of the complete circuit, including the internal resistance of the battery.

We've already calculated the internal resistance of the battery as 0.4 ohm. Add that to the 5.9 ohms of the load and you get the same result for total circuit resistance of 6.3 ohms. It checks out.

That's right. It's appying Ohm's Law to the appropriate part of the circuit:

1. To calculate the resistance of the entire circuit (battery+load), you need the full EMF which is powering that circuit - 12.6V.

2. To calculate the resistance of the load, you need the voltage which appears across the load, i.e. 11.8V.

3. To calculate the internal resistance of the battery, you use the voltage which is dropped internally by the battery's resistance, i.e. 0.8V.