Electronic – Confused by transistor behaviour

You can do this a number of ways. I usually drive transistors in common-emitter mode when using them as switches. The emitter of the NPN transistor goes to the common (ground) rail, the collector to the cathode of the LED and then a current limiting resistor to B+. The transistor will turn on when the base voltage is approximately 0.7V above the emitter voltage. Use a resistor on the base of the transistor (to the I/O pin) since BJTs are current controlled devices.

When the I/O line goes high (say +5V), you will have 5V across the resistor and the B-E junction of the transistor. The transistor B-E junction will drop 0.7V with the remainder (4.3V) across the resistor. Say you use 1k for the base resistor. This will give you 4.3/1000 or 4.3mA of base drive current. This is TONS of current, and since the B-E junction is forward-biased it will amplify the base current by the gain of the transistor (usually between 50 and 200) and limit the C-E current to this level. (When using transistors as switches you usually don't care about the actual collector current, you just want to make sure the transistor is allowing much more current than you will actually draw.) The end result is that the transistor is fully on and your LED will light, limited by the series resistor you chose to limit the LED current to something like 10-20mA.

When the I/O line is low, the B-E junction has no appreciable voltage across it and the transistor is considered "off." I like to include a 4.7k-ish resistor between the base and emitter to make sure that the transistor doesn't accidentally switch on due to noise or a floating I/O pin.

This basic NPN switch circuit works well as long as your C-E voltage is under about 30V and your load is mainly resistive and relatively low current (under a few Amps). When you're trying to control higher voltages you have to start looking at either more specialized (high voltage) transistors or more complex drive circuits. When driving inductive loads (relays, motors, etc.) you need to protect the transistor from the inductive kickback that occurs when current stops flowing through an inductor. When driving high current loads you may have to again look at specialized transistors or more complex drivers to ensure the transistor remains fully switched-on.

If your load must be connected to common (instead of the transistor emitter) then you can use a PNP transistor. Emitter to B+, collector to the LED anode, LED cathode to a current limit resistor and the other end of the resistor to common. Now a logic '1' should turn the LED off and a logic '0' should turn it on, but you've got a problem. The problem is that your I/O line cannot turn the transistor off, because the highest voltage it can reach is 5V (in our example). This would maintain 1V across the B-E junction and the transistor would remain on, even only if partially on. In this case I like to "cheat". You can turn the I/O line into an input to turn the transistor off (remember I like to have a 4.7kish resistor between the base and emitter). This is not ideal because it slows down your turn-off (which may or may not be a problem) but also because you now have (in this example) 6V going to an I/O line. It may not be able to withstand this kind of voltage and you can damage the line or the internal protection circuitry on the input. What I do to mitigate the problem is to use an NPN transistor to turn on the PNP transistor. This doesn't solve the turn-off problem but for most general cases it's nothing to worry about.

You've got the wrong setup: connect the emitter to ground and add a few resistors.

The base-emitter junction is like a diode, and the base will be 0.7 V higher than the emitter. If you would just apply 5 V to it you're kind of creating a short circuit: there's no resistance between 5 V and 0.7 V. Adding a 2 kΩ resistor will limit the current as per Ohm's Law:

\$ I = \dfrac{V}{R} = \dfrac{5 V - 0.7 V}{2 k\Omega} = 2.15 mA \$

Then the collector current will be a multiple of that. If that's 100 times (you can find the value in the BC108's datasheet as \$H_{21E}\$, which is a name nobody uses, everybody talks about \$H_{FE}\$) then the collector current will be 215 mA, 100 times the base current.

But your transistor will be useless: it will always have 12 V at the collector, no matter what current. And it will get hot: 12 V across it and 215 mA through it is 2.58 W!! Too much for the poor thing. So add a resistor between collector and 12 V:

(Here we also have a LED, but we can do with just the 1 kΩ resistor.)

We had a 215 mA collector current, which would cause a voltage drop across the resistor of 215 mA \$\times\$ 1 kΩ = 215 V!, according to Ohm's Law. But that's impossible, we only have 12 V and a 12 V across the resistor will cause 12 mA current, no more than that. So the resistor limits the current, even when the transistor will try to draw more.

If we would increase R2 to 100 kΩ then the base current will be 50 times smaller, or 43 \$\mu\$A, and the collector current would be 100 times that, or 4.3 mA. Then the voltage drop across R1 will be 4.3 mA \$\times\$ 1 kΩ = 4.3 V. So the collector will be 4.3 V lower than the 12 V, or be at 7.7 V.

So by choosing the right base current you can create a certain voltage at the collector, and when the base current is too high the collector voltage will go all the way to zero.

Note
You can make a circuit like you did, with a resistance between emitter and ground, but then the resistance should be much smaller than the multimeter's, which is often 10 MΩ; a value of 100 Ω will often do. Even then it's not a good circuit here, since the emitter voltage should never go higher than 4.3 V (the 5 V in - 0.7 V base-emitter). You'll never have 12 V there, and I can't even explain that you have a higher voltage than 4.3 V.

edit

"I was thinking of multiplexing four of my displays by putting a transistor before each common anode and then connect all 32 segment cathodes to 8 transistors."

This will work fine. What I described is the driver for one segment. Connect all cathodes for the same segments of the different displays together, and use 8 outputs to drive the 8 transistors.

Then you need something to step from one display to the next.

That will be the part of the circuit around Q1 and Q2 (Q3 is the segment driver). Q1 is a PNP transistor, which will source current to the segments of 1 display, so you'll need 4 of those, plus surrounding parts (Q2, R1, R2 and R3). Q1 will source current to its collector if there's a current from the emitter (12 V) to the base. We get that current by activating Q2, an NPN transistor like we saw earlier. So if you make "Display 1" high there will flow a current from 12 V through Q1's emitter-base and R2 to Q2's collector. You can use a BC807 for Q1.
Note: I would ditch the BC108. It's an old beast, and Digikey, which sells everything, doesn't even list it. Alternative: BC337; high \$H_{FE}\$ selections available, and 500 mA maximum current.