No never remove the fuse from the meter.

The meter's insulation is only 1kVDC, if you try to measure 100kV, an arc will jump through the case which will both damage it and you'll get a nasty shock.

Here's a similar project giving an example of how to do this:
http://www.enide.net/webcms/index.php?page=lm3914-batmon

Sorry, we can't help with piracy, it's against the rules.

Please post a schematic.

The NC function generally needs a permanently live powersupply but there are ways round it.

The simplest way is to is to use a depletion mode power MOSFET or a JUGFET driving a BJT being driven by a small pulse transformer.

Another trick involves using a small battery to bias a transistor on and use an opto-coupler to turn it off.

You need to attach the image.

Click additional options, then choose, find the file containg the image and click on open.

That's probably not such a good idea.

It's a fairly low cost part and if you're unlucky enough to find a bad one in an old piece of equipment, you could waste a lot of time fault finding. It's really best practise to use new and known good parts for a prototype.

You're right, it's a pair of diodes. It's actually quite common to get high power diodes in this kind of configuration. They're commonly used to connect two batteries together or as as a full wave rectifier with a centre tapped transformer in a switched mode power supply.

Have you tried typing the part number into a search engine?

The clue lies in the symbol at the bottom of the package.

http://www.electronics-tutorials.ws/logic/logic_6.html
http://www.ustudy.in/node/3036

Why not buy one? The LM317 is a cheap commonly available part. If you see one in an old piece of electronics then use it, but it may not be as reliable as a new one. You should always opt for new parts whenever possible.

The current is not the proble, it's the voltage. You just need a 12VDC power supply capable of supplying more than 380mA, 500mA or 1A would do. I wouldn't recommend going over a couple of amps unless you add a fuse but that's only in case it develops a fault. You could use a 12V 100A PSU if you wanted but you'd need to fit a 1A fuse to protect the wiring.

The LED should be wired from pin 3 to 0V in order to have a duty cycle of greater than 50%.

To answer the origional question. It's definitely better to use a potential divider and a 5V reference. If more current is required, a non-inverting buffer could be used to reduce he output impedance.